Student Solutions Manual for Physics for Scientists and Engineers: A Strategic Approach Vol 2 (Chs 20-43) (v. 2, Chapters 20-43)

1.1.

Solve:

1.2.

Solve:

1.3.

Solve:

1.4. Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated into a single point. The size and shape of the object will not be considered. This is a reasonable approximation of reality if (i) the distance traveled by the object is large in comparison to the size of the object, and (ii) rotations and internal motions are not significant features of the object’s motion. The particle model is important in that it allows us to simplify a problem. Complete reality—which would have to include the motion of every single atom in the object—is too complicated to analyze. By treating an object as a particle, we can focus on the most important aspects of its motion while neglecting minor and unobservable details. (b) The particle model is valid for understanding the motion of a satellite or a car traveling a large distance. (c) The particle model is not valid for understanding how a car engine operates, how a person walks, how a bird flies, or how water flows through a pipe.

1.5. Solve: (a) An operational definition defines a concept or an idea in terms of a procedure, or a set of operations, that is used to identify or measure the concept. G (b) The displacement Δr of an object is a vector found by drawing an arrow from the object’s initial location to G G G G its final location. Mathematically, Δr = rf − ri . The average velocity v of an object is a vector that points in the G G same direction as the displacement Δr and has length, or magnitude, Δr / Δt , where Δt = tf − ti is the time interval during which the object moves from its initial location to its final location.

1.6.

Solve: The player starts from rest and moves faster and faster (accelerates).

1.7. Solve: The particle starts with an initial velocity but as it slides it moves slower and slower till coming to rest. This is a case of negative acceleration because it is an acceleration opposite to the positive direction of motion.

G Solve: The acceleration of an object is a vector formed by finding the ratio of Δv , the change in the G object’s velocity, to Δt , the time in which the change occurs. The acceleration vector a points in the direction G of Δv , which is found by vector subtraction.

1.8.

G (a) Acceleration is found by the method of Tactics Box 1.3. Let v0 be the velocity vector G between points 0 and 1 and v1 be the velocity vector between points 1 and 2.

1.9.

Solve:

(b) Speed v1 is greater than speed v0 because more distance is covered in the same interval of time.

G Solve: (a) Acceleration is found by the method of Tactics Box 1.3. Let v0 be the velocity vector G between points 0 and 1 and v1 be the velocity vector between points 1 and 2.

1.10.

(b) Speed v1 is greater than speed v0 because more distance is covered in the same interval of time.

1.11. (a)

Solve: (b)

1.12. (a)

Solve: (b)

1.13.

Model:

Represent the car as a particle.

Visualize: The dots are equally spaced until brakes are applied to the car. Equidistant dots indicate constant average speed. On braking, the dots get closer as the average speed decreases.

1.14.

Model:

Represent the (child + sled) system as a particle.

Visualize: The dots in the figure are equally spaced until the sled encounters a rocky patch. Equidistant dots indicate constant average speed. On encountering a rocky patch, the average speed decreases and the sled comes to a stop. This part of the motion is indicated by a separation between the dots that becomes smaller and smaller.

1.15.

Model: Represent the tile as a particle. Visualize: The tile falls from the roof with an acceleration equal to a = g = 9.8 m/s2. Starting from rest, its velocity increases until the tile hits the water surface. This part of the motion is represented by dots with increasing separation, indicating increasing average velocity. After the tile enters the water, it settles to the bottom at roughly constant speed.

1.16.

Model: Represent the tennis ball as a particle. Visualize: The particle falls freely for the three stories under the acceleration of gravity. It strikes the ground and very quickly decelerates to zero (while decompresses) and finally travels upward with negative acceleration under gravity to zero velocity at a height of two stories. The downward and upward motions of the ball are shown in the figure. The increasing length between the dots during downward motion indicates increasing average velocity or downward acceleration. On the other hand, the decreasing length between the dots during upward motion indicates acceleration in a direction opposite to its motion; that is, in the downward direction. Assess: For a free-fall motion, acceleration due to gravity is always vertically downward.

1.17.

Model: Represent the toy car as a particle. Visualize: As the toy car rolls down the ramp, its average speed increases. This is indicated by the increasing G length of the velocity arrows. That is, motion down the ramp is under an acceleration a. At the bottom of the ramp, the toy car continues with the speed obtained with no change in velocity.

1.18. Solve: (a)

Dot 1 2 3 4 5 6 7 8 9

Time (s) 0 2 4 6 8 10 12 14 16

x (m) 0 30 95 215 400 510 600 670 720

(b)

1.19. Solve: A forgetful physics professor goes for a walk on a straight country road. Walking at a constant speed, he covers a distance of 300 m in 300 s. He then stops and watches the sunset for 100 s. Finding that it was getting dark, he walks faster back to his house covering the same distance in 200 s.

1.20. Solve: Forty miles into a car trip north from his home in El Dorado, an absent-minded English professor stopped at a rest area one Saturday. After staying there for one hour, he headed back home thinking that he was supposed to go on this trip on Sunday. Absent-mindedly he missed his exit and stopped after one hour of driving at another rest area 20 miles south of El Dorado. After waiting there for one hour, he drove back very slowly, confused and tired as he was, and reached El Dorado in two hours.

Visualize: The bicycle is moving with an acceleration of 1.5 m/s2. Thus, the velocity will increase by 1.5 m/s each second of motion.

1.21.

G Visualize: The particle moves upward with a constant acceleration a . The final velocity is 200 m/s and is reached at a height of 1000 m.

1.22.

⎛ 10−6 s ⎞ −6 Solve: (a) 9.12 μ s = (9.12 μs) ⎜ ⎟ = 9.12 × 10 s ⎝ 1 μs ⎠ ⎛ 103 m ⎞ 3 (b) 3.42 km = (3.42 km) ⎜ ⎟ = 3.42 × 10 m ⎝ 1 km ⎠

1.23.

−2 ⎛ cm ⎞ ⎛ 10 m ⎞ ⎛ 1 ms ⎞ 2 (c) 44 cm/ms = 44 ⎜ ⎟ ⎜ −3 ⎟ = 4.4 × 10 m/s ⎟⎜ ⎝ ms ⎠ ⎝ 1 cm ⎠ ⎝ 10 s ⎠ 3 ⎛ km ⎞ ⎛ 10 m ⎞ ⎛ 1 hour ⎞ (d) 80 km/hour = 80 ⎜ ⎟⎜ ⎟⎜ ⎟ = 22 m/s ⎝ hour ⎠ ⎝ 1 km ⎠ ⎝ 3600 s ⎠

1.24.

−2 ⎛ 2.54 cm ⎞ ⎛ 10 m ⎞ Solve: (a) 8.0 inches = 8.0 (inch) ⎜ ⎟ = 0.20 m ⎟⎜ ⎝ 1 inch ⎠ ⎝ 1 cm ⎠

1m ⎛ feet ⎞⎛ 12 inch ⎞⎛ ⎞ (b) 66 feet /s = 66 ⎜ ⎟⎜ ⎟⎜ ⎟ = 20 m/s ⎝ s ⎠⎝ 1 foot ⎠⎝ 39.37 inch ⎠ 3 ⎛ miles ⎞⎛ 1.609 km ⎞ ⎛ 10 m ⎞ ⎛ 1 hour ⎞ (c) 60 mph = 60 ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = 27 m/s ⎝ hour ⎠⎝ 1 mile ⎠ ⎝ 1 km ⎠ ⎝ 3600 s ⎠

2

1m ⎛ ⎞ −3 (d) 14 square inches = 14 (inches) 2 ⎜ ⎟ = 9.0 × 10 square meter ⎝ 39.37 inches ⎠

1.25.

Solve: (a)

⎛ 3600 s ⎞ 3 1 hour = 1(hour) ⎜ ⎟ = 3600 s = 3.60 × 10 s 1 hour ⎝ ⎠

⎛ 24 hours ⎞ ⎛ 3600 s ⎞ 4 (b) 1 day = 1 (day) ⎜ ⎟⎜ ⎟ = 8.64 × 10 s ⎝ 1 day ⎠ ⎝ 1 hour ⎠ ⎛ 365.25 days ⎞ ⎛ 8.64 × 104 s ⎞ 7 (c) 1 year = 1 (year) ⎜ ⎟ = 3.16 × 10 s ⎟⎜ ⎝ 1 year ⎠⎝ 1 day ⎠ 1m ⎛ ft ⎞⎛ 12 inch ⎞⎛ ⎞ 2 (d) 32 ft /s 2 = 32 ⎜ 2 ⎟⎜ ⎟⎜ ⎟ = 9.75 m/s ⎝ s ⎠⎝ 1 ft ⎠⎝ 39.37 inch ⎠

1.26.

Solve: (a)

⎛1 m ⎞ 20 ft = 20 ( ft ) ⎜ ⎟ = 7.0 m ⎝ 3 ft ⎠

⎛ 1 km ⎞⎛ 1000 m ⎞ 5 (b) 60 miles = 60(miles) ⎜ ⎟⎜ ⎟ = 1.0 × 10 m ⎝ 0.6 miles ⎠⎝ 1 km ⎠

⎛ 1 m/s ⎞ (c) 60 mph = 60(mph) ⎜ ⎟ = 30 m/s ⎝ 2 mph ⎠

⎛ 1 cm ⎞ ⎛ 10−2 m ⎞ (d) 8 in = 8(in) ⎜ ⎟⎜ ⎟ = 0.16 m ⎝ 1/2 in ⎠⎝ 1 cm ⎠

1.27.

Solve: ⎛ 4 in ⎞ (a) (30 cm ) ⎜ ⎟ = 12 in ⎝ 10 cm ⎠

⎛ 2 mph ⎞ (b) ( 25 m/s ) ⎜ ⎟ = 50 mph ⎝ 1 m/s ⎠ ⎛ 0.6 mi ⎞ (c) (5 km ) ⎜ ⎟ = 3 mi ⎝ 1 km ⎠ ⎛1 ⎞ in ⎛1 ⎞⎜ 2 ⎟ 1 (d) ⎜ cm ⎟ ⎜ ⎟ = in ⎝2 ⎠ ⎝ 1 cm ⎠ 4

Solve: (a) 33.3 × 25.4 = 846 (b) 33.3 − 25.4 = 7.9

1.28.

(c) 33.3 = 5.77 (d) 333.3 ÷ 25.4 = 13.1

1.29.

Solve: (a) (33.3) 2 = 1.109 × 103. For numbers starting with “1” an extra digit is kept.

(b) 33.3 × 45.1 = 1.50 × 103 Scientific notation is an easy way to establish significance. (c) 22.2 − 1.2 = 3.5 (d) 1/ 44.4 = 0.0225

1.30.

Solve: The length of a typical car is 15 ft. Or

1m ⎛ 12 inch ⎞⎛ ⎞ 15(ft) ⎜ ⎟⎜ ⎟ = 4.6 m 1 ft 39.37 inch ⎝ ⎠⎝ ⎠ This length of 15 ft is approximately two-and-a-half times my height.

1.31.

Solve: The height of a telephone pole is estimated to be around 50 ft or 15 m. This height is approximately 8 times my height.

1.32.

Solve: I typically take 15 minutes in my car to cover a distance of approximately 6 miles from home to campus. My average speed is ⎛ 0.447 m/s ⎞ 6 miles ⎛ 60 min ⎞ ⎟ = 11 m/s ⎜ ⎟ = 24 mph = 24(mph) ⎜ 15 min ⎝ 1 hour ⎠ ⎝ 1 mph ⎠

1.33.

Solve: My barber trims about an inch of hair when I visit him every month for a haircut. The rate of hair growth is 1(inch) ⎛ 2.54 cm ⎞ ⎛ 10−2 m ⎞⎛ 1 month ⎞ ⎛ 1 day ⎞⎛ 1 h ⎞ −9 ⎟⎜ ⎟⎜ ⎜ ⎟⎜ ⎟⎜ ⎟ = 9.8 × 10 m/s (month) ⎝ 1 inch ⎠ ⎝ 1 cm ⎠ ⎝ 30 days ⎠ ⎝ 24 h ⎠⎝ 3600 s ⎠ 6 ⎛ m ⎞ ⎛ 10 μ m ⎞ ⎛ 3600 s ⎞ = 9.8 × 10−9 ⎜ ⎟ ⎜ ⎟⎜ ⎟ = 35 μ m/h ⎝ s ⎠⎝ 1 m ⎠⎝ 1 h ⎠

1.34.

Model: Visualize:

Represent the Porsche as a particle for the motion diagram.

1.35.

Model: Visualize:

Represent the watermelon as a particle for the motion diagram.

1.36. Model: Visualize:

Represent (Sam + car) as a particle for the motion diagram.

1.37.

Model: Visualize:

Represent the speed skater as a particle for the motion diagram.

1.38.

Model: Visualize:

Represent the wad as a particle for the motion diagram.

1.39.

Model: Visualize:

Represent the ball as a particle for the motion diagram.

1.40.

Model: Visualize:

Represent the ball as a particle for the motion diagram.

1.41.

Model: Visualize:

Represent the motorist as a particle for the motion diagram.

1.42.

Model: Visualize:

Represent Bruce and the puck as particles for the motion diagram.

1.43.

Model: Visualize:

Represent Fred and yourself as particles for the motion diagram.

1.44.

Solve: Rahul was coasting on interstate highway I-35 from Wichita to Kansas City at 65 mph. Seeing an accident at a distance of 200 feet in front of him, he braked his car to a stop with steady deceleration.

1.45. Solve: A car starts coasting at an initial speed of 30.0 m/s up a 10° incline. 230 m up the incline the road levels out to a flat road and the car continues coasting at a reduced speed along the road.

1.46.

Solve: A skier starts from rest down a 25° slope with very little friction. At the bottom of the 100 m slope the skier moves to a flat area and continues at constant velocity.

1.47.

Solve: A ball is dropped from a height to check its rebound properties. It rebounds to 80% of its original height.

1.48.

Solve: Two boards lean against each other at equal angles to the vertical direction. A ball rolls up the incline, over the peak, and down the other side.

1.49.

Solve:

(a)

(b) Sue passes 3rd Street doing 40 mph, slows steadily to the stop sign at 4th Street, stops for 1 s, then speeds up and reaches her original speed as she passes 5th Street. If the blocks are 50 m long, how long does it take Sue to drive from 3rd Street to 5th Street? (c)

1.50.

Solve:

(a)

(b) A train moving at 100 km/hour slows down in 10 s to a speed of 60 km/hour as it enters a tunnel. The driver maintains this constant speed for the entire length of the tunnel that takes the train a time of 20 s to traverse. Find the length of the tunnel. (c)

1.51. Solve: (a)

(b) Jeremy has perfected the art of steady acceleration and deceleration. From a speed of 60 mph he brakes his car to rest in 10 seconds with a constant deceleration. Then he turns into an adjoining street. Starting from rest, Jeremy accelerates with exactly the same magnitude as his earlier deceleration and reaches the same speed of 60 mph over the same distance in exactly the same time. Find the car’s acceleration or deceleration. (c)

1.52. Solve: (a)

(b) A coyote (A) sees a rabbit and begins to run toward it with an acceleration of 3.0 m/s2. At the same instant, the rabbit (B) begins to run away from the coyote with an acceleration of 2.0 m/s2. The coyote catches the rabbit after running 40 m. How far away was the rabbit when the coyote first saw it? (c)

1.53. Solve: Since area equals length × width, the smallest area will correspond to the smaller length and the smaller width. Similarly, the largest area will correspond to the larger length and the larger width. Therefore, the smallest area is (64 m)(100 m) = 6.4 × 103 m2 and the largest area is (75 m)(110 m) = 8.3 × 103 m2.

1.54.

Solve: (a) We need kg/m3. There are 100 cm in 1 m. If we multiply by 3

⎛ 100 cm ⎞ 3 ⎜ ⎟ = (1) ⎝ 1m ⎠ we do not change the size of the quantity, but only the number in terms of the new unit. Thus, the mass density of aluminum is 3

⎛ kg ⎞⎛ 100 cm ⎞ 3 kg 2.7 × 10−3 ⎜ 3 ⎟⎜ ⎟ = 2.7 × 10 3 m ⎝ cm ⎠⎝ 1 m ⎠ (b) Likewise, the mass density of alcohol is

⎛ g ⎞⎛ 100 cm ⎞ 0.81⎜ 3 ⎟⎜ ⎟ ⎝ cm ⎠⎝ 1 m ⎠

3

⎛ 1 kg ⎞ kg ⎜ ⎟ = 810 3 1000 g m ⎝ ⎠

1.55.

Model:

The car is represented by the particle model as a dot.

Solve: (a) Time t (s) 0 10 20 30 40 50 60 70 80 90

Position x (m) 1200 975 825 750 700 650 600 500 300 0

(b)

1.56. Solve: Susan enters a classroom, sees a seat 40 m directly ahead, and begins walking toward it at a constant leisurely pace, covering the first 10 m in 10 seconds. But then Susan notices that Ella is heading toward the same seat, so Susan walks more quickly to cover the remaining 30 m in another 20 seconds, beating Ella to the seat. Susan stands next to the seat for 10 seconds to remove her backpack.

1.57. Solve: A crane operator holds a ton of bricks 30 m above the ground. Four seconds after he is told to lower the bricks, he takes four seconds to lower them 15 m at a constant rate before stopping the bricks to make an eight-second safety check. He then continues lowering the bricks the remaining 15 m, taking four more seconds.

1-1

2.1. Model: We will consider the car to be a particle that occupies a single point in space. Visualize:

Solve:

Since the velocity is constant, we have xf = xi + vx Δt. Using the above values, we get

x1 = 0 m + (10 m/s)(45 s) = 450 m Assess: 10 m/s ≈ 22 mph and implies a speed of 0.4 miles per minute. A displacement of 450 m in 45 s is reasonable and expected.

2.2.

Model: Visualize:

Solve:

We will consider Larry to be a particle.

Since Larry’s speed is constant, we can use the following equation to calculate the velocities:

vs =

sf − si tf − ti

(a) For the interval from the house to the lamppost:

v1 =

200 yd − 600 yd = −200 yd/min 9:07 − 9:05

For the interval from the lamppost to the tree: 1200 yd − 200 yd v2 = = +333 yd/min 9:10 − 9:07 (b) For the average velocity for the entire run: 1200 yd − 600 yd vavg = = +120 yd/min 9:10 − 9:05

2.3.

Model: Visualize:

Solve:

Cars will be treated by the particle model.

Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows: Δx x1 − x0 x −x v= = ⇒ t1 = t0 + 1 0 Δt t1 − t0 v

Using the known values identified in the pictorial representation, we find: x −x 400 mile tAlan 1 = tAlan 0 + Alan 1 Alan 0 = 8:00 AM + = 8:00 AM + 8 hr = 4:00 PM v 50 miles/hour x −x 400 mile tBeth 1 = tBeth 0 + Beth 1 Beth 0 = 9:00 AM + = 9:00 AM + 6.67 hr = 3:40 PM v 60 miles/hour (a) Beth arrives first. (b) Beth has to wait tAlan 1 − tBeth 1 = 20 minutes for Alan. Assess: Times of the order of 7 or 8 hours are reasonable in the present problem.

Solve: (a) The time for each segment is Δt1 = 50 mi/40 mph = 5/4 hr and Δt2 = 50 mi/60 mph = 5/6hr. The average speed to the house is 100 mi = 48 mph 5/6 h + 5/4 h

2.4.

(b) Julie drives the distance Δx1 in time Δt1 at 40 mph. She then drives the distance Δx2 in time Δt2 at 60 mph. She spends the same amount of time at each speed, thus Δt1 = Δt2 ⇒ Δx1 /40 mph = Δx2 /60 mph ⇒ Δx1 = (2/3)Δx2

But Δx1 + Δx2 = 100 miles, so (2 / 3)Δx2 + Δx2 = 100 miles. This means Δx2 = 60 miles and Δx1 = 40 miles. Thus, the times spent at each speed are Δt1 = 40 mi/40 mph = 1.00 h and Δt2 = 60 mi/60 mph = 1.00 h. The total time for her return trip is Δt1 + Δt2 = 2.00 h. So, her average speed is 100 mi/2 h = 50 mph.

2.5.

Model: The bicyclist is a particle. Visualize: Please refer to Figure EX2.5. Solve: The slope of the position-versus-time graph at every point gives the velocity at that point. The slope at t = 10 s is Δs 100 m − 50 m v= = = 2.5 m/s 20 s Δt The slope at t = 25 s is v=

100 m − 100 m = 0 m/s 10 s

v=

0 m − 100 m = −10 m/s 10 s

The slope at t = 35 s is

2.6. Solve:

Visualize: Please refer to Figure EX2.6. (a) We can obtain the values for the velocity-versus-time graph from the equation v = Δs/Δt.

(b) There is only one turning point. At t = 3 s the velocity changes from +5 m/s to −20 m/s, thus reversing the direction of motion. At t = 1 s, there is an abrupt change in motion from rest to +5 m/s, but there is no reversal in motion.

Visualize: Please refer to Figure EX2.7. The particle starts at x0 = 10 m at t0 = 0. Its velocity is initially in the –x direction. The speed decreases as time increases during the first second, is zero at t = 1 s, and then increases after the particle reverses direction. Solve: (a) The particle reverses direction at t = 1 s, when vx changes sign. (b) Using the equation xf = x0 + area of the velocity graph between t1 and tf ,

2.7.

x2 s = 10 m − (area of triangle between 0 s and 1 s) + (area of triangle between 1 s and 2 s) 1 1 (4 m/s)(1 s) + (4 m/s)(1 s) = 10 m 2 2 x3 s = 10 m + area of trapazoid between 2 s and 3 s = 10 m −

1 = 10 m + (4 m/s + 8 m/s)(3 s − 2 s) = 16 m 2 x4 s = x3 s + area between 3 s and 4 s 1 = 16 m + (8 m/s + 12 m/s)(1 s) = 26 m 2

2.8. Visualize: Please refer to Figure EX2.8. Solve: A constant velocity from t = 0 s to t = 2 s means zero acceleration. On the other hand, a linear increase in velocity between t = 2 s and t = 4 s implies a constant positive acceleration.

2.9. Visualize: Please refer to Figure EX2.9. Solve: (a) The acceleration of the train at t = 3.0 s is the slope of the v vs t graph at t = 3 s. Thus a = ( 2 m/s − ( − 2 m/s) ) (8 s ) = 0.5 m/s 2 . (b)

2.10. Solve:

Visualize: Please refer to Figure EX2.10. (a) At t = 2.0 s, the position of the particle is

x2 s = 2.0 m + area under velocity graph from t = 0 s to t = 2.0 s 1 = 2.0 m + (4 m/s)(2.0 s) = 6 m 2 (b) From the graph itself at t = 2.0 s, v = 4 m/s. (c) The acceleration is ax =

Δvx vfx − vix 6 m/s − 0 m/s = = = 2 m/s 2 3s Δt Δt

2.11. Solve:

Visualize: Please refer to Figure EX2.11. (a) Using the equation

xf = xi + area under the velocity-versus-time graph between ti and tf we have

x (at t = 1 s) = x (at t = 0 s) + area between t = 0 s and t = 1 s

= 2.0 m + (4 m/s)(1 s) = 6 m

Reading from the velocity-versus-time graph, vx (at t = 1 s) = 4 m/s. Also, ax = slope = Δv/Δt = 0 m/s 2 . (b) x (at t = 3.0 s) = x(at t = 0 s) + area between t = 0 s and t = 3 s = 2.0 m + 4 m/s × 2 s + 2 m/s × 1 s + (1/2) × 2 m/s × 1 s = 13.0 m

Reading from the graph, vx (t = 3 s) = 2 m/s. The acceleration is ax (t = 3 s) = slope =

vx (at t = 4 s) − vx (at t = 2 s) = −2 m/s 2 2s

2.12.

Model: Visualize:

Solve:

Represent the jet plane as a particle.

(a) Since we don’t know the time of acceleration, we will use

v12 = v02 + 2a ( x1 − x0 ) ⇒a=

v12 − v02 (400 m/s) 2 − (300 m/s) 2 = = 8.75 m/s 2 2 x1 2(4000 m)

(b) The acceleration of the jet is approximately equal to g, the acceleration due to gravity.

2.13.

Model: We are using the particle model for the skater and the kinematics model of motion under constant acceleration. Solve: Since we don’t know the time of acceleration we will use

vf2 = vi2 + 2a ( xf − xi ) ⇒a= Assess:

v −v (6.0 m/s) 2 − (8.0 m/s) 2 = = −2.8 m/s 2 2( xf − xi ) 2(5.0 m) 2 f

2 i

A deceleration of 2.8 m/s2 is reasonable.

2.14.

Model: Visualize:

Solve:

We are assuming both cars are particles.

The Porsche’s time to finish the race is determined from the position equation

1 xP1 = xP0 + vP0 (tP1 − tP0 ) + aP (tP1 − tP0 ) 2 2 1 ⇒ 400 m = 0 m + 0 m + (3.5 m/s 2 )(tP1 − 0 s)2 ⇒ tP1 = 15.1 s 2 The Honda’s time to finish the race is obtained from Honda’s position equation as

1 xH1 = xH0 + vH0 (tH1 − tH0 ) + aH0 (tH1 − tH0 ) 2 2 1 400 m = 50 m + 0 m + (3.0 m/s 2 )(tH1 − 0 s)2 ⇒ tH1 = 15.3 s 2 So, the Porsche wins.

2.15.

Model: Visualize:

Solve:

Represent the spherical drop of molten metal as a particle.

(a) The shot is in free fall, so we can use free fall kinematics with a = − g . The height must be such that

the shot takes 4 s to fall, so we choose t1 = 4 s. Then,

1 1 1 y1 = y0 + v0 (t1 − t0 ) − g (t1 − t0 ) 2 ⇒ y0 = gt12 = (9.8 m/s 2 )(4 s) 2 = 78.4 m 2 2 2 (b) The impact velocity is v1 = v0 − g (t1 − t0 ) = − gt1 = −39.2 m/s. G Assess: Note the minus sign. The question asked for velocity, not speed, and the y-component of v is negative because the vector points downward.

2.16.

Model: Visualize:

Solve:

Assume the ball undergoes free-fall acceleration and that the ball is a particle.

(a) We will use the kinematic equations v = v0 + a (t − t0 ) and

1 y = y0 + v0 (t − t0 ) + a (t − t0 ) 2 2

as follows:

v(at t = 1.0 s) = 19.6 m/s + (−9.8 m/s 2 )(1.0 s − 0 s) = 9.8 m/s y (at t = 1.0 s) = 0 m + (19.6 m/s)(1.0 s − 0 s) + 1/2(−9.8 m/s 2 )(1.0 s − 0 s) 2 = 14.7 m

v(at t = 2.0 s) = 19.6 m/s + (−9.8 m/s 2 )(2.0 s − 0 s) = 0 m/s y (at t = 2.0 s) = 0 m + (19.6 m/s)(2.0 s − 0 s) + 1/2(−9.8 m/s 2 )(2.0 s − 0 s) 2 = 19.6 m v(at t = 3.0 s) = 19.6 m/s + (−9.8 m/s 2 )(3 s − 0 s) = 9.8 m/s

y (at t = 3.0 s) = 0 m + (19.6 m/s)(3.0 s − 0 s) + 1/2(−9.8 m/s 2 )(3.0 s − 0 s)2 = 14.7 m v(at t = 4.0 s) = 19.6 m/s + (−9.8 m/s 2 )(4.0 s − 0 s) = −19.6 m/s y (at t = 4.0 s) = 0 m + (19.6 m/s)(4.0 s − 0 s) + 1/2(−9.8 m/s 2 )(4.0 s − 0 s) 2 = 0 m (b)

Assess: (a) A downward acceleration of 9.8 m/s 2 on a particle that has been given an initial upward velocity of +19.6 m/s will reduce its speed to 9.8 m/s after 1 s and then to zero after 2 s. The answers obtained in this solution are consistent with the above logic.

(b) Velocity changes linearly with a negative uniform acceleration of 9.8 m/s 2 . The position is symmetrical in time around the highest point which occurs at t = 2 s.

2.17.

Model: Visualize:

We represent the ball as a particle.

Solve: Once the ball leaves the student’s hand, the ball undergoes free fall and its acceleration is equal to the acceleration due to gravity that always acts vertically downward toward the center of the earth. According to the constant-acceleration kinematic equations of motion 1 y1 = y0 + v0 Δt + a Δt 2 2

Substituting the known values

−2 m = 0 m + (15 m/s)t1 + (1/2)(−9.8 m/s 2 )t12 The solution of this quadratic equation gives t1 = 3.2 s. The other root of this equation yields a negative value for t1 , which is not valid for this problem. Assess: A time of 3.2 s is reasonable.

2.18.

Model: Visualize:

Solve:

We will use the particle model and the constant-acceleration kinematic equations.

(a) Substituting the known values into y1 = y0 + v0 Δt + 12 a Δt 2 , we get

1 −10 m = 0 m + 20 (m/s)t1 + (−9.8 m/s 2 )t12 2 One of the roots of this equation is negative and is not relevant physically. The other root is t1 = 4.53 s, which is

the answer to part (b). Using v1 = v0 + aΔt , we obtain

v1 = 20(m/s) + (−9.8 m/s 2 )(4.53 s) = −24 m/s (b) The time is 4.5 s. Assess: A time of 4.5 s is a reasonable value. The rock’s velocity as it hits the bottom of the hole has a negative sign because of its downward direction. The magnitude of 24 m/s compared to 20 m/s, when the rock was tossed up, is consistent with the fact that the rock travels an additional distance of 10 m into the hole.

2.19.

Model: Visualize:

We will represent the skier as a particle.

Note that the skier’s motion on the horizontal, frictionless snow is not of any interest to us. Also note that the acceleration parallel to the incline is equal to g sin 10°. Solve: Using the following constant-acceleration kinematic equations,

vf2x = vi2x + 2ax ( xf − xi ) ⇒ (15 m / s) 2 = (3.0 m / s) 2 + 2(9.8 m / s 2 )sin10°( x1 − 0 m) ⇒ x1 = 64 m vfx = vix + ax (tf − ti ) ⇒ (15 m / s) = (3.0 m / s) + (9.8 m / s 2 )(sin10°)t ⇒ t = 7.1 s Assess:

A time of 7.1 s to cover 64 m is a reasonable value.

2.20.

Model: Visualize:

Represent the car as a particle.

Solve: Note that the problem “ends” at a turning point, where the car has an instantaneous speed of 0 m/s before rolling back down. The rolling back motion is not part of this problem. If we assume the car rolls without friction, then we have motion on a frictionless inclined plane with an accleration a = − g sin θ = − g sin 20° = −3.35 m/s 2 . Constant acceleration kinematics gives v12 = v02 + 2a ( x1 − x0 ) ⇒ 0 m 2 /s 2 = v02 + 2ax1 ⇒ x1 = −

v02 (30 m/s) 2 =− = 134 m 2a 2( −3.35 m/s 2 )

Notice how the two negatives canceled to give a positive value for x1. G Assess: We must include the minus sign because the a vector points down the slope, which is in the negative x-direction.

Solve: (a) The position t = 2 s is x2 s = [2(2) 2 − 2 + 1] m = 7 m. (b) The velocity is the derivative v = dx / dt and the velocity at t = 2 s is calculated as follows:

2.21.

v = (4t 2 − 1) m/s ⇒ v2 s = [4(2) − 1] m/s = 7 m/s (c) The acceleration is the derivative a = dv / dt and the acceleration at t = 2 s is calculated as follows:

a = (4) m/s 2 ⇒ a2 s = 4 m/s 2

2.22.

Solve: The formula for the particle’s position along the x-axis is given by tf

xf = xi + ∫ vx dt ti

Using the expression for vx we get 2 xf = xi + ⎡⎣tf3 − ti3 ⎤⎦ 3

ax =

dvx d = (2t 2 m/s) = 4t m/s 2 dt dt

(a) The particle’s position at t = 1 s is x1 s = 1 m + 23 m = 53 m. (b) The particle’s speed at t = 1 s is v1 s = 2 m/s. (c) The particle’s acceleration at t = 1 s is a1 s = 4 m/s 2 .

2.23.

Solve: The formula for the particle’s velocity is given by

vf = vi + area under the acceleration curve between ti and tf For t = 4 s, we get 1 v4 s = 8 m/s + (4 m/s 2 )4 s = 16 m/s 2 Assess: The acceleration is positive but decreases as a function of time. The initial velocity of 8.0 m/s will therefore increase. A value of 16 m/s is reasonable.

2.24.

Solve: (a) Time (s) 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Position (m) −4 −2 0 1.75 3 4 5 6 7

(b)

(c) Δs = s (at t = 1 s) − s (at t = 0 s) = 0 m − (−4 m) = 4 m. (d) Δs = s (at t = 4 s) − s (at t = 2 s) = 7 m − 3 m = 4 m. (e) From t = 0 s to t = 1 s, vs = Δs / Δt = 4 m/s. (f) From t = 2 s to t = 4 s, vs = Δs / Δt = 2 m/s. (g) The average acceleration is a=

Δv 2 m/s − 4 m/s = = −2 m/s 2 Δt 2 s −1 s

2.25.

Solve: (a)

(b) To be completed by student. dx = vx = 2t − 4 ⇒ vx (at t = 1 s) = [2 m/s 2 (1 s) − 4 m/s] = −2 m/s (c) dt (d) There is a turning point at t = 2 s. (e) Using the equation in part (c), vx = 4 m/s = (2t − 4) m/s ⇒ t = 4

Since x = (t 2 − 4t + 2) m, x = 2 m.

(f)

2.26.

Visualize: Please refer to Figure P2.26. Solve: The graph for particle A is a straight line from t = 2 s to t = 8 s. The slope of this line is −10 m/s, which is the velocity at t = 7.0 s. The negative sign indicates motion toward lower values on the x-axis. The velocity of particle B at t = 7.0 s can be read directly from its graph. It is −20 m/s, The velocity of particle C can be obtained from the equation vf = vi + area under the acceleration curve between ti and tf

This area can be calculated by adding up three sections. The area between t = 0 s and t = 2 s is 40 m/s, the area between t = 2 s and t = 5 s is 45 m/s, and the area between t = 5 s and t = 7 s is −20 m/s. We get (10 m/s) + (40 m/s) + (45 m/s) − (20 m/s) = 75 m/s.

2.27. Solve:

Visualize: Please refer to Figure P2.27. (a) We can calculate the position of the particle at every instant with the equation

xf = xi + area under the velocity-versus-time graph between ti and tf The particle starts from the origin at t = 0 s, so xi = 0 m. Notice that the each square of the grid in Figure P2.27 has “area” (5 m/s) × (2 s) = 10 m. We can find the area under the curve, and thus determine x, by counting squares. You can see that x = 35 m at t = 4 s because there are 3.5 squares under the curve. In addition, x = 35 m at t = 8 s because the 5 m represented by the half square between 4 and 6 s is cancelled by the –5 m represented by the half square between 6 and 8 s. Areas beneath the axis are negative areas. The particle passes through x = 35 m at t = 4 s and again at t = 8 s. (b) The particle moves to the right for 0 s ≥ t ≥ 6 s, where the velocity is positive. It reaches a turning point at x = 40 m at t = 6 s. The motion is to the left for t > 6 s. This is shown in the motion diagram below.

2.28. Visualize:

Solve: We will determine the object’s velocity using graphical methods first and then using calculus. Graphically, v(t ) = v0 + area under the acceleration curve from 0 to t. In this case, v0 = 0 m/s. The area at each time t requested is a triangle. t =0 s

v(t = 0 s) = v0 = 0 m/s

t=2s

1 v(t = 2 s) = (2 s)(5 m/s) = 5 m/s 2

t=4s

1 v(t = 4 s) = (4 s)(10 m/s) = 20 m/s 2

t =6 s

1 v(t = 6 s) = (6 s)(10 m/s) = 30 m/s 2

t =8 s

v(t = 8 s) = v (t = 6 s) = 30 m/s

The last result arises because there is no additional area after t = 6 s. Let us now use calculus. The acceleration function a(t) consists of three pieces and can be written: 0 s≤t ≤4 s ⎧ 2.5t ⎪ a (t ) = ⎨−5t + 30 4 s ≤ t ≤ 6 s ⎪ 0 6 s≤t ≤8 s ⎩ These were determined by the slope and the y-intercept of each of the segments of the graph. The velocity function is found by integration as follows: For 0 ≤ t ≤ 4 s, t

t

v(t ) = v(t = 0 s) + ∫ a (t )dt = 0 + 2.5 0

This gives

t2 = 1.25t 2 20

t =0 s

v(t = 0 s) = 0 m / s

t=2s

v(t = 2 s) = 5 m / s

t=4s

v(t = 4 s) = 20 m / s

For 4 s ≤ t ≤ 6 s, t

t ⎡ −5t 2 ⎤ v(t ) = v(t = 4 s) + ∫ a(t )dt = 20 m/s + ⎢ + 30t ⎥ = −2.5t 2 + 30t − 60 4 ⎣ 2 ⎦4

This gives:

t=6s

v(t = 6 s) = 30 m/s

For 6 s ≤ t ≤ 8 s, t

v(t ) = v(t = 6 s) + ∫ a (t )dt = 30 m/s + 0 m/s = 30 m/s 6

This gives: t=8s

v(t = 8 s) = 30 m/s

Assess: graphs.

The same velocities are found using calculus and graphs, but the graphical method is easier for simple

2.29.

Visualize: Please refer to Figure P2.29. Solve: (a) The velocity-versus-time graph is the derivative with respect to time of the distance-versus-time graph. The velocity is zero when the slope of the position-versus-time graph is zero, the velocity is most positive when the slope is most positive, and the velocity is most negative when the slope is most negative. The slope is zero at t = 0, 1 s, 2 s, 3 s, . . . ; the slope is most positive at t = 0.5 s, 2.5 s, . . ; and the slope is most negative at t = 1.5 s, 3.5 s, . . . (b)

2.30. Solve:

Visualize: Please refer to Figure P2.30. (a) We can determine the velocity as follows: vx = vx (at t = ti ) + area under the acceleration-versus-time graph from t = ti to tf

vx (at t = 4 s) = 0 m/s + (−1 m/s 2 )(4 s) = −4 m/s vx (at t = 8 s) = −4 m/s + (3 m/s 2 )(4 s) m/s = 8 m/s vx (at t = 10 s) = 8 m/s + (−2 m/s 2 )(4 s) = 4 m/s

(b) If vx (at t = 0 s) = 2.0 m/s, the entire velocity-versus-time graph will be displaced upward by 2.0 m/s.

2.31.

Solve: (a) The velocity-versus-time graph is given by the derivative with respect to time of the position function: vx =

dx = (6t 2 − 18t )m/s dt

For vx = 0 m/s, there are two solutions to the quadratic equation: t = 0 s and t = 3 s. (b) At the first of these solutions,

x(at t = 0 s) = 2(0 s)3 − 9(0 s) 2 + 12 = 12 m The acceleration is the derivative of the velocity function: ax =

dvx = (12t − 18) m/s 2 ⇒ a (at t = 0 s) = −18 m/s 2 dt

At the second solution,

x(at t = 3 s) = 2(3 s)3 − 9(3 s) 2 + 12 = −15 m

ax (at t = 3 s) = 12(3 s) − 18 = +18 m/s 2

2.32.

Model: Represent the object as a particle. Solve: (a) Known information: x0 = 0 m, v0 = 0 m/s, x1 = 40 m, v1 = 11 m/s, t1 = 5 s. If the acceleration is uniform (constant a), then the motion must satisfy the three equations 1 x1 = at12 ⇒ a = 3.20 m/s 2 2

v1 = at1 ⇒ a = 2.20 m/s 2

v12 = 2ax1 ⇒ a = 1.51 m/s 2

But each equation gives a different value of a. Thus the motion is not uniform acceleration. (b) We know two points on the velocity-versus-time graph, namely at t0 = 0 and t1 = 5 s. What shape does the function have between these two points? If the acceleration was uniform, which it’s not, then the graph would be a straight line. The area under the graph is the displacement Δx. From the figure you can see that Δx = 27.5 m for a straight-line graph. But we know that, in reality, Δx = 40 m. To get a larger Δx, the graph must bulge upward above the straight line. Thus the graph is curved, and it is concave downward.

2.33.

Solve: The position is the integral of the velocity. t1

t1

t1

t0

0

0

x1 = x0 + ∫ vx dt = x0 + ∫ kt 2 dt = x0 + 13 kt 3 = x0 + 13 kt13 We’re given that x0 = −9.0 m and that the particle is at x1 = 9.0 m at t1 = 3.0 s. Thus

9.0 m = ( −9.0 m ) + 13 k (3.0 s)3 = ( −9.0 m ) + k ( 9.0 s3 )

Solving for k gives

k = 2.0 m/s3 .

2.34.

Solve: (a) The velocity is the integral of the acceleration.

v1x = v0 x + ∫ ax dt = 0 m/s + ∫ (10 − t ) dt = (10t − 12 t 2 ) = 10t1 − 12 t12 t1

t1

t1

t0

0

0

The velocity is zero when

v1x = 0 m/s = (10t1 − 12 t12 ) = (10 − 12 t1 ) × t1 ⇒ t1 = 0 s or t1 = 20 s The first solution is the initial condition. Thus the particle’s velocity is again 0 m/s at t1 = 20 s. (b) Position is the integral of the velocity. At t1 = 20 s, and using x0 = 0 m at t0 = 0 s, the position is t1

20

t0

0

x1 = x0 + ∫ vx dt = 0 m + ∫

(10t − t ) dt = 5t 1 2 2

2 20 0

− 16 t 3

20 0

= 667 m

2.35.

Model: Represent the ball as a particle. Visualize: Please refer to Figure P2.35. Solve: In the first and third segments the acceleration as is zero. In the second segment the acceleration is negative and constant. This means the velocity vs will be constant in the first two segments and will decrease linearly in the third segment. Because the velocity is constant in the first and third segments, the position s will increase linearly. In the second segment, the position will increase parabolically rather than linearly because the velocity decreases linearly with time.

2.36.

Model: Represent the ball as a particle. Visualize: Please refer to Figure P2.36. The ball rolls down the first short track, then up the second short track, and then down the long track. s is the distance along the track measured from the left end (where s = 0). Label t = 0 at the beginning, that is, when the ball starts to roll down the first short track. Solve: Because the incline angle is the same, the magnitude of the acceleration is the same on all of the tracks.

Assess: Note that the derivative of the s versus t graph yields the vs versus t graph. And the derivative of the vs versus t graph gives rise to the as versus t graph.

2.37.

Model: Represent the ball as a particle. Visualize: Please refer to Figure P2.37. The ball moves to the right along the first track until it strikes the wall, which causes it to move to the left on a second track. The ball then descends on a third track until it reaches the fourth track, which is horizontal. Solve:

Assess: Note that the time derivative of the position graph yields the velocity graph, and the derivative of the velocity graph gives the acceleration graph.

2.38. Solve:

Visualize:

Please refer to Figure P2.38.

2.39. Solve:

Visualize:

Please refer to Figure P2.39.

2.40.

Visualize: Please refer to Figure P2.40. There are four frictionless tracks. Solve: For the first track, as is negative and constant and vs is decreasing linearly. This is consistent with a ball rolling up a straight track but not so far that vs goes to zero. For the second track, as is zero and vs is constant but greater than zero. This is consistent with a ball rolling on a horizontal track. For the third track, as is positive and constant and vs is increasing linearly. This is consistent with a ball rolling down a straight track. For the fourth track, as is zero and vs is constant. This is again consistent with a ball rolling on a horizontal track. The as on the first track has the same absolute value as the as on the third track. This means the slope of the first track up is the same as the slope of the third track down.

2.41.

Model: The plane is a particle and the constant-acceleration kinematic equations hold. Solve: (a) To convert 80 m/s to mph, we calculate 80 m/s × 1 mi/1609 m × 3600 s/h = 179 mph. (b) Using as = Δv / Δt , we have, as (t = 0 to t = 10 s) =

23 m/s − 0 m/s = 2.3 m/s 2 10 s − 0 s

as (t = 20 s to t = 30 s) =

69 m/s − 46 m/s = 2.3 m/s 2 30 s − 20 s

For all time intervals a is 2.3 m/s2. (c) Using kinematics as follows:

vfs = vis + a (tf − ti ) ⇒ 80 m/s = 0 m/s + (2.3 m/s 2 )(tf − 0 s) ⇒ tf = 35 s (d) Using the above values, we calculate the takeoff distance as follows:

1 1 sf = si + vis (tf − ti ) + as (tf − ti ) 2 = 0 m + (0 m/s)(35 s) + (2.3 m/s 2 )(35 s) 2 = 1410 m 2 2 For safety, the runway should be 3 ×1410 m = 4230 m or 2.6 mi. This is longer than the 2.4 mi long runway, so the takeoff is not safe.

2.42. Solve:

Model: (a)

The automobile is a particle.

The acceleration is not constant because the velocity-versus-time graph is not a straight line. (b) Acceleration is the slope of the velocity graph. You can use a straightedge to estimate the slope of the graph at t = 2 s and at t = 8 s. Alternatively, we can estimate the slope using the two data points on either side of 2 s and 8 s. Either way, you need the conversion factor 1 mph = 0.447 m/s from Table 1.4.

ax ( at 2 s ) ≈

vx ( at 4 s ) − vx ( at 0 s ) mph 0.447 m/s = 11.5 × = 5.1 m/s 2 4 s−0 s s 1 mph

ax ( at 8 s ) ≈

vx ( at 10 s ) − vx ( at 6 s ) mph 0.447 m/s = 4.5 × = 2.0 m/s 2 10 s − 6 s s 1 mph

(c) The displacement, or distance traveled, is

Δx = x ( at 10 s ) − x ( at 0 s ) = ∫

10 s 0s

vx dx = area under the velocity curve from 0 s to 10 s

We can approximate the area under the curve as the area of five rectangular steps, each with width Δt = 2 s and height equal to the average of the velocities at the beginning and end of each step. 0 s to 2 s

vavg = 14 mph = 6.26 m/s

area = 12.5 m

2 s to 4 s

vavg = 37 mph = 16.5 m/s

area = 33.0 m

4 s to 6 s 6 s to 8 s

vavg = 53 mph = 23.7 m/s

area = 47.4 m

vavg = 65 mph = 29.1 m/s

area = 58.2 m

8 s to 10 s

vavg = 74 mph = 33.1 m/s

area = 66.2 m

The total area under the curve is ≈ 217 m , so the distance traveled in 10 s is ≈ 217 m .

2.43. Solve:

Model: Represent the car as a particle. (a) First, we will convert units:

60

miles 1 hour 1610 m × × = 27 m/s hour 3600 s 1 mile

The motion is constant acceleration, so v1 = v0 + aΔt ⇒ a =

v1 − v0 (27 m/s − 0 m/s) = = 2.7 m/s 2 10 s Δt

(b) The fraction is a/g = 2.7 / 9.8 = 0.28. So a is 28% of g. (c) The distance is calculated as follows:

1 1 x1 = x0 + v0 Δt + a (Δt ) 2 = a (Δt ) 2 = 1.3 × 102 m = 4.3 × 102 feet 2 2

2.44.

Model: Represent the spaceship as a particle. Solve: (a) The known information is: x0 = 0 m, v0 = 0 m/s, t0 = 0 s, a = g = 9.8 m/s 2 , and v1 = 3.0 × 108 m/s. Constant acceleration kinematics gives v1 = v0 + aΔt ⇒ Δt = t1 =

v1 − v0 = 3.1 × 107 s a

The problem asks for the answer in days, so we need a conversion: t1 = (3.1× 107 s) ×

1 hour 1 day × = 3.6 × 102 days 3600 s 24 hour

(b) The distance traveled is 1 1 x1 − x0 = v0 Δt + a (Δt ) 2 = at12 = 4.6 × 1015 m 2 2

(c) The number of seconds in a year is

1 year = 365 days ×

24 hours 3600 s × = 3.15 × 107 s 1 day 1 hour

In one year light travels a distance

1 light year = (3.0 × 108m/s)(3.15 × 107 s) = 9.46 × 1015 m The distance traveled by the spaceship is 4.6 × 1015m/9.46 × 1015m = 0.49 of a light year. Assess:

Note that x1 gives “Where is it?” rather than “How far has it traveled?” “How far” is represented by

x1 − x0 . They happen to be the same number in this problem, but that isn’t always the case.

2.45.

Model: Visualize:

Solve:

The car is a particle, and constant-acceleration kinematic equations hold.

This is a two-part problem. During the reaction time, we can use

1 x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 2 = 0 m + (20 m/s)(0.50 s − 0 s) + 0 m = 10 m During deceleration,

v22 = v12 + 2a1 ( x2 − x1 )

0 = (20 m/s) 2 + 2(−6.0 m/s 2 )( x2 − 10 m) ⇒ x2 = 43 m

She has 50 m to stop, so she can stop in time.

2.46.

Model: Visualize:

Solve:

The car is a particle and constant-acceleration kinematic equations hold.

(a) This is a two-part problem. During the reaction time,

x1 = x0 + v0 (t1 − t0 ) + 1/2a0 (t1 − t0 ) 2 = 0 m + (20 m/s)(0.50 s − 0 s) + 0 m = 10 m After reacting, x2 − x1 = 110 m − 10 m = 100 m, that is, you are 100 m away from the intersection. (b) To stop successfully,

v22 = v12 + 2a1 ( x2 − x1 ) ⇒ (0 m/s) 2 = (20 m/s) 2 + 2a1 (100 m) ⇒ a1 = −2 m/s 2 (c) The time it takes to stop once the brakes are applied can be obtained as follows:

v2 = v1 + a1 (t2 − t1 ) ⇒ 0 m/s = 20 m/s + (−2 m/s 2 )(t2 − 0.50 s) ⇒ t2 = 11 s The total time to stop since the light turned red is 11.5 s.

2.47.

Model: Visualize:

Solve:

(a)

To

We will use the particle model and the constant-acceleration kinematic equations.

find

x2 ,

we

first

need

to

determine

x1.

Using

x1 = x0 + v0 (t1 − t0 ),

we

get

x1 = 0 m + (20 m/s)(0.50 s − 0 s) = 10 m. Now,

v22 = v12 + 2a1 ( x2 − x1 ) ⇒ 0 m 2 /s 2 = (20 m/s) 2 + 2(−10 m/s 2 )( x2 − 10 m) ⇒ x2 = 30 m The distance between you and the deer is ( x3 − x2 ) or (35 m − 30 m) = 5 m. (b) Let us find v0 max such that v2 = 0 m/s at x2 = x3 = 35 m. Using the following equation,

v22 − v02 max = 2a1 ( x2 − x1 ) ⇒ 0 m 2 /s 2 − v02 max = 2(−10 m/s 2 )(35 m − x1 ) Also, x1 = x0 + v0 max (t1 − t0 ) = v0 max (0.50 s − 0 s) = (0.50 s)v0 max . Substituting this expression for x1 in the above equation yields

−v02 max = (−20 m/s 2 )[35 m − (0.50 s) v0 max ] ⇒ v02 max + (10 m/s)v0 max − 700 m 2 /s 2 = 0 The solution of this quadratic equation yields v0 max = 22 m/s. (The other root is negative and unphysical for the present situation.) Assess: An increase of speed from 20 m/s to 22 m/s is very reasonable for the car to cover an additional distance of 5 m with a reaction time of 0.50 s and a deceleration of 10 m/s2.

2.48.

Model: Visualize:

The car is represented as a particle.

Solve: (a) This is a two-part problem. First, we need to use the information given to determine the acceleration during braking. Second, we need to use that acceleration to find the stopping distance for a different initial velocity. First, the car coasts at constant speed before braking:

x1 = x0 + v0 (t1 − t0 ) = v0t1 = (30 m/s)(0.5 s) = 15 m Then, the car brakes to a halt. Because we don’t know the time interval, use

v22 = 0 = v12 + 2a1 ( x2 − x1 ) v12 (30 m/s) 2 =− = −10 m/s 2 2( x2 − x1 ) 2(60 m − 15 m) G We used v1 = v0 = 30 m/s. Note the minus sign, because a1 points to the left. ⇒ a1 = −

We can repeat these steps now with v0 = 40 m/s. The coasting distance before braking is x1 = v0t1 = (40 m/s)(0.5 s) = 20 m

The position x2 after braking is found using

v22 = 0 = v12 + 2a1 ( x2 − x1 ) ⇒ x2 = x1 −

v12 (40 m/s) 2 = 20 m − = 100 m 2a1 2(−10 m/s 2 )

(b) The car coasts at a constant speed for 0.5 s, traveling 20 m. The graph will be a straight line with a slope of 40 m/s. For t ≥ 0.5 the graph will be a parabola until the car stops at t2.We can find t2 from

v2 = 0 = v1 + a1 (t2 − t1 ) ⇒ t2 = t1 −

v1 = 4.5 s a1

The parabola will reach zero slope (v = 0 m/s) at t = 4.5 s. This is enough information to draw the graph shown in the figure.

2.49.

Model: Visualize:

The rocket is represented as a particle.

Solve: (a) There are three parts to the motion. Both the second and third parts of the motion are free fall, with a = − g . The maximum altitude is y2.. In the acceleration phase:

1 1 1 y1 = y0 + v0 (t1 − t0 ) + a (t1 − t0 ) 2 = at12 = (30 m/s 2 )(30 s) 2 = 13,500 m 2 2 2 v1 = v0 + a (t1 − t0 ) = at1 = (30 m/s 2 )(30 s) = 900 m/s In the coasting phase,

v22 = 0 = v12 − 2 g ( y2 − y1 ) ⇒ y2 = y1 +

v12 (900 m/s) 2 = 13,500 m + = 54,800 m = 54.8 km 2g 2(9.8 m/s 2 )

The maximum altitude is 54.8 km (≈ 33 miles). (b) The rocket is in the air until time t3 . We already know t1 = 30 s. We can find t2 as follows:

v2 = 0 m/s = v1 − g (t2 − t1 ) ⇒ t2 = t1 +

v1 = 122 s g

Then t3 is found by considering the time needed to fall 54,800 m:

1 1 2 y2 y3 = 0 m = y2 + v2 (t3 − t2 ) − g (t3 − t2 ) 2 = y2 − g (t3 − t2 ) 2 ⇒ t3 = t2 + = 228 s 2 2 g (c) The velocity increases linearly, with a slope of 30 (m/s)/s, for 30 s to a maximum speed of 900 m/s. It then begins to decrease linearly with a slope of −9.8(m/s)/s. The velocity passes through zero (the turning point at

y2 ) at t2 = 122 s. The impact velocity at t3 = 228 s is calculated to be v3 = v2 − g (t3 − t2 ) = −1040 m/s.

Assess: In reality, friction due to air resistance would prevent the rocket from reaching such high speeds as it falls, and the acceleration upward would not be constant because the mass changes as the fuel is burned, but that is a more complicated problem.

2.50.

Model: Visualize:

Solve:

We will model the rocket as a particle. Air resistance will be neglected.

(a) Using the constant-acceleration kinematic equations,

v1 = v0 + a0 (t1 − t0 ) = 0 m/s + a0 (16 s − 0 s) = a0 (16 s) 1 1 y1 = y0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = a0 (16 s − 0 s) 2 = a0 (128 s 2 ) 2 2 1 y2 = y1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 2 1 2 ⇒ 5100 m = 128 s a0 + 16 s a0 (20 s − 16 s) + ( − 9.8 m/s 2 )(20 s − 16 s) 2 ⇒ a0 = 27 m/s 2 2 (b) The rocket’s speed as it passes through a cloud 5100 m above the ground can be determined using the kinematic equation:

v2 = v1 + a1 (t2 − t1 ) = (16 s) a0 + (−9.8 m/s 2 )(4 s) = 4.0 × 102 m/s Assess: 400 m/s ≈ 900 mph, which would be the final speed of a rocket that has been accelerating for 20 s at a rate of approximately 20 m/s2 or 66 ft/s2.

2.51.

Model: equations. Visualize:

We will model the lead ball as a particle and use the constant-acceleration kinematic

Note that the particle undergoes free fall until it hits the water surface. Solve: The kinematics equation y1 = y0 + v0 (t1 − t0 ) + 12 a0 (t1 − t0 ) 2 becomes

1 −5.0 m = 0 m + 0 m + (−9.8 m/s 2 )(t1 − 0) 2 ⇒ t1 = 1.01 s 2 Now, once again,

1 y2 = y1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 2 ⇒ y2 − y1 = v1 (3.0 s − 1.01 s) + 0 m/s = 1.99 v1 v1 is easy to determine since the time t1 has been found. Using v1 = v0 + a0 (t1 − t0 ) , we get

v1 = 0 m/s − (9.8 m/s 2 )(1.01 s − 0 s) = − 9.898 m/s With this value for v1 , we go back to: y2 − y1 = 1.99v1 = (1.99)(−9.898 m/s) = −19.7 m y2 − y1 is the displacement of the lead ball in the lake and thus corresponds to the depth of the lake. The negative sign shows the direction of the displacement vector. Assess: A depth of about 60 ft for a lake is not unusual.

2.52.

Model: Visualize:

Solve:

The elevator is a particle moving under constant-acceleration kinematic equations.

(a) To calculate the distance to accelerate up:

(v1 ) 2 = v02 + 2a0 ( y0 − y0 ) ⇒ (5 m/s) 2 = (0 m/s) 2 + 2(1 m/s 2 )( y1 − 0 m) ⇒ y1 = 12.5 m (b) To calculate the time to accelerate up:

v1 = v0 + a0 (t1 − t0 ) ⇒ 5 m/s = 0 m/s + (1 m/s 2 )(t1 − 0 s) ⇒ t1 = 5 s To calculate the distance to decelerate at the top:

v32 = v22 + 2a2 ( y3 − y2 ) ⇒ (0 m/s) 2 = (5 m/s) 2 + 2(−1 m/s 2 )( y3 − y2 ) ⇒ y3 − y2 = 12.5 m To calculate the time to decelerate at the top: v3 = v2 + a2 (t3 − t2 ) ⇒ 0 m/s = 5 m/s + (−1 m/s 2 )(t3 − t2 ) ⇒ t3 − t2 = 5 s The distance moved up at 5 m/s is y2 − y1 = ( y3 − y0 ) − ( y3 − y2 ) − ( y1 − y0 ) = 200 m − 12.5 m − 12.5 m = 175 m The time to move up 175 m is given by 1 y2 − y1 = v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 ⇒ 175 m = (5 m/s)(t2 − t1 ) ⇒ (t2 − t1 ) = 35 s 2 To total time to move to the top is (t1 − t0 ) + (t2 − t1 ) + (t3 − t2 ) = 5 s + 35 s + 5 s = 45 s Assess: To cover a distance of 200 m at 5 m/s (ignoring acceleration and deceleration times) will require a time of 40 s. This is comparable to the time of 45 s for the entire trip as obtained above.

2.53.

Model: Visualize:

The car is a particle moving under constant-acceleration kinematic equations.

Solve: This is a three-part problem. First the car accelerates, then it moves with a constant speed, and then it decelerates. First, the car accelerates:

v1 = v0 + a0 (t1 − t0 ) = 0 m/s + (4.0 m/s 2 )(6 s − 0 s) = 24 m/s 1 1 x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + (4.0 m/s 2 )(6 s − 0 s) 2 = 72 m 2 2 Second, the car moves at v1: 1 x2 − x1 = v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 = (24 m/s)(8 s − 6 s) + 0 m = 48 m 2 Third, the car decelerates: v3 = v2 + a2 (t3 − t2 ) ⇒ 0 m/s = 24 m/s + (−3.0 m/s 2 )(t3 − t2 ) ⇒ (t3 − t2 ) = 8 s 1 1 x3 = x2 + v2 (t3 − t2 ) + a2 (t3 − t2 ) 2 ⇒ x3 − x2 = (24 m/s)(8 s) + (−3.0 m/s 2 )(8 s) 2 = 96 m 2 2 Thus, the total distance between stop signs is:

x3 − x0 = ( x3 − x2 ) + ( x2 − x1 ) + ( x1 − x0 ) = 96 m + 48m + 72 m = 216 m Assess: A distance of approximately 600 ft in a time of around 10 s with an acceleration/deceleration of the order of 7 mph/s is reasonable.

2.54.

Model: Visualize:

Solve:

The car is a particle moving under constant linear acceleration.

Using the kinematic equation for position:

1 1 x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 ⇒ x1 + 30 m = x1 + v1 (5.0 s − 4.0 s) + (2 m/s 2 )(5.0 s − 4.0 s) 2 2 2 1 ⇒ 30 m = v1 (1.0 s) + (2 m/s 2 )(1.0 s) 2 ⇒ v1 = 29 m/s 2 And 4.0 seconds before:

v1 = v0 + a0 (t1 − t0 ) ⇒ 29 m/s = v0 + (2 m/s 2 )(4.0 s − 0 s) ⇒ v0 = 21 m/s Assess:

21 m/s ≈ 47 mph and is a reasonable value.

2.55.

Model: Santa is a particle moving under constant-acceleration kinematic equations. Visualize: Note that our x-axis is positioned along the incline.

Solve:

Using the following kinematic equation,

v12 = v02 + 2a|| ( x1 − x2 ) = (0 m/s) 2 + 2(4.9 m/s 2 )(10 m − 0 m) ⇒ v1 = 9.9 m/s Assess:

Santa’s speed of 20 mph as he reaches the edge is reasonable.

2.56.

Model: Visualize:

The cars are represented as particles.

Solve: (a) Ann and Carol start from different locations at different times and drive at different speeds. But at time t1 they have the same position. It is important in a problem such as this to express information in terms of positions (that is, coordinates) rather than distances. Each drives at a constant velocity, so using constant velocity kinematics gives xA1 = xA0 + vA (t1 − tA0 ) = vA (t1 − tA0 )

xC1 = xC0 + vC (t1 − tC0 ) = xC0 + vCt1

The critical piece of information is that Ann and Carol have the same position at t1 , so xA1 = xC1. Equating these two expressions, we can solve for the time t1 when Ann passes Carol: vA (t1 − tA0 ) = xC0 + vCt1

⇒ (vA − vC )t1 = xC0 + vAtA0

⇒ t1 =

xC0 + vAtA0 2.4 mi + (50 mph)(0.5 h) = = 2.0 h vA − vC 50 mph − 36 mph

(b) Their position is x1 = xA1 = xC1 = xC0 + vCt1 = 74 miles. (c) Note that Ann’s graph doesn’t start until t = 0.5 hours, but her graph has a steeper slope so it intersects Carol’s graph at t ≈ 2.0 hours.

2.57.

Model: Visualize:

We will use the particle model for the puck.

We can view this problem as two one-dimensional motion problems. The horizontal segments do not affect the motion because the speed does not change. So, the problem “starts” at the bottom of the uphill ramp and “ends” at the bottom of the downhill ramp. At the top of the ramp the speed does not change along the horizontal section. The final speed from the uphill roll (first problem) becomes the initial speed of the downhill roll (second problem). Because the axes point in different directions, we can avoid possible confusion by calling the downhill axis the z-axis and the downhill velocities u. The uphill axis as usual will be denoted by x and the uphill velocities as v. Note that the height information, h = 1 m, has to be transformed into information about positions along the two axes. Solve: (a) The uphill roll has a0 = − g sin 30D = −4.90 m/s 2 . The speed at the top is found from

v12 = v02 + 2a0 ( x1 − x0 )

⇒ v1 = v02 + 2a0 x1 = (5 m/s) 2 + 2(−4.90 m/s 2 )(2.00 m) = 2.32 m/s (b) The downward roll starts with velocity u1 = v1 = 2.32 m/s and a1 = + g sin 20D = 3.35 m/s 2 . Then,

u22 = u12 + 2a1 ( z2 − z1 ) = (2.32 m/s) 2 + 2(3.35 m/s 2 )(2.92 m − 0 m) ⇒ u2 = 5.00 m/s (c) The final speed is equal to the initial speed, so the percentage change is zero! Assess: This result may seem surprising, but can be more easily understood after we introduce the concept of energy. For now, imagine this is a one dimensional vertical problem. The total vertical change in height of the puck is zero. We have already seen how an object with an initial velocity upward has the same velocity in the opposite direction as it passes through that height going down.

2.58.

Model: Visualize:

Solve:

We will model the toy train as a particle.

Using kinematics, 1 x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 2 m + (2.0 m/s)(2.0 s − 0 s) + 0 m = 6.0 m 2

The acceleration can now be obtained as follows:

v22 = v12 + 2a1 ( x2 − x1 ) ⇒ 0 m 2 /s 2 = (2.0 m/s)2 + 2a1 (8.0 m − 6.0 m) ⇒ a1 = −1.0 m/s 2 Assess:

A deceleration of 1 m/s 2 in bringing the toy train to a halt over a distance of 2.0 m is reasonable.

2.59.

Model: Visualize:

Solve:

We will use the particle model and the kinematic equations at constant-acceleration.

To find x2 , let us use the kinematic equation

v22 = v12 + 2a1 ( x2 − x1 ) = (0 m/s) 2 = (50 m/s) 2 + 2(−10 m/s 2 )( x2 − x1 ) ⇒ x2 = x1 + 125 m Since the nail strip is at a distance of 150 m from the origin, we need to determine x1 : x1 = x0 + v0 (t1 − t0 ) = 0 m + (50 m/s)(0.60 s − 0.0 s) = 30 m Therefore, we can see that x2 = (30 + 125) m = 155 m. That is, he can’t stop within a distance of 150 m. He is in jail. Assess: Bob is driving at approximately 100 mph and the stopping distance is of the correct order of magnitude.

2.60.

Model: Visualize:

Solve:

We will use the particle model with constant-acceleration kinematic equations.

The acceleration, being the same along the incline, can be found as

v12 = v02 + 2a ( x1 − x0 ) ⇒ (4.0 m/s)2 = (5.0 m/s)2 + 2a (3.0 m − 0 m) ⇒ a = −1.5 m/s 2 We can also find the total time the puck takes to come to a halt as v2 = v0 + a(t2 − t0 ) ⇒ 0 m/s = (5.0 m/s) + (−1.5 m/s 2 )t2 ⇒ t2 = 3.3 s Using the above obtained values of a and t2 , we can find x2 as follows: 1 1 x2 = x0 + v0 (t2 − t0 ) + a (t2 − t0 ) 2 = 0 m + (5.0 m/s)(3.3 s) + (−1.5 m/s 2 )(3.3 s) 2 = 8.3 m 2 2 That is, the puck goes through a displacement of 8.3 m. Since the end of the ramp is 8.5 m from the starting position x0 and the puck stops 0.2 m or 20 cm before the ramp ends, you are not a winner.

2.61.

Model: Visualize:

We will use the particle model for the skier’s motion ignoring air resistance.

Solve: (a) As discussed in the text, acceleration along a frictionless incline is a = g sin 25°, where g is the acceleration due to gravity. The acceleration of the skier on snow therefore is a = (0.90) g sin 25°. Also since

h/x1 = sin 25°, x = 200 m/ sin 25°. The final velocity can now be determined using kinematics ⎛ 200 m ⎞ v12 = v02 + 2a ( x1 − x0 ) = (0 m/s) 2 + 2(0.90)(9.8 m/s 2 )sin 25° ⎜ − 0 m⎟ sin 25 ° ⎝ ⎠ ⎛ 1 km ⎞⎛ 3600 s ⎞ ⇒ v1 = 59.397 m/s = (59.397 m/s) ⎜ ⎟⎜ ⎟ = 214 km/h ⎝ 1000 m ⎠⎝ 1 h ⎠ (b) The speed lost to air resistance is (214 − 180)/214 × 100% = 16%. Assess: A record of 180 km/h on such a slope in the presence of air resistance makes the obtained speed of 214 km/h (without air resistance) physical and reasonable.

2.62.

Model: We will use the particle model for the motion of the rocks, which move according to constantacceleration kinematic equations. Visualize:

Solve:

(a) For Heather,

1 yH1 = yH0 + vH0 (tH1 − tH0 ) + a0 (tH1 − tH0 ) 2 2 1 ⇒ 0 m = (50 m) + (−20 m/s)(tH1 − 0 s) + (−9.8 m/s 2 )(tH1 − 0 s) 2 2 2 2 ⇒ 4.9 m/s tH1 + 20 m/s tH1 − 50 m = 0 The two mathematical solutions of this equation are −5.83 s and +1.75 s. The first value is not physically acceptable since it represents a rock hitting the water before it was thrown, therefore, tH1 = 1.75 s. For Jerry, 1 yJ1 = yJ0 + vJ0 (tJ1 − tJ0 ) + a0 (tJ1 − tJ0 ) 2 2 1 ⇒ 0 m = (50 m) + (+20 m/s)(tJ1 − 0 s) + (−9.8 m/s 2 )(tJ1 − 0 s) 2 2 Solving this quadratic equation will yield tJ1 = −1.75 s and +5.83 s. Again only the positive root is physically meaningful. The elapsed time between the two splashes is tJ1 − tH1 = 5.83 s − 1.75 s = 4.08 s ≈ 4.1 s. (b) Knowing the times, it is easy to find the impact velocities:

vH1 = vH0 + a0 (tH1 − tH0 ) = (−20 m/s) + (−9.8 m/s)(1.75 s − 0 s) = −37.1 m/s vJ1 = vJ0 + a0 (tJ1 − tJ0 ) = (+20 m/s) + (−9.8 m/s 2 )(5.83 s − 0 s) = −37.1 m/s Assess: The two rocks hit water with equal speeds. This is because Jerry’s rock has the same downward speed as Heather’s rock when it reaches Heather’s starting position during its downward motion.

2.63.

Model: The ball is a particle that exhibits freely falling motion according to the constant-acceleration kinematic equations. Visualize:

Solve: Using the known values, we have v12 = v02 + 2a0 ( y1 − y0 ) ⇒ (−10 m/s) 2 = v02 + 2(−9.8 m/s 2 )(5.0 m − 0 m) ⇒ v0 = 14 m/s

2.64.

Model: Visualize:

Solve:

The car is a particle that moves with constant linear acceleration.

The reaction time is 1.0 s, and the motion during this time is

x1 = x0 + v0 (t1 − t0 ) = 0 m + (20 m/s)(1.0 s) = 20 m During slowing down, 1 x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 = 200 m 2 1 = 20 m + (20 m/s)(15 s − 1.0 s) + a1 (15 s − 1.0 s)2 ⇒ a1 = −1.02 m/s 2 2 The final speed v2 can now be obtained as

v2 = v1 + a1 (t2 − t1 ) = (20 m/s) + (−1.02 m/s 2 )(15 s − 1 s) = 5.7 m/s

2.65. Solve: (a) The quantity

2 P 2(3.6 × 104 W) = = 60 m 2 /s3 . Thus m 1200 kg

vx =

At t = 10 s, vx = (b) With vx =

(60 m s )t 2

3

(60 m s ) (10 s) = 24 m/s ( ≈ 50 mph), and at t = 20 s, v = 35 m/s (≈ 75 mph). 2

3

x

2 P 12 t , we have m

ax = (c) At t = 1 s, ax =

dvx 2 P 1 −12 P = × t = dt m 2 2mt

P (3.6 × 104 W) = = 3.9 m/s 2 . Similarly, at t = 10 s, ax = 1.2 m/s 2 . 2mt 2(1200 kg)(1 s)

(d) Consider the limiting case of very short times. Note that ax → ∞ as t → 0. This is physically impossible for the Alfa Romeo. dx 2 P 12 t and the initial (e) We can use the relationship that vx = and integrate to find x(t ). We have vx = dt m condition xi = 0 at ti = 0. Thus

∫

x

0

and x =

dx =

2 P t 1/ 2 t dt m ∫0

2P t 3 / 2 2 2P 3 / 2 = t m 3/ 2 3 m

(f) Time to travel a distance x is found by solving the above equation for t.

⎡3 m ⎤ t=⎢ x⎥ ⎢⎣ 2 2 P ⎥⎦ For x = 402 m, t = 18.2 s.

2/3

2.66.

Model: Visualize:

Solve:

Both cars are particles that move according to the constant-acceleration kinematic equations.

(a) David’s and Tina’s motions are given by the following equations:

1 xD1 = xD0 + vD0 (tD1 − tD0 ) + aD (tD1 − tD0 ) 2 = vD0tD1 2 1 1 2 xT1 = xT0 + vT0 (tT1 − tT0 ) + aT (tT1 − tT0 ) 2 = 0 m + 0 m + aTtT1 2 2 When Tina passes David the distances are equal and tD1 = tT1 , so we get

1 2 1 2v 2(30 m/s) xD1 = xT1 ⇒ vD0tD1 = aTtT1 ⇒ vD0 = aTtT1 ⇒ tT1 = D0 = = 30 s aT 2 2 2.0 m/s 2 Using Tina’s position equation, 1 2 1 xT1 = aTtT1 = (2.0 m/s 2 )(30 s) 2 = 900 m 2 2

(b) Tina’s speed vT1 can be obtained from

vT1 = vT0 + aT (tT1 − tT0 ) = (0 m/s) + (2.0 m/s 2 )(30 s − 0 s) = 60 m/s Assess: This is a high speed for Tina (~134 mph) and so is David’s velocity (~67 mph). Thus the large distance for Tina to catch up with David (~0.6 miles) is reasonable.

2.67.

Model: Visualize:

We will represent the dog and the cat in the particle model.

Solve: We will first calculate the time tC1 the cat takes to reach the window. The dog has exactly the same time to reach the cat (or the window). Let us therefore first calculate tC1 as follows:

1 xC1 = xC0 + vC0 (tC1 − tC0 ) + aC (tC1 − tC0 ) 2 2 1 2 ⇒ 3.0 m = 1.5 m + 0 m + (0.85 m/s 2 )tC1 ⇒ tC1 = 1.879 s 2 In the time tD1 = 1.879 s, the dog’s position can be found as follows: 1 xD1 = xD0 + vD0 (tD1 − tD0 ) + aD (tD1 − tD0 ) 2 2 1 = 0 m + (1.50 m/s)(1.879 s) + (−0.10 m/s 2 )(1.879 s) 2 = 2.6 m 2 That is, the dog is shy of reaching the cat by 0.4 m. The cat is safe.

2.68.

Model: Visualize:

Solve:

We use the particle model for the large train and the constant-acceleration equations of motion.

Your position after time tY1 is

1 xY1 = xY0 + vY0 (tY1 − tY0 ) + aY (tY1 − tY0 ) 2 2 = 0 m + (8.0 m/s)(tY1 − 0 s) + 0 m = 8.0 tY1 The position of the train, on the other hand, after time tT1 is 1 xT1 = xT0 + vT0 (tT1 − tT0 ) 2 + aT (tT1 − tT0 ) 2 2 1 2 = 30 m + 0 m + (1.0 m/s 2 )(tT1 ) 2 = 30 + 0.5tT1 2 The two positions xY1 and xT1 will be equal at time tY1 ( = tT1 ) if you are able to jump on the back step of the train. That is, 2 2 30 + 0.5tY1 = 8.0tY1 ⇒ tY1 − (16 s)tY1 + 60 s 2 = 0 ⇒ tY1 = 6 s and 10 s

The first time, 6 s, is when you will overtake the train. If you continue to run alongside, the accelerating train will then pass you at 10 s. Let us now see if the first time tY1 = 6.0 s corresponds to a distance before the barrier. From the position equation for you, xY1 = (8.0 m/s)(6.0 s) = 48.0 m. The position equation for the train will yield the same number. Since the barrier is at a distance of 50 m from your initial position, you can just catch the train before crashing into the barrier.

2.69.

Model: Jill and the grocery cart will be treated as particles that move according to the constantacceleration kinematic equations. Visualize:

Solve:

The final position of Jill when the cart is caught is given by 1 1 1 xJ1 = xJ0 + vJ0 (tJ1 − tJ0 ) + aJ0 (tJ1 − tJ0 ) 2 = 0 m + 0 m + aJ0 (tJ1 − 0 s) 2 = (2.0 m/s 2 )tJ12 2 2 2

The cart’s position when it is caught is

1 1 xC1 = xC0 + vC0 (tC1 − tC0 ) + aC0 (tC1 − tC0 ) 2 = 50 m + 0 m + (0.5 m/s 2 )(tC1 − 0 s) 2 2 2 2 = 50 m + (0.25 m/s 2 )tC1 Since xJ1 = xC1 and tJ1 = tC1 , we get

1 2 2 (2.0)tJ12 = 50 s 2 + 0.25tC1 ⇒ 0.75tC1 = 50 s 2 ⇒ tC1 = 8.20 s 2 2 ⇒ xC1 = 50 m + (0.25 m/s 2 )tC1 = 50 m + (0.25 m/s 2 )(8.2 s) 2 = 67.2 m So, the cart has moved 17.2 m.

2.70.

Model: The watermelon and Superman will be treated as particles that move according to constantacceleration kinematic equations. Visualize:

Solve:

The watermelon’s and Superman’s position as they meet each other are

1 yW1 = yW0 + vW0 (tW1 − tW0 ) + aW0 (tW1 − tW0 ) 2 2 1 yS1 = yS0 + vS0 (tS1 − tS0 ) + aS0 (tS1 − tS0 ) 2 2 1 ⇒ yW1 = 320 m + 0 m + (−9.8 m/s 2 )(tW1 − 0 s) 2 2 ⇒ yS1 = 320 m + (−35 m/s)(tS1 − 0 s) + 0 m Because tS1 = t W1 , 2 yW1 = 320 m − (4.9 m/s 2 ) t W1

yS1 = 320 m − (35 m/s) t W1

Since yW1 = yS1 , 2 320 m − (4.9 m/s 2 )tW1 = 320 m − (35 m/s)tW1 ⇒ tW1 = 0 s and 7.1 s

Indeed, t W1 = 0 s corresponds to the situation when Superman arrives just as the watermelon is dropped off the Empire State Building. The other value, tW1 = 7.1 s, is the time when the watermelon will catch up with Superman. The speed of the watermelon as it passes Superman is vW1 = vW0 + aW0 (tW1 − t W0 ) = 0 m/s + (−9.8 m/s 2 )(7.1 s − 0 s) = −70 m/s Note that the negative sign implies a downward velocity. Assess: A speed of 140 mph for the watermelon is understandable in view of the significant distance (250 m) involved in the free fall.

2.71. Model: equations. Visualize:

Treat the car and train in the particle model and use the constant acceleration kinematics

Solve: In the particle model the car and train have no physical size, so the car has to reach the crossing at an infinitesimally sooner time than the train. Crossing at the same time corresponds to the minimum a1 necessary to avoid a collision. So the problem is to find a1 such that x2 = 45 m when y2 = 60 m. The time it takes the train to reach the intersection can be found by considering its known constant velocity.

v0 y = v2 y = 30 m/s =

y2 − y0 60 m = ⇒ t2 = 2.0 s t 2 − t0 t2

Now find the distance traveled by the car during the reaction time of the driver. x1 = x0 + v0 x ( t1 − t0 ) = 0 + ( 20 m/s )( 0.50 s ) = 10 m The kinematic equation for the final position at the intersection can be solved for the minimum acceleration a1. 1 2 x2 = 45 m = x1 + v1x ( t2 − t1 ) + a1 ( t2 − t1 ) 2 1 2 = 10 m + ( 20 m/s )(1.5 s ) + a1 (1.5 s ) 2 ⇒ a1 = 4.4 m/s 2 Assess: The acceleration of 4.4 m/s2 = 2.0 miles/h/s is reasonable for an automobile to achieve. However, you should not try this yourself! Always pay attention when you drive! Train crossings are dangerous locations, and many people lose their lives at one each year.

2.72.

Solve: A comparison of the given equation with the constant-acceleration kinematics equation

1 x1 = x0 + v0 (t1 − t0 ) + ax (t1 − t0 ) 2 2 yields the following information: x0 = 0 m, x1 = 64 m, t0 = 0, t1 = 4 s, and v0 = 32 m/s. (a) After landing on the deck of a ship at sea with a velocity of 32 m/s, a fighter plane is observed to come to a complete stop in 4.0 seconds over a distance of 64 m. Find the plane’s deceleration. (b)

1 (c) 64 m = 0 m + (32 m/s)(4 s − 0 s) + ax (4 s − 0 s) 2 2

64 m = 128 m + (8 s 2 ) ax ⇒ ax = −8 m/s 2

2.73.

Solve: (a) A comparison of this equation with the constant-acceleration kinematic equation 2 (v1y ) 2 = v0y + 2(ay )( y1 − y0 )

yields the following information: y0 = 0 m, y1 = 10 m, a y = −9.8 m/s 2 , and v1 y = 10 m/s. It is clearly a problem of

free fall. On a romantic Valentine’s Day, John decided to surprise his girlfriend, Judy, in a special way. As he reached her apartment building, he found her sitting in the balcony of her second floor apartment 10 m above the first floor. John quietly armed his spring-loaded gun with a rose, and launched it straight up to catch her attention. Judy noticed that the flower flew past her at a speed of 10 m/s. Judy is refusing to kiss John until he tells her the initial speed of the rose as it was released by the spring-loaded gun. Can you help John on this Valentine’s Day? (b)

2 (c) (10 m/s) 2 = v0y − 2(9.8 m/s 2 )(10 m − 0 m) ⇒ v0y = 17.2 m/s

Assess: The initial velocity of 17.2 m/s, compared to a velocity of 10 m/s at a height of 10 m, is very reasonable.

2.74.

Solve: A comparison with the constant-acceleration kinematics equation

(v1x ) 2 = (v0x ) 2 + 2ax ( x1 − x0 ) yields the following quantities: x0 = 0 m, v0 x = 5 m/s, v1x = 0 m/s, and ax = −(9.8 m/s 2 )sin10D. (a) A wagon at the bottom of a frictionless 10° incline is moving up at 5 m/s. How far up the incline does it move before reversing direction and then rolling back down? (b)

(c) (0 m/s) 2 = (5 m/s) 2 − 2(9.8 m/s 2 )sin10°( x1 − 0 m) ⇒ 25(m/s) 2 = 2(9.8 m/s 2 )(0.174) x1 ⇒ x1 = 7.3 m

2.75.

Solve: (a) From the first equation, the particle starts from rest and accelerates for 5 s. The second equation gives a position consistent with the first equation. The third equation gives a subsequent position following the second equation with zero acceleration. A rocket sled accelerates from rest at 20 m/s2 for 5 sec and then coasts at constant speed for an additional 5 sec. Draw a graph showing the velocity of the sled as a function of time up to t = 10 s. Also, how far does the sled move in 10 s? (b)

1 (c) x1 = (20 m/s 2 )(5 s)2 = 250 m 2

v1 = 20 m/s 2 (5 s) = 100 m/s

x2 = 250 m + (100 m/s)(5 s) = 750 m

2.76. Model: Visualize:

Solve:

The masses are particles.

The rigid rod forms the hypotenuse of a right triangle, which defines a relationship between x2 and y1 :

x22 + y12 = L2 . Taking the time derivative of both sides yields 2 x2

dx2 dx + 2 y1 1 = 0 dt dt

dx2 dy and v1y = 1 to write x2v2 x + y1v1 y = 0 . dt dt ⎛ y1 ⎞ y Thus v2 x = − ⎜ ⎟ v1 y . But from the figure, 1 = tan θ ⇒ v2 x = −v1 y tan θ . x 2 ⎝ x2 ⎠

We can now use v2 x =

Assess:

As x2 decreases (v2 x < 0), y1 increases (v1 y > 0), and vice versa.

2.77.

Model: Visualize:

The rocket and the bolt will be represented as particles to investigate their motion.

The initial velocity of the bolt as it falls off the side of the rocket is the same as that of the rocket, that is, vB0 = vR1 and it is positive since the rocket is moving upward. The bolt continues to move upward with a deceleration equal to g = 9.8 m/s 2 before it comes to rest and begins its downward journey. Solve:

To find aR we look first at the motion of the rocket: 1 yR1 = yR0 + vR0 (tR1 − tR0 ) + aR (tR1 − tR0 ) 2 2 1 = 0 m + 0 m/s + aR (4.0 s − 0 s) 2 = 8aR 2

To find aR we must determine the magnitude of yR1 or yB0 . Let us now look at the bolt’s motion: 1 yB1 = yB0 + vB0 (tB1 − tB0 ) + aB (tB1 − tB0 ) 2 2 1 0 = yR1 + vR1 (6.0 s − 0 s) + (−9.8 m/s 2 )(6.0 s − 0 s) 2 2

⇒ yR1 = 176.4 m − (6.0 s) vR1 Since vR1 = vR0 + aR (tR1 − tR0 ) = 0 m/s + 4 aR = 4 aR the above equation for yR1 yields yR1 = 176.4 − 6.0(4aR ). We know from the first part of the solution that yR1 = 8aR . Therefore, 8aR = 176.4 − 24.0aR and hence aR = 5.5 m/s 2 .

2.78.

Model: motion. Visualize:

Solve:

The rocket car is a particle that moves according to the constant-acceleration equations of

This is a two-part problem. For the first part,

1 1 1 x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + a0 (9.0 s − 0 s) 2 = (81 s 2 ) a0 2 2 2 v1 = v0 + a0 (t1 − t0 ) = 0 m/s + a0 (9.0 s − 0 s) = (9.0 s) a0 During the second part of the problem,

1 x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 2 1 1 ⇒ 990 m = (81 s 2 )a0 + (9.0 s)a0 (12 s − 9.0 s) + (−5.0 m/s 2 )(12 s − 9.0 s) 2 2 2 ⇒ a0 = 15 m/s 2 This leads to:

v1 = (9.0 s)a0 = (9.0 s)(15 m/s 2 ) = 135 m/s Using this value of v1 , we can now calculate v2 as follows:

v2 = v1 + a1 (t2 − t1 ) = (135 m/s) + (−5.0 m/s 2 )(12 s − 9.0 s) = 120 m/s That is, the car’s speed as it passes the judges is 120 m/s. Assess: This is a very fast motion (~250 mph), but the acceleration is large and the long burn time of 9 s yields a high velocity.

2.79.

Model: Visualize:

Solve:

Use the particle model.

(a) Substituting into the constant-acceleration kinematic equation

1 10 ⎞ ⎛ x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 ⇒ 100 m = x1 + v1 ⎜ t2 − ⎟ + 0 m 2 3⎠ ⎝ 100 − x1 10 t2 = + v1 3 Let us now find v1 and x1 as follows: ⎛ 10 ⎞ s − 0 s ⎟ = 12 m/s v1 = v0 + a0 (t1 − t0 ) = 0 m/s + (3.6 m/s 2 ) ⎜ 3 ⎝ ⎠ 2

1 1 ⎛ 10 ⎞ x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + (3.6 m/s 2 ) ⎜ s − 0 s ⎟ = 20 m 2 2 ⎝ 3 ⎠ The expression for t2 can now be solved as

t2 =

100 m − 20 m 10 s + = 10 s 12 m/s 3

(b) The top speed = 12 m/s which means v1 = 12 m/s. To find the acceleration so that the sprinter can run the 100-meter dash in 9.9 s, we use

v1 = v0 + a0 (t1 − t0 ) ⇒ 12 m/s = 0 m/s + a0t1 ⇒ t1 =

12 m/s a0

1 1 1 x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + a0t12 = a0t12 2 2 2 Since x2 = x1 + v1 (t2 − t1 ) + 12 a1 (t2 − t1 ) 2 , we get 1 100 m = a0t12 + (12 m/s) (9.9 s − t1 ) + 0 m 2

Substituting the above equation for t1 in this equation, 2

⎛ 12 m/s ⎞ ⎛ 1 ⎞ ⎛ 12 m/s ⎞ 2 100 m = ⎜ ⎟ a0 ⎜ ⎟ + (12 m/s ) ⎜ 9.9 s − ⎟ ⇒ a0 = 3.8 m/s a ⎝ 2 ⎠ ⎝ a0 ⎠ 0 ⎝ ⎠ (c) We see from parts (a) and (b) that the acceleration has to be increased from 3.6 m/s2 to 3.8 m/s2 for the sprint time to be reduced from 10 s to 9.9 s, that is, by 1%. This decrease of time by 1% corresponds to an increase of acceleration by

3.8 − 3.6 × 100% = 5.6% 3.6

2.80.

Solve:

(a) The acceleration is the time derivative of the velocity. dv d ax = x = ⎡⎣ a (1 − e − bt ) ⎤⎦ = abe − bt dt dt With a = 11.81 m/s and b = 0.6887 s −1 , ax = 8.134e −0.6887 t m/s 2 . At the times t = 0 s, 2 s, and 4 s, this has the values 8.134 m/s2, 2.052 m/s2, and 0.5175 m/s2. dx dx (b) Since vx = = a − ae −bt and the initial condition , the position x is the integral of the velocity. With vx = dt dt that xi = 0 m at ti = 0 s ,

∫

x o

t

t

o

o

dx = ∫ a dt − ∫ ae − bt dt

Thus t

a a a t x = at o + e − bt = at + e − bt − b b b o This can be written a little more neatly as a x = ( bt + e− bt − 1) b = 17.15 ( 0.6887t + e−0.6887 t − 1) m (c) By trial and error, t = 9.92 s yields x = 100.0 m. Assess: Lewis’s actual time was 9.93 s.

2.81.

Model: Visualize:

Solve:

We will use the particle-model to represent the sprinter and the equations of kinematics.

Substituting into the constant-acceleration kinematic equations,

1 1 1 1 x1 = x0 + v0 (t1 − t0 ) + a0 (t1 − t0 ) 2 = 0 m + 0 m + a0 (4 s − 0 s) 2 = a0t12 = a0 (4.0 s) 2 2 2 2 2 ⇒ x1 = (8 s 2 ) a0 v1 = v0 + a0 (t1 − t0 ) = 0 m/s + a0 (4.0 s − 0 s) ⇒ v1 = (4.0 s) a0 From these two results, we find that x1 = (2 s)v1. Now,

1 x2 = x1 + v1 (t2 − t1 ) + a1 (t2 − t1 ) 2 2 ⇒ 100 m = (2 s)v1 + v1 (10 s − 4 s) + 0 m ⇒ v1 = 12.5 m/s Assess: Using the conversion 2.24 mph = 1 m/s, v1 = 12.5 m/s = 28 mph. This speed as the sprinter reaches the finish line is physically reasonable.

2.82. Model: Visualize:

Solve:

The balls are particles undergoing constant acceleration.

(a) The positions of each of the balls at t1 is found from kinematics.

1 2 1 gt1 = v0t1 − gt12 2 2 1 1 ( y1 ) B = ( y0 )B + ( v0 y )B t1 − gt12 = h − gt12 2 2 In the particle model the balls have no physical extent, so they meet when ( y1 ) A = ( y1 ) B . This means

( y1 ) A = ( y0 ) A + ( v0 y ) A t1 −

Thus the collision height is ycoll

1 1 h v0t1 − gt12 = h − gt12 ⇒ t1 = 2 2 v0 1 gh 2 . = h − gt12 = h − 2 2v0 2

(b) We need the collision to occur while ycoll ≥ 0 . Thus h−

So hmax =

gh gh 2 2v 2 ⇒ h≤ 0 ≥ 0 ⇒ 1≥ 2 2 2v0 g 2v0

2v0 2 . g

(c) Ball A is at its highest point when its velocity ( v1 y ) = 0 . A

(v ) = (v ) 1y A

0y A

v0 g

h h v0 v2 . Equating these, = ⇒h= 0 . v0 v0 g g Interestingly, the height at which a collision occurs while Ball A is at its highest point is exactly half of

In (a) we found that the collision occurs at t1 = Assess: hmax .

− gt1 ⇒ 0 = v0 − gt1 ⇒ t1 =

2.83.

Model: Visualize:

The space ships are represented as particles.

Solve: The difficulty with this problem is how to describe “barely avoid.” The Klingon ship is moving with constant speed, so its position-versus-time graph is a straight line from xK0 = 100 km. The Enterprise will be decelerating, so its graph is a parabola with decreasing slope. The Enterprise doesn’t have to stop; it merely has to slow quickly enough to match the Klingon ship speed at the point where it has caught up with the Klingon ship. (You do the same thing in your car when you are coming up on a slower car; you decelerate to match its speed just as you come up on its rear bumper.) Thus the parabola of the Enterprise will be tangent to the straight line of the Klingon ship, showing that the two ships have the same speed (same slopes) when they are at the same position. Mathematically, we can say that at time t1 the two ships will have the same position ( xE1 = xK1 ) and the

same velocity (vE1 = vK1 ). Note that we are using the particle model, so the ships have zero length. At time t1, xK1 = xK0 + vK0t1 1 xE1 = vE0t1 + at12 2

vK1 = vK0 vE1 = vE0 + at1

Equating positions and velocities at t1: 1 xK0 + vK0t1 = vE0t1 + at12 2

vK0 = vE0 + at1

We have two simultaneous equations in the two unknowns a and t1. From the velocity equation, t1 = (vK0 − vE0 )/ a Substituting into the position equation gives 2

xK0 = −(vK0 − vE0 ) ⋅

(vK0 − vE0 ) 1 ⎛ (vK0 − vE0 ) ⎞ (vK0 − vE0 ) 2 + a⋅⎜ ⎟ =− a 2 ⎝ a 2a ⎠

⇒a=−

(vK0 − vE0 ) 2 (20,000 m/s − 50,000 m/s) 2 =− = − 4500 m/s 2 2 xK0 2(100,000 m)

Assess: The magnitude of the deceleration is 4500 m/s2, which is a rather extreme ≈460g. Fortunately, the Enterprise has other methods to keep the crew from being killed.

3.1.

Visualize:

G G G G G Solve: (a) To find A + B , we place the tail of vector B on the tip of vector A and connect the tail of vector A G with the tip of vector B. G G G G G G (b) Since A − B = A + (− B) , we place the tail of the vector (− B ) on the tip of vector A and then connect the tail G G of vector A with the tip of vector (− B ) .

3.2.

Visualize:

G G G G G Solve: (a) To find A + B , we place the tail of vector B on the tip of vector A and then connect vector A’s G tail with vector B’s tip. G G G G G G G G (b) To find A − B , we note that A − B = A + (− B) . We place the tail of vector − B on the tip of vector A and G G then connect vector A’s tail with the tip of vector − B.

3.3.

Solve:

Visualize:

G Vector E points to the left and up, so the components Ex and E y are negative and positive,

respectively, according to the Tactics Box 3.1. (a) Ex = − E cosθ

and

E y = E sin θ .

(b) Ex = − E sin φ

and

E y = E cos φ .

Assess: Note that the role of sine and cosine are reversed because we are using a different angle. θ and φ are complementary angles.

G Visualize: The position vector r whose magnitude r is 10 m has an x-component of 6 m. It makes an angle θ with the + x -axis in the first quadrant.

3.4.

Using trigonometry, rx = r cosθ , or 6 m = (10 m)cosθ . This gives θ = 53.1°. Thus the y-component of G the position vector r is ry = r sin θ = (10 m)sin 53.1° = 8 m.

Solve:

Assess: The y-component is positive since the position vector is in the first quadrant.

3.5.

Solve:

Visualize:

The figure shows the components vx and vy, and the angle θ.

We have, v y = −v sin 40°, or −10 m/s = −v sin 40°, or v = 15.56 m/s.

Thus the x-component is vx = v cos 40° = (15.56 m/s ) cos 40° = 12 m/s. Assess: The x-component is positive since the position vector is in the fourth quadrant.

3.6.

Visualize:

We will follow rules in Tactics Box 3.1. G Solve: (a) Vector r points to the right and down, so the components rx and ry are positive and negative, respectively:

rx = r cosθ = (100 m)cos 45° = 70.7 m

ry = − r sin θ = −(100 m)sin 45° = −70.7 m

G (b) Vector v points to the right and up, so the components vx and v y are both positive:

vx = v cosθ = (300 m /s) cos 20° = 282 m/s G (c) Vector a has the following components:

ax = −a cosθ = −(5.0 m/s 2 )cos90° = 0 m/s 2

v y = v sin θ = (300 m/s)sin 20° = 103 m/s

a y = − a sin θ = −(5.0 m/s 2 )sin 90° = −5.0 m/s 2

Assess: The components have same units as the vectors. Note the minus signs we have manually inserted according to Tactics Box 3.1.

3.7.

Visualize:

We will follow the rules given in Tactics Box 3.1. Solve: v y = (5 cm/s)cos90° = 0 cm/s (a) vx = −(5 cm/s)sin 90° = −5 cm/s (b) ax = −(10 m/s 2 )sin 40° = −6.4 m/s 2 (c) Fx = (50 N)sin 36.9° = 30 N

a y = −(10 m/s 2 )cos 40° = −7.7 m/s 2

Fy = (50 N)cos36.9° = 40 N

Assess: The components have the same units as the vectors. Note the minus signs we have manually inserted according to Tactics Box 3.1.

3.8.

Visualize:

G G The components of the vector C and D, and the angles θ are shown. K K Solve: For C we have C x = −(3.15 m)cos15° = −3.04 m and C y = (3.15 m)sin15° = 0.815 m. For D we

have Dx = 25.6sin 30° = 12.8 and Dy = −25.67cos30° = −22.2. G G Assess: The components of the vector C have the same units as C itself. Dx and Dy G are unitless because D is without units. Note the minus signs we have manually inserted

following rules of Tactics Box 3.1.

3.9.

Visualize:

Solve: The magnitude of the vector is E = ( Ex ) 2 + ( E y ) 2 = (125 V/m) 2 + ( −250 V/m) 2 = 280 V/m. In the G K expression for E , the − ˆj and +iˆ means that E is in quadrant IV. The angle θ is below the positive x-axis. We have:

θ = tan −1 Assess:

| Ey | Ex

⎛ 250 V/m ⎞ −1 = tan −1 ⎜ ⎟ = tan 2 = 63.4° 125 V/m ⎝ ⎠

Since | E y | > | Ex | , the angle θ made with the +x-axis is larger than 45°. θ = 45° for | E y | = | Ex | .

3.10.

Visualize:

Solve:

(a) Using the formulas for the magnitude and direction of a vector, we have:

4 4

θ = tan −1 = tan −1 1 = 45°

B = ( − 4) 2 + (4) 2 = 5.7 (b) r = (−2 cm) 2 + (−1 cm) 2 = 2.2 cm

1 2

θ = tan −1 = tan −1 0.5 = 26.6°

100 = tan −1 10 = 84.3° 10 10 (d) a = (10 m/s 2 ) 2 + (20 m/s 2 ) 2 = 22.4 m/s 2 θ = tan −1 = tan −1 0.5 = 26.6° 20 Assess: Note that θ ≤ 45° when | E y | ≤ | Ex |, where θ is the angle made with the x-axis. On the other hand, (c) v = (−10 m/s) 2 + (−100 m/s) 2 = 100.5 m/s

θ > 45° when | E y | > | Ex | .

θ = tan −1

3.11.

Visualize:

Solve:

(a) Using the formulas for the magnitude and direction of a vector, we have: 6 4

θ = tan −1 = tan −1 1.5 = 56.3°

A = (4) 2 + ( −6) 2 = 7.21 (b) r = (50 m) 2 + (80 m) 2 = 94.3 m

⎛ ry ⎞ −1 ⎛ 80 m ⎞ ⎟ = tan ⎜ ⎟ = 58.0° r ⎝ 50 m ⎠ ⎝ x⎠

θ = tan −1 ⎜

40 = tan −1 2 = 63.4° 20 2 (d) a = (2 m/s 2 ) 2 + (−6 m/s 2 ) 2 = 6.3 m/s 2 θ = tan −1 = tan −1 0.33 = 18.4° 6 Assess: Note that the angle θ made with the x-axis is smaller than 45° whenever | E y | < | Ex | , θ = 45° for

(c) v = (−20 m/s) 2 + (40 m/s) 2 = 44.7 m/s

θ = tan −1

| E y | = | Ex |, and θ > 45° for | E y | > | Ex | . In part (d), θ is with the y-axis, where the opposite of this rule applies.

3.12.

Visualize:

G G G G G G G K K We have C = A − B or C = A + (− B), where − B = ( B, direction opposite B ). Look back at Tactics Box 1.2, which shows how to perform vector subtraction graphically. G G G G G Solve: To obtain vector C from A and B, we place the tail of − B on the tip of A, and then use the tip-totail rule of graphical addition.

3.13.

Visualize:

G G G G G The vectors A, B, and C = A + B are shown.

G G G G G (a) We have A = 5iˆ + 2 ˆj and B = −3iˆ − 5 ˆj. Thus, C = A + B = (5iˆ + 2 ˆj ) + (−3iˆ − 5 ˆj ) = 2iˆ − 3 ˆj. G G G (b) Vectors A, B, and C are shown with their tails together. G K (c) Since C = 2iˆ − 3 ˆj = C xiˆ + C y ˆj , Cx = 2, and C y = −3. Therefore, the magnitude and direction of C are Solve:

C = (2) 2 + (−3) 2 = 3.6

θ = tan −1

| Cy |

⎛3⎞ = tan −1 ⎜ ⎟ = 56° below the + x-axis Cx ⎝2⎠

G Assess: The vector C is to the right and down, thus implying a negative y-component and positive xcomponent, as obtained above. Also θ > 45° since | C y | > | Cx | .

3.14.

Visualize:

G G G G G K (a) We have A = 5iˆ + 2 ˆj, B = −3iˆ − 5 ˆj, and − B = +3iˆ + 5 ˆj. Thus, D = A + ( − B) = 8iˆ + 7 ˆj. G G G (b) Vectors A, B and D are shown in the above figure. G G (c) Since D = 8iˆ + 7 ˆj = Dxiˆ + Dy ˆj , Dx = 8 and Dy = 7. Therefore, the magnitude and direction of D are Solve:

D = (8) 2 + (7) 2 = 10.6 Assess:

⎛ Dy ⎞ −1 ⎛ 7 ⎞ ⎟ = tan ⎜ ⎟ = 41° ⎝8⎠ ⎝ Dx ⎠

θ = tan −1 ⎜

Since | Dy | < | Dx | , the angle θ is less than 45°, as it should be.

3.15.

Visualize:

G G G G Solve: (a) We have A = 5iˆ + 2 ˆj and B = −3iˆ − 5 ˆj. This means 2 A = 10iˆ + 4 ˆj and 3B = −9iˆ − 15 ˆj. Hence, G G G E = 2 A + 3B = 1iˆ − 11 ˆj. G G G (b) Vectors A , B, and E are shown in the above figure. G G (c) From the E vector, Ex = 1 and E y = −11 . Therefore, the magnitude and direction of E are

E = (1) 2 + (−11) 2 = 11.05

⎛ Ex ⎞ ⎛1⎞ = tan −1 ⎜ ⎟ = 5.19° right of the − y -axis ⎜ | E | ⎟⎟ ⎝ 11 ⎠ ⎝ y ⎠

φ = tan −1 ⎜

Assess: Note that φ is the angle made with the y-axis, and that is why φ = tan −1 ( Ex /| E y |) rather than

tan −1 (| E y |/ Ex ), which would be the case if φ were the angle made with the x-axis.

3.16.

Visualize:

G G Solve: (a) We have A = 5iˆ + 2 ˆj and B = −3iˆ − 5 ˆj. G G G F = A − 4 B = 17iˆ + 22 ˆj = Fxiˆ + Fy ˆj with Fx = 17 and Fy = 22. G G G (b) The vectors A, B, and F are shown in the above figure. G (c) The magnitude and direction of F are

This

means

F = Fx2 + Fy2 = (17) 2 + (22) 2 = 27.8 ⎛ Fy ⎞ −1 ⎛ 22 ⎞ ⎟ = tan ⎜ ⎟ = 52.3° ⎝ 17 ⎠ ⎝ Fx ⎠

θ = tan −1 ⎜ Assess:

Fy > Fx implies θ > 45°, as is observed.

G − 4 B = +12iˆ + 20 ˆj.

Hence,

3.17.

Solve: A different coordinate system can only mean a different orientation of the grid and a different origin of the grid. (a) False, because the size of a vector is fixed. (b) False, because the direction of a vector in space is independent of any coordinate system. (c) True, because the orientation of the vector relative to the axes can be different.

3.18.

Visualize:

G G In coordinate system I, A = −(4 m) ˆj , so Ax = 0 m and Ay = −4 m. The vector B makes an angle of G 60° counterclockwise from vertical, which makes it have an angle of θ = 30° with the –x-axis. Since B points to the left and up, it has a negative x-component and a positive y-component. That is, G Bx = −(5.0 m)cos30° = −4.3 m and By = + (5.0 m)sin 30° = 2.5 m. Thus, B = −(4.3 m)iˆ + (2.5 m) ˆj. G In coordinate system II, A points to the left and down, and makes an angle of 30° with the –y-axis. Therefore, Ax = −(4.0 m)sin 30° = −2.0 m and Ay = −(4.0)cos30° = −3.5 m. This implies G ˆ ˆ A = −(2.0 m)i − (3.5 m) j. The vector G B makes an angle of 30° with the +y-axis and is to the left and up. This means we have to manually insert a x-component. Bx = − B sin 30° = −(5.0 m)sin 30° = −2.5 m, and minus sign with the G ˆ ˆ By = + B cos30° = (5.0 m)cos30° = 4.3 m. Thus B = −(2.5 m)i + (4.3 m) j. Solve:

3.19.

Visualize:

G Refer to Figure EX3.19. The velocity vector v points west and makes an angle of 30° with

the G –x-axis. v points to the left and up, implying that vx is negative and v y is positive. Solve:

We

have

vx = −v cos30°

v y = + v sin 30° = (100 m/s)sin 30° = 50.0 m/s. G Assess: vx and v y have the same units as v .

= −(100 m/s)cos30°

= −86.6 m/s

and

3.20.

Visualize:

(a)

G G G Solve: (b) The components of the vectors A, B, and C are Ax = ( 3.0 m ) cos 20° = 2.8 m and Ay = − ( 3.0 m ) sin 20°= −1.0 m ; Bx = 0 m and By = 2 m;

Cx = − ( 5.0 m ) cos70° = −1.71 m and C y = − ( 5.0 m ) sin 70°= −4.7 m. This means the vectors can be written, G G G A = (2.8 m)iˆ − (1.0 m) ˆj B = (2.0 m) ˆj C = (−1.71 m)iˆ − (4.7 m) ˆj G G G G (c) We have D = A + B + C = (1.09 m)iˆ − (3.7 m) ˆj. This means D = (1.09 m) 2 + (3.7 m) 2 = 3.9 m

θ = tan −1

G The direction of D is south of east, 7 4 ° b elow the positive x-axis.

3.9 = tan −1 3.58 = 74° 1.09

3.21.

Visualize:

G G G Solve: Using the method of tail-to-tip graphical addition, the diagram shows the resultant for D + E + F in (a), G G G G G the resultant for D + 2 E in (b), and the resultant for D − 2 E + F in (c).

G E = Exiˆ + E y ˆj = 2iˆ + 3 ˆj ,

which means Ex = 2 and E y = 3. Also, G F = Fxiˆ + Fy ˆj = 2iˆ − 2 ˆj , which means Fx = 2 and Fy = −2. G G (a) The magnitude of E is given by E = Ex2 + E y2 = (2) 2 + (3) 2 = 3.6 and the magnitude of F is given by

3.22.

Solve: We

have

F = Fx2 + Fy2 = (2) 2 + (2) 2 = 2.8. G G G G (b) Since E + F = 4iˆ + 1 ˆj , the magnitude of E + F is (4) 2 + (1) 2 = 4.1. G G G G (c) Since − E − 2 F = −(2iˆ + 3 ˆj ) − 2(2iˆ − 2 ˆj ) = −6iˆ + 1 ˆj , the magnitude of − E − 2 F is

(−6) 2 + (1) 2 = 6.1.

G G Solve: We have r = (5iˆ + 4 ˆj )t 2 m. This means that r does not change the ratio of its components as t G G increases, that is, the direction of r is constant. The magnitude of r is given by

3.23.

r = (5t 2 ) 2 + (4t 2 ) 2 m = (6.40t 2 ) m. (a) The particle’s distance from the origin at t = 0 s, t = 2 s, and t = 5 s is 0 m, 25.6 m, and 160 m. G dt 2 G dr (b) The particle’s velocity is v = = (5iˆ + 4 ˆj ) m/s = (5iˆ + 4 ˆj )2t m/s = (10iˆ + 8 ˆj )t m/s dt dt (c) The magnitude of the particle’s velocity is given by v = (10 t ) 2 + (8t ) 2 = 12.8t m/s. The particle’s speed at t = 0 s, t = 2 s, and t = 5 s is 0 m/s, 25.6 m/s , and 64.0 m/s .

3.24.

Visualize:

G G G Solve: (a) Vector C is the sum of vectors A and B, which is obtained using the tip-to-tip rule of graphical addition. Its magnitude is measured to be 4.7 and its angle made with the +x-axis is measured to be 33°. (b) Using the law of cosines, C 2 = A2 + B 2 − 2 AB cos φ , and the geometry of parallelograms, which shows that φ = 180° − (θ B − θ A ) = 180° − (60° − 20°) = 140°, we obtain C = (3) 2 + (2) 2 − 2(3)(2)cos(140°) = 4.71

Using the law of sines: sin α sin140° ⇒ α = 15.8° = 2 4.71 Thus, θC = α + 20° = 35.8°. (c) We have:

Ax = A cos θ A = 3cos 20° = 2.82

Ay = A sin θ A = 3sin 20° = 1.03

Bx = B cos θ B = 2cos60° = 1.00

By = B sin θ B = 2sin 60° = 1.73

This means: Cx = Ax + Bx = 3.82 and C y = Ay + By = 2.76. The magnitude and direction of C are given by

C = Cx2 + C y2 = (3.82)2 + (2.76) 2 = 4.71

⎛ Cy ⎞ −1 ⎛ 2.76 ⎞ ⎟ = tan ⎜ ⎟ = 35.8° C ⎝ 3.82 ⎠ ⎝ x⎠

θC = tan −1 ⎜

Assess: Using the method of vector components and their algebraic addition to find the resultant vector yields the same results as using the graphical addition of vectors.

3.25.

Visualize: Refer to Figure P3.25 in your textbook. G G G G G G G Solve: (a) We are given that A + B + C = −2iˆ with A = 4iˆ, and C = −2 ˆj. This means A + C = 4iˆ − 2 ˆj. Thus, G G G G G G B = ( A + B + C ) − ( A + C ) = (−2iˆ) − (4iˆ − 2 ˆj ) = −6iˆ + 2 ˆj. G (b) We have B = Bxiˆ + By ˆj with Bx = −6 and By = 2. Hence, B = ( −6) 2 + (2) 2 = 6.3

θ = tan −1

By | Bx |

= tan −1

2 = 18° 6

G G Since B has a negative x-component and a positive y-component, the angle θ made by B is with the –x-axis and it is above the –x-axis. Assess: Since | By | < | Bx |, θ < 45° as is obtained above.

3.26.

Visualize:

Solve:

(a) θ E = tan −1 ⎛⎜⎝ 11 ⎞⎟⎠ = 45°

θ F = tan −1 ⎛⎜⎝ 12 ⎞⎟⎠ = 63.4° Thus φ = 180° − θ E − θ F = 71.6° G G (b) From the figure, E = 2 and F = 5. Using

G 2 = E 2 + F 2 − 2 EF cos φ = ( 2) 2 + ( 5) 2 − 2( 2)( 5)cos(180° − 71.6°) ⇒ G = 3.00. sin α sin(180° − 71.6°) = ⇒ α = 45° 2.975 5 G Since θ E = 45° , the angle made by the vector G with the +x-axis is θG = (α + θ E ) = 45° + 45° = 90°. (c) We have

Furthermore, using

Ex = +1.0,

and E y = +1.0

Fx = −1.0,

and Fy = +2.0

Gx = 0.0,

and

⇒G =

( 0.0 )

G y = 3.0 2

+ ( 3.0 ) = 3.0, 2

and θ = tan −1

|G y | |Gx |

⎛ 3.0 ⎞ = tan −1 ⎜ ⎟ = 90° ⎝ 0.0 ⎠

G That is, the vector G makes an angle of 90° with the x-axis. Assess: The graphical solution and the vector solution give the same answer within the given significance of figures.

3.27. Solve:

Visualize: Refer to Figure P3.27. From the rules of trigonometry,

we

have

Ay = 4sin 40° = 2.6. Also, G G G G Since A + B + C = 0,

Ax = 4cos 40° = 3.1 and

Bx = −2cos10° = −1.97 and By = +2sin10° = 0.35. G G G G G ˆ ˆ C = − A − B = (− A) + (− B) = (−3.1i − 2.6 j ) + (+1.97iˆ − 0.35 ˆj ) = −1.1iˆ − 3.0 ˆj.

3.28.

Visualize:

D u

G G In the tilted coordinate system, the vectors A and B are expressed as: G G A = (2sin15° m)iˆ + (2cos15° m) ˆj and B = (4cos15° m)iˆ − (4sin15° m) ˆj. G G G Therefore, D = 2 A + B = (4 m)[(sin15° + cos15°)iˆ + (cos15° − sin15°) ˆj ] = (4.9 m)iˆ + (2.9 m) ˆj. The magnitude of Solve:

this vector is D = 5.7 m, and it makes an angle of θ = tan −1 (2.9 m/4.9 m) = 31° with the +x-axis. Assess: The resultant vector can be obtained graphically by using the rule of tail-to-tip addition.

3.29.

Visualize:

G The magnitude of the unknown vector is 1 and its direction is along iˆ + ˆj . Let A = iˆ + ˆj as shown in the G G diagram. That is, A = 1iˆ + 1 ˆj and the x- and y-components of A are both unity. Since θ = tan −1 ( Ay / Ax ) = 45°, the unknown vector must make an angle of 45° with the +x-axis and have unit magnitude. G Solve: Let the unknown vector be B = Bxiˆ + By ˆj where

Bx = B cos 45° =

1 B 2

and By = B sin 45° =

G We want the magnitude of B to be 1, so we have

1 B 2

B = Bx2 + By2 = 1 2

2

⎛ 1 ⎞ ⎛ 1 ⎞ ⇒ ⎜ B⎟ + ⎜ B ⎟ = 1 ⇒ B2 = 1 ⇒ B = 1 ⎝ 2 ⎠ ⎝ 2 ⎠ Hence, Bx = By = Finally,

1 2

G 1 ˆ 1 ˆ B = Bxiˆ + By ˆj = i+ j 2 2

3.30.

Model: Carlos will be represented as a particle and the particle model will be used for motion under constant acceleration kinetic equations. Visualize:

Solve: Carlos runs at constant speed without changing direction. The total distance he travels is found from kinematics:

r1 = r0 + v0 Δt = 0 m + ( 5 m/s )( 600 s ) = 3000 m G This displacement is north of east, or θ = 25° from the +x-axis. Thus the position r1 becomes G r1 = (3000 m)(cos 25°iˆ + sin 25° ˆj ) = 2.7 km iˆ + 1.27 km ˆj

That is, Carlos ends up 1.27 km north of his starting position. Assess: The choice of our coordinate system is such that the x-component of the displacement is along the east and the y-component is along the north. The displacement of 3.0 km is reasonable for Carlos to run in 10 minutes if he is an athlete.

3.31. Visualize: The coordinate system (x,y,z) is shown here. While +x denotes east and +y denotes north, G G G G the +z-direction is vertically up. The vectors S morning (shortened as S m ), Safternoon (shortened as Sa ), and the total G G G displacement vector S total = Sa + S m are also shown.

G G Solve: S m = (2000iˆ + 3000 ˆj + 200kˆ) m, and Sa = (−1500iˆ + 2000 ˆj − 300kˆ) m. The total displacement is the sum of the individual displacements. (a) The sum of the z-components of the afternoon and morning displacements is

Sza + Szm = −300 m + 200 m = −100 m, that is, 100 m lower. G G G (b) S total = Sa + S m = (500iˆ + 5000 ˆj − 100kˆ) m, that is, (500 m east) + (5000 m north) – (100 m vertical). The magnitude of your total displacement is S total =

( 500 )

2

+ ( 5000 ) + ( −100 ) m = 5.03 km 2

2

3.32.

Visualize:

Only the minute hand is shown in the figure. G G Solve: (a) We have S8:00 = (2.0 cm) ˆj and S8:20 = (2.0 cm)cos30°iˆ − (2.0 cm)sin 30° ˆj. The displacement vector is G G G Δr = S8:20 − S8:00 = (2.0 cm)[cos30°iˆ − (sin 30° + 1) ˆj ]

= (2.0 cm)[0.87iˆ − 1.50 ˆj ] = (1.74 cm)iˆ − (3.00 cm) ˆj

G G G G G (b) We have S8:00 = (2.0 cm) ˆj and S9:00 = (2.0 cm) ˆj. The displacement vector is Δr = S9:00 − S8:00 = 0. Assess: The displacement vector in part (a) has a positive x-component and a negative y-component. The vector thus is to the right and points down, in quadrant IV. This is where the vector drawn from the tip of the 8:00 a.m. arm to the tip of the 8:20 a.m. arm will point.

3.33.

Visualize:

(a)

Note that +x is along the east and +y is along the north. G G Solve: (b) We are given A = −(200 m) ˆj , and can use trigonometry to obtain B = −(283 m)iˆ − (283 m) ˆj and G G G G G C = (100 m)iˆ + (173 m) ˆj. We want A + B + C + D = 0. This means G G G G D = −A − B − C = (200 m ˆj ) + (283 m iˆ + 283 m ˆj ) + (−100 m iˆ − 173 m ˆj ) = 183 m iˆ + 310 m ˆj K The magnitude and direction of D are

D = (183 m) 2 + (310 m) 2 = 360 m and θ = tan −1

Dy Dx

⎛ 310 m ⎞ = tan −1 ⎜ ⎟ = 59.4° ⎝ 183 m ⎠

G This means D = (360 m, 59.4° north of Geast). (c) The measuredGlength of the vector D on the graph (with a ruler) is approximately 1.75 times the measured length of vector A . Since A = 200 m, this gives D = 1.75 × 200 m = 350 m. Similarly, the angle θ measured with the protractor is close to 60°. These answers are in close agreement to part (b).

3.34.

Visualize:

(a) The figure shows Sparky’s individual displacements and his net displacement.

G G G G (b) Dnet = D1 + D2 + D3 , where individual displacements are G D1 = (50cos 45°iˆ + 50sin 45° ˆj ) m = (35.4iˆ + 35.4 ˆj ) m G D2 = −70iˆ m G D3 = −20 ˆj m G Thus Sparky’s displacement is Dnet = ( −35iˆ + 15.4 ˆj ) m. (c) As a magnitude and angle, Solve:

Dnet = ( Dnet ) 2x + ( Dnet ) 2y = 38 m,

⎛ ( Dnet ) y ⎞ ⎟ = 24° ⎝ | ( Dnet ) x | ⎠

θ net = tan −1 ⎜

Sparky’s net displacement is 38 m in a direction 24° north of west.

3.35.

Visualize:

Solve:

G G G We are given A = 5 m iˆ and C = (−1 m)kˆ. Using trigonometry, B = (3cos 45° m)iˆ − (3sin 45° m) ˆj. The G G G G G is displace-ment is r = A + B + C = (7.12 m)iˆ − (2.12 m) ˆj − (1 m)kˆ. The magnitude of r

total

r = (7.12)2 + (2.12) 2 + (1)2 m = 7.5 m. Assess: A displacement of 7.5 m is a reasonable displacement.

3.36.

Visualize:

Solve:

We

have

G v = vxiˆ + v y ˆj

= v||iˆ + v⊥ ˆj

= v cosθ iˆ + v sin θ ˆj.

Thus,

v|| = v cosθ = (100 m/s)cos30° = 86.6 m/s.

Assess:

For the small angle of 30°, the obtained value of 86.6 m/s for the horizontal component is reasonable.

3.37.

Visualize:

⎛ 2.5 m/s ⎞ (a) Since vx = v cosθ , we have 2.5 m/s = (3.0 m/s)cosθ ⇒ θ = cos −1 ⎜ ⎟ = 34°. ⎝ 3.0 m/s ⎠ (b) The vertical component is v y = v sin θ = (3.0 m/s) sin 34° = 1.7 m/s.

Solve:

3.38.

Visualize:

The coordinate system used here is tilted with x-axis along the slope. Solve: The component of the velocity parallel to the x-axis is v|| = −v cos70° = −v sin 20° = −10 m/s (0.34) = −3.4 m/s. This is the speed down the slope. The component of the velocity perpendicular v⊥ = −v sin 70° = −v cos 20° = −10 m/s (0.94) = −9.4 m/s. This is the speed toward the ground. Assess: A speed of approximately 10 m/s implies a fall time of approximately 1 second under free fall. Note that g = –9.8 m/s2. This time is reasonable for a drop of approximately 5 m, or 16 feet.

3.39.

Visualize:

Solve: (a) The river is 100 m wide. If Mary rows due north at a constant speed of 2.0 m/s, it will take her 50 s to row across. But while she’s doing so, the current sweeps her boat sideways a distance 1 m/s × 50 s = 50 m. Mary’s net displacement is the vector sum of the displacement due to her rowing plus the displacement due to the river’s current. She lands 50 m east of the point that was directly across the river from her when she started. (b) Mary’s net displacement is shown on the figure.

3.40. G

Visualize: Establish a coordinate system with origin at the tree and with the x-axis pointing east. Let G G A be a displacement vector directly from the tree to the treasure. Vector A is A = (100iˆ + 500 ˆj ) paces. This describes the displacement you would undergo by walking north 500 paces, then east 100 paces. Instead, you follow the road for 300 paces and undergo displacement G B = (300sin 60°iˆ + 300cos60° ˆj ) paces = (260iˆ + 150 ˆj ) paces

G G G G Solve: Now let C be the displacement vector from your position to the treasure. From the figure A = B + C. G G G So the displacement you need to reach the treasure is C = A − B = (−160iˆ + 350 ˆj ) paces. G If θ is the angle measured between C and the y-axis, ⎛ 160 ⎞ ⎟ = 24.6° ⎝ 350 ⎠

θ = tan −1 ⎜

You should head 24.6° west of north. You need to walk distance C = Cx2 + C y2 = 385 paces to get to the

treasure.

3.41. Visualize: A 3% grade rises 3 m for every 100 m horizontal distance. The angle of the ground is thus α = tan −1 (3/100) = tan −1 (0.03) = 1.72°. Establish a tilted coordinate system with one axis parallel to the ground and the other axis perpendicular to the ground.

G Solve: From the figure, the magnitude of the component vector of v perpendicular to the ground is v⊥ = v sin α = 15.0 m/s. But this is only the size. We also have to note that the direction of vG⊥ is down, so the

component is

v⊥ = −15.0 m/s.

3.42.

Visualize:

G G G G G The resulting velocity is given by v = vfly + vwind , where vwind = 6 m/s iˆ and vfly = −v sin θ iˆ − v cosθ ˆj. G Substituting the known values we get v = −8 m/s sin θ iˆ − 8 m/s cosθ ˆj + 6 m/s iˆ. We need to have vx = 0. This

Solve:

means 0 = −8 m/s sin θ + 6 m/s, so sin θ =

6 and θ = 48.6°. Thus the ducks should head 48.6° west of south. 8

3.43.

Model: Visualize:

Solve:

The car is treated as a particle in this problem.

(a) The tangential component is a|| = a sin 30° = (2.0 m/s 2 )(0.5) = 1.0 m/s 2 .

(b) The perpendicular component is a⊥ = a cos30° = (2.0 m/s 2 )(0.866) = 1.7 m/s 2 . Assess: Magnitudes of the tangential and perpendicular components of acceleration are reasonable.

3.44.

Model: Visualize:

We will treat the knot in the rope as a particle in static equilibrium.

G G Solve: Expressing the vectors using unit vectors, we have F1 = 3.0iˆ and F2 = −5.0sin 30°iˆ + 5.0cos30° ˆj. Since G G G G G G G G F1 + F2 + F3 = 0, we can write F3 = − F1 − F2 = −0.5iˆ − 4.33 ˆj . The magnitude of F3 is given by G F3 = (−0.5) 2 + (−4.33) 2 = 4.4 units. The angle F3 makes is θ = tan −1 (4.33/0.5) = 83° and is below the negative xaxis. Assess: The resultant vector has both components negative, and is therefore in quadrant III. Its magnitude and G direction are reasonable. Note the minus sign that we have manually inserted with the force F2 .

3.45.

Visualize:

Use a tilted coordinate system such that x-axis is down the slope. G G Solve: Expressing all three forces in terms of unit vectors, we have F1 = −(3.0 N)iˆ, F2 = + (6.0 N) ˆj, and G F3 = (5.0 N)sin θ iˆ − (5.0 N)cosθ ˆj. G (a) The component of Fnet parallel to the floor is ( Fnet ) x = −(3.0 N) + 0 N + (5.0 N)sin 30° = −0.50 N, or 0.50 N up the slope. G (b) The component of Fnet perpendicular to the floor is ( Fnet ) y = 0 N + (6.0 N) − (5.0 N)cos30° = 1.67 N. G G (c) The magnitude of Fnet is Fnet = ( Fnet ) x + ( Fnet ) y = (−0.50 N) 2 + (1.67 N) 2 = 1.74 N. The angle Fnet makes

is (F ) y ⎛ 1.67 N ⎞ = tan −1 ⎜ ⎟ = 73° |( Fnet ) x | ⎝ 0.50 N ⎠ G G Fnet is 73° above the floor on the left side of F2 . 00000

φ = tan −1 Gnet

3.46.

Visualize:

Using trigonometry to calculate θ, we get θ = tan −1 (100 cm/141 cm) = 35.3°. Expressing the three forces in unit G G G vectors, FB = −(3.0 N)iˆ, FC = −(6.0 N) ˆj , and FD = + (2.0 N)cos35.3iˆ − (2.0 N)sin 35.3 ˆj = (1.63 N)iˆ − (1.16 N) ˆj. The total G G G G G force is Fnet = FB + FC + FD = −1.37 N iˆ − 7.2 N ˆj. The magnitude of Fnet is Fnet = (1.37 N) 2 + (7.2 N) 2 = 7.3 N. Solve:

θ net = tan −1 G Fnet = (7.3 N,79° below −x in quadrant III).

|( Fnet ) y |

⎛ 7.2 N ⎞ = tan −1 ⎜ ⎟ = 79° |( Fnet ) x | ⎝ 1.37 N ⎠

3-1

4.1.

Solve:

(a)

(b) A race car slows from an initial speed of 100 mph to 50 mph in order to negotiate a tight turn. After making the 90° turn the car accelerates back up to 100 mph in the same time it took to slow down.

4.2.

Solve:

(a)

(b) A car drives up a hill, over the top, and down the other side at constant speed.

4.3.

Solve:

(a)

(b) A ball rolls along a level table at 3 m/s. It rolls over the edge and falls 1 m to the floor. How far away from the edge of the table does it land?

4.4. Solve: (a) The figure shows a motion diagram of a pendulum as it swings from one side to the other. It’s clear that the velocity at the lowest point is not zero. The velocity vector at this point is tangent to the circle. We can use the method of Tactics Box 1.3 to find the acceleration at the lowest point. The acceleration is not zero. Instead, you can see that the acceleration vector points toward the center of the circle.

(b) The end of the arc is like the highest point of a ball tossed straight up. The velocity is zero for an instant as the vector changes from pointing outward to pointing inward. However, the acceleration is not zero at this point. G The velocity is changing at the end point, and this requires an acceleration. The motion diagram shows that Δv , G and thus a , is tangent to the circle at the end of the arc.

4.5.

Model: The boat is treated as a particle whose motion is governed by constant-acceleration kinematic equations in a plane. Visualize:

Solve:

Resolving the acceleration into its x and y components, we obtain G a = ( 0.80 m/s 2 ) cos 40°iˆ + ( 0.80 m/s 2 ) sin 40° ˆj = ( 0.613 m/s 2 ) iˆ + ( 0.514 m/s 2 ) ˆj G G G From the velocity equation v1 = v0 + a ( t1 − t0 ) , G v1 = ( 5.0 m/s ) iˆ + ⎡⎣( 0.613 m/s 2 ) iˆ + ( 0.514 m/s 2 ) ˆj ⎤⎦ ( 6 s − 0 s ) = ( 8.68 m/s ) iˆ + ( 3.09 m/s ) ˆj G The magnitude and direction of v are

( 8.68 m/s )

v=

2

+ ( 3.09 m/s ) = 9.21 m/s 2

⎛ v1 y ⎞ −1 ⎛ 3.09 m/s ⎞ ⎟ = tan ⎜ ⎟ = 20° north of east v ⎝ 8.68 m/s ⎠ ⎝ 1x ⎠

θ = tan −1 ⎜ Assess:

An increase of speed from 5.0 m/s to 9.21 m/s is reasonable.

G Solve: (a) At t = 0 s, x = 0 m and y = 0 m, or r = (0iˆ + 0ˆj ) m. At t = 4 s, x = 0 m and y = 0 m, or G r = (0iˆ + 0ˆj ) m. In other words, the particle is at the origin at both t = 0 s and at t = 4 s. From the expressions for x and y,

4.6.

⎤ dy ˆ ⎡⎛ 3 2 G dx ⎞ v = iˆ + j = ⎢⎜ t − 4t ⎟ iˆ + ( t − 2 ) ˆj ⎥ m/s dt dt ⎠ ⎣⎝ 2 ⎦ G G At t = 0 s, v = −2ˆj m/s, v = 2 m/s. At t = 4 s, v = 8iˆ + 2ˆj m/s, v = 8.3 m/s. G (b) At t = 0 s, v is along − ˆj , or 90° south of + x. At t = 4 s,

(

)

⎛ 2 m/s ⎞ ⎟ = 14° north of +x ⎝ 8 m/s ⎠

θ = tan −1 ⎜

4.7. Solve:

Visualize: Refer to Figure EX4.7. From the figure, identify the following:

x1 = 0 m

y1 = 0 m

x2 = 2000 m

y2 = 1000 m

v1x = 0 m/s

v1 y = 200 m/s

v2 x = 200 m/s

v2 y = −100 m/s

The components of the acceleration can be found by applying v22 = v12 + 2aΔs for the x and y directions. Thus v22x − v12x ( 200 m/s ) − ( 0 m/s ) = = 10.00 m/s 2 2Δx 2 ( 2000 m − 0 m ) 2

ax =

( −100 m/s ) − ( 200 m/s ) 2 (1000 m − 0 m ) 2

ay =

2

2

= −15.00 m/s 2

G So a = (10.00iˆ − 15.00 ˆj ) m/s 2 . Assess: A time of 20 s is needed to change v1x = 0 m/s to v2 x = 200 m/s at ax = 10 m/s 2 . This is the same time

needed to change v1 y to v2 y at a y = −15 m/s 2 .

4.8.

Model: The puck is a particle and follows the constant-acceleration kinematic equations of motion. Visualize: Please refer to Figure EX4.8. G Solve: (a) At t = 2 s, the graphs give vx = 16 cm/s and v y = 30 cm/s. The angle made by the vector v with the

x-axis can thus be found as ⎛ vy ⎞ −1 ⎛ 30 cm/s ⎞ ⎟ = tan ⎜ ⎟ = 62° above the x-axis v ⎝ 16 cm/s ⎠ ⎝ x⎠

θ = tan −1 ⎜

(b) After t = 5 s, the puck has traveled a distance given by: 5s

x1 = x0 + ∫ vx dt = 0 m + area under vx -t curve = 12 (40 cm/s)(5 s) = 100 cm 0

5s

y1 = y0 + ∫ v y dt = 0 m + area under v y -t curve = (30 cm/s)(5 s) = 150 cm 0

⇒ r1 = x12 + y12 =

(100 cm )

2

+ (150 cm ) = 180 cm 2

4.9.

Model: Use the particle model for the puck. Visualize: Please refer to Figure EX4.9 Solve: (a) Since the vx vs t and v y vs t graphs are straight lines, the puck is undergoing constant acceleration

along the x- and y- axes. The components of the puck’s acceleration are ax =

dvx Δvx ( −10 m/s − 10 m/s) = = = −2.0 m/s 2 dt 10 s − 0 s Δt

ay =

(10 m/s − 0 m/s) = 1.0 m/s 2 (10 s − 0 s)

The magnitude of the acceleration is a = a x2 + a y2 = 2.2 m/s 2 . (b) The puck is undergoing constant acceleration in both the x and y directions. Identify from the graphs vix = 10 m/s,

viy = 0 m/s. Since the puck starts at the origin, xi = yi = 0 m, and set ti = 0 s. Using kinematics, x = 0 m + (10 m/s)t + 12 ( −2.0 m/s 2 )t 2

y = 0 m + 0 m/s + 12 (1.0 m/s 2 )t 2

The distance from the origin at time t is r = x 2 + y 2 . The table below shows the values of x, y, and r at the times t = 0, 5, 10 s. t =0 s 5s 10 s

x 0m 25 m 0m

y 0m 12.5 m 50 m

R 0m 28 m 50 m

Assess: The puck turns around at t = 5 s in the x direction, and constantly accelerates in the y direction. Traveling 50 m from the starting point in 10 s is reasonable.

4.10.

Model: Assume the particle model for the ball, and apply the constant-acceleration kinematic equations of motion in a plane. Visualize:

Solve:

G (a) We know the velocity v1 = (2.0iˆ + 2.0jˆ) m/s at t = 1 s. The ball is at its highest point at t = 2 s, so

v y = 0 m/s. The horizontal velocity is constant in projectile motion, so vx = 2.0 m/s at all times. Thus G v2 = 2.0iˆ m/s at t = 2 s. We can see that the y-component of velocity changed by Δv y = −2.0 m/s. between t = 1 s and t = 2 s. Because a y is constant, v y changes by –2.0 m/s in any 1-s interval. At t = 3 s, v y is 2.0 m/s less than its value of 0 at t = 2 s. At t = 0 s, v y must have been 2.0 m/s more than its value of 2.0 m/s at t = 1 s. Consequently, at t = 0 s,

G v0 = (2.0iˆ + 4.0 jˆ ) m/s

At t = 1 s, G v0 = (2.0iˆ + 2.0 jˆ ) m/s

At t = 2 s, G v0 = (2.0iˆ + 0.0 jˆ ) m/s

At t = 3 s, G v0 = (2.0iˆ − 2.0 jˆ ) m/s (b) Because v y is changing at the rate –2.0 m/s per s, the y-component of acceleration is a y = −2.0 m/s 2 . But

a y = − g for projectile motion, so the value of g on Exidor is g = 2.0 m/s 2 . G (c) From part (a) the components of v0 are v0 x = 2.0 m/s and v0 y = 4.0 m/s. This means

⎛ v0 y ⎞ −1 ⎛ 4.0 m/s ⎞ ⎟ = tan ⎜ ⎟ = 63.4° above +x v ⎝ 2.0 m/s ⎠ ⎝ 0x ⎠

θ = tan −1 ⎜

Assess: The y-component of the velocity vector decreases from 2.0 m/s at t = 1 s to 0 m/s at t = 2 s. This gives an acceleration of −2 m/s 2 . All the other values obtained above are also reasonable.

4.11.

Model: Visualize:

Solve:

The ball is treated as a particle and the effect of air resistance is ignored.

Using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) , 2

50 m = 0 m + (25 m/s)(t1 − 0 s) + 0 m ⇒ t1 = 2.0 s Now, using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , 2

y1 = 0 m + 0 m + 12 ( −9.8 m/s 2 )(2.0 s − 0 s) 2 = −19.6 m Assess: The minus sign with y1 indicates that the ball’s displacement is in the negative y direction or downward. A magnitude of 19.6 m for the height is reasonable.

4.12.

Model: neglected. Visualize:

Solve:

The bullet is treated as a particle and the effect of air resistance on the motion of the bullet is

(a) Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , we obtain 2

(−2.0 × 10−2 m) = 0 m + 0 m + 12 (−9.8 m/s 2 )(t1 − 0 s) 2 ⇒ t1 = 0.0639 s (b) Using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) , 2

(50 m) = 0 m + v0 x (0.0639 s − 0 s) + 0 m ⇒ v0 x = 782 m/s Assess: The bullet falls 2 cm during a horizontal displacement of 50 m. This implies a large initial velocity, and a value of 782 m/s is understandable.

4.13.

Model: We will use the particle model for the food package and the constant-acceleration kinematic equations of motion. Visualize:

Solve: For the horizontal motion, x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 = 0 m + (150 m/s)(t1 − 0 s) + 0 m = (150 m/s)t1 We will determine t1 from the vertical y-motion as follows: y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2

⇒ 0 m = 100 m + 0 m + 12 (−9.8 m/s 2 )t12 ⇒ t1 =

200 m = 4.518 s 9.8 m/s 2

From the above x-equation, the displacement is x1 = (150 m/s)(4.518 s) = 678 m. Assess: The horizontal distance of 678 m covered by a freely falling object from a height of 100 m and with an initial horizontal velocity of 150 m/s (≈ 335 mph) is reasonable.

4.14.

Model: Visualize:

Assume the particle model for the spyglass and use the projectile motion equations.

Solve: (a) The spyglass has an initial horizontal velocity equal to that of the ship. As the spyglass falls, it and the ship move forward together at the same velocity, so the spyglass lands at the bottom of the mast directly vertically below on the ship where the sailor dropped it. (b) The time the spyglass takes to fall can be found by considering the vertical motion:

y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ⇒ 0 m = 15 m + 0 m + 12 (−9.8 m/s 2 )(t1 − 0 s)2 ⇒ t1 = 1.749 s Therefore, x1 = 0 m/s + (4.0 m/s)(1.749 s) = 7.0 m. The spyglass lands 7.0 m to the right of the fisherman.

4.15.

Model:

G G G G G The position vectors r and r ′ in frames S and S′ are related by the equation r = r′ + R, where

G R is the position vector of the origin of frame S′ as measured in frame S. S is Ted’s frame and S′ is Stella’s frame. Visualize:

Solve:

G G The relation between R and the velocity of Stella V is G G R = V = (100 m/s ) cos 45°iˆ − (100 m/s ) sin 45° ˆj 5s

G G G Because r = r′ + R,

G ⎡100 ˆ 100 ⇒ R = ( 5.0 m ) ⎢ i− 2 ⎣ 2

ˆj ⎤ = 353.6iˆ − 353.6 ˆj m ⎥ ⎦

(

)

G G G r ′ = r − R = ( 200 m ) iˆ − ⎡⎣( 353.6 m ) iˆ − ( 353.6 m ) ˆj ⎤⎦ = − (154 m ) iˆ + ( 354 m ) ˆj

G The vector r ′ determines the position of the exploding firecracker as seen by Stella.

4.16.

Model: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the water be S′. Frame S′ moves relative to S with a velocity Vx . Visualize:

Solve: Let vx be the velocity of the boat in frame S, and v′x be the velocity of the boat in S′. Then for travel down the river, 30 km vx = v′x + Vx = = 10.0 km/h 3.0 hr For travel up the river,

⎛ 30 km ⎞ −v′x + Vx = − ⎜ ⎟ = −6.0 km/h ⎝ 5.0 hr ⎠ Adding these two equations yields Vx = 2.0 km/h. That is, the velocity of the flowing river relative to the earth is 2.0 km/h.

4.17.

Motion: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the moving sidewalk be S′. Frame S′ moves relative to S with velocity Vx . Solve:

Let vx be your velocity in frame S and v′x be your velocity in S′. In the first case, when the moving

sidewalk is broken, Vx = 0 m/s and vx W =

( x1 − x0 ) 50 s

In the second case, when you stand on the moving sidewalk, v′x = 0 m/s. Therefore, using vx = v′x + Vx , we get vx S = Vx =

x1 − x0 75 s

In the third case, when you walk while riding, v′x = vx W . Using vx = v′x + Vx , we get

x1 − x0 x1 − x0 x1 − x0 = + ⇒ t = 30 s 50 s 75 s t Assess:

A time smaller than 50 s was expected.

4.18. Model: Let the earth frame be S and a frame attached to the water be S′. Frame S′ moves relative to S with velocity V. We define the x-axis along the direction of east and the y-axis along the direction of north for both frames. The frames S and S′ have their origins coincident at t = 0 s. G G G G G Solve: (a) According to the Galilean transformation of position: r = r ′ + R = r′ + tV . We need to find Mary’s G position vector r from the earth frame S. The observer in frame S′ will observe the boat move straight north G G and will find its position as r ′ = ( 2.0t ) ˆj m/s. We also know that V = ( 3.0 ) iˆ m/s. Since r ′ = 100 m = ( 2.0 m/s ) t and t = 50 s, we have G r = ( 2.0 m/s )( 50 s ) ˆj + ( 50 s )( 3.0 m/s ) t iˆ = (150 m ) iˆ + (100 m ) ˆj

Thus she lands 150 m east of the point that was straight across the river from her when she started. (b)

G G G Note that r ′ is the displacement due to rowing, R is the displacement due to the river’s motion, and r is the net displacement.

Model: Let Susan’s frame be S and Shawn’s frame be S′. S′ moves relative to S with velocity V. Both Susan and Shawn are observing the intersection point from their frames. G G G G Solve: The Galilean transformation of velocity is v = v′ + V , where v is the velocity of the intersection point G G from Susan’s reference frame, v′ is the velocity of the intersection point from Shawn’s frame S′, and V is the G velocity of S′ relative to S or Shawn’s velocity relative to Susan. Because v = − ( 60 mph ) ˆj and G G G v′ = − ( 45 mph ) iˆ, we have V = v − v′ = ( 45 mph ) iˆ − ( 60 mph ) ˆj. This means that Shawn’s speed relative to

4.19.

Susan is V=

( 45 mph )

2

+ ( −60 mph ) = 75 mph 2

Solve: (a) From t = 0 s to t = 1 s the particle does not rotate. From t = 1 s to t = 3 s, the particle rotates clockwise from the angular position 0 rad to −2π rad. Therefore, Δθ = −2π rad in two seconds, or

4.20.

ω = −π rad s. From t = 3 s to t = 4 s the particle rotates counterclockwise from the angular position −2π rad to +4π rad. Thus Δθ = 4π − ( −2π ) = 6π rad and ω = +6π rad s. (b)

4.21.

Solve: Since ω = ( dθ dt ) we have

θ f = θ i + area under the ω -versus-t graph between ti and tf From t = 0 s to t = 2 s, the area is

( 20 rad/s )( 2 s ) = 40 rad. (60 rad) × (1 rev/2π rad) = 9.55 revolutions.

1 2

( 20 rad/s )( 2 s ) = 20 rad.

From t = 2 s to t = 4 s, the area is

Thus, the area under the ω -versus-t graph during the total time interval of 4 s is 60 rad or

4.22.

Solve: Since ω = ( dθ dt ) we have

θ f = θ i + area under the ω versus t graph between ti and tf From t = 0 s to s to t = 4 s, the area is 1 2

1 2

( 20 rad/s )( 4 s ) = 40 rad.

( −10 rad/s )( 4 s ) = −20 rad. Thus, the area under the ω

rad or (20 rad) × (1 rev/2π rad) = 3.2 revolutions.

From t = 4 s to t = 8 s, the area is

versus t graph during the total time interval of 8 s is 20

4.23.

Model: Treat the record on a turntable as a particle rotating at 45 rpm. Solve: (a) The angular velocity is

ω = 45 rpm ×

1min 2π rad × = 1.5π rad/s 60 s 1 rev

(b) The period is

T=

2π rad

ω

=

2π rad = 1.33 s 1.5π rad/s

4.24.

Model: Visualize:

Solve:

The airplane is to be treated as a particle.

(a) The angle you turn through is

θ 2 − θ1 =

s 5000 miles 180° = = 1.2500 rad = 1.2500 rad × = 71.62° π rad r 4000 miles

(b) The plane’s angular velocity is

ω=

θ 2 − θ1 t2 − t1

=

1.2500 rad rad 1h = 0.13889 rad/h = 0.13889 × = 3.858 × 10−5 rad/s 9 hr h 3600 s

Assess: An angular displacement of approximately one-fifth of a complete rotation is reasonable because the separation between Kampala and Singapore is approximately one-fifth of the earth’s circumference.

Solve: Let RE be the radius of the earth at the equator. This means RE + 300 m is the radius to the top of the tower. Letting T be the period of rotation, we have

4.25.

vtop − vbottom =

2π ( RE + 300 m) 2π RE 2π (300 m) 600π m − = = = 2.18 × 10−2 m/s T T 24 h 24(3600) s

4.26.

Solve: The plane must fly as fast as the earth’s surface moves, but in the opposite direction. That is, the plane must fly from east to west. The speed is

km km 1 mile ⎛ 2π rad ⎞ 3 v = ωr = ⎜ = 1680 × = 1040 mph ⎟ (6.4 × 10 km) = 1680 h h 1.609 km ⎝ 24 h ⎠

4.27. Solve:

Model: The rider is assumed to be a particle. Since ar = v 2 / r , we have

v 2 = ar r = (98 m/s 2 )(12 m) ⇒ v = 34 m/s Assess:

34 m/s ≈ 70 mph is a large yet understandable speed.

4.28. Solve:

Model: The earth is a particle orbiting around the sun. (a) The magnitude of the earth’s velocity is displacement divided by time:

v=

2π r 2π (1.5 × 1011 m) = = 3.0 × 104 m/s 24 h 3600 s T 365 days × × 1 day 1h

(b) Since v = rω , the angular acceleration is

v r

ω= =

3.0 × 104 m/s = 2.0 × 10−7 rad/s 1.5 × 1011 m

(c) The centripetal acceleration is

ar =

v 2 (3.0 × 104 m/s) 2 = = 6.0 × 10−3 m/s 2 r 1.5 × 1011 m

Assess: A tangential velocity of 3.0 × 104 m/s or 30 km/s is large, but needed for the earth to go through a displacement of 2π (1.5 × 1011 m) ≈ 9.4 × 108 km in 1 year.

Solve: The pebble’s angular velocity ω = (3.0 rev/s)(2π rad/rev) = 18.9 rad/s. The speed of the pebble as it moves around a circle of radius r = 30 cm = 0.30 m is

4.29.

v = ω r = (18.9 rad/s)(0.30 m) = 5.7 m/s

The radial acceleration is v 2 ( 5.7 m/s ) = = 108 m/s 2 0.30 m r 2

ar =

4.30.

Model: The crankshaft is a rotating rigid body. Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up to 200 rad/s from t = 4 s to t = 7 s. The angular acceleration (α ) graph is based on the fact that α is the slope of the ω -versus-t graph.

4.31. Model: The turntable is a rotating rigid body. The angular velocity is the area under the α -versus-t graph:

Solve:

α=

dω ⇒ ω = ∫ α ( x ) dt = ω 0 + area under the α graph dt

The values of ω at selected values of time (t) are: t (s)

0

ω (rad/s)

0.5

0 (5 + 3.75)(0.5)/2 = 2.18

1.0

(5 + 2.5)(1)/2 = 3.75

1.5

(5 + 1.25)(1.5)/2 = 4.68

2.0

(5 + 0)(2)/2 = 5.0

2.5

5.0

3.0

5.0

4.32. Model: The wheel is a rotating rigid body. Solve:

(a)

The angular acceleration (α ) is the slope of the ω -versus-t graph. (b) The car is at rest at t = 0 s. It gradually speeds up for 4 s and then slows down for 4 s. The car is at rest from t = 8 s to t = 12 s, and then speeds up again for 4 s.

4.33. Model: The angular velocity and angular acceleration graphs correspond to a rotating rigid body. Solve:

(a) The α-versus-t graph has a positive slope of 5 rad/s2 from t = 0 s to t = 2 s and a negative slope of −5 rad/s 2 from t = 2 s to t = 4 s. (b) The angular velocity is the area under the α -versus-t graph:

α=

dω ⇒ ω = ∫ α ( x ) dt = ω 0 + area under α graph. dt

4.34.

Model: Visualize:

Model the car as a particle in nonuniform circular motion.

Note that halfway around the curve, the tangent is 45° south of east. The perpendicular component of the acceleration is 45° north of east. Solve: The radial and tangential components of the acceleration are

ar = a cos 25° = ( 3.0 m s 2 ) cos 25° = 2.7 m s 2 at = a sin 25° = ( 3.0 m s 2 ) sin 25° = 1.27 m s 2

4.35.

Model: Visualize:

Model the child on the merry-go-round as a particle in nonuniform circular motion.

Solve: (a) The speed of the child is v0 = rω = (2.5 m)(1.57 rad/s) = 3.9 m/s. (b) The merry-go-round slows from 1.57 rad/s to 0 in 20 s. Thus

ω1 = 0 = ω 0 +

at rω (2.5 m)(1.57 rad/s) = –0.197 m/s 2 t1 ⇒ at = − 0 = − r t1 20 s

During these 20 s, the wheel turns through angle

θ1 = θ 0 + ω 0t1 +

at 2 0.197 m/s 2 t1 = 0 + (1.57 rad/s) (20 s) − (20 s) 2 = 15.6 rad 2r 2(2.5 m)

In terms of revolutions, θ1 = (15.6 rad)(1 rev/2π rad) = 2.49 rev.

4.36.

Model: Visualize:

Model the particle on the crankshaft as being in nonuniform circular motion.

Solve: (a) The initial angular velocity is ω 0 = 2500 rpm × (1 min/60 s) × (2π rad/rev) = 261.8 rad/s. The crankshaft slows from 261.8 rad/s to 0 in 1.5 s. Thus

ω1 = 0 = ω 0 +

at rω (0.015 m)(261.8 rad/s) = –2.618 m/s 2 = −2.6 m/s 2 t1 ⇒ at = − 0 = − r t1 1.5 s

(b) During these 1.5 s, the crankshaft turns through angle

θ1 = θ 0 + ω 0t1 +

at 2 2.618 m/s 2 t1 = 0 + (261.8 rad/s) (1.5 s) − (1.5 s) 2 = 196 rad 2r 2(0.015 m)

In terms of revolutions, θ1 = (196 rad)(1 rev/π rad) = 31.2 rev.

4.37. Model: The fan is in nonuniform circular motion. Visualize:

Solve:

⎛ min ⎞ Note 1800 rev/min ⎜ ⎟ = 30 rev/s. Thus ⎝ 60 s ⎠

ω f = ωi + αΔt ⇒ 30 rev/s = 0 rev/s + α ( 4.0 s ) ⇒ α =7.5 rev/s 2 . This can be expressed as 2π rad ⎞ 2 ⎟ = 47 rad/s . ⎝ rev ⎠ Assess: An increase in the angular velocity of a fan blade by 7.5 rev/s each second seems reasonable.

( 7.5 rev/s ) ⎛⎜

4.38. Model: The wheel is in nonuniform circular motion. Visualize:

Solve:

(a) Express ωi in rad/s:

⎛ min ⎞⎛ 2π rad ⎞ ⎟⎜ ⎟ ⇒ 5.2 rad/s ⎝ 60 s ⎠⎝ rev ⎠

ωi = ( 50 rev/min ) ⎜

After 10 s, ω f = ωi + αΔt ⇒ ω f = 5.2 rad/s + ( 0.50 rad/s 2 ) (10 s ) = 55 rad/s. Converting to rpm, 2

( 55 rad/s ) ⎛⎜

60 s ⎞⎛ rev ⎞ ⎟⎜ ⎟ = 53 rpm ⎝ min ⎠⎝ 2π rad ⎠

(b) In 10 s, the wheel has turned a number of radians

θ f = θi + ω0 Δt + 12 αΔt 2 ⇒ θ f = 0 rad/sec + (5.2 rad/s)(10 s) + 12 (0.50 rad/s 2 )(10 s) 2 = 77 radians. Converting, 77 rad = 12.3 revolutions. Assess: Making a bicycle wheel turn just over 12 revolutions in 10 s when it is initially turning almost one revolution per second to begin with seems attainable by a cyclist.

4.39.

Model: Visualize:

We will assume that constant-acceleration kinematic equations in a plane apply.

G (a) The particle’s position r0 = (9.0 ˆj ) m implies that at t0 the particle’s coordinates are x0 = 0 m and G y0 = 9.0 m. The particle’s position r1 = (20iˆ) m at time t1 implies that x1 = 20 m and y1 = 0 m. This is the

Solve:

position where the wire hoop is located. Let us find the time t1 when the particle crosses the hoop at x1 = 20 m. From the vx -versus-t curve and using the relation x1 = x0 + area of the vx -t graph, we get 20 m = 0 m + area of the vx -t graph = area of the vx -t graph From Figure P4.39 we see that the area of the vx -t graph equals 20 m when t = t1 = 3 s. (b) We can now look at the y-motion to find a y . Note that the slope of the vx -t graph (that is, a y ) is negative

and constant, and we can determine a y by substituting into y3 = y0 + v0 y (t3 − t0 ) + 12 a y (t3 − t0 ) 2 : 0 m = 9 m + 0 m + 12 a y (3 s − 0 s) 2 ⇒ a y = −2 m/s 2 Therefore, v4 y = v0 y + a y (t4 − t0 ) = 0 m/s + (−2 m/s 2 )(4 s − 0 s) = −8 m/s (c)

4.40.

Solve: From the expression for R, Rmax = v02 /g . Therefore,

R= Assess:

Rmax v02 sin 2θ 1 = ⇒ sin 2θ = ⇒ θ = 15° and 75° g 2 2

The discussion of Figure 4.22 explains why launch angles θ and (90° − θ ) give the same range.

4.41.

Model: Visualize:

Solve:

Assume particle motion in a plane and constant-acceleration kinematics for the projectile.

(a) We know that v0 y = v0 sin θ , a y = − g , and v1 y = 0 m/s. Using v1y2 = v02y + 2a y ( y1 − y0 ) ,

0 m 2 /s 2 = v02 sin 2 θ + 2 ( − g ) h ⇒ h =

v02 sin 2 θ 2g

(b) Using Equation 4.19 and the above expression for θ = 30.0° :

h=

( x2 − x0 ) =

( 33.6 m/s )

2

sin 2 30.0°

2 ( 9.8 m/s 2 )

= 14.4 m

( 33.6 m/s ) sin ( 2 × 30.0° ) = 99.8 m v02 sin 2θ = g ( 9.8 m/s2 ) 2

For θ = 45.0° : h=

( 33.6 m/s )

2

sin 2 45.0°

2 ( 9.8 m/s 2 )

( x2 − x0 ) =

( 33.6 m/s )

2

= 28.8 m

sin ( 2 × 45.0° )

( 9.8 m/s2 )

= 115.2 m

For θ = 60.0° : h=

( x2 − x0 ) = Assess:

( 33.6 m/s )

2

sin 2 60.0°

2 ( 9.8 m/s 2 )

( 33.6 m/s )

2

= 43.2 m

sin ( 2 × 60.0° )

2 ( 9.8 m/s 2 )

= 99.8 m

The projectile’s range, being proportional to sin(2θ ), is maximum at a launch angle of 45°, but the

maximum height reached is proportional to sin 2 (θ ). These dependencies are seen in this problem.

4.42.

Model: Visualize:

Solve:

Assume the particle model for the projectile and motion in a plane.

(a) Using y2 = y0 + v0 y ( t2 − t0 ) + 12 a y ( t2 − t0 ) , 2

y2 = 0 m + ( 30 m/s ) sin 60° ( 7.5 s − 0 s ) + 12 ( −9.8 m/s 2 ) ( 7.5 s − 0 s ) = −80.8 m 2

Thus the launch point is 81 m higher than where the projectile hits the ground. (b) Using v12y = v02y + 2a y ( y1 − y0 ) ,

0 m 2 /s 2 = ( 30sin 60° m/s ) + 2 ( −9.8 m/s 2 ) ( y1 − 0 m ) ⇒ y1 = 34.4 m , or y1 = 34 m 2

(c) The x-component is v2 x = v0 x = v0 cos60° = ( 30 m/s ) cos60° = 15 m/s. The y-component is

v2 y = v0 y + a y ( t2 − t0 ) = v0 sin 60° − g ( t2 − t1 ) = ( 30 m/s ) sin 60° − ( 9.8 m/s 2 ) ( 7.5 s − 0 s ) = −47.52 m/s ⇒v=

(15 m/s )

θ = tan −1

v2 y v2 x

2

+ ( −47.52 m/s ) = 50 m/s

= tan −1

2

47.52 = 73° below + x 15

Assess: An angle of 73° made with the ground, as the projectile hits the ground 81 m below its launch point, is reasonable in view of the fact that the projectile was launched at an angle of 60°.

4.43.

Model: plane. Visualize:

Solve:

Assume the particle model and motion under constant-acceleration kinematic equations in a

(a) Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , 2

0 m = 1.80 m + v0 sin 40° ( t1 − 0 s ) + 12 ( −9.8 m/s 2 ) ( t1 − 0 s )

2

= 1.80 m + ( 7.713 m/s ) t1 − ( 4.9 m/s 2 ) t12 ⇒ t1 = − 0.206 s and 1.780 s The negative value of t1 is unphysical for the current situation. Using t1 = 1.780 s and x1 = x0 + v0 x ( t1 − t0 ) , we get

x1 = 0 + ( v0 cos 40° m/s )(1.780 s − 0 s ) = (12.0 m/s ) cos 40° (1.78 s ) = 16.36 m (b) We can repeat the calculation for each angle. A general result for the flight time at angle θ is

)

(

t1 = 12sin θ + 144sin 2 θ + 35.28 / 9.8 s and the distance traveled is x1 = (12.0)cosθ × t1. We can put the results in a table.

θ

t1

x1

40.0° 42.5° 45.0° 47.5°

1.780 s 1.853 s 1.923 s 1.990 s

16.36 m 16.39 m 16.31 m 16.13 m

Maximum distance is achieved at θ ≈ 42.5°. Assess: The well-known “fact” that maximum distance is achieved at 45° is true only when the projectile is launched and lands at the same height. That isn’t true here. The extra 0.03 m = 3 cm obtained by increasing the angle from 40.0° to 42.5° could easily mean the difference between first and second place in a world-class meet.

4.44.

Model: Visualize:

The golf ball is a particle following projectile motion.

(a) The distance traveled is x1 = v0 xt1 = v0 cosθ × t1. The flight time is found from the y-equation, using the fact that the ball starts and ends at y = 0 :

y1 − y0 = 0 = v0 sin θ t1 − 12 gt12 = (v0 sinθ − 12 gt1 ) t1 ⇒ t1 =

2v0 sin θ g

Thus the distance traveled is x1 = v0 cosθ ×

2v0 sin θ 2v02sin θ cosθ = g g

For θ = 30°, the distances are

( x1 )earth =

2v02sin θ cosθ 2(25 m/s) 2sin30°cos30° = = 55.2 m g earth 9.80 m/s2

( x1 ) moon =

2v02sin θ cosθ 2v02sin θ cosθ 2v02sin θ cosθ = = 6 × = 6( x1 )earth = 331.2 m 1 g moon g earth 6 g earth

The golf ball travels 331.2 m − 55.2 m = 276 m farther on the moon than on earth. (b) The flight times are (t1 )earth =

2v0 sin θ = 2.55 s g earth

(t1 ) moon =

2v0 sin θ 2v0 sinθ = 1 = 6(t1 )earth = 15.30 s g moon g 6 earth

The ball spends 15.30 s − 2.55 s = 12.75 s longer in flight on the moon.

4.45.

Model: Visualize:

Solve:

The particle model for the ball and the constant-acceleration equations of motion are assumed.

(a) Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , 2

h = 0 m + ( 30 m/s ) sin 60° ( 4 s − 0 s ) + 12 ( −9.8 m/s 2 ) ( 4 s − 0 s ) = 25.5 m 2

The height of the cliff is 26 m. (b) Using ( v y2 ) = v y2 + 2a y ( ytop − y0 ) , top

0 m /s = ( v0 sin θ ) + 2 ( − g ) ( ytop ) ⇒ ytop 2

2

2

( v sinθ ) = 0 2g

2

⎡( 30 m/s ) sin 60°⎤⎦ =⎣ = 34.4 m 2 ( 9.8 m/s 2 ) 2

The maximum height of the ball is 34 m. (c) The x and y components are v1 y = v0 y + a y ( t1 − t0 ) = v0 sin θ − gt1 = ( 30 m/s ) sin 60° − ( 9.8 m/s 2 ) × ( 4.0 s ) = −13.22 m/s

v1x = v0 y = v0 cos60° = ( 30 m/s ) cos60° = 15.0 m/s ⇒ v1 = v12x + v12y = 20.0 m/s

The impact speed is 20 m/s. Assess: Compared to a maximum height of 34.4 m, a height of 25.5 for the cliff is reasonable.

4.46.

Model: assumed. Visualize:

Solve:

The particle model for the ball and the constant-acceleration equations of motion in a plane are

The initial velocity is

v0 x = v0 cos5.0° = ( 20 m/s ) cos5.0° = 19.92 m/s v0 y = v0 sin 5.0° = ( 20 m/s ) sin 5.0° = 1.743 m/s The time it takes for the ball to reach the net is

x1 = x0 + v0 x ( t1 − t0 ) ⇒ 7.0 m = 0 m + (19.92 m/s )( t1 − 0 s ) ⇒ t = 0.351 s The vertical position at t1 = 0.351 s is y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 )

2

= ( 2.0 m ) + (1.743 m/s )( 0.351 s − 0 s ) + 12 ( −9.8 m/s 2 ) ( 0.351 s − 0 s ) = 2.01 m 2

Thus the ball clears the net by 1.01 m. Assess: The vertical free fall of the ball, with zero initial velocity, in 0.351 s is 0.6 m. The ball will clear by approximately 0.4 m if the ball is thrown horizontally. The initial launch angle of 5° provides some initial vertical velocity and the ball clears by a larger distance. The above result is reasonable.

4.47. Model: The particle model for the ball and the constant-acceleration equations of motion in a plane are assumed. Visualize:

Solve:

(a) The time for the ball to fall is calculated as follows:

y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 )

2

⇒ 0 m = 4 m + 0 m + 12 ( −9.8 m/s 2 ) ( t1 − 0 s ) ⇒ t1 = 0.9035 s 2

Using this result for the horizontal velocity:

x1 = x0 + v0 x ( t1 − t0 ) ⇒ 25 m = 0 m + v0 x ( 0.9035 s − 0 s ) ⇒ v0 x = 27.7 m/s The friend’s pitching speed is 28 m/s. (b) We have v0 y = ± v0 sinθ , where we will use the plus sign for up 5° and the minus sign for down 5°. We can

write y1 = y0 ± v0 sin θ ( t1 − t0 ) −

g g 2 ( t1 − t0 ) ⇒ 0 m = 4 m ± v0 sinθ t1 − t12 2 2

Let us first find t1 from x1 = x0 + v0 x ( t1 − t0 ) : 25 m = 0 m + v0 cosθ t1 ⇒ t1 =

25 m v0 cosθ

Now substituting t1 into the y-equation above yields ⎛ 25 m ⎞ g ⎛ 25 m ⎞ 0 m = 4 m ± v0 sin θ ⎜ ⎟− ⎜ ⎟ ⎝ v0 cosθ ⎠ 2 ⎝ v0 cosθ ⎠

2

g ( 25 m ) ⎧⎪ 1 ⎪⎫ ⎨ ⎬ = 22.3 m/s and 44.2 m/s 2 2cos θ ⎩⎪ 4 m ± ( 25 m ) tan θ ⎭⎪ 2

⇒ v02 =

The range of speeds is 22 m/s to 44 m/s, which is the same as 50 mph to 92 mph. Assess: These are reasonable speeds for baseball pitchers.

4.48.

Model: Visualize:

Solve:

We will use the particle model and the constant-acceleration kinematic equations in a plane.

The x- and y-equations of the ball are

x1B = x0B + ( v0B ) x ( t1B − t0B ) + 12 ( aB ) x ( t1B − t0B ) ⇒ 65 m = 0 m + ( v0B cos30° ) t1B + 0 m 2

2 y1B = y0B + ( v0B ) y ( t1B − t0B ) + 12 ( aB ) y ( t1B − t0B ) ⇒ 0 m = 0 m + ( v0B sin 30° ) t1B + 12 ( − g ) t1B

2

From the y-equation,

v0B =

gt1B

( 2sin 30° )

Substituting this into the x-equation yields 2 g cos30° t1B 2sin 30° ⇒ t1B = 2.77 s

65 m =

For the runner:

t1R =

20 m = 2.50 s 8.0 m/s

Thus, the throw is too late by 0.27 s. Assess: The times involved in running the bases are small, and a time of 2.5 s is reasonable.

4.49.

Model: Visualize:

Use the particle model for the ball and the constant-acceleration kinematic equations.

Solve: (a) The distance from the ground to the peak of the house is 6.0 m. From the throw position this distance is 5.0 m. Using the kinematic equation v12y = v02y + 2a y ( y1 − y0 ) ,

0 m 2 /s 2 = v02y + 2 ( −9.8 m/s 2 ) ( 5.0 m − 0 m ) ⇒ v0 y = 9.899 m/s The time for up and down motion is calculated as follows: y2 = y0 + v0 y ( t2 − t0 ) + 12 a y ( t2 − t0 ) ⇒ 0 m = 0 m + ( 9.899 m/s ) t2 − 12 ( 9.8 m/s 2 ) t22 ⇒ t2 = 0 s and 2.02 s 2

The zero solution is not of interest. Having found the time t2 = 2.02 s, we can now find the horizontal velocity needed to cover a displacement of 18.0 m:

x2 = x0 + v0 x ( t2 − t0 ) ⇒ 18.0 m = 0 m + v0 x ( 2.02 s − 0 s ) ⇒ v0 x = 8.911 m/s ⇒ v0 =

(8.911 m/s )

2

+ ( 9.899 m/s ) = 13.3 m/s 2

G (b) The direction of v0 is given by

θ = tan −1

v0 y v0 x

= tan −1

9.899 = 48° 8.911

Assess: Since the maximum range corresponds to an angle of 45°, the value of 48° corresponding to a range of 18 m and at a modest speed of 13.3 m/s is reasonable.

4.50.

Model: equations. Visualize:

Solve:

We will assume a particle model for the sand, and use the constant-acceleration kinematic

Using the equation x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) , 2

x1 = 0 m + ( v0 cos15° )( t1 − 0 s ) + 0 m = ( 60 m/s ) (cos15°)t1 We can find t1 from the y-equation, but note that v0 y = −v0 sin15° because the sand is launched at an angle below

horizontal. y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ 0 m = 3.0 m − ( v0 sin15° ) t1 − 12 gt12 2

= 3.0 m − ( 6.0 m/s ) (sin15°)t1 − 12 ( 9.8 m/s 2 ) t12

⇒ 4.9t12 + 1.55t1 − 3.0 = 0 ⇒ t1 = 0.6399 s and − 0.956 s (unphysical) Substituting this value of t1 in the x-equation gives the distance

d = x1 = ( 6.0 m/s ) cos15° ( 0.6399 s ) = 3.71 m

4.51.

Model: We will assume a particle model for the cannonball, and apply the constant-acceleration kinematic equations. Visualize:

Solve:

(a) The cannonball that was accidentally dropped can be used to find the height of the wall: 2 y1A = y0A + ( v0A ) y ( t1A − t0A ) + 12 ( aA ) y ( t1A − t0A ) ⇒ 0 m = y0A + 0 m − 12 gt1A 2

⇒ y0A =

1 2

( 9.8 m/s ) (1.5 s ) 2

2

= 11.03 m

For the cannonball that was shot: (v0S ) x = v0S cos30° = (50 m/s)cos30° = 43.30 m/s

(v0S ) y = v0s sin 30° = (50 m/s)sin 30° = 25.0 m/s We can now find the time it takes the cannonball to hit the ground: y2S = y0S + (v0S ) y (t2S − t0S ) + 12 (aS ) y (t2S − t0S ) 2

⇒ 0 m = (11.03 m) + (25.0 m/s)t2S −

(9.8 m/s 2 ) 2 t2S 2

2 ⇒ (4.9 m/s 2 )t2S − (25.0 m/s)t2S − (11.03 m) = 0 ⇒ t2S = 5.51 s

There is also an unphysical root t2S = −0.41 s. Using this time t2S , we can now find the horizontal distance from the wall as follows: x2s = x0s + (v0s ) x (t2s − t0s ) = 0 m + (43.30 m/s)(5.51 s) = 239 m The cannonball hits the ground 2.4 × 10 2 m from the castle wall. (b) At the top of the trajectory (v1S ) y = 0 m/s. Using (v1S ) 2y = (v0S ) 2y − 2 g ( y1S − y0S ),

0 m 2 /s 2 = (25.0 m/s)2 − 2(9.8 m/s 2 )( y1s − 11.03 m) ⇒ y1s = 42.9 m The maximum height above the ground is 43 m. Assess: In view of the fact that the cannonball has a speed of approximately 110 mph, a distance of 239 m for the cannonball to hit the ground is reasonable.

4.52.

Model: Visualize:

Solve:

We will use the particle model and the constant-acceleration kinematic equations for the car.

(a) The initial velocity is

v0 x = v0 cosθ = (20 m/s)cos 20° = 18.79 m/s

v0 y = v0 sin θ = (20 m/s)sin 20° = 6.840 m/s Using y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ,

0 m = 30 m + (6.840 m/s)(t1 − 0 s) + 12 (−9.8 m/s 2 )(t1 − 0 s)2 ⇒ 4.9t12 − 6.840t1 − 30 = 0 The positive root to this equation is t1 = 3.269 s. The negative root is physically unreasonable in the present case. Using x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 )2 , we get

x1 = 0 m + (18.79 m/s)(3.269 s − 0 s) + 0 = 61.4 m The car lands 61 m from the base of the cliff. (b) The components of the final velocity are v1x = v0 x = 18.79 m/s and v1 y = v0 y + a y (t1 − t0 ) = 6.840 m/s − (9.8 m/s 2 )(3.269 s − 0 s) = −25.2 m/s

⇒ v = (18.79 m/s) + (−25.2 m/s) 2 = 31.4 m/s The car’s impact speed is 31 m/s. Assess: A car traveling at 45 mph and being driven off a 30-m high cliff will land at a distance of approximately 200 feet (61.4 m). This distance is reasonable.

4.53.

Model: Visualize:

Solve:

Use the particle model for the cat and apply the constant-acceleration kinematic equations.

The relative velocity of the cat from the mouse’s reference frame is

(4.0cos30° − 1.5) m/s = 1.964 m/s = v0 Thus, v0 x = v0 = 1.964 m/s and v0 y = v0 sin 30° = (4.0 m/s)sin 30° = 2.0 m/s . The time for the cat to land on the floor is found as follows: y1 = y0 + v0 y (t1 − t0 ) + 12 a y (t1 − t0 ) 2 ⇒ 0 m = 0 m + (2.0 m/s)t1 + 12 (−9.8 m/s 2 )t12

⇒ t1 = 0 s (trivial solution) and 0.408 s The horizontal distance covered in time t1 is

x1 = x0 + v0 x (t1 − t0 ) + 12 ax (t1 − t0 ) 2 = 0 m + (1.964 m/s)(0.408 s) = 0.802 m That is, the cat should leap when he is 80 cm behind the mouse.

4.54.

Model: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the staple gun be S′. Frame S′ moves relative to S with velocity Vx . Solve:

The velocity vx of the parts to be stapled in frame S is +3.0 m/s. Also Vx = −1.0 m/s. Using

vx = v′x + Vx , we get v′x = vx − Vx = +3.0 m/s − (−1.0 m/s) = 4.0 m/s That is, the velocity of the parts in the frame of the staple gun is 4.0 m/s. In this frame 10 staples are fired per second. That is, 4.0 m/s =

distance between two staples distance = ⇒ Distance = 0.4 m 1 time between two staples s 10

Assess: Note that the staple gun is in frame S′ and we had to find the velocity of the moving parts in this frame to solve the problem.

G Model: If a frame S′ is in motion with velocity V relative to another frame S and has a displacement G G G G G R relative to S, the positions and velocities (r and v ) in S are related to the positions and velocities (r ′ and v′) G G G G G in S ′ as r = r ′ + R and v = v′ + V . In the present case, ship A is frame S and ship B is frame S′. Both ships have a common origin at t = 0 s. The position and velocity measurements are made in S and S′ relative to their origins.

4.55.

Solve: (a) The velocity vectors of the two ships are:

G

G vA = (20 mph)[cos30°iˆ − sin 30° ˆj ] = (17.32 mph)iˆ − (10.0 mph) ˆj G vB = (25 mph)[cos 20°iˆ + sin 20° ˆj ] = (23.49 mph)iˆ + (8.55 mph) ˆj

G

Since r = v Δt ,

G

G

K

As rA = rB + R,

G G rA = vA (2 h) = (34.64 miles)iˆ − (20.0 miles) ˆj G G rB = vB (2 h) = (46.98 miles)iˆ + (17.10 miles) ˆj G G G R = rA − rB = (−12.34 miles)iˆ − (37.10 miles) ˆj ⇒ R = 39.1 miles

The distance between the ships two hours after they depart is 39 miles. G G G (b) Because vA = vB + V , G G G V = vA − vB = −(6.17 mph)iˆ − (18.55 mph) ˆj ⇒ V = 19.5 mph The speed of ship A as seen by ship B is 19.5 mph. Assess: The value of the speed is reasonable.

4.56. Model: Let the earth frame be S and a frame attached to the water be S′. Frame S′ moves relative to S with velocity V. We define the x-axis along the direction of east and the y-axis along the direction of north for both frames. Solve: (a) The kayaker’s speed of 3.0 m/s is relative to the water. Since he’s being swept toward the east, he needs to point at angle θ west of north. In frame S′, the water frame, his velocity is G v′ = (3.0 m/s, θ west of north) = (−3.0sinθ m/s)iˆ + (3.0cosθ m/s)ˆj G G G G We can find his velocity in earth frame S from the Galilean transformation v = v′ + V , with V = (2.0 m/s)iˆ. Thus G v = ((−3.0sin θ + 2.0) m/s)iˆ + (3.0cosθ m/s)ˆj In order to go straight north in the earth frame, the kayaker needs vx = 0. This will be true if sin θ =

2.0 3.0

⎛ 2.0 ⎞ ⇒ θ = sin −1 ⎜ ⎟ = 41.8° ⎝ 3.0 ⎠

Thus he must paddle in a direction 42° west of north. (b) His northward speed is v y = 3.0 cos(41.8°) m/s = 2.236 m/s. The time to cross is t=

The kayaker takes 45 s to cross.

100 m = 44.7 s 2.236 m/s

Model: Let Mike’s frame be S and Nancy’s frame be S′. Frame S′ moves relative to S with velocity V. x is the horizontal direction and y is the vertical direction for motion. The frames S and S′ coincide at t = 0 s. G G G Solve: (a) According to the Galilean transformation of velocity v = v′ + V . Mike throws the ball with velocity G G v = (22 m/s)cos63°iˆ + (22 m/s)sin63° ˆj, and V = (30 m/s)iˆ. Thus in Nancy’s frame G G G v′ = v − V = (22cos63° − 30)iˆ m/s + (22sin 63°) ˆj m/s = (−20.0iˆ + 19.6ˆj ) m/s

4.57.

θ = tan −1

v′y 19.6 m/s = tan −1 = 44.4° 20.0 m/s v′x

The direction of the angle is 44.4° above the − x′ axis (in the second quadrant). (b) In Nancy’s frame S′ the equation x′ = x0′ + v′x (t ′ − t0′ ) + 12 a′x (t ′ − t0′ ) 2 becomes x′ = −(20.0 m/s)t ′

and the equation y′ = y0′ + v′y (t ′ − t0′ ) + 12 a′y (t ′ − t0′ ) 2 becomes y′ = 0 m + (19.6 m/s)t ′ − 12 (9.8 m/s 2 )t ′2 = (19.6 m/s)t ′ − (4.9 m/s 2 )t ′2

Model: Let the earth frame be S and a frame attached to the sailboat be S′. Frame S′ moves relative to G S with velocity V = 8.0iˆ mph. G G G G Solve: (a) From Equation 6.24, v = v′ + V , where v′ = (12.0 ) cos 45°iˆ mph + (12.0 ) sin 45° ˆj mph is the velocity

4.58.

of the wind in S′, or the apparent wind velocity. Thus, G v = (12.0 ) cos 45°iˆ mph + (12.0 ) sin 45° ˆj mph + ( 8.0 ) iˆ mph

(

)

= ( 8.0 + 12.0cos45° ) iˆ mph + (12.0sin45° ) ˆj mph = 16.48iˆ + 8.48ˆj mph

θ = tan −1

8.48 mph = 27.2° 16.48 mph

The wind speed is v = 18.5 mph and the direction is from 27.2° south of west. G G (b) In this case, v′ = − (12.0 mph ) cos 45°iˆ − (12.0 mph ) sin 45° ˆj and V = ( 8.0 mph ) iˆ. So, G v = ⎡⎣ − (12.0 mph ) cos 45° + 8.0 mph ⎤⎦ iˆ − (12.0 mph ) sin 45° ˆj = −0.485iˆ − 8.485 ˆj

θ = tan −1

8.485 mph = 86.7° 0.485 mph

The speed is 8.5 mph and the direction is from 86.7° north of east or from 3.3° east of north.

Model: Let the ground frame be S and the car frame be S′. S′ moves relative to S with a velocity V along the x-direction. G G G G G Solve: The Galilean transformation of velocity is v = v′ + V where v and v′ are the velocities of the raindrops G in frames S and S′. While driving north, V = ( 25 m/s ) iˆ and v = −vR cosθ ˆj − vR sin θ iˆ. Thus, G G G v′ = v − V = ( −vR sinθ − 25 m/s ) iˆ − vR cosθ ˆj

4.59.

Since the observer in the car finds the raindrops making an angle of 38° with the vertical, we have

vR sin θ + 25 m/s = tan 38° vR cosθ G G While driving south, V = − ( 25 m/s ) iˆ, and v = −vR cosθ ˆj − vR sin θ iˆ. Thus, G v′ = ( −vR sin θ + 25 m/s ) iˆ − vR cosθ ˆj

Since the observer in the car finds the raindrops falling vertically straight, we have −vR sinθ + 25 m/s = tan 0° = 0 ⇒ vR sinθ = 25 m/s vR cosθ Substituting this value of vR sinθ into the expression obtained for driving north yields: 25 m/s + 25 m/s 50 m/s = tan 38° ⇒ vR cosθ = = 64.0 m/s vR cosθ tan 38° Therefore, we have for the velocity of the raindrops:

( vR sinθ )

2

+ ( vR cosθ ) = ( 25 m/s ) + ( 64.0 m/s ) ⇒ vR2 = 4721( m/s ) ⇒ vR = 68.7 m/s 2

2

tan θ =

2

vR sin θ 25 m/s = ⇒ θ = 21.3° vR cosθ 64 m/s

The raindrops fall at 69 m/s while making an angle of 21° with the vertical.

2

4.60. Model: Let the ground be frame S and the wind be frame S′. S′ moves relative to S. The ground frame S has the x-axis along the direction of east and the y-axis along the direction of north. Visualize:

Solve:

G G G (a) The Galilean velocity transformation is v = v′ + V , where G V = ( 50 mph ) cos30°iˆ + ( 50 mph ) sin 30° ˆj G v′ = ( 200 mph ) cosθ iˆ − ( 200 mph ) sin θ ˆj

G G Thus, v = ⎡⎣( 50cos30° + 200cosθ ) iˆ + ( 50sin 30° − 200sinθ ) ˆj ⎤⎦ mph. Because v should have no j-component, 50sin 30° − 200sin θ = 0 ⇒ θ = 7.18°

G (b) The pilot must head 7.18° south of east. Substituting this value of θ in the v equation gives 600 mi G v = ( 242 ) iˆ mph, along the direction of east. At a speed of 242 mph, the trip takes t = = 2.48 h 242 mph

4.61. Solve:

Model: We will use the particle model for the test tube which is in nonuniform circular motion. (a) The radial acceleration is 2

rev 1 min 2π rad ⎞ ⎛ 4 2 × × ar = rω 2 = ( 0.1 m ) ⎜ 4000 ⎟ = 1.75 × 10 m/s min 60 s 1 rev ⎠ ⎝ (b) An object falling 1 meter has a speed calculated as follows:

v12 = v02 + 2a y ( y1 − y0 ) = 0 m + 2 ( −9.8 m/s 2 ) ( −1.0 m ) ⇒ v1 = 4.43 m/s When this object is stopped in 1 × 10−3 s upon hitting the floor, v2 = v1 + a y ( t2 − t1 ) ⇒ 0 m s = −4.43 m s + a y (1× 10−3 s ) ⇒ a y = 4.4 × 103 m/s 2 This result is one-fourth of the above radial acceleration. Assess: The radial acceleration of the centrifuge is large, but it is also true that falling objects are subjected to large accelerations when they are stopped by hard surfaces.

4.62. Solve:

Model: We will use the particle model for the astronaut undergoing nonuniform circular motion. (a) The initial conditions are ω 0 = 0 rad/s, θ 0 = 0 rad, t0 = 0 s, and r = 6.0 m. After 30 s,

ω1 =

1 rev 1 rev 2π rad = × × = 4.83 rad/s 1.3 s 1.3 s rev

Using these values at t1 = 30 s,

ω1 = ω 0 + ( at r )( t1 − t0 ) = 0 + ( at r ) t1 ⎛ 1 ⎞ 2 ⇒ at = ( 6.0 m )( 4.83 rad/s ) ⎜ ⎟ = 0.97 m/s 30 s ⎝ ⎠

(b) The radial acceleration is

ar = rω12 = ( 6.0 m )( 4.83 rad/s ) Assess:

2

g = 14.3g 9.8 ( m/s2 )

The above acceleration is typical of what astronauts experience during liftoff.

4.63.

Model: Visualize:

Model the car as a particle in nonuniform circular motion.

Note that the tangential acceleration stays the same at 1.0 m/s2. As the tangential velocity increases, the radial acceleration increases as well. After a time t1 , as the car goes through an angle θ1 − θ 0 , the total acceleration will increase to 2.0 m/s2. Our objective is to find this angle. Solve: Using v1 = v0 + at ( t1 − t0 ) , we get

v1 = 0 m/s + (1.0 m/s 2 ) ( t1 − 0 s ) = (1.0 m/s 2 ) t1 2 2 rv 2 (1.0 m/s ) t1 t2 ⇒ ar = 21 = = 1 ( m/s 4 ) r 120 m 120 2

⇒ atotal = 2.0 m/s = a + a = 2

2 t

2 r

(1.0 m/s )

2 2

2

⎡ t2 ⎤ + ⎢ 1 ( m/s 4 ) ⎥ ⇒ t1 = 14.4 s 120 ⎣ ⎦

We can now determine the angle θ1 using 2 ⎛ at ⎞ ⎟ ( t1 − t0 ) r ⎝ ⎠

θ1 = θ 0 + ω 0 ( t1 − t0 ) + 12 ⎜ = 0 rad + 0 rad +

2 1 (1.0 m/s ) 2 (14.4 s ) = 0.864 rad = 49.5° 2 (120 m )

The car will have traveled through an angle of 50°.

4.64. Model: The earth is a rigid, rotating, and spherical body. Visualize:

Solve:

At a latitude of θ degrees, the radius is r = Re cos θ with Re = 6400 km = 6.400 × 106 m.

(a) In Miami θ = 26°, and we have r = (6.400 × 106 m)(cos 26°) = 5.752 × 106 m . The angular velocity of the earth is

ω=

2π 2π = = 7.272 × 10−5 rad/s T 24 × 3600 s

Thus, vstudent = rω = (5.752 × 106 m)(7.272 × 10−5 rad/s) = 418 m/s. (b) In Fairbanks θ = 65°, so r = (6.400 × 106 m)cos 65° = 2.705 × 106 m and vstudent = rω = (2.705 × 106 m)(7.272 × 10−5 rad/s) = 197 m/s.

4.65. Model:

The satellite is a particle in uniform circular motion.

Visualize:

Solve: (a) The satellite makes one complete revolution in 24 h about the center of the earth. The radius of the motion of the satellite is r = 6.37 × 106 m + 3.58 × 107 m = 4.22 × 107 m The speed of the satellite is v =

( distance traveled ) = 2π r = 3.07 × 103 m/s. 24 h ( time taken )

(b) The acceleration of the satellite is centripetal, with magnitude 3 v 2 ( 3.07 × 10 m/s ) ar = = = 0.223 m/s 2 r 4.22 × 107 m Assess: The small centripetal acceleration makes sense when realized it is for an object traveling in a circle with radius ≈ 26,400 miles. 2

4.66. Model: The magnetic computer disk is a rigid rotating body. Visualize:

Solve:

Using the rotational kinematic equation ω f = ωi + αΔt , we get

ω1 = 0 rad + (600 rad/s 2 )(0.5 s − 0 s) = 300 rad/s ω 2 = (300 rad/s) + (0 rad/s 2 )(1.0 s − 0.5 s) = 300 rad/s The speed of the painted dot v2 = rω 2 = (0.04 m)(300 rad/s) = 12 m/s. The number of revolutions during the time interval t0 to t2 is 1 2 1 θ 2 = θ1 + ω1 (t2 − t1 ) + α1 (t2 − t1 ) 2 2

1 2

θ1 = θ 0 + ω0 (t1 − t0 ) + α 0 (t1 − t0 ) 2 = 0 rad + 0 rad + (600 rad/s 2 )(0.5 s − 0 s) 2 = 75 rad

⎛ 1 rev ⎞ = 75 rad + (300 rad/s)(1.0 s − 0.5 s) + 0 rad = 225 rad = (225 rad) ⎜ ⎟ = 35.8 rev ⎝ 2π rad ⎠

4.67. Model: The drill is a rigid rotating body. Visualize:

The figure shows the drill’s motion from the top. Solve: (a) The kinematic equation

ωf = ωi + α (tf − ti )

becomes,

ωi = 2400 rpm = (2400)(2π )/60 = 251.3 rad/s, tf − ti = 2.5 s − 0 s = 2.5 s, and wf = 0 rad/s, 0 rad = 251.3 rad/s + α (2.5 s) ⇒ α = −100 rad/s 2

(b) Applying the kinematic equation for angular position yields:

1 2

θ f = θ i + ωi (tf − ti ) + α (tf − ti ) 2 1 = 0 rad + (251.3 rad/s)(2.5 s − 0 s) + (−100 rad/s2 )(2.5 s − 0 s)2 2 = 3.2 × 102 rad = 50 rev

after

using

4.68. Model: The turbine is a rigid rotating body. The known values are ωi = 3600 rpm = (3600)(2π )/60 = 120π rad/s, ti = 0 s, tf = 10 min = 600 s, ω f = 0 rad/s, and θ i = 0 rad Using the rotational kinematic equation ω f = ωi + α (tf − ti ), we get

Solve:

0 rad = (120π rad/s) + α (600 s − 0 s). Thus, α = −0.628 rad/s 2 . Now,

1 2

θ f = θ i + ωi (tf − ti ) + α (tf − ti ) 2 1 = 0 rad + (120π rad/s)(600 s − 0 s) + (−0.628 rad/s 2 )(600 s − 0 s) 2 2 = 113,100 rad = 1.80 × 104 rev Assess:

18,000 revolutions during 10 minutes when the starting angular velocity is 3600 rpm is reasonable.

4.69. Visualize: Please refer to Figure P4.69. Solve: Since ω f = ωi + (area under α vs t curve), at t = 3 s, the angular velocity is ω f = 60 rpm + ( 4.0 rad/s 2 ) ( 2 s − 1 s ) ⎛ 20 rev 60 s ⎞ = 60 rpm + ( 4 rad/s ) ⎜ × ⎟ ⎝ 2π rad 1 min ⎠ = 60 rpm + 38 rpm = 98 rpm

4.69. Visualize: Please refer to Figure P4.69. Solve: Since ω f = ωi + (area under α vs t curve), at t = 3 s, the angular velocity is ω f = 60 rpm + ( 4.0 rad/s 2 ) ( 2 s − 1 s ) ⎛ 20 rev 60 s ⎞ = 60 rpm + ( 4 rad/s ) ⎜ × ⎟ ⎝ 2π rad 1 min ⎠ = 60 rpm + 38 rpm = 98 rpm

4.70. Model: The car’s wheels are in nonuniform circular motion with constant angular acceleration. Solve: The problem must be solved in two steps: first, the angular acceleration α is found for the case of stopping in one revolution, then α is applied to the case of stopping with twice the initial angular velocity ωi . To find α for the case of stopping in 1.0 revolutions, set Δθ = 1.0 rev = 2π rad in ω f2 = ω i2 + 2αΔθ ⇒ ⎛ ω i2 ⎞ 0 m = ωi2 + 2α (2π rad) ⇒ α = − ⎜ ⎟ ⎝ 4π rad ⎠ Now use the same relationship, with that value of α , but this time Δθ is unknown and the initial angular velocity has been doubled.

⎛ −ωi2 ⎞ ωi2 2 2 Δθ ⇒ Δθ = 8π rad = 4.0 rev 0 m = ( 2ωi ) + 2 ⎜ ⎟ Δθ ⇒ 0 m = 4ω i − 2π rad ⎝ 4π rad ⎠ Assess: Doubling the initial velocity quadruples the number of turns the wheels make before locking. This is within reason.

4.71. Model: The bicycle wheel undergoes nonuniform circular motion with constant angular acceleration. Visualize:

Solve:

First find the angular acceleration α , then use it to find θ f . Using kinematics,

ω f = ωi + αΔt ⇒ 0 rpm = 100 rpm + α (1.0 min) ⇒ α = −100 (rpm)/min = −100 rev/min 2 The minus sign indicates the wheel is slowing down. The total number of revolutions the wheel makes while stopping is

θ f = 0 rev + (100 rpm)(1.0 min) + 12 (−100 rev/min 2 )(1.0 min) 2 = 50 rev Assess: A total of 50 revolutions in 60 s is on average less than one revolution per second, which is quite reasonable.

4.72. Model: Treat the rock as a particle in nonuniform circular motion with constant angular acceleration. Visualize:

Solve:

(a) The angular acceleration α is

α=

at −1.00 m/s 2 = = −1.667 rad/s 2 r 0.600 m

The minus sign indicates the angular acceleration is clockwise, as shown in the figure above, opposite to the 3.00 m/s angular velocity. The initial angular velocity is ωi = = 5.00 rad/s. Using angular kinematics, the 0.600 m angular velocity at some time t after braking starts is

ωf ( t ) = ωi + 0αΔt = 5.00 rad/s − 1.667 rad/s ( t − 0 s ) At t = 1.5 s, ω f = 2.50 rad/s while α = −1.667 rad/s. (b) The total acceleration of the rock is a = at2 + ar2 . The value of at = 1.00 m/s 2 is given. The radial

acceleration at a certain time t is ar = multiplying through by r = 0.600 m:

vt2 . Since vt = ω r , we can use the expression for ω f (t ) to find vt (t ) by r

ω f ( t ) r = ( 5.00 rad/s ) r − (1.667 rad/s ) rt ⇒ vt ( t ) = 3.00 m/s − (1.00m/s 2 ) t ⇒ t = The value of vt for a = g is needed. Setting a = g and substituting ar =

3.00 m/s − vt 1.00 m/s 2

vt2 , r

2

⎛ v2 ⎞ v4 g 2 = at2 + ⎜ t ⎟ ⇒ g 2 − at2 = t2 = ( 9.80 m/s 2 ) − (1.00 m/s 2 ) = 95.0 m 2 /s 4 r ⎝ r ⎠ ⇒ vt2 = ( 0.600 m ) 95.0 m 2 /s 4 ⇒ vt = 2.42 m/s 3.00 m/s − ( 2.42 m/s ) = 0.58 s. 1.00 m/s 2 Assess: The time of 0.58 s occurs early during the braking. This is reasonable since an acceleration equal to that supplied by gravity is fairly strong. Finally, t =

4.73. Model: The string is wrapped around the spool in such a way that it does not pile up on itself, and unwinds without slipping. Visualize:

Solve:

Since the string unwinds without slipping, the angular distance the spool turns as the string is pulled 1.0 Δx 1.0 m m is Δθ = = = 33 radians. r 3.0 × 10−2 m The angular acceleration of the spool due to the pull on the string is

α=

at 1.5 m/s 2 = = 50 rad/s 2 r 3.0 × 10−2 m

The angular velocity of the spool after pulling the string is found with kinematics.

ω f2 = ωi2 + 2αΔθ ⇒ ω f2 = 0 rad 2 /s 2 + 2 ( 50 rad/s 2 ) ( 33 rad ) ⇒ ω f2 = 57 rad/s

Converting to revolutions per minute,

( 57 rad/s ) ⎛⎜

rev ⎞⎛ 60 s ⎞ 2 ⎟⎜ ⎟ = 5.5 × 10 rpm π 2 rad min ⎝ ⎠⎝ ⎠

Assess: The angular speed of 57 rad/s ≈ 9 rev/s is reasonable for a medium-sized spool.

4.74.

Solve: (a) A golfer hits an iron shot with a new club as she approaches the green. She is pretty sure, based on past experience, that she hit the ball with a speed of 50 m/s, but she is not sure at what angle the golf ball took flight. She observed that the ball traveled 100 m before hitting the ground. What angle did she hit the ball?

(b) From the second equation, (4.9 m/s 2 )t12 − (50sinθ m/s) t1 = 0 ⇒ t1 = 0 s and t1 =

(50 m/s)sinθ 4.9 m/s 2

Using the above value for t1 in the first equation yields: 100 m =

(50cosθ )(50sinθ ) m 2 /s 2 4.9 m/s 2

⇒ 2cosθ sinθ = sin 2θ =

Assess:

9.8 = 0.392 ⇒ 2θ = 23.1 ⇒ θ = 11.5° 25

Although the original speed is reasonably high (50 m/s = 112 mph), the ball travels a distance of only

100 m, implying either a small launch angle around 10° or an angle closer to 80°. The calculated angle of 11.5° is thus pretty reasonable.

4.75.

Solve: (a) A submarine moving east at 3.0 m/s sees an enemy ship 100 m north of its path. The submarine’s torpedo tube happens to be stuck in a position pointing 45° west of north. The tube fires a torpedo with a speed of 6.0 m/s relative to the submarine. How far east or west of the ship should the sub be when it fires? (b) Relative to the water, the torpedo will have velocity components

vx = −6.0cos 45° m/s + 3.0 m/s = −4.24 m/s + 3 m/s = −1.24 m/s

v y = +6.0cos 45° m/s = +4.2 m/s The time to travel north to the ship is

100 m = ( 4.2 m/s ) t1 ⇒ t1 = 24 s Thus, x = (1.24 m/s )( 24 s ) = −30 m. That is, the ship should be 30 m west of the submarine.

4.76.

Solve: (a) A 1000 kg race car enters a 50 m radius curve and accelerates around the curve for 10.0 s. The forward force provided by the car’s wheels is 1500 N. After 10.0 s the car has moved 125 m around the track. Find the initial and final angular velocities. (b) From Newton’s second law,

Ft = mat ⇒ 1500 N = (1000 kg ) at ⇒ at = 1.5 m/s 2

θ f = θ i + ω it +

Δθ =

Δs 125 m = = 2.5 rad r 50 m

at 2 1.5 m/s 2 2 t ⇒ 2.5 rad = 0 rad + ωi (10 s ) + (10 s ) ⇒ ωi = 0.10 rad/s 2r 2 ( 50 m )

ωf = ωi +

at 1.5 m/s 2 t = 0.1 rad/s + (10 s ) = 0.40 rad/s r 50 m

4.77.

Solve: You decide to test fly your model airplane off of a 125 m tall building. The model’s engine starts fine and gets the airplane moving at 4.0 m/s but quits just as it gets to the edge of the building. The model proceeds to fall “like a rock.” How far from the edge of the building will it crash into the ground? (Assume g = 10 m/s 2 for easier calculation.) Visualize:

Using the equation y1 = y0 + ν 0 yt + 12 a y ( t0 − t1 ) , we get 2

y1 = −(5 m/s 2 ) t12 = −125 m ⇒ t1 =

The distance x1 = ( 4 m/s )( 5 s ) = 20 m.

125 m = 25 s 2 = 5 s 5 m/s 2

4.78.

Model: The ions are particles that move in a plane. They have vertical acceleration while between the acceleration plates, and they move with constant velocity from the plates to the tumor. The flight time will be so small, because of the large speeds, that we’ll ignore any deflection due to gravity. Visualize:

Solve: There’s never a horizontal acceleration, so the horizontal motion is constant velocity motion at vx = 5.0 × 106 m/s. The times to pass between the 5.0-cm-long acceleration plates and from the plates to the tumor are

0.050 m = 1.00 × 10−8 s 5.0 × 106 m/s 1.50 m t2 − t1 = = 3.00 × 10−7 s 5.0 × 106 m/s t1 − t0 = t1 =

Upon leaving the acceleration plates, the ion has been deflected sideways to position y1 and has velocity v1 y . These are y1 = y0 + v0 y t1 + 12 a y t12 = 12 a y t12 v1 y = v0 y + a yt1 = a yt1 In traveling from the plates to the tumor, with no vertical acceleration, the ion reaches position

y2 = y1 + v1 y (t2 − t1 ) = 12 a yt12 + (a yt1 )(t2 − t1 ) = ( 12 t12 + t1 (t2 − t1 ) ) a y We know y2 = 2.0 cm = 0.020 m, so we can solve for the acceleration a y that the ion had while between the plates: ay =

y2 0.020 m =1 = 6.6 × 1012 m/s2 2 −8 t + t1 (t2 − t1 ) 2 (1.00 × 10 s) + (1.00 × 10−8 s)(3.00 × 10−7 s)

1 2 2 1

Assess: This acceleration is roughly 1012 times larger than the acceleration due to gravity. This justifies our assumption that the acceleration due to gravity can be neglected.

4.79.

Model: We will use the particle model for the ball’s motion under constant-acceleration kinematic equations. Note that the ball’s motion on the smooth, flat board is a y = − g sin 20° = −3.352 m/s 2 .

Visualize:

Solve:

The ball’s initial velocity is v0 x = v0 cosθ = ( 3.0 m/s ) cosθ

v0 y = v0 sinθ = ( 3.0 m/s ) sinθ

Using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) , 2

2.5 m = 0 m + ( 3.0 m/s ) cosθ ( t1 − 0 s ) + 0 m ⇒ t1 =

( 2.5 m ) ( 3.0 m/s ) cosθ

=

0.833 s cosθ

Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) and the above equation for t1 , 2

⎛ 0.833 s ⎞ 1 2 ( 0.833 s ) 0 m = 0 m + ( 3.0 m/s ) sinθ ⎜ ⎟ − 2 ( 3.352 m/s ) cos 2 θ ⎝ cosθ ⎠ ⇒ ( 2.5 m )

2

sinθ 1.164 = ⇒ 2.5sin θ cosθ = 1.164 ⇒ 2θ = 68.6° ⇒ θ = 34.3° cosθ cos 2 θ

4.80.

Model: Visualize:

Solve:

Use the particle model for the arrow and the constant-acceleration kinematic equations.

Using v1 y = v0 y + a y ( t1 − t0 ) , we get

v1 y = 0 m/s − gt1 ⇒ v1 y = − gt1 Also using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) , 2

60 m = 0 m + v0 xt1 + 0 m ⇒ v0 x =

60 m = v1x t1

Since v1 y / v1x = − tan 3° = − 0.0524, using the components of v0 gives − gt1

( 60 m/ t1 )

= − 0.0524 ⇒ t1 =

( 0.0524 )( 60 m ) = 0.566 s

( 9.8 m/s ) 2

Having found t1 , we can go back to the x-equation to obtain v0 x = 60 m/0.566 s = 106 m/s. Assess: In view of the fact that the arrow took only 0.566 s to cover a horizontal distance of 60 m, a speed of 106 m/s or 237 mph for the arrow is understandable.

4.81.

Model: Use the particle model for the arrow and the constant-acceleration kinematic equations. We will assume that the archer shoots from 1.75 m above the slope (about 5′ 9′′). Visualize:

Solve:

For the y-motion:

y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ y1 = 1.75 m + ( v0 sin 20° ) t1 − 12 gt12 2

⇒ y1 = 1.75 m + ( 50 m/s ) sin 20°t1 − 12 gt12

For the x-motion: x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) = 0 m + ( v0 cos 20° ) t1 + 0 m = ( 50 m/s ) (cos 20°)t1 2

Because y1 x1 = − tan15° = − 0.268,

1.75 m + ( 50 m/s ) (sin 20°)t1 − 12 gt12 = − 0.268 ⇒ t1 = 6.12 s and −0.058 s (unphysical) ( 50 m/s ) (cos 20°)t1 Using t1 = 6.12 s in the x- and y-equations above, we get y1 = −77.0 m and x1 = 287 m. This means the distance down the slope is

x12 + y12 =

( 287 m )

2

+ ( −77.0 m ) = 297 m. 2

Assess: With an initial speed of 112 mph (50 m/s) for the arrow, which is shot from a 15° slope at an angle of 20° above the horizontal, a horizontal distance of 287 m and a vertical distance of 77.0 m are reasonable numbers.

4.82.

Model: Visualize:

Treat the ball as a particle and apply the constant-acceleration equations of kinematics.

Solve: After the first bounce, the ball leaves the surface at 40° relative to the vertical or 50° relative to the horizontal. We first calculate the time t1 between the second bounce and the first bounce as follows:

x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ⇒ 3.0 m = 0 m + ( v0 cos50° ) t1 + 0 m ⇒ t1 = 2

3.0 m v0 cos50°

In this time, the ball undergoes a vertical displacement of y1 − y0 = −(3.0 m) tan 20° = −1.092 m. Substituting these values in the equation for the vertical displacement yields: y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 )

2

⎛ 3.0 m ⎞ 1 3.0 m ⎞ 2 ⎛ −1.092 m = 0 m + ( v0 sin 50° ) t1 − 12 gt12 = ( v0 sin 50° ) ⎜ ⎟ − 2 ( 9.8 m/s ) ⎜ ⎟ ⎝ v0 cos50° ⎠ ⎝ v0 cos50° ⎠ ⇒ −1.092 m − 3.575 m = Assess:

−106.73 m3 /s 2 ⇒ v0 = 4.78 m/s, or v0 = 4.8 m/s v02

A speed of 4.8 m/s or 10.7 mph on the first bounce is reasonable.

2

4.83.

Model: Treat the skateboarder as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xycoordinates for the second. The skateboarder’s final velocity at the top of the ramp is her initial velocity as she becomes airborne.

Solve: Without friction, the skateboarder’s acceleration on the ramp is a0 = − g sin 30° = −4.90 m/s 2 . The length

of the ramp is s1 = (1.0 m) / sin 30° = 2.0 m. We can use kinematics to find her speed at the top of the ramp:

v12 = v02 + 2a0 ( s1 − s0 ) = v02 + 2a0 s1 ⇒ v1 = (7.0 m/s)2 + 2(−4.90 m/s 2) (2.0 m) = 5.4 m/s This is the skateboarder’s initial speed into the air, giving her velocity components v1x = v1 cos30° = 4.7 m/s and v1 y = v1 cos30° = 2.7 m/s. We can use the y-equation of projectile motion to find her time in the air: y2 = 0 m = y1 + v1 yt2 + 12 a1 yt22 = 1.0 m + (2.7 m/s) t2 − (4.90 m/s 2) t22 This quadratic equation has roots t2 = −0.253 s (unphysical) and t2 = 0.805 s. The x-equation of motion is thus x2 = x1 + v1xt2 = 0 m + (4.7 m/s) t2 = 3.8 m She touches down 3.8 m from the end of the ramp.

4.84.

Model: motion. Visualize:

Solve:

Use the particle model for the motorcycle daredevil and apply the kinematic equations of

We need to find the coordinates of the landing ramp ( x1 , y1 ). We have

x1 = x0 + v0 x ( t1 − t0 ) = 0 m + ( 40 m/s ) t1 y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) = 0 m + 0 m + 12 ( −9.8 m/s 2 ) t12 = − ( 4.9 m/s 2 ) t12 2

G This means we must find t1. Since v1 makes an angle of 20° below the horizontal we can find v1y as follows:

v1 y v1x

= tan 20° ⇒ v1 y = −v1x tan 20° = − ( 40 m/s ) tan 20° = −14.56 m/s

We now use this value of v1y , a y = −9.8 m/s 2 , and v0y = 0 m/s in the following equation to obtain t1: v1 y = v0 y + a y ( t1 − t0 ) ⇒ −14.56 m/s = 0 m/s − ( 9.8 m/s 2 ) t1 ⇒ t1 = 1.486 s Now, we are able to obtain x1 and y1 using the above x- and y-equations:

x1 = ( 40 m/s )(1.486 s ) = 59.4 m

y1 = − ( 4.9 m/s 2 ) (1.486 s ) = −10.82 m 2

That is, the landing ramp should be placed 10.8 m lower and 59 m away from the edge of the horizontal platform.

4.85. Model: The train and projectile are treated in the particle model. The height of the cannon above the tracks is ignored. Visualize:

Solve:

In the ground reference frame, the projectile is launched with velocity components

( v0 x )P = v0 cosθ + vtrain

(v )

0y P

= v0 sinθ

While the projectile is in free fall, vfy = viy − g Δt. The time for the projectile to rise to the highest point is

(with vfy = 0) Δt =

viy g

. So the time to vertically rise and fall is t1 = 2Δt = 2

(v )

0y P

g

=

2v0 sin θ g

During this time the projectile travels a horizontal distance

( x1 )P = ( v0 x )P t1 =

2v0 2 2v v sinθ cosθ + 0 train sinθ g g

During the same time, the train travels a horizontal distance

( x1 )T = ( v0 x )T t1 +

1 2 2v0vtrain 2v 2 a at1 = sin θ + 02 sin 2 θ 2 g g

The range R is the difference between the two horizontal distances:

R = ( x1 )P − ( x1 )T =

2v0 2 ⎛ a 2 ⎞ ⎜ sin θ cosθ − sin θ ⎟ g ⎝ g ⎠

Note that the range is independent of vtrain the train’s steady motion. This makes sense, since the train and projectile share that motion when the projectile is launched. 2v 2 dR = 0. Thus (ignoring the constant 0 ) Maximizing the range R requires g dθ

dR a a = cos 2 θ − sin 2 θ − ( 2sinθ cosθ ) = cos 2θ − sin 2θ = 0 dθ g g Solving for θ ,

sin 2θ g 1 ⎛g⎞ = tan 2θ = ⇒ θ = tan −1 ⎜ ⎟ a cos 2θ 2 ⎝a⎠ Note that

a > 0 (train speeding up) gives θ < 45D and

a < 0 (train slowing down) gives θ > 45D ⎛g⎞ since tan −1 ⎜ ⎟ will be in the 2nd quadrant. ⎝a⎠ Assess: As a check, see what the angle θ is for the limiting case in which the train does not accelerate: 1 1 a = 0 ⇒ θ = tan −1 ( ∞ ) = × 90° = 45° 2 2 This is the expected answer.

4.86.

Model: Let the earth be frame S and the river be frame S′. Assume the river flows toward the east, which is the x and x′-axis. Visualize:

Solve:

In frame S′, the river’s frame, the child is at rest. The boat can go directly to the child at angle

θ = tan (200 /1500) = 7.595°. The boat’s speed is 8.0 m/s, so the components of the boat’s velocity in S′ are −1

v′x = −(8.0 m/s)cos7.595° = −7.93 m/s

v′y = (8.0 m/s)sin 7.595° = 1.06 m/s →

The river flows with velocity V = 2.0iˆ m/s relative to the earth. In the earth’s frame, which is also the frame of the riverbank and the boat dock, the boat’s velocity is vx = v′x = Vx = −5.93 m/s and v y = v′y + Vy = 1.06 m/s Thus the boat’s angle with respect to the riverbank is θ = tan −1 (5.93/1.06) = 10.1°. Assess: The boat, like the child, is being swept downstream. This moves the boat’s angle away from the shore.

4.87. Model: The airplanes are modeled as particles, and are undergoing relative motion according to the Galilean transformations of position and velocity. We designate Uri’s plane as frame S′ and the earth as frame S. G Frame S′ moves relative to frame S with velocity V . Visualize:

G G G G According to the Galilean transformation of velocity v = v′ + V , where v is the velocity of Val’s plane G G relative to the earth, v′ is the velocity of Val’s plane relative to Uri’s plane, and V is the velocity of Uri’s plane relative to the earth. We have G v = − ( 500 mph ) cos30°iˆ + ( 500 mph ) sin 30° ˆj

Solve:

G V = − ( 500 mph ) cos 20°iˆ − ( 500 mph ) sin 20° ˆj G G G v′ = v − V = ( 36.8 mph ) iˆ + ( 421 mph ) ˆj ⎛ 421 ⎞ ⎟ = 85° ⎝ 36.8 ⎠

θ = tan −1 ⎜

The fuselage of Val’s plane points 30° north of west. Val sees her plane moving in a direction 85° north of east. Thus the angle between the fuselage and the direction of motion is

φ = 180° − 30° − 85° = 65°

4.87.

Model: The airplanes are modeled as particles, and are undergoing relative motion according to the Galilean transformations of position and velocity. We designate Uri’s plane as frame S′ and the earth as frame S. G Frame S′ moves relative to frame S with velocity V . Visualize:

G G G G According to the Galilean transformation of velocity v = v′ + V , where v is the velocity of Val’s plane G G relative to the earth, v′ is the velocity of Val’s plane relative to Uri’s plane, and V is the velocity of Uri’s plane relative to the earth. We have G v = − ( 500 mph ) cos30°iˆ + ( 500 mph ) sin 30° ˆj Solve:

G V = − ( 500 mph ) cos 20°iˆ − ( 500 mph ) sin 20° ˆj G G G v′ = v − V = ( 36.8 mph ) iˆ + ( 421 mph ) ˆj

⎛ 421 ⎞ ⎟ = 85° ⎝ 36.8 ⎠

θ = tan −1 ⎜

The fuselage of Val’s plane points 30° north of west. Val sees her plane moving in a direction 85° north of east. Thus the angle between the fuselage and the direction of motion is

4.88. Model: The ball is a particle launched into projectile motion by the wheel. Visualize:

Solve: The initial velocity of the projectile is the tangential velocity at the point of release, and the direction is tangential to the wheel. The strategy is to find the initial velocity vi = ω f r , then use ω f to find the required angular acceleration α .

Since the release point is Δθ =

11 1 of a complete circle, ϕ = ( 2π rad ) = 30° remains. In the coordinate system of the 12 12

figure,

xi = −20sin30°cm = 10.0 cm

yi = 20cos30°cm = 17.3 cm

xf = 100 cm

yf = 0

ay = − g

vi x = vi cosθ

viy = vi sinθ

ax = 0

The launch angle ϕ is identified by calculating the angles in the right triangle highlighted in the figure and realizing the initial velocity is tangential to the circle. Thus the launch angle is also ϕ . We will use Equations 4.17 for the position of a projectile as a function of time, and use the fact that the time to travel the horizontal distance to the cup is the same for the vertical motion. First, find the flight time required for the ball to hit the cup from the horizontal motion: 100 cm = −10.0 cm + ( vi cos30° ) t ⇒ t =

110 cm 127 cm = vi cos30° vi

Substitute this time in the equations for the vertical motion: ⎛ 127 cm ⎞ 1 ⎛ 127 cm ⎞ 0 cm = 17.3 cm + vi sin 30° ⎜ ⎟− g⎜ ⎟ ⎝ vi ⎠ 2 ⎝ vi ⎠ Solving for the required initial velocity, v =

(127 cm ) g . 2 ( 80.8 cm )

2

The centimeter units cancel, and vi = 2.78 m/s.

Now it is time to consider the angular motion. The required final angular velocity

ωf =

vi 2.78 m/s = = 13.9 rad/s 0.20 m r

Using ω f2 = ωi2 + 2αΔθ ,

ω f2 − ωi2 (13.9 rad/s ) = 16.8 rad/s 2 = ⎛ 11 ⎞ 2 Δθ 2 ⎜ ( 2π ) rad ⎟ 2

α=

⎝ 12

Assess:

⎠

An angular acceleration of 16.8 rad/s 2 ≈ 2.7 rev/s 2 seems reasonable for a spring-loaded wheel.

4-1

Kinematics in Two Dimensions

4-2

5.1. Visualize:

Assess: walls.

Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse

5.2. Visualize:

5.3. Visualize:

5.4. Model: Assume friction is negligible compared to other forces. Visualize:

5.5. Visualize:

Assess: The bow and archer are no longer touching the arrow, so do not apply any forces after the arrow is released.

5.6. Model: An object’s acceleration is linearly proportional to the net force.

Solve: (a) One rubber band produces a force F, two rubber bands produce a force 2F, and so on. Because F ∝ a and two rubber bands (force 2F) produce an acceleration of 1.2 m/s2, four rubber bands will produce an acceleration of 2.4 m/s 2 . (b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2m). Using F = ma,

2 F = (2m) a ⇒ a = F/m = 0.6 m/s 2

5.7. Solve: Let the object have mass m and each rubber band exert a force F. For two rubber bands to accelerate the object with acceleration a, we must have a =

acceleration 3a to a mass

2F . We will need N rubber bands to give m

1 m. Find N: 2

3a = Three rubber bands are required.

NF ⎛ 2 F ⎞ 2 NF ⇒ 3⎜ ⇒ N = 3. ⎟= m m ⎝ m ⎠ 2

1

5.8. Visualize: Please refer to Figure EX5.8. Solve:

Mass is defined to be

m=

1 slope of the acceleration-versus-force graph

A larger slope implies a smaller mass. We know m2 = 0.20 kg, and we can find the other masses relative to m2 by comparing their slopes. Thus m1 1/slope 1 slope 2 1 2 = = = = = 0.40 m2 1/slope 2 slope 1 5 2 5 ⇒ m1 = 0.40m2 = 0.40 × 0.20 kg = 0.08 kg Similarly,

m3 1/slope 3 slope 2 1 5 = = = = = 2.50 m2 1/slope 2 slope 3 2 5 2 ⇒ m3 = 2.50m2 = 2.50 × 0.20 kg = 0.50 kg Assess: From the initial analysis of the slopes we had expected m3 > m2 and m1 < m2 . This is consistent with our numerical answers.

5.9. Visualize: Solve:

Please refer to Figure EX5.9. Mass is defined to be

m=

1 slope of the acceleration-versus-force graph

Thus −1

m1 ( slope of line 1) = m2 ( slope of line 2 )−1 −1

⎛ 5a1 ⎞ 3 ⎜ ⎟ 9 3N ⎠ ⎝ = =5= −1 5 25 ⎛ 3a1 ⎞ ⎜ ⎟ 3 ⎝ 5N ⎠

The ratio of masses is

Assess:

m1 9 = m2 25 More rubber bands produce a smaller acceleration on object 2, so it should be more massive.

5.10

Solve:

Use proportional reasoning. Given that distance traveled is proportional to the square of the d time, d ∝ t , so 2 should be constant. We have t 2.0 furlongs x = 2 2 ( 2.0 s ) ( 4.0 s ) Thus the distance traveled by the object in 4.0 s is x = 8.0 furlongs. Assess: A longer time should result in a longer distance traveled. 2

5.11 Solve: Use proportional reasoning. Let T = period of the pendulum, L = length of pendulum. We are T should be constant. We have L 3.0 s x = 2.0 m 3.0 m Solving, the period of the 3.0 m long pendulum is x = 3.7 s. Assess: Increasing the length increases the period, as expected.

told T ∝ L , so

5.12. Force is not necessary for motion. Constant velocity motion occurs in the absence of forces, that is, when the net force on an object is zero. Thus, it is incorrect to say that “force causes motion.” Instead, force causes acceleration. That is, force causes a change in the motion of an object, and acceleration is the kinematic quantity G that measures a change of motion. Newton’s second law quantifies this idea by stating that the net force Fnet on an object of mass m causes the object to undergo an acceleration: G G F a = net m The acceleration vector and the net force vector must point in the same direction.

5.13. Visualize:

Solve:

(a) Newton’s second law is F = ma. When F = 2 N, we have 2 N = (0.5 kg) a, hence a = 4 m/s 2 .

(b) When F = 1 N, we have 1 N = (0.5 kg) a, hence a = 2 m/s 2 . After repeating this procedure at various points, the above graph is obtained.

5.14 Solve: Newton’s second law tells us that F = ma. Compute F for each case: (a) F = (0.200 kg)(5 m/s 2 ) = 1N . (b) F = (0.200 kg)(10 m/s 2 ) = 2 N . Assess: To double the acceleration we must double the force, as expected.

5.15. Visualize: Please refer to Figure EX5.15.

Solve: Newton’s second law is F = ma. We can read a force and an acceleration from the graph, and hence find the mass. Choosing the force F = 1 N gives us a = 4 m/s 2 . Newton’s second law yields m = 0.25 kg.

5.16. Solve: (a) This problem calls for an estimate so we are looking for an approximate answer. Table 5.3 gives us no information on laptops, but does give the weight of a one-pound object. Place a pound weight in one hand and the laptop on the other. The sensation on your hand is the weight of the object. The sensation from the laptop is about five times the sensation from the pound weight. So we conclude the weight of the laptop is about five times the weight of the one-pound object or about 25 N. (b) According to Table 5.3, the propulsion force on a car is 5000 N. A bicycle (including the rider) is about 100 kg. This is about one-tenth of the mass of a car, which is about 1000 kg for a compact model. The acceleration of a bicycle is somewhat less than that of a car, let’s guess about one-fifth. We can write Newton’s second law as follows:

F ( bicycle ) =

1 1 5000 N = 100 N ( mass of car ) × ( acceleration of car ) = 10 5 50

So we would roughly estimate the propulsion force of a bicycle to be 100 N.

5.17. Solve: (a) This problem calls for an estimate so we are looking for an approximate answer. Table 5.3 gives us no information on pencils, but does give us the weight of the U.S. quarter. Put the quarter on one hand and a pencil on the other hand. The sensation on your hand is the weight of the object. The sensation from the quarter is about the same as the sensation from the pencil. So they both have about the same weight. We can estimate the weight of the pencil to be 0.05 N. (b) According to Table 5.3, the propulsion force on a car is 5000 N. The mass of a sprinter is about 100 kg. This is about one-tenth of the mass of a car, which is about 1000 kg for a compact model. The acceleration of a sprinter is somewhat less than that of a car, let’s guess about one-fifth. We can write Newton’s second law as follows: F ( sprinter ) =

1 1 5000 N = 100 N ( mass of car ) × ( acceleration of car ) = 10 5 50

So, we would roughly estimate the propulsion force of a sprinter to be 100 N. Assess: This is the same estimated number as we obtained in Exercise 5.16. This is reasonable since in both the cases the propulsion force comes from a human and it probably does not matter how the human is providing that force.

5.18. Visualize:

G G G Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite direction so that the sum of all the three forces is zero.

5.19. Visualize:

G G G Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite direction so that the sum of all three forces is zero.

5.20. Visualize:

G G G Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite direction so that the sum of all the three forces is zero.

5.21. Visualize:

Solve: The free-body diagram shows two equal and opposite forces such that the net force is zero. The force directed down is labeled as a gravitational force, and the force directed up is labeled as a tension. With zero net force the acceleration is zero. So, a possible description is: “An object hangs from a rope and is at rest.” Or, “An object hanging from a rope is moving up or down with a constant speed.”

5.22. Visualize:

Solve: The free-body diagram shows three forces with a net force (and therefore net acceleration) upward. G G G There is a force labeled FG directed down, a force Fthrust directed up, and a force D directed down. So a possible description is: “A rocket accelerates upward.”

5.23. Visualize:

G The free-body diagram shows three forces. There is a gravitational force FG , which is down. There is a G G G normal force labeled n, which is up. The forces FG and n are shown with vectors of the same length so they are equal in magnitude and the net G vertical force is zero. So we have an object on the ground which is not moving vertically. There is also a force f k to the left. This must be a frictional force and we need to decide whether it is G static or kinetic friction. The frictional force is the only horizontal force so the net horizontal force must be f k . This means there is a net force to the left producing an acceleration to the left. This all implies motion and therefore the frictional force is kinetic. A possible description is: “A baseball player is sliding into second base.” Solve:

5.24. Visualize:

Assess: you.

G G There is a gravitational force FG . You are touching the park bench, so it exerts a contact force n on

5.25. Visualize:

Assess: The problem says that there is no friction and it tells you nothing about any drag; so we do not include either of these forces. The only remaining forces are the weight and the normal force.

5.26. Visualize:

Assess:

Since the velocity is constant, the acceleration is zero, and the net force is zero.

5.27. Visualize:

Assess: The problem uses the word “sliding.” Any real situation involves friction with the surface. Since we are not told to neglect it, we show that force.

5.28. Visualize:

Figure (a) shows velocity as downward, so the object is moving down. The length of the vector increases with each step showing that the speed is increasing (like a dropped ball). Thus, the acceleration is directed down. G G Since F = ma the force is in the same direction as the acceleration and must be directed down. Figure (b), however, shows the velocity as upward, so the object is moving upward. But the length of the vector decreases with each step showing that the speed is decreasing (like a ball thrown up). Thus, the acceleration is also directed down. As in part (a) the net force must be directed down.

5.29. Visualize:

The velocity vector in figure (a) is shown downward and to the left. So movement is downward and to the left. The velocity vectors get successively longer, which means the speed is increasing. Therefore the acceleration is G G downward and to the left. By Newton’s second law F = ma , the net force must be in the same direction as the acceleration. Thus, the net force is downward and to the left. The velocity vector in (b) is shown to be upward and to the right. So movement is upward and to the right. The velocity vector gets successively shorter, which means the speed is decreasing. Therefore the acceleration is downward and to the left. From Newton’s second law, the net force must be in the direction of the acceleration and so it is directed downward and to the left.

5.30. Visualize:

Solve: According to Newton’s second law F = ma, the force at any time is found simply by multiplying the value of the acceleration by the mass of the object.

5.31. Visualize:

Solve: According to Newton’s second law F = ma, the force at any time is found simply by multiplying the value of the acceleration by the mass of the object.

5.32. Visualize:

Solve: According to Newton’s second law F = ma, the acceleration at any time is found simply by dividing the value of the force by the mass of the object.

5.33. Visualize:

Solve: According to Newton’s second law F = ma, the acceleration at any time is found simply by dividing the value of the force by the mass of the object.

5.34. Model: Use the particle model for the object. Solve:

(a) We are told that for an unknown force (call it F0 ) acting on an unknown mass (call it m0 ) the

acceleration of the mass is 10 m/s 2 . According to Newton’s second law, F0 = m0 (10 m/s 2 ). The force then

becomes

1 2

F0 . Newton’s second law gives 1 1 F0 = m0 a = ⎡⎣ m0 (10 m s 2 ) ⎤⎦ 2 2

This means a is 5 m/s 2 . (b) The force is F0 and the mass is now

1 2

m0 . Newton’s second law gives

1 F0 = m0 a = m0 (10 m s 2 ) 2 This means a = 20 m/s 2 . (c) A similar procedure gives a = 10 m/s 2 . (d) A similar procedure gives a = 2.5 m/s 2 .

5.35. Model: Use the particle model for the object. Solve:

(a) We are told that for an unknown force (call it F0 ) acting on an unknown mass (call it m0 ) the

acceleration of the mass is 8 m/s 2 . According to Newton’s second law, F0 = m0 (8 m/s 2 ). The force then becomes

2 F0 . Newton’s second law gives 2 F0 = m0 a = 2 ⎡⎣ m0 ( 8 m s 2 ) ⎤⎦ This means a is 16 m/s 2 . (b) The force is F0 and the mass is now 2m0 . Newton’s second law gives

F0 = 2m0 a = m0 ( 8 m s 2 ) This means a = 4 m/s 2 . (c) A similar procedure gives a = 8 m/s 2 . (d) A similar procedure gives a = 32 m/s 2 .

5.36. Visualize:

Solve: (d) There are a normal force and a gravitational force which are equal and opposite, so this is an object on a horizontal surface. The description could be: “A tow truck pulls a stuck car out of the mud.”

5.37. Visualize:

Solve: (d) There is a normal force and a gravitational force which are equal and opposite, so this is an object on a horizontal surface, or at least balanced in the vertical direction. The description of this free-body diagram could be: “A jet plane is flying at constant speed.”

5.38. Visualize:

Solve: (d) This is an object on a surface because FG = n. It must be moving to the left because the kinetic friction is to the right. The description of the free-body diagram could be: “A compressed spring is shooting a plastic block to the left.”

5.39. Visualize:

Solve: (d) There is only a single force of weight. We are unable to tell the direction of motion. The description could be: “Galileo has dropped a ball from the Leaning Tower of Pisa.”

5.40. Visualize:

Solve: (d) There is an object on an inclined surface. The net force is down the plane so the acceleration is down the plane. The net force includes both the frictional force and the component of the gravitational force. The direction of the force of kinetic friction implies that the object is moving upward. The description could be: “A car is skidding up an embankment.”

5.41. Visualize:

Solve: (d) There is an object on an inclined surface with a tension force down the surface. There is a small frictional force up the surface implying that the object is sliding down the slope. A description could be: “A sled is being pulled down a slope with a rope that is parallel to the slope.”

5.42. Visualize:

Solve: (d) There is a thrust at an angle to the horizontal and a gravitational force. There is no normal force so the object is not on a surface. The description could be: “A rocket is fired at an angle to the horizontal and there is no drag force.”

5.43. Visualize:

Tension is the only contact force. The downward acceleration implies that FG > T .

5.44. Visualize:

5.45. Visualize:

The normal force is perpendicular to the ground. The thrust force is parallel to the ground and in the direction of acceleration. The drag force is opposite to the direction of motion.

5.46. Visualize:

The normal force is perpendicular to the hill. The frictional force is parallel to the hill.

5.47. Visualize:

The normal force is perpendicular to the hill. The kinetic frictional force is parallel to the hill and directed upward opposite to the direction of motion. The wind force is given as horizontal. Since the skier stays on the slope (that is, there is no acceleration away from the slope) the net force must be parallel to the slope.

5.48. Visualize:

As the rock slides there is kinetic friction between it and the rough concrete sidewalk. Since the rock stays on the level surface, the net force must be along that surface, and is equal to the kinetic friction.

5.49. Visualize:

The drag force due to air is opposite the motion.

5.50. Visualize:

The ball rests on the floor of the barrel because the gravitational force is equal to the normal force. There is a force of the spring to the right which causes an acceleration.

5.51. Visualize:

There are no contact forces on the rock. The gravitational force is the only force acting on the rock.

5.52. Visualize:

The gymnast experiences the long range force of gravity. There is also a contact force from the trampoline that is the normal force of the trampoline on the gymnast. The gymnast is moving downward and the trampoline is decreasing her speed, so the acceleration is upward and there is a net force upward. Thus the normal force must be larger than the gravitational force. The actual behavior of the normal force over time will be complicated as it involves the stretching of the trampoline and therefore tensions.

5.53. Visualize:

You can see from the motion diagram that the box accelerates to the right along with the truck. According to G G Newton’s second law, F = ma , there must be a force to the right acting on the box. This is friction, but not kinetic friction. The box is not sliding against the truck. Instead, it is static friction, the force that prevents slipping. Were it not for static friction, the box would slip off the back of the truck. Static friction acts in the direction needed to prevent slipping. In this case, friction must act in the forward (toward the right) direction.

5.54. Visualize:

You can see from the motion diagram that the bag accelerates to the left along with the car as the car slows G G down. According to Newton’s second law, F = ma , there must be a force to the left acting on the bag. This is friction, but not kinetic friction. The bag is not sliding across the seat. Instead, it is static friction, the force that prevents slipping. Were it not for static friction, the bag would slide off the seat as the car stops. Static friction acts in the direction needed to prevent slipping. In this case, friction must act in the backward (toward the left) direction.

5.55. Visualize: (a)

(b)

(c)

(d) The ball accelerates downward until the instant when it makes contact with the ground. Once it makes contact, it begins to compress and to slow down. The compression takes a short but nonzero distance, as shown in the motion diagram. The point of maximum compression is the turning point, where the ball has an instantaneous speed of v = 0 m/s and reverses direction. The ball then expands and speeds up until it loses G contact with the ground. The motion diagram shows that the acceleration vector a points upward the entireG time that the ball is in contact with the ground. An upward acceleration implies that there is a net upward force Fnet on the ball. The only two forces on the ball are the gravitational force downward and the normal force of the ground upward. To have a net force upward requires n > FG . So the ball bounces because the normal force of the ground exceeds the gravitational force, causing a net upward force during the entire time that the ball is in contact with the ground. This net upward force slows the ball, turns it, and accelerates it upward until it loses contact with the ground. Once contact with the ground is lost, the normal force vanishes and the ball is simply in free fall.

5.56. Visualize: (a)

You are sitting on a bench driving along to the right. Both you and the bench are moving with a constant speed. There is a force on you due to gravity, which is directed down. There is a contact force between you and the bench, which is directed up. Since you are not accelerating up or down the net vertical force on you is zero, which means the two vertical forces are equal in magnitude. The statement of the problem gives no indication of any other contact forces. Specifically, we are told that the bench is very slippery. We can take this to mean there is no frictional force. So our force diagram includes only the normal force up, the gravitational force down, and no horizontal force. (b) The above considerations lead to the free-body diagram that is shown. (c) The car (and therefore the bench) slows down. Does this create any new force on you? No. The forces remain the same. This means the pictorial representation and the free-body diagram are unchanged. (d) The car slows down because of some new contact force on the car (maybe the brakes lock the wheels and the road exerts a force on the tires). But there is no new contact force on you. So the force diagram for you remains unchanged. There are no horizontal forces on you. You do not slow down and you continue at an unchanged velocity until something in the picture changes for you (for example, you fall off the bench or hit the windshield). (e) The net force on you has remained zero because the net vertical force is zero and there are no horizontal forces at all. According to Newton’s first law if the net force on you is zero, then you continue to move in a straight line with a constant velocity. That is what happens to you when the car slows down. You continue to move forward with a constant velocity. The statement that you are “thrown forward” is misleading and incorrect. To be “thrown” there would need to be a net force on you and there is none. It might be correct to say that the car has been “thrown backward” leaving you to continue onward (until you part company with the bench). (f) We are now asked to consider what happens if the bench is NOT slippery. That implies there is a frictional force between the bench and you. This force is certainly horizontal (parallel to the surface of the bench). Is the frictional force directed forward (in the direction of motion) or backward? The car is slowing down and you are staying on the bench. That means you are slowing down with the bench. Your velocity to the right is decreasing (you are moving right and slowing down) so you are accelerating to the left. By Newton’s second law that means the force producing the acceleration must be to the left. That force is the force of static friction and it is shown on the free-body diagram below. Of course, when the car accelerates (increases in speed to the right) and you accelerate with it, then your acceleration is to the right and the frictional force must be to the right.

5-1

Force and Motion

5-2

6.1.

Model: Visualize:

Solve:

We can assume that the ring is a single massless particle in static equilibrium.

Written in component form, Newton’s first law is

( Fnet ) x = ΣFx = T1x + T2 x + T3 x = 0 N ( Fnet ) y = ΣFy = T1 y + T2 y + T3 y = 0 N Evaluating the components of the force vectors from the free-body diagram:

T1x = −T1 T2 x = 0 N T3 x = T3 cos30° T1 y = 0 N T2 y = T2 T3 y = −T3 sin 30° Using Newton’s first law: −T1 + T3 cos30° = 0 N T2 − T3 sin 30° = 0 N Rearranging:

T1 = T3 cos30° = (100 N )( 0.8666 ) = 86.7 N T2 = T3 sin 30° = (100 N )( 0.5 ) = 50.0 N G Assess: Since T3 acts closer to the x-axis than to the y-axis, it makes sense that T1 > T2 .

6.2.

Model: Visualize:

We can assume that the ring is a particle.

This is a static equilibrium problem. We will ignore the weight of the ring, because it is “very light,” so the only three forces are the tension forces shown in the free-body diagram. Note that the diagram defines the angle θ . G Solve: Because the ring is in equilibrium it must obey Fnet = 0 N. This is a vector equation, so it has both xand y-components:

( Fnet ) x = T3 cosθ − T2 = 0 N ⇒ T3 cosθ = T2 ( Fnet ) y = T1 − T3 sinθ = 0 N ⇒ T3 sinθ = T1 We have two equations in the two unknowns T3 and θ . Divide the y-equation by the x-equation:

T3 sin θ T 80 N = tanθ = 1 = = 1.6 ⇒ θ = tan −1 (1.6 ) = 58° T3 cosθ T2 50 N Now we can use the x-equation to find T3 =

T2 50 N = = 94 N cosθ cos58°

The tension in the third rope is 94 N directed 58° below the horizontal.

6.3.

Model: We assume the speaker is a particle in static equilibrium under the influence of three forces: gravity and the tensions in the two cables. Visualize:

Solve:

From the lengths of the cables and the distance below the ceiling we can calculate θ as follows:

sinθ =

2m = 0.677 ⇒ θ = sin −1 0.667 = 41.8° 3m

Newton’s first law for this situation is

( Fnet ) x = ΣFx = T1x + T2 x = 0 N ⇒ −T1 cosθ + T2 cosθ = 0 N ( Fnet ) y = ΣFy = T1 y + T2 y + wy = 0 N ⇒ T1 sinθ + T2 sinθ − w = 0 N The x-component equation means T1 = T2 . From the y-component equation:

( 20 kg ) ( 9.8 m/s w mg 2T1 sin θ = w ⇒ T1 = = = 2sin θ 2sinθ 2sin 41.8°

2

) = 196 N = 147 N 1.333

Assess: It’s to be expected that the two tensions are equal, since the speaker is suspended symmetrically from the two cables. That the two tensions add to considerably more than the weight of the speaker reflects the relatively large angle of suspension.

6.4.

Model: We can assume that the coach and his sled are a particle being towed at a constant velocity by the two ropes, with friction providing the force that resists the pullers. Visualize:

Solve:

Since the sled is not accelerating, it is in dynamic equilibrium and Newton’s first law applies:

( Fnet ) x = ΣFx = T1x + T2 x + f kx = 0 N

( Fnet ) y = ΣFy = T1 y + T2 y +

f ky = 0 N

From the free-body diagram: ⎛1 ⎞ ⎛1 ⎞ T1 cos ⎜ θ ⎟ + T2 cos ⎜ θ ⎟ − f k = 0 N ⎝2 ⎠ ⎝2 ⎠

⎛1 ⎞ ⎛1 ⎞ T1 sin ⎜ θ ⎟ − T2 sin ⎜ θ ⎟ + 0 N = 0 N ⎝2 ⎠ ⎝2 ⎠

From the second of these equations T1 = T2 . Then from the first:

2T1 cos10° = 1000 N ⇒ T1 =

1000 N 1000 N = = 508 N 2cos10° 1.970

Assess: The two tensions are equal, as expected, since the two players are pulling at the same angle. The two add up to only slightly more than 1000 N, which makes sense because the angle at which the two players are pulling is small.

6.5. Solve:

Visualize: Please refer to the Figure EX6.5. Applying Newton’s second law to the diagram on the left,

ax =

( Fnet ) x m

=

4 N−2N = 1.0 m/s 2 2 kg

ay =

( Fnet ) y m

=

3 N−3 N = 0 m/s 2 2 kg

For the diagram on the right: ax =

( Fnet ) x m

=

4 N−2 N = 1.0 m/s 2 2 kg

ay =

( Fnet ) y m

=

3 N −1 N − 2 N = 0 m/s 2 2 kg

6.6. Visualize: Please refer to Figure EX6.6. Solve: For the diagram on the left, three of the vectors lie along the axes of the tilted coordinate system. Notice that the angle between the 3 N force and the –y-axis is the same 20° by which the coordinates are tilted. Applying Newton’s second law, ax =

( Fnet ) x m

ay =

=

( Fnet ) y m

5 N − 1 N − ( 3sin 20° ) N = 1.49 m/s 2 2 kg

=

2.82 N − ( 3cos 20° ) N = 0 m/s 2 2 kg

For the diagram on the right, the 2-newton force in the first quadrant makes an angle of 15° with the positive xaxis. The other 2-newton force makes an angle of 15° with the negative y-axis. The accelerations are ax = ay =

( Fnet ) x ( 2cos15° ) N + ( 2sin15° ) N − 3 N = = −0.28 m/s 2 2 kg

m

( Fnet ) y m

=

1.414 N + ( 2sin15° ) N − ( 2cos15° ) N = 0 m/s 2 2 kg

6.7. Solve:

Visualize: Please refer to Figure EX6.7. (a) Apply Newton’s second law in both the x and y directions.

( Fnet ) x = ( 5.0 N ) cos37° − 2.0 N = ( 5.0 kg ) ax ⇒ ax = 0.40 m/s 2

( Fnet ) y = 2.0 N + ( 5.0 N ) sin 37° − 5.0 N = ( 5.0 kg ) a y ⇒ a y = 0.0 m/s 2

(b) The angle that the 5.0 N force makes with the –y-axis is 37°. Apply Newton’s second law for both the x and y direction.

( Fnet ) x = 3.0 N + ( 5.0 N ) sin 37° − 2.0 N = ( 5.0 kg ) ax ⇒ ax = 0.80 m/s 2

( Fnet ) y = 4.0 N − ( 5.0 N ) cos37° = ( 5.0 kg ) a y ⇒ a y = 0.0 m/s 2 Assess: The orientation of the coordinate axes is chosen for convenience, and does not always need to conform to the horizontal and vertical.

6.8. Visualize: Please refer to Figure EX6.8. Solve: We can use the constant slopes of the three segments of the graph to calculate the three accelerations. For t between 0 s and 3 s, ax =

Δvx 12 m/s − 0 s = = 4 m/s 2 3s Δt

For t between 3 s and 6 s, Δvx = 0 m/s, so ax = 0 m/s 2 . For t between 6 s and 8 s,

Δvx 0 m/s − 12 m/s = = −6 m/s 2 Δt 2s From Newton’s second law, at t = 1 s we have ax =

Fnet = max = (2.0 kg)(4 m/s 2 ) = 8 N At t = 4 s, ax = 0 m/s 2 , so Fnet = 0 N. At t = 7 s,

Fnet = max = (2.0 kg)(−6.0 m/s 2 ) = −12 N Assess: The magnitudes of the forces look reasonable, given the small mass of the object. The positive and negative signs are appropriate for an object first speeding up, then slowing down.

6.9. Visualize: Please refer to Figure EX6.9. Positive forces result in the object gaining speed and negative forces result in the object slowing down. The final segment of zero force is a period of constant speed. Solve: We have the mass and net force for all the three segments. This means we can use Newton’s second law to calculate the accelerations. The acceleration from t = 0 s to t = 3 s is ax =

Fx 4N = = 2 m/s 2 m 2.0 kg

ax =

−2 N Fx = = −1 m/s 2 m 2.0 kg

The acceleration from t = 3 s to t = 5 s is

The acceleration from t = 5 s to 8 s is ax = 0 m/s 2 . In particular, ax ( at t = 6 s ) = 0 m/s 2 . We can now use one-dimensional kinematics to calculate v at t = 6 s as follows: v = v0 + a1 ( t1 − t0 ) + a2 ( t2 − t0 )

= 0 + ( 2 m/s 2 ) ( 3 s ) + ( −1 m/s 2 ) ( 2 s ) = 6 m/s − 2 m/s = 4 m/s

Assess: The positive final velocity makes sense, given the greater magnitude and longer duration of the G positive F1. A velocity of 4 m/s also seems reasonable, given the magnitudes and directions of the forces and the mass involved.

6.10.

Model: We assume that the box is a particle being pulled in a straight line. Since the ice is frictionless, the tension in the rope is the only horizontal force. Visualize:

Solve: (a) Since the box is at rest, ax = 0 m/s 2 , and the net force on the box must be zero. Therefore, according to Newton’s first law, the tension in the rope must be zero. (b) For this situation again, ax = 0 m/s 2 , so Fnet = T = 0 N. (c) Here, the velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since ax = 5.0 m/s 2 ,

Fnet = T = max = ( 50 kg ) ( 5.0 m/s 2 ) = 250 N Assess: For parts (a) and (b), the zero acceleration immediately implies that the rope is exerting no horizontal force on the box. For part (c), the 250 N force (the equivalent of about half the weight of a small person) seems reasonable to accelerate a box of this mass at 5.0 m/s 2 .

6.11.

Model: We assume that the box is a point particle that is acted on only by the tension in the rope and the pull of gravity. Both the forces act along the same vertical line. Visualize:

Solve:

(a) Since the box is at rest, a y = 0 m/s 2 and the net force on it must be zero:

Fnet = T − FG = 0 N ⇒ T = FG = mg = ( 50 kg ) ( 9.8 m/s 2 ) = 490 N (b) Since the box is rising at a constant speed, again a y = 0 m/s 2 , Fnet = 0 N, and T = FG = 490 N. (c) The velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since a y = 5.0 m/s 2 ,

Fnet = T − FG = ma y = ( 50 kg ) ( 5.0 m/s 2 ) = 250 N ⇒ T = 250 N + w = 250 N + 490 N = 740 N (d) The situation is the same as in part (c), except that the rising box is slowing down. Thus a y = −5.0 m/s 2 and

we have instead Fnet = T − FG = ma y = ( 50 kg ) ( −5.0 m/s 2 ) = −250 N ⇒ T = −250 N + FG = −250 N + 490 N = 240 N Assess: For parts (a) and (b) the zero accelerations immediately imply that the gravitational force on the box must be exactly balanced by the upward tension in the rope. For part (c) the tension not only has to support the gravitational force on the box but must also accelerate it upward, hence, T must be greater than FG . When the

box accelerates downward, the rope need not support the entire gravitational force, hence, T is less than FG .

6.12.

Model: We assume the rocket is a particle moving in a vertical straight line under the influence of only two forces: gravity and its own thrust. Visualize:

Solve:

(a) Using Newton’s second law and reading the forces from the free-body diagram,

Fthrust − FG = ma ⇒ Fthrust = ma + mg Earth = ( 0.200 kg ) (10 m/s 2 + 9.80 m/s 2 ) = 3.96 N (b) Likewise, the thrust on the moon is (0.200 kg)(10 m/s 2 + 1.62 m/s 2 ) = 2.32 N. Assess: The thrust required is smaller on the moon, as it should be, given the moon’s weaker gravitational pull. The magnitude of a few newtons seems reasonable for a small model rocket.

6.13. Solve:

Model: The astronaut is treated as a particle. The mass of the astronaut is

m=

wearth 800 N = = 81.6 kg g earth 9.80 m/s 2

Therefore, the weight of the astronaut on Mars is wMars = mg Mars = ( 81.6 kg ) ( 3.76 m/s 2 ) = 307 N Assess: The smaller acceleration of gravity on Mars reveals that objects are less strongly attracted to Mars than to the earth, so the smaller weight on Mars makes sense.

6.14. Solve:

Model: Use the particle model for the woman. (a) The woman’s weight on the earth is

wearth = mg earth = ( 55 kg ) ( 9.80 m/s 2 ) = 540 N (b) Since mass is a measure of the amount of matter, the woman’s mass is the same on the moon as on the earth. Her weight on the moon is

wmoon = mg moon = ( 55 kg ) (1.62 m/s 2 ) = 89 N Assess: The smaller acceleration due to gravity on the moon reveals that objects are less strongly attracted to the moon than to the earth. Thus the woman’s smaller weight on the moon makes sense.

6.15.

Model: We assume that the passenger is a particle subject to two vertical forces: the downward pull of gravity and the upward push of the elevator floor. We can use one-dimensional kinematics and Equation 6.10. Visualize:

Solve:

(a) The weight is

⎛ a ⎞ ⎛ 0⎞ w = mg ⎜1 + y ⎟ = mg ⎜1 + ⎟ = mg = ( 60 kg ) ( 9.80 m/s 2 ) = 590 N g⎠ ⎝ g⎠ ⎝ (b) The elevator speeds up from v0 y = 0 m/s to its cruising speed at v y = 10 m/s. We need its acceleration before

we can find the apparent weight: ay =

Δv 10 m/s − 0 m/s = = 2.5 m/s 2 4.0 s Δt

The passenger’s weight is

⎛ ⎛ a ⎞ 2.5 m/s 2 ⎞ w = mg ⎜ 1 + y ⎟ = ( 590 N ) ⎜1 + = (590 N)(1.26) = 740 N 2 ⎟ g ⎠ ⎝ ⎝ 9.80 m/s ⎠ (c) The passenger is no longer accelerating since the elevator has reached its cruising speed. Thus, w = mg = 590 N as in part (a). Assess: The passenger’s weight is the gravitational force on the passenger in parts (a) and (c), since there is no acceleration. In part (b), the elevator must not only support the gravitational force but must also accelerate him upward, so it’s reasonable that the floor will have to push up harder on him, increasing his weight.

6.16.

Model: We assume that the passenger is a particle acted on by only two vertical forces: the downward pull of gravity and the upward force of the elevator floor. Visualize: Please refer to Figure EX6.16. The graph has three segments corresponding to different conditions: (1) increasing velocity, meaning an upward acceleration; (2) a period of constant upward velocity; and (3) decreasing velocity, indicating a period of deceleration (negative acceleration). Solve: Given the assumptions of our model, we can calculate the acceleration for each segment of the graph and then apply Equation 6.10. The acceleration for the first segment is

ay =

v1 − v0 8 m/s − 0 m/s = = 4 m/s 2 t1 − t0 2 s−0 s

⎛ ⎛ a ⎞ 4 m/s 2 ⎞ 4 ⎞ ⎛ ⇒ w = mg ⎜1 + y ⎟ = mg ⎜1 + = ( 75 kg ) ( 9.80 m/s 2 ) ⎜1 + ⎟ = 1035 N 2 ⎟ g 9.80 m/s 9.80 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ For the second segment, a y = 0 m/s 2 and the weight is ⎛ 0 m/s 2 ⎞ 2 w = mg ⎜1 + ⎟ = mg = ( 75 kg ) ( 9.80 m/s ) = 740 N g ⎠ ⎝ For the third segment,

ay =

v3 − v2 0 m/s − 8 m/s = = −2 m/s 2 t3 − t 2 10 s − 6 s

⎛ −2 m/s 2 ⎞ ⇒ w = mg ⎜1 + = ( 75 kg ) ( 9.80 m/s 2 ) (1 − 0.2 ) = 590 N 2 ⎟ 9.80 m/s ⎝ ⎠ Assess: As expected, the weight is greater than the gravitational force on the passenger when the elevator is accelerating upward and lower than normal when the acceleration is downward. When there is no acceleration the weight is the gravitational force. In all three cases the magnitudes are reasonable, given the mass of the passenger and the accelerations of the elevator.

6.17.

Model: We assume that the safe is a particle moving only in the x-direction. Since it is sliding during the entire problem, we can use the model of kinetic friction. Visualize:

Solve: The safe is in equilibrium, since it’s not accelerating. Thus we can apply Newton’s first law in the vertical and horizontal directions:

( Fnet ) x = ΣFx = FB + FC − f k = 0 N

⇒ f k = FB + FC = 350 N + 385 N = 735 N

( Fnet ) y = ΣFy = n − FG = 0 N ⇒ n = FG = mg = ( 300 kg ) ( 9.80 m/s2 ) = 2.94 × 103 N Then, for kinetic friction: fk = μk n ⇒ μk =

fk 735 N = = 0.250 n 2.94 × 103 N

Assess: The value of μ k = 0.250 is hard to evaluate without knowing the material the floor is made of, but it seems reasonable.

6.18.

Model: We assume that the mule is a particle acted on by two opposing forces in a single line: the farmer’s pull and friction. The mule will be subject to static friction until (and if!) it begins to move; after that it will be subject to kinetic friction. Visualize:

Solve: Since the mule does not accelerate in the vertical direction, the free-body diagram shows that n = FG = mg . The maximum friction force is

f smax = μs mg = ( 0.8 )(120 kg ) ( 9.80 m/s 2 ) = 940 N

The maximum static friction force is greater than the farmer’s maximum pull of 800 N; thus, the farmer will not be able to budge the mule. Assess: The farmer should have known better.

6.19.

Model: Visualize:

We will represent the crate as a particle.

G G (a) When the belt runs at constant speed, the crate has an acceleration a = 0 m/s 2 and is in dynamic G G equilibrium. Thus Fnet = 0. It is tempting to think that the belt exerts a friction force on the crate. But if it did, there would be a net force because there are no other possible horizontal forces to balance a friction force. Because there is no net force, there cannot be a friction force. The only forces are the upward normal force and the gravitational force on the crate. (A friction force would have been needed to get the crate moving initially, but no horizontal force is needed to keep it moving once it is moving with the same constant speed as the belt.) (b) If the belt accelerates gently, the crate speeds up without slipping on the belt. Because it is accelerating, the crate must have a net horizontal force. So now there is a friction force, and the force points in the direction of the crate’s motion. Is it static friction or kinetic friction? Although the crate is moving, there is no motion of the crate relative to the belt. Thus, it is a static friction force that accelerates the crate so that it moves without slipping on the belt. (c) The static friction force has a maximum possible value ( f s ) max = μ s n. The maximum possible acceleration of the crate is Solve:

amax =

( fs )max μs n =

m m If the belt accelerates more rapidly than this, the crate will not be able to keep up and will slip. It is clear from the free-body diagram that n = FG = mg . Thus, amax = μs g = (0.5)(9.80 m/s 2 ) = 4.9 m/s 2

6.20.

Model: Visualize:

We assume that the truck is a particle in equilibrium, and use the model of static friction.

Solve: The truck is not accelerating, so it is in equilibrium, and we can apply Newton’s first law. The normal force has no component in the x-direction, so we can ignore it here. For the other two forces:

( Fnet ) x = ΣFx =

fs − ( FG ) x = 0 N ⇒ fs = ( FG ) x = mg sin θ = ( 4000 kg ) ( 9.80 m/s 2 ) ( sin15° ) = 10,145 N

Assess: The truck’s weight (mg) is roughly 40,000 N. A friction force that is ≈ 25% of the truck’s weight seems reasonable.

6.21. Model: The car is a particle subject to Newton’s laws and kinematics. Visualize:

Solve: Kinetic friction provides a horizontal acceleration which stops the car. From the figure, applying Newton’s first and second laws gives

∑F

x

∑F

y

= − f k = max

= n − FG = 0 ⇒ n = FG = mg

Combining these two equations with f k = μ k n yields

ax = − μk g = − ( 0.50 ) ( 9.80 m/s 2 ) = −4.9 m/s Kinematics can be used to determine the initial velocity.

vf2 = vi2 + 2aΔx ⇒ vi2 = −2ax Δx Thus

vi = −2 ( −4.9 m/s 2 ) ( 65 m − 0 m ) = 25 m/s 2

Assess: The initial speed of 25 m/s 2 ≈ 56 mph is a reasonable speed to have initially for a vehicle to leave 65meter-long skid marks.

6.22.

Model: We assume that the plane is a particle accelerating in a straight line under the influence of two forces: the thrust of its engines and the rolling friction of the wheels on the runway. We can use one-dimensional kinematics. Visualize:

Solve:

We can use the definition of acceleration to find a, and then apply Newton’s second law. We obtain: a=

Δv 82 m/s − 0 m/s = = 2.34 m/s 2 Δt 35 s

( Fnet ) = ΣFx = Fthrust − f r = ma ⇒ Fthrust =

f r + ma

For rubber rolling on concrete, μ r = 0.02 (Table 6.1), and since the runway is horizontal, n = FG = mg. Thus:

Fthrust = μ r FG + ma = μ r mg + ma = m ( μ r g + a ) = ( 75,000 kg ) ⎡⎣( 0.02 ) ( 9.8 m/s 2 ) + 2.34 m/s 2 ⎤⎦ = 190,000 N Assess: It’s hard to evaluate such an enormous thrust, but comparison with the plane’s mass suggests that 190,000 N is enough to produce the required acceleration.

6.23.

Model: We treat the train as a particle subject to rolling friction but not to drag (because of its slow speed and large mass). We can use the one-dimensional kinematic equations. Visualize:

Solve: The locomotive is not accelerating in the vertical direction, so the free-body diagram shows us that n = FG = mg . Thus,

f r = μ r mg = ( 0.002 )( 50,000 kg ) ( 9.80 m/s 2 ) = 980 N

From Newton’s second law for the decelerating locomotive, −f −980 N = −0.01960 m/s 2 ax = r = m 50,000 kg

Since we’re looking for the distance the train rolls, but we don’t have the time:

v12 − v02 = 2ax ( Δx ) ⇒ Δx =

v12 − v02 ( 0 m/s ) − (10 m/s ) = = 2.55 × 103 m 2a x 2 ( −0.01960 m/s 2 ) 2

2

Assess: The locomotive’s enormous inertia (mass) and the small coefficient of rolling friction make this long stopping distance seem reasonable.

6.24.

Model: Visualize:

We can treat the sliding player as a particle experiencing kinetic friction.

Solve: We can assume a mass of 80 kg (which corresponds to a gravitational force of about 175 pounds). We have no value for μk of cloth sliding on loose dirt. It is probably greater than μk for wood on wood, due to the roughness of both surfaces. Let’s guess 0.40. From the free-body diagram, n = FG = mg , since there’s no vertical acceleration. Thus,

f k = μ k mg = ( 0.40 )( 80 kg ) ( 9.80 m/s 2 ) = 314 N

Assess: A frictional force of 314 N would produce an acceleration of

a=

Fnet −314 N = = − 4.0 m/s 2 m 80 kg

A player running initially at 8 m/s would thus be brought to a stop in about 2 seconds, which seems somewhat too long. Our estimate of μk is probably a bit low but 8 m/s is a large speed as well.

6.25.

Model: Visualize:

We assume that the skydiver is shaped like a box and is a particle.

The skydiver falls straight down toward the earth’s surface, that is, the direction of fall is vertical. Since the skydiver falls feet first, the surface perpendicular to the drag has the cross-sectional area A = 20 cm × 40 cm. The physical conditions needed to use Equation 6.16 for the drag force are satisfied. The terminal speed corresponds to the situation when the net force acting on the skydiver becomes zero. Solve: The expression for the magnitude of the drag with v in m/s is D≈

1 2 Av = 0.25 ( 0.20 × 0.40 ) v 2 N = 0.020v 2 N 4

The gravitational force on the skydiver is FG = mg = ( 75 kg ) ( 9.8 m/s 2 ) = 735 N. The mathematical form of the

condition defining dynamical equilibrium for the skydiver and the terminal speed is G G G Fnet = FG + D = 0 N 2 ⇒ 0.02vterm N − 735 N = 0 N ⇒ vterm =

735 ≈ 192 m/s 0.02

Assess: The result of the above simplified physical modeling approach and subsequent calculation, even if approximate, shows that the terminal velocity is very high. This result implies that the skydiver will be very badly hurt at landing if the parachute does not open in time.

6.26.

Model: Visualize:

We will represent the tennis ball as a particle.

The tennis ball falls straight down toward the earth’s surface. The ball is subject to a net force that is the resultant of the gravitational and drag force vectors acting vertically, in the downward and upward directions, respectively. Once the net force acting on the ball becomes zero, the terminal velocity is reached and remains constant for the rest of the motion. Solve: The mathematical equation defining the dynamical equilibrium situation for the falling ball is G G G G Fnet = FG + D = 0 N Since only the vertical direction matters, one can write: ΣFy = 0 N ⇒ Fnet = D − FG = 0 N When this condition is satisfied, the speed of the ball becomes the constant terminal speed v = vterm . The magnitudes of the gravitational and drag forces acting on the ball are:

FG = mg = m ( 9.80 m/s 2 )

1 2 2 2 2 = ( 0.25π )( 0.0325 m ) ( 26 m/s ) = 0.56 N ( Avterm ) = 0.25 (π R 2 ) vterm 4 The condition for dynamic equilibrium becomes: 0.56 N = 57 g ( 9.80 m/s2 ) m − 0.56 N = 0 N ⇒ m = 9.80 m/s 2 D≈

Assess: The value of the mass of the tennis ball obtained above seems reasonable.

6.27. Visualize:

We used the force-versus-time graph to draw the acceleration-versus-time graph. The peak acceleration was calculated as follows:

amax =

Fmax 10 N = = 2 m/s 2 m 5 kg

Solve: The acceleration is not constant, so we cannot use constant acceleration kinematics. Instead, we use the more general result that

v(t ) = v0 + area under the acceleration curve from 0 s to t The object starts from rest, so v0 = 0 m/s. The area under the acceleration curve between 0 s and 6 s is

1 2

(4 s)

(2 m/s ) = 4.0 m/s. We’ve used the fact that the area between 4 s and 6 s is zero. Thus, at t = 6 s, vx = 4.0 m/s. 2

6.28. Visualize:

The acceleration is ax = Fx m , so the acceleration-versus-time graph has exactly the same shape as the forceversus-time graph. The maximum acceleration is amax = Fmax m = ( 6 N ) ( 2 kg ) = 3 m/s 2 . Solve: The acceleration is not constant, so we cannot use constant-acceleration kinematics. Instead, we use the more general result that

v(t ) = v0 + area under the acceleration curve from 0 s to t The object starts from rest, so v0 = 0 m/s. The area under the acceleration curve between 0 s and 4 s is a rectangle (3 m/s 2 × 2 s = 6 m/s) plus a triangle

(

1 2

× 3 m/s 2 × 2 s = 3 m/s ). Thus vx = 9 m/s at t = 4 s.

6.29.

Model: Visualize:

Solve:

You can model the beam as a particle in static equilibrium.

Using Newton’s first law, the equilibrium equations in vector and component form are: G G G G G Fnet = T1 + T2 + FG = 0 N

( Fnet ) x = T1x + T2 x + FGx = 0 N ( Fnet ) y = T1 y + T2 y + FGy = 0 N

Using the free-body diagram yields: −T1 sinθ1 + T2 sin θ 2 = 0 N

T1 cosθ1 + T2 cosθ 2 − FG = 0 N

The mathematical model is reduced to a simple algebraic system of two equations with two unknowns, T1 and T2 . Substituting θ1 = 20°, θ 2 = 30°, and FG = mg = 9800 N, the simultaneous equations become −T1 sin 20° + T2 sin 30° = 0 N

T1 cos 20° + T2 cos30° = 9800 N

You can solve this system of equations by simple substitution. The result is T1 = 6397 N and T2 = 4376 N. Assess: The above approach and result seem reasonable. Intuition indicates there is more tension in the left rope than in the right rope.

6.30.

Model: Visualize:

The plastic ball is represented as a particle in static equilibrium.

Solve: The electric force, like the weight, is a long-range force. So the ball experiences the contact force of the string’s tension plus two long-range forces. The equilibrium condition is

( Fnet ) x = Tx + ( Felec ) x = T sin θ − Felec = 0 N ( Fnet ) y = Ty + ( FG ) y = T cosθ − mg = 0 N We can solve the y-equation to get T=

( 0.001 kg ) ( 9.8 m/s mg = cosθ cos 20°

2

) = 0.0104 N

Substituting this value into the x-equation,

Felec = T sin θ = (1.04 × 10−2 N ) sin 20° = 0.0036 N (b) The tension in the string is 0.0104 N.

6.31.

Model: Visualize:

Solve:

The piano is in static equilibrium and is to be treated as a particle.

(a) Based on the free-body diagram, Newton’s second law is

( Fnet ) x = 0 N = T1x + T2 x = T2 cosθ 2 − T1 cosθ1

( Fnet ) y = 0 N = T1 y + T2 y + T3 y + FGy = T3 − T1 sinθ1 − T2 sinθ 2 − mg Notice how the force components all appear in the second law with plus signs because we are adding forces. The negative signs appear only when we evaluate the various components. These are two simultaneous equations in the two unknowns T2 and T3 . From the x-equation we find T2 =

T1 cosθ1 ( 500 N ) cos15° = = 533 N cosθ 2 cos 25°

(b) Now we can use the y-equation to find

T3 = T1 sin θ1 + T2 sin θ 2 + mg = 5.25 × 103 N

6.32.

Model: We will represent Henry as a particle. His motion is governed by constant-acceleration kinematic equations. Visualize: Please refer to the Figure EX6.32. Solve: (a) Henry undergoes an acceleration from 0 s to 2.0 s, constant velocity motion from 2.0 s to 10.0 s, and another acceleration as the elevator brakes from 10.0 s to 12.0 s. The weight is the same as the gravitational force during constant velocity motion, so Henry’s weight w = FG = mg is 750 N. His weight is less than the gravitational force on him during the initial acceleration, so the acceleration is in a downward direction (negative a). Thus, the elevator’s initial motion is down. (b) Because the gravitational force on Henry is 750 N, his mass is m = FG /g = 76.5 kg. (c) The apparent weight (see Equation 6.10) during vertical motion is given by

⎛ w ⎞ ⎛ a⎞ w = mg ⎜1 + ⎟ ⇒ a = g ⎜ − 1⎟ g F ⎝ ⎠ ⎝ G ⎠ During the interval 0 s ≤ t ≤ 2 s, the elevator’s acceleration is ⎛ 600 N ⎞ a = g⎜ − 1⎟ = −1.96 m/s 2 ⎝ 750 N ⎠

At t = 2 s, Henry’s position is 1 1 2 2 y1 = y0 + v0 Δt0 + a ( Δt0 ) = a ( Δt0 ) = −3.92 m 2 2 and his velocity is v1 = v0 + aΔt0 = aΔt0 = −3.92 m/s During the interval 2 s ≤ t ≤ 10 s, a = 0 m/s 2 . This means Henry travels with a constant velocity v1 = −3.92 m/s. At t = 10 s he is at position y2 = y1 + v1Δt1 = −35.3 m and he has a velocity v2 = v1 = −3.92 m/s. During the interval 10 s ≤ t ≤ 12.0 s, the elevator’s acceleration is

⎛ 900 N ⎞ a = g⎜ − 1⎟ = +1.96 m/s 2 ⎝ 750 N ⎠ The upward acceleration vector slows the elevator and Henry feels heavier than normal. At t = 12.0 s Henry is at position 1 y3 = y2 + v2 (Δt2 ) + a (Δt2 ) 2 = −39.2 m 2 Thus Henry has traveled distance 39.2 m.

6.33.

Model: We’ll assume Zach is a particle moving under the effect of two forces acting in a single vertical line: gravity and the supporting force of the elevator. Visualize:

Solve:

(a) Before the elevator starts braking, Zach is not accelerating. His weight (see Equation 6.10) is

⎛ 0 m/s 2 ⎞ ⎛ a⎞ 2 w = mg ⎜1 + ⎟ = mg ⎜1 + ⎟ = mg = ( 80 kg ) ( 9.80 m/s ) = 784 N g g ⎝ ⎠ ⎝ ⎠ Zach’s weight is 7.8 × 102 N. (b) Using the definition of acceleration, a=

Δv v1 − v0 0 − ( −10 ) m/s = = = 3.33 m/s 2 Δt t1 − t0 3.0 s

⎛ 3.33 m/s 2 ⎞ ⎛ a⎞ ⇒ w = mg ⎜1 + ⎟ = ( 80 kg ) ( 9.80 m/s 2 ) ⎜1 + = (784 N)(1 + 0.340) = 1050 N 2 ⎟ ⎝ g⎠ ⎝ 9.80 m/s ⎠ Now Zach’s weight is 1.05 × 103 N. Assess: While the elevator is braking, it not only must support the gravitational force on Zach but must also push upward on him to decelerate him, so his weight is greater than the gravitational force.

6.34.

Model: We can assume your body is a particle moving in a straight line under the influence of two forces: gravity and the support force of the scale. Visualize:

Solve:

The weight (see Equation 6.10) of an object moving in an elevator is

⎛ ⎛ w ⎞ a⎞ w = mg ⎜1 + ⎟ ⇒ a = ⎜ − 1⎟ g g mg ⎝ ⎠ ⎝ ⎠ When accelerating upward, the acceleration is ⎛ 170 lb ⎞ a =⎜ − 1⎟ ( 9.80 m/s 2 ) = 1.3 m/s 2 ⎝ 150 lb ⎠ When braking, the acceleration is ⎛ 120 lb ⎞ a =⎜ − 1⎟ ( 9.80 m/s 2 ) = −2.0 m/s 2 ⎝ 150 lb ⎠

Assess: A 10-20% change in apparent weight seems reasonable for a fast elevator, as the ones in the Empire State Building must be. Also note that we did not have to convert the units of the weights from pounds to newtons because the weights appear as a ratio.

6.35.

Model: We can assume the foot is a single particle in equilibrium under the combined effects of gravity, the tensions in the upper and lower sections of the traction rope, and the opposing traction force of the leg itself. We can also treat the hanging mass as a particle in equilibrium. Since the pulleys are frictionless, the tension is the same everywhere in the rope. Because all pulleys are in equilibrium, their net force is zero. So they do not contribute to T. Visualize:

Solve:

(a) From the free-body diagram for the mass, the tension in the rope is

T = FG = mg = ( 6 kg ) ( 9.80 m/s 2 ) = 58.8 N (b) Using Newton’s first law for the vertical direction on the pulley attached to the foot,

( Fnet ) y = ΣFy = T sin θ − T sin15° − ( FG )foot = 0 N ⇒ sin θ =

T sin15° + ( FG )foot T

= sin15° +

( 4 kg ) ( 9.80 m/s mfoot g = 0.259 + T 58.8 N

2

) = 0.259 + 0.667 = 0.926

⇒ θ = sin −1 0.926 = 67.8° (c) Using Newton’s first law for the horizontal direction,

( Fnet ) x = ΣFx = T cosθ + T cos15° − Ftraction = 0 N ⇒ Ftraction = T cosθ + T cos15° = T ( cos67.8° + cos15° ) = (58.8 N)(0.3778 + 0.9659) = (58.8 N)(1.344) = 79.0 N

Assess: Since the tension in the upper segment of the rope must support the foot and counteract the downward pull of the lower segment of the rope, it makes sense that its angle is larger (a more direct upward pull). The magnitude of the traction force, roughly one-tenth of the gravitational force on a human body, seems reasonable.

6.36.

Model: We can assume the person is a particle moving in a straight line under the influence of the combined decelerating forces of the air bag and seat belt or, in the absence of restraints, the dashboard or windshield. Visualize:

Solve: (a) In order to use Newton’s second law for the passenger, we’ll need the acceleration. Since we don’t have the stopping time:

v12 = v02 + 2a ( x1 − x0 ) ⇒ a =

0 m 2 /s 2 − (15 m/s ) v12 − v02 = = −112.5 m/s 2 2 ( x1 − x0 ) 2 (1 m − 0 m ) 2

⇒ Fnet = F = ma = ( 60 kg ) ( −112.5 m/s 2 ) = −6750 N The net force is 6750 N to the left. (b) Using the same approach as in part (a), 0 m 2 /s 2 − (15 m/s ) v12 − v02 = ( 60 kg ) = −1,350,000 N 2 ( x1 − x0 ) 2 ( 0.005 m ) 2

F = ma = m

The net force is 1,350,000 N to the left. (c) The passenger’s weight is mg = (60 kg)(9.8 m/s 2 ) = 588 N. The force in part (a) is 11.5 times the passenger’s weight. The force in part (b) is 2300 times the passenger’s weight. Assess: An acceleration of 11.5g is well within the capability of the human body to withstand. A force of 2300 times the passenger’s weight, on the other hand, would surely be catastrophic.

6.37.

Model: Visualize:

The ball is represented as a particle that obeys constant-acceleration kinematic equations.

Solve: This is a two-part problem. During part 1 the ball accelerates upward in the tube. During part 2 the ball undergoes free fall (a = − g ). The initial velocity for part 2 is the final velocity of part 1, as the ball emerges from the tube. The free-body diagram for part 1 shows two forces: the air pressure force and the gravitational force. We need only the y-component of Newton’s second law:

ay = a =

( Fnet ) y m

=

Fair − FG Fair 2N = −g= − 9.80 m/s 2 = 30.2 m/s 2 m m 0.05 kg

We can use kinematics to find the velocity v1 as the ball leaves the tube:

v12 = v02 + 2a ( y1 − y0 ) ⇒ v1 = 2ay1 = 2 ( 30.2 m/s 2 ) (1 m ) = 7.77 m/s For part 2, free-fall kinematics v22 = v12 − 2 g ( y2 − y1 ) gives y2 − y1 =

v12 = 3.1 m 2g

6.38.

Model: Visualize:

We will represent the bullet as a particle.

Solve: (a) We have enough information to use kinematics to find the acceleration of the bullet as it stops. Then we can relate the acceleration to the force with Newton’s second law. (Note that the barrel length is not relevant to the problem.) The kinematic equation is

v02 ( 400 m/s ) = −6.67 × 105 m/s 2 =− 2Δx 2 ( 0.12 m ) G Notice that a is negative, in agreement with the vector a in the motion diagram. Turning to forces, the wood exerts two forces on the bullet. First, an upward normal force that keeps the bullet from “falling” through the G G wood. Second, a retarding frictional force f k that stops the bullet. The only horizontal force is f k , which points to the left and thus has a negative x-component. The x-component of Newton’s second law is 2

v12 = v02 + 2aΔx ⇒ a = −

( Fnet ) x = − f k = ma ⇒

f k = − ma = − ( 0.01 kg ) ( −6.67 × 105 m/s 2 ) = 6670 N

Notice how the signs worked together to give a positive value of the magnitude of the force. (b) The time to stop is found from v1 = v0 + aΔt as follows: Δt = −

v0 = 6.00 × 10−4 s = 600 μs a

(c)

Using the above kinematic equation, we can find the velocity as a function of t. For example at t = 60 μs,

vx = 400 m/s + (−6.667 × 105 m/s 2 )(60 × 10−6 s) = 360 m/s

6.39.

Model: Visualize:

Solve:

Represent the rocket as a particle that follows Newton’s second law.

(a) The y-component of Newton’s second law is

ay = a =

( Fnet ) y m

=

Fthrust − mg 3.0 × 105 N = − 9.80 m/s 2 = 5.2 m/s 2 m 20,000 kg

(b) At 5000 m the acceleration has increased because the rocket mass has decreased. Solving the equation of part (a) for m gives

m5000 m =

Fthrust 3.0 × 105 N = = 1.9 × 104 kg a5000 m + g 6.0 m/s 2 + 9.80 m/s 2

The mass of fuel burned is mfuel = minitial − m5000 m = 1.0 × 103 kg.

6.40.

Model: The steel block will be represented by a particle. Steel-on-steel has a static coefficient of friction μ s = 0.80 and a kinetic coefficient of friction μ k = 0.60. Visualize:

Solve:

(a) While the block is at rest, Newton’s second law is

( Fnet ) x = T − fs = 0 N ⇒ T = The static friction force has a maximum value

fs

( Fnet ) y = n − FG ⇒ n = FG = mg

( fs )max = μs mg. The string tension that will cause the block to slip

and start to move is T = μs mg = ( 0.80 )( 2.0 kg ) ( 9.80 m/s 2 ) = 15.7 N Any tension less than this will not be sufficient to cause the block to move, so this is the minimum tension for motion. (b) As the block is moving with a tension of 20 N in the string, we can find its acceleration from the xcomponent of Newton’s second law as follows:

( Fnet ) x = T − f k = max ⇒ ax =

T − fk m

The kinetic friction force f k = μk mg . The acceleration of the block is ax =

20 N − ( 0.60 )( 2 kg ) ( 9.8 m/s 2 ) 2 kg

= 4.12 m/s 2

Using kinematics, the block’s speed after moving 1.0 m will be v12 = 0 m 2 /s 2 + 2 ( 4.12 m/s 2 ) (1.0 m ) ⇒ v1 = 2.9 m/s (c) The only difference in this case is the coefficient of kinetic friction whose value is 0.050 instead of 0.60. The acceleration of the block is

ax =

20 N − ( 0.050 )( 2.0 kg ) ( 9.80 m/s 2 ) 2.0 kg

= 9.51 m/s 2

The block’s speed after moving 1.0 m will be v12 = 0 m 2 /s 2 + 2 ( 9.51 m/s 2 ) (1.0 m ) ⇒ v1 = 4.4 m/s

6.41.

Model: We assume that Sam is a particle moving in a straight horizontal line under the influence of two forces: the thrust of his jet skis and the resisting force of friction on the skis. We can use one-dimensional kinematics. Visualize:

Solve: (a) The friction force of the snow can be found from the free-body diagram and Newton’s first law, since there’s no acceleration in the vertical direction:

n = FG = mg = ( 75 kg ) ( 9.80 m/s 2 ) = 735 N ⇒ f k = μ k n = ( 0.10 )( 735 N ) = 73.5 N Then, from Newton’s second law:

( Fnet ) x = Fthrust − f k = ma0 ⇒ a0 =

Fthrust − f k 200 N − 73.5 N = = 1.687 m/s 2 m 75 kg

From kinematics:

v1 = v0 + a0t1 = 0 m/s + (1.687 m/s 2 ) (10 s ) = 16.9 m/s (b) During the acceleration, Sam travels to

1 1 2 x1 = x0 + v0t1 + a0t12 = (1.687 m/s 2 ) (10 s ) = 84 m 2 2 After the skis run out of fuel, Sam’s acceleration can again be found from Newton’s second law: F −73.5 N = −0.98 m/s 2 ( Fnet ) x = − f k = −73.5 N ⇒ a1 = net = m 75 kg

Since we don’t know how much time it takes Sam to stop: v22 = v12 + 2a1 ( x2 − x1 ) ⇒ x2 − x1 =

2 2 v22 − v12 0 m /s − (16.9 m/s ) = = 145 m 2a1 2 ( −0.98 m/s 2 ) 2

The total distance traveled is ( x2 − x1 ) + x1 = 145 m + 84 m = 229 m. Assess: A top speed of 16.9 m/s (roughly 40 mph) seems quite reasonable for this acceleration, and a coasting distance of nearly 150 m also seems possible, starting from a high speed, given that we’re neglecting air resistance.

6.42.

Model: We assume Sam is a particle moving in a straight line down the slope under the influence of gravity, the thrust of his jet skis, and the resisting force of friction on the snow. Visualize:

Solve:

From the height of the slope and its angle, we can calculate its length: h h 50 m = sin θ ⇒ x1 − x0 = = = 288 m x1 − x0 sin θ sin10°

Since Sam is not accelerating in the y-direction, we can use Newton’s first law to calculate the normal force:

( Fnet ) y = ΣFy = n − FG cosθ = 0 N

⇒ n = FG cosθ = mg cosθ = (75 kg)(9.80 m/s 2 )(cos10°) = 724 N

One-dimensional kinematics gives us Sam’s acceleration: v12 = v02 + 2ax ( x − x0 ) ⇒ ax =

v12 − v02 ( 40 m/s ) − 0 m 2 /s 2 = 2.78 m/s 2 = 2 ( x1 − x2 ) 2 ( 288 m ) 2

Then, from Newton’s second law and the equation f k = μ k n :

( Fnet ) x = ΣFx = FG sinθ + Fthrust − f k = max ⇒ μk = =

mg sin θ + Fthrust − ma n

( 75 kg ) ( 9.80 m/s 2 ) ( sin10° ) + 200 N − ( 75 kg ) ( 2.78 m/s 2 )

724 N Assess: This coefficient seems a bit high for skis on snow, but not impossible.

= 0.165

6.43.

Model: only. Visualize:

We assume the suitcase is a particle accelerating horizontally under the influence of friction

Solve: Because the conveyor belt is already moving, friction drags your suitcase to the right. It will accelerate until it matches the speed of the belt. We need to know the horizontal acceleration. Since there’s no acceleration in the vertical direction, we can apply Newton’s first law to find the normal force:

n = FG = mg = (10 kg ) ( 9.80 m/s 2 ) = 98.0 N The suitcase is accelerating, so we use μ k to find the friction force

f k = μ k mg = ( 0.3)( 98.0 N ) = 29.4 N We can find the horizontal acceleration from Newton’s second law:

( Fnet ) x = ΣFx =

f k = ma ⇒ a =

f k 29.4 N = = 2.94 m/s 2 m 10 kg

From one of the kinematic equations: v12 = v02 + 2a ( x1 − x0 ) ⇒ x1 − x0 =

v12 − v02 ( 2.0 m/s ) − ( 0 m/s ) = = 0.68 m 2a 2 ( 2.94 m/s 2 ) 2

2

The suitcase travels 0.68 m before catching up with the belt and riding smoothly. Assess: If we imagine throwing a suitcase at a speed of 2.0 m/s onto a motionless surface, 0.68 m seems a reasonable distance for it to slide before stopping.

6.44. Model: The box of shingles is a particle subject to Newton’s laws and kinematics. Visualize:

Solve: Newton’s laws can be used in the coordinate system in which the direction of motion of the box of JG shingles defines the +x-axis. The angle that FG makes with the –y-axis is 25°.

(∑ F )

(∑ F )

y

x

= FG sin25° − f k = ma

= n − FG cos25° = 0 ⇒ n = FG cos25°

We have used the observation that the shingles do not leap off the roof, so the acceleration in the y-direction is zero. Combining these equations with f k = μ k n and FG = mg yields mg sin 25° − μ k mg cos 25° = ma ⇒ a = ( sin25° − μ k cos25° ) g = −0.743 m/s 2 where the minus sign indicates the acceleration is directed up the incline. The required initial speed to have the box come to rest after 5.0 m is found from kinematics.

vf2 = vi2 + 2aΔx ⇒ vi2 = −2 ( −0.743 m/s 2 ) ( 5.0 m ) ⇒ vi = 2.7 m/s Assess: To give the shingles an initial speed of 2.7 m/s requires a strong, determined push, but is not beyond reasonable.

6.45.

Model: Visualize:

We will model the box as a particle, and use the models of kinetic and static friction.

The pushing force is along the +x-axis, but the force of friction acts along the –x-axis. A component of the gravitational force on the box acts along the –x-axis as well. The box will move up if the pushing force is at least equal to the sum of the friction force and the component of the gravitational force in the x-direction. Solve: Let’s determine how much pushing force you would need to keep the box moving up the ramp at steady speed. Newton’s second law for the box in dynamic equilibrium is

( Fnet ) x = ΣFx = nx + ( FG ) x + ( f k ) x + ( Fpush ) x = 0 N − mg sinθ − f k + Fpush = 0 N ( Fnet ) y = ΣFy = n y + ( FG ) y + ( f k ) y + ( Fpush ) y = n − mg cosθ + 0 N + 0 N = 0 N The x-component equation and the model of kinetic friction yield: Fpush = mg sinθ + f k = mg sinθ + μ k n Let us obtain n from the y-component equation as n = mg cosθ , and substitute it in the above equation to get Fpush = mg sinθ + μ k mg cosθ = mg (sinθ + μ k cosθ )

= (100 kg)(9.80 m/s 2 )(sin 20° + 0.60 cos 20°) = 888 N The force is less than your maximum pushing force of 1000 N. That is, once in motion, the box could be kept moving up the ramp. However, if you stop on the ramp and want to start the box from rest, the model of static friction applies. The analysis is the same except that the coefficient of static friction is used and we use the maximum value of the force of static friction. Therefore, we have Fpush = mg (sinθ + μ s cosθ ) = (100 kg)(9.80 m/s 2 )(sin 20° + 0.90 cos 20°) = 1160 N Since you can push with a force of only 1000 N, you can’t get the box started. The big static friction force and the weight are too much to overcome.

6.46.

Model: We will represent the wood block as a particle, and use the model of kinetic friction and kinematics. Assume w sin θ > fs , so it does not hang up at the top. Visualize:

The block ends where it starts, so x2 = x0 = 0 m. We expect v2 to be negative, because the block will be moving in the –x-direction, so we’ll want to take v2 as the final speed. Because of friction, we expect to find v2 < v0 . G → Solve: (a) The friction force is opposite to v, so f k points down the slope during the first half of the motion G G and up the slope during the second half. FG and n are the only other forces. Newton’s second law for the upward motion is

ax = a0 =

( Fnet ) x m

a y = 0 m/s 2 =

=

− FG sinθ − f k − mg sinθ − f k = m m

( Fnet ) y m

=

n − FG cosθ n − mg cosθ = m m

The friction model is f k = μ k n. First solve the y-equation to give n = mg cosθ . Use this in the friction model to get f k = μ k mg cosθ . Now substitute this result for f k into the x-equation: a0 =

−mg sin θ − μ k mg cosθ = − g ( sinθ + μ k cosθ ) = − ( 9.8 m/s 2 ) ( sin 30° + 0.20cos30° ) = −6.60 m/s 2 m

Kinematics now gives

v12 = v02 + 2a0 ( x1 − x0 ) ⇒ x1 =

2 2 v12 − v02 0 m /s − (10 m/s ) = = 7.6 m 2a0 2 ( −6.60 m/s 2 ) 2

The block’s height is then h = x1 sinθ = (7.6 m)sin 30° = 3.8 m. G (b) For the return trip, f k points up the slope, so the x-component of the second law is

ax = a1 =

( Fnet ) x m

=

− FG sin θ + f k −mg sin θ + f k = m m

Note the sign change. The y-equation and the friction model are unchanged, so we have a1 = − g ( sin θ − μ k cosθ ) = −3.20 m/s 2

The kinematics for the return trip are

v22 = v12 + 2a1 ( x2 − x1 ) ⇒ v2 = −2a1 x1 = 2 ( −3.20 m/s 2 ) ( −7.6 m ) = −7.0 m/s Notice that we used the negative square root because v2 is a velocity with the vector pointing in the –x-direction. The final speed is v2 = 7.0 m/s.

6.47.

Model: We will model the sled and friend as a particle, and use the model of kinetic friction because the sled is in motion. Visualize:

The net force on the sled is zero (note the constant speed of the sled). That means the component of the pulling force along the +x-direction is equal to the magnitude of the kinetic force of friction in the –x-direction. Also note that ( Fnet ) y = 0 N, since the sled is not moving along the y-axis. Solve:

Newton’s second law is ( Fnet ) x = ΣFx = nx + ( FG ) x + ( f k ) x + ( Fpull ) x = 0 N + 0 N − f k + Fpull cosθ = 0 N ( Fnet ) y = ΣFy = n y + ( FG ) y + ( f k ) y + ( Fpull ) y = n − mg + 0 N + Fpull sin θ = 0 N

The x-component equation using the kinetic friction model f k = μ k n reduces to

μ k n = Fpull cosθ The y-component equation gives

n = mg − Fpull sin θ

We see that the normal force is smaller than the gravitational force because Fpull has a component in a direction opposite to the direction of the gravitational force. In other words, Fpull is partly lifting the sled. From the xcomponent equation, μ k can now be obtained as

μk =

Fpull cosθ mg − Fpull sinθ

=

( 75 N )( cos30° ) = 0.12 ( 60 kg ) ( 9.80 m/s 2 ) − ( 75 N )( sin 30° )

Assess: A quick glance at the various μ k values in Table 6.1 suggests that a value of 0.12 for μ k is reasonable.

6.48.

Model: As long as the static force of friction between the box and the sled is sufficiently large for the box not to slip, the acceleration a of the box is the same as the acceleration of the sled. We will therefore model the box as a particle and use the model of static friction. Visualize:

The force on the box that is responsible for its acceleration is the force of static friction provided by the sled because the box would slide backward without this force. Solve: Newton’s second law for the box is

( Fnet ) x = ΣFx = nx + ( ( FG )box ) + ( fs ) x = 0 N + 0 N + f s max = max = mamax x

( Fnet ) y = ΣFy = n y + ( ( FG )box ) + ( fs ) y = n − mg + 0 N = ma y = 0 N y

The above equations can be combined along with the friction model to get amax =

fs max μs n μ s mg = = = μs g m m m

To find Tmax , we write Newton’s second law for the box-sled system, with total mass m + M , as

( Fnet ) x = ΣFx = N x + ( ( FG )sled+box ) + ( fs ) x + (Tmax ) x = 0 N + 0 N + 0 N + Tmax = (m + M )amax x

( Fnet ) y = ΣFy = N y + ( ( FG )sled+box ) + ( fs ) y + (Tmax ) y = N − (m + M ) g + 0 N + 0 N = 0 N y

Referring to Table 6.1 for the coefficient of friction, the x-component equation for the box-sled system yields

Tmax = (m + M )amax = ( m + M ) μs g = (0.50)(9.80 m/s 2 )(5.0 kg + 10 kg) = 74 N That is, the largest tension force for which the box does not slip is 74 N. Assess: If the box were glued to the sled, there would be no resistance to motion because the ice is frictionless, and the tension force could be as large as one wants. On the other hand, if the friction between the box and the sled were zero, one could not pull at all without causing the box to slide.

6.49.

Model: We will model the steel cabinet as a particle. It touches the truck’s steel bed, so only the steel bed can exert contact forces on the cabinet. As long as the cabinet does not slide, the acceleration a of the cabinet is equal to the acceleration of the truck. Visualize:

Solve: The shortest stopping distance is the distance for which the static friction force has its maximum value ( fs )max . Newton’s second law for the box and the model of static friction are

( Fnet ) x = ΣFx = nx + ( FG ) x + ( fs ) x = 0 N + 0 N − ( fs )max = max = ma ⇒ − ( f s )max = ma ( Fnet ) y = ΣFy = n y + ( FG ) y + ( f s ) y = n − mg + 0 N = 0 N ⇒ n = mg − ( f s )max = − μs n = − μs mg = ma These three equations can be combined together to get a = − μ s g . Because constant-acceleration kinematics

gives v12 = v02 + 2a ( x1 − x0 ) and μ s = 0.80 (Table 6.1), we find

( x1 − x0 ) =

v12 − v02 −v02 (15 m/s ) = = = 14.3 m g 2a 2 − μ 2 0.80 ( s ) ( )( ) ( 9.80 m/s 2 ) 2

Assess: The truck was moving at a speed of 15 m/s or at approximately 34 mph. A stopping distance without making the contents slide of about 14.3 m or approximately 47 feet looks reasonable.

Model: The antiques (mass = m) in the back of your pickup (mass = M ) will be treated as a particle. The antiques touch the truck’s steel bed, so only the steel bed can exert contact forces on the antiques. The pickup-antiques system will also be treated as a particle, and the contact force on this particle will be due to the road. Visualize:

6.50.

Solve: (a) We will find the smallest coefficient of friction that allows the truck to stop in 55 m, then compare that to the known coefficients for rubber on concrete. For the pickup-antiques system, with mass m + M , Newton’s second law is

( Fnet ) x = ΣFx = N x + ( ( FG )PA ) + ( f ) x = 0 N + 0 N − f = ( m + M ) ax = (m + M )a x

( Fnet ) y = ΣFy = N y + ( ( FG )PA ) + ( f ) y = N − (m + M ) g + 0 N = 0 N y

The model of static friction is f = μ N , where μ is the coefficient of friction between the tires and the road. These equations can be combined to yield a = − μ g . Since constant-acceleration kinematics gives

v12 = v02 + 2a( x1 + x0 ), we find v12 − v02 v02 ( 25 m/s ) ⇒ μ min = = = 0.58 2 ( x1 − x0 ) 2 g ( x1 − x0 ) ( 2 ) ( 9.8 m/s 2 ) ( 55 m ) 2

a=

The truck cannot stop if μ is smaller than this. But both the static and kinetic coefficients of friction, 1.00 and 0.80 respectively (see Table 6.1), are larger. So the truck can stop. (b) The analysis of the pickup-antiques system applies to the antiques, and it gives the same value of 0.58 for μ min . This value is smaller than the given coefficient of static friction ( μs = 0.60) between the antiques and the truck bed. Therefore, the antiques will not slide as the truck is stopped over a distance of 55 m.

Assess: The analysis of parts (a) and (b) are the same because mass cancels out of the calculations. According to the California Highway Patrol Web site, the stopping distance (with zero reaction time) for a passenger vehicle traveling at 25 m/s or 82 ft/s is approximately 43 m. This is smaller than the 55 m over which you are asked to stop the truck.

6.51.

Model: The box will be treated as a particle. Because the box slides down a vertical wood wall, we will also use the model of kinetic friction. Visualize:

Solve: The normal force due to the wall, Gwhich is perpendicular to the wall, is here to the right. JG The G box slides G down the wall at constant speed, so a = 0 and the box is in dynamic equilibrium. Thus, Fnet = 0. Newton’s second law for this equilibrium situation is ( Fnet ) x = 0 N = n − Fpush cos 45°

( Fnet ) y = 0 N = f k + Fpush sin 45° − FG = f k + Fpush sin 45° − mg The friction force is f k = μ k n. Using the x-equation to get an expression for n, we see that f k = μ k Fpush cos 45°. Substituting this into the y-equation and using Table 6.1 to find μ k = 0.20 gives,

μ k Fpush cos 45° + Fpush sin 45° − mg = 0 N ⇒ Fpush =

( 2.0 kg ) ( 9.80 m/s 2 ) mg = = 23 N μ k cos 45° + sin 45° 0.20cos 45° + sin 45°

6.52.

Model: Visualize:

Use the particle model for the block and the model of static friction.

Solve: The block is initially at rest, so initially the friction force is static friction. If the 12 N push is too strong, the box will begin to move up the wall. If it is too weak, the box will begin to slide down the wall. And if the pushing force is within the proper range, the box will remain stuck in place. First, let’s evaluate the sum of all the forces except friction:

ΣFx = n − Fpush cos30° = 0 N ⇒ n = Fpush cos30° ΣFy = Fpush sin 30° − FG = Fpush sin 30° − mg = (12 N)sin 30° − (1 kg)(9.8 m/s 2 ) = −3.8 N In the first equation we utilized the fact that any motion is parallel to the wall, so ax = 0 m/s 2 . These three forces add up to −3.8 ˆj N. This means the static friction force will be able to prevent the box from moving if

fs = +3.8 ˆj N. Using the x-equation and the friction model we get

( fs )max = μs n = μs Fpush cos30° = 5.2 N

G where we used μ s = 0.5 for wood on wood. The static friction force fs needed to keep the box from moving is less than ( fs )max . Thus the box will stay at rest.

6.53.

Model: We will model the skier along with the wooden skis as a particle of mass m. The snow exerts a contact force and the wind exerts a drag force on the skier. We will therefore use the models of kinetic friction and drag. Visualize:

We choose a coordinate system such that the skier’s motion is along the +x-direction. While the forces of kinetic G G friction f k and drag D act along the –x-direction opposing the motion of the skier, the gravitational force on the skier has a component in the +x-direction. At the terminal speed, the net force on the skier is zero as the forces along the +x-direction cancel out the forces along the –x-direction. Solve: Newton’s second law and the models of kinetic friction and drag are

( Fnet ) x = ΣFx = nx + ( FG ) x + ( f k ) x + ( D) x = 0 N + mg sin θ − f k

1 2 Av = max = 0 N 4

( Fnet ) y = ΣFy = n y + ( FG ) y + ( f k ) y + ( D) y = n − mg cosθ + 0 N + 0 N = 0 N

fk = μk n These three equations can be combined together as follows:

(1/4) Av 2 = mg sin θ − f k = mg sin θ − μ k n = mg sinθ − μ k mg cosθ 12

⎛ sinθ − μ k cosθ ⎞ ⇒ vterm = ⎜ mg ⎟ 1 A 4 ⎝ ⎠ Using μ k = 0.06 and A = 1.8 m × 0.40 m = 0.72 m 2 , we find

vterm

⎡ ⎛ sin 40° − 0.06cos 40° ⎞ ⎤ ⎟⎥ = ⎢( 80 kg ) ( 9.8 m/s 2 ) ⎜ 1 ⎜ ( 4 kg m3 )( 0.72 m 2 ) ⎟ ⎥ ⎢ ⎝ ⎠⎦ ⎣

12

= 51 m/s

Assess: A terminal speed of 51 m/s corresponds to a speed of ≈100 mph. This speed is reasonable but high due to the steep slope angle of 40° and a small coefficient of friction.

6.54. Model: The ball is a particle experiencing a drag force and traveling at twice its terminal velocity. Visualize:

Solve: (a) An object falling at greater than its terminal velocity will slow down to its terminal velocity. Thus the drag force is greater than the force of gravity, as shown in the free-body diagrams. When the ball is shot straight up,

(∑ F )

y

1 1 2 ⎛ ⎞ ⎛ 4mg ⎞ = ma = − ( FG + D ) = − ⎜ mg + Av 2 ⎟ = − mg − A ( 2vterm ) = − mg − A ⎜ ⎟ = −5mg 4 4 ⎝ ⎠ ⎝ A ⎠

Thus a = −5 g , where the minus sign indicates the downward direction. We have used Equations 6.16 for the drag force and 6.19 for the terminal velocity. (b) When the ball is shot straight down,

(∑ F )

y

= ma = D − FG =

1 2 A ( 2vterm ) − mg = 3mg 4

Thus a = 3 g , this time directed upward. (c)

The ball will slow down to its terminal velocity, slowing quickly at first, and more slowly as it gets closer to the terminal velocity because the drag force decreases as the ball slows.

6.55.

Model: We will model the sculpture as a particle of mass m. The ropes that support the sculpture will be assumed to have zero mass. Visualize:

Solve:

Newton’s first law in component form is

( Fnet ) x = ΣFx = T1x + T2 x + FGx = −T1 sin 30° + T2 sin 60° + 0 N = 0 N

( Fnet ) y = ΣFy = T1 y + T2 y + FGy = −T1 cos30° + T2 cos60° − FG = 0 N Using the x-component equation to obtain an expression for T1 and substituting into the y-component equation yields: T2 =

500 lbs FG = 250 lbs = ( sin 60°)( cos30°) + cos60° 2 sin 30°

Substituting this value of T2 back into the x-component equation, T1 = T2

sin 60° sin 60° = 250 lbs = 433 lbs sin 30° sin 30°

We will now find a rope size for a tension force of 433 lbs, that is, the diameter of a rope with a safety rating of 433 lbs. Since the cross-sectional area of the rope is 14 π d 2 , we have 12

⎡ ⎤ 4 ( 433 lbs ) ⎥ d =⎢ 2 ⎢⎣ π ( 4000 lbs/inch ) ⎥⎦

= 0.371 inch

Any diameter larger than 0.371 inch will ensure a safety rating of at least 433 lbs. The rope size corresponding to a diameter of 3/8 of an inch will therefore be appropriate. Assess: If only a single rope were used to hang the sculpture, the rope would have to support a gravitational force of 500 lbs. The diameter of the rope for a safety rating of 500 lbs is 0.399 inches, and the rope size jumps from a diameter of 3/8 to 4/8 of an inch. Also note that the gravitational force on the sculpture is distributed in the two ropes. It is the sum of the y-components of the tensions in the ropes that will equal the gravitational force on the sculpture.

6.56.

Model: We will model the container as a particle of mass m. The steel cable of the crane will be assumed to have zero mass. Visualize:

Solve: As long as the container is stationary or it is moving with a constant speed (zero acceleration), the net force on the container is zero. In these cases, the tension in the cable is equal to the gravitational force on the container:

T = mg = 44,000 N The cable should safely lift the load. More tension is required to accelerate the load. Newton’s second law is ( Fnet ) y = ΣFy = ( FG ) y + (T ) y = − mg + T = ma y

The crane’s maximum acceleration is amax = 1.0 m/s 2 . So the maximum cable tension is

Tmax = mg + mamax = 48,600 N This is less than the cable’s rating, so the cable must have been defective.

6.57.

Model: Visualize:

Solve:

We will model the skier as a particle, and use the model of kinetic friction.

G G G G Your best strategy, if it’s possible, is to travel at a very slow constant speed a = 0 so Fnet = 0 .

(

)

Alternatively, you want the smallest positive ax . A negative ax would cause you to slow and stop. Let’s find the G G value of μ k that gives Fnet = 0. Newton’s second law for the skier and the model of kinetic friction are

( Fnet ) x = ΣFx = nx + ( FG ) x + ( f k ) x + ( D) x = 0 + mg sin θ − f k − D cosθ = 0 N ( Fnet ) y = ΣFy = n y + ( FG ) y + ( f k ) y + ( D ) y = n − mg cosθ + 0 N − D sin θ = 0 N

fk = μk n The x- and y-component equations are f k = + mg sin θ − D cosθ n = mg cosθ + D sin θ From the model of kinetic friction,

μk =

2 f k mg sin θ − D cosθ 75 kg ( 9.8 m/s ) sin15° − ( 50 N ) cos15° = = = 0.196 n mg cosθ + D sin θ 75 kg ( 9.8 m/s 2 ) cos15° + ( 50 N ) sin15°

Yellow wax with μ k = 0.20 applied to skis will make the skis stick and hence cause the skier to stop. The skier’s next choice is to use the green wax with μ k = 0.15.

6.58.

Model: Visualize:

The ball hanging from the ceiling of the truck by a string is represented as a particle.

Solve: (a) You cannot tell from within the truck. Newton’s first law says that there is no distinction between “at rest” and “constant velocity.” In both cases, the net force acting on the ball is zero and the ball hangs straight down. (b) Now you can tell. If the truck is accelerating, then the ball is tilted back at an angle. (c) The ball moves with the truck, so its acceleration is 5 m/s 2 in the forward direction. G (d) The free-body diagram shows that the horizontal component of T provides a net force in the forward direction. This is the net force that causes the ball to accelerate in the forward direction along with the truck. (e) Newton’s second law for the ball is

ax =

( Fnet ) x m

=

Tx T sin10° = m m

a y = 0 m 2 /s 2 =

( Fnet ) y m

=

Ty + ( FG ) y m

We can solve the second equation for the magnitude of the tension:

T=

(1.0 kg ) ( 9.8 m/s 2 ) mg = = 9.95 N cos10° cos10°

Then the first equation gives the acceleration of the ball and truck: ax =

T sin10° ( 9.95 N ) sin10° = =1.73 m/s 2 m m

The truck’s velocity cannot be determined.

=

T cos10° − mg m

6.59.

Model: You will be treated as a particle. You will experience your weight force and the force (P) of the scale pulling up. Visualize:

G Solve: The weight of an object is the magnitude of the contact force supporting it. Here, the contact force is P, so the weight is w = P. Newton’s second law is

⎛ a⎞ Fnet = ΣFy = Py + ( FG ) y = ma y ⇒ Fnet = P − FG = ma ⇒ w = P = mg + ma = mg ⎜1 + ⎟ ⎝ g⎠

6.60. Solve: Using ax =

dvx , we express Newton’s second law as a differential equation, which we then use dt

to solve for vx . Fx = m

dvx F ct ⇒ dvx = x dt = dt dt m m

Integrating from the initial to final conditions for each variable of integration, vx

∫

v0 x

Thus

t

dvx =

c ct 2 t dt ⇒ v − v = x x 0 2m m ∫0 vx = v0 x +

ct 2 2m

6.61. Model: The astronaut is a particle oscillating on a spring. Solve:

(a) The position versus time function x(t) can be used to find the velocity versus time function v(t ) =

We have v(t ) =

d ( 0.30 m ) sin ( (π rad/s ) t ) = ( 0.30π m/s ) cos ( (π rad/s ) t ) dt

{

}

This can then be used to find the acceleration a (t ) =

a (t ) =

dv . dt

dv = − ( 0.30π 2 m/s 2 ) sin ( (π rad/s ) t ) dt

Newton’s second law yields a general expression for the force on the astronaut.

Fnet (t ) = ma(t ) = −(75 kg) ( 0.30π 2 m/s 2 ) sin ( (π rad/s ) t ) Evaluating this at t = 1.0 s gives Fnet (1.0 s ) = 0 N, since sin(π ) = 0. (b) Evaluating at t = 1.5 s, ⎛ 3π Fnet = −22.5π 2 N sin ⎜ ⎝ 2

Assess:

⎞ 2 ⎟ = 2.2 × 10 N ⎠

The force of 220 N is only one third of the astronaut’s weight on earth, so is easy for her to withstand.

dx . dt

6.62. Solve: (a) The terminal velocity for a falling object is reached when the downward gravitational force is balanced by the upward drag force.

FG = D mg = bvterm = 6πη Rvterm

⇒ vterm =

mg 6πη R

⎛4 ⎞ (b) The mass of the spherical sand grain of density p = 2400 kg/m3 is m = ρ ⎜ π R 3 ⎟ . ⎝3 ⎠ −4 3 2 2 ρ gR 2 2 ( 2400 kg/m )( 9.80 m/s )( 5.0 × 10 m ) vterm = = = 1.3 m/s Thus 9η 9 ⎛ −3 Ns ⎞ 1.0 × 10 ⎜ ⎟ m2 ⎠ ⎝ 50 m The time required for the sand grain to fall 50 m at this speed is t = = 38 s. 1.3 m/s Assess: The speed of 1.3 m/s for a sand grain falling through water seems about right. 2

6.63.

Solve: (a) A 1.0 kg block is pulled across a level surface by a string, starting from rest. The string has a tension of 20 N, and the block’s coefficient of kinetic friction is 0.50. How long does it take the block to move 1.0 m? (b) Newton’s second law for the block is

ax = a =

( Fnet ) x m

=

T − fk T − μk n = m m

a y = 0 m/s 2 =

( Fnet ) y m

=

n − FG n − mg = m m

where we have incorporated the friction model into the first equation. The second equation gives n = mg . Substituting this into the first equation gives

a=

T − μ k mg 20 N − 4.9 N = = 15.1 m/s 2 m 1.0 kg

Constant acceleration kinematics gives 2 (1.0 m ) 1 1 2 x1 2 2 x1 = x0 + v0 Δt + a ( Δt ) = a ( Δt ) ⇒ Δt = = = 0.36 s 2 2 a 15.1 m/s 2

6.64.

Solve: (a) A 15,000 N truck starts from rest and moves down a 15° hill with the engine providing a 12,000 N force in the direction of the motion. Assume the frictional force between the truck and the road is very small. If the hill is 50 m long, what will be the speed of the truck at the bottom of the hill? (b) Newton’s second law is

ΣFy = ny + FGy + f y + E y = ma y = 0 ΣFx = nx + FGx + f x + Ex = max ⇒ 0 N + FG sin θ + 0 N + 12,000 N = ma ⇒a=

mg sinθ + 12,000 N (15,000 N ) sin15° + 12,000 N = = 10.4 m/s 2 m (15,000 N/9.8 m/s2 )

where we have calculated the mass of the truck from the gravitational force on it. Using the constant-acceleration kinematic equation vx2 − v02 = 2ax,

vx2 = 2ax x = 2 (10.4 m/s 2 ) ( 50 m ) ⇒ vx = 32 m/s

6.65.

Solve: (a) A 1.0 kg box is pushed along an assembly line by a mechanical arm. The arm exerts a 20 N force at a downward angle of 30°. The box has a coefficient of kinetic friction on the assembly line of 0.25. What is the speed of the box after traveling 50 cm, starting from rest? (b) Newton’s second law for the box is

ax = a =

( Fnet ) x m

a y = 0 m/s 2 =

=

Fpush cos30° − f k

( Fnet ) y m

=

Fpush cos30° − μ k n

m m n − FG − Fpush sin 30° n − mg − Fpush sin 30° = = m m

The second equation is solved to give an expression for n. Substituting into the first equation:

a=

Fpush cos30° − μ k ( Fpush sin 30° + mg ) m

=

( 20 N ) cos30° − 4.9 N = 12.4 m/s 2 1.0 kg

Using kinematics,

v12 = v02 + 2aΔx = 2aΔx ⇒ v1 = 2aΔx = 2 (12.4 m/s 2 ) ( 0.50 m ) = 3.5 m/s

6.66.

Solve: (a) A driver traveling at 40 m/s in her 1500 kg auto slams on the brakes and skids to rest. How far does the auto slide before coming to rest? (b)

(c) Newton’s second law is ΣFy = n y + ( FG ) y = n − mg = ma y = 0 N

ΣFx = −0.80n = max

The y-component equation gives n = mg = (1500 kg ) ( 9.8 m/s 2 ) . Substituting this into the x-component equation

yields

(1500 kg ) ax = −0.80 (1500 kg ) ( 9.8 m/s 2 ) ⇒ ax = ( −0.80 ) ( 9.8 m/s2 ) = −7.8 m/s 2 Using the constant-acceleration kinematic equation v12 = v02 + 2aΔx, we find

( 40 m/s ) = 102 m v02 =− 2a 2 ( −7.8 m/s 2 ) 2

Δx = −

6.67. Solve: (a) A 20.0 kg wooden crate is being pulled up a 20° wooden incline by a rope that is connected to an electric motor. The crate’s acceleration is measured to be 2.0 m/s 2 . The coefficient of kinetic friction between the crate and the incline is 0.20. Find the tension T in the rope. (b)

(c) Newton’s second law for this problem in the component form is

( Fnet ) x = ΣFx = T − 0.20n − (20 kg)(9.80 m/s 2 )sin 20° = (20 kg)(2.0 m/s 2 ) ( Fnet ) y = ΣFy = n − (20 kg)(9.80 m/s 2 )cos 20° = 0 N Solving the y-component equation, n = 184.18 N. Substituting this value for n in the x-component equation yields T = 144 N.

6.68.

Solve: (a) You wish to pull a 20 kg wooden crate across a wood floor ( μ k = 0.20) by pulling on a rope

attached to the crate. Your pull is 100 N at an angle of 30° above the horizontal. What will be the acceleration of the crate? (b)

(c) Newton’s equations and the model of kinetic friction are

ΣFx = nx + Px + ( FG ) x + f x = 0 N + (100 N ) cos30° + 0 N − f k = (100 N ) cos30° − f k = max ΣFy = n y + Py + ( FG ) y + f y = n + (100 N ) sin 30° − mg − 0 N = ma y = 0 N fk = μk n From the y-component equation, n = 150 N. From the x-component equation and using the model of kinetic friction with μ k = 0.20,

(100 N)cos30° − (0.20)(150 N) = (20 kg) ax ⇒ ax = 2.8 m/s 2

6.69.

Model: Take the coin to be a particle held against the palm of your hand (which is like a vertical wall) by friction. The friction needs to be vertical and equal to the weight of the coin to hold the coin in place. Visualize:

The hand pushes on the coin giving a normal force to the coin and causing a friction force. Solve: (a) We need a force (push) to create a normal force. Since the force accelerates the coin, a minimum acceleration amin is needed. (b) Newton’s second law and the model of friction are

ΣFy = ( f min ) y − ( FG ) y = f min − mg = 0 N

ΣFx = nmin = mamin f min = μ nmin

Since you do not want the coin to slip down the hand you need μs . Combining the above three equations yields

f min = mg ⇒ μ s nmin = mg ⇒ μ s mamin = mg ⇒ amin =

g

μs

=

9.8 m/s 2 = 12.3 m/s 2 0.80

6.70.

Model: We will model the shuttle as a particle and assume the elastic cord to be massless. We will also use the model of kinetic friction for the motion of the shuttle along the square steel rail. Visualize:

Solve:

The upward tension component Ty = T sin 45° = 14.1 N is larger than the gravitational force on the

shuttle. Consequently, the elastic cord pulls the shuttle up against the rail and the rail’s normal force pushes downward. Newton’s second law in component form is ( Fnet ) x = ΣFx = Tx + ( f k ) x + (n) x + ( FG ) x = T cos 45° − f k + 0 N + 0 N = max = max ( Fnet ) y = ΣFy = Ty + ( f k ) y + ( n) y + ( FG ) y = T sin 45° + 0 N − n − mg = ma y = 0 N

The model of kinetic friction is f k = μ k n. We use the y-component equation to get an expression for n and hence f k . Substituting into the x-component equation and using the value of μ k in Table 6.1 gives us

T cos 45° − μ k (T sin 45° − mg ) m ( 20 N ) cos 45° − ( 0.60 ) ⎡⎣ + ( 20 N ) sin 45° − ( 0.800 kg ) ( 9.80 m/s 2 )⎤⎦ = = 13.0 m/s 2 0.800 kg

ax =

Assess: The x-component of the tension force is 14.1 N. On the other hand, the net force on the shuttle in the xdirection is max = (0.800 kg)(13.0 m/s 2 ) = 10.4 N. This value for ma is reasonable since a part of the 14.1 N tension force is used up to overcome the force of kinetic friction.

6.71.

Model: Assume the ball is a particle on a slope, and that the slope increases as the x-displacement increases. Assume that there is no friction and that the ball is being accelerated to the right so that it remains at rest on the slope. Visualize: Although the ball is on a slope, it is accelerating to the right. Thus we’ll use a coordinate system with horizontal and vertical axes.

Solve:

Newton’s second law is

ΣFx = n sin θ = max

ΣFy = n cosθ − FG = ma y = 0 N

Combining the two equations, we get max =

FG sin θ = mg tan θ ⇒ ax = g tan θ cosθ

The curve is described by y = x 2 . Its slope a position x is tanθ, which is also the derivative of the curve. Hence, dy = tan θ = 2 x ⇒ ax = (2 x) g dx

(b) The acceleration at x = 0.20 m is ax = (2)(0.20)(9.8 m/s 2 ) = 3.9 m/s 2 .

6.72.

Model: Visualize:

We will represent the widget as a particle.

Please refer to Figure CP6.72. Solve: (a) There are only two forces on the widget: the normal force of the table and the gravitational force. (b) Newton’s second law along the y-axis is

( Fnet ) y = ny + ( FG ) y = ny − mg = ma y ⇒ ny = m ( a y + g ) We know what the a y -vs-t graph looks like. To get the ny -vs-t graph, we need to add g y to the graph, which amounts to shifting the whole graph up by 9.8 m/s 2 , and multiplying by m = 5 kg. (c) The normal force is negative for t > 0.75 s. Physically, this means that the normal force is pointed in the downward direction. In other words, the table is pulling down on the widget rather than pushing up on the widget. It can do this because the widget is glued to the table rather than simply sitting on the table. (d) The weight is a maximum at t = 0 s, when the upward acceleration is maximum. (e) The weight is zero at t = 0.75 s when a y = −9.8 m/s 2 = − g . (f) If not glued down, the widget will fly off the table at t = 0.75 s, the instant at which its weight becomes zero. The table is accelerating in the negative direction so quickly after t = 0.75 s that the widget can stay on only if the table pulls downward on it. That is the significance of the negative value for ny . If the widget is not glued down, the largest downward acceleration it can achieve is the free-fall acceleration.

6.73. Visualize:

Solve: (a) The horizontal velocity as a function of time is determined by the horizontal net force. Newton’s second law as the x-direction gives ( Fnet ) x = max = − D cosθ = −bv cosθ = −bvx G JJG Note that D points opposite to v, so the angle θ with the x-axis is the same for both vectors, and the x components of both vectors have the same cosθ term. As the particle changes direction as it falls, the evolution of the horizontal motion depends only on the horizontal component of the velocity. Thus m vx ( t )

Separating and integrating,

∫

v0

dvx = −bvx dt

t

dvx b = − ∫ dt vx m0 ⎛ v (t ) ⎞ b ⇒ ln ⎜ x ⎟ = − t v m ⎝ 0 ⎠

Solving, vx ( t ) = v0e

−

bt m

= v0e

−

6πη Rt m

1 (b) The time to reach v ( t ) = v0 is found by solving for the time when 2 6πη Rt − 1 v0 = v0e m 2

Hence t=

m ln ( 2 ) 6πη R

With η = 1.0 × 10−3 Ns/m 2 , R = 2.0 × 10−2 m, and m = 0.033 kg, we get t = 61 s. 6πη R v = 1.1× 10−2 s-1 vx . This is a small fraction of the Assess: The magnitude of the acceleration is ax = m velocity, so a time of about one minute to slow to half the initial speed is reasonable.

(

)

6.74. Visualize:

dvx ⎛ dvx ⎞⎛ dx ⎞ dvx . =⎜ ⎟⎜ ⎟ = vx dt ⎝ dx ⎠⎝ dt ⎠ dx (b) The horizontal motion is determined by using Newton’s second law in the horizontal direction. Using the free-body diagram at a later time t, ( Fnet ) x = max = − D cosθ = −bv cosθ = −bvx G JJG Note that since D points opposite to v, the angle θ with the x-axis is the same for both vectors, and the xcomponents of both vectors have the same cosθ term. Thus

Solve:

(a) Using the chain rule, ax =

max = mvx

dvx = −bvx dx

vx ( x )

∫

Separating and integrating,

dvx = −

v0

b m

x (t )

∫ dx

x0

⇒ vx ( x ) − v0 = −

b ( x ( t ) − x0 ) m

Solving with x0 = 0, vx ( x ) = v0 −

b 6πη R x = v0 − x m m

(c) The marble stops after traveling a distance d when vx ( d ) = 0.

v0 =

Hence

⇒d =

6πη R d m

mv0 6πη R

Using v0 = 10 cm/s, R = 0.50 cm, m = 1.0 × 10−3 kg, and using η = 1.0 × 10−3 Ns/m, d= Assess:

(1.0 ×10

−3

kg ) ( 0.10 m/s )

6π (1.0 × 10 Ns/m 2 )( 5.0 × 10−3 m ) −3

= 1.1 m

The equation for d indicates that a marble with a faster initial velocity travels a further distance.

6.75.

Model: Visualize:

We will model the object as a particle, and use the model of drag.

Solve: (a) We cannot use the constant-acceleration kinematic equations since the drag force causes the acceleration to change with time. Instead, we must use ax = dvx /dt and integrate to find vx . Newton’s second law for the object is

( Fnet ) x = ΣFx = nx + wx + ( f k ) x + ( D) x = 0 N + 0 N + 0 N −

1 2 dv Avx = max = m x 4 dt

This can be written

dvx A = dt 2 vx 4m We can integrate this from the start (v0 x at t = 0) to the end (vx at t ):

∫

vx v0 x

dvx A t 1 1 A = dt ⇒ − + = t 2 ∫ 0 vx vx v0 x 4m 4m

Solving for vx gives vx =

v0 x 1 + Av0 xt 4m

(b) Using A = (1.6 m)(1.4 m) = 2.24 m 2 , v0 x = 20 m/s, and m = 1500 kg, we get

vx =

20 m/s 1 20 ⎛ ⎞ ⎛ 20 m/s ⎞ ⇒t =⎜ − 1⎟ = ⎟⎜ 2.24t ( 20 ) 1 + 0.007467t 0.007467 ⎝ ⎠ ⎝ vx ⎠ 1+ 4 × 1500

where t is in seconds. We can now obtain the time t for v = 10 m/s:

1 ⎛ ⎞ ⎛ 20 m/s ⎞ ⎛ 20 ⎞ t =⎜ − 1⎟ = 133.93⎜ − 1⎟ = 134 s ⎟⎜ ⎝ 0.007467 ⎠ ⎝ 10 m/s ⎠ ⎝ 10 ⎠ When vx = 5 m/s, then t = 402 s.

(c) If the only force acting on the object was kinetic friction with, say, μ k = 0.05, that force would be (0.05)(1500 kg) (9.8 m/s 2 ) = 735 N. The drag force at an average speed of 10 m/s is D = 14 ( 2.24 )(10 )2 N = 56 N. We conclude that it is not reasonable to neglect the kinetic friction force.

7.1.

Visualize:

Solve: (a) The weight lifter is holding the barbell in dynamic equilibrium as he stands up, so the net force on the barbell and on the weight lifter must be zero. The barbells have an upward contact force from the weight lifter and the gravitational force downward. The weight lifter has a downward contact force from the barbells and an upward one from the surface. Gravity also acts on the weight lifter.

(b) The system is the weight lifter and barbell, as indicated in the figure. (c)

7.2.

Visualize:

Solve: (a) Both the bowling ball and the soccer ball have a normal force from the surface and gravitational force on them. The interaction forces between the two are equal and opposite.

(b) The system consists of the soccer ball and bowling ball, as indicated in the figure. (c)

Assess: Even though the soccer ball bounces back more than the bowling ball, the forces that each exerts on the other are part of an action/reaction pair, and therefore have equal magnitudes. Each ball’s acceleration due to the forces on it is determined by Newton’s second law, a = Fnet / m, which depends on the mass. Since the masses of the balls are different, their accelerations are different.

7.3.

Visualize:

Solve: (a) Both the mountain climber and bag of supplies have a normal force from the surface on them, as well as a gravitational force vertically downward. The rope has gravity acting on it, along with pulls on each end from the mountain climber and supply bag. Both the mountain climber and supply bag also experience a frictional force with the surface of the mountain. In the case of the motionless mountain climber it is static friction, but the sliding supply bag experiences kinetic friction.

(b) The system consists of the mountain climber, rope, and bag of supplies, as indicated in the figure. (c)

Assess: Since the motion is along the surface, it is convenient to choose the x-coordinate axis along the surface. The free-body diagram of the rope shows pulls that are slightly off the x-axis since the rope is not massless.

7.4.

Visualize:

Solve: (a) The car and rabbit both experience a normal force and friction from the floor and a gravitational force from the Earth. The push each exerts on the other is a Newton’s third law force pair.

(b) The system consists of the car and stuffed rabbit, as indicated in the figure. (c)

7.5. Visualize: Please refer to Figure EX7.5. Solve: (a) Gravity acts on both blocks, and where Block A is in contact with the floor there is a normal force and friction. The string tension is the same on both blocks since the rope and pulley are massless, and the pulley is frictionless. There are two third law pairs of forces at the surface where the two blocks meet. Block B pushes against Block A with a normal force, while Block A has a reaction force that pushes back against Block B. There is also friction between the two blocks at the surface.

(b) A string that will not stretch constrains the two blocks to accelerate at the same rate but in opposite directions. Block A accelerates down the incline with the same acceleration that Block B has up the incline. The system consists of the two blocks, as indicated in the figure. (c)

Assess: The inclined coordinate systems allow the acceleration a to be purely along the x-axis. This is convenient since the one component of a is zero, simplifying the mathematical expression of Newton’s second law.

7.6. Visualize: Please refer to Figure EX7.6. Solve: (a) For each block, there is a gravitational force with the earth, a normal force and kinetic friction with the surface, and a tension force due to the rope.

(b) The tension in the massless ropes over the frictionless pulley is the same on both blocks. Block A accelerates down the incline with the same acceleration that Block B has up the incline. The system consists of the two blocks, as indicated in the figure. (c)

Assess: The inclined coordinate systems allow the acceleration a to be purely along the x-axis. This is convenient since then one component of a is zero, simplifying the mathematical expression of Newton’s second law.

7.7. Model: We will model the astronaut and the chair as particles. The astronaut and the chair will be denoted by A and C, respectively, and they are separate systems. The launch pad is a part of the environment. Visualize:

Solve:

(a) Newton’s second law for the astronaut is

∑( F )

on A y

= nC on A − ( FG )A = mA aA = 0 N ⇒ nC on A = ( FG )A = mA g

By Newton’s third law, the astronaut’s force on the chair is

nA on C = nC on A = mA g = ( 80 kg ) ( 9.8 m/s 2 ) = 7.8 × 102 N (b) Newton’s second law for the astronaut is:

∑( F )

on A y

= nC on A − ( FG )A = mA aA ⇒ nC on A = ( FG )A + mA aA = mA ( g + aA )

By Newton’s third law, the astronaut’s force on the chair is

nA on C = nC on A = mA ( g + aA ) = ( 80 kg ) ( 9.8 m/s 2 + 10 m/s 2 ) = 1.6 × 103 N Assess:

This is a reasonable value because the astronaut’s acceleration is more than g.

7.8. (a) Visualize: The upper magnet is labeled U, the lower magnet L. Each magnet exerts a long-range magnetic force on the other. Each magnet and the table exert a contact force (normal force) on each other. In addition, the table experiences a normal force due to the surface.

(b) Solve:

Each object is in static equilibrium with ( Fnet ) = 0. Start with the lower magnet. Because

FU on L = 3( FG ) L = 6.0 N, equilibrium requires nT on L = 4.0 N. For the upper magnet, FL on U = FU on L = 6.0 N because these are an action/ reaction pair. Equilibrium for the upper magnet requires nT on U = 8.0 N. For the table, action/reaction pairs are nL on T = nT on L = 4.0 N and nU on T = nT on U = 8.0. The table’s gravitational force is ( FG )T = 20 N, leaving nS on T = 24 N in order for the table to be in equilibrium. Summarizing, Upper magnet

Table

Lower magnet

( FG ) U = 2.0 N nT on U = 8.0 N FL on U = 6.0 N

( FG )T = 20 N nU on T = 8.0 N nL on T = 4.0 N nS on T = 24 N

( FG ) L = 2.0 N nT on L = 4.0 N FU on L = 6.0 N

Assess: The result nS on T = 24 N makes sense. The combined gravitational force on the table and two magnets is 24 N. Because the table is in equilibrium, the upward normal force of the surface has to exactly balance the total gravitational force on the table and magnets.

7.9. Model: The car and the truck will be modeled as particles and denoted by the symbols C and T, respectively. The surface of the ground will be denoted by the symbol S. Visualize:

Solve:

(a) The x-component of Newton’s second law for the car is

∑( F )

on C x

= FS on C − FT on C = mC aC

The x-component of Newton’s second law for the truck is

∑( F )

on T x

= FC on T = mT aT

Using aC = aT = a and FT on C = FC on T , we get

(F

C on S

⎛ 1 ⎞ ⎛ 1 ⎞ − FC on T ) ⎜ ⎟ = a ( FC on T ) ⎜ ⎟=a ⎝ mT ⎠ ⎝ mC ⎠

Combining these two equations,

(F

C on S

⎛ 1 ⎞ ⎛ 1 ⎛ 1 ⎞ ⎛ 1 ⎞ 1 ⎞ − FC on T ) ⎜ + ⎟ = ( FC on T ) ⎜ ⎟ = ( FC on S ) ⎜ ⎟ ⎟ ⇒ FC on T ⎜ m m m m ⎝ T⎠ T ⎠ ⎝ C⎠ ⎝ C ⎝ mC ⎠

⎛ mT ⎞ ⎛ ⎞ 2000 kg ⇒ FC on T = ( FC on S ) ⎜ ⎟ = ( 4500 N ) ⎜ ⎟ = 3000 N ⎝ 1000 kg + 2000 kg ⎠ ⎝ mC + mT ⎠ (b) Due to Newton’s third law, FT on C = 3000 N.

7.10. Model: The blocks are to be modeled as particles and denoted as 1, 2, and 3. The surface is frictionless and along with the earth it is a part of the environment. The three blocks are our three systems of interest. Visualize:

The force applied on block 1 is FA on 1 = 12 N. The acceleration for all the blocks is the same and is denoted by a. Solve: (a) Newton’s second law for the three blocks along the x-direction is

∑( F )

on 1 x

= FA on 1 − F2 on 1 = m1a

∑( F )

on 2 x

= F1 on 2 − F3 on 2 = m2 a

∑( F )

on 3 x

= F2 on 3 = m3a

Adding these three equations and using Newton’s third law ( F2 on 1 = F1 on 2 and F3 on 2 = F2 on 3 ), we get

FA on 1 = ( m1 + m2 + m3 ) a ⇒ (12 N ) = (1 kg + 2 kg + 3 kg ) a ⇒ a = 2 m/s 2 Using this value of a, the force equation on block 3 gives F2 on 3 = m3a = ( 3 kg ) ( 2 m/s 2 ) = 6 N (b) Substituting into the force equation on block 1,

12 N − F2 on 1 = (1 kg ) ( 2 m/s 2 ) ⇒ F2 on 1 = 10 N Assess: Because all three blocks are pushed forward by a force of 12 N, the value of 10 N for the force that the 2 kg block exerts on the 1 kg block is reasonable.

7.11. Model: The block (B) and the steel cable (C), the two objects of interest to us, are treated like particles. The motion of these objects is governed by the constant-acceleration kinematic equations. Visualize:

Solve:

Using v12x = v02x + 2ax ( x1 − x0 ) ,

( 4.0 m/s )

2

= 0 m 2 /s 2 + 2ax ( 2.0 m ) ⇒ ax = 4.0 m/s 2

From the free-body diagram on the block:

∑( F )

on B x

= FC on B = mBax ⇒ FC on B = ( 20 kg ) ( 4.0 m/s 2 ) = 80 N

Also, according to Newton’s third law FB on C = FC on B = 80 N. Newton’s second law on the cable is:

∑( F )

on C x

= Fext − FB on C = mC ax ⇒ 100 N − 80 N = mC ( 4.0 m/s 2 ) ⇒ mC = 5 kg

7.12. Model: The man (M) and the block (B) are interacting with each other through a rope. We will assume the pulley to be frictionless. This assumption implies that the tension in the rope is the same on both sides of the pulley. The system is the man and the block. Visualize:

Solve: Clearly the entire system remains in equilibrium since mB > mM . The block would move downward but it is already on the ground. From the free-body diagrams, we can write down Newton’s second law in the vertical direction as

∑( F )

on M y

= TR on M − ( FG )M = 0 N ⇒ TR on M = ( FG )M = ( 60 kg ) ( 9.8 m/s 2 ) = 588 N

Since the tension is the same on both sides, TB on R = TM on R = T = 588 N.

7.13. Model: Together the carp (C) and the trout (T) make up the system that will be represented through the particle model. The fishing rod line (R) is assumed to be massless. Visualize:

Solve: Jimmy’s pull T2 is larger than the total weight of the fish, so they accelerate upward. They are tied together, so each fish has the same acceleration a. Newton’s second law along the y-direction for the carp and the trout is

∑( F )

on C y

= T2 − T1 − ( FG )C = mC a

∑( F )

on T y

= T1 − ( FG )T = mT a

Adding these two equations gives a=

T2 − ( FG )C − ( FG )T

( mC + mT )

=

60 N − (1.5 kg ) ( 9.8 m/s 2 ) − ( 3 kg ) ( 9.8 m/s 2 ) 1.5 kg + 3.0 kg

= 3.533 m/s 2

Substituting this value of acceleration back into the force equation for the trout, we find that

T1 = mT ( a + g ) = ( 3 kg ) ( 3.533 m/s 2 + 9.8 m/s 2 ) = 40 N ( FG )T = mT g = ( 3 kg ) ( 9.8 m/s 2 ) = 29.4 N Thus, T2 > T1 > ( FG )T > ( FG )C .

( FG )C = mC g = (1.5 kg ) ( 9.8 m/s 2 ) = 14.7 N

7.14. Model: The block of ice (I) is a particle and so is the rope (R) because it is not massless. We must therefore consider both the block of ice and the rope as objects in the system. Visualize:

Solve:

G The force Fext acts only on the rope. Since the rope and the ice block move together, they have the same

acceleration. Also because the rope has mass, Fext on the front end of the rope is not the same as FI on R that acts on the rear end of the rope. Newton’s second law along the x-axis for the ice block and the rope is

∑( F )

= FR on I = mI a = (10 kg ) ( 2.0 m/s 2 ) = 20 N

∑( F )

= Fext − FI on R = mR a ⇒ Fext − FR on I = mR a

on I x

on R x

⇒ Fext = FR on I + mR a = 20 N + ( 0.500 kg ) ( 2.0 m/s 2 ) = 21 N

7.15. Model: The hanging block and the rail car are objects in the systems. Visualize:

Solve: The mass of the rope is very small in comparison to the 2000-kg block, so we will assume a massless G G rope. In that case, the forces T1 and T1′ act as if they are an action/reaction pair. The hanging block is in static G equilibrium, with Fnet = 0 N, so T1′ = mblock g = 19,600 N. The rail car with the pulley is also in static equilibrium:

T2 + T3 − T1 = 0 N Notice how the tension force in the cable pulls both the top and bottom of the pulley to the right. Now, T1 = T1′ = 19,600 N by Newton’s third law. Also, the cable tension is T2 = T3 = T . Thus, T = 12 T1′ = 9800 N.

7.16. Visualize:

Solve: The rope is treated as two 1.0-kg interacting objects. At the midpoint of the rope, the rope has a tension TB on T = TT on B ≡ T . Apply Newton’s first law to the bottom half of the rope to find T.

( Fnet ) y = 0 = T − ( FG ) B

⇒ T = mB g = (1.0 kg ) ( 9.80 m/s 2 ) = 9.8 N Assess: 9.8 N is half the gravitational force on the whole rope. This is reasonable since the top half is holding up the bottom half of the rope against gravity.

7.17. Model: The two hanging blocks, which can be modeled as particles, together with the two knots where rope 1 meets with rope 2 and rope 2 meets with rope 3 form a system. All the four objects in the system are in static equilibrium. The ropes are assumed to be massless. Visualize:

Solve: (a) We will consider both the two hanging blocks and the two knots. The blocks are G in static G equilibrium with Fnet = 0 N. Note that there are three action/reaction pairs. For Block 1 and Block 2, Fnet = 0 N and we have T4′ = ( FG )1 = m1 g T5′ = ( FG )2 = m2 g Then, by Newton’s third law:

T4 = T4′ = m1 g T5 = T5′ = m2 g

The knots are also in equilibrium. Newton’s law applied to the left knot is

( Fnet ) x = T2 − T1 cosθ1 = 0 N ( Fnet ) y = T1 sinθ1 − T4 = T1 sinθ1 − m1g = 0 N The y-equation gives T1 = m1 g sin θ1 . Substitute this into the x-equation to find

T2 =

m1 g cosθ1 m1 g = sin θ1 tanθ1

Newton’s law applied to the right knot is

( Fnet ) x = T3 cosθ 3 − T2′ = 0 N ( Fnet ) y = T3 sinθ 3 − T5 = T3 sinθ 3 − m2 g = 0 N These can be combined just like the equations for the left knot to give

T2′ =

m2 g cosθ 3 mg = 2 sin θ 3 tan θ 3

G G But the forces T2 and T2′ are an action/reaction pair, so T2 = T2′. Therefore,

m1 g mg m = 2 ⇒ tanθ 3 = 2 tanθ1 ⇒ θ 3 = tan −1 ( 2 tan 20° ) = 36° tan θ1 tanθ 3 m1 We can now use the y-equation for the right knot to find T3 = m2 g sinθ 3 = 67 N.

7.18. Visualize: Please refer to Figure P7.18.

Solve: Since the ropes are massless we can treat the tension force they transmit as a Newton’s third law force pair on the blocks. The connection shown in figure P7.18 has the same effect as a frictionless pulley on these massless ropes. The blocks are in equilibrium as the mass of A is increased until block B slides, which occurs when the static friction on B is at its maximum value. Applying Newton’s first law to the vertical forces on block B gives nB = ( FG ) B = mB g. The static friction force on B is thus

( fs )B = μs nB = μs mB g.

( fs )B = TA on B , and the same Since TA on B = TB on A , we have ( f s )B = mA g , so

Applying Newton’s first law to the horizontal forces on B gives vertical forces on A gives TB on A = ( FG )A = mA g .

μs mB g = mA g ⇒ mA = μs mB = ( 0.60 )( 20 kg ) = 12.0 kg

analysis of the

7.19. Model: The astronaut and the satellite, the two objects in our system, will be treated as particles. Visualize:

Solve: The astronaut and the satellite accelerate in opposite directions for 0.50 s. The force on the satellite and the force on the astronaut are an action/reaction pair, so both are 100 N. Newton’s second law for the satellite along the x-direction is

∑( F )

on S x

= FA on S = mSaS ⇒ aS =

FA on S mS

=

− (100 N ) = −0.156 m/s 2 640 kg

Newton’s second law for the astronaut along the x-direction is

∑( F )

on A x

= FS on A = mA aA ⇒ aA =

FS on A mA

=

FA on S mA

=

100 N = 1.25 m/s 2 80 kg

Let us first calculate the positions and velocities of the astronaut and the satellite at t1 = 0.50 s under the

accelerations aA and aS :

x1A = x0A + v0A ( t1 − t0 ) + 12 aA ( t1 − t0 ) = 0 m + 0 m + 12 (1.25 m/s 2 ) ( 0.50 s − 0 s ) = 0.156 m 2

2

x1S = x0S + v0S ( t1 − t0 ) + 12 aS ( t1 − t0 ) = 0 m + 0 m + 12 ( −0.156 m/s 2 ) ( 0.50 s − 0 s ) = −0.020 m 2

2

v1A = v0A + aA ( t1 − t0 ) = 0 m/s + (1.25 m/s 2 ) ( 0.50 s − 0 s ) = 0.625 m/s v1S = v0S + aS ( t1 − t0 ) = 0 m/s + ( −0.156 m/s 2 ) ( 0.5 s − 0 s ) = −0.078 m/s With x1A and x1S as initial positions, v1A and v1S as initial velocities, and zero accelerations, we can now obtain the new positions at ( t2 − t1 ) = 59.5 s :

x2A = x1A + v1A ( t2 − t1 ) = 0.156 m + ( 0.625 m/s )( 59.5 s ) = 37.34 m x2S = x1S + v1S ( t2 − t1 ) = − 0.02 m + ( −0.078 m/s )( 59.5 s ) = −4.66 m Thus the astronaut and the satellite are x2A − x2S = ( 37.34 m ) − ( −4.66 m ) = 42 m apart.

7.20. Model: The block (B) and the steel cable (C), the two objects in the system, are considered particles, and their motion is determined by the constant-acceleration kinematic equations. Visualize:

Solve:

Using v1x = v0 x + ax ( t1 − t0 ) ,

4.0 m/s = 0 m/s + ax ( 2.0 s − 0 s ) ⇒ ax = 2.0 m/s 2 Newton’s second law along the x-direction for the block is

∑( F )

on B x

= FC on B = mB ax = ( 20 kg ) ( 2.0 m/s 2 ) = 40 N

Fext acts on the right end of the cable and FB on C acts on the left end. According to Newton’s third law, FB on C = FC on B = 40 N. The difference in tension between the two ends of the cable is thus Fext − FB on C = 100 N − 40 N = 60 N

7.21. Model: The block (B) and the steel cable (C), the two objects in the system, are treated as particles, and their motion is determined by constant-acceleration kinematic equations. We ignore the hanging shape of the cable. Visualize:

Solve:

Using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) , 2

4.0 m = 0 m + 0 m + 12 ax ( 2.0 s − 0 s ) ⇒ ax = 2.0 m/s 2 2

Newton’s second law along the x-direction for the block is

∑( F )

on B x

= FC on B = mBax = ( 20 kg ) ( 2.0 m/s 2 ) = 40 N

Therefore, the change in tension (T) in the cable from one Fext − FB on C = 100 N − 40 N = 60 N. The tension in the cable as a function of x is

( 60 N ) x

end

to

the

other

= ( 40 + 60 x ) N 1m with x in meters. x = 0 m is where the cable attaches to the box and x = 1.0 m is at the right end of the cable. T ( x ) = 40 N +

is

7.22. Visualize: Consider a segment of the rope of length y, starting from the bottom of the rope. The weight of this segment of rope is a downward force. It is balanced by the tension force at height y.

Solve: The mass m of this segment of rope is the same fraction of the total mass M = 2.2 kg as length y is a fraction of the total length L = 3.0 m. That is, m/M = y/L, from which we can write the mass of the rope segment

m=

M y L

This segment of rope is in static equilibrium, so the tension force pulling up on it is

T = FG = mg =

Mg (2.2 kg)(9.8 m /s 2 ) y= y = 7.19 y N 3.0 m L

where y is in m. The tension increases linearly from 0 N at the bottom ( y = 0 m) to 21.6 N at the top ( y = 3 m). This is shown in the graph.

7.23. Model: Sled A, sled B, and the dog (D) are treated like particles in the model of kinetic friction. Visualize:

Solve:

The acceleration constraint is ( aA ) x = ( aB ) x = ax . Newton’s second law on sled A is G G ∑ Fon A = nA − ( FG )A = 0 N ⇒ nA = ( FG )A = mA g ∑ Fon A = T1 on A − f A = mA ax

(

)

(

y

)

x

Using f A = μ k nA , the x-equation yields

T 1 on A − μ k nA = mA ax ⇒ 150 N − ( 0.1)(100 kg ) ( 9.8 m/s 2 ) = (100 kg ) ax ⇒ ax = 0.52 m/s 2 On sled B:

G

∑( F ) on B

y

= nB − ( FG )B = 0 N ⇒ nB = ( FG )B = mB g

G

∑( F ) on B

x

= T2 − T1 on B − f B = mB ax

T 1 on B and T 1 on A act as if they are an action/reaction pair, so T 1 on B = 150 N. Using f B = μ k nB = ( 0.10 )( 80 kg )

( 9.8 m/s ) = 78.4 N, 2

we get T2 − 150 N − 78.4 N = ( 80 kg ) ( 0.52 m/s 2 ) ⇒ T2 = 270 N

Thus the tension T2 = 2.7 × 102 N.

7.24. Model: The coffee mug (M) is the only object in the system, and it will be treated as a particle. The model of friction and the constant-acceleration kinematic equations will also be used. Visualize:

Solve: The mug and the car have the same velocity. If the mug does not slip, their accelerations will also be the same. Using v12x = v02x + 2ax ( x1 − x0 ) , we get

0 m 2 /s 2 = ( 20 m/s ) + 2ax ( 50 m ) ⇒ ax = −4.0 m/s 2 2

The static force needed to stop the mug is

( Fnet ) x = − fs = max = ( 0.5 kg ) ( −4.0 m/s2 ) = −2.0 N ⇒

f s = 2.0 N

The maximum force of static friction is ( f s ) max = μ s n = μ s FG = μ s mg = ( 0.50 )( 0.50 kg ) ( 9.8 m/s 2 ) = 2.45 N Since ( f s ) max < ( f s ) max , the mug does not slide.

7.25.

Visualize:

The car and the ground are denoted by C and S, respectively. Solve: (a) The car has an internal source of energy–fuel–that allows it to turn the wheels and exert the Gforce G FC on G . As the drive wheels turn they push backward against the ground. This is a static friction force FC on S because the wheels don’t slip against the ground. ByG Newton’s third law, the ground exerts a reaction force G FS on C . This reaction force is opposite in direction to FC on S , hence, is in the forward direction. This is the force that accelerates the car. Houses do not have an internal source of energy that allows them to push sideways against the ground. They also aren’t on wheels, which let the car slide across the ground with minimal friction. (b) The car presses down against the ground at both the drive wheels (assumed to be the front wheels F, although that is not critical) and the nondrive wheels. For this car, two-thirds of the gravitational force rests on the front G wheels. Physically, force FS on C is a static friction force. The maximum acceleration of the car on the ground (or concrete surface) occurs when the static friction reaches its maximum possible value.

F S on C = ( fs )max = μs nF = μs ( FG )F = (1.00 ) ( 23 ) (1500 kg ) ( 9.8 m/s 2 ) = 9800 N ⇒ amax =

FS on C m

=

9800 N = 6.533 m/s 2 1500 kg

7.26. Model: The starship and the shuttlecraft will be denoted as M and m, respectively, and both will be treated as particles. We will also use the constant-acceleration kinematic equations. Visualize:

G (a) The tractor beam is some kind of long-range force FM on m . Regardless of what kind of force it is, by G Newton’s third law there must be a reaction force Fm on M on the starship. As a result, both the shuttlecraft and the starship move toward each other (rather than the starship remaining at rest as it pulls the shuttlecraft in). However, the very different masses of the two crafts means that the distances they each move will also be very different. The pictorial representation shows that they meet at time t1 when xM1 = xm1. There’s only one force on each craft, so Newton’s second law is very simple. Furthermore, because the forces are an action/reaction pair, Solve:

FM on m = Fm on M = Ftractor beam = 4.0 × 104 N The accelerations of the two craft are aM =

Fm on M M

=

G FM on m −4.0 × 104 N 4.0 × 104 N 2 a = = = −2.0 m/s 2 = 0.020 m/s m 2.0 × 106 kg m 2.0 × 104 kg

Acceleration am is negative because the force and acceleration vectors point in the negative x-direction. The accelerations are very different even though the forces are the same. Now we have a constant-acceleration problem in kinematics. At a later time t1 the positions of the crafts are xM1 = xM0 + vM0 ( t1 − t0 ) + 12 aM ( t1 − t0 ) = 12 aM t12 2

xm1 = xm0 + vm0 ( t1 − t0 ) + 12 am ( t1 − t0 ) = xm0 + 12 amt12 2

The craft meet when xM1 = xm1 , so 1 2

aMt12 = xm0 + 12 amt12 ⇒ t1 =

2 (10,000 m ) 2 xm0 2 xm0 = = = 99.5 s aM − am aM + am 2.02 m/s 2

Knowing t1 , we can now find the starship’s position as it meets the shuttlecraft: xM1 = 12 aMt12 = 99 m

The starship moves 99 m as it pulls in the shuttlecraft from 10 km away.

7.27. Model: The rock (R) and Bob (B) are the two objects in our system, and will be treated as particles. We will also use the constant-acceleration kinematic equations. Visualize:

G Solve: (a) Bob exerts a forward force FB on R on the rock to accelerate it forward. The rock’s acceleration is calculated as follows: v2 ( 30 m/s ) = 450 m/s 2 v = v + 2a0R Δx ⇒ aR = 1R = 2Δx 2 (1.0 m ) 2

2 1R

2 0R

The force is calculated from Newton’s second law:

FB on R = mR aR = (.500 kg ) ( 450 m/s 2 ) =225 N Bob exerts a force of 2.3 × 102 N on the rock.

G G G (b) Because Bob pushes on the rock, the rock pushes back on Bob with a force FR on B . Forces FR on B and FB on R are an action/reaction pair, so FR on B = FB on R = 225 N. The force causes Bob to accelerate backward with an acceleration equal to

aB =

(F

)

net on B x

mB

=−

FR on B mB

=−

225 N = −3.0 m/s 2 75 kg

This is a rather large acceleration, but it lasts only until Bob releases the rock. We can determine the time interval by returning to the kinematics of the rock:

v1R = v0R + aR Δt = aR Δt ⇒ Δt =

v1R = 0.0667 s aR

At the end of this interval, Bob’s velocity is v1B = v0B + aBΔt = aBΔt = −0.20 m/s Thus his recoil speed is 0.20 m/s.

7.28. Model: The boy (B) and the crate (C) are the two objects in our system, and they will be treated in the particle model. We will also use the static and kinetic friction models. Visualize:

Solve: The fact that the boy’s feet occasionally slip means that the maximum force of static friction must exist between the boy’s feet and the sidewalk. That is, f sB = μ sB nB . Also f kC = μ kC nC . Newton’s second law for the crate is

∑( F )

on C y

= nC − ( FG )C = 0 N ⇒ nC = mC g

∑( F )

on C x

= FB on C − f kC = 0 N ⇒ FB on C = f kC = μ kC nC = μ kC mC g

∑( F )

= fsB − FC on B = 0 N ⇒ FC on B = f sB = μsB nB = μsB mB g

Newton’s second law for the boy is

∑( F )

on B y

= nB − ( FG )B = 0 N ⇒ nB = mB g

on B x

G G FC on B and FB on C are an action/reaction pair, so FC on B = FB on C ⇒ μsB mB g = μ kC mC g ⇒ mC =

μsB mB ( 0.8 )( 50 kg ) = = 2.0 × 102 kg μ kC ( 0.2 )

7.29. Model: Assume package A and package B are particles. Use the model of kinetic friction and the constant-acceleration kinematic equations. Visualize:

Solve: Package B has a smaller coefficient of friction. It will try to overtake package A and push against it. Package A will push back on B. The acceleration constraint is ( aA ) x = ( aB ) x = a.

Newton’s second law for each package is

∑( F )

on A x

= FB on A + ( FG )A sin θ − f kA = mA a

⇒ FB on A + mA g sin θ − μ kA ( mA g cosθ ) = mA a

∑( F )

on B x

= − FA on B − f kB + ( FG )B sinθ = mBa

⇒ − FA on B − μ kB ( mB g cosθ ) + mB g sinθ = mB a where we have used nA = mA cosθ g and nB = mB cosθ g . Adding the two force equations, and using

FA on B = FB on A because they are an action/reaction pair, we get a = g sin θ −

( μ kA mA + μ kBmB )( g cosθ ) mA + mB

= 1.82 m/s 2

Finally, using x1 = x0 + v0x ( t1 − t0 ) + 12 a ( t1 − t0 ) , 2

2.0 m = 0 m + 0 m + 12 (1.82 m/s 2 ) ( t1 − 0 s ) ⇒ t1 = 1.48 s 2

7.30. Model: The two blocks form a system of interacting objects. Visualize:

Please refer to Figure P7.30.

Solve: It is possible that the left-hand block (Block L) is accelerating down the slope faster than the right-hand block (Block R), causing the string to be slack (zero tension). If that were the case, we would get a zero or negative answer for the tension in the string. Newton’s first law applied to the y-direction on Block L yields

(∑ F )

L y

= 0 = nL − ( FG )L cos 20° ⇒ nL = mL g cos 20°

Therefore

( f k )L = ( μ k )L mL g cos 20° = ( 0.20 )(1.0 kg ) ( 9.80 m/s2 ) cos 20° = 1.84 N A similar analysis of the vertical forces on Block R gives ( f k )R = 1.84 N as well. Using Newton’s second law in the x-direction for Block L, ( ∑ FL ) x = mLa = TR on L − ( f k )L + ( FG )L sin 20° ⇒ mLa = TR on L − 1.84 N + mL g sin 20°. For Block R,

(∑ F )

R x

= mR a = ( FG )R sin 20° − 1.84 N − TL on R ⇒ mR a = mR g sin 20° − 1.84 N − TL on R

These are two equations in the two unknowns a and TL on R = TR on L ≡ T . Solving them, we obtain a = 2.12 m/s 2 and T = 0.61 N. Assess: The tension in the string is positive, and is about 1/3 of the kinetic friction force on each of the blocks, which is reasonable.

7.31. Model: The two ropes and the two blocks (A and B) will be treated as particles. Visualize:

Solve:

(a) The two blocks and two ropes form a combined system of total mass M = 2.5 kg. This combined

system is accelerating upward at a = 3.0 m/s 2 under the influence of a force F and the gravitational force − Mg ˆj. Newton’s second law applied to the combined system is

( Fnet ) y = F − Mg = Ma ⇒ F = M ( a + g ) = 32 N (b) The ropes are not massless. We must consider both the blocks and the ropes as systems. The force F acts only on block A because it does not contact the other objects. We can proceed to apply the y-component of Newton’s second law to each system, starting at the top. Each has an acceleration a = 3.0 m/s 2 . For block A:

(F

)

net on A y

= F − mA g − T1 on A = mA a ⇒ T1 on A = F − mA ( a + g ) = 19.2 N

(c) Applying Newton’s second law to rope 1: ( Fnet on 1 ) y = TA on 1 − m1g − TB on 1 = m1a G G G TA on 1 and T1 on A are an action/reaction pair. But, because the rope has mass, the two tension forces TA on 1 and G TB on 1 are not the same. The tension at the lower end of rope 1, where it connects to B, is

TB on 1 = TA on 1 − m1 ( a + g ) = 16.0 N (d) We can continue to repeat this procedure, noting from Newton’s third law that

T1 on B = TB on 1 and T2 on B = TB on 2 Newton’s second law applied to block B is

(F

)

net on B y

= T1 on B − mB g − T2 on B = mB a ⇒ T2 on B = T1 on B − mB ( a + g ) = 3.2 N

7.32. Model: The two blocks (1 and 2) are the systems of interest and will be treated as particles. The ropes are assumed to be massless, and the model of kinetic friction will be used. Visualize:

Solve: (a) The separate free-body diagrams for the two blocks show that there are two action/reaction pairs. G G Notice how block 1 both pushes down on block 2 (force n1′ ) and exerts a retarding friction force f 2 top on the top surface of block 2. Block 1 is in static equilibrium (a1 = 0 m/s 2 ) but block 2 is accelerating. Newton’s second law for block 1 is

(F

)

net on 1 x

= f1 − Trope = 0 N ⇒ Trope = f1

(F

)

net on 1 y

= n1 − m1 g = 0 N ⇒ n1 = m1 g

Although block 1 is stationary, there is a kinetic force of friction because there is motion between block 1 and block 2. The friction model means f1 = μ k n1 = μ k m1 g . Substitute this result into the x-equation to get the tension in the rope: Trope = f1 = μ k m1 g = 3.92 N (b) Newton’s second law for block 2 is ax = a =

(F

)

net on 2 x

m2

=

Tpull − f 2 top − f 2 bot m2

a y = 0 m/s 2 =

(F

)

net on 2 y

m2

=

n2 − n1′ − m2 g m2

G G Forces n1 and n1′ are an action/reaction pair, so n1′ = n1 = m1 g . Substituting into the y-equation gives

n2 = ( m1 + m2 ) g . This is not surprising because the combined weight of both objects presses down on the

surface. The kinetic friction on the bottom surface of block 2 is then f 2 bot = μ k n2 = μ k ( m1 + m2 ) g G G The forces f1 and f 2 top are an action/reaction pair, so f 2 bot = f1 = μ k m1 g . Inserting these friction results into

the x-equation gives

a=

(F

)

net on 2 x

m2

=

Tpull − μ k m1 g − μ k ( m1 + m2 ) g m2

= 2.16 m/s 2

7.33. Model: The 3-kg and 4-kg blocks are to be treated as particles. The models of kinetic and static friction and the constant-acceleration kinematic equations will be used. Visualize:

Solve: Minimum time will be achieved when static friction is at its maximum possible value. Newton’s second law for the 4-kg block is

∑( F )

on 4 y

= n3 on 4 − ( FG )4 = 0 N ⇒ n3 on 4 = ( FG )4 = m4 g = ( 4.0 kg ) ( 9.8 m/s 2 ) = 39.2 N

⇒ f s4 = ( f s ) max = μs n3 on 4 = ( 0.60 )( 39.2 N ) = 23.52 N Newton’s second law for the 3-kg block is

∑( F )

on 3 y

= n3 − n4 on 3 − ( FG )3 = 0 N ⇒ n3 = n4 on 3 + ( FG )3 = 39.2 N + ( 3.0 kg ) ( 9.8 m/s 2 ) = 68.6 N

Friction forces f and fs4 are an action/reaction pair. Thus

∑( F )

on 3 x

= fs3 − f k3 = m3a3 ⇒ f s4 − μ k n3 = m3a3 ⇒ 23.52 N − ( 0.20 )( 68.6 N ) = ( 3.0 kg ) a3 ⇒ a3 = 3.267 m/s 2

Since block 3 does not slip, this is also the acceleration of block 4. The time is calculated as follows: x1 − x0 + v0 x ( t1 − t0 ) + 12 a ( t1 − t0 ) ⇒ 5.0 m = 0 m + 0 m + 12 ( 3.267 m/s 2 ) ( t1 − 0 s ) ⇒ t1 = 1.75 s 2

2

7.34. Model: Blocks 1 and 2 make up the system of interest and will be treated as particles. Assume a massless rope and frictionless pulley. Visualize:

Solve: The blocks accelerate with the same magnitude but in opposite directions. Thus the acceleration constraint is a2 = a = −a1 , where a will have a positive value. There are two real action/reaction pairs. The two tension forces will act as if they are action/reaction pairs because we are assuming a massless rope and a frictionless pulley. Make sure you understand why the friction forces point in the directions shown in the freeG body diagrams, especially force f1′ exerted on block 2 by block 1. We have quite a few pieces of information to

include. First, Newton’s second law for blocks 1 and 2: G Fnet on 1 = f1 − T1 = μ k n1 − T1 = m1a1 = − m1a

(

)

x

(F

(F

)

net on 1 y

= n1 − m1 g = 0 N ⇒ n1 = m1 g

) = T − f ′− f − T = T − f ′− μ n − T = m a ( F ) = n − n′ − m g = 0 N ⇒ n = n′ + m g

net on 2 x

pull

net on 2 y

1

2

2

2

1

pull

2

1

k 2

2

2

1

2 2

= m2 a

2

We’ve already used the kinetic friction model in both x-equations. Next, Newton’s third law:

n1′ = n1 = m1 g f1′ = f1 = μ k n1 = μ k m1 g T1 = T2 = T Knowing n1′, we can now use the y-equation of block 2 to find n2 . Substitute all these pieces into the two xequations, and we end up with two equations in two unknowns:

μ k m1 g − T = −m1a Tpull − T − μ k m1 g − μ k ( m1 + m2 ) g = m2 a Subtract the first equation from the second to get

Tpull − μ k ( 3m1 + m2 ) g = ( m1 + m2 ) a ⇒ a =

Tpull − μ k ( 3m1 + m2 ) g m1 + m2

= 1.77 m/s 2

7.35. Model: The sled (S) and the box (B) will be treated in the particle model, and the model of friction will be used. Visualize:

In the sled’s free-body diagram nS is the normal (contact) force on the sled due to the snow. Similarly f kS is the force of kinetic friction on the sled due to snow. Solve: Newton’s second law on the box in the y-direction is

nS on B − ( FG )B cos 20° = 0 N ⇒ nS on B = (10 kg ) ( 9.8 m/s 2 ) cos 20° = 92.09 N G The static friction force fS on B accelerates the box. The maximum acceleration occurs when static friction reaches its maximum possible value.

( fs ) max = μSnS on B = ( 0.50 )( 92.09 N ) = 46.05 N Newton’s second law along the x-direction thus gives the maximum acceleration fS on B − ( FG )B sin 20° = mB a ⇒ 46.05 N − (10 kg ) ( 9.8 m/s 2 ) sin 20° = (10 kg ) a ⇒ a = 1.25 m/s 2 Newton’s second law for the sled along the y-direction is

nS − nB on S − ( FG )S cos 20° = 0 N ⇒ nS = nB on S + mS g cos 20° = ( 92.09 N ) + ( 20 kg ) ( 9.8 m/s 2 ) cos 20° = 276.27 N Therefore, the force of friction on the sled by the snow is f kS = ( μ k )nS = (0.06)(276.27 N) = 16.58 N Newton’s second law along the x-direction is

Tpull − wS sin 20° − f kS − f B on S = mSa The friction force f B on S = fS on B because these are an action/reaction pair. We’re using the maximum acceleration, so the maximum tension is

Tmax − ( 20 kg ) ( 9.8 m/s 2 ) sin 20° − 16.58 N − 46.05 N = ( 20 kg ) (1.25 m/s 2 ) ⇒ Tmax = 155 N

7.36. Model: The masses m and M are to be treated in the particle model. We will also assume a massless rope and frictionless pulley, and use the constant-acceleration kinematic equations for m and M. Visualize:

Solve:

Using y1 = y0 + v0 y ( t1 − t0 ) + 12 aM ( t1 − t0 ) , 2

( −1 m ) = 0 m + 0 m + 12 aM ( 6.0 s − 0 s )

2

⇒ aM = −0.0556 m/s 2

Newton’s second law for m and M is:

∑( F )

on m y

= TR on m − ( FG )m = mam

∑( F )

on M y

= TR on M − ( FG )M = MaM

The acceleration constraint is am = − aM . Also, the tensions are an action/reaction pair, thus TR on m = TR on M . With these, the second law equations are TR on M − Mg = MaM TR on M − mg = −maM

Subtracting the second from the first gives

⎡ g + aM ⎤ − Mg + mg = MaM + maM ⇒ m = M ⎢ ⎥ ⎣ g − aM ⎦ ⎡ 9.8 − 0.556 ⎤ = (100 kg ) ⎢ ⎥ = 99 kg ⎣ 9.8 + 0.556 ⎦ Assess: Note that am = − aM = 0.0556 m/s 2 . For such a small acceleration, a mass of 99 kg for m compared to M = 100 kg is understandable.

7.37. Model: Use the particle model for the block of mass M and the two massless pulleys. Additionally, the

G rope is massless and the pulleys are frictionless. The block is kept in place by an applied force F . Visualize:

Solve:

Since there is no friction on the pulleys, T2 = T3 and T2 = T5 . Newton’s second law for mass M is

T1 − FG = 0 N ⇒ T1 = Mg = (10.2 kg ) ( 9.8 m/s 2 ) = 100 N Newton’s second law for the small pulley is T2 + T3 − T1 = 0 N ⇒ T2 = T3 =

T1 = 50 N = T5 = F 2

Newton’s second law for the large pulley is T4 − T2 − T3 − T5 = 0 N ⇒ T4 = T2 + T3 + T5 = 150 N

7.38. Model: Assume the particle model for m1 , m2 , and m3 , and the model of kinetic friction. Assume the ropes to be massless, and the pulleys to be frictionless and massless. Visualize:

Solve:

Newton’s second law for m1 is T1 − ( FG )1 = m1a1. Newton’s second law for m2 is

∑( F ) on m2

∑ (F

y

= n2 − ( FG )2 = 0 N ⇒ n2 = ( 2.0 kg ) ( 9.8 m/s 2 ) = 19.6 N ) = T2 − f k2 − T = m2 a2 ⇒ T2 − μ k n2 − T1 = (2.0 kg)a2

on m2 x

Newton’s second law for m3 is

∑( F ) on m3

y

= T2 − ( FG )3 = m3a3

Since m1 , m2 , and m3 move together, a1 = a2 = −a3 = a. The equations for the three masses thus become

T1 − ( FG )1 = m1a = (1.0 kg ) a T2 − μ k n2 − T1 = m2 a = ( 2.0 kg ) a T2 − ( FG )3 = −m3a = − ( 3.0 kg ) a Subtracting the third equation from the sum of the first two equations yields: − ( FG )1 − μ k n2 + ( FG )3 = ( 6.0 kg ) a ⇒ − (1.0 kg ) ( 9.8 m/s 2 ) − ( 0.30 )(19.6 N ) + ( 3.0 kg ) ( 9.8 m/s 2 ) = ( 6.0 kg ) a ⇒ a = 2.3 m/s 2

7.39. Model: Assume the particle model for the two blocks, and the model of kinetic and static friction. Visualize:

Solve: (a) If the mass m is too small, the hanging 2.0 kg mass will pull it up the slope. We want to find the smallest mass that will stick because of friction. The smallest mass will be the one for which the force of static friction is at its maximum possible value: f s = ( f s )max = μs n. As long as the mass m is stuck, both blocks are at G rest with Fnet = 0 N. Newton’s second law for the hanging mass M is

( Fnet ) y = TM − Mg = 0 N ⇒ TM = Mg = 19.6 N For the smaller mass m,

( Fnet ) x = Tm − fs − mg sinθ = 0 N ( Fnet ) y = n − mg cosθ ⇒ n = mg cosθ

G G For a massless string and frictionless pulley, forces Tm and TM act as if they are an action/reaction pair. Thus Tm = TM . Mass m is a minimum when

( fs )max = μs n = μs mg cosθ .

Substituting these expressions into the x-

equation, TM − μ s mg cosθ − mg sinθ = 0 N ⇒ m =

TM

( μs cosθ + sinθ ) g

= 1.83 kg

(b) Because μ k < μS the 1.83 kg block will begin to slide up the ramp, and the 2.0 kg mass will begin to fall, if the block is nudged ever so slightly. Now the net force and the acceleration are not zero. Notice how, in the pictorial representation, we chose different coordinate systems for the two masses. This gives block M an acceleration with only a y-component and block m an acceleration with only an x-component. The magnitudes of the accelerations are the same because the blocks are tied together. But block M has a negative acceleration G component a y (vector a points down) whereas block m has a positive ax . Thus the acceleration constraint is

(am ) x = −(aM ) y = a, where a will have a positive value. Newton’s second law for block M is

( Fnet ) y = T − Mg = M ( aM ) y = − Ma For block m we have

( Fnet ) x = T − f k − mg sinθ = T − μ k mg cosθ − mg sinθ = m ( am ) x = ma In writing these equations, we used Newton’s third law to write Tm = TM = T . Also, we noticed that the yequation and the friction model for block m don’t change, except for μ s becoming μ k , so we already know f k from part (a). Notice that the tension in the string is not the gravitational force Mg. We have two equations in the two unknowns T and a: T − Mg = − Ma T − ( μ k cosθ + sin θ ) mg = ma

Subtracting the second equation from the first to eliminate T, − Mg + ( μ k cosθ + sinθ ) mg = − Ma − ma = −( M + m)a

⇒a=

M − ( μ k cosθ + sinθ ) m g = 1.32 m/s 2 M +m

7.40. Model: Assume the particle model for the two blocks. Visualize:

Solve: (a) The slope is frictionless, so the blocks stay in place only if held. Once m is released, the blocks will G move one way or the other. As long as m is held, the blocks are in static equilibrium with Fnet = 0 N. Newton’s second law for the hanging block M is ( Fnet on M ) = TM − Mg = 0 N ⇒ TM = Mg = 19.6 N y

By Newton’s third law, TM = Tm = T = 19.6 N is the tension in the string. (b) The free-body diagram shows box m after it is released. Whether it moves up or down the slope depends on whether the acceleration a is positive or negative. The acceleration constraint is (am ) x = a = −(aM ) y . Newton’s

second law for each system gives ( Fnet on m ) = T − mg sin 35° = m ( am )x = ma x

(F

)

net on M y

= T − Mg = M ( aM ) y = − Ma

We have two equations in two unknowns. Subtract the second from the first to eliminate T: M − m sin 35° − mg sin 35° + Mg = ( m + M ) a ⇒ a = g = −0.481 m/s 2 M +m Since a < 0 m/s 2 , the box accelerates down the slope. (c) It is now straightforward to compute T = Mg − Ma = 21 N. Notice how the tension is larger than when the blocks were motionless.

7.41. Model: Assume the particle model for the book (B) and the coffee cup (C), the models of kinetic and static friction, and the constant-acceleration kinematic equations. Visualize:

Solve:

(a) Using v12x = v02x + 2a ( x1 − x0 ) ,

0 m 2 /s 2 = ( 3.0 m/s ) + 2a ( x1 ) ⇒ ax1 = −4.5 m 2 /s 2 2

To find x1 , we must first find a. Newton’s second law for the book and the coffee cup is

∑( F )

on B y

= nB − ( FG )B cos 20° = 0 N ⇒ nB = (1.0 kg ) ( 9.8 m/s 2 ) cos 20° = 9.21 N

∑( F )

on B x

= −T − f k − ( FG )B sin 20° = mB aB

∑( F )

on C y

= T − ( FG )C = mC aC

The last two equations can be rewritten, using aC = aB = a, as −T − μ k nB − mB g sin 20° = mB a T − mC g = mC a Adding the two equations,

a ( mC + mB ) = − g ( mC + mB sin 20° ) − μ k ( 9.21 N )

⇒ (1.5 kg ) a = − ( 9.8 m/s 2 ) ⎣⎡0.500 kg + (1.0 kg ) sin 20° ⎦⎤ − ( 0.20 )( 9.21 N ) ⇒ a = −6.73 m/s 2 Using this value for a, we can now find x1 as follows: x1 =

−4.5 m 2 /s 2 −4.5 m 2 /s 2 = = 0.67 m −6.73 m/s 2 a

(b) The maximum static friction force is ( f s ) max = μs nB = ( 0.50 )( 9.21 N ) = 4.60 N. We’ll see if the force f s

needed to keep the book in place is larger or smaller than ( f s ) max . When the cup is at rest, the string tension is T = mC g . Newton’s first law for the book is

∑( F )

on B x

= fs − T − wB sin 20° = fs − mC g − mB g sin 20° = 0

⇒ fs = ( M C + M B sin 20° ) g = 8.25 N Because fs > ( f s ) max , the book slides back down.

7.42. Model: Use the particle model for the cable car and the counterweight. Assume a massless cable. Visualize:

Solve: (a) Notice the separate coordinate systems for the cable car (object 1) and the counterweight (object 2). G G G G Forces T1 and T2 act as if they are an action/reaction pair. The braking force FB works with the cable tension T1 G to allow the cable car to descend at a constant speed. Constant speed means dynamic equilibrium, so Fnet = 0 N for both systems. Newton’s second law for the cable car is

(F

)

net on 1 x

Newton’s second law for the counterweight is

(F

(F

= T1 + FB − m1 g sin θ1 = 0 N

)

net on 2 x

= m2 g sinθ 2 − T2 = 0 N

)

net on 1 y

(F

)

net on 2 y

= n1 − m1 g cosθ1 = 0 N

= n2 − m2 g cosθ 2 = 0 N

From the x-equation for the counterweight, T2 = m2 g sin θ 2 . By Newton’s third law, T1 = T2 . Thus the x-equation for the cable car becomes FB = m1 g sinθ1 − T1 = m1 g sin θ1 − m2 g sin θ 2 = 3770 N (b) If the brakes fail, then FB = 0 N. The car will accelerate down the hill on one side while the counterweight accelerates up the hill on the other side. Both will have negative accelerations because of the direction of the acceleration vectors. The constraint is a1x = a2 x = a, where a will have a negative value. Using T1 = T2 = T , the two x-equations are

( Fnet on 1 ) x = T − m1g sinθ1 = m1a1x = m1a

(F

)

net on 2 x

= m2 g sinθ 2 − T = m2 a2x = m2 a

Note that the y-equations aren’t needed in this problem. Add the two equations to eliminate T: m sin θ1 − m2 sinθ 2 g = −0.991 m/s 2 − m1 g sin θ1 + m2 g sin θ = ( m1 + m2 ) a ⇒ a = − 1 m1 + m2

Now we have a problem in kinematics. The speed at the bottom is calculated as follows: v12 = v02 + 2a ( x1 − x0 ) = 2ax1 ⇒ v1 = 2ax1 = 2 ( −0.991 m/s 2 ) ( −400 m ) = 28.2 m/s Assess: A speed of approximately 60 mph as the cable car travels a distance of 2000 m along a frictionless slope of 30° is reasonable.

7.43. Model: Assume the cable mass is negligible compared to the car mass and that the pulley is frictionless. Use the particle model for the two cars. Visualize: Please refer to Figure P7.43.

Solve: (a) The cars are moving at constant speed, so they are in dynamic equilibrium. Consider the descending car D. We can find the rolling friction force on car D, and then find the cable tension by applying Newton’s first law. In the y-direction for car D, ( Fnet ) y = 0 = nD − ( FG )D cos35°

⇒ nD = mD g cos35° So the rolling friction force on car D is

( f R )D = μ R nD = μ R mD g cos35°

Applying Newton’s first law to car D in the x-direction, ( Fnet ) x = TA on D + ( f R )D − ( FG )D sin 35° = 0 Thus

TA on D = mD g sin 35° − μ R mD g cos35°

= (1500 kg ) ( 9.80 m/s 2 ) ( sin 35° − 0.020cos35° )

= 8.2 × 103 N (b) Similarly, we find that for car A, ( f R )A = μ R mA g cos35°. In the x-direction for car A,

( Fnet ) x = Tmotor + TD on A − ( f R )A − ( FG )A sin 35° = 0 ⇒ Tmotor = mA g sin 35° + μ R mA g cos35° − ( mD g sin 35° − μ R mD g cos35° )

Here, we have used TA on D = TD on A . If we also use m A = m D , then Tmotor = 2μ R mA g cos35° = 4.8 × 102 N. Assess: Careful examination of the free-body diagrams for cars D and A yields the observation that Tmotor = 2( FR ) A in order for the cars to be in dynamic equilibrium. It is a tribute to the design that the motor must only provide such a small force compared to the tension in the cable connecting the two cars.

7.44. Model: The painter and the chair are treated as a single object and represented as a particle. We assume that the rope is massless and that the pulley is massless and frictionless. Visualize:

Solve: If the painter pulls down on the rope with force F, Newton’s third law requires the rope to pull up on the painter with force F. This is just the tension in the rope. With our model of the rope and pulley, the same tension force F also pulls up on the painter’s chair. Newton’s second law for (painter + chair) is

2F − FG = ( mP + mC ) a ⇒F =(

1 2

) ⎡⎣( mP + mC ) a + ( mP + mC ) g ⎤⎦ = 12 ( mP + mC )( a + g )

= ( 12 ) ( 70 kg + 10 kg ) ( 0.20 m/s 2 + 9.8 m/s 2 ) = 4.0 × 102 N Assess: A force of 400 N, which is approximately one-half the total gravitational force, is reasonable since the upward acceleration is small.

7.45. Model: Use the particle model for the tightrope walker and the rope. The rope is assumed to be massless, so the tension in the rope is uniform. Visualize:

Solve:

Newton’s second law for the tightrope walker is

FR on W − FG = ma ⇒ FR on W = m ( a + g ) = ( 70 kg ) ( 8.0 m/s 2 + 9.8 m/s 2 ) = 1.25 × 103 N Now, Newton’s second law for the rope gives

∑( F ) on R

y

= T sin θ + T sinθ − FW on R = 0 N ⇒ T =

FW on R 2sin10°

We used FW on R = FR on W because they are an action/reaction pair.

=

FR on W 2sin10°

=

1.25 × 103 N = 3.6 × 103 N 2sin10°

7.46. Model: Use the particle model for the wedge and the block. Visualize:

The block will not slip relative to the wedge if they both have the same horizontal acceleration a. Note: n1 on 2 = n2 on 1. Solve:

The y-component of Newton’s second law for block m2 is

∑( F )

on 2 y

= n1 on 2 cosθ − ( FG )2 = 0 N ⇒ n1 on 2 =

m2 g cosθ

Combining this equation with the x-component of Newton’s second law yields:

∑( F )

on 2 x

= n1 on 2 sin θ = m2 a ⇒ a =

n1 on 2 sin θ = g tan θ m2

Now, Newton’s second law for the wedge is

∑( F )

on 1 x

= F − n2 on 1 sinθ = m1a

⇒ F = m1a + n2 on 1 sin θ = m1a + m2 a = ( m1 + m2 )a = (m1 + m2 ) g tan θ

7.47. Model: Treat the basketball player (P) as a particle, and use the constant-acceleration kinematic equations.

Visualize:

Solve: (a) While in the process of jumping, the basketball player is pressing down on the floor as he straightens G his legs. He exerts a force FP on F on the floor. The player experiences a gravitational force FG as well as a P G G normal force by the floor nF on P . The only force that the floor experiences is the one exerted by the player FP on F . G G G (b) The player standing at rest exerts a force FP on F on the floor. The normal force nF on P is the reaction force to FP on F . G But nF on P = FP on F , so Fnet = 0 N. When the basketball player accelerates upward by straightening his legs, his

( )

speed has to increase from zero to v1y with which he leaves the floor. Thus, according to Newton’s second law, there must be a net upward force on him during this time. This can be true only if nF on P > ( FG )P . In other words, the player presses on the floor with a force FP on F larger than the gravitational force on him, which is equal to his G weight. The reaction force nF on P then exceeds his weight and accelerates him upward until his feet leave the floor. (c) The height of 80 cm = 0.80 m is sufficient to determine the speed v1y with which he leaves the floor. Once his feet are off the floor, he is simply in free fall, with a1 = − g. From kinematics, v22 y = v12y + 2a1 ( y2 − y1 ) ⇒ 0 m 2 /s 2 = v12y + 2 ( − g )( 0.80 m )

⇒ v1 y = 2g ( 0.80 m ) = 3.96 m/s The basketball player reached v1 y = 4.0 m/s by accelerating from rest through a distance of 0.60 m. (d) Assuming a0 to be constant during the jump, we find

v12y = v02y + 2a0 ( y1 − y0 ) = 0 m 2 /s 2 + 2a0 ( y1 − 0 m ) ⇒ a0 =

v12y 2 y1

( 3.96 m/s ) 2 ( 0.60 m )

2

=

= 13.1 m/s 2

(e) The scale reads the value of nF on P , the force exerted by the scale on the player. Before jumping,

nF on P − ( FG )P = 0 N ⇒ nF on P = ( FG )P = mg = (100 kg ) ( 9.8 m/s 2 ) = 980 N While accelerating up,

⎛ a ⎞ ⎛ 13.1 ⎞ nF on P − mg = ma0 ⇒ nF on P = ma0 + mg = mg ⎜ 1 + 0 ⎟ = ( 980 N ) ⎜1 + ⎟ = 2290 N 9.8 ⎠ g⎠ ⎝ ⎝ After leaving the scale, nF on P = 0 N because there is no contact with the scale.

7.48.

A 1.0 kg wood block is placed on top of a 2.0 kg wood block. A horizontal rope pulls the 2.0 kg block across a frictionless floor with a force of 21.0 N. Does the 1.0 kg block on top slide? Visualize:

Solve:

The 1.0 block is accelerated by static friction. It moves smoothly with the lower block if f s < ( fs ) max . It

slides if the force that would be needed to keep it in place exceeds ( f s ) max . Begin by assuming the blocks move together with common acceleration a. Newton’s second is

∑ ( Fon 1 ) x = fs = m1a Bottom block: ∑ ( Fon 2 ) x = Tpull − fs = m2 a

Top block:

Adding these two equations gives Tpull = (m1 + m2 )a, or a = (21.0 N)/(1.0 kg + 2.0 kg) = 7.0 m/s 2 . The static

friction force needed to accelerate the top block at 7.0 m/s2 is

fs m1a = (1.0 kg)(7.0 m/s 2 ) = 7.0 N To find the maximum possible static friction force ( f s ) max = μ s n1 , the y-equation of Newton’s second law for the top block shows that n1 = m1 g . Thus ( f s ) max = μs m1 g = (0.50)(1.0 kg)(9.80 m/s 2 ) = 4.9 N Because 7.0 N > 4.9 N, static friction is not sufficient to accelerate the top block. It slides.

7.49.

A 1.0 kg wood block is placed behind a 2.0 kg wood block on a horizontal table. The coefficients of kinetic friction with the table are 0.3 for the 1.0 kg block and 0.5 for the 2.0 kg block. The 1.0 kg block is pushed forward, against the 2.0 block, and released with a speed of 2.0 m/s. How far do the blocks travel before stopping? Visualize:

Solve: The 2.0 kg block in front has a larger coefficient of friction. Thus the 1.0 kg block pushes against the rear of the 2.0 kg block and, in reaction, the 2.0 kg block pushes backward against the 1.0 kg block. There’s no vertical acceleration, so n1 = m1 g and n2 = m2 g , leading to f1 = μ1m1 g and f 2 = μ 2 m2 g . Newton’s second law along the x-axis is

1 kg block: 2 kg block:

∑ (F

) = − F2 on 1 − f1 = − F2 on 1 − μ1m1 g = m1a

on 1 x

∑ ( Fon 2 ) x = F1 on 2 − f 2 = F2 on 1 − μ 2m2 g = m2a

where we used a1 = a2 = a. Also, F1 on 2 = F2 on 1 because they are an action/reaction pair. Adding these two equations gives − ( μ1m1 + μ 2 m2 ) g = ( m1 + m2 )a a=−

μ1m1 + μ 2 m2 m1 + m2

g =−

(0.3)(1.0 kg) + (0.5)(2.0 kg) × 9.80 m/s2 = −4.25 m/s2 1.0 kg + 2.0 kg

We can now use constant-acceleration kinematics to find v12x = 0 = v02x + 2a ( x1 − x0 ) ⇒ x1 = −

v02x (2.0 m/s) 2 =− = 0.47 m 2a 2( −4.25 m/s2)

7.50. Model: Visualize:

Treat the ball of clay and the block as particles.

G G Solve: (a) Forces FC on B and FB on C are an action/reaction pair, so FB on C = FC on B . Note that aB ≠ aC because the clay is decelerating while the block is accelerating. Newton’s second law in the x-direction is

∑ ( Fon C ) x = − FB on C = mC aC Block: ∑ ( Fon B ) x = FC on B = FB on C = mB aB

Clay:

Equating the two expressions for FB on C gives

aC = −

mB aB mC

Turning to kinematics, the velocity of each after Δt is (vC )1 = (vC )0 + aC Δt (vB )1 = (vB )0 + aB Δt = aBΔt But (vC )1 = (vB )1 because the clay and the block are moving together after Δt has elapsed. Equating these two expressions gives (vC )0 + aC Δt = aBΔt , from which we find aC = aB −

(vC )0 Δt

We can now equate the two expressions for aC : −

mB (v ) (vC )0 / Δt (10 m/s)(0.01 s) aB = aB − C 0 ⇒ aB = = = 100 m/s2 1 + mB / mC 1 + (900 g)(100 g) mC Δt

Then aC = −9aB = −900 m/s 2 . With the acceleration now known, we can use either kinematic equation to find

(vC )1 = (vB )1 = (100 m/s 2 )(0.010 s) = 1.0 m/s (b) FC on B = mB aB = (0.90 kg)(100 m/s 2 ) = 90 N. (c) FB on C = mC aC = (0.10 kg)(900 m/s 2 ) = 90 N. Assess: The two forces are of equal magnitude, as expected from Newton’s third law.

7.51. Model: Use the particle model for the two blocks. Assume a massless rope, and massless, frictionless pulleys. Visualize:

Note that for every meter block 1 moves forward, one meter is provided to block 2. So each rope on m2 has to be lengthened by one-half meter. Thus the acceleration constraint is a2 = − 12 a1. Solve: Newton’s second law for block 1 is T = m1a1. Newton’s second law for block 2 is 2T − ( FG ) 2 = m2 a2 . Combining these two equations gives

2 ( m1a1 ) − m2 g = m2 ( − 12 a1 ) ⇒ a1 [ 4m1 + m2 ] = 2m2 g ⇒ a1 =

2m2 g 4m1 + m2

where we have used a2 = − 12 a1. Assess:

If m1 = 0 kg, then a2 = − g . This is what is expected for a freely falling object.

7.52. Model: Use the particle model for the two blocks. Assume a massless rope and massless, frictionless pulleys. Visualize:

For every one meter that the 1.0-kg block goes down, each rope on the 2.0-kg block will be shortened by onehalf meter. Thus the acceleration constraint is a1 = −2a2 . Solve: Newton’s second law for the two blocks is

2T = m2 a2 T − ( FG )1 = m1a1 Since a1 = −2a2 , the above equations become 2T = m2 a2 T − m1 g = m1 ( −2a2 ) ⇒ m2

Assess:

2 (1.0 kg ) ( 9.8 m/s 2 ) a2 2m1 g + m1 ( 2a2 ) = m1 g ⇒ a2 = = = 3.3 m/s 2 2 m2 + 4m1 ( 2.0 kg + 4.0 kg )

If m1 = 0 kg, then a2 = 0 m/s 2 , which is expected.

7.53. Model: The hamster of mass m and the wedge with mass M will be treated as objects 1 and 2, respectively. They will be treated as particles. Visualize:

The scale is denoted by the letter s. G Solve: (a) The reading of the scale is the magnitude of the force n2 that the scale exerts upward. There are two G action/reaction pairs. Initially the hamster of mass m is stuck in place and is in static equilibrium with Fnet = 0 N. G Because of the shape of the blocks, it is not clear whether the scale has to exert a horizontal friction force fs on 2 to prevent horizontal motion. We’ve included one just in case. Newton’s second law for the hamster is

(F

)

net on 1 x

= mg sinθ − f 2 on 1 = 0 N ⇒ f 2 on 1 = mg sinθ

(F

)

net on 1 y

= n1 − mg cosθ = 0 N ⇒ n1 = mg cosθ

For the wedge, we see from Newton’s third law that n1′ = n1 = mg cosθ and that f 2 on 1 = f1 on 2 = mg sin θ . Using

these equations, Newton’s second law for the wedge is

( Fnet on 2 ) x =

(F

f1 on 2 cosθ + fs on 2 − n1′ sin θ = mg sinθ cosθ + fs on 2 − mg cosθ sinθ = 0 N ⇒ f s on 2 = 0 N

)

net on 2 y

= n2 − n1′ cosθ − f1 on 2 sin θ − Mg = n2 − mg cos 2 θ − mg sin 2 θ − Mg = 0 N

⇒ n2 = mg ( cos 2 θ + sin 2 θ ) + Mg = ( M + m ) g = ( 0.800 kg + 0.200 kg ) ( 9.8 m/s 2 ) = 9.80 N First we find that f s on 2 = 0 N, so no horizontal static friction is needed to prevent motion. More interesting, the scale reading is ( M + m ) g which is the total gravitational force resting on the scale. This is the expected result.

(b) Now suppose that the hamster is accelerating down the wedge. The total mass is still M + m, but is the reading still ( M + m) g ? The frictional forces between the systems 1 and 2 have now vanished, and system 1 now

has an acceleration. However, the acceleration is along the hamster’s x-axis, so a1 y = 0 m/s 2 . The hamster’s yequation is still

( Fnet on 1 ) y = n1 − mg cosθ = 0 N ⇒ n1 = mg cosθ We still have n1′ = n1 = mg cosθ , so the y-equation for block 2 (with a2 y = 0 m/s 2 ) is

(F

)

net on 2 y

= n2 − n1′ cosθ − Mg = n2 − mg cos 2 θ − Mg = 0 N

⇒ n2 = mg cos 2 θ + Mg = ( M + m cos 2 θ ) g = 8.99 N

Assess: The scale reads less than it did when the hamster was at rest. This makes sense if you consider the limit θ → 90°, in which case cosθ → 0. If the face of the wedge is vertical, then the hamster is simply in free fall and can have no effect on the scale (at least until impact!). So for θ = 90° we expect the scale to record Mg only, and that is indeed what the expression for n2 gives.

7.54. Model: The hanging masses m1 , m2 , and m3 are modeled as particles. Pulleys A and B are massless and frictionless. The strings are massless. Visualize:

Solve:

(a) The length of the string over pulley B is constant. Therefore,

( yB − y3 ) + ( yB − yA ) = LB ⇒ yA = 2 yB − y3 − LB The length of the string over pulley A is constant. Thus,

( yA − y2 ) + ( yA − y1 ) = LA = 2 yA − y1 − y2 ⇒ 2 ( 2 yB − y3 − LB ) − y1 − y2 = LA ⇒ 2 y3 + y2 + y1 = constant This constraint implies that 2

dy3 dy2 dy1 + + = 0 m/s = 2v3 y + v2 y + v1 y dt dt dt

Also by differentiation, 2a3 y + a2 y + a1 y = 0 m/s 2 . (b) Newton’s second law for the masses m3 , m2 , m1 , and pulley A is

TB − m3 g = m3a3 y TA − m2 g = m2 a2 y TA − m1 g = m1a1 y TB − 2TA = 0 N

The pulley equation is zero because the pulley is massless. These four equations plus the acceleration constraint are five equations for the five unknowns (two tensions and three accelerations). To solve for TA , multiply the m3 equation by 2, substitute 2TB = 4TA , then divide each of the mass equations by the mass. This gives the three equations 4TA /m3 − 2 g = 2a3 y TA /m 2 − g = a2 y TA /m 1 − g = a1 y If these three equations are added, the right side adds to zero because of the acceleration constraint. Thus

( 4/m3 + 1/m2 + 1/m2 )TA − 4 g = 0 ⇒ TA =

4g ( 4/m3 + 1/m2 + 1/m2 )

(c) Using numerical values, we find TA = 18.97 N. Then

a1 y = TA /m1 − g = −2.2 m/s 2 a2 y = TA /m 2 − g = 2.9 m/s 2 a3 y = 2TA /m 3 − g = −0.32 m/s 2 (d) m3 = m1 + m2 , so it looks at first like m3 should hang in equilibrium. For it to do so, tension TB would need

to equal m3 g . However, TB is not (m1 + m2 ) g because masses m1 and m2 are accelerating rather than hanging at rest. Consequently, tension TB is not able to balance the weight of m3 .

8.1.

Model: The model rocket and the target will be treated as particles. The kinematics equations in two dimensions apply. Visualize:

Solve:

For the rocket, Newton’s second law along the y-direction is

( Fnet ) y = FR − mg = maR ⇒ aR =

1 1 ⎡15 N − ( 0.8 kg ) ( 9.8 m/s 2 ) ⎤ = 8.95 m/s 2 ( FR − mg ) = ⎦ 0.8 kg ⎣ m

Using the kinematic equation y1R = y0R + (v0R ) y ( t1R − t0R ) + 12 aR ( t1R − t0R ) , 2

30 m = 0 m + 0 m + 12 ( 8.95 m/s 2 ) ( t1R − 0 s ) ⇒ t1R = 2.589 s 2

For the target (noting t1T = t1R ), x1T = x0T + (v0T ) x ( t1T − t0T ) + 12 aT ( t1T − t0T ) = 0 m + (15 m/s )( 2.589 s − 0 s ) + 0 m = 39 m 2

You should launch when the target is 39 m away. Assess: The rocket is to be fired when the target is at x0T . For a net acceleration of approximately 9 m/s 2 in the vertical direction and a time of 2.6 s to cover a vertical distance of 30 m, a horizontal distance of 39 m is reasonable.

8.2.

Model: The model rocket will be treated as a particle. Kinematic equations in two dimensions apply. Air resistance is neglected. Visualize:

The horizontal velocity of the rocket is equal to the speed of the car, which is 3.0 m/s. Solve: For the rocket, Newton’s second law along the y-direction is: 1 ⎡( 8.0 N ) − ( 0.5 kg ) ( 9.8 m/s 2 ) ⎤ = 6.2 m/s 2 ( Fnet ) y = FR − mg = maR ⇒ a y = ⎦ 0.5 kg ⎣

Thus using y1 = y0 + (v0 ) y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , 2

( 20 m ) = 0 m + 0 m + 12 ( 6.2 m/s2 ) ( t1R − 0 s )

2

⇒ ( 20 m ) = ( 3.1 m/s 2 ) t12 ⇒ t1 = 2.54 s

Since t1 is also the time for the rocket to move horizontally up to the hoop, x1 = x0 + (v0 ) x ( t1 − t0 ) + 12 ax ( t1 − t0 ) = 0 m + ( 3.0 m/s )( 2.54 s − 0 s ) + 0 m = 7.6 m 2

Assess: In view of the rocket’s horizontal speed of 3.0 m/s and its vertical thrust of 8.0 N, the above-obtained value for the horizontal distance is reasonable.

8.3.

Model: The asteroid and the giant rocket will be treated as particles undergoing motion according to the constant-acceleration equations of kinematics. Visualize:

Solve:

(a) The time it will take the asteroid to reach the earth is

displacement 4.0 × 106 km = = 2.0 × 105 s = 56 h velocity 20 km/s (b) The angle of a line that just misses the earth is

tanθ =

⎛R⎞ R ⎛ 6400 km ⎞ ⇒ θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ = 0.092° 6 y0 ⎝ 4.0 × 10 km ⎠ ⎝ y0 ⎠

(c) When the rocket is fired, the horizontal acceleration of the asteroid is

ax =

5.0 × 109 N = 0.125 m/s 2 4.0 × 1010 kg

(Note that the mass of the rocket is much smaller than the mass of the asteroid and can therefore be ignored completely.) The velocity of the asteroid after the rocket has been fired for 300 s is vx = v0 x + ax ( t − t0 ) = 0 m/s + ( 0.125 m/s 2 ) ( 300 s − 0 s ) = 37.5 m/s After 300 s, the vertical velocity is v y = 2 × 104 m/s and the horizontal velocity is vx = 37.5 m/s. The deflection due to this horizontal velocity is tan θ = That is, the earth is saved.

⎛ 37.5 m/s ⎞ vx ⇒ θ = tan −1 ⎜ ⎟ = 0.107° 4 vy ⎝ 2 × 10 m/s ⎠

8.4.

Model: We are using the particle model for the car in uniform circular motion on a flat circular track. There must be friction between the tires and the road for the car to move in a circle. Visualize:

Solve:

The centripetal acceleration is v 2 ( 25 m/s ) = = 6.25 m/s 2 100 m r 2

ar =

The acceleration points to the center of the circle, so the net force is G G Fr = ma = (1500 kg ) ( 6.25 m/s 2 , toward center ) = ( 9380 N, toward center )

This force is provided by static friction

f s = Fr = 9.4 kN

8.5.

Model: Visualize:

Solve:

We will use the particle model for the car which is in uniform circular motion.

The centripetal acceleration of the car is

v 2 (15 m/s ) = = 4.5 m/s 2 50 m r The acceleration is due to the force of static friction. f s = mar = (1500 kg ) ( 4.5m s 2 ) = 6750 N = 6.8 kN. 2

ar =

Assess:

The model of static friction is

The

( fs )max = nμs = mg μs ≈ mg ≈ 15,000 N

surface. We see that f s < ( fs )max , which is reasonable.

force

of

friction

since μ s ≈ 1 for a dry road

is

8.6.

Model: Treat the block as a particle attached to a massless string that is swinging in a circle on a frictionless table. Visualize:

Solve:

(a) The angular velocity and speed are

ω = 75

rev 2π rad 1min × = 471.2 rad/min vt = rω = ( 0.50 m )( 471.2 rad min ) × = 3.93 m/s min 1 rev 60 s

The tangential velocity is 3.9 m/s. (b) The radial component of Newton’s second law is

∑F

r

=T =

mv 2 r

Thus

T = ( 0.20 kg )

( 3.93 m/s ) 0.50 m

2

= 6.2 N

8.7.

Solve:

Newton’s second law is Fr = mar = mrω 2 . Substituting into this equation yields:

ω=

Fr = mr

8.2 × 10−8 N (9.1×10 kg )(5.3 × 10−11 m ) −31

= 4.37 × 1016 rad/s = 4.37 × 1016

Assess:

rad 1 rev × = 6.6 × 1015 rev/s s 2π rad

This is a very high number of revolutions per second.

8.8.

Model: Visualize:

The vehicle is to be treated as a particle in uniform circular motion.

On a banked road, the normal force on a vehicle has a horizontal component that provides the necessary centripetal acceleration. The vertical component of the normal force balances the gravitational force. Solve: From the physical representation of the forces in the r-z plane, Newton’s second law can be written

∑F

r

= n sin θ =

mv 2 r

∑F

z

= n cosθ − mg = 0 ⇒ n cosθ = mg

Dividing the two equations and making the conversion 90 km h = 25 m/s yields:

( 25 m/s ) v2 = = 0.128 ⇒ θ = 7.3° rg ( 9.8 m/s 2 ) 500 m 2

tanθ =

Assess: Such a banking angle for a speed of approximately 55 mph is clearly reasonable and within our experience as well.

8.9.

Model: The motion of the moon around the earth will be treated through the particle model. The circular motion is uniform. Visualize:

Solve:

The tension in the cable provides the centripetal acceleration. Newton’s second law is

⎛ 2π ⎞ ∑ Fr = T = mrω 2 = mr ⎜ T ⎟ ⎝ moon ⎠

2

2

⎡ 2π 1 day 1h ⎤ 20 = ( 7.36 × 1022 kg )( 3.84 × 108 m ) ⎢ × × ⎥ = 2.01 × 10 N 27.3 days 24 h 3600 s ⎣ ⎦ Assess: This is a tremendous tension, but clearly understandable in view of the moon’s large mass and the large radius of circular motion around the earth.

8.10.

Model: Visualize:

Solve:

Model the ball as a particle in uniform circular motion. Rolling friction is ignored.

The track exerts both an upward normal force and an inward normal force. From Newton’s second law, 2

⎡ 60 rev 2π rad 1min ⎤ Fnet = n2 = mrω 2 = ( 0.030 kg )( 0.20 m ) ⎢ × × = 0.24 N 1 rev 60 s ⎥⎦ ⎣ min

8.11.

Model: Visualize:

The satellite is considered to be a particle in uniform circular motion around the moon.

Solve: The radius of the moon is 1.738 × 106 m and the satellite’s distance from the center of the moon is the same quantity. The angular velocity of the satellite is

ω=

2π 2π rad 1min = × = 9.52 × 10−4 rad/s T 110 min 60 s

and the centripetal acceleration is

ar = rω 2 = (1.738 × 106 m )( 9.52 × 10−4 rad/s ) = 1.58 m/s 2 2

The acceleration of a body in orbit is the local “g” experienced by that body.

8.12.

Model: The earth is considered to be a particle in uniform circular motion around the sun. Solve: The earth orbits the sun in 365 days and is 1.5 × 1011 m from the sun. The angular velocity and centripetal acceleration are

ω=

2π rad 1 day 1h × × = 2.0 × 10−7 rad/s 365 days 24 h 3600 s

ar = g = rω 2 = (1.5 × 1011 m )( 2.0 × 10−7 rad/s ) = 6.0 × 10−3 m/s 2 2

Assess:

The smallness of this acceleration due to gravity is essentially due to the large earth-sun distance.

8.13.

Model: Visualize:

Use the particle model for the car which is undergoing circular motion.

Solve: The car is in circular motion with the center of the circle below the car. Newton’s second law at the top of the hill is

∑F = (F ) r

G r

− nr = mg − n = mar =

mv 2 n⎞ ⎛ ⇒ v2 = r ⎜ g − ⎟ m r ⎝ ⎠

Maximum speed is reached when n = 0 and the car is beginning to lose contact with the road.

vmax = rg = Assess:

( 50 m ) ( 9.8 m/s 2 ) = 22 m/s

A speed of 22 m/s is equivalent to 49 mph, which seems like a reasonable value.

8.14.

Model: Visualize:

The passengers are particles in circular motion.

Solve: The center of the circle of motion of the passengers is directly above them. There must be a net force pointing up that provides the needed centripetal acceleration. The normal force on the passengers is their weight. Ordinarily their weight is FG , so if their weight increases by 50%, n = 1.5 FG . Newton’s second law at the bottom of the dip is mv 2 ∑ Fr = n − FG = (1.5 − 1) FG = 0.5mg = r

⇒ v = 0.5 gr = 0.5 ( 9.8 m/s 2 ) ( 30 m ) = 12.1 m/s Assess:

A speed of 12.1 m/s is 27 mph, which seems very reasonable.

8.15.

Model: Model the roller coaster car as a particle at the top of a circular loop-the-loop undergoing uniform circular motion. Visualize:

Notice that the r-axis points downward, toward the center of the circle. G K Solve: The critical speed occurs when n goes to zero and FG provides all the centripetal force pulling the car in the vertical circle. At the critical speed mg = mvc2 r , therefore vc = rg . Since the car’s speed is twice the

critical speed, vt = 2vc and the centripetal force is

∑ Fr = n + FG =

2 mv 2 m ( 4vc ) m ( 4rg ) = = = 4mg r r r

Thus the normal force is n = 3 mg . Consequently, n FG = 3.

8.16.

Model: Visualize:

Model the roller coaster car as a particle undergoing uniform circular motion along a loop.

Notice that the r-axis points downward, toward the center of the circle. Solve: In this problem the normal force is equal to the gravitational force: n = FG = mg . We have

∑F

r

= n + FG =

mv 2 = mg + mg ⇒ v = 2rg = 2 ( 20 m ) ( 9.8 m/s 2 ) = 19.8 m/s r

8.17.

Model: Visualize:

Model the bucket of water as a particle in uniform circular motion.

Solve: Let us say the distance from the bucket handle to the top of the water in the bucket is 35 cm. This makes the shoulder to water distance 65 cm + 35 cm = 1.00 m. The minimum angular velocity for swinging a bucket of water in a vertical circle without spilling any water corresponds to the case when the speed of the bucket is critical. In this case, n = 0 N when the bucket is in the top position of the circular motion. We get

∑F

r

⇒ ωc = g / r =

= n + FG = 0 N + mg =

mvc 2 = mrω c2 r

9.8 m/s 2 1 rev 60 s = 3.13 rad/s = 3.13 rad s × × = 30 rpm 1.00 m 2π rad 1 min

8.18.

Model: Visualize:

Solve:

Use the particle model for the car, which is undergoing nonuniform circular motion.

The car is in circular motion with radius r =

d = 100 m. We require 2

1.5 m/s 2 1.5 m/s 2 = = 0.122 s −1 100 m r The definition of the angular velocity can be used to determine the time Δt using the angular acceleration a 1.5 m/s 2 α= t = = 1.5 × 10−2 s −2 . r 100 m ω = ωi + αΔt ar = ω 2 r = 1.5 m/s 2 ⇒ ω =

⇒ Δt =

ω − ωi 0.122 s −1 − 0 s −1 = = 8.2 s α 0.015 s −2

8.19.

Model: Visualize:

Solve:

The train is a particle undergoing nonuniform circular motion.

(a) Newton’s second law in the vertical direction is ( Fnet ) y = n − FG = 0

from which n = mg . The rolling friction is f R = μ R n = μ R mg . This force provides the tangential acceleration at = −

The angular acceleration is

fR = −μR g m

2 at − μ R g − ( 0.10 ) ( 9.8 m/s ) = = = −1.96 rad/s 2 r r 0.50 m ⎛ rev ⎞⎛ 1 min ⎞⎛ 2π rad ⎞ (b) The initial angular velocity is 30 ⎜ ⎟⎜ ⎟⎜ ⎟ = 3.14 rad/s. The time to come to a stop due to ⎝ min ⎠⎝ 60 sec ⎠⎝ rev ⎠ the rolling friction is ω − ωi 0 − 3.14 rad/s Δt = f = = 1.60 s −1.96 rad/s 2 α Assess: The original angular speed of π rad/s means the train goes around the track one time every 2 seconds, so a stopping time of less than 2 s is reasonable.

α=

8.20.

Model: The object is treated as a particle in the model of kinetic friction with its motion governed by constant-acceleration kinematics. Visualize:

The velocity v1x as the object sails off the edge is related to the initial velocity v0 x by

Solve:

v = v + 2ax ( x1 − x0 ) . Using Newton’s second law to determine ax while sliding gives 2 1x

2 0x

∑ Fx = − f k = max ⇒ ∑ Fy = n − mg = 0 N ⇒ n = mg

Using this result and the model of kinetic friction

( fk = μk n ) ,

the x-component equation can be written as

− μ k mg = max . This implies ax = − μ k g = − ( 0.50 ) ( 9.8 m/s 2 ) = −4.9 m/s 2 Kinematic equations for the object’s free fall can be used to determine v1x : y2 = y1 + v1 y ( t2 − t1 ) + 12 ( − g )( t2 − t1 ) ⇒ 0 m = 1.0 m + 0 m − 2

g 2 ( t2 − t1 ) ⇒ ( t2 − t1 ) = 0.4518 s 2

x2 = x1 + v1x ( t2 − t1 ) = 2.30 m = 2.0 m + v1x ( 0.4518 s ) ⇒ v1x = 0.664 m/s Having determined v1x and ax , we can go back to the velocity equation v12x = v02x + 2ax ( x1 − x0 ) :

( 0.664 m/s ) Assess:

2

= v02x + 2 ( −4.9 m/s 2 ) ( 2.0 m ) ⇒ v0 x = 4.5 m/s

v0 x = 4.5 m/s is about 10 mph and is a reasonable speed.

8.21.

Model: The rocket and puck together make a particle moving on frictionless ice. The thrust of the rocket motor is assumed to be constant. Visualize:

Solve The kinematics equations can be used to examine the motion for each coordinate axis independently as a function of time t, then combined to eliminate t. In the x-direction, x f = xi + v0 xt = 0 m + ( 2.0 m/s ) t

So t =

xf 2.0 m/s

. The acceleration of the rocket and puck in the y-direction is

ay =

( Fnet ) y mrocket + mpuck

=

8.0 N = 13.3 m/s 2 0.600 kg

In the y–direction, 1 1 yf = yi + v0 yt + a yt 2 = 0 m + ( 0 m/s ) t + (13.3 m/s 2 ) t 2 2 2 = ( 6.67 m/s 2 ) t 2 Substituting t from the equation for the x-direction, 2

⎛ x ⎞ 2 yf = ( 6.67 m/s 2 ) ⎜ ⎟ = 1.67 xf 2.0 m/s ⎝ ⎠ A graph of this equation is shown in the figure. Assess: When the puck has traveled 9 m in the x-direction it has traveled 15 m in the y-direction.

8.22.

Model: Treat Sam as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xycoordinates for the second. Sam’s final velocity at the top of the slope is his initial velocity as he becomes airborne.

Solve:

Sam’s acceleration up the slope is given by Newton’s second law:

( Fnet ) s = F − mg sin10° = ma0 a0 =

F 200 N − g sin10° = − (9.8 m/s 2 ) sin10° = 0.965 m/s 2 m 75 kg

The length of the slope is s1 = (50 m)/ sin10° = 288 m. His velocity at the top of the slope is v12 = v02 + 2a0 ( s1 − s0 ) = 2a0 s1 ⇒ v1 = 2(0.965 m/s 2) (288 m) = 23.6 m/s This is Sam’s initial speed into the air, giving him velocity components v1x = v1 cos10° = 23.2 m/s and

v1 y = v1 sin10° = 410 m/s. This is not projectile motion because Sam experiences both the force of gravity and the thrust of his skis. Newton’s second law for Sam’s acceleration is a1x = a1 y =

( Fnet ) y m

=

( Fnet ) x (200 N)cos10° = = 2.63 m/s 2 m 75 kg

(200 N)sin10° − (75 kg)(9.80 m/s 2 ) = −9.34 m/s 2 75 kg

The y-equation of motion allows us to find out how long it takes Sam to reach the ground: y2 = 0 m = y1 + v1 yt2 + 12 a1 yt22 = 50 m + (4.10 m/s) t2 − (4.67 m/s 2) t22

This quadratic equation has roots t2 = −2.86 s (unphysical) and t2 = 3.74 s. The x-equation of motion—this time with an acceleration—is x2 = x1 + v1xt2 + 12 a1xt22 = 0 m + (23.2 m/s) t2 − 12 (2.63 m/s 2) t22 = 105 m Sam lands 105 m from the base of the cliff.

8.23. Model: Treat the motorcycle and rider as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xycoordinates for the second. The motorcycle’s final velocity at the top of the ramp is its initial velocity as it becomes airborne.

Solve:

The motorcycle’s acceleration on the ramp is given by Newton’s second law:

( Fnet ) s = − f r − mg sin 20° = − μ r n − mg sin 20° = − μ r mg cos 20° − mg sin 20° = ma0 a0 = − g ( μ r cos 20° + sin 20°) = −(9.8 m/s 2 )((0.02)cos 20° + sin 20°) = −3.536 m/s 2 The length of the ramp is s1 = (2.0 m)/ sin 20° = 5.85 m. We can use kinematics to find its speed at the top of the ramp: v12 = v02 + 2a0 ( s1 − s0 ) = v02 + 2a0 s1 ⇒ v1 = (11.0 m/s) 2 + 2(−3.536 m/s 2) (5.85 m) = 8.92 m/s This is the motorcycle’s initial speed into the air, with velocity components v1x = v1 cos 20° = 8.38 m/s and

v1 y = v1 sin 20° = 3.05 m/s. We can use the y-equation of projectile motion to find the time in the air: y2 = 0 m = y1 + v1 yt2 + 12 a1 yt22 = 2.0 m + (3.05 m/s) t2 − (4.90 m/s 2) t22

This quadratic equation has roots t2 = −0.399 s (unphysical) and t2 = 1.021 s. The x-equation of motion is thus x2 = x1 + v1xt2 = 0 m + (8.38 m/s) t2 = 8.56 m

8.56 m < 10.0 m, so it looks like crocodile food.

8.24.

Model: Use the particle model and the constant-acceleration equations of kinematics for the rocket. Solve: (a) The acceleration of the rocket in the launch direction is obtained from Newton’s second law G G F = ma :

140,700 N = ( 5000 kg ) a ⇒ a = 28.14 m/s 2 Therefore, ax = a cos 44.7° = 20.0 m/s 2 and a y = a sin 44.7° = 19.8 m/s 2 . The net acceleration in the y-direction is thus

( anet ) y = a y − g = (19.8 − 9.8) m/s 2

= 10.0 m/s 2

With this acceleration, we can write the equations for the x- and y-motions of the rocket.

y = y0 + v0 y ( t − t0 ) + 12 ( anet ) y ( t − t0 ) = 0 m + 0 m + 12 (10.0 m/s 2 ) t 2 = ( 5.00 m/s 2 ) t 2 2

x = x0 + v0 x ( t − t0 ) + 12 ( anet ) x ( t − t0 ) = 0 m + 0 m + 12 ( 20.0 m/s 2 ) t 2 = (10.0 m/s 2 ) t 2 2

From these two equations, 2 2 x (10.0 m/s ) t = =2 y ( 5.00 m/s 2 ) t 2

The equation that describes the rocket’s trajectory is y = 12 x. (b) It is a straight line with a slope of 12 . (c) In general, v y = v0 y + ( anet ) y ( t1 − t0 ) = 0 + (10.0 m/s 2 ) t1

vx = v0 x + ( anet ) x ( t1 − t0 ) = 0 + ( 20.0 m/s 2 ) t1 v=

(10.0 m/s ) t + ( 20.0 m/s ) t = ( 22.36 m/s ) t 2 2 2 1

2 2 2 1

2

1

The time required to reach the speed of sound is calculated as follows: 330 m/s = ( 22.36 m/s 2 ) t1 ⇒ t1 = 14.76 s We can now obtain the elevation of the rocket. From the y-equation, y = ( 5.00 m/s 2 ) t12 = ( 5.00 m/s 2 ) (14.76 s ) = 1090 m 2

8.25.

Model: The hockey puck will be treated as a particle whose motion is determined by constantacceleration kinematic equations. We break this problem in two parts, the first pertaining to motion on the table and the second to free fall.

Visualize: Solve:

Newton’s second law is:

Fx = max ⇒ ax =

Fx 2.0 N = = 2.0 m/s 2 m 1.0 kg

The kinematic equation v12x = v02x + 2ax ( x1 − x0 ) yields:

v12x = 0 m 2 /s 2 + 2 ( 2.0 m/s 2 ) ( 4.0 m ) ⇒ v1x = 4.0 m/s Let us now find the time of free fall ( t2 − t1 ) :

y2 = y1 + v1 y ( t2 − t1 ) + 12 a1 y ( t2 − t1 )

2

⇒ 0 m = 2.0 m + 0 m + 12 ( −9.8 m/s 2 ) ( t2 − t1 ) ⇒ ( t2 − t1 ) = 0.639 s 2

Having obtained v1x and ( t2 − t1 ) , we can now find ( x2 − x1 ) as follows: x2 = x1 + v1x ( t2 − t1 ) + 12 ax ( t2 − t1 )

2

⇒ x2 − x1 = ( 4.0 m/s )( 0.639 s ) + 12 ( 2.0 m/s 2 ) ( 0.639 s ) = 3.0 m 2

Assess:

For a modest horizontal thrust of 2.0 N, a landing distance of 3.0 m is reasonable.

8.26.

Model: The model rocket is treated as a particle and its motion is determined by constant-acceleration kinematic equations. Visualize:

Solve:

As the rocket is accidentally bumped v0 x = 0.5 m/s and v0 y = 0 m/s. On the other hand, when the

engine is fired

Fx = max ⇒ ax =

Fx 20 N = = 40 m/s 2 m 0.500 kg

(a) Using y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , 2

0 m = 40 m + 0 m + 12 ( −9.8 m/s 2 ) t12 ⇒ t1 = 2.857 s The distance from the base of the wall is x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) = 0 m + ( 0.5 m/s )( 2.857 s ) + 12 ( 40 m/s 2 ) ( 2.857 s ) = 165 m 2

2

(b) The x- and y-equations are y = y0 + v0 y ( t − t0 ) + 12 a y ( t − t0 ) = 40 − 4.9t 2 2

x = x0 + v0 x ( t − t0 ) + 12 ax ( t − t0 ) = 0.5t + 20t 2 2

Except for a brief interval near t = 0, 20t 2  0.5t. Thus x ≈ 20t 2 , or t 2 = x / 20. Substituting this into the yequation gives y = 40 − 0.245 x This is the equation of a straight line, so the rocket follows a linear trajectory to the ground.

8.27.

Model: Visualize:

Assume the particle model for the satellite in circular motion.

To be in a geosynchronous orbit means rotating at the same rate as the earth, which is 24 hours for one complete 3.58 × 107 m, r = 3.58 × 107 m, rotation. Because the altitude of the satellite is

re = 3.58 × 107 m + 6.37 × 106 m = 4.22 × 107 m. Solve: (a) The period (T ) of the satellite is 24.0 hours. (b) The acceleration due to gravity is 2

2

1 hr ⎞ ⎛ 2π ⎞ ⎛ 2π 7 2 × g = ar = rω 2 = r ⎜ ⎟ = ( 4.22 × 10 m ) ⎜ ⎟ = 0.223 m/s ⎝ T ⎠ ⎝ 24.0 hr 3600 s ⎠ (c) There is no normal force on a satellite, so the weight is zero. It is in free fall.

8.28.

Model: earth rotates. Visualize:

Treat the man as a particle. The man at the equator undergoes uniform circular motion as the

Solve: The scale reads the man’s weight FG = n, the force of the scale pushing up against his feet. At the north pole, where the man is in static equilibrium,

nP = FG = mg = 735 N At the equator, there must be a net force toward the center of the earth to keep the man moving in a circle. The raxis points toward the center, so

∑F = F r

G

− nE = mω 2 r ⇒ nE = mg − mω 2 r = nP − mω 2 r

The equator scale reads less than the north pole scale by the amount mω 2 r. The man’s angular velocity is that of the equator, or 2π 2π rad ω= = = 7.27 × 10−5 rad/s T 24 hours × (3600 s/1 h)

Thus the north pole scale reads more than the equator scale by

Δw = (75 kg)(7.27 × 10−5 rad/s) 2 (6.37 × 106 m) = 2.5 N Assess:

The man at the equator appears to have lost Δm = Δw/g ≈ 0.25 kg, or the equivalent of ≈

1 2

lb.

8.29.

Model: Visualize:

Solve:

Use the particle model for the (cart + child) system which is in uniform circular motion.

Newton’s second law along r and z directions can be written:

∑F

r

= T cos 20° − n sin 20° = mar

∑F

z

= T sin 20° − n cos 20° − mg = 0

The cart’s centripetal acceleration is 2

rev 1 min 2π rad ⎞ ⎛ 2 × × ar = rω 2 = ( 2.0cos20° m ) ⎜14 ⎟ = 4.04 m/s 1 rev ⎠ ⎝ min 60 s The above force equations can be rewritten as

0.94T − 0.342n = ( 25 kg ) ( 4.04 m/s 2 ) = 101 N 0.342T + 0.94n = ( 25 kg ) ( 9.8 m/s 2 ) = 245 N Solving these two equations yields T = 179 N for the tension in the rope. Assess: In view of the child + cart weight of 245 N, a tension of 179 N is reasonable.

8.30.

Model: Visualize:

Solve:

Model the ball as a particle which is in a vertical circular motion.

At the bottom of the circle,

∑ Fr = T − FG =

( 0.500 kg ) v ⇒ v = 5.5 m/s mv 2 ⇒ (15 N ) − ( 0.500 kg ) ( 9.8 m/s 2 ) = r (1.5 m ) 2

8.31.

Model: We will use the particle model for the car, which is undergoing uniform circular motion on a banked highway, and the model of static friction. Visualize:

Note that we need to use the coefficient of static friction μ s , which is 1.0 for rubber on concrete. Solve: Newton’s second law for the car is

mv 2 ∑ Fz = n cosθ − fs sinθ − FG = 0 N r Maximum speed is when the static friction force reaches its maximum value ( fs )max = μs n. Then

∑F

r

= f s cosθ + n sinθ =

mv 2 n ( cos15° − μs sin15° ) = mg r Dividing these two equations and simplifying, we get n ( μ s cos15° + sin15° ) =

μs + tan15° v 2 μ + tan15° = ⇒ v = gr s 1 − μs tan15° gr 1 − μ s tan15° = Assess:

( 9.80 m/s ) ( 70 m ) ( (1 − 0.268) ) = 34 m/s 2

1.0 + 0.268

The above value of 34 m/s ≈ 70 mph is reasonable.

8.32.

Model: Visualize:

Solve:

Use the particle model for the rock, which is undergoing uniform circular motion.

Newton’s second law is

mv 2 ∑ Fz = T sin10° − mg = 0 N r where the radius of the circular motion is r = (1.0 m ) cos10° = 0.985 m. Dividing these two equations, we get

∑F

r

tan10° = ⇒ω =

= T cos10° =

gr ⇒v= v2

gr = tan10°

( 9.8 m/s ) ( 0.985 m ) = 7.40 m/s 2

tan10°

v 7.40 m/s rad 1 rev 60 s = = 7.51 rad/s = 7.51 × × = 72 rpm r 0.985 m s 2π rad 1 min

8.33.

Model: motion. Visualize:

Use the particle model and static friction model for the coin, which is undergoing circular

Solve: The force of static friction is f s = μ s n = μs mg . This force is equivalent to the maximum centripetal force that can be applied without sliding. That is,

μs mg = m

vt 2 2 = m ( rω max ) ⇒ ωmax = r = 7.23

μs g r

=

( 0.80 ) ( 9.8 m/s 2 ) 0.15 m

= 7.23 rad/s

rad 1 rev 60 s × × = 69 rpm s 2π rad 1 min

So, the coin will stay still on the turntable. Assess: A rotational speed of approximately 1 rev per second for the coin to stay stationary seems reasonable.

8.34.

Model: Visualize:

Solve:

Use the particle model for the car, which is in uniform circular motion.

Newton’s second law is

∑ Fr = T sin 20° = mar =

mv 2 r

∑F

z

= T cos 20° − FG = 0 N

These equations can be written as T sin 20° =

mv 2 T cos 20° = mg r

Dividing these two equations gives

tan 20° = v 2 rg ⇒ v = rg tan 20° =

( 4.55 m ) ( 9.8 m/s 2 ) tan 20° = 4.03 m/s

8.35.

Model: Visualize:

Use the particle model for the ball in circular motion.

G Solve: (a) The mass moves in a horizontal circle of radius r = 20 cm. The acceleration a and the net force G vector point to the center of the circle, not along the string. The only two forces are the string tension T , which G does point along the string, and the gravitational force FG . These are shown in the free-body diagram. Newton’s second law for circular motion is mv 2 ∑ Fz = T cosθ − FG = T cosθ − mg = 0 N ∑ Fr = T sinθ = mar = r From the z-equation, T=

( 0.500 kg ) ( 9.8 m/s mg = cosθ cos11.54°

2

) = 5.00 N

(b) We can find the tangential speed from the r-equation:

v=

rT sin θ = 0.63 m/s m

The angular speed is v r

ω= =

0.63 ms = 3.15 rad/s = 30 rpm 0.20 m

8.36.

Model: Visualize:

Consider the passenger to be a particle and use the model of static friction.

Solve: The passengers stick to the wall if the static friction force is sufficient to support the gravitational force on them: fs = FG The minimum angular velocity occurs when static friction reaches its maximum possible value

( fs )max = μs n.

Although clothing has a range of coefficients of friction, it is the clothing with the smallest

coefficient ( μ s = 0.60) that will slip first, so this is the case we need to examine. Assuming that the person is stuck to the wall, Newton’s second law is

∑ F = n = mω r 2

r

The minimum frequency occurs when

∑ Fz = fs − w = 0 ⇒ fs = mg

2 f s = ( f s )max = μ s n = μ s mrω min

Using this expression for fs in the z-equation gives 2 fs = μ s mrω min = mg

⇒ ω min =

g 9.80 m/s 2 1 rev 60 s = = 2.56 rad/s = 2.56 rad/s × × = 24 rpm 0.60(2.5 m) 2π rad 1 min μs r

Assess: Note the velocity does not depend on the mass of the individual. Therefore, the minimum mass sign is not necessary.

8.37.

Model: Visualize:

Use the particle model for the marble in uniform circular motion.

Solve: The marble will roll in a horizontal circle if the static friction force is sufficient to support the gravitational on it: f s = FG If mg > ( fs )max then static friction is not sufficient and the marble will slip down the side as it rolls around the circumference. The r-equation of Newton’s second law is

∑

2

2π rad 1 min ⎞ ⎛ × Fr = n = mrω 2 = (0.010 kg)(0.060 m) ⎜150 rpm × ⎟ = 0.148 N 1 rev 60 s ⎠ ⎝

( fs )max = μs n = (0.80)(0.148 N) = 0.118 N. The friction force f s = mg = 0.098 N. We see that f s < ( fs )max , therefore friction is sufficient

Thus the maximum possible static friction is needed to support a 10 g marble is

and the marble spins in a horizontal circle. Assess: In reality, rolling friction will cause the marble to gradually slow down until point, it will begin to slip down the inside wall.

( fs )max < mg.

At that

8.38.

Model: Visualize:

Use the particle model for the car and the model of kinetic friction.

Solve: We will apply Newton’s second law to all three cars. Car A:

∑ F = n + ( f ) + ( F ) = 0 N − f + 0 N = ma ∑ F = n + ( f ) + y = n + 0 N − mg = 0 N x

k x

x

y

G x

k y

y

k

x

y

The y-component equation means n = mg . Since f k = μ k n, we have f k = μ k mg . From the x-component equation, −f − μ k mg ax = k = = − μ k g = −9.8 m/s 2 m m Car B: Car B is in circular motion with the center of the circle above the car. mv 2 ∑ Fr = nr + ( f k )r + ( FG )r = n + 0 N − mg = mar = r

∑F = n + ( f ) + (F ) t

k t

t

G t

= 0 N − f k + 0 N = + mat

From the r-equation

n = mg +

⎛ mv 2 v2 ⎞ ⇒ fk = μk n = μk m ⎜ g + ⎟ r r ⎠ ⎝

Substituting back into the t-equation,

at = −

2 ⎛ fk μ m⎛ v2 ⎞ ( 25 m/s ) ⎞⎟ = −12.9 m/s 2 = − k ⎜ g + ⎟ = − μ k ⎜ 9.8 m/s 2 + ⎜ 200 m ⎟⎠ m m ⎝ r ⎠ ⎝

Car C: Car C is in circular motion with the center of the circle below the car.

∑F

r

= n r + ( f k )r + ( FG )r = − n + 0 N + mg = mar =

∑F = n + ( f ) + (F ) t

t

k t

G t

mv 2 r

= 0 N − f k + 0 N = mat

From the r-equation n = m ( g − v 2 r ) . Substituting this into the t-equation yields at =

− fk −μk n = = − μ k ( g − v 2 r ) = −6.7 m/s 2 m m

8.39.

Model: Visualize:

Solve:

Model the ball as a particle that is moving in a vertical circle.

(a) The ball’s gravitational force FG = mg = ( 0.500 kg ) ( 9.8 m/s 2 ) = 4.9 N.

(b) Newton’s second law at the top is

∑F

r

= T1 + FG = mar = m

v2 r

⎡ ( 4.0 m/s )2 ⎤ ⎛ v2 ⎞ ⇒ T1 = m ⎜ − g ⎟ = ( 0.500 kg ) ⎢ − 9.8 m/s 2 ⎥ = 2.9 N ⎢⎣ 1.02 m ⎥⎦ ⎝ r ⎠

(c) Newton’s second law at the bottom is

∑ Fr = T2 − FG =

mv 2 r

2 ⎡ ⎛ ( 7.5 m/s ) ⎤ = 32 N v2 ⎞ ⇒ T2 = m ⎜ g + ⎟ = ( 0.500 kg ) ⎢9.8 m/s 2 + ⎥ 1.02 m ⎥⎦ r ⎠ ⎢⎣ ⎝

8.40.

Model: Visualize:

Solve:

Use the particle model for yourself while in uniform circular motion.

(a) The speed and acceleration are

2π r 2π (15 m ) v 2 ( 3.77 m/s ) = 0.95 m/s 2 = = 3.77 m/s ar = = 15 m r T 25 s 2

v=

So the speed is 3.8 m/s and the centripetal acceleration is 0.95 m/s 2 . (b) The weight w = m, the normal force. On the ground, your weight is the same as the gravitational force FG . Newton’s second law at the top is mv 2 ∑ Fr = FG − n = mar = r 2 ⎛ ⎛ ( 3.77 m/s ) ⎞⎟ = m 8.85 m/s 2 v2 ⎞ ⇒ n = w = m ⎜ g − ⎟ = m ⎜ 9.8 m/s 2 − ( ) ⎜ ⎟ 15 m r ⎠ ⎝ ⎝ ⎠

⇒

w 8.85 m s 2 = = 0.90 FG 9.8 m s 2

(c) Newton’s second law at the bottom is

∑ Fr = n − FG = mar =

mv 2 r

⎛ (3.77 m/s2 ) ⎟⎞ = 10.75 m/s2 m ⎛ v2 ⎞ ⇒ n = w = m ⎜ g + ⎟ = m ⎜ 9.8 m/s 2 + ( ) ⎜ ⎟ 15 m r ⎠ ⎝ ⎝ ⎠ ⇒

w 10.75 m s 2 = = 1.10 FG 9.8 m s 2

8.41.

Model: Visualize:

Solve:

Model a passenger as a particle rotating in a vertical circle.

(a) Newton’s second law at the top is

∑F

r

= nT + FG = mar =

mv 2 mv 2 ⇒ nT + mg = r r

The speed is

v=

2π r 2π ( 8.0 m ) = = 11.17 m/s T 4.5 s

⎡ (11.17 m/s )2 ⎤ ⎛ v2 ⎞ ⇒ nT = m ⎜ − g ⎟ = ( 55 kg ) ⎢ − 9.8 m/s 2 ⎥ = 319 N ⎢⎣ 8.0 m ⎥⎦ ⎝ r ⎠

That is, the ring pushes on the passenger with a force of 3.2 × 102 N at the top of the ride. Newton’s second law at the bottom: ⎛ v2 ⎞ mv 2 mv 2 ⇒ nB = + mg = m ⎜ + g ⎟ r r r ⎝ ⎠ ⎡ (11.17 m/s )2 ⎤ = ( 55 kg ) ⎢ + 9.8 m/s 2 ⎥ = 1397 N ⎢⎣ 8.0 m ⎥⎦ Thus the force with which the ring pushes on the rider when she is at the bottom of the ring is 1.4 kN. (b) To just stay on at the top, nT = 0 N in the r-equation at the top in part (a). Thus,

∑F

r

= nB − FG = mar =

2

⎛ 2π ⎞ mv 2 r 8.0 m = mrω 2 = mr ⎜ mg = = 2π = 5.7 s ⎟ ⇒ Tmax = 2π r g 9.8 m/s 2 ⎝ Tmax ⎠

8.42.

Model: Visualize:

Solve:

Model the chair and the rider as a particle in uniform circular motion.

Newton’s second law along the r-axis is

∑F

r

= Tr + ( FG )r = mar ⇒ T sin θ + 0 N = mrω 2

Since r = L sin θ , this equation becomes 2

⎡ 2π rad ⎤ T = mLω 2 = (150 kg )( 9.0 m ) ⎢ ⎥ = 3330 N ⎣ 4.0 s ⎦ Thus, the 3000 N chain is not strong enough for the ride.

8.43.

Model: Visualize:

Model the ball as a particle in motion in a vertical circle.

G Solve: If the ball moves in a complete circle, then there is a tension force T when the ball is at the top of the circle. The tension force adds to the gravitational force to cause the centripetal acceleration. The forces are along the r-axis, and the center of the circle is below the ball. Newton’s second law at the top is

( Fnet )r = T + FG = T + mg = ⇒ vtop = rg +

mv 2 r

rT m

The tension T can’t become negative, so T = 0 N gives the minimum speed vmin at which the ball moves in a

circle. If the speed is less than vmin , then the string will go slack and the ball will fall out of the circle before it reaches the top. Thus, vmin = rg ⇒ ω min =

rg vmin = = r r

g = r

( 9.8 m/s ) = 3.13 rad / s = 30 rpm 2

(1.0 m )

8.44.

Model: The ball is a particle on a massless rope in circular motion about the point where the rope is attached to the ceiling. Visualize:

Solve:

Newton’s second law in the radial direction is

(∑ F ) = T − F r

G

= T − mg =

mv 2 r

Solving for the tension in the rope and evaluating, 2 ⎛ ⎛ ( 5.5 m/s ) ⎞⎟ = 168 N v2 ⎞ T = m ⎜ g + ⎟ = (10.2 kg ) ⎜ 9.8 m/s 2 + ⎜ 4.5 m ⎟⎠ r ⎠ ⎝ ⎝ Assess: The tension in the rope is greater than the gravitational force on the ball in order to keep the ball moving in a circle.

8.45.

Model: Visualize:

Model the person as a particle in uniform circular motion.

Solve: The only force acting on the passengers is the normal force of the wall. Newton’s second law along the r-axis is:

∑F

r

= n = mrω 2

To create “normal” gravity, the normal force by the inside surface of the space station equals mg. Therefore,

mg = mrω 2 ⇒ ω = Assess:

2π = T

500 m g r ⇒ T = 2π = 2π = 45.0 s 9.8 m/s 2 r g

This is a fast rotation. The tangential speed is v=

2π r 2π ( 500 m ) = = 70 m/s ≈ 140 mph T 45 s

8.46. Model: Visualize:

Masses m1 and m2 are considered particles. The string is assumed to be massless.

Solve: The tension in the string causes the centripetal acceleration of the circular motion. If the hole is smooth, G G it acts like a pulley. Thus tension forces T1 and T2 act as if they were an action/reaction pair. Mass m1 is in circular motion of radius r, so Newton’s second law for m1 is

∑ Fr = T1 =

m1v 2 r

Mass m2 is at rest, so the y-equation of Newton’s second law is

∑F

y

= T2 − m2 g = 0 N ⇒ T2 = m2 g

Newton’s third law tells us that T1 = T2 . Equating the two expressions for these quantities:

m1v 2 m2 rg = m2 g ⇒ v = r m1

8.47.

Model: Visualize:

Model the ball as a particle swinging in a vertical circle, then as a projectile.

Solve: Initially, the ball is moving in a circle. Once the string is cut, it becomes a projectile. The final circularmotion velocity is the initial velocity for the projectile. The free-body diagram for circular motion is shown at the bottom of the circle. Since T > FG , there is a net force toward the center of the circle that causes the centripetal acceleration. The r-equation of Newton’s second law is

( Fnet )r = T − FG = T − mg =

mv 2 r

r 0.60 m ⎡5.0 N − ( 0.10 kg ) ( 9.8 m/s 2 ) ⎤ = 4.91 m/s (T − mg ) = ⎦ m 0.100 kg ⎣ G As a projectile the ball starts at y0 = 1.4 m with v0 = 4.91iˆ m/s. The equation for the y-motion is ⇒ vbottom =

y1 = 0 m = y0 + v0 y Δt − 12 g ( Δt ) = y0 − 12 gt12 2

This is easily solved to find that the ball hits the ground at time

t1 =

2 y0 = 0.535 s g

During this time interval it travels a horizontal distance x1 = x0 + v0 xt1 = ( 4.91 m/s )( 0.535 s ) = 2.63 m So the ball hits the floor 2.6 m to the right of the point where the string was cut.

8.48.

Model: Visualize:

Solve:

Use the particle model for a ball in motion in a vertical circle and then as a projectile.

For the circular motion, Newton’s second law along the r-direction is

∑F

r

= T + FG =

mvt2 r

Since the string goes slack as the particle makes it over the top, T = 0 N. That is,

FG = mg =

mvt2 ⇒ vt = gr = r

( 9.8 m/s ) ( 0.5 m ) = 2.21 m/s 2

The ball begins projectile motion as the string is released. The time it takes for the ball to hit the floor can be found as follows:

y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ 0 m = 2.0 m + 0 m + 12 ( −9.8 m/s2 ) ( t1 − 0 s ) ⇒ t1 = 0.639 s 2

2

The place where the ball hits the ground is

x1 = x0 + v0 x ( t1 − t0 ) = 0 m + ( +2.21 m/s )( 0.639 s − 0 s ) = +1.41 m The ball hits the ground 1.41 m to the right of the point beneath the center of the circle.

8.49.

Model: Visualize:

Model the ball as a particle undergoing circular motion in a vertical circle.

Solve: Initially, the ball is moving in circular motion. Once the string breaks, it becomes a projectile. The final circular-motion velocity is the initial velocity for the projectile, which we can find by using the kinematic equation

v12 = v02 + 2a y ( y1 − y0 ) ⇒ 0 m 2 s 2 = ( v0 ) + 2 ( −9.8 m/s 2 ) ( 4.0 m − 0 m ) ⇒ v0 = 8.85 m/s 2

This is the speed of the ball as the string broke. The tension in the string at that instant can be found by using the r-component of the net force on the ball: ⎛ v2 ⎞ ( 8.85 m/s ) ∑ Fr = T = m ⎜⎜ r0 y ⎟⎟ ⇒ T = ( 0.100 kg ) 0.60 m = 13.1 N ⎝ ⎠ 2

8.50.

Model: Visualize:

Solve:

Model the car as a particle on a circular track.

(a) Newton’s second law along the t-axis is

∑ F = F = ma t

t

t

⇒ 1000 N = (1500 kg ) at ⇒ at = 2 3 m/s 2

With this tangential acceleration, the car’s tangential velocity after 10 s will be

v1t = v0t + at ( t1 − t0 ) = 0 m s + ( 2 3 m/s 2 ) (10 s − 0 s ) = 20 3 m/s The radial acceleration at this instant is

v12t ( 20 3 m/s ) 16 m/s 2 = = 25 m 9 r 2

ar = The car’s acceleration at 10 s has magnitude

a1 = at2 + ar2 =

( 2 3 m/s ) + (16 9 m/s ) 2 2

2 2

= 1.90 m/s 2 θ = tan −1

⎛ 23⎞ at = tan −1 ⎜ ⎟ = 21° ar ⎝ 16 9 ⎠

where the angle is measured from the r-axis. (b) The car will begin to slide out of the circle when the static friction reaches its maximum possible value ( fs )max = μs n. That is,

mv22t ⇒ v2t = rg = ( 25 m ) ( 9.8 m/s 2 ) = 15.7 m/s r In the above equation, n = mg follows from Newton’s second law along the z-axis. The time when the car begins to slide can now be obtained as follows:

∑F = ( f ) r

s max

= μ s n = μ s mg =

v2t = v0t + at ( t2 − t0 ) ⇒ 15.7 m/s = 0 m s + ( 2 3 m/s 2 ) ( t2 − 0 ) ⇒ t2 = 24 s

8.51.

Model: Visualize:

Solve:

Model the steel block as a particle and use the model of kinetic friction.

G (a) The components of thrust ( F ) along the r-, t-, and z-directions are

Fr = F sin 20° = ( 3.5 N ) sin 20° = 1.20 N Ft = F cos 20° = ( 3.5 N ) cos 20° = 3.29 N Fz = 0 N Newton’s second law is

( Fnet )r = T + Fr = mrω 2 ( Fnet )t = Ft − f k = mat

( Fnet ) z = n − mg = 0 N The z-component equation means n = mg . The force of friction is

f k = μ k n = μ k mg = ( 0.60 )( 0.500 kg ) ( 9.8 m/s 2 ) = 2.94 N Substituting into the t-component of Newton’s second law

( 3.29 N ) − ( 2.94 N ) = ( 0.500 kg ) at ⇒ at = 0.70 m/s 2 Having found at , we can now find the tangential velocity after 10 revolutions = 20π rad as follows: 1⎛ a ⎞

θ1 = ⎜ t ⎟ t12 ⇒ t1 = 2⎝ r ⎠

2rθ1 = 18.95 s at

⎛ at ⎞ ⎟ t1 = 6.63 rad/s ⎝r⎠

ω1 = ω 0 + ⎜

The block’s angular velocity after 10 s is 6.6 rad/s. (b) Substituting ω1 into the r-component of Newton’s second law yields:

T1 + Fr = mrω12 ⇒ T1 + (1.20 N ) = ( 0.500 kg )( 2.0 m )( 6.63 rad/s ) ⇒ T1 = 44 N 2

8.52.

Model: Visualize:

Solve:

Assume the particle model for a ball in vertical circular motion.

(a) Newton’s second law in the r- and t-directions is

( Fnet )r = T + mg cosθ = mar =

mvt2 r

( Fnet )t = −mg sinθ = mat

Substituting into the r-component,

( 20 N ) + ( 2.0 kg ) ( 9.8 m/s 2 ) cos30° = ( 2.0 kg )

vt2 ⇒ vt = 3.85 m/s ( 0.80 m )

The tangential velocity is 3.8 m/s. (b) Substituting into the t-component, − ( 9.8 m/s 2 ) sin 30° = at ⇒ at = −4.9 m/s 2 The radial acceleration is vt2 ( 3.85 m/s ) = = 18.5 m/s 2 0.80 m r 2

ar = Thus, the magnitude of the acceleration is a = ar2 + at2 =

(18.5 m/s ) + ( −4.9 m/s ) 2 2

2 2

The angle of the acceleration vector from the r-axis is

φ = tan −1 The angle is below the r-axis.

at 4.9 = tan −1 = 14.8° 18.5 ar

= 19.1 m/s 2

8.53.

Solve: (a) You are spinning a lead fishing weight in a horizontal 1.0 m diameter circle on the ice of a pond when the string breaks. You know that the test weight (breaking force) of the line is 60 N and that the lead weight has a mass of 0.30 kg. What was the weight’s angular velocity in rad/s and in rpm? 60 N rev 60 s ω2 = ⇒ ω = 20 rad/s × × = 191 rpm (b) 2π rad min ( 0.3 kg )( 0.5 m )

8.54.

Solve: (a) At what speed does a 1500 kg car going over a hill with a radius of 200 m have a weight of 11,760 N? (b) The weight is the normal force. 1500 kg v 2 2940 N = ⇒ v = 19.8 m/s 200 m

8.55.

Model: Visualize:

Assume the particle model and apply the constant-acceleration kinematic equations.

Solve:

(a) Newton’s second law for the projectile is G −F Fnet = − Fwind = max ⇒ ax = x m where Fwind is shortened to F. For the y-motion:

( )

y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ 0 m = 0 m + ( v0 sin θ ) t1 − 12 gt12 ⇒ t1 = 0 s and t1 = 2

2v0 sin θ g

Using the above expression for t1 and defining the range as R we get from the x motion: x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 )

2

⎛ 2v sin θ ⎛ F⎞ ⇒ x1 − x0 = R = v0 xt1 + 12 ⎜ − ⎟ t12 = ( v0 cosθ ) ⎜ 0 g ⎝ m⎠ ⎝ =

⎞ F ⎛ 2v0 sin θ ⎞ ⎟− ⎜ ⎟ g ⎠ 2m ⎝ ⎠

2

2v02 2v 2 F cosθ sinθ − 0 2 sin 2 θ g mg

We will now maximize R as a function of θ by setting the derivative equal to 0: dR 2v02 2 Fv02 cos 2 θ − sin 2 θ ) − 2sinθ cosθ = 0 = ( dθ g mg 2

⎛ 2 Fv02 ⎞ ⎛ g ⎞ mg ⇒ cos 2 θ − sin 2 θ = cos 2θ = ⎜ sin 2θ ⇒ tan 2θ = 2 ⎟⎜ 2 ⎟ F ⎝ mg ⎠⎝ 2v0 ⎠ Thus the angle for maximum range is θ = 12 tan −1 ( mg / F ) (b) We have

2 mg ( 0.50 kg ) ( 9.8 m/s ) = = 8.167 ⇒ θ = 12 tan −1 ( 8.167 ) = 41.51° F 0.60 N The maximum range without air resistance is

R′ =

2v02 sin 45° cos 45° v02 = g g

Therefore, we can write the equation for the range R as

R = 2 R′ sin 41.51° cos 41.51° −

⇒

2F R′ sin 2 41.51° = R′ ( 0.9926 − 0.1076 ) = 0.885R′ mg

R R′ − R = 0.8850 ⇒ = 1 − 0.8850 = 0.115 R′ R′

Thus R is reduced from R′ by 11.5%. Assess: The condition for maximum range ( tan 2θ = mg / F ) means 2θ → 90° as F → 0. That is, θ = 45° when F = 0, as is to be expected.

8.56.

Visualize:

G ⎛1 ⎞ From Chapter 6 the drag on a projectile is D = ⎜ Av 2 , direction opposite to motion ⎟ , where A is the 4 ⎝ ⎠ cross- sectional area. Using the free-body diagram above, apply Newton’s second law to each direction. In the x-direction, 1 ( Fnet ) x D cosθ Av 2 cosθ ax = =− =−4 m n m 1 ( Fnet ) y Av 2 sin θ D sin θ − FG ay = =− =−4 −g m m m Solve:

Since v = vx2 + v y2 , vx = v cosθ , and v y = v sinθ , we can rewrite these as ax = −

Avx vx2 + v y2 A ( v cosθ ) v =− 4m 4m

ay = −

Av y vx2 + v y2 A ( v sin θ ) v − g = −g − 4m 4m

8.57.

Model: Visualize:

Solve:

Use the particle model and apply the constant-acceleration kinematic equations of motion.

(a) The thrust of the rocket burn must be such that vx decreases from 2.0 km/s to 0 during the same

interval that v y increases from 0 to 1.0 km/s. The forward distance in which to accomplish this is

x1 = (500 km)cos30° = 433 km. We can find the burn time by combining two kinematic equations: v1x = 0 = v0 x + axt1 ⇒ axt1 = −v0 x x1 = 0 + v0 xt1 + 12 axt12 = v0 xt1 + 12 (−v0 x )t1 = 12 v0 xt1 ⇒ t1 =

2 x1 2(433 km) = = 433 s 2.0 km/s v0 x

The required acceleration is ax = −v0 y /t1 = −(2000 m/s)/(433 s) = −4.62 m/s. During this interval, v y increases

from 0 to 1.0 km/s, thus v1 y = 0 + a yt1 ⇒ a y =

1000 m/s = 2.31 m/s 2 433 s

The thrust force is given by Newton’s second law: K G Fthrust = ma = (20,000 kg)(–4.62iˆ + 2.31 ˆj m/s 2 ) = (−92,400iˆ + 46,200 ˆj ) N The magnitude of the thrust force is 103,300 N and it is at angle θ = tan −1 (46,200/92,400) = 26.6° below the +xaxis (i.e., to the right and down). In order to point the thrusters in this direction, you must rotate the rocket counterclockwise 180° − θ = 153.4°. To make it home with one rocket burn, you need to rotate your rocket to 153.4° and fire with a thrust of 103,300 N for 433 s. (b) The x- and y-positions during the burn are x = (2000 m/s) t − 12 (4.62 m/s 2) t 2 y = 12 (2.31 m/s 2) t 2

The y-position at t1 = 433 s is 216,500 m = 216.5 km, 33.5 km away from the entrance. With the rocket off, you’re now coasting straight toward the entrance at 1.0 km/s and will cover this distance in 33.5 s. Thus you pass through the entrance at 433 s + 33.5 s = 466.5 s. The following table calculates your position at intervals of 50 s and, for accuracy, at t1 = 433 s when you end the burn. The trajectory is a parabola that intersects the x = 433 km line at y = 216.5 km, then a vertical line into the entrance.

8.58.

Model: Visualize:

Solve:

Use the particle model for the ball, which is in uniform circular motion.

From Newton’s second law along r and z directions,

∑F

r

= n cosθ =

mv 2 r

∑F

z

= n sin θ − mg = 0 ⇒ n sin θ = mg

Dividing the two force equations gives tan θ =

gr v2

From the geometry of the cone, tanθ = r y . Thus r gr = ⇒ v = gy y v2

8.59.

Model: Visualize:

Model the block as a particle and use the model of kinetic friction.

Solve: The only radial force is tension, so we can use Newton’s second law to find the angular velocity ω max at which the tube breaks:

∑ F = T = mω r ⇒ ω 2

r

max

=

Tmax 50 N = = 9.12 rad/s mr (0.50 kg)(1.2 m)

The compressed air and friction exert tangential forces, and the second law along the tangential direction is

∑F = F − f t

at =

t

k

= Ft − μ k n = Ft − μ k mg = mat

4.0 N Ft − μk g = − (0.60)(9.80 m/s 2 ) = 2.12 m/s 2 0.50 kg m

The time needed to accelerate to 9.12 rad/s is given by rω max (1.2)(9.12 rad/s) ⎛ at ⎞ = = 5.16 s ⎟ t1 ⇒ t1 = at 2.12 m/s 2 ⎝r⎠

ω1 = ω max = 0 + ⎜

During this interval, the block turns through angle

⎛ 2.12 m/s 2 ⎞ 1 rev ⎛a ⎞ 2 Δθ = θ1 − θ 0 = ω 0t1 + 12 ⎜ t ⎟ t12 = 0 + 12 ⎜ = 3.7 rev ⎟ (5.16 s) = 23.52 rad × π rad 1.2 m 2 ⎝r⎠ ⎝ ⎠

8.60.

Model: Visualize:

Solve:

Assume the particle model for a sphere in circular motion at constant speed.

(a) Newton’s second law along the r and z axes is:

mvt2 ∑ Fz = T1 cos30° + T 2 cos60° − FG = 0 N r Since we want T1 = T2 = T , these two equations become

∑ Fr = T1 sin 30° + T2 sin 60° =

T ( sin 30° + sin 60° ) =

mvt2 T ( cos30° + cos60° ) = mg r

Since sin 30° + sin 60° = cos30° + cos 60°, mvt2 ⇒ vt = rg r The triangle with sides L1 , L2 , and 1.0 m is isosceles, so L2 = 1.0 m and r = L2 cos30°. Thus mg =

L2 cos 30° g = (b) The tension is

T=

(1.0 m ) cos 30° g = ( 0.866 m ) ( 9.8 m/s2 ) = 2.9 m/s

( 2.0 kg ) ( 9.8 m/s mg = cos30° + cos60° 0.866 + 0.5

2

) = 14.3 N

8.61.

Model: Visualize:

Solve:

Use the particle model for a sphere revolving in a horizontal circle.

Newton’s second law in the r- and z-directions is

mvt2 ∑ ( F ) z = T1 sin 30° − T2 sin 30° − FG = 0 N r r Using r = (1.0 m ) cos30° = 0.886 m, these equations become

∑( F )

= T1 cos30° + T2 cos30° =

mvt2 ( 0.300 kg )( 7.5 m/s ) = 22.5 N = r cos30° ( 0.866 m )( 0.866 ) 2

T1 + T2 =

T1 − T2 =

( 0.300 kg ) ( 9.8 m/s 2 ) mg = = 5.88 N sin 30° ( 0.5)

Solving for T1 and T2 yields T1 = 14.2 N and T2 = 8.3 N.

8.62.

Model: Visualize:

Solve:

Use the particle model for the ball.

(a) Newton’s second law along the r- and z-directions is

∑F

r

= n cosθ = mrω 2

∑F

z

= n sin θ − FG = 0 N

Using FG = mg and dividing these equations yields:

tan θ =

g R− y = rω 2 r

where you can see from the figure that tanθ = ( R − y ) r . Thus ω =

g . R− y

(b) ω will be minimum when ( R − y) is maximum or when y = 0 m. Then ω min = g / R . (c) Substituting into the above expression,

ω=

g 9.8 m/s 2 rad 60 s 1 rev = = 9.9 × × = 95 rpm R−y 0.20 m − 0.10 m s 1 min 2π rad

8.63.

Model: Visualize:

Use the particle model for the airplane.

G Solve: In level flight, the lift force L balances the gravitational force. When turning, the plane banks so that the radial component of the lift force can create a centripetal acceleration. Newton’s second law along the r- and z-directions is

∑F

r

= L sin θ =

mvt2 r

∑F

z

= L cosθ − mg = 0 N

These can be written: sin θ =

mv 2 mg cosθ = L rL

Dividing the two equations gives: 2

miles 1 hr 1610 m ⎤ ⎡ ⎢⎣ 400 hour × 3600 s × 1 mile ⎥⎦ v2 v2 ⇒r= = = 18.5 km tanθ = ⎡⎣9.8 m s 2 ⎤⎦ tan10° gr g tan θ The diameter of the airplane’s path around the airport is 2 × 18.5 km = 37 km.

8.64.

Model: Visualize:

Use the particle model for a small volume of water on the surface.

Solve: Consider a particle of water of mass m at point C on the surface. Newton’s second law along the r- and zdirections is

( Fnet )r = n cosθ = mrω 2 ⇒ cosθ =

mrω 2 n

( Fnet ) z = n sinθ − mg = 0 N ⇒ sinθ =

mg n

Dividing both equations gives tanθ = g rω 2 . For a parabola z = ar 2 . This means dz 1 1 = 2ar = slope of the curve at C = tan φ = tan ( 90° − θ ) = ⇒ tan θ = dr tan θ 2ar Equating the two equations for tanθ , we get

1 g ω2 = 2 ⇒a= 2ar rω 2g Thus the surface is described by the equation z=

ω2 2g

r2

which is the equation of a parabola.

8-1

9.1. Model: Model the car and the baseball as particles. Solve:

(a) The momentum p = mv = (1500 kg )(10 m/s ) = 1.5 × 104 kg m/s.

(b) The momentum p = mv = ( 0.2 kg )( 40 m/s ) = 8.0 kg m/s.

9.2. Model: Model the bicycle and its rider as a particle. Also model the car as a particle. Solve:

From the definition of momentum,

pcar = pbicycle ⇒ mcar v car = mbicyclevbicycle ⇒ vbicycle =

⎛ 1500 kg ⎞ mcar v car = ⎜ ⎟ ( 5.0 m/s ) = 75 m/s mbicycle ⎝ 100 kg ⎠

Assess: This is a very high speed (≈ 168 mph). This problem shows the importance of mass in comparing two momenta.

9.3. Solve:

Visualize: Please refer to Figure EX9.3. The impulse J x is defined in Equation 9.6 as tf

J x = ∫ Fx ( t ) dt = area under the Fx (t ) curve between ti and tf ti

Jx =

1 ( 4 ms )(1000 N ) + ( 6 − 4 ms )(1000 N ) = 4 Ns 2

9.4. Model: The particle is subjected to an impulsive force. Visualize: Please refer to Figure EX9.4. Solve: Using Equation 9.6, the impulse is the area under the curve. From 0 s to 2 ms the impulse is

∫ Fdt = ( −500 N ) ( 2 × 10 s ) = −0.5 N s 1 2

−3

From 2 ms to 8 ms the impulse is

∫ Fdt = ( +2000 N )(8 ms − 2 ms ) = +6.0 N s 1 2

From 8 ms to 10 ms the impulse is

∫ Fdt = ( −500 N )(10 ms − 8 ms ) = −0.5 N s 1 2

Thus, from 0 s to 10 ms the impulse is ( −0.5 + 6.0 − 0.5 ) N s = 5.0 N s.

9.5. Visualize: Please refer to Figure EX9.5. Solve:

The impulse is defined in Equation 9.6 as tf

J x = ∫ Fx ( t ) dt = area under the Fx (t ) curve between ti and tf ti

⇒ 6.0 N s =

1 2

( Fmax )(8 ms ) ⇒ Fmax = 1.5 × 103 N

9.6. Model: Model the object as a particle and the interaction as a collision. Visualize: Please refer to Figure EX9.6. Solve: The object is initially moving to the right (positive momentum) and ends up moving to the left (negative momentum). Using the impulse-momentum theorem pfx = pix + J x ,

−2 kg m/s = +6 kg m/s + J x ⇒ J x = −8 kg m/s = −8 N s Since J x = Favg Δt , we have

Favg Δt = −8 N s ⇒ Favg = G The force is F = ( 8 × 102 N, left ) .

−8 N s = −8 × 102 N 10 ms

9.7. Model: Model the object as a particle and the interaction with the force as a collision. Visualize: Please refer to Figure EX9.7. Solve: Using the equations tf

pfx = pix + J x and J x = ∫ Fx (t ) dt = area under force curve ti

( 2.0 kg ) vfx = ( 2.0 kg )(1.0 m/s ) + (area under the force curve) ⇒ vfx = (1.0 m/s ) +

1 (1.0 s )( 2.0 N ) = 2.0 m/s 2.0 kg

Becaue vfx is positive, the object moves to the right at 2.0 m/s. Assess: For an object with positive velocity, a positive impulse increases the object’s speed. The opposite is true for an object with negative velocity.

9.8. Model: Model the object as a particle and the interaction with the force as a collision. Visualize: Please refer to Figure EX9.8. Solve: Using the equations tf

pfx = pix + J x and J x = ∫ Fx (t ) dt = area under force curve ti

( 2.0 kg ) vfx = ( 2.0 kg )(1.0 m/s ) + (area under the force curve) ⎛ 1 ⎞ ⇒ vfx = (1.0 m/s ) + ⎜ ⎟ ( −2.0 N )( 0.50 s ) = 0.50 m/s ⎝ 2.0 kg ⎠ Assess: For an object with positive velocity, a negative impulse slows the object. The opposite is true for an object with negative velocity.

9.9. Model: Use the particle model for the sled, the model of kinetic friction, and the impulse-momentum theorem. Visualize:

Note that the force of kinetic friction f k imparts a negative impulse to the sled. Solve:

Using Δpx = J x , we have tf

tf

ti

ti

pfx − pix = ∫ Fx (t ) dt = − f k ∫ dt = − f k Δt ⇒ mvfx − mvix = − μ k nΔt = − μ k mg Δt We have used the model of kinetic friction f k = μ k n, where μ k is the coefficient of kinetic friction and n is the normal (contact) force by the surface. The force of kinetic friction is independent of time and was therefore taken out of the impulse integral. Thus,

Δt =

1

μk g

( vix − vfx )

=

1

( 0.25) ( 9.8 m/s 2 )

( 8.0 m/s − 5.0 m/s ) = 1.22 s

9.10. Model: Use the particle model for the falling object and the impulse-momentum theorem. Visualize:

Note that the object is acted on by the gravitational force, whose magnitude is mg. Solve: Using the impulse-momentum theorem, tf

pfy − piy = J y = ∫ Fy (t ) dt ⇒ mvfy − mviy = −mg Δt ti

⇒ Δt = Assess:

viy − vfy g

=

−5.5 m/s − ( −10.4 m/s ) 9.8 m/s 2

= 0.50 s

Since Fy = − mg is independent of time, we have taken it out of the impulse integral.

9.11. Model: Model the tennis ball as a particle, and its interaction with the wall as a collision. Visualize:

The force increases to Fmax during the first two ms, stays at Fmax for two ms, and then decreases to zero during the last two ms. The graph shows that Fx is positive, so the force acts to the right. Solve: Using the impulse-momentum theorem pfx = pix + J x , 6 ms

( 0.06 kg )( 32 m/s ) = ( 0.06 kg )( −32 m/s ) + ∫

Fx ( t ) dt

0

The impulse is 6 ms

1 1 ∫ F ( t ) dx = area under force curve = 2 F ( 0.002 s ) + F ( 0.002 s ) + 2 F ( 0.002 s ) = ( 0.004 s ) F max

x

0

⇒ Fmax =

max

max

( 0.06 kg )( 32 m/s ) + ( 0.06 kg )( 32 m/s ) = 9.6 × 102 N 0.004 s

max

9.12. Model: Model the ball as a particle, and its interaction with the wall as a collision in the impulse approximation. Visualize: Please refer to Figure EX9.12. Solve: Using the equations tf

pfx = pix + J x and J x = ∫ Fx (t ) dt = area under force curve ti

( 0.250 kg ) vfx = ( 0.250 kg )( −10 m/s ) +

(500 N)(8.0 ms)

⎛ 4.0 N ⎞ ⇒ vfx = ( −10 m/s ) + ⎜ ⎟ = 6 m/s ⎝ 0.250 kg ⎠ Assess:

The ball’s final velocity is positive, indicating it has turned around.

9.13. Model: Model the glider cart as a particle, and its interaction with the spring as a collision. Visualize:

Solve:

Using the impulse-momentum theorem pfx − pix = ∫ Fdt ,

( 0.6 kg )( 3 m/s ) − ( 0.6 kg )( −3 m/s ) = area under force curve = 12 ( 36 N )( Δt ) ⇒ Δt = 0.20 s

9.14. Model: Visualize:

Solve:

Choose car + gravel to be the system. Ignore friction in the impulse approximation.

There are no external forces on the car + gravel system, so the horizontal momentum is conserved. This

means pfx = pix . Hence,

(10,000 kg + 4000 kg ) vfx = (10,000 kg )( 2.0 m/s ) + ( 4000 kg )( 0.0 m/s )

⇒ vfx = 1.43 m/s

9.15. Model: Visualize:

Choose car + rainwater to be the system.

There are no external horizontal forces on the car + water system, so the horizontal momentum is conserved. Solve: Conservation of momentum is pfx = pix . Hence,

( mcar + mwater )( 20 m/s ) = ( mcar )( 22 m/s ) + ( mwater )( 0 m/s ) ⇒ ( 5000 kg + mwater )( 20 m/s ) = ( 5000 kg )( 22 m/s ) ⇒ mwater = 5.0 × 102 kg

9.16. Model: Visualize:

Choose skydiver + glider to be the system in the impulse approximation.

Note that there are no external forces along the x-direction (ignoring friction in the impulse approximation), implying conservation of momentum along the x-direction. Solve: The momentum conservation equation pfx = pix is

( 680 kg − 60kg )( vG ) x + ( 60 kg )( vD ) x = ( 680 kg )( 30 m/s ) Immediately after release, the skydiver’s horizontal velocity is still ( vD ) x = 30 m/s. Thus

( 620 kg )( vG ) x + ( 60 kg )( 30 m/s ) = ( 680 kg )( 30 m/s ) ⇒ ( vG ) x = 30 m/s Assess:

The skydiver’s motion in the vertical direction has no influence on the glider’s horizontal motion.

9.17. Model: We will define our system to be bird + bug. This is the case of an inelastic collision because the bird and bug move together after the collision. Horizontal momentum is conserved because there are no external forces acting on the system during the collision in the impulse approximation. Visualize:

Solve:

The conservation of momentum equation pfx = pix is

( m1 + m2 ) vfx = m1 ( vix )1 + m2 ( vix )2

⇒ ( 300 g + 10 g ) vfx = ( 300 g )( 6.0 m/s ) + (10 g )( −30 m/s ) ⇒ vfx = 4.8 m/s

Assess: We left masses in grams, rather than convert to kilograms, because the mass units cancel out from both sides of the equation. Note that (vix ) 2 is negative.

9.18. Model: The two cars are not an isolated system because of external frictional forces. But during the collision friction is not going to be significant. Within the impulse approximation, the momentum of the Cadillac + Volkswagen system will be conserved in the collision. Visualize:

Solve:

The momentum conservation equation pfx = pix is

( mC + mVW ) vfx = mC ( vix )C + mVW ( vix )VW ⇒ 0 kg mph = ( 2000 kg )(1.0 mph ) + (1000 kg )( vix )VW ⇒ ( vix )VW = −2.0 mph You need a speed of 2.0 mph.

9.19. Model: Because of external friction and drag forces, the car and the blob of sticky clay are not exactly an isolated system. But during the collision, friction and drag are not going to be significant. The momentum of the system will be conserved in the collision, within the impulse approximation. Visualize:

Solve:

The conservation of momentum equation pfx = pix is

( mC + mB )( vf ) x = mB ( vix )B + mC ( vix )C ⇒ 0 kg m/s = (10 kg )( vix )B + (1500 kg )( −2.0 m/s ) ⇒ (vix ) B = 3.0 × 102 m/s

Assess: This speed of the blob is around 600 mph, which is very large. However, we must point out that a very large speed is expected in order to stop a car with only 10 kg of clay.

9.20. Model: We will define our system to be archer + arrow. The force of the archer (A) on the arrow (a) is equal to the force of the arrow on the archer. These are internal forces within the system. The archer G is Gstanding on frictionless ice, and the normal force by ice on the system balances the weight force. Thus Fext = 0 on the system, and momentum is conserved. Visualize:

The initial momentum pix of the system is zero, because the archer and the arrow are at rest. The final moment pfx must also be zero. Solve:

We have M A vA + ma va = 0 kg m/s. Therefore, vA =

−ma va − ( 0.100 kg )(100 m/s ) = = − 0.20 m/s 50 kg mA

The archer’s recoil speed is 0.20 m/s. Assess: It is the total final momentum that is zero, although the individual momenta are nonzero. Since the arrow has forward momentum, the archer will have backward momentum.

9.21. Model: We will define our system to be Bob + rock. Bob’s (B) force on the rock (R) is equal to the rock’s force on Bob. These are internal forces within the Bob is standing on frictionless ice, and the G G system. normal force by ice on the system balances the weight. Fext = 0 on the system, and thus momentum is conserved. Visualize:

The initial momentum pix of the system is zero because Bob and the rock are at rest. Thus pfx = 0 kg m/s. Solve: We have mBvB + mR vR = 0 kg m/s. Hence,

vB = −

⎛ 0.500 kg ⎞ mR vR = − ⎜ ⎟ ( 30 m/s ) = −0.20 m/s mB ⎝ 75 kg ⎠

Bob’s recoil speed is 0.20 m/s. Assess: Since the rock has forward momentum, Bob’s momentum is backward. This makes the total momentum zero.

9.22. Model: We will define our system to be Dan + skateboard, and their interaction as an explosion. While friction is present between the skateboard and the ground, it is negligible in the impulse approximation. Visualize:

The system has nonzero initial momentum pix . As Dan (D) jumps backward off the gliding skateboard (S), the skateboard will move forward in such a way that the final total momentum of the system pfx is equal to pix . Solve: We have mS ( vfx )S + mD ( vfx )D = ( mS + mD ) vix . Hence,

( 5.0 kg )( 8.0 m/s ) + ( 50 kg )( vfx )D = ( 5.0 kg + 50 kg )( 4.0 m/s ) ⇒ ( vfx )D = 3.6 m/s

9.23. Model: We assume that the momentum is conserved in the collision. Visualize: Please refer to Figure EX9.23. Solve: The conservation of momentum equation yields

( pfx )1 + ( pfx )2 = ( pix )1 + ( pix )2 ⇒ ( pfx )1 + 0 kg m/s = 2 kg m/s − 4 kg m/s ⇒ ( pfx )1 = −2 kg m/s

(p ) +(p ) =(p ) +(p ) fy 1

fy 2

iy 1

iy 2

⇒ ( pfy ) − 1 kg m/s = 2 kg m/s + 1 kg m/s ⇒ ( pfy ) = 4 kg m/s 1

Thus, the final momentum of particle 1 is ( −2iˆ + 4 ˆj ) kg m/s.

1

9.24. Model: This problem deals with the conservation of momentum in two dimensions in an inelastic collision. Visualize:

Solve:

G G The conservation of momentum equation pbefore = pafter is m1 ( vix )1 + m2 ( vix )2 = ( m1 + m2 ) vfx m1 ( viy ) + m2 ( viy ) = ( m1 + m2 ) vfy 1

2

Substituting in the given values,

(.02 kg )( 3.0 m/s ) + 0 kg m/s = (.02 kg + .03 kg ) vf cosθ 0 kg m/s + (.03 kg )( 2.0 m/s ) = (.02 kg + .03 kg ) vf sin θ

⇒ vf cosθ = 1.2 m/s vf sinθ = 1.2 m/s ⇒ vf =

(1.2 m/s )

2

+ (1.2 m/s ) = 1.7 m/s θ = tan −1

The ball of clay moves 45° north of east at 1.7 m/s.

2

vy vx

= tan −1 (1) = 45°

9.25. Model: Model the ball as a particle. We will also use constant-acceleration kinematic equations.

Solve:

(a) The momentum just after throwing is

p0 x = p0 cos30° = mv0 cos30° = ( 0.050 kg )( 25 m/s ) cos30° = 1.083 kg m/s p0 y = p0 sin 30° = mv0 sin 30° = ( 0.050 kg )( 25 m/s ) sin 30° = 0.625 kg m/s The momentum just after throwing is (1.08, 0.63) kg m/s. The momentum at the top is p1x = mv1x = mv0 x = 1.08 kg m/s p1 y = mv1 y = 0 kg m/s Just before hitting the ground, the momentum is p2 x = mv2 x = mv0 x = 1.08 kg m/s p2 y = mv2 y = − mv0 y = −0.63 kg m/s (b) px is constant because no forces act on the ball in the x-direction. Mathematically,

Fx =

dpx = 0 N ⇒ px = constant dt

(c) The change in the y-component of the momentum during the ball’s flight is

Δp y = −0.625 kg m/s − 0.625 kg m/s = −1.250 kg m/s From kinematics, the time to reach the top is obtained as follows:

v1 y = v0 y + a y ( t1 − t0 ) ⇒ t1 =

−v0 y ay

=

− (12.5 m/s ) −9.8 m/s 2

= 1.276 s

The time of flight is thus Δt = 2t1 = 2 (1.276 s ) = 2.552 s. Multiplying this time by –mg, the y-component of the weight, yields –1.25 kg m/s. This follows from the impulse-momentum theorem: Δp y = − mg Δt

9.26. Model: Model the rocket as a particle, and use the impulse-momentum theorem. The only force acting on the rocket is due to its own thrust. Visualize: Please refer to Figure P9.26. Solve: (a) The impulse is

J x = ∫ Fx dt = area of the Fx (t ) graph between t = 0 s and t = 30 s =

1 2

(1000 N )( 30 s ) = 1.5 × 104 N s

(b) From the impulse-momentum theorem, pfx = pix + 1.5 × 104 N s. That is, the momentum or velocity increases as

long as J x increases. When J x increases no more, the speed will be a maximum. This happens at t = 30 s. At this time, mvfx = mvix + 1.5 × 104 N s ⇒ ( 425 kg ) vfx = ( 425 kg )( 75 m/s ) + 1.5 × 10 4 N s ⇒ vfx = 110 m/s

9.27.

Solve: Using the equation

area under the force curve m 2.0 s 1 2π t ⎞ = 0 m/s + (10 N ) sin ⎛⎜ ⎟ dt 0.250 kg ∫0 ⎝ 4.0 s ⎠

v fx = vix +

2.0 s ⎞ ⎛ 4.0 s ⎞ ⎛ ⎛ 2π t ⎞ ⎜ ⎟ = ( 40 N/m ) ⎜ − cos ⎟⎜ ⎜ ⎟ ⎝ 2π ⎠ ⎝ ⎝ 4.0 s ⎠ 0 ⎟⎠

⎛ 80 ⎞ = −⎜ m/s ⎟ ( cos (π ) − cos ( 0 ) ) ⎝π ⎠ = 25 m/s ⎛ 2π t ⎞ The force is applied for half the period of 4.0 s. During that time, sin ⎜ ⎟ is positive, so an object ⎝ 4.0 s ⎠ initially at rest acquires a positive velocity. Assess:

9.28. Model: Let the system be ball + racket. During the collision of the ball and racket, momentum is conserved because all external interactions are insignificantly small. Visualize:

Solve:

(a) The conservation of momentum equation pfx = pix is

mR ( vfx )R + mB ( vfx )B = mR ( vix )R + mB ( vix )B

(1.000 kg )( vfx )R + ( 0.060 kg )( 40 m/s ) = (1.000 kg )(10 m/s ) + ( 0.060 kg )( −20 m/s ) ⇒ ( vfx )R = 6.4 m/s (b) The impulse on the ball is calculated from ( pfx ) B = ( pix ) B + J x as follows:

( 0.060 kg )( 40 m/s ) = ( 0.060 kg )( −20 m/s ) + J x ⇒ Favg =

⇒ J x = 3.6 N s = ∫ Fdt = Favg Δt

3.6 Ns = 3.6 × 102 N 10 ms

Let us now compare this force with the gravitational ball ( FG )B = mB g = ( 0.060 kg ) ( 9.8 m/s 2 ) = 0.588 N. Thus, Favg = 612( FG ) B .

force

on

the

Assess: This is a significant force and is reasonable because the impulse due to this force changes the direction as well as the speed of the ball from approximately 45 mph to 90 mph.

9.29. Model: Model the ball as a particle that is subjected to an impulse when it is in contact with the floor. We will also use constant-acceleration kinematic equations. Ignore any forces other than the interaction between the floor and the ball during the collision in the impulse approximation. Visualize:

Solve:

To find the ball’s velocity just before and after it hits the floor:

v12y = v02y + 2a y ( y1 − y0 ) = 0 m 2 /s 2 + 2 ( −9.8 m/s 2 ) ( 0 − 2.0 m ) ⇒ v1 y = −6.261 m/s v32y = v22 y + 2a y ( y3 − y2 ) ⇒ 0 m 2 /s 2 = v22 y + 2 ( −9.8 m/s 2 ) (1.5 m − 0 m ) ⇒ v2 y = 5.422 m/s The force exerted by the floor on the ball can be found from the impulse-momentum theorem:

mv2 y = mv1 y + ∫ Fdt = mv1 y + area under the force curve

⇒ ( 0.200 kg )( 5.422 m/s ) = − ( 0.200 kg )( 6.261 m/s ) + 12 Fmax ( 5.0 × 10−3 s )

⇒ Fmax = 9.3 × 102 N Assess:

A maximum force of 9.3 × 102 N exerted by the floor is reasonable.

9.30. Model: Model the rubber ball as a particle that is subjected to an impulsive force when it comes in contact with the floor. We will also use constant-acceleration kinematic equations and the impulse-momentum theorem. Visualize:

Solve: (a) To find the magnitude and direction of the impulse that the floor exerts on the ball, we use the impulse-momentum theorem: pfy = piy + J y ⇒ J y = p2 y − p1 y = m ( v2 y − v1 y ) Let us now find v1y and v2 y by using kinematics. For the falling and rebounding ball,

v12y = v02y + 2a y ( y1 − y0 ) = 0 m 2 /s 2 + 2 ( −9.8 m/s 2 ) ( 0 m − 1.8 m ) ⇒ v1 y = −5.940 m/s v32y = v22 y + 2a y ( y3 − y2 ) ⇒ 0 m 2 /s 2 = v22 y + 2 ( −9.8 m/s 2 ) (1.2 m − 0 m ) ⇒ v2 y = 4.850 m/s Going back to the impulse-momentum equation, we find J y = ( 0.040 kg ) ⎡⎣ 4.850 m/s − ( −5.940 m/s ) ⎤⎦ = 0.432 N s

The impulse is 0.43 N s upward. (b) As the ball compresses, the force of contact increases and the ball slows to v y = 0 m/s. Then in

decompression the ball is accelerated upward. To a good approximation, the force due to the floor as a function of time is shown in the figure. (c) For a rubber ball, Δt is likely in the range 5 to 10 ms. For 10 ms, J y = Favg Δt = 0.432 N s ⇒ Favg ≈ 40 N

The force is ≈80 N if Δt = 5 ms. Altogether, 40 to 80 N is a reasonable estimate.

9.31. Model: Model the cart as a particle rolling down a frictionless ramp. The cart is subjected to an impulsive force when it comes in contact with a rubber block at the bottom of the ramp. We will use the impulsemomentum theorem and the constant-acceleration kinematic equations. Visualize:

Solve:

From the free-body diagram on the cart, Newton’s second law “before the collision” is

∑( F )

x

= FG sin θ = max ⇒ ax =

mg sin θ g = g sin 30° = m 2

Using this acceleration, we can find the cart’s speed just before its contact with the rubber block: v12x = v02x + 2ax ( x1 − x0 ) = 0 m 2 /s 2 + 2 ( 12 g )(1.0 m − 0 m) ⇒ v1x = 3.13 m/s

Now we can use the impulse-momentum theorem to obtain the velocity just after the collision:

mv2 x = mv1x + ∫ Fx dt = mv1x + area under the force graph ⇒ ( 0.500 kg ) v2 x = ( 0.500 kg )( 3.13 m/s ) − 12 ( 200 N ) ( 26.7 × 10−3 s ) ⇒ v2 x = −2.21 m/s Note that the given force graph is positive, but in this coordinate system the impulse of the force is to the left up the slope. That is the reason to put a minus sign while evaluating the ∫ Fx dt integral. We can once again use a kinematic equation to find how far the cart will roll back up the ramp: v32x = v22x + 2ax ( x3 − x2 ) ⇒ ( 0 m/s ) = ( −2.21 m/s ) + 2 ( − 12 g ) ( x3 − x2 ) ⇒ ( x3 − x2 ) = 0.50 m 2

2

9.32.

Solve: Using Newton’s second law for the x–direction, Fx =

Fx =

dpx . Therefore, dt

d ( 6t 2 kg m/s ) = 12t N dt

Assess: The x-component of the net force on an object is equal to the time rate of change of the x-component of the object’s momentum.

9.33.

Visualize:

Solve:

Using Newton’s second law for the y-direction and the chain rule,

( Fnet ) y =

dp y dt

=

⎛ dv ⎞ d dm mv y ) = vy ) + m ⎜ y ⎟ ( ( dt dt ⎝ dt ⎠

= ( −0.50 kg/s )(120 m/s ) + ( 48 kg ) (18 m/s 2 ) = 8.0 × 102 N Assess:

Since the rocket is losing mass,

dm < 0. The time derivative of the velocity is the acceleration. dt

9.34. Model: Model the train cars as particles. Since the train cars stick together, we are dealing with perfectly inelastic collisions. Momentum is conserved in the collisions of this problem in the impulse approximation, in which we ignore external forces during the time of the collision. Visualize:

Solve:

In the collision between the three-car train and the single car:

mv1x + ( 3m ) v2 x = 4mv3 x ⇒ v1x + 3v2 x = 4v3 x ⇒ ( 4.0 m/s ) + 3( 2.0 m/s ) = 4v3 x ⇒ v3 x = 2.5 m/s In the collision between the four-car train and the stationary car:

( 4m ) v3 x + mv4 x = ( 5m ) v5 x

⇒ 4v3 x + 0 m/s = 5v5 x ⇒ v5 x =

4v3 x = ( 0.8 )( 2.5 m/s ) = 2.0 m/s 5

9.35.

Model: The dart and cork are particles in free fall until they have a perfectly inelastic head-on collision. Ignore air friction. Visualize:

Solve:

The positions of the dart and cork just before the collision are 1 ( y1 )D = ( y0 )D + ( v0 )D t1 − gt12 2 1 2 ( y1 )C = ( y0 )C − gt1 2 In the particle model, the dart and cork have no physical size, so ( y1 )D = ( y1 )C . Hence

1 1 1 9.8 m/s 2 ) t12 = 3 m − ( 9.8 m/s 2 ) t12 ⇒ t1 = s ( 2 2 3 At this time, the velocities of the dart and cork are 1 ( v1 )D = ( v0 )D − gt1 = 9.0 m/s − ( 9.8 m/s 2 ) ⎛⎜ s ⎞⎟ = 5.73 m/s ⎝3 ⎠ 1 ( v1 )C = ( v0 )C − gt1 = − ( 9.8 m/s 2 ) ⎜⎛ s ⎟⎞ = −3.27 m/s ⎝3 ⎠ These are the initial velocities to use with momentum conservation in a perfectly inelastic collision. pfy = piy 0 m + ( 9.0 m/s ) t1 −

( mD + mC ) v = mD ( v1 )D + mC ( v1 )C Thus v=

( 0.030 kg )( 5.73 m/s ) + ( 0.020 kg )( −3.27 m/s ) = 2.13 m/s, up ( 0.030 kg + 0.20 kg )

Assess: The heavier, faster upward-going dart has more momentum than the falling cork, so the total momentum is upwards.

9.36. Model: Model the earth (E) and the asteroid (A) as particles. Earth + asteroid is our system. Since the two stick together during the collision, this is a case of a perfectly inelastic collision. Momentum is conserved in the collision since no significant external force acts on the system. Visualize:

Solve:

(a) The conservation of momentum equation pfx = pix is

mA ( vix )A + mE ( vix )E = ( mA + mE ) vfx ⇒ (1.0 × 1013 kg )( 4.0 × 104 m s ) + 0 kg m/s = (1.0 × 1013 kg + 5.98 × 1024 kg ) vfx ⇒ vfx = 6.7 × 10−8 m/s (b) The speed of the earth going around the sun is

vE =

11 2π r 2π (1.50 × 10 m ) = = 3.0 × 104 m/s T 3.15 × 107 s

Hence, vfx vE = 2 × 10−12 = 2 × 10−10%. Assess: The earth’s recoil speed is insignificant compared to its orbital speed because of its large mass.

9.37. Model: Model the skaters as particles. The two skaters, one traveling north (N) and the other traveling west (W), are a system. Since the two skaters hold together after the “collision,” this is a case of a perfectly inelastic collision in two dimensions. Momentum is conserved since no significant external force in the x-y plane acts on the system in the “collision.” Visualize:

Solve:

(a) The x-component of the conservation of momentum is

( mN + mW ) vfx = mN ( vix ) N + mW ( vix )W

⇒ ( 75 kg + 60 kg ) vfx = 0 kg m/s + ( 60 kg )( −3.5 m/s ) ⇒ vfx = −1.556 m/s

The y-component of the conservation of momentum is

( mN + mW ) vfy = mN ( viy )N + mW ( viy )W

⇒ ( 75 kg + 60 kg ) vfy = ( 75 kg )( 2.5 m/s ) + 0 kg m/s

⇒ vfy = 1.389 m/s ⇒ vf =

( vfx )

2

+ ( vfy ) = 2.085 m/s 2

The time to glide to the edge of the rink is radius of the rink 25 m = = 12.0 s vf 2.085 m/s (b) The location is θ = tan −1 vfy vfx = 42° north of west. Assess:

A time of 12.0 s in covering a distance of 25 m at a speed of ≈ 2 m/s is reasonable.

9.38. Model: This problem deals with a case that is the opposite of a collision. The two ice skaters, heavier and lighter, will be modeled as particles. The skaters (or particles) move apart after pushing off against each other. During the “explosion,” the total momentum of the system is conserved. Visualize:

Solve:

The initial momentum is zero. Thus the conservation of momentum equation pfx = pix is

mH ( vfx )H + mL ( vfx )L = 0 kg m/s ⇒ ( 75 kg )( vfx )H + ( 50 kg )( vfx )L = 0 kg m/s Using the observation that the heavier skater takes 20 s to cover a distance of 30 m, we find ( vfx )H = 30 m 20 s = 1.5 m/s. Thus,

( 75 kg )(1.5 m/s ) + ( 50 kg )( vfx )L = 0 kg m/s

⇒ ( vfx )L = −2.25 m/s

Thus, the time for the lighter skater to reach the edge is 30 m 30 m = = 13.3 s v 2.25 m/s ( fx )L Assess: Conservation of momentum leads to a higher speed for the lighter skater, and hence a shorter time to reach the edge of the ice rink.

9.39. Model: This problem deals with a case that is the opposite of a collision. Our system is comprised of three coconut pieces that are modeled as particles. During the blow up or “explosion,” the total momentum of the system is conserved in the x-direction and the y-direction. Visualize:

Solve:

The initial momentum is zero. From pfx = pix , we get

+ m1 ( vfx )1 + m3 ( vf )3 cosθ = 0 kg m/s ⇒ ( vf )3 cosθ =

−m1 ( vfx )1 m3

=

−m ( −20 m/s ) = 10 m/s 2m

=

− m ( −20 m/s ) = 10 m/s 2m

From pfx = pix , we get

+ m2 ( vfy ) + m3 ( vf )3 sin θ = 0 kg m/s ⇒ ( vf )3 sin θ = 2

⇒ ( vf )3 =

(10 m/s )

The velocity is 14.1 m/s at 45° east of north.

2

− m2 ( vfy ) m3

2

+ (10 m/s ) = 14.1 m/s θ = tan −1 (1) = 45° 2

9.40. Model: The billiard balls will be modeled as particles. The two balls, m1 (moving east) and m2 (moving west), together are our system. This is an isolated system because any frictional force during the brief collision period is going to be insignificant. Within the impulse approximation, the momentum of our system will be conserved in the collision. Visualize:

Note that m1 = m2 = m. Solve:

The equation pfx = pix yields:

m1 ( vfx )1 + m2 ( vfx )2 = m1 ( vix )1 + m2 ( vix )2 ⇒ m1 ( vf )1 cosθ + 0 kg m/s = m1 ( vix )1 + m2 ( vix )2 ⇒ ( vf )1 cosθ = ( vix )1 + ( vix )2 = 2.0 m/s − 1.0 m/s = 1.0 m/s The equation pfy = piy yields:

+ m1 ( vfy ) sin θ + m2 ( vfy ) = 0 kg m/s ⇒ ( vf )1 sinθ = − ( vfy ) = −1.41 m/s 1

2

⇒ ( vf )1 =

2

(1.0 m/s )

2

+ ( −1.41 m/s ) = 1.73 m/s 2

⎛ 1.41 m/s ⎞ ⎟ = 55° ⎝ 1.0 m/s ⎠

θ = tan −1 ⎜ The angle is below +x axis, or south of east.

9.41. Model: This is a two-part problem. First, we have an inelastic collision between the wood block and the bullet. The bullet and the wood block are an isolated system. Since any external force acting during the collision is not going to be significant (the impulse approximation), the momentum of the system will be conserved. The second part involves the dynamics of the block + bullet sliding on the wood table. We treat the block and the bullet as particles. Visualize:

Solve:

The equation pfx = pix gives

( mB + mW ) vfx = mB ( vix )B + mW ( vix )W ⇒ ( 0.010 kg + 10 kg ) vfx = ( 0.010 kg )( vix )B + (10 kg )( 0 m/s ) ⇒ vfx =

1 ( vix )B 1001

From the model of kinetic friction, f k = − μ k n = − μ k ( mB + mW ) g = ( mB + mW ) ax ⇒ ax = − μ k g Using the kinematic equation v12x = v22x + 2ax ( x1 − x0 ) , 2

2 ⎛ 1 ⎞ v12x = v22x − 2 μ k g ( x1 − x0 ) ⇒ 0 m 2 s 2 = vf2x − 2μ k g x1 ⇒ ⎜ ⎟ ( vix )B = 2 μ k g x1 ⎝ 1001 ⎠

⇒ ( vix )B = 1001 2μ k g x1 = 1001 2 ( 0.20 ) ( 9.8 m/s 2 ) ( 0.050 m ) = 443 m/s The bullet’s speed is 4.4 × 102 m/s. Assess: The bullet’s speed is reasonable (≈ 900 mph).

9.42. Model: This is a two-part problem. First, we have an inelastic collision between Fred (F) and Brutus (B). Fred and Brutus are an isolated system. The momentum of the system during collision is conserved since no significant external force acts on the system. The second part involves the dynamics of the Fred + Brutus system sliding on the ground. Visualize:

Note that the collision is head-on and therefore one-dimensional. Solve: The equation pfx = pix is

( mF + mB ) vfx = mF ( vix )F + mB ( vix )B ⇒ ( 60 kg + 120 kg ) vfx = ( 60 kg )( −6.0 m/s ) + (120 kg )( 4.0 m/s ) ⇒ vfx = 0.667 m/s The positive value indicates that the motion is in the direction of Brutus. The model of kinetic friction yields: f k = − μ k n = − μ k ( mF + mB ) g = ( mF + mB ) ax ⇒ ax = − μ k g Using the kinematic equation v12x = v02x + 2ax ( x1 − x0 ) , we get v12x = v02x − 2 μ k g x1 ⇒ 0 m 2 s 2 = vf2x − 2 ( 0.30 ) ( 9.8 m/s 2 ) x1 ⇒ 0 m 2 s 2 = ( 0.667 m/s ) − ( 5.9 m/s 2 ) x1 ⇒ x1 = 7.6 cm 2

They slide 7.6 cm in the direction Brutus was running. Assess: After the collision, Fred and Brutus slide with a small speed but with a good amount of kinetic friction. A stopping distance of 7.6 cm is reasonable.

9.43. Model: Model the package and the rocket as particles. This is a two-part problem. First we have an inelastic collision between the rocket (R) and the package (P). During the collision, momentum is conserved since no significant external force acts on the rocket and the package. However, as soon as the package + rocket system leaves the cliff they become a projectile motion problem. Visualize:

Solve: The minimum velocity after collision that the package + rocket must have to reach the explorer is v0 x , which can be found as follows:

y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ −200 m = 0 m + 0 m + 12 ( −9.8 m/s 2 ) t12 ⇒ t1 = 6.389 s 2

With this time, we can now find v0 x using x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) . We obtain 2

30 m = 0 m + v0 x ( 6.389 s ) + 0 m ⇒ v0 x = 4.696 m/s = vfx We now use the momentum conservation equation pfx = pix which can be written

( mR + mP ) vfx = mR ( vix )R + mP ( vix )P ⇒ (1.0 kg + 5.0 kg )( 4.696 m/s ) = (1.0 kg )( vix )R + ( 5.0 kg )( 0 m/s ) ⇒ ( vix )R = 28 m/s

9.44.

Model: Visualize:

Solve:

The clay balls undergo a perfectly inelastic collision in the impulse approximation.

(a) The total momentum in reference frame S is P = ( px )1 + ( px )2 = m1 ( vx )1 + m2 ( vx )2

= ( 0.020 kg )(12 m/s ) + 0 kg m/s

= 0.24 kg m/s (b) In the S′ frame, the total momentum is zero. P′ = ( px )1′ + ( px )2′ = 0 ⇒ m1 ( vx )1′ = − m2 ( vx )2′ Equation 4.24 relates velocities in the two reference frames. ( vx )1′ = ( vx )1 − Vx

( vx )2′ = ( vx )2 − Vx Thus

m1 ( ( vx )1 − Vx ) = − m2 ( ( vx )2 − Vx ) ⇒ Vx =

m1 ( vx )1 + m2 ( vx )2 m1 + m2

= 4.0 m/s

(c) Since the total momentum is zero, the final velocity of the resulting 60 g clay ball must be ( vx )f ′ = 0 m/s. (d) Using Equation 4.24 again,

( vx )f = ( vx )f ′ + Vx = 0 m/s + 4.0 m/s = 4.0 m/s Assess: The final velocity of both balls in the S frame is less than the original velocity of the smaller ball, and is in the same direction. The S′ frame is called the “center of mass frame.” Some collisions are more easily studied in the center of mass frame, such as high energy particle collisions at particle accelerator facilities.

9.45.

Model: This is a two-part problem. First, we have an explosion that creates two particles. The momentum of the system, comprised of two fragments, is conserved in the explosion. Second, we will use kinematic equations and the model of kinetic friction to find the displacement of the lighter fragment. Visualize:

The initial momentum is zero. Using momentum conservation pfx = pix during the explosion,

Solve:

mH ( v1x )H + mL ( v1x )L = mH ( v0 x )H + mL ( v0 x )L ⇒ 7m ( v1x )H + m ( v1x )L = 0 kg m/s ⇒ ( v1x )H = − ( 17 ) ( v1x )L Because mH slides to x2H = −8.2 m before stopping, we have f k = μ k nH = μ k wH = μ k mH g = mH aH ⇒ aH = μ k g

Using kinematics,

( v2 x )H = ( v1x )H + 2aH ( x2H − x1H ) 2

2

⇒ 0 m 2 s 2 = ( 17 ) ( v1x )L + 2μ k g ( −8.2 m − 0 m ) ⇒ ( v1x )L = −88.74 μ k m/s 2

2

How far does mL slide? Using the information obtained above in the following kinematic equation,

( v2 x )L = ( v1x )L + 2aL ( x2L − x1L ) ⇒ 0 2

Assess:

2

m 2 s 2 = μ k ( 88.74 ) − 2μ k gx2L ⇒ x2L = 4.0 × 102 m 2

Note that aH is positive, but aL is negative, and both are equal in magnitude to μ k g . Also, x2H is

negative but x2L is positive.

9.46. Model: We will model the two fragments of the rocket after the explosion as particles. We assume the explosion separates the two parts in a vertical manner. This is a three-part problem. In the first part, we will use kinematic equations to find the vertical position where the rocket breaks into two pieces. In the second part, we will apply conservation of momentum to the system (that is, the two fragments) in the explosion. In the third part, we will again use kinematic equations to find the velocity of the heavier fragment just after the explosion. Visualize:

Solve:

The rocket accelerates for 2.0 s from rest, so

v1 y = v0 y + a y ( t1 − t0 ) = 0 m/s + (10 m/s 2 ) ( 2 s − 0 s ) = 20 m/s

y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) = 0 m + 0 m + 12 (10 m/s 2 ) ( 2 s ) = 20 m 2

2

At the explosion the equation pfy = piy is

mL ( v2 y ) + mH ( v2 y ) = ( mL + mH ) v1 y ⇒ ( 500 kg ) ( v2 y ) + (1000 kg ) ( v2 y ) = (1500 kg )( 20 m/s ) L

H

L

H

To find (v2 y ) H we must first find (v2 y ) L , the velocity after the explosion of the upper section. Using kinematics,

(v ) = (v ) 2

2

3y L

2y L

+ 2 ( −9.8 m/s 2 ) ( y3L − y2L ) ⇒ ( v2 y ) = 2 ( 9.8 m/s 2 ) ( 530 m − 20 m ) = 99.98 m/s L

Now, going back to the momentum conservation equation we get

( 500 kg )( 99.98 m/s ) + (1000 kg ) ( v2 y )H = (1500 kg )( 20 m/s ) ⇒ ( v2 y )H = −20 m/s The negative sign indicates downward motion.

9.47. Model: Let the system be bullet + target. No external horizontal forces act on this system, so the horizontal momentum is conserved. Model the bullet and the target as particles. Since the target is much more massive than the bullet, it is reasonable to assume that the target undergoes no significant motion during the brief interval in which the bullet passes through. Visualize:

Solve: (a) By assuming that the target has negligible motion during the interval in which the bullet passes through, the time is that needed to slow from 1200 m/s to 900 m/s in a distance of 30 cm. We’ll use kinematics to first find the acceleration, then the time.

(v1x ) B2 = (v0 x ) B2 + 2ax Δx ⇒ ax =

⇒ Δt =

(v1x ) B2 − (v0 x ) B2 (900 m/s)2 − (1200 m/s)2 = = −1.05 × 106 m/s 2 2Δx 2(0.30 m) (v1x ) B = (v0 x ) B + ax Δt (v1x ) B − (v0 x ) B 900 m/s − 1200 m/s = = 2.86 × 10−4 s = 286 μ s ax −1.05 × 106 m/s 2

The average force on the bullet is Favg = m ax = 26 kN. (b) Now we can use the conservation of momentum equation p1x = p0 x to find

mT (v1x )T + mB (v1x ) B = mT (v0 x )T + mB (v0 x ) B = 0 + mB (v0 x ) B ⇒ (v1x )T =

mB (v0 x ) B − mB (v1x ) B 0.025 kg = ( (1200 m/s) − (900 m/s) ) = 0.021 m/s mT 350 kg

9.48. Model: Model the two blocks (A and B) and the bullet (L) as particles. This is a two-part problem. First, we have a collision between the bullet and the first block (A). Momentum is conserved since no external force acts on the system (bullet + block A). The second part of the problem involves a perfectly inelastic collision between the bullet and block B. Momentum is again conserved for this system (bullet + block B). Visualize:

Solve:

For the first collision the equation pfx = pix is

mL ( v1x )L + mA ( v1x )A = mL ( v0 x )L + mA ( v0 x )A ⇒ ( 0.010 kg )( v1x )L + ( 0.500 kg )( 6.0 m/s ) = ( 0.010 kg )( 400 m/s ) + 0 kg m/s ⇒ ( v1x )L = 100 m/s The bullet emerges from the first block at 100 m/s. For the second collision the equation pfx = pix is

( mL + mB ) v2 x = mL ( v1x )L ⇒ ( 0.010 kg + 0.500 kg ) v2 x = ( 0.010 kg )(100 m/s ) ⇒ v2 x = 1.96 m/s

9.49. Model: Model Brian (B) along with his wooden skis as a particle. The “collision” between Brian and

Ashley lasts for a short time, and during this time no significant external forces act on the Brian + Ashley system. Within the impulse approximation, we can then assume momentum conservation for our system. After finding the velocity of the system immediately after the collision, we will apply constant-acceleration kinematic equations and the model of kinetic friction to find the final speed at the bottom of the slope. Visualize:

Solve:

Brian skiing down for 100 m:

( v1x )B = ( v0 x )B + 2ax ( x1B − x0B ) 2

2

= 0 m 2 s 2 + 2ax (100 m − 0 m ) ⇒ ( v1x )B =

( 200 m ) ax

To obtain ax , we apply Newton’s second law to Brian in the x and y directions as follows:

∑( F )

on B x

= wB sinθ − f k = mBax

∑( F )

on B y

= n − wB cosθ = 0 N ⇒ n = w cosθ

From the model of kinetic friction, f k = μ k n = μ k wB cosθ . The x-equation thus becomes wB sin θ − μ k wB cosθ = mB ax

⇒ ax = g ( sin θ − μ k cosθ ) = ( 9.8 m/s 2 ) ⎡⎣sin 20° − ( 0.060 ) cos 20°⎤⎦ = 2.80 m/s 2 Using this value of ax ,

( v1x )B = ( 200 m ) ( 2.80 m/s 2 ) = 23.7 m/s.

In the collision with Ashley the conservation

of momentum equation pfx = pix is

( mB + mA ) v2 x = mB ( v1x )B ⇒ v2 x =

mB 80 kg ( v1x )B = ( 23.66 m/s ) = 14.56 m/s mB + mA 80 kg + 50 kg

Brian + Ashley skiing down the slope:

v32x = v22x + 2ax ( x3 − x2 ) = (14.56 m/s ) + 2 ( 2.80 m/s 2 ) (100 m ) ⇒ v3 x = 27.8 m/s 2

That is, Brian + Ashley arrive at the bottom of the slope with a speed of 27.8 m/s. Note that we have used the same value of ax in the first and the last parts of this problem. This is because ax is independent of mass. Assess:

A speed of approximately 60 mph on a ski slope of 200 m length and 20° slope is reasonable.

9.50. Model: Model the spy plane (P) and the rocket (R) as particles. The plane and the rocket undergo a

perfectly inelastic collision. During the brief collision time, momentum for the plane + rocket system will be conserved because no significant external forces act on this system during the collision. After finding the velocity of the system immediately after the collision, we will apply projectile equations to find where the system will hit the ground. Visualize:

Solve:

For the collision we have pfx = pix and pfy = piy . This means the magnitude of the final momentum is

pf = ⇒ vf =

( pfx )

2

+ ( pfy ) = 2

1

( pix )

2

+ ( piy ) ⇒ ( mR + mP ) vf = ⎡⎣ mR ( vix )R ⎤⎦ + ⎡ mP ( viy )P ⎤ ⎣ ⎦ 2

2

⎡⎣(1280 kg )( 725 m/s ) ⎤⎦ + ⎡⎣( 575 kg )( 450 m/s ) ⎤⎦ = 519.4 m/s 2

(1280 kg + 575 kg )

2

2

⎛ ( 575 kg )( 450 m/s ) ⎞ ⎛ pfy ⎞ −1 ⎟⎟ = 15.58° north of east ⎟ = tan ⎜⎜ ⎝ pfx ⎠ ⎝ (1280 kg )( 725 m/s ) ⎠

θ = tan −1 ⎜

We call this direction the x′ axis in the x-y plane. Let us now look at the falling plane + rocket system in the x′-z plane where z is the vertical axis perpendicular to the x-y plane. We have z0 = 2700 m

z1 = 0 m

v0 z = 0 m/s

x0′ = 0 m

t0 = 0 s

v0 x′ = 519.4 m/s

Using the kinematic equation z1 = z0 + v0 z ( t1 − t0 ) + 12 az ( t1 − t0 ) , we can find the time to fall: 2

0 m = 2700 m + 0 m + 12 ( −9.8 m/s 2 ) t12 ⇒ t1 = 23.47 s In this time t1 , the wreckage travels horizontally to

x1′ = x0′ + v0 x′ ( t1 − t0 ) = 0 m + ( 519.4 m/s )( 23.47 s ) = 12,192 m = 12.19 km The enmeshed plane + rocket system lands 12.19 km from the collision point at an angle of 15.58° north of east.

9.51. Model: This is an isolated system, so momentum is conserved in the explosion. Momentum is a vector

G quantity, so the direction of the initial velocity vector v1 establishes the direction of the momentum vector. The

final momentum vector, after the explosion, must still point in the +x-direction. The two known pieces continue to move along this line and have no y-components of momentum. The missing third piece cannot have a ycomponent of momentum if momentum is to be conserved, so it must move along the x-axis—either straight forward or straight backward. We can use conservation laws to find out. Visualize:

Solve:

From the conservation of mass, the mass of piece 3 is

m3 = mtotal − m1 − m2 = 7.0 × 105 kg To conserve momentum along the x-axis, we require

[ pi = mtotalvi ] = [ pf

= p1f + p2f + p3f = m1v1f + m2v2f + p3f ]

⇒ p3f = mtotalvi − m1v1f − m2v2f = +1.02 × 1013 kg m/s Because p3f > 0, the third piece moves in the +x-direction, that is, straight forward. Because we know the mass m3 , we can find the velocity of the third piece as follows: v3f =

p3f 1.02 × 1013 kg m/s = = 1.46 × 107 m/s m3 7.0 × 105 kg

9.52.

Model: The two railcars make up a system. The impulse approximation is used while the spring is expanding, so friction can be ignored. The spring is massless. Visualize:

Solve: holds.

Since the cars are at rest initially, the total momentum of the system is zero. Conservation of momentum

0 = m1 ( vfx )1 + m2 ( vfx )2

We are only told that the relative velocity of the two cars after the spring expands is 4.0 m/s, so

( vfx )2 − ( vfx )1 = 4.0 m/s The positive-going car 2 velocity is written first to ensure that the relative velocity is positive. Substitute ( vfx )2 = ( vfx )1 + 4.0 m/s into the conservation of momentum equation, then solve for ( vfx )1 .

0 = m1 ( vfx )1 + m2 ( ( vfx )1 + 4.0 m/s ) ⇒ ( vfx )1 = −

Assess:

m2 ( 4.0 m/s ) ( 90 tons )( 4.0 m/s ) = −3.0 m/s =− ( m1 + m2 ) ( 30 tons + 90 tons )

The other more massive railcar has a velocity ( vfx )2 = ( vfx )1 + 4.0 m/s = 1.0 m/s. A slower speed for the

more massive car makes sense.

9.53. Model: This is a three-part problem. In the first part, the shell, treated as a particle, is launched as a projectile and reaches its highest point. We will use constant-acceleration kinematic equations for this part. The shell, which is our system, then explodes at the highest point. During this brief explosion time, momentum is conserved. In the third part, we will again use the kinematic equations to find the horizontal distance between the landing of the lighter fragment and the origin. Visualize:

Solve:

The initial velocity is

v0 x = v cosθ = (125 m/s ) cos55° = 71.7 m/s v0 y = v sinθ = (125 m/s ) sin 55° = 102.4 m/s

At the highest point, v1 y = 0 m/s and v1x = 71.7 m/s. The conservation of momentum equation pfx = pix is mL ( v1x )L + mH ( v1x )H = ( mL + mH ) v1x The heavier particle falls straight down, so (v1x ) H = 0 m/s. Thus,

(15 kg )( v1x )L + 0 kg m s = (15 kg + 60 kg )( 71.7 m/s ) ⇒ ( v1x )L = 358 m/s That is, the velocity of the smaller fragment immediately after the explosion is 358 m/s and this velocity is in the horizontal x-direction. Note that (v1 y ) L = 0 m/s. To find x2 , we will first find the displacement x1 − x0 and then x2 − x1. For x1 − x0 ,

v1 y = v0 y + a y ( t1 − t0 ) ⇒ 0 m s = (102.4 m/s ) + ( −9.8 m/s 2 ) ( t1 − 0 s ) ⇒ t1 = 10.45 s x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ⇒ x1 − x0 = ( 71.7 m/s )(10.45 s ) + 0 m = 749 m 2

For x2 − x1 : x2 = x1 + ( v1x ) L ( t2 − t1 ) + 12 ax ( t2 − t1 ) ⇒ x2 − x1 = ( 358 m/s )(10.45 s ) + 0 m=3741 m 2

⇒ x2 = ( x2 − x1 ) + ( x1 − x0 ) = 3741 m + 749 m = 4490 m = 4.5 km Assess: Note that the time of ascent to the highest point is equal to the time of descent to the ground, that is, t1 − t0 = t2 − t1.

9.54. Model: Model the proton (P) and the gold atom (G) as particles. The two constitute our system, and momentum is conserved in the collision between the proton and the gold atom. Visualize:

Solve:

The conservation of momentum equation pfx = pix is

mG ( vfx )G + mP ( vfx )P = mP ( vix )P + mG ( vix )G ⇒ (197 u )( vfx )G + (1 u ) ( −0.90 × 5.0 × 107 m/s ) = (1 u ) ( 5.0 × 107 m/s ) + 0 u m/s ⇒ ( vfx )G = 4.8 × 105 m/s

9.55. Model: Model the proton (P) and the target nucleus (T) as particles. The proton and the target nucleus make our system and in the collision between them momentum is conserved. This is due to the impulse approximation because the collision lasts a very short time and the external forces acting on the system during this time are not significant. Visualize:

Solve:

The conservation of momentum equation pfx = pix is

mT ( vfx )T + mP ( v fx ) = mT ( vix )T + mP ( vix )P P

⇒ mT ( 3.12 × 10 m/s ) + (1 u ) ( −0.75 × 2.5 × 106 m/s ) = 0 u m/s + (1 u ) ( 2.5 × 106 m/s ) 5

⇒ mT = 14.0 u Assess:

This is the mass of the nucleus of a nitrogen atom.

9.56. Model: This problem deals with a case that is the opposite of a collision. It is a case of an “explosion” in which a 214 Po nucleus (P) decays into an alpha-particle (A) and a daughter nucleus (N). During the “explosion” or decay, the total momentum of the system is conserved. Visualize:

Solve: Conservation of mass requires the daughter nucleus to have mass mN = 214 u − 4 u = 210 u. The conservation of momentum equation pfx = pix is

mN ( vfx ) N + mA ( vfx )A = ( mN + mA ) vP ⇒ ( 210 u )( vfx ) N + ( 4 u ) ( −1.92 × 107 m / s ) = 0 u m/s ⇒ ( vfx ) N = 3.66 × 105 m/s

9.57. Model: The neutron’s decay is an “explosion” of the neutron into several pieces. The neutron is an isolated system, so its momentum should be conserved. The observed decay products, the electron and proton, move in opposite directions. Visualize:

Solve:

(a) The initial momentum is pix = 0 kg m/s. The final momentum pfx = meve + mpvp is

2.73 × 10−23 kg m/s − 1.67 × 10−22 kg m/s = −1.40 × 10−22 kg m/s No, momentum does not seem to be conserved. (b) and (c) If the neutrino is needed to conserve momentum, then pe + pP + pneutrino = 0 kg m/s. This requires pneutrino = − ( pe + pP ) = +1.40 × 10−22 kg m/s The neutrino must “carry away” 1.40 × 10−22 kg m/s of momentum in the same direction as the electron.

9.58. Model: Model the two balls of clay as particles. Our system comprises these two balls. Momentum is conserved in the perfectly inelastic collision. Visualize:

Solve:

The x-component of the final momentum is

pfx = pix = m1 ( vix )1 + m2 ( vix )2 = ( 0.020 kg )( 2.0 m/s ) − ( 0.030 kg )(1.0 m/s ) cos30° = 0.0140 kg m/s The y-component of the final momentum is

pfy = piy = m1 ( viy ) + m2 ( viy ) 1

2

= ( 0.02 kg )( 0 m/s ) − ( 0.03 kg )(1.0 m/s ) sin 30° = −0.0150 kg m/s

⇒ pf =

( 0.014 kg m/s )

2

+ ( −0.015 kg m/s ) = 0.0205 kg m/s 2

Since pf = ( m1 + m2 ) vf = 0.0205 kg m/s, the final speed is

vf =

0.0205 kg m/s = 0.41 m/s ( 0.02 + 0.03) kg

and the direction is

θ = tan −1

pfy pfx

= tan −1

0.015 = 47° south of east 0.014

9.59. Model: Model the three balls of clay as particle 1 (moving north), particle 2 (moving west), and particle 3 (moving southeast). The three stick together during their collision, which is perfectly inelastic. The momentum of the system is conserved. Visualize:

Solve:

The three initial momenta are G G pi1 = m1vi1 = ( 0.020 kg )( 2.0 m/s ) ˆj = 0.040ˆj kg m/s

(

)

G G pi2 = m2vi2 = ( 0.030 kg ) −3.0 m/s iˆ = −0.090iˆ kg m/s

(

)

G G pi3 = m3vi3 = ( 0.040 kg ) ⎡⎣( 4.0 m/s ) cos 45°iˆ − ( 4.0 m/s ) sin 45° ˆj ⎤⎦ = 0.113iˆ − 0.113 ˆj kg m/s G G G G Since pf = pi = pi1 + pi2 + pi3 , we have G

( m1 + m2 + m3 ) vf

(

)

(

)

G = 0.023iˆ − 0.073 ˆj kg m/s ⇒ vf = 0.256iˆ − 0.811 ˆj m/s

⇒ vf =

( 0.256 m/s )

θ = tan −1

vfy vfx

2

+ ( −0.811 m/s ) = 0.85 m/s

= tan −1

2

0.811 = 72° below +x 0.256

9.60. Model: Model the truck (T) and the two cars (C and C′) as particles. The three forming our system stick together during their collision, which is perfectly inelastic. Since no significant external forces act on the system during the brief collision time, the momentum of the system is conserved. Visualize:

Solve:

The three momenta are G G piT = mT viT = ( 2100 kg )( 2.0 m/s ) iˆ = 4200iˆ kg m/s G G piC = mCviC = (1200 kg )( 5.0 m/s ) ˆj = 6000ˆj kg m/s G G piC′ = mC′viC′ = (1500 kg )(10 m/s ) iˆ = 15,000iˆ kg m/s

(

)

G G G G G pf = pi = piT + piC + piC′ = 19,200iˆ + 6000ˆj kg m/s ⇒ pf = ( mT + mC + mC′ ) vf = ⇒ vf = 4.2 m/s θ = tan −1

(19,200 kg m/s ) py px

= tan −1

2

+ ( 6000 kg m/s )

2

6000 = 17.4° above +x 19,200

Assess: A speed of 4.2 m/s for the entangled three vehicles is reasonable since the individual speeds of the cars and the truck before entanglement were of the same order of magnitude.

9.61. Model: The 14 C atom undergoes an “explosion” and decays into a nucleus, an electron, and a neutrino. Momentum is conserved in the process of “explosion” or decay. Visualize:

Solve:

G G The conservation of momentum equation pf = pi = 0 kg m/s is G G G G G G G G pe + pn + pN = 0 N ⇒ pN = − ( pe + pn ) = − meve − mn vn

(

)

= − ( 9.11 × 10−31 kg )( 5.0 × 107 m/s ) iˆ − ( 8.0 × 10−24 kg m/s ) ˆj = − 45.55 × 10−24iˆ + 8.0 × 10−24 ˆj kg m/s ⇒ pN = mN vN =

( 45.55 × 10 ) + (8.0 × 10 ) −24 2

−24

2

kg m/s

⇒ ( 2.34 × 10−26 kg ) vN = 4.62 × 10−23 kg m/s ⇒ vN = 1.97 × 103 m/s

9.62. (a) A 100 g ball traveling to the left at 30 m/s is batted back to the right at 40 m/s. The force curve for the force of the bat on the ball can be modeled as a triangle with a maximum force of 1400 N. How long is the ball in contact with the bat? (b)

(c) The solution is Δt = 0.100 s = 10 ms.

9.63. (a) A 200 g ball of clay traveling to the right overtakes and collides with a 400 g ball of clay traveling to the right at 3.0 m/s. The balls stick and move forward at 4.0 m/s. What was the speed of the 200 g ball of clay? (b)

(c) The solution is ( vix )2 = 6.0 m/s.

9.64. (a) A 2000 kg auto traveling east at 5.0 m/s suffers a head-on collision with a small 1000 kg auto traveling west at 4.0 m/s. They lock bumpers and stick together after the collision. What will be the speed and direction of the combined wreckage after the collision? (b)

(c) The solution is vfx = 2.0 m/s along +x direction.

9.65. (a) A 150 g spring-loaded toy is sliding across a frictionless floor at 1.0 m/s. It suddenly explodes into two pieces. One piece, which has twice the mass of the second piece, continues to slide in the forward direction at 7.5 m/s. What is the speed and direction of the second piece? (b)

(c) The solution is (vfx )1 = −12 m/s. The minus sign tells us that the second piece moves backward at 12 m/s.

9.66. Model: Visualize:

Solve:

The cart + man (C + M) is our system. It is an isolated system, and momentum is conserved.

The conservation of momentum equation pfx = pix is

mM ( vfx )M + mC ( vfx )C = mM ( vix )M + mC ( vix )C Note that (vfx ) M and (vfx )C are the final velocities of the man and the cart relative to the ground. What is given in this problem is the velocity of the man relative to the moving cart. The man’s velocity relative to the ground is

( vfx )M = ( vfx )C − 10 m/s With this form for (vfx ) M, we rewrite the momentum conservation equation as

mM ⎡⎣( vfx )C − 10 m/s ⎤⎦ + mC ( vfx )C = mM ( 5.0 m/s ) + mC ( 5.0 m/s )

⇒ ( 70 kg ) ⎡⎣( vfx )C − 10 m/s ⎤⎦ + (1000 kg )( vfx )C = (1000 kg + 70 kg )( 5.0 m/s ) ⇒ ( vfx )C [1000 kg + 70 kg ] = (1070 kg )( 5.0 m/s ) + ( 70 kg )(10 m/s ) ⇒ ( vfx )C = 5.7 m/s

9.67. Model: Model Ann and cart as particles. The initial momentum is pi = 0 kg m/s in a coordinate system attached to the ground. As Ann begins running to the right, the cart will have to recoil to the left to conserve momentum. Visualize:

Solve: The difficulty with this problem is that we are given Ann’s velocity of 5.0 m/s relative to the cart. If the cart is also moving with velocity vcart then Ann’s velocity relative to the ground is not 5.0 m/s. Using the Galilean transformation equation for velocity, Ann’s velocity relative to the ground is

( vfx )Ann = ( vfx )cart + 5.0 m/s Now, the momentum conservation equation pix = pfx is

0 kg m/s = mAnn ( vfx )Ann + mcart ( vfx )cart ⇒ 0 kg m/s = ( 50 kg ) ⎡⎣( vfx )cart + 5.0 m/s ⎤⎦ + ( 500 kg )( vfx )cart ⇒ ( vfx )cart = −0.45 m/s Using the recoil velocity (vfx )cart relative to the ground, we find Ann’s velocity relative to the ground to be (vfx ) Ann = 5.00 m/s − 0.45 m/s = 4.55 m/s The distance Ann runs relative to the ground is Δx = (vfx ) Ann Δt , where Δt is the time it takes to reach the end of the cart. Relative to the cart, which is 15 m long, Ann’s velocity is 5 m/s. Thus Δt = (15 m ) ( 5 m/s ) = 3.00 s. Her

distance over the ground during this interval is

Δx = ( vfx )Ann Δt = ( 4.55 m/s )( 3.00 s ) = 13.6 m

9.68. Model: Visualize:

Solve:

The projectile + wood ball are our system. In the collision, momentum is conserved.

The momentum conservation equation pfx = pix is

( mP + mB ) vfx = mP ( vix )P + mB ( vix )B

⇒ (1.0 kg + 20 kg ) vfx = (1.0 kg )( vix )P + 0 kg m/s ⇒ ( vix )P = 21vfx

We therefore need to determine vfx . Newton’s second law for circular motion is

T − FG = T − ( mP + mB ) g =

( mP + mB ) vf2x r

Using Tmax = 400 N, this equation gives

400 N − (1.0 kg + 20 kg )( 9.8 m/s ) =

(1.0 kg + 20 kg ) vf2x 2.0 m

⇒ ( vfx )max = 4.3 m/s

Going back to the momentum conservation equation,

( vix )P = 21vfx = ( 21)( 4.3 m/s ) = 90 m/s That is, the largest speed this projectile can have without causing the cable to break is 90 m/s.

9.69. Model: This is an “explosion” problem and momentum is conserved. The two-stage rocket is our system. Visualize:

Solve:

Relative to the ground, the conservation of momentum equation pfx = pix is

m1 ( vfx )1 + m2 ( vfx )2 = ( m1 + m2 ) v1x ⇒ 3 m 2 ( vfx )1 + m2 ( vfx )2 = ( 4 m 2 )(1200 m/s ) ⇒ 3( vfx )1 + ( vfx )2 = 4800 kg m/s The fact that the first stage is pushed backward at 35 m/s relative to the second can be written

( vfx )1 = −35 m/s + ( vfx )2 Substituting this form of (vfx )1 in the conservation of momentum equation, 3 ⎡⎣ −35 m/s + ( vfx )2 ⎤⎦ + ( vfx )2 = 4800 kg m/s ⇒ ( vfx )2 = 1226 m/s

9.70. Model: Model the bullet and the vehicle as particles, and use the impulse-momentum theorem to find the impulse provided to the vehicle by bullet(s). Because the final speed of the vehicle is small compared to the bullet speed, and the mass of the bullet is so much smaller than the mass of the vehicle, we will assume that each bullet exerts the same impulse on the vehicle. Visualize:

Solve: For one bullet, the impulse-momentum theorem Δpx = J x allows us to find that the impulse exerted on the bullet by the vehicle’s sail is

( J x ) B = mB (Δv) B = (0.020 kg) ( ( −200 m/s) − (400 m/s) ) = −12.0 kg m/s From Newton’s third law, the impulse that the bullet exerts on the vehicle is ( J x ) V = −( J x ) B = 12.0 kg m/s. A second use of the impulse momentum theorem allows us to find the vehicle’s increase in velocity due to one bullet: (Δv) V =

( J x ) V 12.0 kg m/s = = 0.120 m/s mV 100 kg

This increase in velocity is independent of the vehicle’s speed, so as long as the impulse per bullet stays essentially constant (which it does because the bullet speeds are so much larger than the vehicle speed), the number of impacts needed to increase the vehicle speed to 12 m/s is N bullets =

12 m/s = 100 bullets 0.12 m/s per bullet

To reach this speed in 20 s requires the bullets to be fired at a rate of 5 bullets per second.

9.71. Model: Let the system be rocket + bullet. This is an isolated system, so momentum is conserved. Visualize: The fact that the bullet’s velocity relative to the rocket is 139,000 can be written (vf ) B = (vf ) R + 139,000 m/s.

Solve:

Consider the firing of one bullet when the rocket has mass M and velocity vi . The conservation of

momentum equation pf = pi is

( M − 5kg)vf + (5 kg)(vf + 139,000 m/s) = Mvi ⇒ Δv = vf − vi = −

5 kg (139,000 m/s) M

The rocket starts with mass M = 2000 kg, which is much larger than 5 kg. If only a few bullets are needed, M will not change significantly as the rocket slows. If we assume that M remains constant at 2000 kg, the loss of speed per bullet is Δv = −347.5 m/s = −1250 km/h. Thus exactly 8 bullets will reduce the speed by 10,000 km/h, from 25,000 km/h to 15,000 km/h. If you’re not sure that treating M as a constant is valid, you can calculate Δv for each bullet and reduce M by 5 kg after each shot. The loss of mass causes Δv to increase slightly for each bullet. An eight-step calculation then finds that 8 bullets will slow the rocket to 14,900 km/h. Seven bullets wouldn’t be enough, and 9 would slow the rocket far too much.

9.72.

Visualize:

Solve: Ladies and gentlemen of the jury, how far would the chair slide if it was struck with a bullet from my client’s gun? We know the bullet’s velocity as it leaves the gun is 450 m/s. The bullet travels only a small distance to the chair, so we will neglect any speed loss due to air resistance. The bullet and chair can be considered an isolated system during the brief interval of the collision. The bullet embedded itself in the chair, so this was a perfectly inelastic collision. Momentum conservation allows us to calculate the velocity of the chair immediately after the collision as follows:

pix = pfx ⇒ mB ( vi )B = ( mB + mC ) vf ⇒ vf =

mB ( vi )B mB + mC

=

( 450 m/s )( 0.010 kg ) = 0.225 m/s 20.01 kg

This is the velocity immediately after the collision when the chair starts to slide but before it covers any distance. For the purpose of the problem in dynamics, call this the initial velocity v0 . The free-body diagram of the chair shows three forces. Newton’s second law applied to the chair (with the embedded bullet) is

ax = a =

( Fnet ) x mtot

=

( Fnet ) y n − mtot g − fk μn = = − k a y = 0 m/s 2 = mtot mtot mtot mtot

where we’ve used the friction model in the x-equation. The y-equation yields n = mtot g , and the x-equation yields

a = − μ k g = −1.96 m/s 2 . We know the coefficient of kinetic friction because it is a wood chair sliding on a wood floor. Finally, we have to determine the stopping distance of the chair. The motion of the chair ends with v1 = 0 m/s after sliding a distance Δx, so

( 0.225 m/s ) = 0.013 m = 1.3 cm v02 =− 2a 2 ( −1.96 m/s 2 ) 2

v12 = 0 m 2 /s 2 = v02 + 2aΔx ⇒ Δx = −

If the bullet lost any speed in the air before hitting the chair, the sliding distance would be even less. So you can see that the most the chair could slide if it had been struck by a bullet from my client’s gun, would be 1.3 cm. But in actuality, the chair slid 3 cm, more than twice as far. The murder weapon, ladies and gentlemen, was a much more powerful gun than the one possessed by my client. I rest my case.

9-1

10.1. Model: We will use the particle model for the bullet (B) and the running student (S). Visualize:

Solve:

For the bullet,

1 1 K B = mBvB2 = (0.010 kg)(500 m/s) 2 = 1250 J 2 2 For the running student, 1 1 KS = mSvS2 = (75 kg)(5.5 m/s) 2 = 206 J 2 2 Thus, the bullet has the larger kinetic energy. Assess: Kinetic energy depends not only on mass but also on the square of the velocity. The above calculation shows this dependence. Although the mass of the bullet is 7500 times smaller than the mass of the student, its speed is more than 90 times larger.

10.2. Model: Model the hiker as a particle. Visualize:

The origin of the coordinate system chosen for this problem is at sea level so that the hiker’s position in Death Valley is y0 = −8.5 m. Solve: The hiker’s change in potential energy from the bottom of Death Valley to the top of Mt. Whitney is

ΔU = U gf − U gi = mgyf − mgyi = mg ( yf − yi ) = (65 kg)(9.8 m/s 2 )[4420 m − (−85 m)] = 2.9 × 106 J Assess:

Note that ΔU is independent of the origin of the coordinate system.

10.3. Model: Model the compact car (C) and the truck (T) as particles. Visualize:

Solve:

For the kinetic energy of the compact car and the kinetic energy of the truck to be equal,

1 1 20,000 kg mT (25 km/hr) = 112 km/hr K C = K T ⇒ mCvC2 = mT vT2 ⇒ vC = vT = 2 2 1000 kg mC Assess:

A smaller mass needs a greater velocity for its kinetic energy to be the same as that of a larger mass.

10.4.

Model: Model the car (C) as a particle. This is an example of free fall, and therefore the sum of kinetic and potential energy does not change as the car falls. Visualize:

Solve:

(a) The kinetic energy of the car is

KC =

1 1 mCvC2 = (1500 kg)(30 m/s) 2 = 6.75 × 105 J 2 2

The car’s kinetic energy is 6.8 × 105 J. (b) Let us relabel K C as K f and place our coordinate system at yf = 0 m so that the car’s potential energy U gf

is zero, its velocity is vf , and its kinetic energy is K f . At position yi , vi = 0 m/s or K i = 0 J, and the only energy the car has is U gi = mgyi . Since the sum K + U g is unchanged by motion, K f + U gf = K i + U gi . This

means K f + mgyf = K i + mgyi ⇒ K f + 0 = K i + mgyi ⇒ yi =

( K f − Ki ) (6.75 × 105 J − 0 J) = = 46 m mg (1500 kg)(9.8 m / s 2 )

(c) From part (b),

1 2 1 2 2 2 ( K f − K i ) 2 mvf − 2 mvi ( vf − vi ) yi = = = mg mg 2g Free fall does not depend upon the mass.

10.5.

Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the ball rises and falls. Visualize:

The figure shows a ball’s before-and-after pictorial representation for the three situations in parts (a), (b) and (c). Solve: The quantity K + U g is the same during free fall: K f + U gf = Ki + U gi . We have (a)

1 2 1 mv1 + mgy1 = mv02 + mgy0 2 2 ⇒ y1 = ( v02 − v12 ) 2 g = [(10 m/s) 2 − (0 m/s) 2 ]/(2 × 9.8 m / s 2 ) = 5.10 m

5.1 m is therefore the maximum height of the ball above the window. This is 25.1 m above the ground. 1 2 1 (b) mv2 + mgy2 = mv02 + mgy0 2 2 Since y2 = y0 = 0, we get for the magnitudes v2 = v0 = 10 m / s. (c)

1 2 1 mv3 + mgy3 = mv02 + mgy0 ⇒ v32 + 2 gy3 = v02 + 2 gy0 ⇒ v32 = v02 + 2 g ( y0 − y3 ) 2 2 ⇒ v32 = (10 m / s) 2 + 2(9.8 m / s 2 )[0 m − (−20 m)] = 492 m 2 / s 2

This means the magnitude of v3 is equal to 22 m/s. Assess: Note that the ball’s speed as it passes the window on its way down is the same as the speed with which it was tossed up, but in the opposite direction.

10.6. Model: This is a problem of free fall. The sum of the kinetic and gravitational potential energy for the ball, considered as a particle, does not change during its motion. Visualize:

The figure shows the ball’s before-and-after pictorial representation for the two situations described in parts (a) and (b). Solve: The quantity K + U g is the same during free fall. Thus, K f + U gf = Ki + U gi . 1 2 1 mv1 + mgy1 = mv02 + mgy0 ⇒ v02 = v12 + 2 g ( y1 − y0 ) (a) 2 2 ⇒ v02 = (0 m/s) 2 + 2(9.8 m/s 2 )(10 m − 1.5 m) = 166.6 m 2 /s 2 ⇒ v0 = 12.9 m/s (b)

1 2 1 mv2 + mgy2 = mv02 + mgy0 ⇒ v22 = v02 + 2 g ( y0 − y2 ) 2 2 2 2 ⇒ v2 = 166.6 m /s 2 + 2(9.8 m/s 2 )(1.5 m − 0 m) ⇒ v2 = 14.0 m/s

Assess: An increase in speed from 12.9 m/s to 14.0 m/s as the ball falls through a distance of 1.5 m is reasonable. Also, note that mass does not appear in the calculations that involve free fall.

10.7. Model: Model the oxygen and the helium atoms as particles. Visualize: We denote the oxygen and helium atoms by O and He, respectively. Note that the oxygen atom is four times heavier than the helium atom, so mO = 4 mHe . Solve:

The energy conservation equation K O = K He is

1 1 v 2 2 mOvO2 = mHevHe ⇒ (4 mHe )vO2 = mHevHe ⇒ He = 2.0 2 2 vO Assess: The result vHe = 2vO , combined with the fact that mHe = 14 mO , is a consequence of the way kinetic energy is defined: It is directly proportional to the mass and to the square of the speed.

10.8. Model: Model the ball as a particle undergoing rolling motion with zero rolling friction. The sum of the ball’s kinetic and gravitational potential energy, therefore, does not change during the rolling motion. Visualize:

Solve:

Since the quantity K + U g does not change during rolling motion, the energy conservation equations

apply. For the linear segment the energy conservation equation K 0 + U g0 = K1 + U g1 is 1 2 1 1 1 1 mv1 + mgy1 = mv02 + mgy0 ⇒ m v12 + mg (0 m) = m(0 m/s) 2 + mgy0 ⇒ mv12 = mgy0 2 2 2 2 2

For the parabolic part of the track, K1 + U g1 = K 2 + U g 2 is 1 2 1 1 1 1 mv1 + mgy1 = mv22 + mgy2 ⇒ mv12 + mg (0 m) = m(0 m / s) 2 + mgy2 ⇒ mv12 = mgy2 2 2 2 2 2

Since from the linear segment we have 12 m v12 = mgy0 , we get mgy0 = mgy2 or y2 = y0 = 1.0 m. Thus, the ball rolls up to exactly the same height as it started from. Assess: Note that this result is independent of the shape of the path followed by the ball, provided there is no rolling friction. This result is an important consequence of energy conservation.

10.9. Model: Model the skateboarder as a particle. Assuming that the track offers no rolling friction, the sum of the skateboarder’s kinetic and gravitational potential energy does not change during his rolling motion. Visualize:

The vertical displacement of the skateboarder is equal to the radius of the track. Solve: The quantity K + U g is the same at the upper edge of the quarter-pipe track as it was at the bottom. The energy conservation equation K f + U gf = K i + U gi is

1 2 1 mvf + mgyf = mvi2 + mgyi ⇒ vi2 = vf2 + 2 g ( yf − yi ) 2 2 vi2 = (0 m/s) 2 + 2(9.8 m/s 2 )(3.0 m − 0 m) = 58.8 m 2 /s 2 ⇒ vi = 7.7 m/s Assess:

Note that we did not need to know the skateboarder’s mass, as is the case with free-fall motion.

10.10. Model: Model the puck as a particle. Since the ramp is frictionless, the sum of the puck’s kinetic and gravitational potential energy does not change during its sliding motion. Visualize:

Solve:

The quantity K + U g is the same at the top of the ramp as it was at the bottom. The energy conservation

equation K f + U gf = K i + U gi is

1 2 1 mvf + mgyf = mvi2 + mgyi ⇒ vi2 = vf2 + 2 g ( yf − yi ) 2 2 ⇒ vi2 = (0 m/s) 2 + 2(9.8 m/s 2 )(1.03 m − 0 m) = 20.2 m 2 /s 2 ⇒ vi = 4.5 m/s Assess: An initial push with a speed of 4.5 m/s ≈ 10 mph to cover a distance of 3.0 m up a 20° ramp seems reasonable.

10.11.

Model: In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change as the pendulum swings from one side to the other. Visualize:

The figure shows the pendulum’s before-and-after pictorial representation for the two situations described in parts (a) and (b). Solve: (a) The quantity K + U g is the same at the lowest point of the trajectory as it was at the highest point. Thus, K1 + U g1 = K 0 + U g0 means

1 2 1 mv1 + mgy1 = mv02 + mgy0 ⇒ v12 + 2 gy1 = v02 + 2 gy0 2 2 ⇒ v12 + 2 g (0 m) = (0 m/s) 2 + 2 gy0 ⇒ v1 = 2 gy0 From the pictorial representation, we find that y0 = L − L cos30°. Thus,

v1 = 2 gL(1 − cos30°) = 2(9.8 m/s 2 )(0.75 m)(1 − cos30°) = 1.403 m/s The speed at the lowest point is 1.40 m/s. (b) Since the quantity K + U g does not change, K 2 + U g 2 = K1 + U g1. We have 1 2 1 mv2 + mgy2 = mv12 + mgy1 ⇒ y2 = ( v12 − v22 ) 2 g 2 2 ⇒ y2 = [(1.403 m/s) 2 − (0 m/s) 2 ]/(2 × 9.8 m/s 2 ) = 0.100 m Since y2 = L − L cosθ , we obtain

cosθ =

L − y2 (0.75 m) − (0.10 m) = = 0.8667 ⇒ θ = cos −1 (0.8667) = 30° L (0.75 m)

That is, the pendulum swings to the other side by 30°.

Assess: The swing angle is the same on either side of the rest position. This result is a consequence of the fact that the sum of the kinetic and gravitational potential energy does not change. This is shown as well in the energy bar chart in the figure.

10.12. Model: Model the child and swing as a particle, and assume the chain to be massless. In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change during the swing’s motion. Visualize:

Solve:

The quantity K + U g is the same at the highest point of the swing as it is at the lowest point. That is,

K 0 + U g0 = K1 + U g1. It is clear from this equation that maximum kinetic energy occurs where the gravitational potential energy is the least. This is the case at the lowest position of the swing. At this position, the speed of the swing and child will also be maximum. The above equation is 1 2 1 mv0 + mgy0 = mv12 + mgy1 ⇒ v12 = v02 + 2 g ( y0 − y1 ) 2 2 2 ⇒ v1 = (0 m/s) 2 + 2 g ( y0 − 0 m) ⇒ v1 = 2 gy0 We see from the pictorial representation that

y0 = L − L cos 45° = (3.0 m) − (3.0 m)cos 45° = 0.879 m ⇒ v1 = 2 gy0 = 2(9.8 m/s 2 )(0.879 m) = 4.2 m/s Assess: We did not need to know the swing’s or the child’s mass. Also, a maximum speed of 4.2 m/s is reasonable.

10.13. Model: Model the car as a particle with zero rolling friction. The sum of the kinetic and gravitational potential energy, therefore, does not change during the car’s motion. Visualize:

Solve:

The initial energy of the car is

1 1 K 0 + U g0 = mv02 + mgy0 = (1500 kg)(10.0 m/s) 2 + (1500 kg)(9.8 m/s 2 )(10 m) = 2.22 × 105 J 2 2 The car increases its height to 15 m at the gas station. The conservation of energy equation K 0 + U g0 = K1 + U g1 is

1 1 2.22 × 105 J = mv12 + mgy1 ⇒ 2.22 × 105 J = (1500 kg)v12 + (1500 kg)(9.8 m/s 2 )(15 m) 2 2 ⇒ v1 = 1.41 m/s Assess: A lower speed at the gas station is reasonable because the car has decreased its kinetic energy and increased its potential energy compared to its starting values.

10.14.

Model: Assume that the spring is ideal and obeys Hooke’s law. Also model the sled as a particle. G Visualize: The only horizontal force acting on the sled is Fsp .

Solve:

Applying Newton’s second law to the sled gives

∑ (F

) = Fsp = max ⇒ k Δy = max

on sled x

⇒ ax = k Δ x / m = (150 N/m)(0.20 m)/20 kg = 1.50 m/s 2

10.15.

Model: Assume that the spring is ideal and obeys Hooke’s law. Visualize: According to Hooke’s law, the spring force acting on a mass (m) attached to the end of a spring is given as Fsp = k Δ x, where Δx is the change in length of the spring. If the mass m is at rest, then Fsp is also equal

to the gravitational force FG = mg . Solve:

We have Fsp = k Δ x = mg . We want a 0.100 kg mass to give Δ x = 0.010 m. This means k = mg / Δx = (0.100 kg)(9.8 N/m)/(0.010 m) = 98 N/m

10.16. Model: Assume an ideal spring that obeys Hooke’s law. Visualize:

Solve:

(a) The spring force on the 2.0 kg mass is Fsp = − k Δy. Notice that Δy is negative, so Fsp is positive.

This force is equal to mg, because the 2.0 kg mass is at rest. We have − k Δy = mg . Solving for k:

k = −( mg / Δy ) = −(2.0 kg)(9.8 m/s 2 )/(−0.15 m − (−0.10 m)) = 392 N/m The spring constant is 3.9 × 102 N/m. (b) Again using − k Δy = mg: Δy = − mg / k = −(3.0 kg)(9.8 m/s 2 )/(392 N/m) y′ − ye = −0.075 m ⇒ y′ = ye − 0.075 m = −0.10 m − 0.075 m = −0.175 m = −17.5 cm The length of the spring is 17.5 cm when a mass of 3.0 kg is attached to the spring. The position of the end of the spring is negative because it is below the origin, but length must be a positive number.

10.17. Model: Assume that the spring is ideal and obeys Hooke’s law. We also model the 5.0 kg mass as a particle. Visualize:

Solve:

We will use the subscript s for the scale and sp for the spring.

(a) The scale reads the upward force Fs on m that it applies to the mass. Newton’s second law gives

∑ (F

) = Fs on m − FG = 0 ⇒ Fs on m = FG = mg = (5.0 kg)(9.8 m/s 2 ) = 49 N

on m y

(b) In this case, the force is

∑ (F

) = Fs on m + Fsp − FG = 0 ⇒ 20 N + k Δy − mg = 0

on m y

⇒ k = (mg − 20 N)/ Δy = (49 N − 20 N)/0.02 m = 1450 N/m The spring constant for the lower spring is 1.45 × 103 N/m. (c) In this case, the force is ∑ ( Fon m ) y = Fsp − FG = 0 ⇒ k Δy − mg = 0 ⇒ Δy = mg / k = (49 N)/(1450 N/m) = 0.0338 m = 3.4 cm

10.18. Model: Model the student (S) as a particle and the spring as obeying Hooke’s law. Visualize:

Solve:

According to Newton’s second law the force on the student is

∑ (F

) = Fspring on S − FG = ma y

on S y

⇒ Fspring on S = FG + ma y = mg + ma y = (60 kg)(9.8 m/s 2 + 3.0 m/s 2 ) = 768 N Since Fspring on S = FS on spring = k Δy, k Δy = 768 N. This means Δy = (768 N)/(2500 N/m) = 0.31 m.

10.19. Model: Assume an ideal spring that obeys Hooke’s law. Solve: The elastic potential energy of a spring is defined as U s = 12 k (Δs ) 2 , where Δs is the magnitude of the stretching or compression relative to the unstretched or uncompressed length. We have Δs = 20 cm = 0.20 m and k = 500 N/m. This means

1 1 U s = k (Δs ) 2 = (500 N/m)(0.20 m) = 10 J 2 2 Assess: Since Δs is squared, U s is positive for a spring that is either compressed or stretched. U s is zero when the spring is in its equilibrium position.

10.20. Model: Assume an ideal spring that obeys Hooke’s law. Solve:

The elastic potential energy of a spring is defined as U s = 12 k (Δs ) 2 , where Δs is the magnitude of the

stretching or compression relative to the unstretched or uncompressed length. ΔU s = 0 when the spring is at its equilibrium length and Δs = 0. We have U s = 200 J and k = 1000 N/m. Solving for Δs : Δs =

2U s / k = 2(200 J) / 1000 N/m = 0.632 m

10.21. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, so the mechanical energy K + U s is conserved. Also model the book as a particle. Visualize:

The figure shows a before-and-after pictorial representation. The compressed spring will push on the book until the spring has returned to its equilibrium length. We put the origin of our coordinate system at the equilibrium position of the free end of the spring. The energy bar chart shows that the potential energy of the compressed spring is entirely transformed into the kinetic energy of the book. Solve: The conservation of energy equation K 2 + U s2 = K1 + U s1 is

1 2 1 1 1 mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe ) 2 2 2 2 2 Using x2 = xe = 0 m and v1 = 0 m/s, this simplifies to 1 2 1 (1250 N/m)(0.040 m) 2 kx12 mv2 = k ( x1 − 0 m) 2 ⇒ v2 = = = 2.0 m/s 2 2 (0.500 kg) m

Assess: This problem cannot be solved using constant-acceleration kinematic equations. The acceleration is not a constant in this problem, since the spring force, given as Fs = − k Δx, is directly proportional to Δx or | x − xe |.

10.22. Model: Assume an ideal spring that obeys Hooke’s law. Since there is no friction, the mechanical energy K + U s is conserved. Also, model the block as a particle. Visualize:

The figure shows a before-and-after pictorial representation. We have put the origin of our coordinate system at the equilibrium position of the free end of the spring. This gives us x1 = xe = 0 cm and x2 = 2.0 cm. Solve:

The conservation of energy equation K 2 + U s2 = K1 + U s1 is

1 2 1 1 1 mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe ) 2 2 2 2 2 Using v2 = 0 m/s, x1 = xe = 0 m, and x2 − xe = 0.020 m, we get

m 1 1 k ( x2 − xe ) 2 = mv12 ⇒ Δx = ( x2 − xe ) = v1 k 2 2 That is, the compression is directly proportional to the velocity v1. When the block collides with the spring with twice the earlier velocity (2v1 ), the compression will also be doubled to 2( x2 − xe ) = 2(2.0 cm) = 4.0 cm. Assess: This problem shows the power of using energy conservation over using Newton’s laws in solving problems involving nonconstant acceleration.

10.23. Model: Model the grocery cart as a particle and the spring as an ideal that obeys Hooke’s law. We will also assume zero rolling friction during the compression of the spring, so that mechanical energy is conserved. Visualize:

The figure shows a before-and-after pictorial representation. The “before” situation is when the cart hits the spring in its equilibrium position. We put the origin of our coordinate system at this equilibrium position of the free end of the spring. This give x1 = xe = 0 and ( x2 − xe ) = 60 cm. Solve:

The conservation of energy equation K 2 + U s2 = K1 + U s1 is

1 2 1 1 1 mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe ) 2 2 2 2 2 Using v2 = 0 m/s,( x2 − xe ) = 0.60 m, and x1 = xe = 0 m gives:

1 1 k 250 N/m k ( x2 − xe ) 2 = m v12 ⇒ v1 = ( x2 − xe ) = (0.60 m) = 3.0 m/s 2 2 m 10 kg

10.24. Model: Model the jet plane as a particle, and the spring as an ideal that obeys Hooke’s law. We will also assume zero rolling friction during the stretching of the spring, so that mechanical energy is conserved. Visualize:

The figure shows a before-and-after pictorial representation. The “before” situation occurs just as the jet plane lands on the aircraft carrier and the spring is in its equilibrium position. We put the origin of our coordinate system at the right free end of the spring. This gives x1 = xe = 0 m. Since the spring stretches 30 m to stop the

plane, x2 − xe = 30 m. Solve:

The conservation of energy equation K 2 + U s2 = K1 + U s1 for the spring-jet plane system is 1 2 1 1 1 mv2 + k ( x2 − xe ) 2 = mv12 + k ( x1 − xe ) 2 2 2 2 2

Using v2 = 0 m/s, x1 = xe = 0 m, and x2 − xe = 30 m yields

1 1 k 60,000 N/m k ( x2 − xe ) 2 = mv12 ⇒ v1 = ( x2 − x1 ) = (30 m) = 60 m/s 2 2 m 15,000 kg Assess:

A landing speed of 60 m/s or ≈120 mph is reasonable.

10.25. Model: We assume this is a one-dimensional collision that obeys the conservation laws of momentum and mechanical energy. Visualize:

Note that momentum conservation alone is not sufficient to solve this problem because the two final velocities (vfx )1 and (vfx ) 2 are unknowns and can not be determined from one equation. Solve:

Momentum conservation: m1 (vix )1 + m2 (vix ) 2 = m1 (vfx )1 + m2 (vfx ) 2 Energy conservation:

1 1 1 1 m1 (vix )12 + m2 (vix ) 22 = m1 (vfx )12 + m2 (vfx ) 22 2 2 2 2

These two equations can be solved for (vfx )1 and (vfx ) 2 , as shown by Equations 10.39 through 10.43, to give

(vfx )1 =

m1 − m2 50 g − 20 g (vix )1 = (2.0 m/s) = 0.86 m/s m1 + m2 50 g + 20 g

(vfx ) 2 =

2m1 2(50 g) (vix )1 = (2.0 m/s) = 2.9 m/s m1 + m2 50 g + 20 g

Assess: These velocities are of a reasonable magnitude. Since both these velocities are positive, both balls move along the +x-direction.

10.26. Model: This is a case of a perfectly elastic collision between a proton and a carbon atom. The collision obeys the momentum as well as the energy conservation law. Visualize:

Solve:

Momentum conservation: mP (vix ) P + mC (vix ) C = mP (vfx ) P + mC (vfx ) C

1 1 1 1 mP (vix ) 2P + mC (vix )C2 = mP (vfx ) P2 + mC (vfx ) C2 2 2 2 2 These two equations can be solved, as described in the text through Equations 10.39 to 10.43: Energy conservation:

⎛ m − 12mP ⎞ mP − mC 7 7 mP + mC (vix ) P = ⎜ P ⎟ (2.0 × 10 m/s) = −1.69 × 10 m/s + mP + mC m 12 m ⎝ P P ⎠ ⎛ 2mP ⎞ 2mP 7 6 (vfx )C = (vix ) P = ⎜ ⎟ (2.0 × 10 m/s) = 3.1 × 10 m/s mP + mC ⎝ mP + 12mP ⎠ (vfx ) P =

After the elastic collision the proton rebounds at 1.69 × 107 m/s and the carbon atom moves forward at

3.08 × 106 m/s.

10.27. Model: This is the case of a perfectly inelastic collision. Momentum is conserved because no external force acts on the system (clay + brick). We also represent our system as a particle. Visualize:

Solve:

(a) The conservation of momentum equation pfx = pix is

(m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2 Using (vix )1 = v0 and (vix ) 2 = 0, we get

vfx =

m1 0.050 kg (vix )1 = (vix )1 = 0.0476(vix )1 = 0.0476 v0 m1 + m2 (1.0 kg + 0.050 kg)

The brick is moving with speed 0.048v0 . (b) The initial and final kinetic energies are given by 1 1 1 1 2 K i = m1 (vix )12 + m2 (vix ) 22 = (0.050 kg)v02 + (1.0 kg) ( 0 m/s ) = (0.025 kg)v02 2 2 2 2 1 1 K f = (m1 + m2 )vf2x = (1.0 kg + 0.050 kg)(0.0476) 2 v02 = 0.00119 v02 2 2 ⎛ K − Kf The percent of energy lost = ⎜ i ⎝ Ki

⎞ ⎛ 0.00119 ⎞ ⎟ × 100% = ⎜1 − ⎟ × 100% = 95% 0.025 ⎠ ⎝ ⎠

10.28. Model: In this case of a one-dimensional collision, the momentum conservation law is obeyed whether the collision is perfectly elastic or perfectly inelastic. Visualize:

Solve: In the case of a perfectly elastic collision, the two velocities (vfx )1 and (vfx ) 2 can be determined by combining the conservation equations of momentum and mechanical energy. By contrast, a perfectly inelastic collision involves only one final velocity vfx and can be determined from just the momentum conservation equation. (a)

Momentum conservation: m1 (vix )1 + m2 (vix ) 2 = m1 (vfx )1 + m2 (vfx ) 2

1 1 1 1 m1 (vix )12 + m2 (vix ) 22 = m1 (vfx )12 + m2 (vfx ) 22 2 2 2 2 These two equations can be solved as shown in Equations 10.39 through 10.43: m − m2 (100 g) − (300 g) (vfx )1 = 1 (vix )1 = (10 m/s) = −5.0 m/s m1 + m2 (100 g) + (300 g) Energy conservation:

(vfx ) 2 =

2m1 2(100 g) (vix )1 = (10 m/s) = +5.0 m/s m1 + m2 (100 g) + (300 g)

(b) For the inelastic collision, both balls travel with the same final speed vfx . The momentum conservation equation

pfx = pix is (m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2 ⎛ ⎞ 100 g ⇒ vfx = ⎜ ⎟ (10 m/s) + 0 m/s = 2.5 m/s ⎝ 100 g + 300 g ⎠ Assess: In the case of the perfectly elastic collision, the two balls bounce off each other with a speed of 5.0 m/s. In the case of the perfectly inelastic collision, the balls stick together and move together at 2.5 m/s.

10.29. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize:

The particle is released from rest at x = 1.0 m. That is, K = 0 at x = 1.0 m. Since the total energy is given by E = K + U , we can draw a horizontal total energy (TE) line through the point of intersection of the potential energy curve (PE) and the x = 1.0 m line. The distance from the PE curve to the TE line is the particle’s kinetic energy. These values are transformed as the position changes, causing the particle to speed up or slow down, but the sum K + U does not change. Solve: (a) We have E = 4.0 J and this energy is a constant. For x < 1.0, U > 4.0 J and, therefore, K must be negative to keep E the same (note that K = E − U or K = 4.0 J − U ). Since negative kinetic energy is unphysical, the particle can not move to the left. That is, the particle will move to the right of x = 1.0 m. (b) The expression for the kinetic energy is E − U . This means the particle has maximum speed or maximum kinetic energy when U is minimum. This happens at x = 4.0 m. Thus,

K max = E − U min = (4.0 J) − (1.0 J) = 3.0 J

1 2 2(3.0 J) 8.0 J mvmax = 3.0 J ⇒ vmax = = = 17.3 m/s 2 m 0.020 kg

The particle possesses this speed at x = 4.0 m. (c) The total energy (TE) line intersects the potential energy (PE) curve at x = 1.0 m and x = 6.0 m. These are the turning points of the motion.

10.30. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize:

The particle with a mass of 500 g is released from rest at A. That is, K = 0 at A. Since E = K + U = 0 J + U , we can draw a horizontal TE line through U = 5.0 J. The distance from the PE curve to the TE line is the particle’s kinetic energy. These values are transformed as the position changes, causing the particle to speed up or slow down, but the sum K + U does not change. Solve: The kinetic energy is given by E − U , so we have

1 2 mv = E − U ⇒ v = 2( E − U )/ m 2 Using U B = 2.0 J, U C = 3.0 J, and U D = 0 J, we get vB = 2(5.0 J − 2.0 J)/0.500 kg = 3.5 m/s vD = 2(5.0 J − 0 J)/0.500 kg = 4.5 m/s

vC = 2(5.0 J − 3.0 J)/0.500 kg = 2.8 m/s

10.31. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize:

Since the particle oscillates between x = 2.0 mm and x = 8.0 mm, the speed of the particle is zero at these points. That is, for these values of x, E = U = 5.0 J, which defines the total energy (TE) line. The distance from the potential energy (PE) curve to the TE line is the particle’s kinetic energy. These values are transformed as the position changes, but the sum K + U does not change. Solve: The equation for total energy E = U + K means K = E − U , so that K is maximum when U is minimum. We have

⇒ vmax

1 2 K max = mvmax = 5.0 J − U min 2 = 2(5.0 J − U min )/ m = 2(5.0 J − 1.0 J)/0.0020 kg = 63 m/s

10.32. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize:

For the speed of the particle at A that is needed to reach B to be a minimum, the particle’s kinetic energy as it reaches the top must be zero. Similarly, the minimum speed at B for the particle to reach A obtains when the particle just makes it to the top with zero kinetic energy. Solve: (a) The energy equation K A + U A = K top + U top is

1 2 mvA + U A = 0 J + U top 2 ⇒ vA = 2(U top − U A )/ m = 2(5.0 J − 2.0 J)/0.100 kg = 7.7 m/s (b) To go from point B to point A, K B + U B = K top + U top is

1 2 mvB + U B = 0 J + U top 2 ⇒ vB = 2(U top − U B )/ m = 2(5.0 J − 0 J)/0.100 kg = 10.0 m/s Assess:

The particle requires a higher kinetic energy to reach A from B than to reach B from A.

10.33. Model: Model your vehicle as a particle. Assume zero rolling friction, so that the sum of your kinetic and gravitational potential energy does not change as the vehicle coasts down the hill. Visualize:

The figure shows a before-and-after pictorial representation. Note that neither the shape of the hill nor the angle of the downward slope is given, since these are not needed to solve the problem. All we need is the change in potential energy as you and your vehicle descend to the bottom of the hill. Also note that

35 km / hr = (35,000 m / 3600 s) = 9.722 m / s Solve:

Using yf = 0 and the equation K i + U gi = K f + U gf we get

1 2 1 mvi + mgyi = mvf2 + mgyf ⇒ vi2 + 2 gyi = vf2 2 2 ⇒ vf = vi2 + 2 gyi = (9.722 m/s) 2 + 2(9.8 m/s 2 )(15 m) = 19.7 m/s = 71 km/h You are driving over the speed limit. Yes, you will get a ticket. Assess: A speed of 19.7 m/s or 71 km/h at the bottom of the hill, when your speed at the top of the hill was 35 km/s, is reasonable. From the energy bar chart, we see that the initial potential energy is completely transformed into the final kinetic energy.

10.34. Model: This is case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the cannon ball falls. Visualize:

The figure shows a before-and-after pictorial representation. To express the gravitational potential energy, we put the origin of our coordinate system on the ground below the fortress. Solve: Using yf = 0 and the equation K i + U gi = K f + U gf we get

1 2 1 mvi + mgyi = mvf2 + mgyf ⇒ vi2 + 2 gyi = vf2 2 2 vf = vi2 + 2 gyi = (80 m/s) 2 + 2(9.8 m/s 2 )(10 m) = 81 m/s Assess: Note that we did not need to use the tilt angle of the cannon, because kinetic energy is a scalar. Also note that using the energy conservation equation, we can find only the magnitude of the final velocity, not the direction of the velocity vector.

10.35. Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the rock is thrown. Model the Frisbee and the rock as particles. Visualize:

The coordinate system is put on the ground for this system, so that yf = 16 m. The rock’s final velocity vf must be at least 5.0 m/s to dislodge the Frisbee. Solve: (a) The energy conservation equation for the rock K f + U gf = K i + U gi is

1 2 1 mvf + mgyf = mvi2 + mgyi 2 2 This equation involves only the velocity magnitudes and not the angle at which the rock is to be thrown to dislodge the Frisbee. This equation is true for all angles that will take the rock to the Frisbee. (b) Using the above equation we get

vf2 + 2 gyf = vi2 + 2 gyi

vi = vf2 + 2 g ( yf − yi ) = (5.0 m/s)2 + 2(9.8 m/s 2 )(16 m − 2.0 m) = 17.3 m/s

Assess: Kinetic energy is defined as K = 12 mv 2 and is a scalar quantity. Scalar quantities do not have directional properties.

10.36. Model: For the ice cube sliding around the inside of a smooth pipe, the sum of the kinetic and gravitational potential energy does not change. Visualize:

We use a coordinate system with the origin at the bottom of the pipe, that is, y1 = 0. The radius (R) of the pipe is 10 cm, and therefore ytop = y2 = 2 R = 0.20 m. At an arbitrary angle θ , measured counterclockwise from the bottom of the circle, y = R − R cosθ . Solve: (a) The energy conservation equation K 2 + U g2 = K1 + U g1 is

1 1 ⇒ mv22 + mgy2 = mv12 + mgy1 2 2 ⇒ v2 = v12 + 2 g ( y1 − y2 ) = (3.0 m/s) 2 + 2(9.8 m/s 2 )(0 m − 0.20 m) = 2.25 m/s (b) Expressing the energy conservation equation as a function of θ :

1 1 K (θ ) + U g (θ ) = K1 + U g1 ⇒ mv 2 (θ ) + mgy (θ ) = mv12 + 0 J 2 2 ⇒ v (θ ) = v12 − 2 gy (θ ) = v12 − 2 gR (1 − cosθ ) Using v1 = 3.0 m/s, g = 9.8 m/s 2 , and R = 0.10 m, we get v(θ ) = 9 − 1.96(1 − cosθ ) (m/s) (c) The accompanying figure shows a graph of v for a complete revolution (0° ≤ θ ≤ 360°).

Assess: Beginning with a speed of 3.0 m/s at the bottom, the marble’s potential energy increases and kinetic energy decreases as it gets toward the top of the circle. At the top, its speed is 2.25 m/s. This is reasonable since some of the kinetic energy has been transformed into the marble’s potential energy.

10.37. Model: Assume that the rubber band behaves similar to a spring. Also, model the rock as a particle. Visualize:

Please refer to Figure P10.37. Solve: (a) The rubber band is stretched to the left since a positive spring force on the rock due to the rubber band results from a negative displacement of the rock. That is, ( Fsp ) x = − kx, where x is the rock’s displacement

from the equilibrium position due to the spring force Fsp . (b) Since the Fsp versus x graph is linear with a negative slope and can be expressed as Fsp = − kx, the rubber

band obeys Hooke’s law. (c) From the graph, |ΔFsp | = 20 N for |Δx| = 10 cm. Thus, k=

|ΔFsp | |Δx|

=

20 N = 200 N/m = 2.0 × 102 N/m 0.10 m

(d) The conservation of energy equation K f + U sf = Ki + U si for the rock is

1 2 1 2 1 2 1 2 1 1 1 1 mvf + kxf = mvi + kxi ⇒ mvf2 + k (0 m) 2 = m(0 m/s) 2 + kxi2 2 2 2 2 2 2 2 2 k 200 N/m vf = xi = (0.30 m) = 19.0 m/s m 0.050 kg Assess: Note that xi is Δx, which is the displacement relative to the equilibrium position, and xf is the equilibrium position of the rubber band, which is equal to zero.

10.38. Model: Model the block as a particle and the springs as ideal springs obeying Hooke’s law. There is no friction, hence the mechanical energy K + U s is conserved. Visualize:

Note that xf = xe and xi − xe = Δx. The before-and-after pictorial representations show that we put the origin of the coordinate system at the equilibrium position of the free end of the springs. Solve: The conservation of energy equation K f + U sf = Ki + U si for the single spring is

1 2 1 1 1 mvf + k ( xf − xe ) 2 = mvi2 + k ( xi − xe ) 2 2 2 2 2 Using the value for vf given in the problem, we get 1 2 1 1 1 mv0 + 0 J = 0 J + k (Δx) 2 ⇒ mv02 = k (Δx ) 2 2 2 2 2

Conservation of energy for the two-spring case: 1 1 1 mVf2 + 0 J = 0 J + k ( xi − xe ) 2 + k ( xi − xe ) 2 2 2 2 Using the result of the single-spring case,

1 mVf2 = k (Δx ) 2 2

1 mVf2 = mv02 ⇒ Vf = 2v 0 2 Assess:

The block separates from the spring at the equilibrium position of the spring.

10.39. Model: Model the block as a particle and the springs as ideal springs obeying Hooke’s law. There is no friction, hence the mechanical energy K + U s is conserved. Visualize:

The springs in both cases have the same compression Δx. We put the origin of the coordinate system at the equilibrium position of the free end of the spring for the single-spring case (a), and at the free end of the two connected springs for the two-spring case (b). Solve: The conservation of energy for the single-spring case: 1 1 1 1 K f + U sf = K i + U si ⇒ mvf2 + k ( xf − xe ) 2 = mvi2 + k ( xi − xe ) 2 2 2 2 2 Using xf = xe = 0 m, vi = 0 m/s, and vf = v0, this equation simplifies to

1 2 1 mv0 = k (Δx) 2 2 2 Conservation of energy in the case of the two springs in series, where each spring compresses by Δx /2, is 1 1 1 1 K f + U sf = K i + U si ⇒ mVf2 + 0 = mvi2 + k (Δx /2) 2 + k (Δx /2) 2 2 2 2 2 Using xf = xe′ = 0 m and vi = 0 m/s, we get 1 1 ⎡1 ⎤ mVf2 = ⎢ k (Δx) 2 ⎥ 2 2 ⎣2 ⎦ Comparing the two results we see that Vf = v0 / 2. Assess: The block pushes on the spring until the spring has returned to its equilibrium length.

10.40. Model: Model the ball and earth as particles. An elastic collision between the ball and earth conserves both momentum and mechanical energy. Visualize:

The before-and-after pictorial representation of the collision is shown in the figure. The ball as it is dropped from a height of 10 m has zero velocity. Its speed just before it hits earth can be found using kinematics. We can then apply the conservation laws to find earth’s recoil velocity. Solve: (a) To get the speed of the ball at collision:

(v1 ) 2B = (v0 ) B2 − 2 g ( y1 − y0 ) B = (0 m/s) − 2(9.8 m/s 2 )(−10.0 m) ⇒ (v1 ) B = 2(9.8 m/s 2 )(10 m) = 14.0 m/s In an elastic collision, the velocity of the object that is struck is

(v2 ) E =

2mB 2(0.500 kg) (v1 ) B = (14.0 m / s) = 2.3 × 10−24 m/s mE + mB 5.98 × 1024 kg

(b) The time it would take earth to move 1.0 mm at this speed is given by speed =

distance 1.0 × 10−3 m = = 4.3 × 1020 s = 1.36 × 1013 years velocity 2.3 × 10−24 m/s

10.41. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, and therefore the mechanical energy K + U s + U g is conserved. Visualize:

The figure shows a before-and-after pictorial representation. We have chosen to place the origin of the coordinate system at the position where the ice cube has compressed the spring 10 cm. That is, y0 = 0. Solve:

The energy conservation equation K 2 + U s2 + U g2 = K 0 + U s0 + U g0 is 1 2 1 1 1 mv2 + k ( xe − xe ) 2 + mgy0 = mv02 + k ( x − xe ) 2 + mgy0 2 2 2 2

Using v2 = 0 m / s, y0 = 0 m, and v0 = 0 m / s,

k ( x − xe ) 2 1 (25 N/m)(0.10 m) 2 = = 25.5 cm mgy2 = k ( x − xe ) 2 ⇒ y2 = 2 2 mg 2(0.050 kg)(9.8 m/s 2 ) The distance traveled along the incline is y2 / sin 30° = 25.5 cm/ sin 30° = 51 cm. Assess: The net effect of the launch is to transform the potential energy stored in the spring into gravitational potential energy. The block has kinetic energy as it comes off the spring, but we did not need to know this energy to solve the problem.

10.42. Model: Model the two packages as particles. Momentum is conserved in both inelastic and elastic collisions. Kinetic energy is conserved only in a perfectly elastic collision. Visualize:

Solve:

For a package with mass m the conservation of energy equation is

1 1 K1 + U g1 = K 0 + U g0 ⇒ m(v1 ) 2m + mgy1 = m(v0 ) m2 + mgy0 2 2

Using (v0 ) m = 0 m / s and y1 = 0 m, 1 m(v1 ) 2m = mgy0 ⇒ (v1 ) m = 2 gy0 = 2(9.8 m/s 2 )(3.0 m) = 7.668 m/s 2 (a) For the perfectly inelastic collision the conservation of momentum equation is

pfx = pix ⇒ ( m + 2m)(v2 )3m = m(v1 ) m + (2m)(v1 ) 2m Using (v1 ) 2m = 0 m / s, we get (v2 )3m = (v1 ) m / 3 = 2.56 m/s The packages move off together at a speed of 2.6 m/s. (b) For the elastic collision, the mass m package rebounds with velocity m − 2m 1 (v1 ) m = − ( 7.668 m/s ) = −2.56 m/s m + 2m 3 The negative sign with (v3 ) m shows that the package with mass m rebounds and goes to the position y4 . We can (v3 ) m =

determine y4 by applying the conservation of energy equation as follows. For a package of mass m: 1 1 K f + U gf = K i + U gi ⇒ m(v4 ) 2m + mgy4 = m(v3 ) m2 + mgy3 2 2 Using (v3 )m = −2.55 m/s, y3 = 0 m, and (v4 ) m = 0 m/s, we get 1 mgy4 = m(−2.56 m/s) 2 ⇒ y4 = 33 cm 2

10.43. Model: Model the marble and the steel ball as particles. We will assume an elastic collision between the marble and the ball, and apply the conservation of momentum and the conservation of energy equations. We will also assume zero rolling friction between the marble and the incline. Visualize:

This is a two-part problem. In the first part, we will apply the conservation of energy equation to find the marble’s speed as it exits onto a horizontal surface. We have put the origin of our coordinate system on the horizontal surface just where the marble exits the incline. In the second part, we will consider the elastic collision between the marble and the steel ball. Solve: The conservation of energy equation K1 + U g1 = K 0 + U g0 gives us:

1 1 2 2 mM (v1 ) M + mM gy1 = mM (v0 ) M + mM gy0 2 2 1 2 (v1 ) M = gy0 ⇒ (v1 )M = 2 gy0 . When the marble collides with the 2 steel ball, the elastic collision gives the ball velocity Using (v0 ) M = 0 m / s and y1 = 0 m, we get

(v2 )S =

2mM 2mM (v1 ) M = 2 gy0 mM + mS mM + mS

Solving for y0 gives 2

⎤ 1 ⎡ mM + mS y0 = (v2 )S ⎥ = 0.258 m = 25.8 cm ⎢ 2 g ⎣ 2mM ⎦

10.44. Model: Assume an ideal spring that obeys Hooke’s law. Since this is a free-fall problem, the mechanical energy K + U g + U s is conserved. Also, model the safe as a particle. Visualize:

We have chosen to place the origin of our coordinate system at the free end of the spring, which is neither stretched nor compressed. The safe gains kinetic energy as it falls. The energy is then converted into elastic potential energy as the safe compresses the spring. The only two forces are gravity and the spring force, which are both conservative, so energy is conserved throughout the process. This means that the initial energy—as the safe is released—equals the final energy—when the safe is at rest and the spring is fully compressed. Solve: The conservation of energy equation K1 + U g1 + U s1 = K 0 + U g0 + U s0 is

1 2 1 1 1 mv1 + mg ( y1 − ye ) + k ( y1 − ye ) 2 = mv02 + mg ( y0 − ye ) + k ( ye − ye ) 2 2 2 2 2 Using v1 = v0 = 0 m/s and ye = 0 m, the above equation simplifies to 1 mgy1 + ky12 = mgy0 2

⇒k =

2mg ( y0 − y1 ) 2(1000 kg)(9.8 m/s 2 )(2.0 m − ( −0.50 m)) = = 1.96 × 105 N/m y12 ( −0.50 m) 2

Assess: By equating energy at these two points, we do not need to find how fast the safe was moving when it hit the spring.

10.45. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction and hence the mechanical energy K + U s + U g is conserved. Visualize:

Solve:

(a) When releasing the block suddenly, K 2 + U s2 + U g 2 = K1 + U s1 + U g1 1 2 1 1 1 mv2 + k ( y2 − ye ) 2 + mgy2 = mv12 + k ( y1 − ye ) 2 + mgy1 2 2 2 2

Using v2 = 0 m/s, v1 = 0 m/s, and y1 = ye , we get

1 0 J + (490 N/m)( y2 − y1 ) 2 + mgy2 = 0 J + 0 J + mgy1 ⇒ (245 N/m)( y2 − y1 ) 2 = mg ( y1 − y2 ) 2 ⇒ (245 N/m)( y1 − y2 ) 2 = (5.0 kg)(9.8 m/s 2 )( y1 − y2 ) ⇒ ( y1 − y2 ) = 0.20 m (b) When lowering the block gently until it rests on the spring, the block reaches a point of static equilibrium. Fnet = k Δy − mg = 0 ⇒ Δy =

mg (5.0 kg)(9.8 m/s 2 ) = = 0.10 m k 490 N/m

(c) In part (b), at a point 0.10 m down, the forces balance. But in part (a) the block has kinetic energy as it reaches 0.10 m. So the block continues on past the equilibrium point until all the gravitational potential energy is stored in the spring.

10.46. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, and thus the mechanical energy K + U s + U g is conserved. Visualize:

We place the origin of our coordinate system at the spring’s compressed position y1 = 0. The rock leaves the

spring with velocity v2 as the spring reaches its equilibrium position. Solve: (a) The conservation of mechanical energy equation is 1 2 1 1 1 K 2 + U s 2 + U g 2 = K1 + U s1 + U g1 mv2 + k ( y2 − ye ) 2 + mgy2 = mv12 + k ( y1 − ye ) 2 + mgy1 2 2 2 2 Using y2 = ye , y1 = 0 m, and v1 = 0 m/s, this simplifies to 1 2 1 mv2 + 0 J + mgy2 = 0 J + k ( y1 − ye ) 2 + 0 2 2 1 1 (0.400 kg)v22 + (0.400 kg)(9.8 m/s 2 )(0.30 m) = (1000 N/m)(0.30 m) 2 ⇒ v2 = 14.8 m/s 2 2 (b) Let us use the conservation of mechanical energy equation once again to find the highest position ( y3 ) of the rock where its speed (v3 ) is zero:

1 1 K 3 + U g3 = K 2 + U g 2 ⇒ mv32 + mgy3 = mv22 + mgy2 2 2 1 v 2 (14.8 m/s) 2 ⇒ 0 + g ( y3 − y2 ) = v22 ⇒ ( y3 − y2 ) = 2 = = 11.2 m 2 2 g 2(9.8 m/s 2 ) If we assume the spring’s length to be 0.5 m, then the distance between ground and fruit is 11.2 m + 0.5 m = 11.7 m. This is much smaller than the distance of 15 m between fruit and ground. So, the rock does not reach the fruit, and the contestants go hungry.

10.47. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, so the mechanical energy K + U g + U s is conserved. Visualize:

Place the origin of the coordinate system at the end of the unstretched spring, making ye = 0 m. Solve: The clay is in static equilibrium while resting in the pan. The net force on it is zero. We can start by using this to find the spring constant.

Fsp = FG ⇒ − k ( y1 − ye ) = − ky1 = mg ⇒ k = −

mg (0.10 kg)(9.8 m/s2 ) =− = 9.8 N/m y1 −0.10 m

Now apply conservation of energy. Initially, the spring is unstretched and the clay ball is at the ceiling. At the end, the spring has maximum stretch and the clay is instantaneously at rest. Thus K 2 + (U g ) 2 + (U s ) 2 = K 0 + (U g )0 + (U s )0 ⇒ 12 mv22 + mgy2 + 12 ky22 = 12 mv02 + mgy0 + 0 J

Since v0 = 0 m/s and v2 = 0 m/s, this equation becomes

2mg 2mgy0 y2 − =0 k k y22 + 0.20 y2 − 0.10 = 0

mgy2 + 12 ky22 = mgy0 ⇒ y22 +

The numerical values were found using known values of m, g, k, and y0 . The two solutions to this quadratic equation are y2 = 0.231 m and y2 = −0.432 m. The point we’re looking for is below the origin, so we need the negative root. The distance of the pan from the ceiling is

L = y2 + 50 cm = 93 cm

10.48. Model: Assume an ideal spring that obeys Hooke’s law. Also assume zero rolling friction between the roller coaster and the track, and a particle model for the roller coaster. Since no friction is involved, the mechanical energy K + U s + U g is conserved. Visualize:

We have chosen to place the origin of the coordinate system on the end of the spring that is compressed and touches the roller coaster car. Solve: (a) The energy conservation equation for the car going to the top of the hill is K 2 + U g 2 + U s2 = K 0 + U g0 + U s0

1 2 1 1 1 mv2 + mgy2 + k ( xe − xe ) 2 = mv02 + mgy0 + k ( x0 − xe ) 2 2 2 2 2

Noting that y0 = 0 m, v2 = 0 m/s, v1 = 0 m/s, and x0 − xe = 2.0 m, we obtain 1 0 J + mgy2 + 0 J = 0 J + 0 J + k (2.0 m) 2 2 2mgy2 2(400 kg)(9.8 m/s 2 )(10 m) ⇒k = = = 1.96 × 104 N/m (2.0 m) 2 (2.0 m) 2 We now increase this value for k by 10% for safety, giving a value of 2.156 × 104 N/m ≈ 2.2 × 104 N/m. (b) The energy conservation equation K 3 + U g3 + U s3 = K 0 + U g 0 + U s0 is 1 2 1 1 1 mv3 + mgy3 + k ( xe − xe ) 2 = mv02 + mgy0 + k ( x0 − xe ) 2 2 2 2 2

Using y3 = −5.0 m, v0 = 0 m / s, y0 = 0 m, and | x0 − xe | = 2.0 m, we get 1 2 1 mv3 + mg (−5.0 m) + 0 J = 0 J + 0 J + k ( x0 − xe ) 2 2 2 1 1 (400 kg)v32 − (400 kg)(9.8 m/s 2 )(5.0 m) = (2.156 × 104 N/m)(2.0 m) 2 2 2 ⇒ v3 = 17.7 m/s

10.49. Model: Since there is no friction, the sum of the kinetic and gravitational potential energy does not change. Model Julie as a particle. Visualize:

We place the coordinate system at the bottom of the ramp directly below Julie’s starting position. From geometry, Julie launches off the end of the ramp at a 30º angle. 1 1 Solve: Energy conservation: K1 + U g1 = K 0 + U g0 ⇒ mv12 + mgy1 = mv02 + mgy0 2 2 Using v0 = 0 m/s, y0 = 25 m, and y1 = 3 m, the above equation simplifies to

1 2 mv1 + mgy1 = mgy0 ⇒ v1 = 2 g ( y0 − y1 ) = 2(9.8 m/s 2 )(25 m − 3 m) = 20.77 m/s 2 We can now use kinematic equations to find the touchdown point from the base of the ramp. First we’ll consider the vertical motion: 1 1 y2 = y1 + v1 y (t2 − t1 ) + a y (t2 − t1 ) 2 0 m = 3 m + (v1 sin 30°)(t2 − t1 ) + (−9.8 m / s 2 )(t2 − t1 ) 2 2 2 (20.77 m/s)sin 30° (3 m) 2 ⇒ (t2 − t1 ) − =0 (t2 − t1 ) − (4.9 m/s 2 ) (4.9 m/s 2 ) (t2 − t1 ) 2 − (2.119 s)(t2 − t1 ) − (0.6122 s 2 ) = 0 ⇒ (t2 − t1 ) = 2.377 s For the horizontal motion: 1 x2 = x1 + v1x (t2 − t1 ) + ax (t2 − t1 ) 2 2 x2 − x1 = (v1 cos30°)(t2 − t1 ) + 0 m = (20.77 m/s)(cos30°)(2.377 s) = 43 m Assess: Note that we did not have to make use of the information about the circular arc at the bottom that carries Julie through a 90° turn.

10.50. Model: We assume the spring to be ideal and to obey Hooke’s law. We also treat the block (B) and the ball (b) as particles. In the case of an elastic collision, both the momentum and kinetic energy equations apply. On the other hand, for a perfectly inelastic collision only the equation of momentum conservation is valid. Visualize:

Place the origin of the coordinate system on the block that is attached to one end of the spring. The before-andafter pictorial representations of the elastic and perfectly inelastic collision are shown in figures (a) and (b), respectively. Solve: (a) For an elastic collision, the ball’s rebound velocity is

(vf ) b =

−80 g mb − mB (vi ) b = (5.0 m/s) = −3.33 m/s mb + mB 120 g

The ball’s speed is 3.3 m/s. (b) An elastic collision gives the block speed

(vf ) B =

2mB 40 g (vi ) b = (5.0 m/s) = 1.667 m/s mb + mB 120 g

To find the maximum compression of the spring, we use the conservation equation of mechanical energy for the block + spring system. That is K1 + U s1 = K 0 + U s0 : 1 1 1 1 mB (vf′ ) 2B + k ( x1 − x0 ) 2 = mB (vf ) B2 + k ( x0 − x0 ) 2 2 2 2 2

0 + k ( x1 − x0 ) 2 = mB (vf )B2 + 0

(x1 − x0 ) = (0.100 kg )(1.667 m/s )2 /(20 N/m ) = 11.8 cm (c) Momentum conservation pf = pi for the perfectly inelastic collision means (mB + mB )vf = mb (vi ) b + mB (vi ) B (0.100 kg + 0.020 kg)vf = (0.020 kg)(5.0 m/s) + 0 m/ v ⇒ vf = 0.833 m/s

The maximum compression in this case can now be obtained using the conservation of energy equation K1 + U s1 = K 0 + U s0 :

0 J + (1/ 2)k (Δx) 2 = (1/ 2)(mB + mb )vf 2 + 0 J ⇒ Δx =

mB + mb 0.120 kg vf = (0.833m/s) = 0.0645 m = 6.5 cm k 20 N/m

10.51.

Model: Assume an ideal spring that obeys Hooke’s law. We treat the bullet and the block in the particle model. For a perfectly inelastic collision, the momentum is conserved. Furthermore, since there is no friction, the mechanical energy of the system (bullet + block + spring) is conserved.

Visualize:

We place the origin of our coordinate system at the end of the spring that is not anchored to the wall. Solve: (a) Momentum conservation for perfectly inelastic collision states pf = pi . This means ⎛ m (m + M )vf = m(vi ) m + M (vi ) M ⇒ (m + M )vf = mvB + 0 kg m/s ⇒ vf = ⎜ ⎝m+M

⎞ ⎟ vB ⎠

where we have used vB for the initial speed of the bullet. The mechanical energy conservation equation K1 + U s1 = K e + U se as the bullet embedded block compresses the spring is: 1 1 1 1 m(v′f ) 2 + k ( x1 − xe ) 2 = (m + M )(vf ) 2 + k ( xe − xe ) 2 2 2 2 2 2

1 1 (m + M )kd 2 ⎛ m ⎞ 2 0 J + kd 2 = (m + M ) ⎜ 0 J v + ⇒ v = B ⎟ B 2 2 m2 ⎝m+M ⎠ (b) Using the above formula with m = 5.0 g, M = 2.0 kg, k = 50 N/m, and d = 10 cm,

vB = (0.0050 kg + 2.0 kg)(50 N/m)(0.10 m) 2 /(0.0050) 2 = 2.0 × 102 m/s (c) The fraction of energy lost is 1 2 1 2 2 mvB − (m + M )vf2 m+M ⎛ vf ⎞ m+M ⎛ m ⎞ 2 2 =1− = − 1 ⎜ ⎟ ⎜ ⎟ 1 2 m ⎝ vB ⎠ m ⎝m+M ⎠ mvB 2 m 0.0050 kg =1− =1− = 0.9975 m+M (0.0050 kg + 2.0 kg)

That is, during the perfectly inelastic collision 99.75% of the bullet’s energy is lost. The energy is dissipated inside the block. Although it is common to say, “The energy is lost to heat,” in the next chapter we’ll see that it is more accurate to say, “The energy is transformed to thermal energy.”

10.52. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanical energy K + U g + U s is conserved. Visualize:

We have chosen to place the origin of the coordinate system on the free end of the spring that is neither stretched nor compressed, that is, at the equilibrium position of the end of the unstretched spring. The bullet’s mass is m and the block’s mass is M. Solve: (a) The energy conservation equation K 2 + U s2 + U g2 = K1 + U s1 + U g1 becomes

1 1 1 1 (m + M )v22 + k ( y2 − ye )2 + (m + M ) g ( y2 − ye ) = (m + M )v12 + k ( y1 − ye )2 − (m + M ) g ( y1 − ye ) 2 2 2 2 Noting v2 = 0 m/s, we can rewrite the above equation as

k (Δy2 ) 2 + 2( m + M ) g ( Δy2 + Δy1 ) = (m + M )v12 + k (Δy1 ) 2

Let us express v1 in terms of the bullet’s initial speed vB by using the momentum conservation equation pf = pi which is (m + M )v1 = mvB + Mvblock . Since vblock = 0 m/s, we have ⎛ m ⎞ v1 = ⎜ ⎟ vB ⎝m+M ⎠

We can also find the magnitude of y1 from the equilibrium condition k ( y1 − ye ) = Mg . Δy1 =

Mg k

With these substitutions for v1 and Δy1 , the energy conservation equation simplifies to k (Δy2 ) 2 + 2(m + M ) g (Δy1 + Δy2 ) =

m 2vB2 ⎛ Mg ⎞ + k⎜ ⎟ (m + M ) ⎝ k ⎠

2

2

(m + M ) M 2 g 2 ⎛m+M ⎞ ⎛m+M ⎞ 2 ⇒ vB2 = 2 ⎜ Δ + Δ − + k⎜ g y y ( ) 1 2 ⎟ ⎟ ( Δy 2 ) 2 2 m k ⎝ m ⎠ ⎝ m ⎠ We still need to include the spring’s maximum compression (d) into this equation. We assume that d = Δy1 + Δy2 , that is, maximum compression is measured from the initial position ( y1 ) of the block. Thus, using Δy2 = d − Δy1 = (d − Mg / k ), we have ⎡ ⎛ m + M ⎞2 ⎛m+M vB = ⎢ 2 ⎜ ⎟ gd − ⎜ 2 ⎝ m ⎢⎣ ⎝ m ⎠

2 2 ⎞M g ⎛m+M + k⎜ ⎟ 2 ⎠ k ⎝ m

12

⎤ ⎞ 2 ⎟ ( d − Mg / k ) ⎥ ⎠ ⎥⎦

(b) Using m = 0.010 kg, M = 2.0 kg, k = 50 N/m, and d = 0.45 m, 2

2

⎛ 2.0 kg ⎞ ⎛ 2.010 kg ⎞ 2 2 2 vB2 = 2 ⎜ ⎟ (9.8 m/s ) /(50 N/m) ⎟ (9.8 m/s )(0.45 m) − (2.010 kg) ⎜ 0.010kg 0.010 kg ⎝ ⎠ ⎝ ⎠ 1 + (50 N/m)(2.010 kg) [0.45 m − (2.0 kg) × (9.8 m/s 2 )/50 N/m]2 (0.010 kg) 2 ⇒ vB = 453 m/s The bullet has a speed of 4.5 × 10 2 m/s. Assess: This is a reasonable speed for the bullet.

10.53. Model: Because the track is frictionless, the sum of the kinetic and gravitational potential energy does not change during the car’s motion. Visualize:

We place the origin of the coordinate system at the ground level directly below the car’s starting position. This is a two-part problem. If we first find the maximum speed at the top of the hill, we can use energy conservation to find the maximum initial height. Solve: Because its motion is circular, at the top of the hill the car has a downward-pointing centripetal accelG eration ac = −(mv 2 / r ) ˆj. Newton’s second law at the top of the hill is

( Fnet ) y = n y + ( FG ) y = n − mg = m( ac ) y = −

⎛ mv 2 mv 2 v2 ⎞ ⇒ n = mg − = m⎜ g − ⎟ R R R⎠ ⎝

If v = 0 m / s, n = mg as expected in static equilibrium. As v increases, n gets smaller—the weight of the car and passengers decreases as they go over the top. But n has to remain positive for the car to be on the track, so the maximum speed vmax occurs when n → 0. We see that vmax = gR . Now we can use energy conservation to relate the top of the hill to the starting height:

1 1 1 K f + U f = Ki + U i ⇒ mvf2 + mgyf = mv12 + mgy1 ⇒ mgR + mgR = 0 J + mghmax 2 2 2 where we used vf = vmax and yf = R. Solving for hmax gives hmax = 32 R. (b) If R = 10 m, then hmax = 15 m.

10.54. Model: This is a two-part problem. In the first part, we will find the critical velocity for the block to go over the top of the loop without falling off. Since there is no friction, the sum of the kinetic and gravitational potential energy is conserved during the block’s motion. We will use this conservation equation in the second part to find the minimum height the block must start from to make it around the loop. Visualize:

We place the origin of our coordinate system directly below the block’s starting position on the frictionless track. Solve: The free-body diagram on the block implies mv 2 FG + n = c R For the block to just stay on track, n = 0. Thus the critical velocity vc is

FG = mg =

mvc2 ⇒ vc2 = gR R

The block needs kinetic energy 12 mvc2 = 12 mgR to go over the top of the loop. We can now use the conservation of mechanical energy equation to find the minimum height h. 1 1 K f + U gf = K i + U gi ⇒ mvf2 + mgyf = mvi2 + mgyi 2 2

Using vf = vc = gR , yf = 2 R, vi = 0 m/s, and yi = h, we obtain

1 gR + g (2 R ) = 0 + gh ⇒ h = 2.5R 2

10.55. Model: Model Lisa (L) and the bobsled (B) as particles. We will assume the ramp to be frictionless,

so that the mechanical energy of the system (Lisa + bobsled + spring) is conserved. Furthermore, during the collision, as Lisa leaps onto the bobsled, the momentum of the Lisa + bobsled system is conserved. We will also assume the spring to be an ideal one that obeys Hooke’s law. Visualize:

We place the origin of our coordinate system directly below the bobsled’s initial position. Solve: (a) Momentum conservation in Lisa’s collision with bobsled states p1 = p0 , or (mL + mB )v1 = mL (v0 ) L + mB (v0 ) B ⇒ (mL + mB )v1 = mL (v0 ) L + 0 ⎛ mL ⎞ ⎛ ⎞ 40 kg ⇒ v1 = ⎜ ⎟ (v0 ) L = ⎜ ⎟ (12 m/s) = 8.0 m/s m + m 40 kg + 20 kg ⎝ ⎠ B ⎠ ⎝ L The energy conservation equation: K 2 + U s2 + U g 2 = K1 + U s1 + U g1 is 1 1 1 1 (mL + mB )v22 + k ( x2 − xe ) 2 + ( mL + mB ) gy2 = (mL + mB )v12 + k ( xe − xe ) 2 + (mL + mB ) gy1 2 2 2 2 Using v2 = 0 m/s, k = 2000 N/m, y2 = 0 m, y1 = (50 m)sin 20° = 17.1 m, v1 = 8.0 m/s, and (mL mB ) = 60 kg, we get 1 1 0 J + (2000 N / m)( x2 − xe ) 2 + 0 J = (60 kg)(8.0 m / s) 2 + 0 J + (60 kg)(9.8 m / s 2 )(17.1 m) 2 2 Solving this equation yields ( x2 − xe ) = 3.5 m. (b) As long as the ice is slippery enough to be considered frictionless, we know from conservation of mechanical energy that the speed at the bottom depends only on the vertical descent Δy. Only the ramp’s height h is important, not its shape or angle.

10.56. Model: We can divide this problem into two parts. First, we have an elastic collision between the 20 g ball (m) and the 100 g ball (M). Second, the 100 g ball swings up as a pendulum. Visualize:

The figure shows three distinct moments of time: the time before the collision, the time after the collision but before the two balls move, and the time the 100 g ball reaches its highest point. We place the origin of our coordinate system on the 100 g ball when it is hanging motionless. Solve: For a perfectly elastic collision, the ball moves forward with speed

(v1 ) M =

2mm 1 (v0 ) m = (v0 ) m mm + mM 3

In the second part, the sum of the kinetic and gravitational potential energy is conserved as the 100 g ball swings up after the collision. That is, K 2 + U g 2 = K1 + U g1. We have 1 1 M (v2 ) 2M + Mgy2 = M (v1 ) 2M + Mgy1 2 2 Using (v2 ) M = 0 J, (v1 ) M =

(v0 ) m , y1 = 0 m, and y2 = L − L cosθ , the energy equation simplifies to 3

g ( L − L cosθ ) =

1 (v0 ) 2m 2 9

⇒ (v0 ) m = 18 g L(1 − cosθ ) = 18(9.8 m/s 2 )(1.0 m)(1 − cos50°) = 7.9 m/s

10.57. Model: Model the balls as particles. We will use the Galilean transformation of velocities (Equation 10.44) to analyze the problem of elastic collisions. We will transform velocities from the lab frame S to a frame S′ in which one ball is at rest. This allows us to apply Equations 10.43 to the case of a perfectly elastic collision in S′, find the final velocities of the balls in S′, and then transform these velocities back to the lab frame S. Visualize: Let S′ be the frame of the 200 g ball. Denoting masses as m1 = 100 g and m2 = 200 g, the initial velocities in the S frame are (vix )1 = 4 m/s and (vix ) 2 = −3 m/s.

Figure (a) shows the before-collision situation as seen in frame S, and figure (b) shows the before-collision situation as seen in frame S′. The after-collision velocities in S′ are shown in figure (c), and figure (d) indicates velocities in S after they have been transformed to frame S from S′. Solve: (a) In the S frame, (vix )1 = 4 m/s and (vix ) 2 = −3 m/s. S′ is the reference frame of the 200 g ball, so V = −3 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of velocities v′ = v − V . So, (vix )1′ = (vix )1 − V = 4 m/s − ( − 3 m/s) = 7 m/s

(vix )′2 = (vix ) 2 − V = −3 m/s − ( − 3 m/s) = 0 m/s

Figure (b) shows the “before” situation, where ball 2 is at rest. Now we can use Equations 10.43 to find the after-collision velocities in frame S′: (vfx )1′ =

m1 − m2 100 g − 200 g 7 (vix )1′ = (7 m/s) = − m/s m1 + m2 100 g + 200 g 3

(vfx )′2 =

2m1 2(100) g 14 (vix )1′ = (7 m/s) = m/s m1 + m2 100 g + 200 g 3

Finally, we need to apply the reverse Galilean transformation v = v′ + V , with the same V, to transform the aftercollision velocities back to the lab frame S: 7 (vfx )1 = (vfx )1′ + V = − m/s − 3 m/s = −5.33 m/s 3 14 (vfx ) 2 = (vfx )′2 + V = m/s − 3 m/s = 1.67 m/s 3 Figure (d) shows the “after” situation in the lab frame. The 100 g ball is moving left at 5.3 m/s; the 200 g ball is moving right at 1.7 m/s.

(b) The momentum conservation equation pfx = pix for a perfectly inelastic collision is

(m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2 (0.100 kg + 0.200 kg)vfx = (0.100 kg)(4.0 m/s) + (0.200 kg)(−3.0 m/s) ⇒ vfx = −0.667 m/s Both balls are moving left at 0.67 m/s.

10.58. Model: Model the balls as particles. We will use the Galilean transformation of velocities (Equation 10.44) to analyze the problem of elastic collisions. We will transform velocities from the lab frame S to a frame S′ in which one ball is at rest. This allows us to apply Equations 10.43 to a perfectly elastic collision in S′. After finding the final velocities of the balls in S′, we can then transform these velocities back to the lab frame S. Visualize: Let S′ be the frame of the 400 g ball. Denoting masses as m1 = 100 g and m2 = 400 g, the initial velocities in the S frame are (vix )1 = + 4.0 m/s and (vix ) 2 = +1.0 m/s.

Figures (a) and (b) show the before-collision situations in frames S and S′, respectively. The after-collision velocities in S′ are shown in figure (c). Figure (d) indicates velocities in S after they have been transformed to S from S′. Solve: In frame S, (vix )1 = 4.0 m/s and (vix ) 2 = 1.0 m/s. Because S′ is the reference frame of the 400 g ball, V = 1.0 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of velocities v′ = v − V . So, (vix )1′ = (vix )1 − V = 4.0 m/s − 1.0 m/s = 3.0 m/s (vix )′2 = (vix ) 2 − V = 1.0 m/s − 1.0 m/s = 0 m/s

Figure (b) shows the “before” situation in frame S′ where the ball 2 is at rest. Now we can use Equations 10.43 to find the after-collision velocities in frame S′: (vfx )1′ =

m1 − m2 100 g − 400 g (vix )1′ = (3.0 m/s) = −1.80 m/s m1 + m2 100 g + 400 g

(vfx )′2 =

2m1 2(100 g) (vix )1′ = (3.0 m/s) = 1.20 m/s m1 + m2 100 g + 400 g

Finally, we need to apply the reverse Galilean transformation v = v′ + V , with the same V, to transform the aftercollision velocities back to the lab frame S. (vfx )1 = (vfx )1′ + V = −1.80 m/s + 1.0 m/s = −0.80 m/s (vfx ) 2 = (vfx )′2 + V = 1.20 m/s + 1.0 m/s = 2.20 m/s

Figure (d) shows the “after” situation in frame S. The 100 g ball moves left at 0.80 m/s, the 400 g ball right at 2.2 m/s. Assess: The magnitudes of the after-collision velocities are similar to the magnitudes of the before-collision velocities.

10.59. Model: Use the model of the conservation of mechanical energy. Visualize:

Solve: (a) The turning points occur where the total energy line crosses the potential energy curve. For E = 12 J, this occurs at the points x = 1 m and x = 7 m. (b) The equation for kinetic energy K = E − U gives the distance between the potential energy curve and total energy line. U = 8 J at x = 2 m, so K = 12 J − 8 J = 4 J. The speed corresponding to this kinetic energy is

v=

2K 2(4 J) = = 4.0 m/s m 0.5 kg

(c) Maximum speed occurs for minimum U. This occurs at x = 1 m and x = 4 m, where U = 0 J and K = 12 J. The speed at these two points is

v=

2K 2(12 J) = = 6.9 m/s m 0.500 kg

(d) The particle leaves x = 1 m with v = 6.9 m/s. It gradually slows down, reaching x = 2 m with a speed of 4.0 m/s. After x = 2 m, it speeds up again, returning to a speed of 6.9 m/s as it crosses x = 4 m. Then it slows again, coming instantaneously to a halt (v = 0 m/s) at the x = 7 m turning point. Now it will reverse direction and move back to the left. (e) If the particle has E = 4 J it cannot cross the 8 J potential energy “mountain” in the center. It can either oscillate back and forth over the range 1.0 m ≤ x ≤ 1.5 m or over the range 3 m ≤ x ≤ 5 m.

10.60. Model: We will use the conservation of mechanical energy. Visualize:

The potential energy (U ) of the nitrogen atom as a function of z exhibits a double-minimum behavior; the two minima correspond to the nitrogen atom’s position on both sides of the plane containing the three hydrogens. Solve: (a) At room temperature, the total energy line is below the “hill” in the center of the potential energy curve. That is, the nitrogen atom does not have sufficient energy to pass from one side of the molecule to the other. There’s a stable equilibrium position on either side of the hydrogen-atom plane at the points where U = 0. Since E > 0, the nitrogen atom will be on one side of the plane and will make small vibrations back and forth along the z-axis—that is, toward and away from the hydrogen-atom plane. In the figure above, the atom oscillates between points A and B. (b) The total energy line is now well above the “hill,” and the turning points of the nitrogen atom’s motion (where the total energy line crosses the potential curve) are at points C and D. In other words, the atom oscillates from one side of the H3 plane to the other. It slows down a little as it passes through the plane of hydrogen atoms, but it has sufficient energy to get through.

dU = 0. dx dU 1 = 0 = 1 + 2cos ( 2 x ) ⇒ cos ( 2 x ) = − dx 2 1 ⎛ 1⎞ ⇒ x = cos −1 ⎜ − ⎟ 2 ⎝ 2⎠

10.61. Solve: (a) The equilibrium positions are located at points where

1 2π 4π ⎛ 1⎞ is in radians and x is in meters. The function cos −1 ⎜ − ⎟ may have values and . Thus 2 3 3 2 ⎝ ⎠ there are two values of x, π 2π x1 = and x2 = 3 3 within the interval 0 m ≤ x ≤ π m. (b) A point of stable equilibrium corresponds to a local minimum, while a point of unstable equilibrium corresponds to a local maximum. Compute the concavity of U(x) at the equilibrium positions to determine their stability. d 2U = −4sin ( 2 x ) dx 2 ⎛ 3⎞ d 2U π π d 2U π At x1 = , x < 0, x1 = is a local maximum, so x1 = is a point x = −4 ⎜⎜ ⎟⎟ = −2 3. Since 2 ( 1) 2 ( 1) 3 3 dx dx 3 2 ⎝ ⎠ of unstable equilibrium. ⎛ 2π d 2U 3⎞ d 2U 2π 2π − x2 ) = −4 ⎜⎜ − At x2 = > 0, x2 = is a local minimum, so x2 = is a , ⎟⎟ = +2 3. Since 2 ( 2 3 dx 2 dx 3 3 ⎝ ⎠ point of stable equilibrium.

Note that −

10.62. Model: The potential energy of two nucleons interacting via the strong force is U = U 0 [1 − e − x / x0 ] where x is the distance between the centers of the two nucleons, x0 = 2.0 × 10−15 m, and U 0 = 6.0 × 10−11 J. Visualize: Nucleons are protons and neutrons, and they are held together in the nucleus by a force called the strong force. This force exists between nucleons at very small separations. Solve: (a)

(b) For x = 5.0 × 10−15 m, U = 55.1 × 10−12 J. This energy is represented by a total energy line. (c) Due to conservation of total energy, the potential energy when x = 5.0 × 10−15 J is transformed into kinetic energy until x = twice the radius = 1.0 × 10−15 m. At this separation, u = 15.7 × 10−12 J. Thus,

1 2 1 2 mv + mv + 15.7 × 10−12 J = 55.1 × 10 −12 J ⇒ v = 1.53 × 108 m/s 2 2 Assess: A speed of 1.53 × 108 m/s is approximately 0.5 c where c is the speed of light. This speed is understandable for the present model.

10.63. A 2.5 kg ball is thrown upward at a speed of 4.0 m/s from a height of 82 cm above a vertical spring. When the ball comes down it lands on and compresses the spring. If the spring has a spring constant of k = 600 N/m, by how much is it compressed?

10.64. (a) A 1500 kg auto coasts up a 10.0 m high hill and reaches the top with a speed of 5.0 m/s. What initial speed must the auto have had at the bottom of the hill? (b)

We place the origin of our coordinate system at the bottom of the hill. (c) The solution of the equation is vi = 14.9 m/s. This is approximately 30 mph and is a reasonable value for the speed at the bottom of the hill.

10.65. (a) A spring gun is compressed 15 cm to launch a 200 g ball on a horizontal, frictionless surface. The ball has a speed of 2.0 m/s as it loses contact with the spring. Find the spring constant of the gun. (b)

We place the origin of our coordinate system on the free end of the spring in the equilibrium position. Because the surface is frictionless, the mechanical energy for the system (ball + spring) is conserved. (c) The conservation of energy equation is

K f + U sf = K i + U si 1 2 1 1 1 mv1 + k (0 m) 2 = m(0 m/s) 2 + k ( −0.15 m) 2 2 2 2 2 (0.200 kg)(2.0 m/s) 2 = k ( −0.15 m) 2 k = 36 N/m

10.66. (a) A 100 g lump of clay traveling at 3.0 m/s strikes and sticks to a 200 g lump of clay at rest on a

frictionless surface. The combined lumps smash into a horizontal spring with k = 3.0 N/m. The other end of the spring is firmly anchored to a fixed post on the surface. How far will the spring compress? (b)

(c) Solving the conservation of momentum equation we get v1x = 1.0 m/s. Substituting this value into the conservation of energy equation yields Δx2 = 32 cm.

10.67. (a) A spring with spring constant 400 N/m is anchored at the bottom of a frictionless 30° incline. A 500 g block is pressed against the spring, compressing the spring by 10 cm, then released. What is the speed with which the block is launched up the incline? (b) The origin is placed at the end of the uncompressed spring. This is the point from which the block is launched as the spring expands.

(c) Solving the energy conservation equation, we get vf = 2.6 m/s.

10.68. Model: Assume an ideal spring that obeys Hooke’s law. The mechanical energy K + U s + U g is conserved during the launch of the ball. Visualize:

This is a two-part problem. In the first part, we use projectile equations to find the ball’s velocity v2 as it leaves the spring. This will yield the ball’s kinetic energy as it leaves the spring. Solve: Using the equations of kinematics,

1 x3 = x2 + v2 x (t3 − t2 ) + ax (t3 − t2 ) 2 ⇒ 5.0 m = 0 m + (v2 cos30°)(t3 − 0 s) + 0 m 2 (v2 cos30°)t3 = 5.0 m ⇒ t3 = (5.0 m/ v2 cos30°) 1 y3 = y2 + v2 y (t3 − t2 ) + a y (t3 − t2 ) 2 2 1 −1.5 m = 0 + (v2 sin 30°)(t3 − 0 s) + (9.8 m / s 2 )(t3 − 0 s) 2 2 ⎛ 5.0 m ⎞ 5.0 m ⎞ 2 ⎛ Substituting the value for t3 , (−1.5 m) = (v2 sin 30°) ⎜ ⎟ − ( 4.9 m/s ) ⎜ ⎟ ⎝ v2 cos30° ⎠ ⎝ v2 cos30° ⎠

⇒ (−1.5 m) = + (2.887 m) −

2

163.33 ⇒ v2 = 6.102 m/s v22

The conservation of energy equation K 2 + U s2 + U g 2 = K1 + U s1 + U g1 is 1 2 1 1 1 mv2 + k (0 m) 2 + mgy2 = mv12 + k ( Δs ) 2 + mgy1 2 2 2 2

Using y2 = 0 m, v1 = 0 m/s, Δs = 0.20 m, and y1 = −(Δs )sin g 30°, we get 1 2 1 mv2 + 0 J + 0 J = 0 J + k ( Δs ) 2 − mg (Δs )sin 30° (Δs ) 2 k = mv22 + 2mg (Δs )sin 30° 2 2 (0.20 m) 2 k = (0.020 kg)(6.102 m/s)2 + 2(0.020 kg)(9.8 m/s 2 )(0.20)(0.5) ⇒ k = 19.6 N/m Assess: Note that y1 = −( Δs )sin 30° is with a minus sign and hence the gravitational potential energy mgy1 is − mg (Δs )sin 30°.

10.69. Model: This is a two-part problem. In the first part, we will find the critical velocity for the ball to go over the top of the peg without the string going slack. Using the energy conservation equation, we will then obtain the gravitational potential energy that gets transformed into the critical kinetic energy of the ball, thus determining the angle θ . Visualize:

We place the origin of our coordinate system on the peg. This choice will provide a reference to measure gravitational potential energy. For θ to be minimum, the ball will just go over the top of the peg. Solve: The two forces in the free-body force diagram provide the centripetal acceleration at the top of the circle. Newton’s second law at this point is mv 2 FG + T = r where T is the tension in the string. The critical speed vc at which the string goes slack is found when T → 0. In this case, mg =

mvC2 ⇒ vC2 = gr = gL 3 r

The ball should have kinetic energy at least equal to 1 2 1 ⎛ L⎞ mvC = mg ⎜ ⎟ 2 2 ⎝3⎠ for the ball to go over the top of the peg. We will now use the conservation of mechanical energy equation to get the minimum angle θ . The equation for the conservation of energy is 1 1 K f + U gf = K i + U gi ⇒ mvf2 + mgyf = mvi2 + mgyi 2 2

Using vf = vc , yf = 13 L, vi = 0, and the above value for vC2 , we get 1 L L L mg + mg = mgyi ⇒ yi = 2 3 3 2 That is, the ball is a vertical distance

1 2

L above the peg’s location or a distance of ⎛ 2L L ⎞ L − ⎟= ⎜ ⎝ 3 2⎠ 6

below the point of suspension of the pendulum, as shown in the figure on the right. Thus, cosθ =

L /6 1 = ⇒ θ = 80.4° L 6

10.70. Model: Choose yourself + spring + earth as the system. There are no forces from outside this system, so it is an isolated system. The interaction forces within the system are the spring force of the bungee cord and the gravitational force. These are both conservative forces, so mechanical energy is conserved. Visualize:

We can equate the system’s initial energy, as you step off the bridge, to its final energy when you reach the lowest point. We do not need to compute your speed at the point where the cord starts to stretch. We do, however, need to note that the end of the unstretched cord is at y0 = y1 − 30 m = 70 m, so U 2s = 12 k ( y2 − y0 ) 2 .

Also note that U1s = 0, since the cord is not stretched. The energy conservation equation is 1 K 2 + U 2g + U 2s = K1 + U1g + U1s ⇒ 0 J + mgy2 + k ( y2 − y0 ) 2 = 0 J + mgy1 + 0 J 2 Multiply out the square of the binomial and rearrange: 1 1 mgy2 + ky22 − ky0 y2 + ky02 = mgy1 2 2 ⎛ 2mg ⎞ ⎛ 2 2mgy1 ⎞ 2 2 ⇒ y2 + ⎜ − 2 y0 ⎟ y2 + ⎜ y0 − ⎟ = y2 − 100.8 y2 + 980 = 0 k k ⎝ ⎠ ⎝ ⎠ This is a quadratic equation with roots 89.9 m and 10.9 m. The first is not physically meaningful because it is a height above the point where the cord started to stretch. So we find that your distance from the water when the bungee cord stops stretching is 10.9 m.

10.71. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanical energy K + U g + U s is conserved. Visualize:

We have chosen to place the origin of the coordinate system at the point of maximum compression. We will use lengths along the ramp with the variable s rather than x. Solve: (a) The conservation of energy equation K 2 + U g2 + U s2 = K1 + U g1 + U s1 is

1 2 1 1 mv2 + mgy2 + k ( Δs ) 2 = mv12 + mgy1 + k (0 m) 2 2 2 2 1 1 1 m(0 m/s) 2 + mg (0 m) + k ( Δs ) 2 = m(0 m/s) 2 + mg (4.0 m + Δs )sin 30° + 0 J 2 2 2 1 ⎛1⎞ (250 N/m)( Δs ) 2 = (10 kg)(9.8 m/s 2 )(4.0 m + Δs ) ⎜ ⎟ 2 ⎝ 2⎠ This gives the quadratic equation: (125 N/m)(Δs ) 2 − (49 kg ⋅ m/s 2 ) Δs − 196 kg ⋅ m 2 /s 2 = 0 ⇒ Δs = 1.46 m and − 1.07 m (unphysical)

The maximum compression is 1.46 m. (b) We will now apply the conservation of mechanical energy to a point where the vertical position is y and the block’s velocity is v. We place the origin of our coordinate system on the free end of the spring when the spring is neither compressed nor stretched.

1 2 1 1 1 mv + mgy + k ( Δs ) 2 = mv12 + mgy1 + k (0 m) 2 2 2 2 2 1 2 1 mv + mg (−Δs sin 30°) + k ( Δs ) 2 = 0 J + mg (4.0 m sin 30°) + 0 J 2 2 1 1 2 k (Δs ) − ( mg sin 30°) Δs + mv 2 − mg sin 30°(4.0 m) = 0 2 2 To find the compression where v is maximum, take the derivative of this equation with respect to Δs : 1 1 dv k 2(Δs ) − (mg sin 30°) + m 2v −0=0 2 2 d Δs

Since

dv = 0 at the maximum, we have d Δs Δs = (mg sin 30°)/ k = (10 kg)(9.8 m/s 2 )(0.5)/(250 N/m) = 19.6 cm

10.72. Model: Assume an ideal, massless spring that obeys Hooke’s law. Let us also assume that the cannon (C) fires balls (B) horizontally and that the spring is directly behind the cannon to absorb all motion. Visualize:

The before-and-after pictorial representation is shown, with the origin of the coordinate system located at the spring’s free end when the spring is neither compressed nor stretched. This free end of the spring is just behind the cannon. Solve: The momentum conservation equation pfx = pix is

mB (vfx ) B + mC (vfx )C = mB (vix ) B + mC (vix )C Since the initial momentum is zero,

(vfx ) B = −

⎛ 200 kg ⎞ mC (vfx )C = − ⎜ ⎟ (vfx )C = −20(vfx )C mB ⎝ 10 kg ⎠

The mechanical energy conservation equation for the cannon + spring K f + U sf = Ki + U si is 1 1 1 1 1 m(vf )C2 + k (Δx) 2 = m(vi )C2 + 0 J ⇒ 0 J + k (Δx) 2 = m(vfx )C2 2 2 2 2 2 k (20,000 N/m) ⇒ (vfx )C = ± Δx = ± (0.50 m) = ±5.0 m/s m 200 kg To make this velocity physically correct, we retain the minus sign with (vfx )C . Substituting into the momentum conservation equation yields: (vfx ) B = −20(−5.0 m/s) = 100 m/s

10.73. Model: This is a collision between two objects, and momentum is conserved in the collision. In addition, because the interaction force is a spring force and the surface is frictionless, energy is also conserved. Visualize:

Let part 1 refer to the time before the collision starts, part 2 refer to the instant when the spring is at maximum compression, and part 3 refer to the time after the collision. Notice that just for an instant, when the spring is at maximum compression, the two blocks are moving side by side and have equal velocities: vA2 = vB2 = v2 . This is an important observation. Solve: First relate part 1 to part 2. Conservation of energy is 1 1 1 1 1 1 2 2 2 mA vA1 = mA vA2 + mBvB2 + k (Δxmax ) 2 = ( mA + mB )v22 + k (Δxmax ) 2 2 2 2 2 2 2 where Δxmax is the spring’s compression. U g is not in the equation because there are no elevation changes. Also note

that K 2 is the sum of the kinetic energies of all moving objects. Both v2 and Δxmax are unknowns. Now add the conservation of momentum: mA vA1 mA vA1 = mA vA 2 + mBvB2 = (mA + mB )v2 ⇒ v2 = = 2.667 m/s mA + mB

Substitute this result for v2 into the energy equation to find: 1 1 1 2 k (Δxmax ) 2 = mA vA1 − (mA + mB )v22 2 2 2 2 mAvA1 − (mA + mB )v22 = 0.046 m = 4.6 cm k Notice how both conservation laws were needed to solve this problem. (b) Again, both energy and momentum are conserved between “before” and “after.” Energy is 1 1 1 1 2 2 2 2 2 mA vA1 = mA vA3 + mBvB3 ⇒ 16 = vA3 + vB3 2 2 2 2 The spring is no longer compressed, so the energies are purely kinetic. Momentum is mA vA1 = mA vA3 + mBvB3 ⇒ 8 = 2vA3 + vB3

⇒ Δxmax =

We have two equations in two unknowns. From the momentum equation, we can write vB3 = 2(4 − vA3 ) and use this in the energy equation to obtain: 2 1 2 2 2 2 16 = vA3 + ⋅ 4 ( 4 − vA3 = 3vA3 − 16vA3 + 32 ⇒ 3vA3 − 16vA3 + 16 = 0 ) 2 This is a quadratic equation for vA3 with roots vA3 = (4 m/s, 1.33 m/s). Using vB3 = 2(4 − vA3 ), these two roots give vB3 = (0 m/s, 5.333 m/s). The first pair of roots corresponds to a “collision” in which A misses B, so each keeps its initial speed. That’s not the situation here. We want the second pair of roots, from which we learn that the blocks’ speeds after the collision are vA3 = 1.33 m/s and vB3 = 5.3 m/s.

10.74. Model: Mechanical energy and momentum are conserved during the expansion of the spring. Visualize: Please refer to Figure CP10.74. Solve: Example 10.8 is a very similar problem, except that the objects are initially at rest. We can use the solution from Example 10.8 for this problem in a reference frame S′ in which the two carts are initially at rest, then transform the answer to the frame S in which the carts are initially moving. Thus in the S′ frame,

( vfx )′2 =

k ( Δxi ) m m2 1 + 2

( vfx )1′ = −

(

2

m1

)

m2 ( vfx )2′ m1

Let the 100 g cart be Cart 1 and the 300 g cart be Cart 2. With k = 120 N/m and Δxi = 4.0cm,

( vfx )2′ = 0.40 m/s, ( vfx )1′ = −1.2 m/s An object at rest in the S′ frame is traveling to the right at 1.0 m/s in the S frame. The equation of transformation is therefore vx = vx′ + 1.0 m/s In the S frame, the velocities of the carts are ( vfx )1 = −1.2 m/s + 1.0 m/s = −0.2 m/s

( vfx )2 = 0.40 m/s + 1.0 m/s = 1.4 m/s

Assess:

Cart 1 is moving slowly to the left while the heavier Cart 2 is moving quickly to the right.

10.75. Model: Model the balls as particles, and assume a perfectly elastic collision. After the collision is over, the balls swing out as pendulums. The sum of the kinetic energy and gravitational potential energy does not change as the balls swing out. Visualize:

In the pictorial representation we have identified before-and-after quantities for both the collision and the pendulum swing. We have chosen to place the origin of the coordinate system at a point where the two balls at rest barely touch each other. Solve: As the ball with mass m1 , whose string makes an angle θ il with the vertical, swings through its equilibrium position, it lowers its gravitational energy from m1 gy0 = m1 g ( L − L cosθ i1 ) to zero. This change in potential energy transforms into a change in kinetic energy. That is, 1 m1 g ( L − L cosθ i1 ) = m1 (v1 )12 ⇒ (v1 )1 = 2 gL(1 − cosθ i1 ) 2

Similarly, (v1 )2 = 2 gL(1 − cosθ i2 ). Using θ i1 = θ i2 = 45°, we get (v1 )1 = 2.396 m/s = (v1 ) 2 . Both balls are moving at the point where they have an elastic collision. Since our analysis of elastic collisions was for a situation in which ball 2 is initially at rest, we need to use the Galilean transformation to change to a frame S′ in which ball 2 is at rest. Ball 2 is at rest in a frame that moves with ball 2, so choose S′ to have V = −2.396 m/s, with the minus sign because this frame (like ball 2) is moving to the left. In this frame, ball 1 has velocity (v1′ )1 = (v1 )1 − V = 2.396 m/s + 2.396 m/s = 4.792 m/s and ball 2 is at rest. The elastic collision causes the balls to move with velocities m − m2 1 (v2′ )1 = 1 (v1′ )1 = – (4.792 m/s)= − 1.597 m/s 3 m1 + m2 (v′2 ) 2 =

2m1 2 (v1′ )1 = (2.396 m/s)=3.194 m/s m1 + m2 3

We can now use v = v′ + V to transform these back into the laboratory frame: (v2 )1 = −1.597 m/s − 2.396 m/s = −3.99 m/s (v2 ) 2 = 3.195 m/s − 2.396 m/s = 0.799 m/s

Having determined the velocities of the two balls after collision, we will once again use the conservation equation Kf + U gf = Ki + U gi for each ball to solve for the θ f1 and θ f2 . 1 1 m1 (v3 )12 + m1 gL(1 − cosθ f 1 ) = m1 (v2 )12 + 0 J 2 2 Using (v3 )1 = 0, this equation simplifies to

1 1 gL(1 − cosθ f 1 ) = (−3.99 m/s) 2 ⇒ cosθ f 1 = 1 − (−3.99 m/s) 2 ⇒ θ f 1 = 79.3° 2 2 gL The 100 g ball rebounds to 79º. Similarly, for the other ball: 1 1 m2 (v3 ) 22 + m2 gL(1 − cosθ f 2 ) = m2 (v2 ) 22 + 0 J 2 2 Using (v3 ) 2 = 0, this equation becomes

⎛ 1 ⎞ 2 cosθ f 2 = 1 − ⎜ ⎟ (0.799) ⇒ θ f 2 = 14.7° ⎝ 2gL ⎠ The 200 g ball rebounds to 14.7°.

10.76. Model: Model the sled as a particle. Because there is no friction, the sum of the kinetic and gravitational potential energy is conserved during motion. Visualize:

Place the origin of the coordinate system at the center of the hemisphere. Then y0 = R and, from geometry,

y1 = R cos φ . Solve:

The energy conservation equation K1 + U1 = K 0 + U 0 is

1 2 1 1 mv1 + mgy1 = mv02 + mgy0 ⇒ mv12 + mgR cos φ = mgR ⇒ v1 = 2 gR (1 − cos φ ) 2 2 2 G (b) If the sled is on the hill, it is moving in a circle and the r-component of Fnet has to point to the center with magnitude Fnet = mv 2 / R. Eventually the speed gets so large that there is not enough force to keep it in a circular trajectory, and that is the point where it flies off the hill. Consider the sled at angle φ . Establish an r-axis pointing toward the center of the circle, as we usually do for circular motion problems. Newton’s second law along this axis requires:

mv 2 R ⎛ mv 2 v2 ⎞ ⇒ n = mg cos φ − = m ⎜ g cos φ − ⎟ R R⎠ ⎝

( Fnet ) r = FG cos φ − n = mg cos φ − n = mar =

The normal force decreases as v increases. But n can’t be negative, so the fastest speed at which the sled stays on the hill is the speed vmax that makes n → 0. We can see that vmax = gR cos φ . (c) We now know the sled’s speed at angle φ , and we know the maximum speed it can have while remaining on the hill. The angle at which v reaches vmax is the angle φmax at which the sled will fly off the hill. Combining the two expressions for v1 and vmax gives: 2 gR (1 − cos φ ) = gR cos φ ⇒ 2 R (1 − cos φmax ) = R cos φmax

⇒ cos φmax =

2 ⎛ 2⎞ ⇒ φmax = cos −1 ⎜ ⎟ = 48° 3 ⎝3⎠

11.1. Visualize: Please refer to Figure EX11.1. G G Solve: (a) A ⋅ B = AB cosα = (4)(5)cos 40° = 15.3. (b) (c)

G G C ⋅ D = CD cosα = (2)(4)cos120° = −4.0. G G E ⋅ F = EF cosα = (3)(4)cos90° = 0.

11.2. Visualize: Please refer to Figure EX11.2. G G Solve: (a) A ⋅ B = AB cosα = (3)(4)cos110° = −4.1. (b) (c)

G G C ⋅ D = CD cosα = (4)(5)cos180° = −20. G G E ⋅ F = EF cosα = (4)(3)cos30° = 10.4.

11.3. Solve: (a) (b)

G G A ⋅ B = Ax Bx + Ay By = (3)( −2) + ( −4)(6) = −30.

G G A ⋅ B = Ax Bx + Ay By = (2)(6) + (3)(−4) = 0.

11.4. Solve: (a) (b)

G G A ⋅ B = Ax Bx + Ay By = (4)(−3) + (−2)( −2) = −16.

G G A ⋅ B = Ax Bx + Ay By = (−4)(−1) + (2)(−2) = 0.

G

G

G G

G

11.5. Solve: (a) The length of A is A = A = A ⋅ A = ( 3) + ( −4 ) = 25 = 5. The length of B is B=

( −2 )

2

2

2

G G 2 + ( 6 ) = = 40 = 2 10. Using the answer A ⋅ B = −30 from Ex11.3(a),

G G A ⋅ B = AB cosα

(

)

⇒ −30 = ( 5 ) 2 10 cosα G G (b) From EX11.3 (b), A ⋅ B = 0. Thus

⇒ α = cos

−1

( −0.9487 ) = 162°

cosα = 0 α = 90°

Assess: In part (a) the vectors are nearly pointing in opposite directions, while in part (b) the vectors are perpendicular.

G

G G G G 2 2 A = A = A ⋅ A = ( 4 ) + ( 2 ) = 20. The length of B is G G 13. Using the answer A ⋅ B = −16 from EX11.4(a), G G A ⋅ B = AB cosα

11.6. Solve: (a) The length of A is B=

( −3)

2

+ ( −2 ) = 2

⇒ −16 =

(

⇒ α = cos

20 −1

)( 13 ) cosα

( −0.9923) = 173°

G G (b) From EX11.4(b), A ⋅ B = 0. Thus

cosα = 0 α = 90° Assess: In part (a) the vectors are nearly pointing in opposite directions, while in part (b) the vectors are perpendicular.

G

G G G W = F ⋅ Δr = (6.0iˆ − 3.0 ˆj ) ⋅ (2.0 ˆj ) N ⋅ m = (12.0iˆ ⋅ ˆj − 6.0 ˆj ⋅ ˆj ) J = −6.0 J.

11.7. Solve: (a) W = F ⋅ Δr = (6.0iˆ − 3.0 ˆj ) ⋅ (2.0iˆ) N ⋅ m = (12.0iˆ ⋅ iˆ − 3.0 ˆj ⋅ iˆ) J = 12.0 J. (b)

G

G

G G G G W = F ⋅ Δr = ( −4.0iˆ − 6.0 ˆj ) N ⋅ (−3.0iˆ + 2.0 ˆj ) m = (12.0 − 12.0) J=0 J.

11.8. Solve: (a) W = F ⋅ Δr = (−4.0iˆ − 6.0 ˆj ) N ⋅ (3.0iˆ) m = ( −12.0iˆ ⋅ iˆ + 12.0 ˆj ⋅ iˆ) J= − 12.0 J. (b)

11.9. Model: Use the work-kinetic energy theorem to find the net work done on the particle. Visualize:

Solve:

From the work-kinetic energy theorem, 1 1 1 1 W = ΔK = mv12 − mv02 = m ( v12 − v02 ) = (0.020 kg)[(30 m/s) 2 − (−30 m/s) 2 ] = 0 J 2 2 2 2

Assess: Negative work is done in slowing down the particle to rest, and an equal amount of positive work is done in bringing the particle to the original speed but in the opposite direction.

G

G

G

G

G

11.10. Model: Work done by a force F on a particle is defined as W = F ⋅ Δr , where Δr is the particle’s displacement. Visualize:

Solve:

(a) The work done by gravity is G G Wg = FG ⋅ Δr = (− mgjˆ) N ⋅ (2.25 − 0.75) ˆj m = −(2.0 kg)(9.8 m/s 2 )(1.50 m) J = −29 J

G G (b) The work done by hand is WH = Fhand on book ⋅ Δr . As long as the book does not accelerate, G G Fhand on book = − Fearth on book = −(− mgjˆ) = mgjˆ

⇒ WH = ( mgjˆ) ⋅ (2.25 − 0.75) ˆj m = (2.0 kg)(9.8 m/s 2 )(1.50 m)=29 J

G

G

G

11.11. Model: Model the piano as a particle and use W = F ⋅ Δr , where W is the work done by the force F G through the displacement Δr . Visualize:

G For the force FG : G G G G W = F ⋅ Δr = FG ⋅ Δr = ( Fg )(Δr )cos0° = (2500 N)(5.00 m)(1) = 1.250 × 104 J G For the tension T1 : G G W = T1 ⋅ Δr = (T1 )(Δr )cos(150°) = (1830 N)(5.00 m)(−0.8660) = −7.92 × 103 J G For the tension T2 : G G W = T2 ⋅ Δr = (T2 )(Δr )cos(135°) = (1295 N)(5.00 m)(−0.7071) = −4.58 × 103 J

Solve:

Assess:

G Note that the displacement Δr in all the above cases is directed downwards along − ˆj.

G

G

G

11.12. Model: Model the crate as a particle and use W = F ⋅ Δr , where W is the work done by a force F on G a particle and Δr is the particle’s displacement. Visualize:

Solve:

G For the force f k :

G For the tension T1 :

G For the tension T2 :

G G W = f k ⋅ Δr = f k (Δr )cos(180°) = (650 N)(3.0 m)(−1) = −1.95 kJ

G G W = T1 ⋅ Δr = (T1 )( Δr )cos 20° = (600 N)(3.0 m)(0.9397) =1.69 kJ

G G W = T2 ⋅ Δr = (T2 )(Δr )cos30° = (410 N)(3.0 m)(0.866) = 1.07 kJ G Assess: Negative work done by the force of kinetic friction ( f k ) means that 1.95 kJ of energy has been transferred out of the crate.

11.13. Model: Model the 2.0 kg object as a particle, and use the work-kinetic energy theorem. Visualize: Please refer to Figure EX11.13. For each of the five intervals the velocity-versus-time graph gives the initial and final velocities. The mass of the object is 2.0 kg. Solve: According to the work-kinetic energy theorem:

1 1 1 W = ΔK = mvf2 = mvi2 = m ( vf2 − vi2 ) 2 2 2 1 Interval AB: vi = 2 m/s, vf = −2 m/s ⇒ W = (2.0 kg)[(−2 m/s) 2 − (2 m/s) 2 ] = 0 J 2 1 Interval BC: vi = −2 m/s, vf = −2 m/s ⇒ W = (2.0 kg)[(−2 m/s) 2 − (−2 m/s) 2 ] = 0 J 2 1 Interval CD: vi = −2 m/s, vf = 0 m/s ⇒ W = (2.0 kg)[(0 m/s) 2 − (−2 m/s) 2 ] = −4.0 J 2 1 Interval DE: vi = 0 m/s, vf = 2 m/s ⇒ W = (2.0 kg)[(2 m/s) 2 − (0 m/s)2 ] = +4.0 J 2 1 Interval EF: vi = 2 m/s, vf = 1 m/s ⇒ W = (2.0 kg)[(1 m/s)2 − (2 m/s)2 ] = −3.0 J 2 Assess: The work done is zero in intervals AB and BC. In the interval CD + DE the total work done is zero. It is not whether v is positive or negative that counts because K ∝ v 2 . What is important is the magnitude of v and how v changes.

11.14. Model: Use the definition of work. Visualize: Please refer to Figure EX11.14. Solve: Work is defined as the area under the force-versus-position graph: sf

W = ∫ Fs ds = area under the force curve si

Interval 0–1 m: W = (4.0 N)(1.0 m − 0.0 m) = 4.0 J Interval 1–2 m: W = (4.0 N)(0.5 m) + (− 4.0 N)(0.5 m) = 0 J 1 Interval 2–3 m: W = ( −4.0 N)(1 m) = −2.0 J 2

11.15. Model: Use the work-kinetic energy theorem to find velocities. Visualize: Please refer to Figure EX11.15. Solve: The work-kinetic energy theorem is x

f 1 1 ΔK = mvf2 − mvi2 = W = ∫ Fx dx = area under the force curve from xi to xf 2 2 xi

x

f 1 1 1 ⇒ mvf2 − (0.500 kg)(2.0m/s) 2 = mvf2 − 1.0 J = ∫ Fx dx = area from 0 to x 2 2 2 0m

1 (0.500 kg)vf2 − 1.0 J = 12.5 J ⇒ vf = 7.35 m/s 2 1 At x = 2 m: (0.500 kg)vf2 − 1.0 J = 20 J ⇒ vf = 9.17 m/s 2 1 At x = 3 m: (0.500 kg)vf2 − 1.0 J = 22.5 J ⇒ vf = 9.70 m/s 2 At x =1 m:

11.16. Model: Use the work-kinetic energy theorem. Visualize: Please refer to Figure EX11.16. Solve: The work-kinetc energy theorem is x

f 1 1 ΔK = mvf2 − mvi2 = W = ∫ Fx dx = area under the force curve from xi to xf 2 2 xi

x

f 1 1 1 ⇒ mvf2 − (2.0 kg)(4.0 m/s) 2 = mvf2 − 16.0 J = ∫ Fx dx = area from 0 to x 2 2 2 0m

1 1 (2.0 kg)vf2 − 16.0 J = (10 N)(2 m) = 10 J ⇒ vf = 5.1 m/s 2 2 1 2 At x = 4 m : (2.0 kg)vf − 16.0 J = 0 J ⇒ vf = 4.0 m/s 2

At x = 2 m :

11.17. Model: Use the work-kinetic energy theorem. Visualize: Please refer to Figure EX11.17. Solve: The work-kinetic energy theorem is xf

ΔK = W = ∫ Fx dx = area of the Fx -versus- x graph between xi and xf xi

1 2 1 2 1 mvf − mvi = ( Fmax )(2 m) 2 2 2 Using m = 0.500 kg, vf = 6.0 m/s, and vi = 2.0 m/s, the above equation yields Fmax = 8.0 N. Assess: Problems in which the force is not a constant can not be solved using constant-acceleration kinematic equations.

11.18. Model: Use the definition Fs = − dU / ds. Visualize: Please refer to Figure EX11.18. Solve: Fx is the negative of the slope of the potential energy graph at position x. Between x = 0 cm and x = 10 cm the slope is

slope = (U f − U i ) / ( xf − xi ) = (0 J − 10 J) / (0.10 m − 0.0 m) = −100 N Thus, Fx = 100 N at x = 5 cm. The slope between x = 10 cm and x = 20 cm is zero, so Fx = 0 N at x = 15 cm. Between 20 cm and 40 cm, slope = (10 J − 0 J) / (0.40 m − 0.20 m) = 50 N At x = 25 cm and x = 35 cm, therefore, Fx = −50 N.

11.19. Model: Use the definition Fs = − dU ds . Visualize: Please refer to Figure EX11.19. Solve: Fx is the negative of the slope of the potential energy graph at position x.

⎛ dU ⎞ Fx = − ⎜ ⎟ ⎝ dx ⎠

Between x = 0 m and x = 3 m, the slope is slope = (U f − U i ) / ( xf − xi ) = (60 J − 0 J) / (3 m − 0 m) = 20 N Thus, Fx = −20 N at x = 1 m. Between x = 3 m and x = 5 m, the slope is

slope = (U f − U i ) / ( xf − xi ) = ( 0 J − 60 J ) / ( 5 m − 3 m ) = −30 N

Thus, Fx = 30 N at x = 4 m.

11.20. Model: Use the negative derivative of the potential energy to determine the force acting on a particle. Solve:

(a)

The graph of the potential energy U = 4 y 3 is shown. (b) The y-component of the force is

Fy = −

dU d = − (4 y 3 ) = −12 y 2 dy dy

At y = 0 m, Fy = 0 N; at y = 1 m, Fy = −12 N; and at y = 2 m, Fy = −48 N.

11.21. Model: Use the negative derivative of the potential energy to determine the force acting on a particle. Solve:

(a)

The graph of the potential energy U = 10 / x is shown. (b) The x-component of the force is

dU d ⎛ 10 ⎞ 10 =− ⎜ ⎟= dx dx ⎝ x ⎠ x 2 10 10 = 2 = 2.5 N Fx = 5 m = 2 x x=2 m x

Fx = − Fx = 2 m

= 0.40 N x =5 m

Fx =8 m =

10 x2

= 0.156 N x =8 m

11.22. Model: Assume the carbon-carbon bond acts like an ideal spring that obeys Hooke’s law. Visualize:

The quantity ( x − xe ) is the stretching relative to the spring’s equilibrium length. In the present case, bond stretching is analogous to spring stretching. Solve: (a) The kinetic energy of the carbon atom is

1 1 K = mv 2 = (2.0 × 10−26 kg)(500 m/s) 2 = 2.5 × 10−21 J 2 2 (b) The energy of the spring is given by

1 U s = k ( x − xe ) 2 = K 2 2K 2(2.5 × 10−21 J) N ⇒k = = = 2.0 2 ( x − xe ) (0.050 × 10−9 m) 2 m

11.23. Visualize: One mole of helium atoms in the gas phase contains N A = 6.02 × 1023 atoms. Solve:

If each atom moves with the same speed v, the microscopic total kinetic energy will be

2 K micro 2(3700 J) ⎛1 ⎞ = = 1360 m/s K micro = N A ⎜ mv 2 ⎟ = 3700 J ⇒ v = −27 × 2 (6.68 10 kg)(6.02 × 1023 ) mN ⎝ ⎠ A

11.24. Solve: (a) The car has an initial kinetic energy K i . That energy is transformed into thermal energy of the car and the road surface. The gravitational potential energy does not change and no work is done by external forces, so during the skid K i transforms entirely into thermal energy Eth . This energy transfer and transformation is shown on the energy bar chart. Note that 1 1 K i = mv 2 = (1500 kg)(20 m/s) 2 = 3.0 × 105 J 2 2

(b) The change in the thermal energy of the car and the road surface is 3.0 × 105 J.

11.25. Visualize:

1 K i = K 0 = mv02 = 0 J U i = U g0 = mgy0 = (20 kg)(9.8 m/s 2 )(3.0 m) = 5.9 × 102 J 2 1 1 Wext = 0 J K f = K1 = mv12 = (20 kg)(2.0 m/s) 2 = 40 J U f = U g1 = mgy1 = 0 J 2 2 At the top of the slide, the child has gravitational potential energy of 5.9 × 102 J. This energy is transformed into thermal energy of the child’s pants and the slide and the kinetic energy of the child. This energy transfer and transformation is shown on the energy bar chart. (b)

Solve:

(a)

The change in the thermal energy of the slide and of the child’s pants is 5.9 × 10 2 J − 40 J = 5.5 × 10 2 J.

11.26. Visualize: The system loses 400 J of potential energy. In the process of losing this energy, it does 400 J of work on the environment, which means Wext = −400 J. Since the thermal energy increases 100 J, we have ΔEth = 100 J, which must have been 100 J of kinetic energy originally. This is shown in the energy bar chart.

11.27. Visualize:

Note that the conservation of energy equation K i + U i + Wext = K f + U f + ΔEth requires that Wext be equal to +400 J.

11.28. Solve: Please refer to Figure EX11.28. The energy conservation equation yields K i + U i + Wext = K f + U f + ΔEth ⇒ 4 J + 1 J + Wext = 1 J + 2 J + 1 J ⇒ Wext = −1 J Thus, the work done to the environment is −1 J. In other words, 1 J of energy is transferred from the system into the environment. This is shown in the energy bar chart.

11.29. Visualize: The tension of 20.0 N in the cable is an external force that does work on the block Wext = (20.0 N)(2.00 m) = 40.0 J, increasing the gravitational potential energy of the block. We placed the origin of our

coordinate system on the initial resting position of the block, so we have U i = 0 J and U f = mgyf =

(1.02 kg)(9.8 m/s 2 )(2.00 m) = 20.0 J. Also, K i = 0 J, and ΔEth = 0 J. The energy bar chart shows the energy transfers and transformations.

Solve:

The conservation of energy equation is 1 K i + U i + Wext = K f + U f + ΔEth ⇒ 0 J + 0 J + 40.0 J = mvf2 + 20.0 J + 0 J 2 ⇒ vf = (20.0 J)(2) /1.02 kg = 6.26 m/s

11.30. Model: Model the elevator as a particle, and apply the conservation of energy equation. Solve: The tension in the cable does work on the elevator to lift it. Because the cable is pulled by the motor, we say that the motor does the work of lifting the elevator. (a) The energy conservation equation is K i + U i + Wext = K f + U f + ΔEth . Using K i = 0 J, K f = 0 J, and

ΔEth = 0 J gives

Wext = (U f − U i ) = mg ( yf − yi ) = (1000 kg)(9.8 m/s 2 )(100 m) = 9.80 × 105 J (b) The power required to give the elevator this much energy in a time of 50 s is

P=

Wext 9.8 × 105 J = = 1.96 × 104 W Δt 50 s

Assess: Since 1 horsepower (hp) is 746 W, the power of the motor is 26 hp. This is a reasonable amount of power to lift a mass of 1000 kg to a height of 100 m in 50 s.

11.31. Model: Model the steel block as a particle subject to the force of kinetic friction and use the energy conservation equation. Visualize:

G G G Solve: (a) The work done on the block is Wnet = Fnet ⋅ Δr where Δr is the displacement. We will find the displacement using kinematic equations and the force using Newton’s second law of motion. The displacement in the x-direction is 1 Δx = x1 = x0 + v0 x (t1 − t0 ) + ax (t1 − t0 ) 2 = 0 m + (1.0 m/s)(3.0 s − 0 s) + 0 m = 3.0 m 2

G Thus Δr = 3.0iˆ m. The equations for Newton’s second law along the x and y components are

( Fnet ) y = n − FG = 0 N ⇒ n = FG = mg = (10 kg)(9.8 m/s2 ) = 98.0 N G

G

( Fnet ) x = F − f k = 0 N ⇒ F = ⇒ Wnet

f k = μ k n = (0.6)(98.0 N) = 58.8 N G G = Fnet ⋅ Δr = F Δ x cos0° = (58.8 N)(3.0 m)(1) = 176 J

(b) The power required to do this much work in 3.0 s is P=

W 176 J = = 59 W 3.0 s t

11.32. Solve: The power of the solar collector is the solar energy collected divided by time. The intensity of the solar energy striking the earth is the power divided by area. We have ΔE 150 × 106 J W = = 41,667 W and intensity = 1000 2 3600 s m Δt 41,667 W 2 ⇒ Area of solar collector = = 41.7 m 1000 W/m 2

P=

11.33. Solve: The night light consumes more energy than the hair dryer. The calculations are 1.2 kW × 10 min = 1.2 × 103 × 10 × 60 J = 7.2 × 105 J 10 W × 24 hours = 10 × 24 × 60 × 60 J = 8.64 × 105 J

11.34. Solve: (a) A kilowatt hour is a kilowatt multiplied by 3600 seconds. It has the dimensions of energy. (b)

One kilowatt hour is energy

1 kwh = (1000 J/s)(3600 s) = 3.6 × 106 J Thus

⎛ 3.6 × 106 J ⎞ 9 500 kwh = (500 kwh) ⎜ ⎟ = 1.8 × 10 J ⎝ 1 kwh ⎠

11.35. Model: Model the sprinter as a particle, and use the constant-acceleration kinematic equations and the definition of power in terms of velocity. Visualize:

Solve: (a) We can find the acceleration from the kinematic equations and the horizontal force from Newton’s second law. We have

1 1 x = x0 + v0 x (t1 − t0 ) + ax (t1 − t0 ) 2 ⇒ 50 m = 0 m + 0 m + ax (7.0 s − 0 s) 2 ⇒ ax = 2.04 m/s 2 2 2 2 ⇒ Fx = max = (50 kg)(2.04 m/s ) = 102 N G G G (b) We obtain the sprinter’s power output by using P = F ⋅ v , where v is the sprinter’s velocity. At t = 2.0 s the power is

P = ( Fx )[v0 x + ax (t − t0 )] = (102 N)[0 m/s + (2.04 m/s 2 )(2.0 s − 0 s)] = 0.42 kW The power at t = 4.0 s is 0.83 kW, and at t = 6.0 s the power is 1.25 kW.

G

G

G

G

11.36. Model: Use the definition of work for a constant force F , W = F ⋅ Δs , where Δs is the displacement. G Visualize: Please refer to Figure P11.36. The force F = (6iˆ + 8 ˆj ) N on the particle is a constant. G G G G Solve: (a) WABD = WAB + WBD = F ⋅ (Δs ) AB + F ⋅ (Δs ) BD = (6iˆ + 8ˆ j ) N ⋅ (3iˆ) m + (6iˆ + 8 ˆj ) N ⋅ (4 ˆj ) m = 18 J + 32 J = 50 J G G G G (b) WACD = WAC + WCD = F ⋅ (Δs ) AC + F ⋅ (Δs )CD = (6iˆ + 8ˆ j ) N ⋅ (4 ˆj ) m + (6iˆ + 8 ˆj ) N ⋅ (3 ˆj ) m = 32 J + 18 J = 50 J G G (c) WAD = F ⋅ (Δs ) AD = (6iˆ + 8ˆ j ) N ⋅ (3iˆ + 4 ˆj ) m = 18 J + 32 J = 50 J The force is conservative because the work done is independent of the path.

11.37. Model:

The force is conservative, so it has a potential energy. Visualize: Please refer to Figure P11.37 for the graph of the force. Solve: The definition of potential energy is ΔU = −W (i → f ). In addition, work is the area under the force-versus-

displacement graph. Thus ΔU = U f − U i = − (area under the force curve). Since U i = 0 at x = 0 m, the potential energy at position x is U ( x) = − (area under the force curve from 0 to x). From 0 m to 3 m, the area increases linearly from 0 N m to −60 N m, so U increases from 0 J to 60 J. At x = 4 m, the area is −70 J. Thus U = 70 J at x = 4 m, and U doesn’t change after that since the force is then zero. Between 3 m and 4 m, where F changes linearly, U must have a quadratic dependence on x (i.e., the potential energy curve is a parabola). This information is shown on the potential energy graph below.

(b) Mechanical energy is E = K + U . From the graph, U = 20 J at x = 1.0 m. The kinetic energy is K = 12 mv 2 = 12 (0.100 kg) (25 m/s) 2 = 31.25 J. Thus E = 51.25 J. (c) The total energy line at 51.25 J is shown on the graph above. (d) The turning point occurs where the total energy line crosses the potential energy curve. We can see from the graph that this is at approximately 2.5 m. For a more accurate value, the potential energy function is U = 20x J. The TE line crosses at the point where 20 x = 51.25, which is x = 2.56 m.

11.38. Model: Use the relationship between force and potential energy and the work-kinetic energy theorem. Visualize:

Please refer to Figure P11.38. We will find the slope in the following x regions: 0 cm < x < 1 cm, 1 < x < 3 cm, 3 < x < 5 cm, 5 < x < 7 cm, and 7 < x < 8 cm. Solve: (a) Fx is the negative slope of the U-versus-x graph, for example, for 0 m < x < 2 m dU −4 J = = −400 N ⇒ Fx = +400 N dx 0.01 m Calculating the values of Fx in this way, we can draw the force-versus-position graph as shown. (b) Since W = ∫ xxif Fx dx = area of the Fx-versus-x graph between xi and xf, the work done by the force as the

particle moves from xi = 2 cm to xf = 6 cm is −2 J. (c) The conservation of energy equation is K f + U f = Ki + U i . We can see from the graph that U i = 0 J and

U f = 2 J in moving from x = 2 cm to x = 6 cm. The final speed is vf = 10 m/s, so

2 J + 12 (0.010 kg)(10.0 m/s) 2 = 0 J + 12 (0.010 kg)vi 2 ⇒ vi = 22.4 m/s

11.39. Model: Use the relationship between a conservative force and potential energy. Visualize: Please refer to Figure P11.39. We will obtain U as a function of x and Fx as a function of x by using the calculus techniques of integration and differentiation. Solve: (a) For the interval 0 m < x < 0.5 m, Fx = (4 x) N, where x is in meters. This means

dU = − Fx = −4 x ⇒ U = −2 x 2 + C1 = −2 x 2 dx where we have used U = 0 J at x = 0 m to obtain C1 = 0. For the interval 0.5 m < x < 1 m, Fx = ( −4 x + 4) N. Likewise, dU = 4 x − 4 ⇒ U = 2 x 2 − 4 x + C2 dx Since U should be continuous at the junction, we have the continuity condition

(−2 x 2 ) x = 0.5 m = (2 x 2 − 4 x + C2 ) x = 0.5 m ⇒ −0.5 = 0.5 − 2 + C2 ⇒ C2 = 1 U remains constant for x ≥ 1 m. (b) For the interval 0 m < x < 0.5 m, U = +4 x, and for the interval 0.5 m < x < 1.0 m, U = −4 x + 4, where x is in meters. The derivatives give Fx = −4 N and Fx = +4 N, respectively. The slope is zero for x ≥ 1 m.

dvx , x = ∫ vx dt , K = 12 mvx2 , and F = max . dt Visualize: Please refer to Figure P11.40. We know ax = slope of the vx-versus-t graph and x = area under the vx-versus-x graph between 0 and x. Solve: Using the above definitions and methodology, we can generate the following table: t(s) ax (m/s2) x(m) K(J) F(N)

11.40. Model: Use ax =

0 0.5 1.0 1.5 2.0 2.5

10 10 10 10 +10 or −10 −10 −10 or 0 0 0

3.0 3.5 4.0

0 1.25 5 11.25 20 28.75

0 6.25 25 56.25 100 56.25

35 40 45

25 25 25

5 5 5 5 −5 or +5 −5 −5 or 0 0 0

(e) Let J1 be the impulse from t = 0 s to t = 2 s and J 2 be the impulse from t = 2 s to t = 4 s. We have 2s

J1 =

∫F

x

4s

dt = (5 N)(2 s) = 10 N ⋅ s and J 2 =

0s

∫F

x

dt = (−5 N)(1 s) = −5 N ⋅ s

0s

(f) J = Δp = mvf − mvi ⇒ vf = vi + J / m

At t = 2 s, vx = 0 m/s + (10 N ⋅ s) / 0.5 kg = 0 m/s + 20 m/s = 20 m/s At t = 4 s, vx = 20 m/s + ( −5 N ⋅ s) / 0.5 kg = 20 m/s − 10 m/s = 10 m/s The vx-versus-t graph also gives vx = 20 m/s at t = 2 s and vx = 10 m/s at t = 4 s. (g)

(h) From t = 0 s to t = 2 s, W = ∫ Fx dx = (5 N)(20 m) = 100 J

From t = 2 s to t = 4 s, W = ∫ Fx dx = (−5 N)(15 m) = −75 J (i) At t = 0 s, vx = 0 m/s so the work-kinetic energy theorem for calculating vx at t = 2 s is 1 1 1 1 W = ΔK = mvf2 − mvi2 ⇒ 100 J = (0.5 kg)vx2 − (0.5 kg)(0 m/s) 2 ⇒ vx = 20 m/s 2 2 2 2

To calculate vx at t = 4 s, we use vx at t = 2 s as the initial velocity: 1 1 −75 J = (0.5 kg)vx2 − (0.5 kg)(20 m/s) 2 ⇒ vx = 10 m/s 2 2 Both of these values agree with the values on the velocity graph.

11.41. Model: Model the elevator as a particle. Visualize:

Solve:

(a) The work done by gravity on the elevator is

Wg = −ΔU g = mgy0 − mgy1 = − mg ( y1 − y0 ) = −(1000 kg)(9.8 m/s 2 )(10 m) = −9.8 × 104 J (b) The work done by the tension in the cable on the elevator is

WT = T (Δy )cos0° = T ( y1 − y0 ) = T (10 m) To find T we write Newton’s second law for the elevator:

∑F

y

= T − FG = ma y ⇒ T = FG + ma y = m( g + a y ) = (1000 kg)(9.8 m/s 2 + 1.0 m/s 2 ) = 1.08 × 104 N ⇒ WT = (1.08 × 104 N)(10 m) = 1.08 × 105 J

(c) The work-kinetic energy theorem is

1 1 Wnet = Wg + WT = ΔK = K f − K i = K f − mv02 ⇒ K f = Wg + WT + mv02 2 2 1 ⇒ K f = ( −9.8 × 104 J) + (1.08 × 105 J) + (1000 kg)(0 m/s) 2 = 1.0 × 104 J 2 (d)

1 1 K f = mvf2 ⇒ 1.0 × 104 J = (1000 kg)v 2f ⇒ vf = 4.5 m/s 2 2

11.42. Model: Model the rock as a particle, and apply the work-kinetic energy theorem. Visualize:

Solve:

(a) The work done by Bob on the rock is

1 1 1 1 WBob = ΔK = mv12 − mv02 = mv12 = (0.500 kg)(30 m / s) 2 = 225 J = 2.3 × 102 J 2 2 2 2 (b) For a constant force, WBob = FBob Δx ⇒ FBob = WBob / Δx = 2.3 × 102 N. (c) Bob’s power output is PBob = FBob vrock and will be a maximum when the rock has maximum speed. This is just

as he releases the rock with vrock = v1 = 30 m/s. Thus, Pmax = FBob v1 = ( 225 J )( 30 m/s ) = 6750 W = 6.8 kW.

11.43. Model: Model the crate as a particle, and use the work-kinetic energy theorem. Visualize:

Solve:

(a) The work-kinetic energy theorem is ΔK = 12 mv12 − 12 mv02 = 12 mv12 = Wtotal . Three forces act on the box,

so Wtotal = Wgrav + Wn + Wpush . The normal force is perpendicular to the motion, so Wn = 0 J. The other two forces do the following amount of work: G G Wpush = Fpush ⋅ Δr = Fpush Δx cos 20° = 137.4 J

G G Wgrav = FG ⋅ Δr = ( FG ) x Δ x = (− mg sin 20°)Δ x = −98.0 J

Thus, Wtotal = 39.4 J, leading to a speed at the top of the ramp equal to

v1 =

2Wtotal 2(39.4 J) = = 4.0 m/s 5.0 kg m

(b) The x-component of Newton’s second law is ax = a =

( Fnet ) x Fpush cos 20° − FG sin 20° Fpush cos 20° − mg sin 20° = = = 1.35 m/s 2 m m m

Constant-acceleration kinematics with x1 = h / sin 20° = 5.85 m gives the final speed v12 = v02 + 2a ( x1 − x0 ) = 2ax1 ⇒ v1 = 2ax1 = 2(1.35 m / s 2 )(5.85 m) = 4.0 m/s

11.44. Model: Model Sam strapped with skis as a particle, and apply the law of conservation of energy. Visualize:

Solve:

(a) The conservation of energy equation is

K1 + U g1 + Δ Eth = K 0 + U g0 + Wext The snow is frictionless, so ΔEth = 0 J. However, the wind is an external force doing work on Sam as he moves down the hill. Thus,

Wext = Wwind = ( K1 + U g1 ) − ( K 0 + U g0 ) 1 ⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ = ⎜ mv12 + mgy1 ⎟ − ⎜ mv02 + mgy0 ⎟ = ⎜ mv12 + 0 J ⎟ − (0 J + mgy0 ) = mv12 − mgy0 2 ⎝2 ⎠ ⎝2 ⎠ ⎝2 ⎠ ⇒ v1 = 2 gy0 +

2Wwind m

We compute the work done by the wind as follows: G G Wwind = Fwind ⋅ Δr = Fwind Δr cos160° = (200 N)(146 m)cos160° = −27,400 J

where we have used Δr = h / sin 20° = 146 m. Now we can compute

v1 = 2(9.8 m / s 2 )(50 m) +

2( −27,400 J) = 15.7 m / s 75 kg

(b) We will use a tilted coordinate system, with the x-axis parallel to the slope. Newton’s second law for Sam is

( Fnet ) x FG sin 20° − Fwind cos 20° mg sin 20° − Fwind cos 20° = = m m m (75 kg)(9.8 m/s 2 )sin 20° − (200 N)cos 20° = = 0.846 m/s 2 75 kg

ax = a =

Now we can use constant-acceleration kinematics as follows:

v12 = v02 + 2a( x1 − x0 ) = 2ax1 ⇒ v1 = 2ax1 = 2(0.846 m/s 2 )(146 m) = 15.7 m/s Assess:

We used a vertical y-axis for energy analysis, rather than a tilted coordinate system, because U g is

determined by its vertical position.

11.45. Model: Model Paul and the mat as a particle, assume the mat to be massless, use the model of kinetic friction, and apply the work-kinetic energy theorem. Visualize:

We define the x-axis along the floor and the y-axis perpendicular to the floor. fk . Newton’s second Solve: We need to first determine

law

in

the

y-direction

is

n + T sin 30° = FG = mg ⇒ n = mg − T sin 30° = (10 kg)(9.8 m/s ) − (30 N)(sin 30°) = 83.0 N. Using n and the model of 2

kinetic friction,

f k = μ k n = (0.2)(83.0 N) = 16.60 N. The net force on Paul and the mat is therefore

Fnet = T cos30° − f k = ( 30 N ) cos30° − 16.6 N = 9.4 N . Thus,

Wnet = Fnet Δr = (9.4 N)(3.0 m) = 28 J

G G G The other forces n and FG make an angle of 90° with Δr and do zero work. We can now use the work-kinetic energy theorem to find the final velocity as follows: 1 Wnet = K f − K i = K f − 0 J = mvf2 ⇒ vf = 2Wnet / m = 2(28 J)/10 kg = 2.4 m/s 2

Assess:

A speed of 2.4 m/s or 5.4 mph is reasonable for the present problem.

11.46. Model: Assume an ideal spring that obeys Hooke’s law. Model the box as a particle and use the model of kinetic friction. Visualize:

Solve:

When the horizontal surface is frictionless, conservation of energy means

1 1 1 k ( x0 − xe ) 2 = mv12x = K1 ⇒ K1 = (100 N/m)(0.20 m − 0 m) 2 = 2.0 J 2 2 2 That is, the box is launched with 2.0 J of kinetic energy. It will lose 2.0 J of kinetic energy on the rough surface. G G The net force on the box is Fnet = − f k = − μ k mgiˆ. The work-kinetic energy theorem is G G Wnet = Fnet ⋅ Δr = K 2 − K1 = 0 J − 2.0 J = −2.0 J

⇒ (− μ k mg )( x2 − x1 ) = −2.0 J ⇒ ( x2 − x1 ) =

2.0 J 2.0 J = = 54 cm ( μ k )( mg ) (0.15)(2.5 kg)(9.8 m/s 2 )

Assess: Because the force of friction transforms kinetic energy into thermal energy, energy is transferred out of the box into the environment. In response, the box slows down and comes to rest.

11.47. Model: Model the suitcase as a particle, use the model of kinetic friction, and use the work-kinetic energy theorem. Visualize:

G G The net force on the suitcase is Fnet = f k . Solve: The work-kinetic energy theorem is

G 1 1 1 1 G G G Wnet = ΔK = mv12 − mv02 ⇒ Fnet ⋅ Δr = f k ⋅ Δr = 0 J − mv02 ⇒ ( f k )( x1 − x0 )cos180° = − mv02 2 2 2 2 1 2 v02 (1.2 m/s) 2 ⇒ − μ k mg ( x1 − x0 ) = − mv0 ⇒ μ k = = = 0.037 2 2 g ( x1 − x0 ) 2(9.8 m/s 2 )(2.0 m − 0 m) Assess: Friction transforms kinetic energy of the suitcase into thermal energy. In response, the suitcase slows down and comes to rest.

11.48. Model: Identify the truck and the loose gravel as the system. We need the gravel inside the system because friction increases the temperature of the truck and the gravel. We will also use the model of kinetic friction and the conservation of energy equation. Visualize:

We place the origin of our coordinate system at the base of the ramp in such a way that the x-axis is along the ramp and the y-axis is vertical so that we can calculate potential energy. The free-body diagram of forces on the truck is shown. Solve: The conservation of energy equation is K1 + U g1 + Δ Eth = K 0 + U g0 + Wext . In the present case,

Wext = 0 J, v1x = 0 m/s, U g 0 = 0 J, v0 x = 35 m/s. The thermal energy created by friction is

ΔEth = ( f k )( x1 − x0 ) = ( μ k n)( x1 − x0 ) = μ k ( mg cos6.0°)( x1 − x0 ) = (0.40)(15,000 kg)(9.8 m/s 2 )(cos6.0°)( x1 − x0 ) = (58,478 J/m)( x1 − x0 ) Thus, the energy conservation equation simplifies to 1 0 J + mgy1 + (58,478 J/m)( x1 − x0 ) = mv02x + 0 J + 0 J 2 1 (15,000 kg)(9.8 m/s 2 )[( x1 − x0 )sin 6.0°] + (58,478 J/m)( x1 − x0 ) = (15,000 kg)(35 m/s) 2 2 ⇒ ( x1 − x0 ) = 124 m Assess:

A length of 124 m at a slope of 6º seems reasonable.

11.49. Model: We will use the spring, the package, and the ramp as the system. We will model the package as a particle. Visualize:

We place the origin of our coordinate system on the end of the spring when it is compressed and is in contact with the package to be shot. Model: (a) The energy conservation equation is

K1 + U g1 + U s1 + Δ Eth = K 0 + U g 0 + U s0 + Wext 1 2 1 1 1 mv1 + mgy1 + k ( xe − xe ) 2 + Δ Eth = mv02 + mgy0 + k (Δx) 2 + Wext 2 2 2 2 Using y1 = 1 m, ΔEth = 0 J (note the frictionless ramp), v0 = 0 m/s, y0 = 0 m, Δx = 30 cm, and Wext = 0 J, we get

1 2 1 mv1 + mg (1 m) + 0 J + 0 J = 0 J + 0 J + k (0.30 m)2 + 0 J 2 2 1 1 2 2 (2.0 kg)v1 + (2.0 kg)(9.8 m/s )(1 m) = (500 N/m)(0.30 m) 2 2 2 ⇒ v1 = 1.70 m/s (b) How high can the package go after crossing the sticky spot? If the package can reach y1 ≥ 1.0 m before

stopping (v1 = 0), then it makes it. But if y1 < 1.0 m when v1 = 0, it does not. The friction of the sticky spot generates thermal energy

ΔEth = ( μ k mg )Δx = (0.30)(2.0 kg)(9.8 m/s 2 )(0.50 m) = 2.94 J The energy conservation equation is now 1 2

mv12 + mgy1 + ΔEth = 12 k ( Δx) 2

If we set v1 = 0 m/s to find the highest point the package can reach, we get

y1 = ( 12 k (Δx) 2 − ΔEth ) / mg = ( 12 (500 N/m)(0.30 m) 2 − 2.94 J ) /(2.0 kg)(9.8 m/s 2 ) = 0.998 m The package does not make it. It just barely misses.

11.50. Model: Model the two blocks as particles. The two blocks make our system. Visualize:

We place the origin of our coordinate system at the location of the 3.0 kg block. Solve: (a) The conservation of energy equation is K f + U gf + ΔEth = K i + U gi + Wext . Using ΔEth = 0 J and

Wext = 0 J we get 1 1 1 1 m2 (vf ) 22 + m3 (vf )32 + m2 g ( yf ) = m2 (vi ) 22 + m3 (vi )32 + m2 g ( yi ) 2 2 2 2 Noting that (vf ) 2 = (vf )3 = vf and (vi ) 2 = (vi )3 = 0 m / s, this becomes

1 ( m2 + m3 )vf2 = −m2 g ( yf − yi ) 2 vf =

2m2 g ( yi − yf ) 2(2.0 kg)(9.8 m/s 2 )(1.50 m) = = 3.4 m/s m2 + m3 (2.0 kg + 3.0 kg)

(b) We will use the same energy conservation equation. However, this time

ΔEth = μ k (m3 g )(Δ x) = (0.15)(3.0 kg)(9.8 m/s 2 )(1.50 m) = 6.615 J The energy conservation equation is now 1 1 1 1 m2vf2 + m3vf2 + m2 gyf + 6.615 J = m2 (vi ) 22 + m3 (v1 )32 + m2 gyi + 0 J 2 2 2 2 ⎛ 1 2 ⎞ (m2 + m3 )vf2 + 6.615 J = m2 g ( yi − yf ) ⇒ vf = ⎜ ⎟[ m2 g ( yi − yf ) − 6.615 J] m 2 + ⎝ 2 m3 ⎠ ⎛ 2 ⎞ 2 = ⎜ ⎟[(2.0 kg)(9.8 m/s )(1.50 m) − 6.615 J] = 3.0 m/s ⎝ 5.0 kg ⎠ Assess:

A reduced speed when friction is present compared to when there is no friction is reasonable.

G

G

11.51. Model: Use the particle model, the definition of work W = F ⋅ Δs , and the model of kinetic friction. Visualize: We place the coordinate frame on the incline so that its x-axis is along the incline.

Solve:

(a)

G G WT = T ⋅ Δr = (T )(Δx)cos18° = (120 N)(5.0 m)cos18° = 0.57 kJ

G G Wg = ( FG ⋅ Δr ) = mg (Δx)cos(120°) = (8 kg)(9.8 m/s 2 )(5.0 m)cos120° = −196 J G G Wn = (n ⋅ Δr ) = (n)(Δ x)cos90° = 0 J (b) The amount of energy transformed into thermal energy is ΔEth = ( f k )(Δx) = ( μ k n)(Δx). To find n, we write Newton’s second law as follows:

∑F

y

= n − FG cos30° + T sin18° = 0 N ⇒ n = FG cos30° − T sin18°

⇒ n = mg cos30° − T sin18° = (8.0 kg)(9.8 m/s 2 )cos30° − (120 N)sin18° = 30.814 N Thus, ΔEth = (0.25)(30.814 N)(5.0 m) = 38.5 J. Assess: Any force that acts perpendicular to the displacement does no work.

11.52. Model: Assume the spring is ideal so that Hooke’s law is obeyed, and model the weather rocket as a particle. Visualize:

The origin of the coordinate system is placed on the free end of the spring. Note that the bottom of the spring is anchored to the ground. Solve: (a) The rocket is initially at rest. The free-body diagram on the rocket helps us write Newton’s second law as

(∑ F ) = 0 N ⇒ F y

sp

⇒ Δy =

= FG = mg ⇒ k Δy = mg

mg (10.2 kg)(9.8 m/s 2 ) = = 20.0 cm k (500 N/m)

(b) The thrust does work. Using the energy conservation equation when y2 − y0 = 40 cm = 0.40 m:

K 2 + U g2 + U sp2 = K1 + U g1 + U sp1 + Wext 1 1 1 1 Wext = F ( y2 − y1 ) ⋅ mv22 + mgy2 + k ( y2 − ye ) 2 = mv12 + mgy1 + k ( y1 − ye ) 2 + (200 N)(0.60 m) 2 2 2 2 2 (5.10 kg)v2 + 40.0 J + 40.0 J = 0 − 20.0 J + 10.0 J + 120 J ⇒ v2 = 2.43 m/s If the rocket were not attached to the spring, the energy conservation equation would not involve the spring energy term U sp2 . That is,

K 2 + U g 2 = K1 + U g1 + U sp1 + Wext 1 (10.2 kg)v22 + (10.2 kg)(9.8 m/s 2 )(0.40 m) = 0 J − (10.2 kg)(9.8 m/s 2 )(0.20 m) 2 1 + (500 N/m)(0.20 m) 2 + (200 N)(0.60 m) 2

⇒ (5.10 kg)v22 = 70.0 J ⇒ v2 = 3.70 m/s

11.53. Model: Use the particle model for the ice skater, the model of kinetic/static friction, and the workkinetic energy theorem. Visualize:

Solve:

(a) The work-kinetic energy theorem is 1 1 ΔK = mv12 − mv02 = Wnet = Wwind 2 2

There is no kinetic friction along her direction of motion. Static friction acts to prevent her skates from slipping sideways on the ice, but this force is perpendicular to the motion and does not contribute to a change in thermal G G energy. The angle between Fwind and Δr is θ = 135°, so G G Wwind = Fwind ⋅ Δr = Fwind Δy cos135° = (4 N)(100 m)cos135° = −282.8 J

Thus, her final speed is v1 = v02 +

2Wwind = 2.16 m/s m

(b) If the skates don’t slip, she has no acceleration in the x-direction and so ( Fnet ) x = 0 N. That is:

f s − Fwind cos 45° = 0 N ⇒ fs = Fwind cos 45° = 2.83 N Now there is an upper limit to the static friction: f s ≤ ( fs ) max = μ s mg . To not slip requires

μs ≥

fs 2.83 N = = 0.0058 mg (50 kg)(9.8 m/s 2 )

Thus, the minimum value of μs is 0.0058. Assess: The work done by the wind on the ice skater is negative, because the wind slows the skater down.

11.54. Model: Model the ice cube as a particle, the spring as an ideal that obeys Hooke’s law, and the law of conservation of energy. Visualize:

Solve: (a) The normal force does no work and the slope is frictionless, so mechanical energy is conserved. We’ve drawn two separate axes: a vertical y-axis to measure potential energy and a tilted s-axis to measure distance along the slope. Both have the same origin which is at the point where the spring is not compressed. Thus, the two axes are related by y = s sin θ . Also, this choice of origin makes the elastic potential energy simply

U s = 12 k ( s − s0 ) 2 = 12 ks 2 . Because energy is conserved, we can relate the initial point—with the spring compressed—to the final point where the ice cube is at maximum height. We do not need to find the speed with which it leaves the spring. We have K 2 + U g 2 + U s2 = K1 + U g1 + U s1 1 2 1 1 1 mv2 + mgy2 + ks02 = mv12 + mgy1 + ks12 2 2 2 2 It is important to note that at the final point, when the ice cube is at y2, the end of the spring is only at s0. The spring does not stretch to s2 , so U s2 is not 12 ks22 . Three of the terms are zero, leaving 1 ks 2 mgy2 = + mgy1 + ks12 ⇒ y2 − y1 = Δy = height gained = 1 = 0.255 m = 25.5 cm 2 2mg

The distance traveled is Δs = Δy / sin 30° = 51.0 cm. (b) Using the energy equation and the expression for thermal energy:

K 2 + U g 2 + U s2 + ΔEth = K1 + U g1 + U s1 + Wext

ΔEth = f k Δs = μ k nΔs

From the free-body diagram,

G ( Fnet ) y = 0 N = n − mg cos30° ⇒ n = mg cos30° Now, having found ΔEth = μ k (mg cos30°) Δs, the energy equation can be written 1 0 J + mgy2 + 0 J + μ k (mg cos30°)Δs = 0 J + mgy1 + ks12 + 0 J 2 1 2 ⇒ mg ( y2 − y1 ) − ks1 + μ k mg cos30°Δs = 0 2 Using Δy = ( Δs )sin 30°, the above equation simplifies to

ks12 1 mg Δs sin 30° + μ k mg cos30°Δs = ks12 ⇒ Δs = = 0.379 m = 37.9 cm 2 2mg (sin 30° + μ k cos30°)

11.55. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the box as a particle and apply the energy conservation law. Box, spring, and the ground make our system, and we also use the model of kinetic friction. Visualize: We place the origin of the coordinate system on the ground directly below the box’s starting position.

Solve:

(a) The energy conservation equation is

K1 + U g1 + U s1 + ΔEth = K 0 + U g0 + U s0 + Wext 1 2 1 1 mv1 + mgy1 + 0 J + 0 J = mv02 + mgy0 + 0 J + 0 J ⇒ mv12 + 0 J = 0 J + mgy0 2 2 2 ⇒ v1 = 2 gy0 = 2(9.8m/s 2 )(5.0 m) = 9.9 m/s (b) The friction creates thermal energy. The energy conservation equation for this part of the problem is

1 2 1 mv2 + 0 J + 0 J + μ k mg ( x2 − x1 ) = mv12 + 0 J + 0 J + 0 J 2 2 1 2 1 2 1 2 1 2 mv2 + μ k n( x2 − x1 ) = mv1 ⇒ mv2 + μ k mg ( x2 − x1 ) = mv1 2 2 2 2

K 2 + U g 2 + U s 2 + ΔEth = K1 + U g1 + U s1 + Wext

⇒ v2 = v12 − 2μ k g ( x2 − x1 ) = (9.9 m/s) 2 − 2(0.25)(9.8 m/s 2 )(2.0 m) = 9.4 m/s (c) To find how much the spring is compressed, we apply the energy conservation once again:

K 3 + U g3 + U s3 + ΔEth = K 2 + U g 2 + U s2 + Wext

1 1 0 J + 0 J + k ( x3 − x2 ) 2 + 0 J = mv22 + 0 J + 0 J + 0 J 2 2

Using v2 = 9.4 m/s, k = 500 N/m and m = 5.0 kg, the above equation yields ( x3 − x2 ) = Δx = 94 cm. (d) The initial energy = mgy0 = (5.0 kg)(9.8 m / s 2 )(5.0 m) = 254 J. The energy transformed to thermal energy during each passage is

μ k mg ( x2 − x1 ) = (0.25)(5.0 kg)(9.8 m / s 2 )(2.0 m) = 24.5 J The number of passages is equal to 245 J / 24.5 J or 10.

11.56. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the physics student as a particle and apply the law of conservation of energy. Our system is comprised of the spring, the student, and the ground. We also use the model of kinetic friction. Visualize: We place the origin of the coordinate system on the ground directly below the end of the compressed spring that is in contact with the student.

Solve:

(a) The energy conservation equation is

K1 + U g1 + U s1 + ΔEth = K 0 + U g0 + U s0 + Wext 1 2 1 1 1 mv1 + mgy1 + k ( x1 − xe ) 2 + 0 J = mv02 + mgy0 + k ( x1 − x0 ) 2 + 0 J 2 2 2 2 Since y1 = y0 = 10 m, x1 = xe , v0 = 0 m/s, k = 80,000 N/m, m = 100 kg, and ( x1 − x0 ) = 0.5 m, k 1 2 1 mv1 = k ( x1 − x0 ) 2 ⇒ v1 = ( x1 − x0 ) = 14.14 m/s m 2 2 (b) Friction creates thermal energy. Applying the conservation of energy equation once again:

K 2 + U g 2 + U s2 + ΔEth = K 0 + U g 0 + U s0 + Wext 1 2 1 mv2 + mgy2 + 0 J + f k Δs = 0 J + mgy0 + k ( x1 − x0 ) 2 + 0 J 2 2 With v2 = 0 m/s and y2 = (Δs )sin 30°, the above equation is simplified to 1 mg (Δs )sin 30° + μ k nΔs = mgy0 + k ( x1 − x0 ) 2 2

From the free-body diagram for the physics student, we see that n = FG cos30° = mg cos30°. Thus, the conservation of energy equation gives 1 Δs (mg sin 30° + μ k mg cos30°) = mgy0 + k ( x1 − x0 ) 2 2

Using m = 100 kg, k = 80,000 N/m, ( x1 − x0 ) = 0.50 m, y0 = 10 m, and μ k = 0.15, we get

1 mgy0 + k ( x1 − x0 ) 2 2 Δs = = 32.1 m mg (sin 30° + μ k cos30°) Assess: y2 = (Δs )sin 30° = 16.05 m, which is greater than y0 = 10 m. The higher value is due to the transformation of the spring energy into gravitational potential energy.

11.57. Model: Treat the block as a particle, use the model of kinetic friction, and apply the energy conservation law. The block and the incline comprise our system. Visualize: We place the origin of the coordinate system directly below the block’s starting position at the same level as the horizontal surface. On the horizontal surface the model of kinetic friction applies.

Solve:

(a) For the first incline, the conservation of energy equation gives 1 2 K1 + U g1 + ΔEth = K 0 + U g0 + Wext mv1 + 0 J + 0 J = 0 J + mgy0 + 0 J ⇒ v1 = 2 gy0 = 2 gh 2 (b) The friction creates thermal energy. Applying once again the conservation of energy equation, we have K 3 + U g3 + ΔEth = K1 + U g1 + Wext

1 2 1 mv3 + mgy3 + μ k mg ( x2 − x1 ) = mv12 + mgy1 + Wext 2 2

Using v3 = 0 m/s, y1 = 0 m, Wext = 0 J, v1 = 2 gh , and ( x2 − x1 ) = L, we get 1 mgy3 + μ k mgL = m(2 gh) ⇒ y3 = h − μ k L 2

Assess:

For μ k = 0, y3 = h which is predicted by the law of the conservation of energy.

11.58. Model: Assume an ideal spring, so Hooke’s law is obeyed. Visualize:

Solve: For a conservative force the work done on a particle as it moves from an initial to a final position is independent of the path. We will show that WA →C → B = WA → B for the spring force. Work done by a spring force F = − kx is given by xf

W = ∫ Fdx = − ∫ kx dx xi

This means xB

WA → B = − ∫ kx dx = − xA

x

x

C B k 2 k k xB − xA2 ) , WA → C = − ∫ kx dx = − ( xC2 − xA2 ) , and WC → B = − ∫ kx dx = − ( xB2 − xC2 ) ( 2 2 2 xA xC

Adding the last two:

WA → C → B = WA → C + WC → B = −

k 2 ( xC − xA2 + xB2 − xC2 ) = WA → B 2

11.59. Model: A “sprong” that obeys the force law Fx = − q( x − xe )3 , where q is the sprong constant and xe is the equilibrium position. Visualize: We place the origin of the coordinate system on the free end of the sprong, that is, xe = xf = 0 m.

Solve: (a) The units of q are N/m3 . (b) A cubic curve rises more steeply than a parabola. The force increases by a factor of 8 every time x increases by a factor of 2.

x dU qx 4 , we have U ( x) = − ∫ Fx dx = − ∫ (− qx3 )dx = . 0 dx 4 (d) Applying the energy conservation equation to the ball and sprong system: K f + U f = Ki + U i

(c) Since Fx = −

1 2 qx 4 mvf + 0 J = 0 J + i 2 4 ⇒ vf =

q x4 (40,000 N/m3 ) ( −0.10 m) 4 = ⋅ = 10 m/s m 2 (0.020 kg) 2

11.60. Solve: (a) Because sin (cx) is dimensionless, F0 must have units of force in newtons. (b) The product cx is an angle because we are taking the sine of it. An angle has no real physical units. If x has units of m and the product cx is unitless, then c has to have units of m -1. (c) The force is a maximum when sin(cx) = 1. This occurs when cx = π /2, or for xmax = π /2c. (d) The graph is the first quarter of a sine curve.

(e) We can find the velocity vf at xf = xmax from the work-kinetic energy theorem:

1 1 1 1 2W ΔK = mvf2 − mvi2 = mvf2 − mv02 = W ⇒ vf = v02 + m 2 2 2 2 This is a variable force. As the particle moves from xi = 0 m to xf = xmax = π /2c, the work done on it is xf

π / 2c

xi

0

W = ∫ F ( x) dx = F0 ∫

sin(cx) dx = −

F0 F ⎛ π ⎞ F cos(cx)|π0 / 2 c = − 0 ⎜ cos − cos0 ⎟ = 0 2 c c ⎝ ⎠ c

Thus, the particle’s speed at xf = xmax = π /2c is vf = v02 + 2 F0 / mc .

11.61. Visualize: We place the origin of the coordinate system at the base of the stairs on the first floor.

Solve: (a) We might estimate y2 − y1 ≈ 4.0 m ≈ 12 ft ≈ y3 − y2 , thus, y3 − y1 ≈ 8.0 m. (b) We might estimate the time to run up these two flights of stairs to be 20 s. (c) Estimate your mass as m ≈ 70 kg ≈ 150 lb. Your power output while running up the stairs is

work done by you change in potential energy mg ( y3 − y1 ) = = time time time (70 kg)(9.8 m/s 2 )(8.0 m) ⎛ 1 hp ⎞ = ≈ 270 W = (270 W) ⎜ ⎟ ≈ 0.35 hp 20 s ⎝ 746 W ⎠ Assess:

Your estimate may vary, depending on your mass and how fast you run.

11.62. Solve: Power output during the push-off period is equal to the work done by the cat divided by the time the cat applied the force. Since the force on the floor by the cat is equal in magnitude to the force on the cat by the floor, work done by the cat can be found using the work-kinetic energy theorem during the push-off period, Wnet = Wfloor = ΔK . We do not need to explicitly calculate Wcat , since we know that the cat’s kinetic energy is transformed into its potential energy during the leap. That is, ΔU g = mg ( y2 − y1 ) = (5.0 kg)(9.8 m / s 2 )(0.95 m) = 46.55 J Thus, the average power output during the push-off period is P=

Wnet 46.55 J = = 0.23 kW 0.20 s t

11.63. Solve: Using the conversion 746 W = 1 hp, we have a power of 1492 J/s. This means W = Pt = (1492 J/s)(1 h) = 5.3712 × 106 J is the total work done by the electric motor in one hour. Furthermore,

Wmotor = −Wg = U gf − U gi = mg ( yf − yi ) = mg (10 m) m=

Wmotor 5.3712 × 106 J 1 liter = = 5.481 × 104 kg = 5.481 × 104 kg × = 5.5 × 104 liters g (10 m) (9.8 m/s 2 )(10 m) 1 kg

11.64. Solve: (a) The change in the potential energy of 1.0 kg of water in falling 25 m is ΔU g = − mgh = −(1.0 kg)(9.8 m/s 2 )(25 m) = −245 J ≈ −0.25 kJ (b) The power required of the dam is P=

W W = = 50 × 106 Watts ⇒ W = 50 × 106 J t 1s

That is, 50 × 106 J of energy is required per second for the dam. Out of the 245 J of lost potential energy, (245 J)(0.80) = 196 J is converted to electrical energy. Thus, the amount of water needed per second is (50 × 106 J)(1 kg /196 J) = 255,000 kg ≈ 2.6 × 105 kg.

11.65. Solve: The force required to tow a water skier at a speed v is Ftow = Av. Since power P = Fv, the power required to tow the water skier is Ptow = Ftow v = Av 2 . We can find the constant A by noting that a speed of v = 2.5 mph requires a power of 2 hp. Thus,

(2 hp) = A(2.5 mph) 2 ⇒ A = 0.32

hp (mph) 2

Now, the power required to tow a water skier at 7.5 mph is Ptow = Av 2 = 0.32 Assess:

hp ⋅ (7.5 mph) 2 = 18 hp (mph) 2

Since P ∝ v 2 , a three-fold increase in velocity leads to a nine-fold increase in power.

11.66. Solve: By definition, the maximum power output of a horse is P ≈ 1 hp = 746 W. At maximum G G speed, when the horse is running at constant speed, Fnet = 0. The propulsion force Fhorse , provided by the horse

pushing against the ground, is balanced by the drag force of air resistance: Fhorse = D. We learned in Chapter 6 that a reasonable model for drag is D ≈ 14 Av 2 , where A is the cross section area. Since the power needed for force F to push an object at velocity v is P = Fv, we have

P ≈ 1 hp = 746 W = Fhorsev = ( 14 Av 2 ) v = 14 Av3 1/ 3

⎛ 4P ⎞ v≈⎜ ⎟ ⎝ A⎠ Assess:

1/ 3

⎛ 4(746 W) ⎞ =⎜ ⎟ ⎝ (0.5 m)(1.8 m) ⎠

= 15 m/s

15 m/s ≈ 30 mph is a reasonable top speed for a well-trained horse.

11.67. Solve: The net force on a car moving at a steady speed is zero. The motion is opposed both by rolling friction and by air resistance. Thus the propulsion force provided by the drive wheels must be Fcar = μ r mg + 14 Av 2 , where μ r is the rolling friction, m is the mass, A is the cross-section area, and v is the car’s velocity. The power required to move the car at speed v is 1 3 Av 4 Since the maximum power output is 200 hp and 75% of the power reaches the drive wheels, P = (200 hp)(0.75) = 150 hp. Thus, P = Fcar v = μ r mgv +

⎛ 746 W ⎞ 1 2 3 (150 hp) ⎜ ⎟ = (0.02)(1500 kg)(9.8 m/s )v + (1.6 m)(1.4 m)v 1 hp 4 ⎝ ⎠ ⇒ 0.56 v3 + 294 v − 111,900 = 0 ⇒ v = 55.5 m/s The easiest way to solve this equation is through iterations by trial and error. Assess: A speed of 55.5 m/s ≈ 110 mph is very reasonable.

11.68. Model: Use the model of static friction, kinematic equations, and the definition of power.

Solve: (a) The rated power of the Porsche is 217 hp = 161,882 W and the gravitational force on the car is

(1480 kg)(9.8 m/s 2 ) = 14,504 N. The amount of that force on the drive wheels is (14,504)(2/3) = 9670 N. Because the static friction of the tires on road pushes the car forward, Fmax = fs max = μs n = μ s mg = (1.00)(9670 N) = mamax ⇒ amax =

9670 N = 6.53 m/s 2 1480 kg

(b) Only 70% of the power generated by the motor is applied at the wheels. P = Fvmax ⇒ vmax =

P (0.70)(161,882 W) = = 11.7 m/s 9670 N F

(c) Using the kinematic equation, vmax = v0 + amax (tmin − t0 ) with v0 = 0 m/s and t0 = 0 s, we obtain

tmin = Assess:

vmax 11.7 m/s = = 1.79 s amax 6.53 m/s 2

An acceleration time of 1.79 s for the Porsche to reach a speed of ≈ 26 mph from rest is reasonable.

11.69. (a) A student uses a string to pull her 2.0 kg physics book, starting from rest, across a 2.0-m-long lab bench. The coefficient of kinetic friction between the book and the lab bench is 0.15. If the book’s final speed is 4.0 m/s, what is the tension in the string? (b)

(c) The tension does external work Wext . This work increases the book’s kinetic energy and also causes an

increase ΔEth in the thermal energy of the book and the lab bench. Solving the equation gives T = 10.9 N.

11.70. (a) A 20 kg chicken crate slides down a 2.5-m-high, 40° ramp from the back of a truck to the ground. The coefficient of kinetic friction between the crate and the ramp bench is 0.15. How fast are the chickens going at the bottom of the ramp? (b)

(c) v1 = 6.34 m/s.

11.71. (a) If you expend 75 W of power to push a 30 kg sled on a surface where the coefficient of kinetic friction between the sled and the surface is μ k = 0.20, what speed will you be able to maintain? (b)

(c) Fpush = (0.20)(30 kg)(9.8 m/s 2 ) = 58.8 N ⇒ 75 W = (58.8 N)v ⇒ v =

75 W = 1.28 m/s 58.8 N

11.72. (a) A 1500 kg object is being accelerated upward at 1.0 m/s2 by a rope. How much power must the motor supply at the instant when the velocity is 2.0 m/s? (b)

(c) T = (1500 kg)(9.8 m/s 2 ) + 1500 kg(1.0 m/s 2 ) = 16,200 N=16.2 kN

P = T (2m/s) = (16,200 N)(2.0 m/s) = 32,400 W = 32 kW

11.73. Model: Model the water skier as a particle, apply the law of conservation of mechanical energy, and use the constant-acceleration kinematic equations. Visualize:

We placed the origin of the coordinate system at the base of the frictionless ramp. Solve: We’ll start by finding the smallest speed v1 at the top of the ramp that allows her to clear the shark tank. From the vertical motion for jumping the shark tank, 1 y2 = y1 + v1 y (t2 − t1 ) + a y (t2 − t1 ) 2 2 1 ⇒ 0 m = (2.0 m) + 0 m + (−9.8 m/s 2 )(t2 − t1 ) 2 ⇒ (t2 − t1 ) = 0.639 s 2 From the horizontal motion, 1 x2 = x1 + v1x (t2 − t1 ) + ax (t2 − t1 ) 2 2 5.0 m ⇒ ( x1 + 5.0 m) = x1 + v1 (0.639 s) + 0 m ⇒ v1 = = 7.825 m/s 0.639 s Having found the v1 that will take the skier to the other side of the tank, we now use the energy equation to find the minimum speed v0. We have 1 1 K1 + U g1 = K 0 + U g0 ⇒ mv12 + mgy1 = mv02 + mgy0 2 2

v0 = v12 + 2 g ( y1 − y0 ) = (7.825 m/s) 2 + 2(9.8 m/s 2 )(2.0 m) = 10.0 m/s

11.74. Model: Assume the spring to be ideal that obeys Hooke’s law, and model the block as a particle. Visualize: We placed the origin of the coordinate system on the free end of the compressed spring which is in contact with the block. Because the horizontal surface at the bottom of the ramp is frictionless, the spring energy appears as kinetic energy of the block until the block begins to climb up the incline.

Solve: Although we could find the speed v1 of the block as it leaves the spring, we don’t need to. We can use energy conservation to relate the initial potential energy of the spring to the energy of the block as it begins projectile motion at point 2. However, friction requires us to calculate the increase in thermal energy. The energy equation is K 2 + U g2 + ΔEth = K 0 + U g0 + Wext ⇒ 12 mv22 + mgy2 + f k Δs = 12 k ( x0 − xe ) 2 The distance along the slope is Δs = y2 / sin 45°. The friction force is f k = μ k n, and we can see from the freebody diagram that n = mg cos 45°. Thus 1/ 2

⎡k ⎤ v2 = ⎢ ( x0 − xe ) 2 − 2 gy2 − 2 μ k gy2 cos 45°/ sin45° ⎥ ⎣m ⎦

1/ 2

⎡1000 N/m ⎤ =⎢ (0.15 m) 2 − 2(9.8 m/s 2 )(2.0 m) − 2(0.20)(9.8 m/s 2 )(2.0 m) cos 45° / sin45° ⎥ = 8.091 m/s 0.20 kg ⎣ ⎦ Having found the velocity v2 , we can now find ( x3 − x2 ) = d using the kinematic equations of projectile motion:

1 y3 = y2 + v2 y (t3 − t2 ) + a2 y (t3 − t2 ) 2 2 1 2.0 m = 2.0 m + (v2 sin 45°)(t3 − t2 ) + (−9.8 m / s 2 )(t3 − t2 ) 2 2 ⇒ (t3 − t2 ) = 0 s and 1.168 s Finally, 1 x3 = x2 + v2 x (t3 − t2 ) + a2 x (t3 − t2 ) 2 2 d = ( x3 − x2 ) = (v2 cos 45°)(1.168 s) + 0 m = 6.68 m

11.75. Model: Assume an ideal spring that obeys Hooke’s law, the particle model for the ball, and the model of kinetic friction. The ball, the spring, and the barrel comprise our system. Visualize:

We placed the origin of the coordinate system on the free end of the spring which is in contact with the ball. G Solve: Force F does external work Wext . This work compresses the spring, gives the ball kinetic energy, and is partially dissipated to thermal energy by friction. (a) The energy equation for our system is K1 + U s1 + ΔEth = K 0 + U s0 + Wext

1 2 1 1 1 mv1 + k ( x1 − x0 ) 2 + μ k mg ( x1 − x0 ) = mv02 + k ( xe − xe ) 2 + F ( x1 − x0 ) 2 2 2 2 1 1 (1.0 kg)(2.0 m/s) 2 + (3000 N/m)(0.3 m) 2 + μ k mg (0.30 m) = 0 J + 0 J + F (0.30 m) 2 2 With μ k = 0.30 and m = 1.0 kg, we can solve this equation to obtain F = 460 N. (b) Using the energy equation for our system once again, we have 1 2 1 1 1 mv2 + k ( xe − xe ) 2 + μ k mg ( x2 − x1 ) = mv12 + k ( x1 − x0 ) 2 + 0 J 2 2 2 2 1 2 1 mv2 + μ k mg ( x2 − x1 ) = 0 J+ k ( x1 − x0 ) 2 2 2 1 1 2 2 (1.0 kg)v2 + (0.30)(1.0 kg)(9.8 m/s )(1.5 m) = (3000 N/m)(0.30 m) 2 2 2 (0.5 kg)v22 + 4.41 J=135 J ⇒ v2 = 16.2 m/s

K 2 + U s2 + ΔEth = K1 + U s1 + Wext

11.76. Solve: (a)

The graph is a hyperbola. (b) The separation for zero potential energy is r = ∞, since U =−

Gm1m2 → 0 J as r → ∞ r

This makes sense because two masses don’t interact at all if they are infinitely far apart. (c) Due to the absence of nonconservative forces in our system of two particles, the mechanical energy is conserved.

The equations of energy and momentum conservation are

K f + U gf = K i + U gi

⎛ Gm1m2 ⎞ 1 ⎛ Gm1m2 ⎞ 1 1 1 2 2 m1v1f2 + m2v2f2 + ⎜ − ⎟ = m1v1i + m2v2i + ⎜ − ⎟ 2 2 rf ⎠ 2 2 ri ⎠ ⎝ ⎝ ⎛ 1 1⎞ 1 1 m1v1f2 + m2v2f2 = Gm1m2 ⎜ − ⎟ 2 2 ⎝ rf ri ⎠

pf = pi ⇒ m1v1f + m2v2f = 0 kg m / s ⇒ v2f = −

m1 v1f m2

Substituting this expression for v2f into the energy equation, we get 2

⎞ ⎛ 1 1⎞ 1 1 ⎛m 2Gm2 ⎛ 1 1 ⎞ m1v1f2 + m2 ⎜ 1 v1f ⎟ = Gm1m2 ⎜ − ⎟ ⇒ v1f2 = ⎜ − ⎟ 2 2 ⎝ m2 ⎠ r r (1 m1 / m2 ) ⎝ rf ri ⎠ + i ⎠ ⎝ f With G = 6.67 × 10−11 N ⋅ m(kg) −2 , rf = R1 + R2 = 18 × 108 m, ri = 1.0 × 1014 m, m1 = 8.0 × 1030 kg, and m2 = 2.0 × 1030 kg, the above equation can be simplified to yield

v1f = 1.72 × 105 m/s, and v2f = −

⎛ 8.0 × 1030 kg ⎞ m1 5 5 v1f = ⎜ ⎟ × (1.72 × 10 m/s ) = 6.89 × 10 m/s 30 m2 ⎝ 2.0 × 10 kg ⎠

The speed of the heavier star is 1.7 × 105 m/s. That of the lighter star is 6.9 × 105 m/s.

11.77. Model: Model the lawnmower as a particle and use the model of kinetic friction. Visualize:

We placed the origin of our coordinate system on the lawnmower and drew the free-body diagram of forces. G G G Solve: The normal force n, which is related to the frictional force, is not equal to FG . This is due to the presence of F . G G The rolling friction is f r = μ r n, or n = f r μ r . The lawnmower moves at constant velocity, so Fnet = 0. The two components of Newton’s second law are

( ∑ F ) = n − F − F sin 37° = ma = 0 N ⇒ f / μ − mg − F sin 37° = 0 N ⇒ f ( ∑ F ) = F cos37° − f = 0 N ⇒ F cos37° − μ mg − μ F sin 37° = 0 N G

y

y

r

x

⇒F=

r

r

r

r

= μ r mg + μ r F sin 37°

r

μ r mg (0.15)(12 kg)(9.8 m/s 2 ) = = 29.4 N cos37° − μ r sin 37° (0.7986) − (0.15)(0.6018)

G G Thus, the power supplied by the gardener in pushing the lawnmower at a constant speed of 1.2 m/s is P = F ⋅ v = Fv cosθ = (24.9 N)(1.2 m/s)cos37° = 24 W.

11-1

12.1.

Model: A spinning skater, whose arms are outstretched, is a rigid rotating body.

Visualize:

Solve: The speed v = rω , where r = 140 cm/2 = 0.70 m. Also, 180 rpm = (180)2π /60 rad/s = 6π rad/s. Thus, v = (0.70 m)(6π rad/s) = 13.2 m/s. Assess: A speed of 13.2 m/s ≈ 26 mph for the hands is a little high, but reasonable.

12.2.

Model:

Assume constant angular acceleration.

⎛ 2π rad ⎞⎛ min ⎞ (a) The final angular velocity is ω f = ( 2000 rpm ) ⎜ ⎟⎜ ⎟ = 209.4 rad/s. The definition of ⎝ rev ⎠⎝ 60 s ⎠ angular acceleration gives us Solve:

α=

Δω ω f − ωi 209.4 rad/s − 0 rad/s = = = 419 rad/s Δt Δt 0.50 s

The angular acceleration of the drill is 4.2 × 102 rad/s. (b) 1 2

θ f = θ i + ωi Δt + α ( Δt ) = 0 rad + 0 rad + 2

⎛ rev ⎞ The drill makes (52.4 rad) ⎜ ⎟ = 8.3 revolutions. ⎝ 2π rad ⎠

1 2 ( 419 rad/s )( 0.50 s ) = 52.4 rad 2

12.3.

Model: Visualize:

Solve:

Assume constant angular acceleration.

(a) Since at = rα , find α first. With 90 rpm = 9.43 rad/s and 60 rpm = 6.28 rad/s,

α=

Δω 9.43 rad − 6.28 rad/s = = 0.314 rad/s 2 Δt 10 s

The angular acceleration of the sprocket and pedal are the same. So

at = rα = ( 0.18 m ) ( 0.314 rad/s 2 ) = 0.057 m/s2 (b) The length of chain that passes over the sprocket during this time is L = r Δθ . Find Δθ :

1 2

θ f = θ i + ωi Δt + α ( Δt )

2

θ f − θ i = Δθ = ( 6.28 rad/s )(10 s ) +

1 ( 0.314 rad/s2 ) (10 s )2 = 78.5 rad 2

The length of chain which has passed over the top of the sprocket is L = (0.10 m)(78.5 rad) = 7.9 m

12.4.

Model: Visualize:

Assume constant angular acceleration.

⎛ 2π rad ⎞⎛ min ⎞ The initial angular velocity is ωi = ( 60 rpm ) ⎜ ⎟⎜ ⎟ = 2π rad/s. ⎝ rev ⎠⎝ 60 s ⎠ The angular acceleration is Solve:

α=

ωf − ωi Δt

=

0 rad/s − 2π rad/s = −0.251 rad/s 2 25 s

The angular velocity of the fan blade after 10 s is

ω f = ωi + α ( t − t0 ) = 2π rad/s+ ( −0.251 rad/s 2 ) (10 s − 0 s ) = 3.77 rad/s The tangential speed of the tip of the fan blade is

vt = rω = ( 0.40 m )( 3.77 rad/s ) = 1.51 m/s (b) 1 2 2 The fan turns 78.6 radians = 12.5 revolutions while coming to a stop.

θ f = θ i + ω i Δt + α ( Δt ) = 0 rad + ( 2π rad/s )( 25 s ) +

1 ( −0.251 rad/s 2 ) ( 25 s )2 = 78.6 rad 2

12.5.

Model: Visualize:

The earth and moon are particles.

Choosing xE = 0 m sets the coordinate origin at the center of the earth so that the center of mass location is the distance from the center of the earth. Solve: 24 22 8 m x + mM xM ( 5.98 × 10 kg ) ( 0 m ) + ( 7.36 × 10 kg )( 3.84 × 10 m ) xcm = E E = mE + mM 5.98 × 1024 kg + 7.36 × 10 22 kg

= 4.67 × 106 m Assess: The center of mass of the earth-moon system is called the barycenter, and is located beneath the surface of the earth. Even though xE = 0 m the earth influences the center of mass location because mE is in the denominator of the expression for xcm .

12.6. Visualize: Please refer to Figure EX12.6. The coordinates of the three masses mA , mB , and mC are (0 cm, 0 cm), (0 cm, 10 cm), and (10 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are

xcm =

mA xA + mB xB + mC xC (100 g)(0 cm) + (200 g)(0 cm) + (300 g)(10 cm) = = 5.0 cm mA + mB + mC (100 g + 200 g + 300 g)

ycm =

mA yA + mB yB + mC yC (100 g)(0 cm) + (200 g)(10 cm) + (300 g)(0 cm) = = 3.3 cm mA + mB + mC (100 g + 200 g + 300 g)

12.7. Visualize: Please refer to Figure EX12.7. The coordinates of the three masses mA , mB , and mC are (0 cm, 10 cm), (10 cm, 10 cm), and (10 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are

xcm =

mA xA + mB xB + mC xC (200 g)(0 cm) + (300 g)(10 cm) + (100 g)(10 cm) = = 6.7 cm mA + mB + mC (200 g + 300 g + 100 g)

ycm =

mA yA + mB yB + mC yC (200 g)(10 cm) + (300 g)(10 cm) + (100 g)(0 cm) = = 8.3 cm mA + mB + mC (200 g + 300 g + 100 g)

12.8.

Model: Visualize:

Solve:

The balls are particles located at the ball’s respective centers.

The center of mass of the two balls measured from the left hand ball is xcm =

(100 g )( 0 cm ) + ( 200 g )( 30 cm ) = 20 cm 100 g + 200 g

The linear speed of the 100 g ball is ⎛ 2π rad ⎞⎛ min ⎞ v1 = rω = xcmω = ( 0.20 m )(120 rev/min ) ⎜ ⎟⎜ ⎟ = 2.5 m/s ⎝ rev ⎠⎝ 60 s ⎠

12.9.

Model: The earth is a rigid, spherical rotating body.

Solve:

The rotational kinetic energy of the earth is K rot = 12 Iω 2 . The moment of inertia of a sphere about its

diameter (see Table 12.2) is I = 52 M earth R 2 and the angular velocity of the earth is

ω=

2π rad = 7.27 × 10 −5 rad/s 24 × 3600 s

Thus, the rotational kinetic energy is 1⎛ 2 ⎞ K rot = ⎜ M earth R 2 ⎟ω 2 2⎝ 5 ⎠ 1 = (5.98 × 1024 kg)(6.37 × 106 m) 2 (7.27 × 10−5 rad/s) 2 = 2.57 × 1029 J 5

12.10. Model: The triangle is a rigid body rotating about an axis through the center. Visualize: Please refer to Figure EX12.10. Each 200 g mass is a distance r away from the axis of rotation, where r is given by 0.20 m 0.20 m = cos30° ⇒ r = = 0.2309 m r cos30° Solve: The moment of inertia of the triangle is I = 3 × mr 2 = 3(0.200 kg)(0.2309 m) 2 = 0.0320 kg m 2 . The frequency of rotation is given as 5.0 revolutions per s or 10π rad/s. The rotational kinetic energy is

K rot. =

1 2 1 Iω = (0.0320 kg m 2 )(10.0π rad/s) 2 = 15.8 J 2 2

12.11.

Model: The disk is a rigid body rotating about an axis through its center.

Visualize:

Solve:

The speed of the point on the rim is given by vrim = Rω . The angular velocity ω of the disk can be

determined from its rotational kinetic energy which is K = 12 Iω 2 = 0.15 J. The moment of inertia I of the disk about its center and perpendicular to the plane of the disk is given by 1 1 I = MR 2 = (0.10 kg)(0.040 m) 2 = 8.0 × 10−5 kg m 2 2 2 2(0.15 J) 0.30 J 2 ⇒ω = = ⇒ ω = 61.237 rad/s I 8.0 × 10−5 kg m 2 Now, we can go back to the first equation to find vrim . We get vrim = Rω = (0.040 m)(61.237 rad/s) = 2.4 m/s.

12.12.

Model:

The baton is a thin rod rotating about a perpendicular axis through its center of mass. 1 Solve: The moment of inertia of a thin rod rotating about its center is I = ML2 . For the baton, 12 1 2 I = ( 0.400 kg )( 0.96 m ) = 0.031 kg m 2 12 The rotational kinetic energy of the baton is 2

K rot =

1 2 1 ⎛ ⎛ 2π rad ⎞⎛ min ⎞ ⎞ Iω = ( 0.031 kg m 2 ) ⎜ (100 rev/min ) ⎜ ⎟⎜ ⎟ ⎟ = 1.68 J 2 2 ⎝ rev ⎠⎝ 60 s ⎠ ⎠ ⎝

12.13.

Model: The structure is a rigid body rotating about its center of mass.

Visualize:

We placed the origin of the coordinate system on the 300 g ball. Solve: First, we calculate the center of mass:

xcm =

(300 g)(0 cm) + (600 g)(40 cm) = 26.67 cm 300 g + 600 g

Next, we will calculate the moment of inertia about the structure’s center of mass: I = (300 g)( xcm ) 2 + (600 g)(40 cm − xcm ) 2 = (0.300 kg)(0.2667 m) 2 + (0.600 kg)(0.1333 m) 2 = 0.032 kg m 2 Finally, we calculate the rotational kinetic energy: 2 1 1 ⎛ 100 × 2π ⎞ K rot = Iω 2 = (0.032 kg m 2 ) ⎜ rad/s ⎟ = 1.75 J 2 2 ⎝ 60 ⎠

12.14. Model: The moment of inertia of any object depends on the axis of rotation. In the present case, the rotation axis passes through mass A and is perpendicular to the page. Visualize: Please refer to Figure EX12.14. ∑ mi xi = mA xA + mB xB + mC xC + mD xD Solve: (a) xcm = mA + mB + mC + mD ∑ mi (100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m) = 0.057 m 100 g + 200 g + 200 g + 200 g m y + mB yB + mC yC + mD yD = A A mA + mB + mC + mD

= ycm

(100 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 cm) + (200 g )(0 m) = 0.057 m 700 g (b) The distance from the axis to mass C is 14.14 cm. The moment of inertia through A and perpendicular to the page is =

I A = ∑ mi ri 2 = mA rA2 + mB rB2 + mC rC2 + mD rD2 i

= (0.100 kg)(0 m) 2 + (0.200 kg)(0.10 m) 2 + (0.200 kg)(0.1414 m) 2 + (0.200 kg)(0.10 m) 2 = 0.0080 kg m 2

12.15.

Model: The moment of inertia of any object depends on the axis of rotation.

Visualize:

Solve:

(a) xcm = ∑

mi xi

∑m

i

ycm

=

mA xA + mB xB + mC xC + mD xD mA + mB + mC + mD

(100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m) = = 0.057 m 100 g + 200 g + 200 g + 200 g m y + mB yB + mC yC + mD yD = A A mA + mB + mC + mD

(100 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 cm) + (200 g )(0 m) = 0.057 m 700 g (b) The moment of inertia about a diagonal that passes through B and D is =

I BD = mA rA2 + mC rC2 where rA = rC = (0.10 m)cos 45° = 7.07 cm and are the distances from the diagonal. Thus, I BD = (0.100 kg) rA2 + (0.200 kg)rC2 = 0.0015 kg m 2 Assess:

Note that the masses B and D, being on the axis of rotation, do not contribute to the moment of inertia.

12.16.

Model: The three masses connected by massless rigid rods is a rigid body.

Visualize: Solve:

(a)

Please refer to Figure EX12.16.

xcm =

∑m x ∑m

i i i

ycm =

∑m y ∑m

i i

=

=

(0.100 kg)(0 m) + (0.200 kg)(0.06 m) + (0.100 kg)(0.12 m) = 0.060 m 0.100 kg + 0.200 kg + 0.100 kg

(0.100 kg)(0 m) + (0.200 kg)

(

)

(0.10 m) 2 − (0.06 m) 2 + (0.100 kg)(0 m)

0.100 kg + 0.200 kg + 0.100 kg

i

= 0.040 m

(b) The moment of inertia about an axis through A and perpendicular to the page is

I A = ∑ mi ri 2 = mB (0.10 m) + mC (0.10 m) 2 = (0.100 kg)[(0.10 m) 2 + (0.10 m) 2 ] = 0.0020 kg m 2 2

(c) The moment of inertia about an axis that passes through B and C is

I BC = mA Assess:

(

(0.10 m) 2 − (0.06 m) 2

)

2

= 0.00128 kg m 2

Note that mass mA does not contribute to I A , and the masses mB and mC do not contribute to I BC .

12.17.

Model: The door is a slab of uniform density. (a) The hinges are at the edge of the door, so from Table 12.2, 1 2 I = ( 25 kg )( 0.91 m ) = 6.9 kg m 2 3 (b) The distance from the axis through the center of mass along the height of the door is ⎛ 0.91 m ⎞ d =⎜ − 0.15 m ⎟ = 0.305 m. Using the parallel–axis theorem, 2 ⎝ ⎠ 1 2 2 I = I cm + Md 2 = ( 25 kg )( 0.91 m ) + ( 25 kg )( 0.305 cm ) = 4.1 kg m 2 12 Assess: The moment of inertia is less for a parallel axis through a point closer to the center of mass. Solve:

12.18.

Model: The CD is a disk of uniform density. (a) The center of the CD is its center of mass. Using Table 12.2, 1 1 2 I cm = MR 2 = ( 0.021 kg )( 0.060 m ) = 3.8 × 10−5 kg m 2 2 2 (b) Using the parallel–axis theorem with d = 0.060 m, Solve:

I = I cm + Md 2 = 3.8 × 10−5 kg m 2 + ( 0.021 kg )( 0.060 m ) = 1.14 × 10−4 kg m 2 2

12.19. Visualize:

G Solve: Torque by a force is defined as τ = Fr sin φ where φ is measured counterclockwise from the r vector to G the F vector. The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N force:

(30 N)r1 sinφ1 + (20 N)r2 sinφ2 = (30 N)(0.02 m) sin ( − 90°) + (20 N)(0.02 m) sin (90°) = ( − 0.60 N m) + (0.40 N m) = −0.20 N m Assess:

A negative torque causes a clockwise acceleration of the pulley.

12.20. Visualize:

The two equal but opposite 50 N forces, one acting at point P and the other at point Q, make a couple that causes a net torque. Solve: The distance between the lines of action is l = d cos30°. The net torque is given by

τ = lF = (d cos30°)F = (0.10 m)(0.866)(50 N) = 4.3 N m

12.21. Visualize:

Solve:

The net torque on the spark plug is τ = Fr sin φ = −38 N m = F (0.25 m)sin(−120°) ⇒ F = 176 N

That is, you must pull with a force of 176 N to tighten the spark plug. Assess: The force applied on the wrench leads to its clockwise motion. That is why we have used a negative sign for the net torque.

12.22.

Model: The disk is a rotating rigid body.

Visualize:

The radius of the disk is 10 cm and the disk rotates on an axle through its center. Solve: The net torque on the axle is

τ = FA rA sin φA + FBrB sin φB + FC rC sin φC + FD rD sin φD = (30 N)(0.10 m)sin(−90°) + (20 N)(0.050 m)sin 90° + (30 N)(0.050 m)sin135° + (20 N)(0.10 m)sin 0° = −3 N m + 1 N m + 1.0607 N m = −0.94 N m Assess: A negative torque means a clockwise rotation of the disk.

12.23.

Model: The beam is a solid rigid body.

Visualize:

G The steel beam experiences a torque due to the gravitational force on the construction worker FG and the C G gravitational force on the beam FG . The normal force exerts no torque since the net torque is calculated about the

( )

( )

B

point where the beam is bolted into place. G Solve: The net torque on the steel beam about point O is the sum of the torque due to FG G FG . The gravitational force on the beam acts at the center of mass.

( )

( )

C

and the torque due to

B

τ = (( FG )C )(4.0 m)sin( −90°) + (( FG ) B )(2.0 m)sin(−90°) = −(70 kg)(9.80 m/s 2 )(4.0 m) − (500 kg)(9.80 m/s 2 )(2.0 m) = −12.5 kN m The negative torque means these forces would cause the beam to rotate clockwise. The magnitude of the torque is 12.5 kN m.

12.24.

Model: Model the arm as a uniform rigid rod. Its mass acts at the center of mass.

Visualize:

Solve:

(a) The torque is due both to the gravitational force on the ball and the gravitational force on the arm:

τ = τ ball + τ arm = ( mb g )rb sin 90° + (ma g )ra sin 90° = (3.0 kg)(9.8 m/s 2)(0.70 m)+(4.0 kg)(9.8 m/s 2)(0.35 m) = 34 N m (b) The torque is reduced because the moment arms are reduced. Both forces act at φ = 45° from the radial line, so

τ = τ ball + τ arm = (mb g )rb sin 45° + ( ma g ) ra sin 45° = (3.0 kg)(9.8 m/s 2 )(0.70 m)(0.707) + (4.0 kg)(9.8 m/s 2)(0.35 m)(0.707) = 24 N m

12.25. Solve: τ = Iα is the rotational analog 2 τ = (2.0 kg m )(4.0 rad/s 2 ) = 8.0 kg m 2 /s 2 = 8.0 N m.

of

Newton’s

second

law

F = ma.

We

have

12.26.

Visualize:

Since

α = τ /I , a graph of the angular acceleration looks just like the torque graph with

the numerical values divided by I = 4.0 kg m 2 .

Solve: From the discussion about Figure 4.47

ω f = ωi + area under the angular acceleration α curve between ti and tf The area under the curve between t = 0 s and t = 3 s is 0.75 rad/s. With ω1 = 0 rad/s, we have

ω f = 0 rad/s + 0.75 rad/s = 0.75 rad/s

12.27. Model: Two balls connected by a rigid, massless rod are a rigid body rotating about an axis through the center of mass. Assume that the size of the balls is small compared to 1 m. Visualize:

We placed the origin of the coordinate system on the 1.0 kg ball. Solve: The center of mass and the moment of inertia are

xcm =

(1.0 kg)(0 m) + (2.0 kg)(1.0 m) = 0.667 m and ycm = 0 m (1.0 kg + 2.0 kg)

I about cm = ∑ mi ri 2 = (1.0 kg)(0.667 m) 2 + (2.0 kg)(0.333 m) 2 = 0.667 kg m 2 We have ω f = 0 rad/s, tf − ti = 5.0 s, and ωi = −20 rpm = −20(2π rad/60 s) = − 32 π rad/s, so ω f = ωi + α (tf − ti ) becomes 2π ⎛ 2π ⎞ 0 rad/s = ⎜ − rad/s ⎟ + α (5.0 s) ⇒ α = rad/s 2 15 ⎝ 3 ⎠ Having found I and α , we can now find the torque τ that will bring the balls to a halt in 5.0 s: ⎛2 ⎝3

⎞⎛ 2π ⎞ 4π rad/s 2 ⎟ = N m = 0.28 N m 15 ⎠⎝ ⎠ 45

τ = I about cmα = ⎜ kg m 2 ⎟⎜

The magnitude of the torque is 0.28 N m, applied in the counterclockwise direction.

12.28. Model: A circular plastic disk rotating on an axle through its center is a rigid body. Assume axis is perpendicular to the disk. Solve: To determine the torque (τ) needed to take the plastic disk from ωi = 0 rad/s

to

ω f = 1800 rpm = (1800)(2π ) / 60 rad/s = 60π rad/s in tf − ti = 4.0 s, we need to determine the angular acceleration (α ) and the disk’s moment of inertia (I ) about the axle in its center. The radius of the disk is R = 10.0 cm. We have 1 1 I = MR 2 = (0.200 kg)(0.10 m) 2 = 1.0 × 10−3 kg m 2 2 2 ω − ωi 60π rad/s − 0 rad/s ω f = ωi + α (tf − ti ) ⇒ α = f = = 15π rad/s 2 t f − ti 4.0 s Thus, τ = Iα = (1.0 × 10−3 kg m 2 )(15π rad/s 2 ) = 0.047 N m.

12.29.

Model: The compact disk is a rigid body rotating about its center.

Visualize:

Solve:

(a) The rotational kinematic equation ω1 = ω 0 + α (t1 − t0 ) gives

200π ⎛ 2π ⎞ (2000 rpm) ⎜ rad/s 2 ⎟ rad/s = 0 rad + α (3.0 s − 0 s) ⇒ α = 9 ⎝ 60 ⎠ The torque needed to obtain this operating angular velocity is ⎛ 200π ⎞ rad/s 2 ⎟ = 1.75 × 10−3 N m ⎝ 9 ⎠

τ = Iα = (2.5 × 10−5 kg m 2 ) ⎜ (b) From the rotational kinematic equation,

1

1 ⎛ 200π

θ1 = θ 0 + ω 0 (t1 − t0 ) + α (t1 − t0 ) 2 = 0 rad + 0 rad + ⎜ 2 2⎝ 9 100π revolutions = 50 rev = 100π rad = 2π Assess:

Fifty revolutions in 3 seconds is a reasonable value.

2 ⎞ rad/s 2 ⎟ ( 3.0 s − 0 s ) ⎠

12.30. Model: The rocket attached to the end of a rigid rod is a rotating rigid body. Assume the rocket is small compared to 60 cm. Visualize: Please refer to Figure EX12.30. Solve: We can determine the rocket’s angular acceleration from the relationship τ = Iα . The torque τ can be found from the thrust (F) using τ = Fr sin φ . The moment of inertia (I) can be calculated from equations given in Table 12.2. Specifically, I = I rod about one end + I rocket becomes 1 1 M rod L2 + ML2 = (0.100 kg)(0.60 m) 2 + (0.200 kg)(0.60 m) 2 3 3 = 0.012 kg m 2 + 0.072 kg m 2 = 0.0840 kg m 2

⇒α = Assess:

τ I

=

Fr sin φ (4.0 N)(0.60 m)sin(45°) = = 20 rads/s 2 I 0.0840 kg m 2

The rocket will accelerate counterclockwise since α is positive.

12.31.

Model: The rod is in rotational equilibrium, which means that τ net = 0.

Visualize:

As the gravitational force on the rod and the hanging mass pull down (the rotation of the rod is exaggerated in the figure), the rod touches the pin at two points. The piece of the pin at the very end pushes down on the rod; the right end of the pin pushes up on the rod. To understand this, hold a pen or pencil between your thumb and forefinger, with your thumb on top (pushing down) and your forefinger underneath (pushing up). Solve: Calculate the torque about the left end of the rod. The downward force exerted by the pin acts through this G point, so it exerts no torque. To prevent rotation, the pin’s normal force npin exerts a positive torque (ccw about the left end) to balance the negative torques (cw) of the gravitational force on the mass and rod. The gravitational force on the rod acts at the center of mass, so

τ net = 0 N m = τ pin − (0.40 m)(2.0 kg)(9.8 m/s 2) − (0.80 m)(0.50 kg)(9.8 m/s 2) ⇒ τ pin = 11.8 N m

12.32.

Model: The massless rod is a rigid body.

Visualize:

Solve:

G To be in equilibrium, the object must be in both translational equilibrium ( Fnet = 0 N) and rotational

equilibrium (τ net = 0 Nm). We have ( Fnet ) y = (40 N) − (100 N) + (60 N) = 0 N, so the object is in translational equilibrium. Measuring τ net about the left end,

τ net = (60 N)(3.0 m)sin( +90°) + (100 N)(2.0 m)sin(−90°) = −20 N m The object is not in equilibrium.

12.33.

Model: The object balanced on the pivot is a rigid body.

Visualize:

Since the object is balanced on the pivot, it is in both translational equilibrium and rotational equilibrium. G Solve: There are three forces acting on the object: the gravitational force FG acting through the center of 1 G mass of the long rod, the gravitational force FG acting through the center of mass of the short rod, and the 2 G normal force P on the object applied by the pivot. The translational equilibrium equation ( Fnet ) y = 0 N is

( )

( )

− ( FG )1 − ( FG )2 + P = 0 N ⇒ P = ( FG )1 + ( FG )2 = (1.0 kg)(9.8 m/s 2 ) + (4.0 kg)(9.8 m/s 2 ) = 49 N

Measuring torques about the left end, the equation for rotational equilibrium τ net = 0 Nm is

Pd − w1 (1.0 m) − w2 (1.5 m) = 0 Nm ⇒ (49 N)d − (1.0 kg)(9.8 m/s 2 )(1.0 m) − (4.0 kg)(9.8 m/s2 )(1.5 m) = 0 N ⇒ d = 1.40 m Thus, the pivot is 1.40 m from the left end.

12.34.

Model: Visualize:

Solve:

The see-saw is a rigid body. The cats and bowl are particles.

The see-saw is in rotational equilibrium. Calculate the net torque about the pivot point.

τ net = 0 = ( FG )1 ( 2.0 m ) − ( FG )2 ( d ) − ( FG ) B ( 2.0 m )

m2 gd = m1 g ( 2.0 m ) − mB g ( 2.0 m ) d=

( m1 − mB )( 2.0 m ) = ( 5.0 kg − 2.0 kg )( 2.0 m ) = 1.5 m m2

4.0 kg

Assess: The smaller cat is close but not all the way to the end by the bowl, which makes sense since the combined mass of the smaller cat and bowl of tuna is greater than the mass of the larger cat.

12.35.

Solve:

(a) According to Equation 12.35, the speed of the center of mass of the tire is

vcm = Rω = 20 m/s ⇒ ω =

vcm 20 m/s ⎛ 60 ⎞ 2 = = 66.67 rad/s = ( 66.7 ) ⎜ ⎟ rpm = 6.4 × 10 rpm R 0.30 m 2 π ⎝ ⎠

(b) The speed at the top edge of the tire relative to the ground is vtop = 2vcm = 2(20 m/s) = 40 m/s. (c) The speed at the bottom edge of the tire relative to ground is vbottom = 0 m/s.

12.36. Model: The can is a rigid body rolling across the floor. Assume that the can has uniform mass distribution. Solve: The rolling motion of the can is a translation of its center of mass plus a rotation about the center of mass. The moment of inertia of the can about the center of mass is 12 MR 2, where R is the radius of the can. Also vcm = Rω , where ω is the angular velocity of the can. The total kinetic energy of the can is K = K cm + K rot =

1 1 1 1⎛ 1 ⎞⎛ v ⎞ 2 2 Mvcm + I cmω 2 = Mvcm + ⎜ MR 2 ⎟⎜ cm ⎟ 2 2 2 2⎝ 2 ⎠⎝ R ⎠

3 3 2 = Mvcm = (0.50 kg)(1.0 m/s) 2 = 0.38 J 4 4

2

12.37.

Model: The sphere is a rigid body rolling down the incline without slipping.

Visualize:

The initial gravitational potential energy of the sphere is transformed into kinetic energy as it rolls down. Solve: (a) If we choose the bottom of the incline as the zero of potential energy, the energy conservation equation will be K f = U i . The kinetic energy consists of both translational and rotational energy. This means

Kf =

1 1 1⎛ 2 1 ⎞ 2 I cmω 2 + Mvcm = Mgh ⇒ ⎜ MR 2 ⎟ω 2 + M ( Rω ) 2 = Mgh 2 2 2⎝ 5 2 ⎠ 7 ⇒ MR 2ω 2 = Mg (2.1 m)sin 25° 10

⇒ω =

10 7

g (2.1 m)(sin 25°) = R2

10 7

g (2.1 m)(sin 25°) = 88 rad/s (0.04 m) 2

(b) From part (a)

K total =

1 1 7 1 1⎛ 2 1 ⎞ 2 = MR 2ω 2 and K rot = I cmω 2 = ⎜ MR 2 ⎟ω 2 = MR 2ω 2 I cmω 2 + Mvcm 2 2 10 2 2⎝ 5 5 ⎠ 2 2 1 MR ω K 1 10 2 ⇒ rot = 75 = × = K total 10 MR 2ω 2 5 7 7

12.38. Visualize: Please refer to Figure EX12.38. To determine angle α , put the tails of the vectors together. G G G G Solve: (a) The magnitude of A × B is AB sin α = (6)(4)sin 45° = 17. The direction of A × B, using the right G G hand rule, is out of the page. Thus, A × B = (17, out of the page). G G G G G (b) The magnitude of C × D is CD sin α = (6)(4)sin180° = 0. Thus C × D = 0.

12.39.

Visualize: Please refer to Figure EX12.39. G G G G (a) The magnitude of A × B is AB sin α = (6)(4)sin 60° = 20.78. The direction of A × B is given by the G G G G right hand rule. To curl our fingers from A to B, we have to point our thumb out of the page. Thus, A × B = (21, out of the page). G G (b) C × D = ((6)(4)sin 90°, into the page) = (24, into the page).

Solve:

(a) (iˆ × ˆj ) × iˆ = kˆ × iˆ = ˆj (b) iˆ × ( ˆj × iˆ) = iˆ × (− kˆ) = −iˆ × kˆ = −(− ˆj ) = ˆj

12.40.

Solve:

(a) iˆ × (iˆ × ˆj ) = iˆ × kˆ = − ˆj G (b) (iˆ × ˆj ) × kˆ = kˆ × kˆ = 0

12.41.

Solve:

12.42.

(b)

Solve:

G G (a) A × B = (3iˆ + ˆj ) × (3iˆ − 2 ˆj + 2kˆ) = 9iˆ × iˆ − 6iˆ × ˆj + 6iˆ × kˆ + 3 ˆj × iˆ − 2 ˆj × ˆj + 2 ˆj × kˆ = 0 − 6kˆ + 6(− ˆj ) + 3(−kˆ) − 0 + 2iˆ = 2iˆ − 6 ˆj − 9kˆ

G

G

G

G

12.43. Solve: (a) C × D = G0 implies that D must also be in the same or opposite direction as the C vector or zero, because iˆ × iˆ = 0. Thus D = niˆ, where n could be any real number. G G G G (b) C × E = 6kˆ implies that E must be along the ˆj vector, because iˆ × ˆj = kˆ. Thus E = 2 ˆj. G G G G (c) C × F = −3 ˆj implies that F must be along the kˆ vector, because iˆ × kˆ = − ˆj. Thus F = 1kˆ.

12.44.

Solve:

G

G

G

τ = r × F = (5iˆ + 5 ˆj ) × (−10 ˆj ) N m

G = [−50(iˆ × ˆj ) − 50( ˆj × ˆj )] N m = [−50(+ kˆ) − 0] N m = −50kˆ N m

12.45.

Solve:

G

G

G

τ = r × F = (5 ˆj ) × (−10iˆ + 10 ˆj ) N m = (−5 ˆj × iˆ + 50 ˆj × ˆj ) N m G = [−50( −kˆ) + 0] N m = 50kˆ N m

12.46. Solve:

Visualize: Please refer to Figure EX12.46. G G G L = r × mv = (1.0iˆ + 2.0 ˆj ) m × (0.200 kg)(3.0 m/s) cos45°iˆ − sin 45° ˆj

(

)

= (0.42iˆ × iˆ − 0.42iˆ × ˆj + 0.85 ˆj × iˆ − 0.85 ˆj × ˆj ) kg m 2 /s = −(1.27 kˆ) kg m 2 /s or (1.27 kg m 2 /s, into page)

12.47.

Solve:

Visualize: Please refer to Figure EX12.47. G G G L = r × mv = (3.0iˆ + 2.0 ˆj ) m × (0.1 kg)(4.0ˆj ) m/s = 1.20(iˆ × ˆj ) kg m 2 /s + 0.8( ˆj × ˆj ) kg m 2 /s = 1.20kˆ kg m 2 /s + 0 kg m 2 /s = 1.20kˆ kg m 2 /s or (1.20 kg m 2 /s, out of page)

12.48. Model: The bar is a rotating rigid body. Assume that the bar is thin. Visualize: Please refer to Figure EX12.48. Solve: The angular velocity ω = 120 rpm = (120)(2π ) / 60 rad/s = 4π rad/s. From Table 12.2, the moment of inertial of a rod about its center is I = 121 ML2 . The angular momentum is ⎛1⎞ L = Iω = ⎜ ⎟ (0.50 kg)(2.0 m) 2 (4π rad/s) = 2.1 kg m 2 /s ⎝ 12 ⎠

If we wrap our fingers in the direction of the rod’s rotation, our thumb will point in the z direction or out of the page. Consequently, G L = (2.1 kg m 2 /s, out of the page)

12.49. Model: The disk is a rotating rigid body. Visualize: Please refer to Figure EX12.49. Solve: From Table 12.2, the moment of inertial of the disk about its center is

The

angular

1 1 I = MR 2 = (2.0 kg)(0.020 m) 2 = 4.0 × 10−4 kg m 2 2 2 ω is 600 rpm = 600 × 2π /60 rad/s = 20π rad/s. velocity

Thus,

L = Iω = (4.0 × 10 kg m )(20π rad/s) = 0.025 kg m /s. If we wrap our right fingers in the direction of the disk’s rotation, our thumb will point in the − x direction. Consequently, G L = −0.025 iˆ kg m 2 /s = (0.025 kg m 2 /s, into page) −4

2

2

12.50.

Model: The beach ball is a spherical shell. Solve: From Table 12.2, the moment of inertia about a diameter of a spherical shell is 2 2 1 2 I = MR 2 = ( 0.100 kg )( 0.50 m ) = kg m 2 3 3 60 Require ⎛ 1 ⎞ kg m 2 ⎟ω L = 0.10 kg m 2 /s = Iω = ⎜ 60 ⎝ ⎠ ⇒ ω = ( 60 rad/s )

⎛ rev ⎞ ⎛ 60 s ⎞ In rpm, this is ( 60 rad/s ) ⎜ ⎟⎜ ⎟ = 57 rpm. ⎝ 2π rad ⎠ ⎝ min ⎠

12.51.

Model: The wheel is a rigid rolling body.

Visualize:

Solve:

The front of the disk is moving forward at velocity vcm . Also, because of rotation the point is moving

downward at velocity vrel = Rω = vcm . So, this point has a speed 2 2 v = vcm + vcm = 2vcm = 2(20 m/s) = 28 m/s

Assess:

The speed v is independent of the radius of the wheel.

12.52

Model: The triangle is a rigid body rotating about its center of mass perpendicular to the plane of the triangle. The center of mass of any symmetrical object of uniform density is at the physical center of the object. Visualize:

The distance to one tip of the triangle from the center of mass is given by r cos30° = (5.0/2) cm, which yields r = (2.5 cm)/ cos30° = 2.9 cm. Solve: The speed of the tip is v = rω = (2.9 cm)(120 rpm) = (2.9 cm)(4π rad/s) = 36 cm/s

12.53. Visualize:

Solve: We will consider a vertical strip of width dx and of mass dm at a position x from the origin. The formula for the x component of the center of mass is 1 xcm = x dm M∫

The area of the steel plate is A = 12 (0.2 m)(0.3 m) = 0.030 m 2 . Mass dm in the strip is the same fraction of M as dA is of A. Thus dm dA M ⎛ 0.800 kg ⎞ = ⇒ dm = dA = ⎜ dA = (26.67 kg/m 2 )l dx 2 ⎟ M A A ⎝ 0.030 m ⎠

The relationship between l and x is l x 2 = ⇒l = x 0.20 m 0.30 m 3 Therefore, xcm

1 (17.78 kg/m 2 ) x3 2 ⎛ 2⎞ 2 (26.67 kg/m ) x dx = = ⎜ ⎟ M∫ M 3 ⎝ 3⎠

Due to symmetry ycm = 0 cm.

0.3 m

= 0m

(17.78 kg/m 2 ) (0.3 m)2 = 20 cm 0.8 kg 3

12.54.

Visualize:

Solve: Build the plate from the three shapes 1, 2, and 3. The center of mass of the complete shape is the center of mass of the center of masses of the three smaller shapes. m1 ( xcm )1 + m2 ( xcm )2 + m3 ( xcm )3 xcm = m1 + m2 + m3

ycm =

m1 ( ycm )1 + m2 ( ycm )2 + m3 ( ycm )3

m1 + m2 + m3 The masses of each of the smaller shapes are a fraction of the larger shape’s mass by the ratio of areas. The following table will be useful in the solution. Shape A mass xcm ycm 2 All M xcm ycm 32 cm 1

2 3

24 32

M

−1.0 cm

0 cm

2.0 cm

2

2.0 32

M

2.0 cm

−2.5 cm

6.0 cm

2

6.0 32

M

2.0 cm

1.5 cm

24 cm 2

Thus, ⎛ 24 ⎞ ⎛ 2.0 ⎞ ⎛ 6.0 ⎞ M ⎟ ( −1.0 cm ) + ⎜ M ⎟ ( 2.0 cm ) + ⎜ M ⎟ ( 2.0 cm ) ⎜ 32 ⎠ 32 32 ⎝ ⎝ ⎠ ⎝ ⎠ = −0.25 cm xcm = M Similarly, ycm = 0.125 cm. Assess: The plate’s center of mass of (−0.25 cm, 0.125 cm) is reasonable. The cutout means more of the mass is left of center and above the center.

12.55.

Model:

The disk is a rigid rotating body. The axis is perpendicular to the plane of the disk.

Visualize:

Solve:

(a) From Table 12.2, the moment of inertia of a disk about its center is

1 1 I = MR 2 = (2.0 kg)(0.10 m) 2 = 0.010 kg m 2 2 2 (b) To find the moment of inertia of the disk through the edge, we can make use of the parallel axis theorem:

I = I center + Mh 2 = (0.010 kg m 2 ) + (2.0 kg)(0.10 m) 2 = 0.030 kg m 2 Assess: The larger moment of inertia about the edge means there is more inertia to rotational motion about the edge than about the center.

12.56.

Model: The object is a rigid rotating body. Assume the masses m1 and m2 are small and the rod is thin. Visualize: Please refer to P12.56. Solve: The moment of inertia of the object is the sum of the moment of inertia of the rod, mass m1 , and mass m2 . Using Table 12.2 for the moment of inertia of the rod, we get 2

I rod = I rod about center + I m1 + I m2 = = Assess:

1 ⎛ L⎞ ⎛ L⎞ ML2 + m1 ⎜ ⎟ + m2 ⎜ ⎟ 12 2 ⎝ ⎠ ⎝ 4⎠

1 1 1 L2 ⎛ M m ⎞ ML2 + m1L2 + m2 L2 = ⎜ + m1 + 2 ⎟ 12 4 16 4⎝ 3 4 ⎠

With m1 = m2 = 0 kg, I rod 121 ML2 , as expected.

2

12.57. Visualize:

We chose the origin of the coordinate system to be on the axis of rotation, that is, at a distance d from one end of the rod. Solve: The moment of inertia can be calculated as follows: x2

I = ∫ x 2 dm x1

⇒I =

M L

L−d

∫

−d

3 ⎛M ⎞x x 2 dx = ⎜ ⎟ ⎝ L⎠3

L−d

−d

and

dm dx M = ⇒ dm = dx M L L

1⎛ M = ⎜ 3⎝ L

M ⎞ 3 3 3 3 ⎟ [( L − d ) − (− d ) ] = [( L − d ) + d ] 3L ⎠

For d = 0 m, I = 13 ML , and for d = 12 L, 2

I=

3 3 M ⎡⎛ L ⎞ ⎛ L ⎞ ⎤ 1 2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ = ML 3L ⎣⎢⎝ 2 ⎠ ⎝ 2 ⎠ ⎦⎥ 12

Assess: The special cases d = 0 m and d = L/2 of the general formula give the same results that are found in Table 12.2.

12.58. Visualize:

Solve: We solve this problem by dividing the disk between radii r1 and r2 into narrow rings of mass dm. Let dA = 2π rdr be the area of a ring of radius r. The mass dm in this ring is the same fraction of the total mass M as dA is of the total area A. (a) The moment of inertia can be calculated as follows:

I disk = ∫ r 2 dm and dm =

M M dA = ( 2π r ) dr A π ( r22 − r12 )

r

⇒ I disk =

r

2 M 2M 2 2M r 4 r 2 (2π r ) dr = 2 2 ∫ r 3dr = 2 2 2 2 ∫ π ( r2 − r1 ) r1 ( r2 − r1 ) r1 ( r2 − r1 ) 4

r2

r1

M 2M r 4 − r14 ) = ( r22 + r12 ) = 2 2 ( 2 2 4 ( r2 − r1 ) Replacing r1 with r and r2 with R, the moment of inertia of the disk through its center is I disk = 12 M ( R 2 + r 2 ). (b) For r = 0 m, I disk = 12 MR 2 . This is the moment of inertia for a solid disk or cylinder about the center.

Additionally, for r ≅ R, we have I = MR 2 . This is the expression for the moment of inertia of a cylindrical hoop or ring about the center. (c) The initial gravitational potential energy of the disk is transformed into kinetic energy as it rolls down. If we choose the bottom of the incline as the zero of potential energy, and use vcm = ω R, the energy conservation equation K f = U i is 1 2 1 1⎛ M ⎞ v2 1 2 2 Iω + Mvcm = Mgh ⇒ ⎜ ⎟ ( R 2 + r 2 ) cm2 + Mvcm = Mgyi = Mg (0.50 m)sin 20° 2 2 2⎝ 2 ⎠ R 2 2 2 1 2 1 r2 ⎞ 2 ⎛ R +r ⎞ 2 ⎛1 v v ⇒ vcm + = + + 2 ⎟ = 1.6759 m 2 /s 2 ⎜ ⎟ ⎜ cm cm 2 ⎝ 4R ⎠ 2 ⎝ 2 4 4R ⎠ (0.015 m) 2 ⎞ 2 ⎛3 vcm = 1.6759 m 2 /s 2 ⇒ vcm = 1.37 m/s ⎜ + 2 ⎟ 4 4(0.020 m) ⎝ ⎠ For a sliding particle on a frictionless surface K f = U i , so

1 2 v mvf = mgyi ⇒ vf = 2 gyi = 2 g (0.50 m)sin 20° = 1.83 m/s ⇒ cm = 0.75 2 vf That is, vcm is 75% of the speed of a particle sliding down a frictionless ramp.

12.59.

Model: Visualize:

Solve:

The plate has uniform density.

The moment of inertia is

I = ∫ r 2 dm. Let the mass of the plate be M. Its area is L2 . A region of area dA located at (x,y) has mass M M dm = dA = 2 dx dy. The distance from the axis of rotation to the point (x, y) is r = x 2 + y 2 . With A L L L L L − ≤ x ≤ and − ≤ y ≤ , 2 2 2 2 I=

L 2

L 2

∫ ∫ (x

L L − − 2 2

M = 2 L

2

⎛M + y2 )⎜ 2 ⎝L

M ⎞ ⎟ dx dy = 2 L ⎠

L 2

L 2

⎛ L3 y 2 L ⎛ − L3 y 2 L ⎞ ⎞ M ∫L ⎜⎜ 24 + 2 − ⎜⎝ 24 − 2 ⎟⎠ ⎟⎟dy = L2 ⎠ − ⎝

⎛ L3 L3 ⎞ ⎞ 1 2 M ⎛ L4 ⎜ + L ⎜ + ⎟ ⎟⎟ = ML 2 ⎜ L ⎝ 12 ⎝ 24 24 ⎠ ⎠ 6

⎞

∫ ⎜⎝ 3 + y x ⎟⎠ 2

L − 2

2

=

⎛ x3

L 2

dy −

L 2

L 2

⎛ ⎛ L3 M ⎜ L4 y3 2⎞ Ly dy L + = + ⎜ ⎟ ∫L ⎝ 12 L2 ⎜⎜ 12 3 ⎠ − ⎝ 2

⎞ ⎟ L ⎟ − ⎟ 2 ⎠ L 2

12.60.

Solve:

From Equation 12.16,

I = ∫ ( x 2 + y 2 )dm

A small region of area dA has mass dm, and M M dA = dx dy A A The area of the plate is 12 (0.20 m)(0.30 m) = 0.030 m 2 . So dm =

M ⎛ 0.800 kg ⎞ 2 =⎜ ⎟ = 26.67 kg m A ⎝ 0.030 m 2 ⎠ 1 1 1 The limits for x are 0 ≤ x ≤ 30 cm. For a particular value of x, − x ≤ y ≤ x. Note that ± is the slope of the 3 3 3 top and bottom edges of the triangle. Therefore, 30 cm

I=

1 x 3

∫ ∫ (x 0

2

+ y 2 )( 26.67 kg/m 2 ) dx dy

1 − x 3

1

= ( 26.67 kg/m

30 cm

2

)∫ 0

= ( 26.67 kg/m 2 )

30 cm

∫ 0

x

⎛ 2 y3 ⎞ 3 ⎜ x y + ⎟ dx 3 ⎠ −1 x ⎝ 3

⎛ 2 3 2 x3 ⎞ ⎜ x + ⎟dx 81 ⎠ ⎝3 30 cm

⎛ 56 ⎞⎛ 1 ⎞ = ( 26.67 kg/m 2 ) ⎜ ⎟⎜ x 4 ⎟ ⎝ 81 ⎠⎝ 4 ⎠ 0

= 0.037 kg m 2

12.61. G Model: The ladder is a rigid rod of length L. To not slip, it must be in both translational equilibrium G ( Fnet = 0 N) and rotational equilibrium (τ net = 0 N m). We also apply the model of static friction. Visualize:

G Since the wall is frictionless, the only force from the wall on the ladder is the normal force n2 . On the other G G G hand, the floor exerts both the normal force n1 and the static frictional force f s . The gravitational force FG on the ladder acts through the center of mass of the ladder. G G Solve: The x- and y-components of Fnet = 0 N are

∑F

x

= n2 − fs = 0 N ⇒ fs = n2

∑F

y

= n1 − FG = 0 N ⇒ n1 = FG

The minimum angle occurs when the static friction is at its maximum value f s max = μ s n1. Thus we have n2 = f s = μs n1 = μ s mg . We choose the bottom corner of the ladder as a pivot point to obtain τ net , because two forces pass through this point and have no torque about it. The net torque about the bottom corner is

τ net = d1mg − d 2 n2 = (0.5L cosθ min )mg − ( L sinθ min ) μ s mg = 0 N m ⇒ 0.5cosθ min = μs sin θ min ⇒ tanθ min =

0.5

μs

=

0.5 = 1.25 ⇒ θ min = 51° 0.4

12.62.

Model: The beam is a rigid body of length 3.0 m and the student is a particle.

Visualize:

Solve:

G G To stay in place, the beam must be in both translational equilibrium ( Fnet = 0 N) and rotational

equilibrium (τ net = 0 Nm). The first condition is

∑F

y

= − ( FG )beam − ( FG )student + F1 + F2 = 0 N

⇒ F1 + F2 = ( FG ) beam + ( FG )student = (100 kg + 80 kg)(9.80 m/s 2 ) = 1764 N Taking the torques about the left end of the beam, the second condition is −( FG ) beam (1.5 m) − ( FG )student (2.0 m) + F2 (3.0 m) = 0 N m −(100 kg)(9.8 m/s 2 )(1.5 m) − (80 kg)(9.8 m/s 2 )(2.0 m) + F2 (3.0 m) = 0 N m ⇒ F2 = 1013 N From F1 + F2 = 1764 N, we get F1 = 1764 N − 1013 N = 0.75 kN. Assess: To establish rotational equilibrium, the choice for the pivot is arbitrary. We can take torques about any point on the body of interest.

12.63.

Model: The structure is a rigid body.

Visualize:

Solve:

We pick the left end of the beam as our pivot point. We don’t need to know the forces Fh and Fv

because the pivot point passes through the line of application of Fh and Fv and therefore these forces do not exert a torque. For the beam to stay in equilibrium, the net torque about this point is zero. We can write

τ about left end = −( FG ) B (3.0 m) − ( FG ) W (4.0 m) + (T sin150°)(6.0 m) = 0 N m Using ( FG ) B = (1450 kg)(9.8 m/s 2 ) and ( FG ) W = (80 kg)(9.8 m/s 2 ), the torque equation can be solved to yield T = 15,300 N. The tension in the cable is slightly more than the cable rating. The worker should be worried.

12.64. Model: G Model the beam as a rigid body. For the beam not to fall over, it must be both in translational G equilibrium ( Fnet = 0 N) and rotational equilibrium (τ net = 0 N m). Visualize:

The boy walks along the beam a distance x, measured from the left end of the beam. There are four forces acting G G on the beam. F1 and F2 are from the two supports, FG is the gravitational force on the beam, and FG is

( )

b

( )

B

the gravitational force on the boy. Solve: We pick our pivot point on the left end through the first support. The equation for rotational equilibrium is −( FG ) b (2.5 m) + F2 (3.0 m) − ( FG ) B x = 0 N m −(40 kg)(9.80 m/s 2 )(2.5 m) + F2 (3.0 m) − (20 kg)(9.80 m/s 2 ) x = 0 N m The equation for translation equilibrium is

∑F

y

= 0 N = F1 + F2 − ( FG )b − ( FG )B

⇒ F1 + F2 = ( FG )b + ( FG )B = (40 kg + 20 kg)(9.8 m/s 2 ) = 588 N Just when the boy is at the point where the beam tips, F1 = 0 N. Thus F2 = 588 N. With this value of F2 , we can simplify the torque equation to: −(40 kg)(9.80 m/s 2 )(2.5 m) + (588 N)(3.0 m) − (20 kg)(9.80 m/s 2 ) x = 0 N m ⇒ x = 4.0 m Thus, the distance from the right end is 5.0 m − 4.0 m = 1.0 m.

12.65.

Visualize:

Please refer to Figure P12.65.

Solve: The bricks are stable when the net gravitational torque on each individual brick or combination of bricks is zero. This is true as long as the center of gravity of each individual brick and any combination is over a base of support. To determine the relative positions of the bricks, work from the top down. The top brick can extend past the second brick by L 2. For maximum extension, their combined center of gravity will be at the edge of the third brick, and the combined center of gravity of the three upper bricks will be at the edge of the fourth brick. The combined center of gravity of all four bricks will be over the edge of the table. Measuring from the left edge of the brick 2, the center of gravity of the top two bricks is ⎛L⎞ m ⎜ ⎟ + mL m1 x1 + m2 x2 3 2 ( x12 )com = = ⎝ ⎠ = L. 2m 4 m1 + m2

Thus the top two bricks can extend L 4 past the edge of the third brick. The top three bricks have a center of mass ⎛L⎞ ⎛ 3L ⎞ ⎛ 5L ⎞ m⎜ ⎟ + m⎜ ⎟ + m⎜ ⎟ m1 x1 + m2 x2 + m3 x3 2⎠ 4 ⎠ ⎝ ⎝ ⎝ 4 ⎠ = 5 L. = ( x123 )com = 3m 6 m1 + m2 + m3 Thus the top three bricks can extend past the edge of the fourth brick by L 6. Finally, the four bricks have a combined center of mass at ⎛L⎞ ⎛ 4L ⎞ ⎛ 11L ⎞ ⎛ 17 L ⎞ m⎜ ⎟ + m⎜ ⎟ + m⎜ ⎟ + m⎜ ⎟ 2⎠ 6 ⎠ 12 ⎠ ⎝ ⎝ ⎝ ⎝ 12 ⎠ = 7 L. ( x1234 )com = 4m 8 The center of gravity of all four bricks combined is 7 L 8 from the left edge of the bottom brick, so brick 4 can extend L 8 past the table edge. Thus the maximum distance to the right edge of the top brick from the table edge is L L L L 25 d max = + + + = L. 8 6 4 2 24 Thus, yes, it is possible that no part of the top brick is directly over the table because d max > L. Assess: As crazy as this seems, the center of gravity of all four bricks is stably supported, so the net gravitational torque is zero, and the bricks do not fall over.

12.66.

Model: Visualize:

The pole is a uniform rod. The sign is also uniform.

Solve:

The geometry of the rod and cable give the angle that the cable makes with the rod. ⎛ 250 ⎞ θ = tan −1 ⎜ ⎟ = 51.3° ⎝ 200 ⎠ The rod is in rotational equilibrium about its left-hand end. ⎛1⎞ ⎛1⎞ τ net = 0 = − (100 cm )( FG )P − ( 80 cm ) ⎜ ⎟ ( FG )S − ( 200 cm ) ⎜ ⎟ ( FG )S + ( 200 cm ) T sin 51.3° ⎝ 2⎠ ⎝ 2⎠

= − (100 cm )( 5.0 kg ) ( 9.8 m/s 2 ) − mS ( 9.8 m/s 2 ) (140 cm ) + (156 cm ) T

With T = 300 N , mS = 30.6 kg. Assess: A mass of 30.6 kg is reasonable for a sign.

12.67.

Model: The hollow cylinder is a rigid rotating body.

Visualize:

We placed the origin of the coordinate system on the ground. Solve: (a) Newton’s second law for the block is −( FG ) B + T = mB a y , where T is the tension in the string, ( FG ) B = mB g is the gravitational force on the block, and a y is the acceleration of block. The string tension exerts a negative (cw) torque on the cylinder, so the rotational form of Newton’s second law for the hollow cylinder is

τ = −TR = Iα = I

ay R

⇒T = −

Ia y R2

where we used the acceleration constraint a y = α R. With this expression for T, Newton’s second law for the block becomes

− mB g −

−mB g Ia = mB a y ⇒ a y = 2 R (mB + I / R 2 )

The moment of inertia of a hollow cylinder is I = mC R 2 , so the equation for a y simplifies to ay =

−mB g −(3.0 kg)(9.8 m/s 2 ) = = −5.88 m/s 2 mB + mC 3.0 kg + 2.0 kg

The speed of the block just before it hits the ground can now be found using kinematics: v12 = v02 + 2a y ( y1 − y0 ) = 0 m 2 /s 2 + 2(−5.88 m/s 2 )(0 m − 1.0 m) ⇒ v1 = 3.4 m/s

(b) The conservation of energy equation K1 + U g1 = K 0 + U g0 for the system (block + cylinder + earth) is

1 1 1 1 mBv12 + Iω12 + mB gy1 = mBv02 + Iω 02 + mB gy0 2 2 2 2 2 v m ⎞ 1 1 ⎛m mBv12 + ( mC R 2 ) 12 + 0 J = 0 J + 0 J + mB gy0 ⇒ v12 ⎜ B + C ⎟ = mB gy0 R 2 2 2 ⎠ ⎝ 2 ⇒ v12 =

2mB gy0 2(3.0 kg)(9.8 m/s 2 )(1.0 m) = = 11.76 m 2 /s 2 ⇒ v1 = 3.4 m/s mB + mC (3.0 kg) + (2.0 kg)

Assess: Newton’s second law and the conservation of energy method give the same result for the block’s final velocity.

12.68.

Model: The bar is a solid body rotating through its center.

Visualize:

Solve:

(a) The two forces form a couple. The net torque on the bar about its center is

Iα L where F is the force produced by one of the air jets. We can find I and α as follows:

τ net = LF = Iα ⇒ F =

1 1 ML2 = (0.50 kg)(0.60 m) 2 = 0.015 kg m 2 12 12 ω1 = ω 0 + α (t1 − t0 ) ⇒ 150 rpm = 5.0π rad/s = 0 rad + α (10 s − 0 s) ⇒ α = 0.50π rad/s 2 I=

⇒F=

(0.015 kg m 2 )(0.5π rad/s 2 ) = 0.0393 N (0.60 m)

The force F = 39 mN. (b) The torque of a couple is the same about any point. It is still τ net = LF . However, the moment of inertia has changed.

LF 1 1 where I = ML2 = (0.500 kg)(0.6 m) 2 = 0.060 kg m 2 I 3 3 (0.0393 N) × (0.60 m) ⇒α = = 0.393 rad/s 2 0.060 kg m 2

τ net = LF = Iα ⇒ α =

Finally,

ω1 = ω 0 + α (t1 − t0 ) = 0 rad/s + (0.393 rad/s 2 )(10 s − 0 s) = 3.93 rad/s =

(3.93)(60) rpm = 37.5 rpm 2π

The angular speed is 38 rpm. Assess: Note that ω ∝ α and α ∝ 1/ I . Thus, ω ∝ 1/ I . I about the center of the rod is 4 times smaller than I about one end of the rod. Consequently, ω is 4 times larger.

12.69.

Model: The flywheel is a rigid body rotating about its central axis.

Visualize:

Solve: (a) The radius of the flywheel is R = 0.75 m and its mass is M = 250 kg. The moment of inertia about the axis of rotation is that of a disk: 1 1 I = MR 2 = (250 kg)(0.75 m) 2 = 70.31 kg m 2 2 2 The angular acceleration is calculated as follows:

τ net = Iα ⇒ α = τ net / I = (50 N m)/(70.31 kg m 2 ) = 0.711 rad/s 2 Using the kinematic equation for angular velocity gives

ω1 = ω 0 + α (t1 − t0 ) = 1200 rpm = 40 π rad/s = 0 rad/s + 0.711 rad/s 2 (t1 − 0 s) ⇒ t1 = 177 s (b) The energy stored in the flywheel is rotational kinetic energy: 1 1 K rot = Iω12 = (70.31 kg m 2 )(40π rad/s) 2 = 5.55 × 105 J 2 2 5 The energy stored is 5.6 × 10 J. energy delivered (5.55 × 105 J)/2 = = 1.39 × 105 W = 139 kW (c) Average power delivered = time interval 2.0 s (d) Because τ = Iα , ⇒ τ = I

− ω half energy ⎞ ⎛ω Δω = I ⎜ full energy ⎟ . ω full energy =ω1 (from part (a)) = 40π rad/s. ω half energy Δt Δt ⎝ ⎠

can be obtained as: 1 2 1 Iω half energy = K rot ⇒ ω half energy = 2 2

5.55 × 105 J K rot = = 88.85 rad/s 70.31 kg m 2 I

Thus ⎛ 40 π rad/s − 88.85 rad/s ⎞ ⎟ = 1.30 kN m 2.0 s ⎝ ⎠

τ = (70.31 kg m 2 ) ⎜

12.70. Model: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a disk and that the string does not slip. Visualize:

Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same. Solve: Applying Newton’s second law to m1 , m2 , and the pulley yields the three equations:

T1 − ( FG )1 = m1a1

− ( FG )2 + T2 = m2 a2

T2 R − T1R − 0.50 N m = Iα

Noting that − a2 = a1 = a, I = 12 mp R 2 , and α = a/R, the above equations simplify to T1 − m1 g = m1a

m2 g − T2 = m2 a

0.50 N m ⎛1 ⎞⎛ a ⎞ 1 0.50 N m 1 T2 − T1 = ⎜ mp R 2 ⎟⎜ ⎟ + = mp a + R 2 0.060 m ⎝2 ⎠⎝ R ⎠ R

Adding these three equations, 1 ⎞ ⎛ (m2 − m1 ) g = a ⎜ m1 + m2 + mp ⎟ + 8.333 N 2 ⎠ ⎝ (m − m1 ) g − 8.333 N (4.0 kg − 2.0 kg)(9.8 m/s 2 ) − 8.333 N ⇒a= 2 = = 1.610 m/s 2 2.0 kg + 4.0 kg + (2.0 kg/2) m1 + m2 + 12 mp We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor: 1 1 y1 = y0 + v0 (t1 − t0 ) + a2 (t1 − t0 ) 2 ⇒ 0 = 1.0 m + 0 + (−1.610 m/s 2 )(t1 − 0 s) 2 2 2 2(1.0 m) ⇒ t1 = = 1.11 s (1.610 m/s 2 )

12.71.

Model: Assume the string does not slip on the pulley.

Visualize:

The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward force on the block m2 , but a downward force on the outer edge of the pulley. Similarly the string exerts a force on block m1 to the right, but a leftward force on the outer edge of the pulley. Solve: (a) Newton’s second law for m1 and m2 is T = m1a1 and T − m2 g = m2 a2 . Using the constraint − a2 = + a1 = a, we have T = m1a and −T + m2 g = m2 a. Adding these equations, we get m2 g = (m1 + m2 )a, or

a=

m2 g mm g ⇒ T = m1a = 1 2 m1 + m2 m1 + m2

(b) When the pulley has mass m, the tensions (T1 and T2 ) in the upper and lower portions of the string are different. Newton’s second law for m1 and the pulley are:

T1 = m1a

and

T1R − T2 R = − Iα

We are using the minus sign with α because the pulley accelerates clockwise. Also, a = Rα . Thus, T1 = m1a and T2 − T1 =

I a aI = R R R2

Adding these two equations gives I ⎞ ⎛ T2 = a ⎜ m1 + 2 ⎟ R ⎝ ⎠

Newton’s second law for m2 is T2 − m2 g = m2 a2 = − m2 a. Using the above expression for T2 , I ⎞ m2 g ⎛ a ⎜ m1 + 2 ⎟ + m2 a = m2 g ⇒ a = R ⎠ m1 + m2 + I / R 2 ⎝ Since I = 12 mp R 2 for a disk about its center, a=

m2 g m1 + m2 + 12 mp

With this value for a we can now find T1 and T2 :

T1 = m1a = Assess:

m1m2 g m1 + m2 + 12 mp

T2 = a (m1 + I / R 2 ) =

1 m2 g 1 ⎞ m2 ( m1 + 2 mp ) g ⎛ + = m m 1 p ⎜ ⎟ ( m1 + m2 + 12 mp ) ⎝ 2 ⎠ m1 + m2 + 12 mp

For m = 0 kg, the equations for a, T1 , and T2 of part (b) simplify to a=

m2 g m1 + m2

These agree with the results of part (a).

and T1 =

m1m2 g m1 + m2

and

T2 =

m1m2 g m1 + m2

12.72. Model: The disk is a rigid spinning body. Visualize: Please refer to Figure P12.72. The initial angular velocity is 300 rpm or (300)(2π )/60 = 10π rad/s. After 3.0 s the disk stops. Solve: Using the kinematic equation for angular velocity, ω1 = ω 0 + α (t1 − t0 ) ⇒ α =

ω1 − ω 0 t1 − t0

=

(0 rad/s − 10π rad/s) −10π = rad/s 2 (3.0 s − 0 s) 3

Thus, the torque due to the force of friction that brings the disk to rest is

τ = Iα = − fR ⇒ f = −

( 12 mR 2 )α = − 1 (mR)α = − 1 (2.0 kg)(0.15 m) ⎛ −10 π rad/s2 ⎞ = 1.57 N Iα =− ⎜ ⎟ R R 2 2 3 ⎝ ⎠

The minus sign with τ = − fR indicates that the torque due to friction acts clockwise.

12.73.

Model: The entire structure is a rigid rotating body. The two thrust forces are a couple that exerts a torque on the structure about its center of mass. We will assume the thrust forces are perpendicular to the connecting tunnel. Visualize: Please refer to Figure 12.30. We chose a coordinate system in which m1 and m2 are on the x-axis

and m1 = 100,000 kg is at the origin. m3 is the mass of the tunnel, whose center is at x3 = 45 m. Solve: (a) Assuming the center of mass of the tunnel is at the center of the tunnel, we get

xcm =

m1 x1 + m2 x2 + m3 x3 m1 + m2 + m3

xcm =

(1.0 × 105 kg)(0 m) + (2.0 × 105 kg)(90 m) + (5.0 × 104 kg)(45 m) = 57.9 m 1.0 × 105 kg + 2.0 × 105 kg + 5.0 × 104 kg

The center of mass of the entire structure is 58 m from the 100,000 kg rocket. (b) Initially, the angular velocity is zero. The structure’s angular velocity after 30 s is

ω1 = ω 0 + α (t1 − t0 ) = 0 rad/s + α (30 s − 0 s) = α (30 s) The angular acceleration α can be found from τ = Iα , where τ is the net torque on the structure and I is its moment of inertia. The two thrusts form a couple with torque

τ = lF = (90 m)(50,000 N) = 4.50 × 106 N m 2 2⎞ ⎛1 I = I m1 + I m2 + I tunnel = m1 x12 + m2 x22 + ⎜ m3 ( 90 m ) + m3 ( 58 m − 45 m ) ⎟ 12 ⎝ ⎠ 1 2 5 2 5 2 4 = (1.0 × 10 kg)(57.9 m) + (2.0 × 10 kg)(32.1 m) + (5 × 10 kg)(90 m) 2 + ( 5 × 104 kg ) (13 m ) 12 = 5.83 × 108 kg m 2

⇒α =

τ I

=

4.5 × 106 N m = 7.71 × 10−3 rad/s 2 5.83 × 108 kg m 2

⇒ ω1 = (30 s)(7.71 × 10−3 rad/s 2 ) = 0.23 rad/s

Assess: Note that the parallel axis theorem was used in finding the moment of inertia of the tunnel.

12.74. Model: Assume that the hollow sphere is a rigid rolling body and that the sphere rolls up the incline without slipping. We also assume that the coefficient of rolling friction is zero. Visualize:

The initial kinetic energy, which is a combination of rotational and translational energy, is transformed in gravitational potential energy. We chose the bottom of the incline as the zero of the gravitational potential energy. Solve: The conservation of energy equation K f + U gf = K i + U gi is

1 1 1 1 2 2 + I cm (ω1 ) 2 + Mgy1 = M (v0 )cm + I cm (ω 0 ) 2 + Mgy0 M (v1 )cm 2 2 2 2 1 1 2 1 1 (v ) 2 ⎛ ⎞ 2 2 2 0 J + 0 J + Mgy1 = M (v0 )cm + ⎜ MR 2 ⎟ (ω 0 )cm + 0 J ⇒ Mgy1 = M (v0 )cm + MR 2 0 2cm R 2 2⎝ 3 2 3 ⎠ 5 (v ) 2 5 5 (5.0 m/s) 2 2 ⇒ gy1 = (v0 )cm ⇒ y1 = 6 0 cm = = 2.126 m g 6 6 9.8 m/s 2

The distance traveled along the incline is 2.126 m y1 = = 4.3 m sin 30° 0.5 Assess: This is a reasonable stopping distance for an object rolling up an incline when its speed at the bottom of the incline is approximately 10 mph. s=

12.75.

Model: Visualize:

Solve:

The masses are particles.

(a) The moment of inertia of the barbell is 2 I = I M + I m = Mx 2 + m ( L − x )

The rotational kinetic energy is therefore K rot =

(

)

1 2 1 2 Iω = Mx 2 + m ( L − x ) ω 2 2 2

To find x such that K rot is a minimum, set

1 dK rot = 0 = ( 2 Mx − 2m ( L − x ) )ω 2 dx 2 (b) This is the center of mass location measured from the mass M.

⇒x=

m L m+M

12.76. Model: The disk is a rigid body rotating on an axle passing through one edge. The gravitational potential energy is transformed into rotational kinetic energy as the disk is released. Visualize:

We placed the origin of the coordinate system at a distance R just below the axle. In the initial position, the center of mass of the disk is at the same level as the axle. The center of mass of the disk in the final position is coincident with the origin of the coordinate system. Solve: (a) The torque is due to the gravitational force on the disk acting at the center of mass. Thus

τ = (mg ) R = (5.0 kg)(9.8 m/s 2 )(0.30 m) = 14.7 N m The moment of inertia about the disk’s edge is obtained using the parallel-axis theorem: 1 3 ⎛ 3⎞ I = I cm + mR 2 = mR 2 + mR 2 = mR 2 = ⎜ ⎟ (5.0 kg)(0.30 m)2 = 0.675 kg m 2 2 2 ⎝ 2⎠ τ 14.7 N m ⇒α = = = 22 rad/s 2 I 0.675 kg m 2 (b) The energy conservation equation K f + U gf = K i = U gi is

1 2 1 1 Iω1 + mgy1 = Iω 02 + mgy0 ⇒ Iω12 + 0 J = 0 J + mgR 2 2 2

ω1 =

2mgR 2(5.0 kg)(9.8 m/s 2 )(0.30 m) = = 6.6 rad/s I 0.675 kg m 2

Assess: An angular velocity of 6.6 rad/s (or 1.05 revolutions/s) as the center of mass of the disk reaches below the axle is reasonable.

12.77. Model: The hoop is a rigid body rotating about an axle at the edge of the hoop. The gravitational torque on the hoop causes it to rotate, transforming the gravitational potential energy of the hoop’s center of mass into rotational kinetic energy. Visualize:

We placed the origin of the coordinate system at the hoop’s edge on the axle. In the initial position, the center of mass is a distance R above the origin, but it is a distance R below the origin in the final position. Solve: (a) Applying the parallel-axis theorem, I edge = I cm + mR 2 = mR 2 + mR 2 = 2mR 2 . Using this expression in

the energy conservation equation K f + U gf = K i + U gi yields:

1 1 I edgeω12 + mgy1 = I edgeω 02 + mgy0 2 2

1 2g (2mR 2 )ω12 − mgR = 0 J + mgR ⇒ ω1 = 2 R

(b) The speed of the lowest point on the hoop is

2g (2 R ) = 8 gR R Note that the speed of the lowest point on the loop involves a distance of 2R instead of R. v = (ω1 )(2 R) =

Assess:

12.78. Model: The long, thin rod is a rigid body rotating about a frictionless pivot on the end of the rod. The gravitational torque on the rod causes it to rotate, transforming the gravitational potential energy of the rod’s center of mass into rotational kinetic energy. Visualize:

We placed the origin of the coordinate system at the pivot point. In the initial position, the center of mass is a distance 1 L above the origin. In the final position, the center of mass is at y = 0 m and thus has zero gravitational potential 2 energy. Solve: (a) The energy conservation equation for the rod K f + U gf = K i + U gi is

1 2 1 Iω1 + mgy1 = Iω 02 + mgy0 2 2

1⎛1 2⎞ 2 ⎜ mL ⎟ω1 + 0 J = 0 J + mg ( L /2) ⇒ ω1 = 3 g / L 2⎝3 ⎠

(b) The speed at the tip of the rod is vtip = (ω1 ) L = 3 gL .

12.79. Model: The sphere attached to a thin rod is a rigid body rotating about the rod. Assume the rod is vertical and the sphere solid. Visualize: Please refer to Figure P12.79. The sphere rotates because the string wrapped around the rod exerts a torque τ. Solve: The torque exerted by the string on the rod is τ = Tr. From the parallel-axis theorem, the moment of inertia of the sphere about the rod’s axis is 2

MR 2 13 ⎛R⎞ 2 I off center = I cm + M ⎜ ⎟ = MR 2 + = MR 2 4 20 ⎝2⎠ 5 From Newton’s second law,

α=

τ I

=

Tr 20Tr = (13MR 2 /20) 13MR 2

12.80.

Model: The pulley is a rigid rotating body.

Visualize:

We placed the origin of the coordinate system on the floor. The pulley rotates about its center. Solve: Using kinematics for the physics book (mass = m1 ), 1 1 y1 = y0 + v0 (t1 − t0 ) + a (t1 − t0 ) 2 0 m = 1.0 m + 0 m + a(0.71 s − 0 s) 2 ⇒ a = −3.967 m/s 2 2 2 Since the torque acting on the pulley is τ = −TR = Iα , we have

I =−

TR

α

=−

TR TR 2 =− a /R a

We can compare the measured value of I for the pulley with the theoretical value. We first must find the tension T. From the free-body diagram, Newton’s second law of motion is T − m1 g = m1a. This means T = m1 ( g + a ) = (1.0 kg)(9.80 m/s 2 − 3.967 m/s 2 ) = 5.833 N With these values of T and a, we can now find I as: I =−

TR 2 (5.833 N)(0.06 m) 2 =− = 5.3 × 10−3 kg m 2 a (−3.967 m/s 2 )

Let us now calculate the theoretical values of I:

Since I disk

hoop about center: I = MR 2 = (2.0 kg)(0.06 m) 2 = 7.2 × 10 −3 kg m 2 1 1 disk about center: I disk = MR 2 = (2.0 kg)(0.06 m) 2 = 3.6 × 10 −3 kg m 2 2 2 < I < I hoop , the mass of the disk is not uniformly distributed. The mass is concentrated near the rim.

12.81.

Model: The angular momentum of the satellite in the elliptical orbit is a constant.

Visualize:

Solve: (a) Because the gravitational force is always along the same direction as the direction of the moment G G G arm vector, the torque τ = r × Fg is zero at all points on the orbit. (b) The angular momentum of the satellite at any point on the elliptical trajectory is conserved. The velocity is G perpendicular to r at points a and b, so β = 90° and L = mvr. Thus

⎛r ⎞ Lb = La ⇒ mvb rb = mva ra ⇒ vb = ⎜ a ⎟ va ⎝ rb ⎠ 30,000 km 30,000 km − 9000 km = 6000 km and rb = + 9000 km = 24,000 km ra = 2 2 ⎛ 6000 km ⎞ ⇒ vb = ⎜ ⎟ (8000 m/s) = 2000 m/s ⎝ 24,000 km ⎠ (c) Using the conservation of angular momentum Lc = La , we get

⎛r ⎞ mvc rc sin β c = mva ra ⇒ vc = ⎜ a ⎟ va /sin β c ⎝ rc ⎠

rc = (9000 km)2 + (12,000 km) 2 = 1.5 × 107 m

From the figure, we see that sin β c = 12,000 15,000 = 0.80. Thus ⎛ 6000 km ⎞ (8000 m/s) vc = ⎜ = 4000 m/s ⎟ 0.80 ⎝ 15,000 km ⎠

12.82.

Model: The clay balls are particles and undergo a totally inelastic collision. Linear momentum is conserved during the collision. Visualize:

Solve:

(a)

The

angle

is

measured counterclockwise ⎛1⎞ β1 = 180° + tan −1 (1) = 225°, and β 2 = tan −1 ⎜ ⎟ = 26.6°. So ⎝ 2⎠ L = L1 + L2 + m1r1v1 sin β1 + m2 r2v2 sin β 2

= ( 0.015 kg )

(

)

2 m ( 2 m/s ) sin 225° + ( 0.025 kg )

G r

from

(

to

G v.

)

5 m ( 2.0 m/s ) sin 26.6°

= 0.020 kg m /s. 2

Note that the signs of L1 and L2 agree with those determined by the right-hand rule. (b) At the instant before the clay balls collide they are located at (1.5 m, 1.0 m). Here, ⎛ 2⎞ β 2 = tan −1 ⎜ ⎟ = 33.7° β1 = 180° + β 2 = 146.3° ⎝ 3⎠ So

( (1.0 m) + (1.5 m) )( 2.0 m/s) sin 213.7 + ( 0.025 kg ) ( (1.0 m ) + (1.5 m ) ) ( 2.0 m/s ) sin 33.7°

L = ( 0.015 kg )

2

2

2

2

= 0.020 kg m 2 /s (c) The clay balls have a final speed v after the collision. Linear momentum is conserved. p1i + p2i = p(1+ 2) f

( 0.015 kg )( 2.0 m/s ) + ( 0.025 kg )( −2.0 m/s ) = ( 0.015 kg + 0.025 kg ) v ⇒ v = −0.50 m/s. The balls are moving to the left. The angle β = β 2 from part (b). The angular momentum after the collision is L = ( 0.040 kg ) Assess:

(

(1.0 m )

2

+ (1.5 m )

2

)(0.50 m/s )sin 33.7° = 0.020 kg m /s

G Angular momentum is also conserved since τ net = 0.

From

2

geometry,

Model: For the (bullet + block + rod) system, angular momentum is conserved. After the bullet is stuck in the block, the mechanical energy of the system is conserved. Assume the block is small. Visualize:

12.83.

The origin of the coordinate system was placed at the center-of-mass of the block as it freely hangs from the bottom of the rod. Solve: The initial angular momentum of the system about the pivot is due only to the bullet:

Li = mb vb L = (0.010 kg)vb (1.0 m) = (0.010 kg m)vb The angular momentum immediately after the bullet hits and sticks in the block is equal to Iω . The moment of inertia is 1 ⎛1 ⎞ I = I rod + I block + I bullet = mR L2 + mB L2 + mb L2 = ⎜ mR + mB + mb ⎟ L2 3 ⎝3 ⎠ ⎡1 ⎤ = ⎢ (1.0 kg) + 2.0 kg + 0.010 kg ⎥ (1.0 m) 2 = 2.343 kg m 2 ⎣3 ⎦ Angular momentum is conserved in the collision:

Lf = (2.343 kg m 2 )ω = Li = (0.010 kg m)vb We need to determine ω before we can find vb . To find ω we use the conservation of mechanical energy equation K f + U gf = K i + U gi as the pendulum swings out, which is 1 2 Iω + 0 J 2 1 L (0.010 kg + 2.0 kg)(9.8 m/s 2 ) L(1 − cos30°) + (1.0 kg)(9.8 m/s 2 ) (1 − cos30°) = (2.343 kg m 2 )ω 2 2 2 0 J + (mb + mB ) gyB + mR gyR =

The energy equation can be further simplified to

2.6390 kg m 2 /s 2 + 0.6565 kg m 2 /s 2 = (1.1715 kg m 2 )ω 2 ⇒ ω = 1.677 rad/s Finally, we can use the conservation of angular momentum equation to obtain the speed of the bullet: vb =

(2.343 kg m 2 )(1.677 rad/s) = 3.9 × 102 m/s 0.010 kg m

Assess: A speed of 390 m/s for a bullet is reasonable.

12.84.

Model: For the (bullet + door) system, the angular momentum is conserved in the collision.

Visualize:

G G Solve: As the bullet hits the door, its velocity v is perpendicular to r . Thus the initial angular momentum about the rotation axis, with r = L, is

Li = mBvB L = (0.010 kg)(400 m/s)(1.0 m) = 4.0 kg m 2 /s After the collision, with the bullet in the door, the moment of inertia about the hinges is 1 1 I = I door + I bullet = mD L2 + mB L2 = (10.0 kg)(1.0 m) 2 + (0.010 kg)(1.0 m) 2 = 3.343 kg m 2 3 3

Therefore,

Lf = Iω = (3.343 kg m 2 )ω .

Using

the

angular

Lf = Li (3.343 kg m )ω = 4.0 kg m /s and thus ω = 1.20 rad/s. 2

2

momentum

conservation

equation

12.85.

Model: The mechanical energy of both the hoop (h) and the sphere (s) is conserved. The initial gravitational potential energy is transformed into kinetic energy as the objects roll down the slope. The kinetic energy is a combination of translational and rotational kinetic energy. We also assume no slipping of the hoop or of the sphere. Visualize:

The zero of gravitational potential energy is chosen at the bottom of the slope. Solve: (a) The energy conservation equation for the sphere or hoop K f + U gf = K i + U gi is 1 1 1 1 I (ω1 ) 2 + m(v1 ) 2 + mgy1 = I (ω 0 ) 2 + m(v0 ) 2 + mgy0 2 2 2 2

For the sphere, this becomes 2 1⎛ 2 1 2 ⎞ (v1 )s 2 ⎜ mR ⎟ 2 + m(v1 )s + 0 J = 0 J + 0 J + mghs 2⎝ 5 2 ⎠ R

⇒

7 (v1 )s2 = gh ⇒ (v1 )s = 10 gh /7 = 10(9.8 m/s 2 )(0.30 m)/7 = 2.05 m/s 10

For the hoop, this becomes 1 (v ) 2 1 (mR 2 ) 1 2 h + m(v1 ) 2h + 0 J = 0 J + 0 J + mghhoop 2 R 2 (v ) 2 ⇒ hhoop = 1 h g For the hoop to have the same velocity as that of the sphere, hhoop =

(v1 )s2 (2.05 m/s) 2 = = 42.9 cm g 9.8 m/s 2

The hoop should be released from a height of 43 cm. (b) As we see in part (a), the speed of a hoop at the bottom depends only on the starting height and not on the mass or radius. So the answer is No.

12.86.

Model: Model the turntable as a rigid disk rotating on frictionless bearings. As the blocks fall from above and stick on the turntable, the turntable slows down due to increased rotational inertia of the (turntable + blocks) system. Any torques between the turntable and the blocks are internal to the system, so angular momentum of the system is conserved. Visualize: The initial moment of inertia is I1 and the final moment of inertia is I 2 .

Solve: The initial moment of inertia is I1 = I disk = 12 mR 2 = 12 (2.0 kg)(0.10 m) 2 = 0.010 kg m 2 and the final moment of inertia is

I 2 = I1 + 2mR 2 = 0.010 kg m 2 + 2(0.500 kg) × (0.10 m) 2 = 0.010 kg m2 + 0.010 kg m 2 = 0.020 kg m2 Let ω1 and ω 2 be the initial and final angular velocities. Then Lf = Li ⇒ ω 2 I 2 = ω1I1 ⇒ ω 2 =

I1ω1 (0.010 kg m 2 )(100 rpm) = = 50 rpm I2 0.020 kg m 2

12.87.

Model: Model the turntable as a rigid disk rotating on frictionless bearings. For the (turntable + block) system, no external torques act as the block moves outward towards the outer edge. Angular momentum is thus conserved. Visualize: The initial moment of inertia of the turntable is I1 and the final moment of inertia is I 2 . Solve: The initial moment of inertia is I1 = I disk = 12 mR 2 = 12 (0.2 kg)(0.2 m) 2 = 0.0040 kg m 2 . As the block reaches the outer edge, the final moment of inertia is

I 2 = I1 + mB R 2 = 0.0040 kg m 2 + (0.020 kg)(0.20 m) 2 = 0.0040 kg m 2 + 0.0008 kg m 2 = 0.0048 kg m 2 Let ω1 and ω 2 be the initial and final angular velocities, then the conservation of angular momentum equation is Lf = Li ⇒ ω 2 I 2 = ω1I1 ⇒ ω 2 =

I1ω1 (0.0040 kg m 2 )(60 rpm) = = 50 rpm I2 (0.0048 kg m 2 )

Assess: A change of angular velocity from 60 rpm to 50 rpm with an increase in the value of the moment of inertia is reasonable.

12.88. Model: Model the merry-go-round as a rigid disk rotating on frictionless bearings about an axle in the center and John as a particle. For the (merry-go-round + John) system, no external torques act as John jumps on the merry-go-round. Angular momentum is thus conserved. Visualize: The initial angular momentum is the sum of the angular momentum of the merry-go-round and the angular momentum of John. The final angular momentum as John jumps on the merry-go-round is equal to I finalω final . Solve: John’s initial angular momentum is that of a particle: LJ = mJ vJ R sin β = mJ vJ R. The angle β = 90° since John runs tangent to the disk. The conservation of angular momentum equation Lf = Li is ⎛1 ⎞ I finalω final = Ldisk + LJ = ⎜ MR 2 ⎟ωi + mJ vJ R ⎝2 ⎠ 2π ⎛ rad ⎞ ⎛1⎞ 2 = ⎜ ⎟ (250 kg)(1.5 m) 2 (20 rpm) ⎜ ⎟ + (30 kg)(5.0 m/s)(1.5 m) = 814 kg m /s 60 ⎝ rpm ⎠ ⎝ 2⎠ ⇒ ω final =

814 kg m 2 /s I final

1 1 I final = I disk + I J = MR 2 + mJ R 2 = (250 kg)(1.5 m) 2 + (30 kg)(1.5 m)2 = 349 kg m 2 2 2 814 kg m 2 /s ω final = = 2.33 rad/s = 22 rpm 349 kg m 2

12.89.

Model: The toy car is a particle located at the rim of the track. The track is a cylindrical hoop rotating about its center, which is an axis of symmetry. No net torques are present on the track, so the angular momentum of the car and track is conserved. Visualize:

Solve:

The toy car’s steady speed of 0.75 m/s relative to the track means that vc − vt = 0.75 m/s ⇒ v c = vt + 0.75 m/s,

where vt is the velocity of a point on the track at the same radius as the car. Conservation of angular momentum implies that Li = Lf 0 = I cω c + I tω t = ( mr 2 )ω c + ( Mr 2 )ω t = mω c + M ω t

The initial and final states refer to before and after the toy car was turned on. Table 12.2 was used for the track. v v Since ω c = c , ω t = t , we have r r 0 = mvc + Mvt ⇒ m ( vt + 0.75 m/s ) + Mvt = 0 ⇒ vt = −

M ( 0.200 kg ) ( 0.75 m/s ) = − ( 0.75 m/s ) = −0.125 m/s m+M ( 0.200 kg + 1.0 kg )

The minus sign indicates that the track is moving in the opposite direction of the car. The angular velocity of the track is v ( 0.125 m/s ) ωt = t = = 0.417 rad/s clockwise. r 0.30 m In rpm, ⎛ rev ⎞⎛ 60 s ⎞ ω t = ( 0.417 rad/s ) ⎜ ⎟⎜ ⎟ ⎝ 2π rad ⎠⎝ min ⎠ = 4.0 rpm Assess: The speed of the track is less than that of the car because it is more massive.

12.90.

Model: Model the skater as a cylindrical torso with two rod-like arms that are perpendicular to the axis of the torso in the initial position and collapse into the torso in the final position. Visualize:

Solve: For the initial position, the moment of inertia is I1 = I Torso + 2 I Arm . The moment of inertia of each arm is that of a 66 cm long rod rotating about a point 10 cm from its end, and can be found using the parallel-axis theorem. In the final position, the moment of inertia is I 2 = 12 MR 2 . The equation for the conservation of angular

momentum Lf = Li can be written I 2ω 2 = I1ω1 ⇒ ω 2 = ( I1 / I 2 )ω1. Calculating I1 and I 2 , 1 ⎡1 ⎤ I1 = M T R 2 + 2 ⎢ M A L2A + M A d 2 ⎥ 2 12 ⎣ ⎦ 1 2⎤ ⎡1 2 = (40 kg)(0.10 m) + 2 ⎢ (2.5 kg)(0.66 m) 2 + (2.5 kg) ( 0.33 m + 0.10 m ) ⎥ = 1.306 kg m 2 2 ⎣12 ⎦ 1 1 (1.306 kg m 2 ) I 2 = MR 2 = (45 kg)(0.10 m) 2 = 0.225 kg m 2 ⇒ ω 2 = (1.0 rev/s) = 5.8 rev /s 2 2 (0.225 kg m 2 )

12.91. Model: Assume that the marble does not slip as it rolls down the track and around a loop-the-loop. The mechanical energy of the marble is conserved. Visualize:

Solve: The ball’s center of mass moves in a circle of radius R − r. The free-body diagram on the marble at its highest position shows that Newton’s second law for the marble is

mv12 R−r The minimum height (h) that the track must have for the marble to make it around the loop-the-loop occurs when the normal force of the track on the marble tends to zero. Then the weight will provide the centripetal acceleration needed for the circular motion. For n → 0 N, mg + n =

mg =

mv 2 ⇒ v12 = g ( R − r ) (R − r)

Since rolling motion requires v12 = r 2ω12 , we have

ω12 r 2 = g ( R − r ) ⇒ ω12 =

g (R − r) r2

The conservation of energy equation is 1 1 ( K f + U gf ) top of loop = ( K i + U gi )initial ⇒ mv12 + Iω12 + mgy1 = mgy0 = mgh 2 2 Using the above expressions and I = 52 mr 2 the energy equation simplifies to 1 1⎛ 2⎞ ⎛ g(R − r) ⎞ mg ( R − r ) + ⎜ ⎟ mr 2 ⎜ ⎟ + mg 2( R − r ) = mgh ⇒ h = 2.7 ( R − r ) 2 2 2⎝ 5⎠ ⎝ r ⎠

12.92.

Model: The Swiss cheese wedge is of uniform density—or at least uniform enough that its center of mass is at the same location as that of a solid piece. To find the angle at which the cheese starts sliding, the cheese will be treated as a particle, and the model of static friction will be used. Visualize:

Solve: The angle at which the cheese starts sliding, θS , will be compared to the critical angle θ c for stability. Use Newton’s second law with the free body diagram. ( Fnet ) x = 0 = fs − FG sinθ s

( Fnet ) y = 0 = n − FG cosθ s

With FG = mg , the y-direction equation gives n = mg cosθ . The cheese starts sliding when μ s is at its maximum value. Combining that with the x-direction equation and fs = μ s n, 0 = μs ( mg cosθ s ) − mg sin θ s ⇒ θ s = tan −1 ( μ s ) = tan −1 ( 0.90 ) = 42° The cheese will start sliding at an angle of 42°. The center of mass of the cheese wedge can be found using the result of problem 12.53. There, the center of mass of a triangle with the same proportions as the cheese wedge was found. So xcm is at the center of the cheese wedge (by symmetry). The ycm can be found by proportional reasoning.

ycm ( 30 cm − 20 cm ) ⇒ y = 4.0 cm = cm 12 cm 30 cm Note that here we have measured ycm from the base of the wedge. Stability considerations require that the center of mass be no further than the left corner of the wedge. At the critical angle geometry shown in the figure above, the right triangle formed by the wedge’s center of mass, lower left corner, and center point of the base is a 45°-45°-90° triangle. So θ c = 45°. Assess: The cheese will slide first as the incline reaches 42°. It would not topple until the angle reaches 45°. So Emily is correct.

12.93.

Model: Define the system as the rod and cube. Energy and angular momentum are conserved in a perfectly elastic collision in the absence of a net external torque. The rod is uniform. Visualize: Please refer to Figure CP12.93. Solve: Let the final speed of the cube be vf , and the final angular velocity of the rod be ω . Energy is conserved, and angular momentum around the rod’s pivot point is conserved. 1 1 1 Ei = Ef ⇒ mv02 = mvf2 + I rodω 2 2 2 2 ⎛d⎞ ⎛d⎞ Li = Lf ⇒ mv0 ⎜ ⎟ = mvf ⎜ ⎟ + I rodω ⎝ 2⎠ ⎝ 2⎠ This is two equations in the two unknowns vf and ω . From Table 12.2,

I rod =

1 1 1 Md 2 = ( 2m ) d 2 = md 2 12 12 6

From the angular momentum equation, d v0 = vf + ω 3

⇒ω =

3 ( v0 − vf ) d

Substituting into the energy equation, 1 1 1⎛1 2 ⎞⎛ 9 ⎞ mv0 2 = mvf 2 + ⎜ md 2 ⎟⎜ 2 ⎟ ( v0 − vf ) 2 2 2⎝ 6 ⎠⎝ d ⎠ 3 2 v0 2 = vf 2 + ( v0 − vf ) 2 6 1 0 = vf 2 − v0vf + v0 2 5 5 This is a quadratic equation in vf . The roots are 2

⎛v 2⎞ 6 ⎛6 ⎞ v0 ± ⎜ v0 ⎟ − 4 ⎜ 0 ⎟ 5 ⎝5 ⎠ 2 ⎝ 5 ⎠ 3 vf = = v0 ± v0 2 5 5 ⎧1 ⎪ v = ⎨5 0 ⎪⎩ v0 1 The answer vf = v0 means the ice cube missed the rod. So vf = v0 to the right. 5

12.94.

Model: The clay ball is a particle. The rod is a uniform thin rod rotating about its center. Angular momentum is conserved in the collision. Visualize:

Solve: This is a two-part problem. Angular momentum is conserved in the collision, and energy is conserved as the ball rises like a pendulum. The angular momentum conservation equation about the rod’s pivot point is Li = Lf ⇒ mv0 r = ( I ball+rod )ω

L 1 = 15 cm. The rod and ball are a composite object. From Table 12.2, I rod = ML2 , so 2 12 1 L2 1 L2 ⎛ M⎞ I ball+rod = I ball + I rod = mr 2 + ML2 = m + ML2 = ⎜ m + ⎟ 12 4 12 4⎝ 3 ⎠ v 2v If vf is the final velocity of the clay ball, ω = f = f since the ball sticks to the rod. Thus r L 2 mv0 L L ⎛ M ⎞⎛ 2v ⎞ = ⎜ m + ⎟⎜ f ⎟ 2 4⎝ 3 ⎠⎝ L ⎠ Note r =

mv0 ( 0.010 kg )( 2.5 m/s ) = 0.714 m/s = M ( 0.075 kg ) m+ ( 0.010 kg ) + 3 3 Energy is conserved as the clay ball rises. Compare the energy of the ball-rod system just after the collision to when the ball reaches the maximum height. Note that the center of mass of the rod does not change position. 1 Ei = Ef ⇒ ( I rod + ball )ω 2 = mgh 2 Thus ⇒ vf =

2

1 L2 ⎛ M ⎞⎛ 2vf ⎞ M⎞ 2⎛ ⎜ m + ⎟⎜ ⎟ = mgh ⇒ vf ⎜ m + ⎟ = mgL (1 − cosθ ) 2 4⎝ 3 ⎠⎝ L ⎠ 3 ⎠ ⎝ 2 v ⎛ M⎞ ⇒ 1 − f ⎜ m + ⎟ = cosθ mgL ⎝ 3 ⎠ Using the various values, cosθ = 0.393 ⇒ θ = 67°. L Assess: The clay ball rises h = (1 − cosθ ) = 9.1 cm. This is about 2/3 of the height of the pivot point, and is 2 reasonable.

12.95. Model: Because no external torque acts on the star during gravitational collapse, its angular momentum is conserved. Model the star as a solid rotating sphere. Solve: (a) The equation for the conservation of angular momentum is ⎛2 ⎞ ⎛2 ⎞ Li = Lf ⇒ I iωi = I f ω f ⇒ ⎜ mRi2 ⎟ωi = ⎜ mRf2 ⎟ω f 5 5 ⎝ ⎠ ⎝ ⎠ ⇒ Rf = Ri

ωi ωf

The angular velocity is inversely proportional to the period T. We can write

Rf = Ri

0.10 s Tf = ( 7.0 × 108 m ) = 1.3749 × 105 m = 137 km 2.592 × 106 s Ti

(b) A point on the equator rotates with r = Rf . Its speed is v=

2π Rf 2π (137,490 m) = = 8.6 × 106 m/s 0.10 s T

Model: For the (turntable + bicycle wheel + professor) system the angular momentum is conserved because the turntable is frictionless and no external torques act on the system. Solve: (a) Nothing happens. The bicycle wheel already has an angular momentum and nothing changes for the wheel when it is handed to the professor. So, nothing happens to the professor. (b) The initial angular momentum is 12.96.

⎛ 180 × 2π rad ⎞ 2 ( Lwheel )i = Iω = (mR 2 )ω = (4.0 kg)(0.32 m) 2 ⎜ ⎟ = 7.72 kg m /s s ⎠ ⎝ 60 When the wheel is turned upside down, the angular momentum of the wheel becomes

( Lwheel )f = −7.72 kg m 2 /s Since the initial and final total angular momentum should be equal, the professor must acquire some angular momentum. From the angular momentum conservation equation, Lprof + ( Lwheel ) f = ( Lwheel )i ⇒ Lprof = ( Lwheel )i − ( Lwheel )f = 7.72 kg m 2 /s − (−7.72 kg m 2 /s) = 15.44 kg m 2 /s

Let us now calculate the angular velocity of the turntable by using Lprof = I totalω , where I total = I turntable + I body + I arms +

I wheel in hand . We assume that the axis of the professor and turntable also become the axis of the arms and wheel (i.e., no leaning). 1 1 I turntable = mT RT2 = (5.0 kg)(0.25 m) 2 = 0.156 kg m 2 2 2 1 2 I body = mbody R = (70 kg)(0.125 m) 2 = 0.547 kg m 2 2 1 ⎛ ⎞ 2 I arms = 2 ⎜ marms R 2 ⎟ = (2.5 kg)(0.45 m) 2 = 0.338 kg m 2 ⎝3 ⎠ 3 I wheel in hand = mwheel r 2 = (4.0 kg)(0.45 m) 2 = 0.810 kg m 2 Thus, I total = (0.156 + 0.547 + 0.338 + 0.810) kg m 2 = 1.851 kg m 2 . Using this value for I total , we can now find ω to be

ω=

Lprof I total

=

15.44 kg m 2 /s = 8.34 rad/s = 79.7 rpm 1.851 kg m 2

12-1

13.1.

Model: Model the sun (s) and the earth (e) as spherical masses. Due to the large difference between your size and mass and that of either the sun or the earth, a human body can be treated as a particle. GM s M y GM e M y and Fe on you = Solve: Fs on you = 2 rs − e re2 Dividing these two equations gives

Fs on y Fe on y

2

2

⎛ M ⎞⎛ r ⎞ ⎛ 1.99 × 1030 kg ⎞⎛ 6.37 × 106 m ⎞ −4 = ⎜ s ⎟⎜ e ⎟ = ⎜ ⎟⎜ ⎟ = 6.00 × 10 24 11 M r × × 5.98 10 kg 1.50 10 m ⎠⎝ ⎠ ⎝ e ⎠⎝ s − e ⎠ ⎝

13.2.

Model:

Assume the two lead balls are spherical masses. Gm1m2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(10 kg)(0.100 kg) = = 6.7 × 10−9 N Solve: (a) F1 on 2 = F2 on 1 = r2 (0.10 m) 2 (b) The ratio of the above gravitational force to the gravitational force on the 100 g ball is

6.67 × 10−9 N = 6.81 × 10−9 (0.100 kg)(9.8 m/s 2 ) Assess: The answer in part (b) shows the smallness of the gravitational force between two lead balls separated by 10 cm compared to the gravitational force on the 100 g ball.

13.3.

Model:

Model the sun (s), the moon (m), and the earth (e) as spherical masses. GM s M m GM e M m and Fe on m = Solve: Fs on m = rs2− m re2− m Dividing the two equations and using the astronomical data from Table 13.2, 2

2

Fs on m ⎛ M s ⎞⎛ re − m ⎞ ⎛ 1.99 × 1030 kg ⎞⎛ 3.84 × 108 m ⎞ =⎜ ⎟⎜ ⎟ =⎜ ⎟⎜ ⎟ = 2.18 Fe on m ⎝ M e ⎠⎝ rs − m ⎠ ⎝ 5.98 × 1024 kg ⎠⎝ 1.50 × 1011 m ⎠ Note that the sun-moon distance is not noticeably different from the tabulated sun-earth distance.

13.4.

Solve:

Fsphere on particle =

GM s M p rs2− p

and

Fearth on particle =

GM e M p re2

Dividing the two equations,

Fsphere on particle Fearth on particle

2

2

⎛ M ⎞ ⎛ r ⎞ ⎛ 5900 kg ⎞ ⎛ 6.37 × 106 m ⎞ −7 = ⎜ s ⎟⎜ e ⎟ = ⎜ ⎟ = 1.60 × 10 ⎟⎜ 24 ⎜ ⎟ M r 5.98 10 kg 0.50 m × ⎠⎝ ⎠ ⎝ e ⎠ ⎝ s−p ⎠ ⎝

13.5.

Model:

Solve:

Fw on m

Model the woman (w) and the man (m) as spherical masses or particles. GM w M m (6.67 × 10−11 N ⋅ m 2 /kg 2 )(50 kg)(70 kg) = Fm on w = = = 2.3 × 10−7 N rm2 − w (1.0 m) 2

13.6.

Model: Visualize:

Model the earth (e) as a sphere.

The space shuttle or a 1.0 kg sphere (s) in the space shuttle is Re + rs = 6.37 × 106 m + 0.30 × 106 m = 6.67 × 106 m away from the center of the earth. GM e M s (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(1.0 kg) = = 9.0 N Solve: (a) Fe on s = ( Re + rs ) 2 (6.67 × 106 m) 2 (b) Because the sphere and the shuttle are in free fall with the same acceleration around the earth, there cannot be any relative motion between them. That is why the sphere floats around inside the space shuttle.

13.7. Solve:

Model:

Model the sun (s) as a spherical mass. GM (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) = 274 m/s 2 (a) gsun surface = 2 s = Rs (6.96 × 108 m) 2

(b) gsun at earth =

GM s (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) = = 5.90 × 10−3 m/s 2 rs2− e (1.50 × 1011 m) 2

13.8. Solve:

Model:

Model the moon (m) and Jupiter (J) as spherical masses. GM m (6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.36 × 1022 kg) = = 1.62 m/s 2 (a) g moon surface = Rm2 (1.74 × 106 m) 2

(b) g Jupiter surface =

GM J (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.90 × 1027 kg) = = 25.9 m/s 2 RJ2 (6.99 × 107 m) 2

13.9.

Model: Model the earth (e) as a spherical mass. Visualize: The acceleration due to gravity at sea level is 9.83 m/s 2 (see Table 13.1) and Re = 6.37 × 106 m (see Table 13.2).

Solve:

g observatory =

GM e = ( Re + h) 2

GM e ⎛ h⎞ Re2 ⎜1 + ⎟ ⎝ Re ⎠

2

=

g earth ⎛ h⎞ ⎜1 + ⎟ ⎝ Re ⎠

2

= (9.83 − 0.0075) m/s 2

Here g earth = GM e Re2 is the acceleration due to gravity on a non-rotating earth, which is why we’ve used the value 9.83 m/s 2 . Solving for h, ⎛ 9.83 ⎞ h = ⎜⎜ − 1⎟⎟ Re = 2.43 km ⎝ 9.8225 ⎠

13.10. Model: Model the earth (e) as a spherical mass. Solve:

Let the acceleration due to gravity be 3gsurface when the earth is shrunk to a radius of x. Then,

gsurface = ⇒3

GM e Re2

and

3 gsurface =

GM e x2

GM e GM e R = 2 ⇒ x = e = 0.577 Re Re2 x 3

The earth’s radius would need to be 0.577 times its present value.

13.11.

Model:

Model Planet Z as a spherical mass. GM Z (6.67 × 10−11 N ⋅ m 2 /kg 2 ) M Z ⇒ 8.0 m/s 2 = ⇒ M Z = 3.0 × 1024 kg Solve: (a) g Z surface = 2 RZ (5.0 × 106 m) 2 (b) Let h be the height above the north pole. Thus,

g above N pole =

GM Z = ( RZ + h) 2

GM Z ⎛ h ⎞ RZ2 ⎜1 + ⎟ ⎝ RZ ⎠

2

=

g Z surface ⎛ h ⎞ ⎜1 + ⎟ ⎝ RZ ⎠

2

=

8.0 m/s 2 ⎛ 10.0 × 106 m ⎞ ⎜1 + ⎟ 5.0 × 106 m ⎠ ⎝

2

= 0.89 m/s 2

13.12.

Model: Model Mars (M) as a spherical mass. Ignore air resistance. Also consider Mars and the ball as an isolated system, so mechanical energy is conserved. Visualize:

Solve: A height of 15 m is very small in comparison with the radius of earth or Mars. We can use flat-earth gravitational potential energy to find the speed with which the astronaut can throw the ball. On earth, with yi = 0 m and vf = 0 m/s, energy conservation is 1 2

mvf2 + mgyf = 12 mvi2 + mgyi ⇒ vi = 2 g e yf = 2(9.80 m/s 2)(15 m) = 17.1 m/s

Energy is also conserved on Mars, but the acceleration due to gravity is different. GM M (6.67 × 10−11 N ⋅ m 2 /kg 2 )(6.42 × 1023 kg) g Mar’s surface = = = 3.77 m/s 2 RM2 (3.37 × 106 m) 2 On Mars, with yi = 0 m and vf = 0 m/s, energy conservation is 1 2

mvf2 + mgyf = 12 mvi2 + mgyi ⇒ yf =

vi2 (17.1 m/s) 2 = = 39 m 2 g m 2(3.77 m/s 2)

13.13.

Model: Model Jupiter as a spherical mass and the object as a point particle. The object and Jupiter form an isolated system, so mechanical energy is conserved. The minimum launch speed for escape, which is called the escape speed, causes an object to stop only as the distance approaches infinity. Visualize:

Solve:

The energy conservation equation K 2 + U 2 = K1 + U1 is

1 GM J mo 1 GM J mo mo v22 − = mo v12 − 2 r2 2 RJ where RJ and M J are the radius and mass of Jupiter. Using the asymptotic condition v2 = 0 m/s as r2 → ∞, 1 2GM J 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.90 × 1027 kg) GM J mo 0 J = mo v12 − ⇒ v1 = = = 6.02 × 104 m/s 2 6.99 × 107 m RJ RJ

Thus, the escape velocity from Jupiter is 60.2 km/s.

13.14.

Model: Model the earth (e) as a spherical mass. Compared to the earth’s size and mass, the rocket (r) is modeled as a particle. This is an isolated system, so mechanical energy is conserved. Visualize:

Solve:

The energy conservation equation K 2 + U 2 = K1 + U1 is

1 GM e mr 1 GM e mr mr v22 − = mr v12 − 2 r2 2 Re In the present case, r2 → ∞, so 1 1 GM e mr 2GM e mr v22 = mr v12 − ⇒ v22 = v12 − Re Re 2 2 ⇒ v2 = (1.50 × 104 m/s) 2 − = 9.99 × 103 m/s

2(6.67 × 10−11 N m 2 /kg 2 )(5.98 × 1024 kg) 6.37 × 106 m

13.15. Model: The probe and the sun form an isolated system, so mechanical energy is conserved. The minimum launch speed for escape, which is called the escape speed, causes an object to stop only as the distance approaches infinity. Visualize:

We denote by mp the mass of the probe. M S is the sun’s mass, and RS − p is the separation between the centers of the sun and the probe. Solve: The conservation of energy equation K 2 + U 2 = K1 + U1 is

GM Smp 1 GM Smp 1 mp v22 − = mp v12 − 2 ( r2 ) 2 RS− p Using the condition v2 = 0 m/s asymptotically as r2 → ∞, GM Smp 1 2GM S 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) 2 mp vescape = ⇒ vescape = = = 4.21× 104 m/s 2 (1.50 × 1011 m) RS − p RS − p

13.16.

Model: Model the distant planet (p) and the earth (e) as spherical masses. Because both are isolated, the mechanical energy of the object on both the planet and the earth is conserved. Visualize:

Let us denote the mass of the planet by M p and that of the earth by M e . Your mass is m0 . Also, acceleration due to gravity on the surface of the planet is g p and on the surface of the earth is g e . Rp and Re are the radii of the planet and the earth, respectively. Solve: (a) We are given that M P = 2M e and g p = 14 g e .

Since g p =

GM p Rp2

and g e = GM p 2 p

R

=

GM e , we have Re2

1 GM e 1 G ( M p /2) = ⇒ Rp = 8Re = 8(6.37 × 106 m) = 1.80 × 107 m Re2 4 Re2 4

(b) The conservation of energy equation K 2 + U 2 = K1 + U1 is

GM p mo 1 GM p mo 1 mov22 − = mo v12 − 2 r2 2 Rp Using v2 = 0 m/s as r2 → ∞, we have GM p mo 1 2 = mo vescape 2 Rp ⇒ vescape =

2GM p Rp

=

2G (2M e ) 4(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) = = 9.41 km/s 1.80 × 107 m Rp

13.17.

Model: Model the sun (s) as a spherical mass and the asteroid (a) as a point particle. Visualize: The asteroid, having mass ma and velocity va , orbits the sun in a circle of radius ra . The asteroid’s

time period is Ta = 5.0 earth years = 1.5779 × 108 s. Solve: The gravitational force between the sun (mass = M S ) and the asteroid provides the centripetal acceleration required for circular motion. 2

1/ 3

⎛ GM sTa2 ⎞ GM s ma ma va2 GM s ⎛ 2π ra ⎞ = ⇒ =⎜ ⎟ ⇒ ra = ⎜ ⎟ 2 2 ra ra ra ⎝ 4π ⎠ ⎝ Ta ⎠

Substituting G = 6.67 × 10−11 N ⋅ m 2 /kg 2 , M s = 1.99 × 1030 kg, and the time period of the asteroid, we obtain

ra = 4.37 × 1011 m. The velocity of the asteroid in its orbit will therefore be va =

2π ra (2π )(4.37 × 1011 m) = = 1.74 × 104 m/s Ta 1.5779 × 108 s

13.18.

Model: Model the sun (s) and the earth (e) as spherical masses. Visualize: The earth orbits the sun with velocity ve in a circular path with a radius denoted by rs-e . The sun’s

and the earth’s masses are denoted by M s and me . Solve: The gravitational force provides the centripetal acceleration required for circular motion. GM s me meve2 me (2π rs − e ) 2 = = rs2− e rs − e rs − eTe2 ⇒ Ms =

4π 2 rs3− e 4π 2 (1.50 × 1011 m)3 = = 2.01 × 1030 kg 2 −11 GTe (6.67 × 10 N ⋅ m 2 /kg 2 )(365 × 24 × 3600 s) 2

Assess: The tabulated value is 1.99 × 1030 kg. The slight difference can be ascribed to the fact that the earth’s orbit isn’t exactly circular.

13.19. Solve:

Model: Model the star (s) and the planet (p) as spherical masses. A planet’s acceleration due to gravity is

gp = ⇒ Mp =

g p Rp2 G

=

GM p Rp2

(12.2 m/s 2 )(9.0 × 106 m) 2 = 1.48 × 1025 kg 6.67 × 10−11 N ⋅ m 2 /kg 2

(b) A planet’s orbital period is

⇒ Ms =

4π 2 r 3 = GT 2 (6.67 × 10−11

⎛ 4π 2 ⎞ 3 T2 =⎜ ⎟r ⎝ GM s ⎠ 4π 2 (2.2 × 1011 m)3 = 5.2 × 1030 kg N ⋅ m 2 /kg 2 )(402 × 24 × 3600 s) 2

13.20.

Model: Model the planet and satellites as spherical masses. Visualize: Please refer to Figure EX13.20. Solve: (a) The period of a satellite in a circular orbit is T = [(4π 2 /GM ) r 3 ]1/ 2 . This is independent of the satellite’s mass, so we can find the ratio of the periods of two satellites a and b:

⎛r ⎞ Ta = ⎜ a⎟ Tb ⎝ rb ⎠

3

Satellite 2 has r2 = r1 , so T2 = T1 = 250 min. Satellite 3 has r3 = (3/2) r1 , so T3 = (3/2)3/ 2 T1 = 459 min. (b) The force on a satellite is F = GMm/r 2 . Thus the ratio of the forces on two satellites a and b is 2

Fa ⎛ rb ⎞ ⎛ ma ⎞ =⎜ ⎟ ⎜ ⎟ Fb ⎝ ra ⎠ ⎝ mb ⎠ Satellite 2 has r2 = r1 and m2 = 2m1 , so F2 = (1) 2(2) F1 = 20,000 N. Similarly, satellite 3 has r3 = (3/2) r1 and m3 = m1 , so F3 = (2/3) 2 (1) F1 = 4440 N. (c) The speed of a satellite in a circular orbit is v = (GM/r ) 2 , so its kinetic energy is K = 12 mv 2 = GMm/2r. Thus the ratio of the kinetic energy of two satellites a and b is

K a ⎛ rb ⎞⎛ ma ⎞ = ⎜ ⎟⎜ ⎟ K b ⎝ ra ⎠⎝ mb ⎠ Satellite 3 has r3 = (3/2) r1 and m3 = m1 , so K1/K 3 = (3/2)(1/1) = 3/2 = 1.50.

13.21.

Model: Model the sun (S) as a spherical mass and the satellite (s) as a point particle. Visualize: The satellite, having mass ms and velocity vs , orbits the sun with a mass M S in a circle of radius

rs . Solve: The gravitational force between the sun and the satellite provides the necessary centripetal acceleration for circular motion. Newton’s second law is GM Sms ms vs2 = rs2 rs

Because vs = 2π rs / Ts where Ts is the period of the satellite, this equation simplifies to GM S (2π rs ) 2 GM STs2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)(24 × 3600 s) 2 3 r = ⇒ = = ⇒ rs = 2.9 × 109 m s rs2 Ts2 rs 4π 2 4π 2

13.22.

Model: Model the earth (e) as a spherical mass and the shuttle (s) as a point particle. Visualize: The shuttle, having mass ms and velocity vs , orbits the earth in a circle of radius rs . We will

denote the earth’s mass by M e . Solve: The gravitational force between the earth and the shuttle provides the necessary centripetal acceleration for circular motion. Newton’s second law is

GM e ms msvs2 GM e GM e = ⇒ vs2 = ⇒ vs = rs2 rs rs rs Because rs = Re + 350 km = 6.72 × 106 m, vs = Ts =

(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) = 7704 m/s (6.72 × 106 m) 2π rs 2π (6.72 × 106 m) = = 5580 s = 91.3 minutes vs 7.704 × 103 m/s

13.23.

Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle. Visualize: The satellite has a mass is ms and orbits the earth with a velocity vs . The radius of the circular orbit

is denoted by rs and the mass of the earth by M e . Solve: The satellite experiences a gravitational force that provides the centripetal acceleration required for circular motion: (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) GM e ms ms vs2 GM = ⇒ rs = 2 e = = 1.32 × 107 m 2 (5500 m/s) 2 rs rs vs

⇒ Ts =

2π Rs (2π )(1.32 × 107 m) = = 1.51 × 104 s = 4.2 hr (5500 m/s) vs

13.24.

Model: Model Mars (m) as a spherical mass and the satellite (s) as a point particle. Visualize: The geosynchronous satellite whose mass is ms and velocity is vs orbits in a circle of radius rs

around Mars. Let us denote mass of Mars by M m . Solve: The gravitational force between the satellite and Mars provides the centripetal acceleration needed for circular motion: 1/ 3

⎛ GM mTs2 ⎞ GM m ms ms vs2 ms (2π rs ) 2 = = ⇒ = r ⎜ ⎟ s 2 rs2 rs rs (Ts ) 2 ⎝ 4π ⎠ Using G = 6.67 × 10−11 N ⋅ m 2 /kg 2 ,

M m = 6.42 × 1023 kg, and Ts = (24.8 hrs) = (24.8)(3600) s = 89,280 s, we

obtain rs = 2.05 × 107 m. Thus, altitude = rs − Rm = 1.72 × 107 m.

13.25.

Solve: We are given M 1 + M 2 = 150 kg which means M 1 = 150 kg − M 2 . We also have

GM 1M 2

( 0.20 m ) Thus,

2

= 8.00 × 10−6 N ⇒ M 1M 2 =

(8.00 × 10−6 N)(0.20 m) 2 = 4798 kg 2 6.67 × 10−11 N ⋅ m 2 /kg 2

(150 kg − M 2 ) M 2 = 4798 kg 2 or M 22 − (150 kg) M 2 + (4798 kg 2 ) = 0.

M 2 = 103.75 kg and 46.25 kg. The two masses are 104 kg and 46 kg.

Solving

this

equation

gives

13.26. Visualize: Because of the gravitational force of attraction between the lead spheres, the cables will make an angle of θ with the vertical. The distance between the sphere centers is therefore going to be less than 1 m. The free-body diagram shows the forces acting on the lead sphere.

Solve: We can see from the diagram that the distance between the centers is d = 1.000 m − 2 L sin θ . Each sphere is in static equilibrium, so Newton’s second law is

∑F = F x

grav

− T sin θ = 0 ⇒ T sin θ = Fgrav

∑F

y

= T cosθ − mg = 0 ⇒ T cosθ = mg

Dividing these two equations to eliminate the tension T yields

F Gmm / d 2 Gm sin θ = tan θ = grav = = 2 cosθ mg mg d g We know that d is going to be only very, very slightly less than 1.00 m. The very slight difference is not going to be enough to affect the value of Fgrav , the gravitational attraction between the two masses, so we’ll evaluate the right side of this equation by using 1.00 m for d. This gives tan θ =

(6.67 × 10−11 N m 2 /kg 2 ) (100 kg) = 6.81 × 10−10 ⇒ θ = (3.90 × 10−8 )° (1.00 m) 2 (9.80 m/s 2 )

This small angle causes the two spheres to move closer by 2 L sin θ = 1.4 × 10−7 m = 0.00000014 m. Consequently, the distance between their centers is d = 0.99999986 m.

13.27. Visualize:

We placed the origin of the coordinate system on the 20 kg sphere (m1 ). The sphere (m2 ) with a mass of 10 kg is 20 cm away on the x-axis. The point at which the net gravitational force is zero must lie between the masses m1

and m2 . This is because on such a point, the gravitational forces due to m1 and m2 are in opposite directions. As the gravitational force is directly proportional to the two masses and inversely proportional to the square of distance between them, the mass m must be closer to the 10-kg mass. The small mass m, if placed either to the left of m1 or to the right of m2 , will experience gravitational forces from m1 and m2 pointing in the same direction, thus always leading to a nonzero force. mm m2 m 20 10 Solve: Fm1 on m = Fm2 on m ⇒ G 12 = G ⇒ 2 = 2 x x (0.20 − x ) (0.20 − x) 2 ⇒ 10 x 2 − 8 x + 0.8 = 0 ⇒ x = 0.683 m and 0.117 m The value x = 68.3 cm is unphysical in the current situation, since this point is not between m1 and m2 . Thus, the point ( x, y ) = (11.7 cm, 0 cm) is where a small mass is to be placed for a zero gravitational force.

13.28.

We placed the origin of the coordinate system on the 20.0 kg mass (m1 ) so that the 5.0 kg

Visualize:

mass (m3 ) is on the x-axis and the 10.0 kg mass (m2 ) is on the y-axis.

Solve:

(a) The forces acting on the 20 kg mass (m1 ) are

G Gm1m2 ˆ (6.67 × 10−11 N ⋅ m 2 /kg 2 )(20.0 kg)(10.0 kg) ˆ Fm2 on m1 = j= j = 3.335 × 10−7 ˆj N r122 (0.20 m) 2 G Gm1m3 ˆ (6.67 × 10−7 N ⋅ m 2 /kg 2 )(20.0 kg)(5.0 kg) ˆ Fm3 on m1 = i= i = 6.67 × 10−7 iˆ N r132 (0.10 m) 2 G Fon m = 6.67 × 10−7 iˆ N + 3.335 × 10−7 ˆj N ⇒ Fon m = 7.46 × 10−7 N 1

1

⎛F θ = tan −1 ⎜ m3 on m1 ⎜ ⎝ Fm2 on m1

⎞ ⎛ 6.67 × 10−7 N ⎞ ⎟ = tan −1 ⎜ ⎟ = 63.4° −7 ⎟ ⎝ 3.335 × 10 N ⎠ ⎠

G Thus the force is Fon m1 = ( 7.5 × 10−7 N, 63° cw from the y -axis ). (b) The forces acting on the 5 kg mass (m3 ) are G G Fm1 on m3 = − Fm3 on m1 = −6.67 × 10−7 iˆ N

Gm2 m3 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(10.0 kg)(5.0 kg) = = 6.67 × 10−8 N r232 [(0.20) 2 + (0.10) 2 ] m = −(6.67 × 10−8 N)cos φ iˆ + (6.67 × 10−8 N)sin φ ˆj

Fm2 on m3 = G Fm2 on m3

G Fon m3

⎛ 10 cm ⎞ ˆ ⎛ 20 cm ⎞ ˆ −8 = −(6.67 × 10−8 N) ⎜ ⎟ i + (6.67 × 10 N) ⎜ ⎟j ⎝ 22.36 cm ⎠ ⎝ 22.36 cm ⎠ = −(2.983 × 10−8 N)iˆ + (5.966 × 10−8 N) ˆj = −6.968 × 10−7 iˆ N + 5.966 × 10−8 ˆj N

Fon m3 = (−6.968 × 10−7 N) 2 + (5.966 × 10−8 N)2 = 6.99 × 10−7 N ⎛ 6.968 × 10−7 N ⎞ ⎟ = 85.1° −8 ⎝ 5.966 × 10 N ⎠

θ ′ = tan −1 ⎜

G Thus Fon m3 = ( 7.0 × 10−7 N, 85° ccw from the y -axis ).

13.29. Visualize:

⎛ 5 ⎞ The angle θ = tan −1 ⎜ ⎟ = 14.04°. The distance r1 = r2 = ⎝ 20 ⎠ forces on the 20.0 kg mass are G Mm F1 = G 2 1 − sinθ iˆ + cosθ ˆj r1 G Mm F2 = G 2 2 sin θ iˆ + cosθ ˆj r2

( 0.050 m )

Solve:

(

(

2

+ ( 0.200 m ) = 0.206 m. The 2

)

)

Note m1 = m2 and r1 = r2 . Thus the net force on the 20.0 kg mass is

G G G Mm Fnet = F1 + F2 = 2G 2 1 cosθ ˆj r1 =

2 ( 6.67 × 10−11 N ⋅ m 2 /kg 2 ) ( 20.0 kg )( 5.0 kg ) cos14.04°

= 3.0 × 10−7 ˆj N

( 0.206 m )

2

ˆj

13.30. Visualize:

Solve: The total gravitational potential energy is the sum of the potential energies due to the interactions of the pairs of masses. U = U12 + U13 + U 23

= −G

m1m2 mm mm −G 1 3 −G 2 3 r12 r13 r23

With m1 = 20.0 kg, m2 = 10.0 kg, m3 = 5.0 kg, r12 = 0.20 m, r13 = 0.10 m, and r23 = 0.2236 m, U = −1.48 × 10 −7 J

Assess: The gravitational potential energy is negative because the masses attract each other. It is a scalar, so there are no vector calculations required.

13.31. Visualize:

Solve: The total gravitational potential energy is the sum of the potential energies due to the interactions of the pairs of masses.

U = U12 + U13 + U 23 = −G With m1 = 20.0 kg, m2 = m3 = 10.0 kg, r12 = r13 =

m1m2 mm mm −G 1 3 −G 2 3 r12 r13 r23

( 0.050 m )

2

+ ( 0.200 m ) = 0.206 m, and r23 = 0.100 m, 2

U = −1.96 × 10−7 J Assess: The gravitational potential energy is negative because the masses attract each other. It is a scalar, so there are no vector calculations required.

13.32.

Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle. Visualize: Let h be the height from the surface of the earth where the acceleration due to gravity ( g altitude ) is

10% of the surface value ( gsurface ). Solve:

(a) Since g altitude = (0.10) gsurface , we have

GM e GM = (0.10) 2 e ⇒ ( Re + h) 2 = 10Re2 2 ( Re + h) Re ⇒ h = 2.162 Re ⇒ h = (2.162)(6.37 × 106 m) = 1.377 × 107 m= 1.38 × 107 m (b) For a satellite orbiting the earth at a height h above the surface of the earth, the gravitational force between the earth and the satellite provides the centripetal acceleration necessary for circular motion. For a satellite orbiting with velocity vs ,

GM e ms ms vs2 = ⇒ vs = 2 ( Re + h) ( Re + h)

GM e (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) = = 4.45 km/s Re + h (6.37 × 106 m + 1.377 × 107 m)

13.33.

Model: Model the earth as a spherical mass and the object (o) as a point particle. Ignore air resistance. This is an isolated system, so mechanical energy is conserved. Visualize:

Solve:

(a) The conservation of energy equation K 2 + U g2 = K1 + U g1 is

1 GM e mo 1 GM e mo mov22 − = mo v12 − Re 2 2 ( Re + y1 ) ⎛ 1 1 ⎞ ⇒ v2 = 2GM e ⎜ − ⎟ ⎝ Re Re + y1 ⎠ 1 1 ⎛ ⎞ = 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) ⎜ − ⎟ = 3.02 km/s 6 6 ⎝ 6.37 × 10 m 6.87 × 10 m ⎠ (b) In the flat-earth approximation, U g = mgy. The energy conservation equation thus becomes

1 1 mov22 + mo gy2 = mov12 + mo gy1 2 2 ⇒ v2 = v12 + 2 g ( y1 − y2 ) = 2(9.80 m/s 2 )(5.00 × 105 m − 0 m) = 3.13 km/s (c) The percent error in the flat-earth calculation is

3130 m/s − 3020 m/s ≈ 3.6% 3020 m/s

13.34. Model: Model the earth and the projectile as spherical masses. Ignore air resistance. This is an isolated system, so mechanical energy is conserved. Visualize:

A pictorial representation of the before-and-after events is shown. Solve: After using v2 = 0 m/s, the energy conservation equation K 2 + U 2 = K1 + U1 is

0 J−

GM e mp

GM e mp 1 = mpv12 − Re + h 2 Re

The projectile mass cancels. Solving for h, we find −1

⎡1 v2 ⎤ h = ⎢ − 1 ⎥ − Re ⎣ Re 2GM e ⎦ = 4.18 × 105 m = 418 km

13.35.

Model: Model the planet (p) as a spherical mass and the projectile as a point mass. This is an isolated system, so mechanical energy is conserved. Visualize: The projectile of mass m was launched on the surface of the planet with an initial velocity v0 .

Solve:

(a) The energy conservation equation K1 + U1 = K 0 + U 0 is

1 2 GM p m 1 2 GM p m = mv0 − mv1 − 2 Rp + y1 2 Rp −1

⎡1 v02 ⎤ 6 ⇒ y1 = ⎢ − ⎥ − Rp = 2.8 × 10 m 2 R GM ⎥ p⎦ ⎣⎢ p (b) Using the energy conservation equation K1 + U1 = K 0 + U 0 with y1 = 1000 km = 1.000 × 106 m:

1 2 GM p m 1 2 GM p m mv1 − = mv0 − 2 Rp + y1 2 Rp 12

⎡ ⎛ 1 1 ⎞⎤ ⇒ v1 = ⎢v02 + 2GM p ⎜ − ⎟⎥ ⎜ ⎟ ⎢⎣ ⎝ Rp + y1 Rp ⎠ ⎥⎦

⎡ ⎛ ⎞⎤ 1 1 = ⎢(5000 m/s) 2 + 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(2.6 × 1024 kg) ⎜ − ⎟⎥ 6 6 6 ⎝ (5.0 × 10 m + 1.000 × 10 m) 5.0 × 10 m ⎠ ⎦ ⎣ = 3.7 km/s

12

13.36. Model: The object is a particle. The planet is a spherical mass. Solve:

Conservation of mechanical energy of the object gives

−G

Mm 1 2 Mm = mv − G R+h 2 R

The object’s mass drops out. Solving for the speed as it hits the ground,

⎛ R+h−R⎞ 1 ⎞ ⎛1 v = 2GM ⎜ − ⎟⎟ = ⎟ = 2GM ⎜⎜ ⎝ R R+h⎠ ⎝ R ( R + h) ⎠ Assess:

Compare this to v = 2 gh =

2GMh R ( R + h)

2GMh , which is the result if the potential U = mgh is used. R2

13.37.

Model: Visualize:

Solve:

Model the earth as a spherical mass and the meteoroids as point masses.

(a) The energy conservation equation K 2 + U 2 = K1 + U1 is 1/ 2

1 2

⎡ ⎛ 1 1 ⎞⎤ GM e m 1 2 GM e m = 2 mv1 − ⇒ v2 = ⎢ v12 + 2GM e ⎜ − ⎟ ⎥ mv − Re rm ⎢⎣ ⎝ Re rm ⎠ ⎦⎥ = 1.13 × 104 m/s =11.3 km/s 2 2

The speed does not depend on the meteoroid’s mass. (b) This part differs in that r2 = Re + 5000 km = 1.137 × 107 m. The shape of the meteoroid’s trajectory is not important for using energy conservation. Thus 1 2

⎡ ⎛ GM e m GM e m 1 1 ⎞⎤ mv − = 1 mv12 − ⇒ v2 = ⎢v12 + 2GM e ⎜ − ⎟⎥ + 5000 km Re + 5000 km 2 rm R r ⎢⎣ ⎥ m ⎠⎦ ⎝ e 3 = 8.94 × 10 m/s = 8.94 km/s 2 2

1/ 2

13.38.

Model: Model the two stars as spherical masses, and the comet as a point mass. This is an isolated system, so mechanical energy is conserved. Visualize:

In the initial state, the comet is far away from the two stars and thus it has neither kinetic energy nor potential energy. In the final state, as the comet passes through the midpoint connecting the two stars, it possesses both kinetic energy and potential energy. Solve: The conservation of energy equation K f + U f = K i + U i is

1 2 GMm GMm mvf − − =0 J +0 J 2 rf1 rf2 ⇒ vf = Assess:

4GM 4(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) = = 32,600 m/s rf (0.50 × 1012 m)

Note that the final velocity of 33 km/s does not depend on the mass of the comet.

13.39.

Model: Model the asteroid as a spherical mass and yourself as a point mass. This is an isolated system, so mechanical energy is conserved. Visualize: The radius of the asteroid is M a and its mass is Ra .

Solve:

The conservation of energy equation K f + U f = K i + U i for the asteroid is

1 2 GM a m 1 2 GM a m mvf − = mvi − 2 Ra + r 2 Ra The minimum speed for escape is the one that will cause you to stop only when the separation between you and the asteroid becomes very large. Noting that vf → 0 m/s as r → ∞, we have vi2 =

2GM a 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1014 kg) = ⇒ vi = 2.58 m/s Ra (2.0 × 103 m)

That is, you need a speed of 2.58 m/s to escape from the asteroid. We can now calculate your jumping speed on the earth. The conservation of energy equation is 1 0 J − mvi2 = − mg (0.50 m) ⇒ vi = 2(9.8 m/s 2 )(0.50 m) = 3.13 m/s 2

This means you can escape from the asteroid.

13.40. Model: The projectile is a particle. The earth and moon are spherical masses. Solve: The projectile is attracted to both the moon and earth. Its final velocity and potential energy are zero. Since the projectile is fired from the far side of the moon, its initial distance from the center of the earth is the earth-moon distance Re-m plus the radius of the moon Rm . Let the projectile have mass m. The conservation of mechanical energy equation is

1 2 M em M m mvescape − G −G m = 0 J +0 J R R Rm 2 + ( e-m m ) ⎛ Me M ⎞ 2 ⇒ vescape = 2G ⎜ + m⎟ R R Rm ⎠ + m ⎝ e-m From

Table

13.2,

Re-m = 3.84 × 108 m, Rm = 1.74 × 106 m, M e = 5.98 × 1024 kg,

and

M m = 7.36 × 1022 kg.

vescape = 2.78 km/s. Assess:

The escape velocity does not depend on the mass of the object which is trying to escape.

So

13.41. Model: The earth and sun are spherical masses. The earth is in a circular orbit around the sun. The projectile is a particle. Visualize:

Solve: At the earth’s distance from the sun, to escape the sun’s gravitational pull a projectile must have speed vescape . The energy conservation equation for the projectile is

M m 1 2 mvescape − G 2 = 0 J Re − s 2 ⇒ vescape = 2G

2 ( 6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg ) Ms = = 4.207 × 104 m/s Re − s (1.50 ×1011 m )

The earth’s speed in its orbit is found by considering its period T = 1 year × 365.25 × 24 × 3600 = 3.156 × 107 s.

ve =

11 2π Re − s 2π (1.50 × 10 m ) = = 2.987 × 104 m/s T 3.156 × 107 s

The projectile’s total speed is the sum of its launch speed and the earth’s speed. vescape = ve + vlaunch ⇒ vlaunch = vescape − ve = 1.22 × 104 m/s Assess: The projectile need only attain a launch speed of 12.2 km/s if launched in the direction of earth’s motion. This is about a factor of 3.5 times less than what is required from rest.

13.42.

Model: Visualize:

Gravity is a conservative force, so we can use conservation of energy.

The planets will be pulled together by gravity and each will have speed v2 as they crash and the separation between their centers will be 2R. Solve: The planets begin with only gravitational potential energy. When they crash, they have both potential and kinetic energy. Thus,

⎡ 1 1 GMM ⎤ ⎡ GMM ⎤ 2 2 ⎢ K 2 + U 2 = Mv2 + Mv2 − ⎥ = ⎢ K1 + U1 = 0 J − ⎥ 2 2 r2 ⎦ ⎣ r1 ⎦ ⎣ ⎛ 1 1⎞ ⇒ v2 = GM ⎜ − ⎟ ⎝ r2 r1 ⎠ Because the planet is “Jupiter-size,” we’ll use M = M Jupiter = 1.9 × 1027 kg and r2 = 2 RJupiter = 1.4 × 108 m. The

crash speed of each planet is v2 = 3.0 × 104 m/s. Assess: Note that the force is not constant, because it varies with distance, so the motion is not constant acceleration motion. The formulas from constant-acceleration kinematics do not work for problems such as this.

13.43.

Model: The two asteroids make an isolated system, so mechanical energy is conserved. We will also use the law of conservation of momentum for our system. Visualize:

Solve:

The conservation of momentum equation pfx = pix is

M (vfx )1 + 2 M (vfx ) 2 = 0 kg m/s ⇒ (vfx )1 = −2(vfx ) 2 The equation for mechanical energy conservation K f + U f = K i + U i is

1 1 G ( M )(2M ) G ( M )(2M ) 1 0.8GM M (vfx )12 + (2M )(vfx ) 22 − =− ⇒ [2(vfx ) 2 ]2 + (vfx ) 22 = 2 2 2R 10 R 2 R GM GM ⇒ (vfx ) 2 = −0.516 ⇒ (vfx )1 = −2(vfx ) 2 = 1.032 R R The heavier asteroid has a speed of 0.516(GM/R)1/ 2 and the lighter one a speed of 1.032(GM/R)1/ 2 .

13.44. Solve:

Model: Model the distant planet (P) as a spherical mass. The acceleration at the surface of the planet and at the altitude h are

gsurface =

GM P R2

and g altitude =

1 MP 1 GM P = g surface ⇒ G 2 2 ( R + h) 2 R2

⇒ ( R + h) 2 = 2 R 2 ⇒ R + h = 2 R ⇒ h = ( 2 − 1) R = 0.414 R That is, the starship is orbiting at an altitude of 0.414R.

13.45. Model: The stars are spherical masses. Visualize:

Solve: The starts are identical, so their final speeds vf are the same. They collide when their centers are 2R apart. From energy conservation,

⎛ ⎞ ⎛ M2 ⎞ M2 ⎛1 2⎞ 3 ⎜ −G ⎟ = 3 ⎜ Mvf ⎟ − 3⎜ G ⎟ 9 (5.0 × 10 m) ⎠ ⎝2 ⎠ ⎝ 2R ⎠ ⎝ 1 ⎛ 1 ⎞ vf2 = 2GM ⎜ − ⎟ 9 ⎝ 2 R 5.0 × 10 m ⎠ ⇒ vf = 3.71 × 105 m/s

13.46.

Model: Model the moon (m) as a spherical mass and the lander (l) as a particle. This is an isolated system, so mechanical energy is conserved. Visualize: The initial position of the lunar lander (mass = m1 ) is at a distance r1 = Rm + 50 km from the center of the moon. The final position of the lunar lander is the orbit whose distance from the center of the moon is r2 = Rm + 300 km.

Solve:

The external work done by the thrusters is

Wext = ΔEmech = 12 ΔU g where we used Emech = 12 U g for a circular orbit. The change in potential energy is from the initial orbit at r1 = Rm + 50 km to the final orbit rf = Rm + 300 km. Thus

1 ⎛ −GM m m −GM m m ⎞ GM m m ⎛ 1 1 ⎞ Wext = ⎜ − ⎟= ⎜ − ⎟ 2⎝ rf ri 2 ⎝ ri rf ⎠ ⎠ (6.67 × 10−11 N m 2 / kg 2 )(7.36 × 1022 kg)(4000 kg) ⎛ 1 1 ⎞ = − ⎜ ⎟ 6 6 2 1.79 10 m 2.04 10 m × × ⎝ ⎠ = 6.72 × 108 J

13.47.

Model: Model the earth (e) as a spherical mass and the space shuttle (s) as a point particle. This is an isolated system, so the mechanical energy is conserved. Visualize: The space shuttle (mass = ms ) is at a distance of r1 = Re + 250 km.

Solve:

The external work done by the thrusters is

Wext = ΔEmech = 12 ΔU g where we used Emech = 12 U g for a circular orbit. The change in potential energy is from the initial orbit at

r1 = Re + 250 km to the final orbit rf = Re + 610 km. Thus 1 ⎛ −GM e m −GM e m ⎞ GM e m ⎛ 1 1 ⎞ Wext = ⎜ − ⎟= ⎜ − ⎟ 2 ⎝ rf ri 2 ⎝ ri rf ⎠ ⎠ (6.67 × 10−11 N m 2 / kg 2 )(5.98 × 1024 kg)(75,000 kg) ⎛ 1 1 ⎞ = − ⎜ ⎟ 6 6 2 ⎝ 6.62 × 10 m 6.98 × 10 m ⎠ = 1.17 × 1011 J This much energy must be supplied by burning the on-board fuel.

13.48.

Solve:

(a) Using Equation 13.31 for a satellite in a circular orbit,

⎛ ⎞ 1 1 1 1 ΔEmech = ΔU g = (−GM e m) ⎜ − ⎟ + + 2 2 R 300 km R 500 km e ⎝ e ⎠ =

−1 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(5.00 × 104 kg) 2

1 1 ⎛ ⎞ 10 ×⎜ − ⎟ = −4.35 × 10 J 6 5 6 5 ⎝ 6.37 × 10 m + 3 × 10 m 6.37 × 10 m + 5 × 10 m ⎠ The negative sign indicates that 4.35 × 1010 J of energy is lost. (b)

The shuttle would fire retro-rockets to lose enough energy to go into an elliptical orbit. When halfway around, at rmin , the shuttle would fire retro-rockets again to lose the rest of the energy to stay at rmin .

13.49. Model: Planet Physics is a spherical mass. The cruise ship is in a circular orbit. Solve:

(a) At the surface, the free-fall acceleration is g = G

M . From kinematics, R2

y f = yi + v0 Δt − 2 g ( Δt ) ⇒ 0 m = 0 m + (11 m/s )( 2.5 s ) − 2 g ( 2.5 s ) 2

2

⇒ g = 2.20 m/s 2 The period of the cruise ship’s orbit is 230 × 60 = 13,800 s. For the circular orbit of the cruise ship, ⎛ 4π 2 ⎞ ⎛ R2 ⎞ ⎛1⎞ T2 3 T2 =⎜ R = ⎟ ( 2R ) ⇒ ⎜ ⎟ = R⎜ ⎟ 2 32π ⎝g⎠ ⎝ GM ⎠ ⎝ GM ⎠

( 2.20 m/s ) (13,800 s ) 2

⇒R=

R2 g = 5.8 × 1022 kg. G (b) From part (a), R = 1.33 × 106 m. The mass is thus M =

32π 2

2

= 1.327 × 106 m.

13.50.

Model: Assume a spherical asteroid and a point mass model for the satellite. This is an isolated system, so mechanical energy is conserved. Visualize:

The orbital radius of the satellite is

r = Ra + h = 8,800 m + 5,000 m = 13,800 m Solve:

(a) The speed of a satellite in a circular orbit is 12

GM ⎡ (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1016 kg) ⎤ =⎢ v= ⎥ = 7.0 m/s r (13,800 m) ⎣ ⎦ (b) The minimum launch speed for escape (vi ) will cause the satellite to stop asymptotically (vf = 0 m/s) as rf → ∞. Using the energy conservation equation K 2 + U 2 = K1 + U1 , we get GM a ms 1 GM a ms GM a 1 1 2 ms vf2 − = msvi2 − ⇒ 0 J − 0 J = vescape − rf Ra Ra 2 2 2 ⇒ vescape =

2GM a 2(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.0 × 1016 kg) = = 12.3 m/s Ra 8800 m

13.51.

Model: Model the moon as a spherical mass and the satellite as a point mass. Visualize: The rotational period of the satellite is the same as the rotational period of the moon around its own axis. This time happens to be 27.3 days. Solve: The gravitational force between the moon and the satellite provides the centripetal acceleration necessary for circular motion around the moon. Therefore, 2

GM m m ⎛ 2π ⎞ = mrω 2 = mr ⎜ ⎟ r2 ⎝ T ⎠ GM mT 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.36 × 1022 kg)(27.3 × 24 × 3600 s) 2 ⇒ r3 = = 4π 2 4π 2 ⇒ r = 8.84 × 107 m Since r = Rm + h, then h = r − Rm = 8.84 × 107 m − 1.74 × 106 m = 8.67 × 107 m.

13.52.

Model: Model the earth as a spherical mass and the satellite as a point mass. Visualize: The satellite is directly over a point on the equator once every two days. Thus, T = 2Te = 2 × 24 ×

3600 s = 1.728 × 105 s. Solve: A satellite’s period is

⎛ 4π 2 ⎞ 3 T2 = ⎜ ⎟r ⎝ GM e ⎠ GM eT 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)(1.728 × 105 ) 2 ⇒ r3 = = 4π 2 4π 2 7 ⇒ r = 6.71 × 10 m Assess:

The radius of the orbit is larger than the geosynchronous orbit.

13.53.

Solve:

(a) Taking the logarithm of both sides of v p = Cu q gives

[log(v p ) = p log v] = [log(Cu q ) = log C + q log u ] ⇒ log v =

q log C log u + p p

But x = log u and y = log v, so x and y are related by ⎛q⎞ log C y = ⎜ ⎟x + p p ⎝ ⎠ (b) The previous result shows there is a linear relationship between x and y, hence there is a linear relationship between log u and log v. The graph of a linear relationship is a straight line, so the graph of log v-versus-log u will be a straight line. (c) The slope of the straight line represented by the equation y = (q / p ) x + log C / p is q/p. Thus, the slope of the log v-versus-log u graph will be q/p. (d) From Newton’s theory, the period T and radius r of an orbit around the sun are related by

⎛ 4π 2 ⎞ 3 T2 =⎜ ⎟r ⎝ GM ⎠ This equation is of the form T p = Cr q , with p = 2, q = 3, and C = 4π 2 /GM . If the theory is correct, we expect a graph of log T-versus-log r to be a straight line with slope q/p = 3/2 = 1.500. The experimental measurements of actual planets yield a straight line graph whose slope is 1.500 to four significant figures. Note that the graph has nothing to do with theory—it is simply a graph of measured values. But the fact that the shape and slope of the graph agree precisely with the prediction of Newton’s theory is strong evidence for its correctness. (e) The predicted y-intercept of the graph is log C/p, and the experimentally determined value is 9.264. Equating these, we can solve for M. Because the planets all orbit the sun, the mass we are finding is M = M sun .

⎛ 4π 2 ⎞ 1 1 4π 2 1 log C = log ⎜ = 10−18.528 = 18.528 ⎟ = −9.264 ⇒ 2 2 GM sun 10 ⎝ GM sun ⎠ ⇒ M sun =

4π 2 ⋅ 1018.528 = 1.996 × 1030 kg G

The tabulated value, to three significant figures, is M sum = 1.99 × 1030 kg. We have used the orbits of the planets to “weigh the sun!”

13.54. Solve:

Visualize: Please refer to Figure P13.54. The gravitational force on one of the masses is due to the star and the other planet. Thus 2

G

Mm Gmm mv 2 m ⎛ 2π r ⎞ GM Gm 4π 2 r 2 + = = ⇒ + = ⎜ ⎟ r2 (2r ) 2 r r⎝ T ⎠ r 4r T2 12

⎡ 4π 2 r 3 ⎤ G⎛ m ⎞ 4π 2 r 2 1 T ⇒ = ⎢ ⎥ ⎜M + ⎟ = 2 r⎝ 4⎠ T ⎣ G ( M + m /4) ⎦

13.55.

Solve:

(a) Dividing the circumference of the orbit by the period,

2π Rs 2π (1.0 × 104 m) = = 6.3 × 104 m/s T 1.0 s (b) Using the formula for the acceleration at the surface, v=

gsurface =

GM s (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) = = 1.33 × 1012 m/s 2 Rs2 (1 × 104 m) 2

(c) The mass of an object on the earth will be the same as its mass on the star. The gravitational force is

( FG )star = mg surface = 1.33 × 1012 N (d) The radius of the orbit of the satellite is r = 1 × 104 m + 1.0 × 103 m = 1.1 × 104 m. The period is

T2 =

4π 2 r 3 4π 2 (1.1 × 104 m)3 = ⇒ T = 6.29 × 10−4 s −11 GM s (6.67 × 10 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)

This means there are 1589 revolutions per second or 9.5 × 104 orbits per minute. (e) Applying Equation 13.25 for a geosynchronous orbit, r3 =

GM s 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)(1.0 s) T = ⇒ r = 1.50 × 106 m 4π 2 4π 2

13.56. Solve:

Model: Assume the solar system is a point particle. (a) The radius of the orbit of the solar system in the galaxy is 25,000 light years. This means

r = 25,000 light years = 2500(3 × 108 )(365)(24)(3600) m = 2.36 × 1020 m T=

2π r (2π )(2.36 × 1020 m) = = 6.64 × 1015 s = 2.05 × 108 years v (2.30 × 105 m/s)

5.0 × 109 years = 24 orbits. 2.05 × 108 years (c) Applying Newton’s second law yields GM g center mss mss v 2 v 2 r (2.30 × 105 m/s) 2 (2.36 × 1020 m) = ⇒ M = = = 1.87 × 1041 kg g center r2 r G 6.67 × 10−11 N ⋅ m 2 /kg 2 (b) The number of orbits =

(d) The number of stars in the center of the galaxy is

1.87 × 1041 kg = 9.4 × 1010 1.99 × 1030 kg

13.57.

Model: Assume the two stars are spherical masses. Visualize: The gravitational force between the two stars provides the centripetal acceleration required for circular motion about the center of mass.

Solve:

Newton’s second law is 2

Fgravitation

1

⎛ GT 2 M ⎞ 3 GMM ⎛ 2π ⎞ = = MRω 2 = MR ⎜ ⎟ ⇒ R=⎜ 2 2 ⎟ (2 R ) ⎝ T ⎠ ⎝ 16π ⎠

Using T = 90 days = 90 × 24 × 3600 s and M = 2M sun = 3.98 × 1030 kg, we get R = 4.667 × 1010 m. Thus the star separation is 2 R = 9.33 × 1010 m.

13.58.

Model: Visualize:

Assume the three stars are spherical masses.

The stars rotate about the center of mass, which is the center of the triangle and equal distance r from all three stars. The gravitational force between any two stars is the same. On a given star the two forces from the other stars make an angle of 60°. Solve: The value of r can be found as follows:

L /2 L 1.0 × 1012 m = cos30° ⇒ r = = = 0.577 × 1012 m r 2cos30° 2cos30° The gravitational force between any two stars is Fg =

GM 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) 2 = = 2.64 × 1026 N L2 (1.0 × 1012 m) 2

The component of this force toward the center is Fc = Fg cos30° = (2.64 × 10 26 N)cos30° = 2.29 × 1026 N

The net force on a star toward the center is twice this force, and that force equals MRω 2 . This means ⎛ 2π ⎞ 2 × 2.29 × 1026 N = MRω 2 = MR ⎜ ⎟ ⎝ T ⎠ ⇒T =

2

4π 2 MR 4π 2 (1.99 × 1030 kg)(0.577 × 1012 m) = = 3.15 × 108 s = 10 years 26 4.58 × 10 N 4.58 × 1026 N

13.59.

Model: Visualize:

Angular momentum is conserved for a particle following a trajectory of any shape.

For a particle in an elliptical orbit, the particle’s angular momentum is L = mrvt = mrv sin β , where v is the G G velocity tangent to the trajectory and β is the angle between r and v . Solve: At the distance of closest approach (rmin ) and also at the most distant point, β = 90°. Since there is no tangential force (the only force being the radial force), angular momentum must be conserved:

mPluto v1rmin = mPlutov2 rmax ⇒ v2 = v1 (rmin / rmax ) = (6.12 × 103 m/s)(4.43 × 1012 m/7.30 × 1012 m) = 3.71 km/s

13.60.

Model: Visualize:

Angular momentum is conserved for a particle following a trajectory of any shape.

For a particle in an elliptical orbit, the particle’s angular momentum is L = mrvt = mrv sin β , where v is the velocity G G tangent to the trajectory, and β is the angle between r and v . Solve: At the distance of closest approach (rmin ) and also at the most distant point, β = 90°. Since there is no tangential force (the only force being the radial force), angular momentum must be conserved:

mMercuryv1rmin = mMercury v2 rmax

⇒ rmin = rmax (v2 / v1 ) =

(6.99 × 1010 m)(38.8 km/s) = 4.60 × 1010 m 59.0 km/s

13.61. Model: Visualize:

Solve:

For the sun + comet system, the mechanical energy is conserved.

The conservation of energy equation K f + U f = K i + U i is

1 GM s M c 1 GM s M c M cv22 − = M cv12 − 2 r2 2 r1 Using G = 6.67 × 10−11 Nm 2 /kg 2 , M s = 1.99 × 1030 kg, r1 = 8.79 × 1010 m, r2 = 4.50 × 1012 m, and v1 = 54.6 km/s, we get v2 = 4.49 km/s.

13.62.

Model: Visualize:

Solve:

Model the planet (p) as a spherical mass and the spaceship (s) as a point mass.

(a) For the circular motion of the spaceship around the planet, GM p ms 2 0

r

=

GM p mv02 ⇒ v0 = r0 r0

Immediately after the rockets were fired v1 = v0 /2 and r1 = r0 . Therefore,

v1 =

1 GM p 2 r0

(b) The spaceship’s maximum distance is rmax = r0 . Its minimum distance occurs at the other end of the ellipse. The energy at the firing point is equal to the energy at the other end of the elliptical trajectory. That is,

GM p ms 1 GM p ms 1 ms v12 − = ms v22 − 2 r1 2 r2 Since the angular momentum at these two ends is conserved, we have mv1r1 = mv2 r2 ⇒ v2 = v1 (r1 / r2 ) With this expression for v2 , the energy equation simplifies to GM p 1 2 GM p 1 2 v1 − = v1 (r1 / r2 ) 2 − 2 r1 2 r2 Using r1 = r0 and v1 = v0 /2 =

1 GM p , 2 r0

1 ⎛ 1 GM p ⎞ GM p 1 ⎛ 1 GM p ⎞ r02 GM p 1 1 r 1 = ⎜ ⇒ − = 0 − ⎜ ⎟− ⎟ − 2 ⎝ 4 r0 ⎠ r0 2 ⎝ 4 r0 ⎠ r22 r2 8r0 r0 8r22 r2 ⎛ 7 ⎞ 2 7 r 1 r0 ⇒ + 02 − = 0 ⇒ ⎜ ⎟ r2 − r2 + = 0 8r0 8r2 r2 8 ⎝ 8r0 ⎠ The solutions are r2 = r0 (the initial distance) and r2 = r0 / 7. Thus the minimum distance is rmin = r0 / 7.

1 2

13.63. Solve: (a) The satellite is in a circular orbit if K = − Ug. Calculate K and U g : 2 1 1 K = mv 2 = m ( 5.5 × 103 m/s ) = 1.51 × 107 m. 2 2 ( 5.98 × 1024 kg ) m Mm U g = −G e = − ( 6.67 × 10−11 N ⋅ m 2 /kg 2 ) 1.10 × 107 m r 7 = −3.63 × 10 m

1 The quantity − U g = 1.81× 107 m ≠ K , so no, the satellite is not in a circular orbit. 2 (b) The orbit is bound if the total mechanical energy is negative. Emech = K + U g = 1.51 × 107 m − 3.63 × 107 m < 0.

Yes, the orbit is bound. Assess: Even without knowing the mass of the satellite we could answer questions about its orbit.

13.64. Visualize:

Solve:

We choose two equal time intervals tb − ta and tc − tb . A constant velocity and equal time intervals

means that xb − xa = xc − xb . The area swept from t = 0 s to t = ta is Rxa /2 and the area swept from t = 0 s to t = tb is Rxb /2. Thus, the area swept between t = ta and t = tb is R( xb − xa ) /2. In the same way, the area

swept between t = tb and t = t is R ( xc − xb )/2. Since xc − xb = xb − xa , the area swept during time (tb − ta ) is the same as the area swept during an equal time (tc − tb ). Kepler’s second law is obeyed.

13.65. Solve: (a) At what distance from the center of Saturn is the acceleration due to gravity the same as on the surface of the earth? (b)

(c) The distance is 6.21 × 107 m. This is 1.06RSaturn

13.66.

Solve:

(a) A 1000 kg satellite orbits the earth with a speed of 1997 m/s. What is the radius of the

orbit? (b)

(c) The radius of the orbit is

r=

GM E 6.67 × 10−11 N ⋅ m 2 /kg 2 (5.98 × 1024 kg) = = 1.00 × 108 m (vpayload ) 2 (1997 m/s) 2

13.67. Solve: (a) A 100 kg object is released from rest at an altitude above the moon equal to the moon’s radius. At what speed does it impact the moon’s surface? (b)

(c) The speed is v2 = 1680 m/s.

13.68.

Solve:

(a) A 2.0 × 1030 kg star and a 4.0 × 1030 kg star, each 1.0 × 109 m in diameter, are at rest

1.0 × 1012 m apart. What are their speeds as they crash together? (b)

(c) The first equation is the conservation of momentum. It can be used in the conservation of energy equation to give vf1 = 596 km/s and vf2 = −298 km/s. That is, the speeds of the two stars are 6.0 × 102 km/s and

3.0 × 102 km/s.

13.69. Solve: (a) Kepler’s third law for circular orbits is ⎛ 4π 2 ⎞ 3 4π 2 32 T2 =⎜ r ⎟r ⇒ T = GM ⎝ GM ⎠

Letting a =

4π 2 3 , the first satellite obeys T = ar 2 . For the second satellite, which orbits the same mass, GM 3

T + ΔT = a ( r + Δr ) Since

3 2

⎛ Δr ⎞ 2 = ar ⎜1 + ⎟ r ⎠ ⎝ 3 2

Δr n << 1, we can use the approximation (1 ± x ) ≈ 1 ± nx. r

Thus 3 ⎛ 3 Δr ⎞ T + ΔT ≈ ar 2 ⎜ 1 + ⎟ ⎝ 2 r ⎠

3

Subtracting the equation T = ar 2 for the first satellite from this, 3 3 ⎛ Δr ⎞ ΔT = ar 2 ⎜ ⎟ 2 ⎝ r ⎠

Dividing this by the equation for the first satellite, ΔT 3 Δr = T 2 r (b) The satellites orbit the earth. The fractional difference in their periods is

ΔT 3 1 km = = 2.24 × 10−4 T 2 6700 km After

1 = 4467 periods they will meet again. For the inner satellite, 2.24 × 10−4

T=

3 4π 2 6.700 × 106 m ) 2 ( 24 G ( 5.98 × 10 kg )

= 5456 s = 1.52 hrs So the satellites will meet again in 4467 × 1.52 hrs = 6770 hrs = 282 days. Assess: A communications satellite has an orbital period of around 1.5 h. The surprising length of time between the two satellites meeting is due to the small differences in their periods.

13.70.

Model: Model the earth and sun as spherical masses, the satellite as a point mass. Assume the satellite’s distance from the earth is very small compared to the earth’s (and satellite’s) distance to the sun. Visualize:

Solve: The net force on the satellite is the sum of the gravitational force toward the sun and the gravitational force toward the earth. This net force is responsible for circular motion around the sun. We want to chose the distance d to make the period T match the period Te with which the earth orbits the sun. The earth’s orbital period is given by T e2 = (4π 2 / GM s ) Re3 . Thus 2

Fnet =

GM s m GM e m mv 2 m ⎛ 2π r ⎞ 4π 2 GM ma mr − = = = = = mr 3 s ⎜ ⎟ centripetal 2 2 2 r d r r ⎝ Te ⎠ Te Re

Using r = Re − d and canceling the Gm term gives

Ms M M − 2e = 3s ( Re − d ) 2 ( Re − d ) d Re This equation can’t be solved exactly, but we can make use of the fact that d  Re to use the binomial approximation. Factor the Re out of the expressions ( Re − d ) to get Ms M M − e = 2s (1 − d / Re ) Re2 (1 − d / Re ) 2 d 2 Re If we think of d/Re = x  1, we can simplify the first term by using (1 ± x )n ≈ 1 ± nx. Here n = −2, so we get

Ms M M M M (1 − (−2)d / Re ) − 2e = 2s (1 − d / Re ) ⇒ 2e = 3 2s Re2 d Re d Re Thus d = ( M e /3M s )1/3 Re = 1.50 × 109 m. Assess:

d/Re = 0.010, so our assumption that d/Re  1 is justified.

13.71.

Model: Model the earth as a spherical mass and the shuttle and payload as point masses. We’ll assume mpayload  mshuttle .

Visualize:

Solve: (a) The payload, in steady state, is undergoing uniform circular motion. This means that the net force is directed toward the center of the earth. There is no tangential force component, since such a force would cause the payload to speed up or slow down and the motion wouldn’t be uniform. Since the net force is due to gravity and tension, and the gravitational force is radial, the tension force cannot have any tangential component. Thus the rope is radially outward at angle 0°. (b) To move with the shuttle, the payload’s period Tp in an orbit of radius rp must exactly match the period Ts

of the shuttle in an orbit of radius rs . The shuttle’s period around the earth is given by T s2 = (4π 2 /GM e ) rs3 , where we’ve used the assumption mp  ms to infer that the rope’s tension will be too small to have any influence on the shuttle’s motion. Newton’s second law for the payload’s motion is ( Fnet ) r =

GM e mp rp 2

− T = mp ar =

mp v 2 rp

2

m ⎛ 2π rp ⎞ 4π 2 = p⎜ = mp rp 2 ⎟ rp ⎜⎝ Tp ⎟⎠ Tp

Using Tp = Ts and the above expression for Ts , this becomes

GM e mp r 2p ⇒T =

− T = mp rp

GM e mp ⎡ ⎛ rp ⎞ ⎢1 − ⎜ ⎟ r 2p ⎢ ⎝ rs ⎠ ⎣

3

3 GM e GM e mp r p = rs3 r p2 rs3

⎤ GM e mp ⎡ ⎛ R + 290 km ⎞3 ⎤ ⎥= ⎢1 − ⎜ e ⎟ ⎥ = 4.04 N r p2 ⎢ ⎝ Re + 300 km ⎠ ⎥ ⎥⎦ ⎣ ⎦

Assess: The fact that the tension is so small justifies our assumption that it will have no effect on the motion of the shuttle.

13.72.

Model: Model the 400 kg satellite and the 100 kg satellite as point masses and model the earth as a spherical mass. Momentum is conserved during the inelastic collision of the two satellites. Visualize:

Solve:

For the given orbit, r0 = Re + 1× 106 m = 7.37 × 106 m. The speed of a satellite in this orbit is

GM e = 7357 m/s r0

v0 =

The two satellites collide, stick together, and move with velocity v1. The equation for momentum conservation for the perfectly inelastic collision is (400 kg + 100 kg)v1 = (400 kg)(7357 m/s) − (100 kg)(7357 m/s) ⇒ v1 = 4414 m/s The new satellite’s radius immediately after the collision is still r1 = r0 = 7.37 × 106 m. Now it is moving in an elliptical orbit. We need to determine if the minimum distance r2 is larger or smaller than the earth’s radius Re = 6.37 × 106 m. The combined satellites will continue moving in an elliptical orbit. The momentum of the combined satellite is L = mrv sin β (see Equation 13.26) and is conserved in a trajectory of any shape. The angle β is 90° when v1 = 4414 m/s and when the satellite is at its closest approach to the earth. From the conservation of angular momentum, we have r1v1 = r2v2 = (7.37 × 106 )(4414 m/s) = 3.253 × 1010 m 2 /s

⇒ r2 =

3.253 × 1010 m 2 /s v2

Using the conservation of energy equation at positions 1 and 2,

GM e (500 kg) 1 GM e (500 kg) 1 = (500 kg)v22 − (500 kg)(4414 m/s) 2 − r2 2 7.37 × 106 m 2 Using the above expression for r2 , we can simplify the energy equation to v22 − (2.452 × 104 m/s)v2 + (8.876 × 107 m 2 /s 2 ) = 0 m 2 /s 2 ⇒ v2 = 20,107 m/s and 4414 m/s A velocity of 20,107 m/s for v2 yields r2 =

3.253 × 1010 m 2 /s = 1.62 × 106 m 20,107 m/s

Since r2 < Re = 6.37 × 106 m, the combined mass of the two satellites will crash into the earth.

13.73.

Model: The stars are spherical masses. They each rotate about the system’s center of mass.

Visualize:

Solve: (a) The stars rotate about the system’s center of mass with the same period: T1 = T2 = T . We can locate the center of mass by letting the origin be at the smaller-mass star. Then r1 = rcm =

(2.0 × 1030 kg)(0 m) + (6.0 × 1030 kg)(2.0 × 1012 m) = 1.5 × 1012 m 2.0 × 1030 kg + 6.0 × 1030 kg

Mass m2 undergoes uniform circular motion with radius r2 = 0.5 × 1012 m due to the gravitational force of mass m1 at distance R = 2.0 × 1012 m. The gravitational force is responsible for the centripetal acceleration, so 2

Fgrav = 1/ 2

⎡ 4π 2 ⎤ ⇒ T =⎢ r2 R 2 ⎥ ⎣ Gm1 ⎦

Gm1m2 m2v22 m2 ⎛ 2π r2 ⎞ 4π 2 m2 r2 = = = m a 2 centripetal ⎜ ⎟ = R2 r2 r2 ⎝ T ⎠ T2 1/ 2

⎡ 4π 2 (0.5 × 1012 m)(0.5 × 1012 m) 2 ⎤ =⎢ ⎥ −11 2 2 30 ⎣ (6.67 × 10 N m /kg ) (2.0 × 10 kg) ⎦

(b) The speed of each star is v = (2π r )/T . Thus

2π r1 2π (1.5 × 1012 m) = = 12.3 km/s T 7.693 × 108 s 2π r2 2π (0.5 × 1012 m) v2 = = = 4.1 km/s T 7.693 × 108 s

v1 =

= 7.693 × 108 s = 24 years

13.74. Model: The moon is a spherical mass. The moon lander is originally in a circular orbit. Visualize:

Solve: Energy and momentum are conserved between points 1 and 2 in the elliptical orbit. Also, at both points β = 90°, so L = mrv sin β = mrv. Let h = 1000 km. The original speed of the lander is

GM m v r R +h . The = 1338 m/s. Conservation of angular momentum requires mr1v1 = mr2v2 ⇒ 2 = 1 = m Rm + h v1 r2 Rm energy conservation equation is v0 =

1 M m 1 M m mv12 − G m = mv22 − G m 2 Rm + h 2 Rm ⎛ R +h⎞ Substituting v2 = ⎜ m ⎟ v1 , ⎝ Rm ⎠ 2

1 2 Mm 1 ⎛ R +h⎞ Mm v1 − G = v12 ⎜ m ⎟ −G 2 Rm + h 2 ⎝ Rm ⎠ Rm ⎛ ⎛ R +h⎞ v12 ⎜1 − ⎜ m ⎟ ⎜ Rm ⎠ ⎝ ⎝

2

⎞ ⎛ 1 1 ⎞ ⎟ = 2GM m ⎜ − ⎟ ⎟ ⎝ Rm + h Rm ⎠ ⎠

2GM m h 1 2GM m ⎛ Rm ⎞ 6 2 2 = ⎜ ⎟ = 1.392 × 10 m /s Rm ( Rm + h ) ⎛ R + h ⎞ 2 Rm + h ) ⎝ 2 Rm + h ⎠ ( m ⎜ ⎟ −1 ⎝ Rm ⎠ v1 = 1180 m/s v12 =

The fractional change in speed required to just graze the moon at point 2 is 1338 m/s − 1180 m/s = 11.8% 1338 m/s Assess:

A reduction in speed by almost 12% is reasonable.

13.75.

Model: Model the earth as a spherical mass and the satellite as a point mass. This is an isolated system, so mechanical energy is conserved. Also, the angular momentum of the satellite is conserved. Visualize: Please refer to Figure CP13.75. Solve: (a) Angular momentum is L = mrv sin β . The angle β = 90° at points 1 and 2, so conservation of angular momentum requires

⎛r ⎞ mr1v1′ = mr2v′2 ⇒ v1′ = ⎜ 2 ⎟ v′2 ⎝ r1 ⎠ The energy conservation equation is

⎛ 1 1⎞ 1 GMm 1 GMm m(v2′ ) 2 − = m(v1′ ) 2 − ⇒ (v′2 ) 2 − (v1′ ) 2 = 2GM ⎜ − ⎟ 2 r2 2 r1 ⎝ r2 r1 ⎠ Using the angular momentum result for v1′ gives 2

⎛r ⎞ ⎛r −r ⎞ ⎡ r2 − r2 ⎤ ⎛r −r ⎞ (v2′ ) 2 − (v2′ ) 2 ⎜ 2 ⎟ = 2GM ⎜ 1 2 ⎟ ⇒ (v′2 ) 2 ⎢ 1 2 2 ⎥ = 2GM ⎜ 1 2 ⎟ r r r r ⎝ 1⎠ ⎝ 12 ⎠ ⎣ 1 ⎦ ⎝ r1r2 ⎠ v′2 =

2GM ( r1 / r2 ) r1 + r2

⎛r ⎞ 2GM (r2 / r1 ) and v1′ = ⎜ 2 ⎟ v′2 = r r1 + r2 ⎝ 1⎠

(b) For the circular orbit,

v1 =

GM (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) = = 7730 m/s r1 (6.37 × 106 m + 3 × 105 m)

For the elliptical orbit, r1 = Re + 300 km = 6.37 × 106 m + 3 × 105 m = 6.67 × 106 m r2 = Re + 35,900 km = 6.37 × 106 m + 3.59 × 107 m = 4.23 × 107 m v1′ =

2GM ( r2 / r1 ) ⇒ v1′ = 10,160 m/s ( r1 + r2 )

1 1 (c) From the work-kinetic energy theorem, W = ΔK = mv1′2 − mv12 = 2.17 × 1010 J 2 2 (d) v2′ =

2GM (r1 / r2 ) r1 + r2

Using the same values of r1 and r2 as in (b), v′2 = 1600 m/s. For the circular orbit, v2 =

GM = 3070 m/s r2

1 1 W = mv22 − mv2′2 = 3.43 × 109 J 2 2 (f) The total work done is 2.513 × 1010 J. This is the same as in Example 13.6, but here we’ve learned how the work has to be divided between the two burns. (e)

13.76.

Model: The rod is thin and uniform.

Visualize:

Solve: (a) The rod is not spherical so must be divided into thin sections each dr wide and having mass dm. Since the rod is uniform,

dm dr M = ⇒ dm = dr M L L The width of the rod is small enough so that all of dm is distance r away from m. The gravitational potential energy of dm and m is

m dm dU = −G = −G r

⎛M ⎞ m ⎜ ⎟ dr ⎝ L⎠ r

The total potential energy of the rod and mass m is found by adding the contributions dU from every point along the rod in an integral:

GMm U = ∫ dU = − L

L⎞ ⎛ x+ ⎟ dr GMm ⎜ ∫L r = − L ln ⎜⎜ L2 ⎟⎟ x− x− 2 2⎠ ⎝

x+

L 2

Note r is increasing with the limits chosen as they are. (b) The force on m when at x is ⎛ ⎞ 1 ⎟ dU GMm d ⎛ ⎛ L⎞ L ⎞ ⎞ GMm ⎜ 1 ⎛ = − F =− ⎜ ⎟ ⎜ ln ⎜ x + ⎟ − ln ⎜ x − ⎟ ⎟ = dx L dx ⎝ ⎝ L ⎜ x+ L x− L⎟ 2⎠ 2 ⎠⎠ ⎝ ⎝ 2 2⎠ L 4 ⎛ ⎞ = −GMm ⎜ 2 , x≥ 2 ⎟ 2 ⎝ 4x − L ⎠ Assess: The direction of the force is towards the –x direction, as expected. The force magnitude approaches ∞ 1 as the mass m approaches the end of the rod, but goes to zero like 2 as x gets large. This is expected since from x far away the rod looks like a point mass.

13.77.

Model: The ring is uniform and is so thin that every point on it is the same distance r from m.

Visualize:

Solve: (a) We must determine the gravitational potential (dU) between m and an arbitrary part of the ring dm, then add using an integral all the contributions to U. Since the ring is uniform,

dm dl M dl = ⇒ dm = M 2π r 2π r The distance from m to dm is r = x 2 + R 2 . The gravitational potential between m and dm is

dU = −G

m dm mM = −G r 2π r

dl x + R2 2

The total gravitational potential is

U = ∫ dU = −

GmM

∫ dl

2π r x 2 + R 2 ring

Note that x and R do not change for any location of dm. The integral is just the length of the ring.

∫ dl = 2π R

ring

Thus U =−

GmM x2 + R2

(b) The force on m when at x is

(

)

−1 −3 dU d ⎛ 1⎞ x 2 + R 2 ) 2 = GmM ⎜ − ⎟ ( x 2 + R 2 ) 2 ( 2 x ) = GmM ( dx dx ⎝ 2⎠ x = −GmM 3 2 ( x + R2 ) 2

Fx = −

Thus the magnitude of the force is

x . ( x 2 + R 2 )3 2 Assess: The force is zero at the center of the ring. Elsewhere its direction is toward the origin. As x gets large, the force 1 decreases like 2 . This is expected since from far away the ring looks like a point mass. x F = GmM

13-1

14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, hence

T=

1 1 = = 2.27 × 10−3 s = 2.27 ms f 440 Hz

14.2. Model: The air-track glider oscillating on a spring is in simple harmonic motion. Solve: The glider completes 10 oscillations in 33 s, and it oscillates between the 10 cm mark and the 60 cm mark. 33 s (a) T= = 3.3 s oscillation = 3.3 s 10 oscillations 1 1 (b) f = = = 0.303 Hz ≈ 0.30 Hz T 3.3 s ω = 2π f = 2π ( 0.303 Hz ) = 1.90 rad s (c) (d) The oscillation from one side to the other is equal to 60 cm − 10 cm = 50 cm = 0.50 m. Thus, the amplitude is A = 12 ( 0.50 m ) = 0.25 m. (e) The maximum speed is

⎛ 2π ⎞ vmax = ω A = ⎜ ⎟ A = (1.90 rad s )( 0.25 m ) = 0.48 m s ⎝ T ⎠

14.3. Model: The air-track glider attached to a spring is in simple harmonic motion. Visualize: The position of the glider can be represented as x (t ) = A cos ω t. Solve: The glider is pulled to the right and released from rest at t = 0 s. It then oscillates with a period T = 2.0 s and a maximum speed vmax = 40 cm s = 0.40 m s.

2π 2π v 0.40 m s = = π rad s ⇒ A = max = = 0.127 m = 12.7 cm T 2.0 s ω π rad s (b) The glider’s position at t = 0.25 s is (a) vmax = ω A and ω =

x0.25 s = ( 0.127 m ) cos ⎡⎣(π rad s )( 0.25 s ) ⎤⎦ = 0.090 m = 9.0 cm

14.4. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure EX14.4. Solve: (a) The amplitude A = 10 cm. (b) The time to complete one cycle is the period, hence T = 2.0 s and

f = (c)

The

position

of

an

object

1 1 = = 0.50 Hz T 2.0 s

undergoing

simple

harmonic

motion

is

At t = 0 s, x0 = −5 cm, thus

−5 cm = (10 cm ) cos ⎣⎡ω ( 0 s ) + φ0 ⎦⎤ ⇒ cos φ0 =

−5 cm 1 2π ⎛ 1⎞ = − ⇒ φ0 = cos −1 ⎜ − ⎟ = ± rad or ± 120° 10 cm 2 3 ⎝ 2⎠

Since the oscillation is originally moving to the left, φ0 = +120°.

x ( t ) = A cos (ω t + φ0 ) .

14.5. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure EX14.5. Solve: (a) The amplitude A = 20 cm. (b) The period T = 4.0 s, thus f =

1 1 = = 0.25 Hz T 4.0 s

(c) The position of an object undergoing simple harmonic motion is

x ( t ) = A cos (ω t + φ0 ) . At

t = 0 s, x0 = 10 cm. Thus, 10 cm =

10 cm ⎞ π −1 ⎛ 1 ⎞ ⎟ = cos ⎜ ⎟ = ± rad = ±60° 3 ⎝ 20 cm ⎠ ⎝ 2⎠

( 20 cm ) cos φ0 ⇒ φ0 = cos −1 ⎛⎜

Because the object is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram and thus must have a phase constant between π and 2π radians. Therefore, φ0 = −

π 3

rad = −60°.

14.6. Visualize: The phase constant 32 π has a plus sign, which implies that the object undergoing simple harmonic motion is in the second quadrant of the circular motion diagram. That is, the object is moving to the left. Solve: The position of the object is

x ( t ) = A cos (ω t + φ0 ) = A cos ( 2π ft + φ0 ) = ( 4.0 cm ) cos ⎡⎣( 4π rad s ) t + 32 π rad ⎤⎦ The amplitude is A = 4 cm and the period is T = 1 f = 0.50 s. A phase constant φ0 = 2π 3 rad =120° (second

quadrant) means that x starts at − 12 A and is moving to the left (getting more negative).

Assess:

We can see from the graph that the object starts out moving to the left.

14.7. Visualize: A phase constant of −

π

implies that the object that undergoes simple harmonic motion is 2 in the lower half of the circular motion diagram. That is, the object is moving to the right. Solve: The position of the object is given by the equation

⎡⎛ π ⎤ ⎞ π x ( t ) = A cos (ω t + φ0 ) = A cos ( 2π ft + φ0 ) = ( 8.0 cm ) cos ⎢⎜ rad s ⎟ t − rad ⎥ 2 2 ⎠ ⎣⎝ ⎦ The amplitude is A = 8.0 cm and the period is T = 1 f = 4.0 s. With φ0 = −π 2 rad, x starts at 0 cm and is moving to the right (getting more positive).

Assess:

As we see from the graph, the object starts out moving to the right.

14.8. Solve: The position of the object is given by the equation x ( t ) = A cos (ω t + φ0 ) = A cos ( 2π ft + φ0 ) We can find the phase constant φ0 from the initial condition: 0 cm = ( 4.0 cm ) cos φ0 ⇒ cos φ0 = 0 ⇒ φ0 = cos −1 ( 0 ) = ± 12 π rad Since the object is moving to the right, the object is in the lower half of the circular motion diagram. Hence, φ0 = − 12 π rad. The final result, with f = 4.0 Hz, is x ( t ) = ( 4.0 cm ) cos ⎡⎣( 8.0π rad s ) t − 12 π rad ⎤⎦

14.9. Solve: The position of the object is given by the equation x ( t ) = A cos (ω t + φ0 ) The amplitude A = 8.0 cm. The angular frequency ω = 2π f = 2π ( 0.50 Hz ) = π rad/s. Since at t = 0 it has its most negative velocity, it must be at the equilibrium point x = 0 cm and moving to the left, so φ0 = x ( t ) = ( 8.0 cm ) cos[(π rad/s)t +

π 2

rad]

π 2

. Thus

14.10. Model: The air-track glider is in simple harmonic motion. Solve:

(a) We can find the phase constant from the initial conditions for position and velocity:

x0 = A cos φ0

v0 x = −ω A sin φ0

Dividing the second by the first, we see that

sin φ0 v = tan φ0 = − 0 x cos φ0 ω x0 The glider starts to the left ( x0 = −5.00 cm) and is moving to the right (v0 x = +36.3 cm/s). With a period of 1.5 s =

3 2

s, the angular frequency is ω = 2π /T = 34 π rad/s. Thus ⎛

⎞ 1 36.3 cm/s 2 ⎟ = 3 π rad (60°) or – 3 π rad (–120°) π (4 /3 rad/s)( 5.00 cm) − ⎝ ⎠

φ0 = tan −1 ⎜ −

The tangent function repeats every 180°, so there are always two possible values when evaluating the arctan function. We can distinguish between them because an object with a negative position but moving to the right is in the third quadrant of the corresponding circular motion. Thus φ0 = − 23 π rad, or −120°. (b) At time t, the phase is φ = ω t + φ0 = ( 34 π rad/s) t − 32 π rad. This gives φ = − 23 π rad, 0 rad, 4 3

π rad at, respectively, t = 0 s, 0.5 s, 1.0 s, and 1.5 s. This is one period of the motion.

2 3

π rad, and

14.11. Model: The block attached to the spring is in simple harmonic motion. Solve:

The period of an object attached to a spring is

T = 2π

m = T0 = 2.0 s k

where m is the mass and k is the spring constant. (a) For mass = 2m, T = 2π

(b) For mass

1 2

2m = k

( 2 )T

0

= 2.8 s

m, T = 2π

1 2

m = T0 k

2 = 1.41 s

(c) The period is independent of amplitude. Thus T = T0 = 2.0 s (d) For a spring constant = 2k ,

T = 2π

m = T0 2k

2 = 1.41 s

14.12. Model: The air-track glider attached to a spring is in simple harmonic motion. Solve:

Experimentally, the period is T = (12.0 s ) (10 oscillations ) = 1.20 s. Using the formula for the period, 2

T = 2π

2

m ⎛ 2π ⎞ ⎛ 2π ⎞ ⇒k =⎜ ⎟ m=⎜ ⎟ ( 0.200 kg ) = 5.48 N m k ⎝ T ⎠ ⎝ 1.20 s ⎠

14.13. Model: The mass attached to the spring oscillates in simple harmonic motion. Solve: (a) The period T = 1 f = 1 2.0 Hz = 0.50 s. (b) The angular frequency ω = 2π f = 2π (2.0 Hz) = 4π rad/s. (c) Using energy conservation 1 2

kA2 = 12 kx02 + 12 mv02x

Using x0 = 5.0 cm, v0 x = −30 cm/s and k = mω 2 = (0.200 kg)(4π rad/s) 2 , we get A = 5.54 cm. (d) To calculate the phase constant φ0 ,

A cos φ0 = x0 = 5.0 cm ⎛ 5.0 cm ⎞ ⇒ φ0 = cos −1 ⎜ ⎟ = 0.45 rad ⎝ 5.54 cm ⎠ (e) The maximum speed is vmax = ω A = ( 4π rad s )( 5.54 cm ) = 70 cm s. (f) The maximum acceleration is

amax = ω 2 A = ω (ω A ) = ( 4π rad s )( 70 cm s ) = 8.8 m s 2 2 (g) The total energy is E = 12 mvmax =

1 2

( 0.200 kg )( 0.70 m s )

2

= 0.049 J.

(h) The position at t = 0.40 s is

x0.4 s = ( 5.54 cm ) cos ⎡⎣( 4π rad s )( 0.40 s ) + 0.45 rad ⎤⎦ = +3.8 cm

14.14. Model: The oscillating mass is in simple harmonic motion. Solve: (a) The amplitude A = 2.0 cm. (b) The period is calculated as follows:

ω=

2π 2π = 10 rad s ⇒ T = = 0.63 s T 10 rad s

(c) The spring constant is calculated as follows:

ω=

k 2 ⇒ k = mω 2 = ( 0.050 kg )(10 rad s ) = 5.0 N m m

(d) The phase constant φ0 = − 14 π rad. (e) The initial conditions are obtained from the equations

x ( t ) = ( 2.0 cm ) cos (10t − 14 π ) and vx ( t ) = − ( 20.0 cm s ) sin (10t − 14 π ) At t = 0 s, these equations become

x0 = ( 2.0 cm ) cos ( − 14 π ) = 1.41 cm and v0 x = − ( 20 cm s ) sin ( − 14 π ) = 14.1 cm s In other words, the mass is at +1.41 cm and moving to the right with a velocity of 14.1 cm/s. (f) The maximum speed is vmax = Aω = ( 2.0 cm )(10 rad s ) = 20 cm s. (g) The total energy E = 12 kA2 = 12 ( 5.0 N m )( 0.020 m ) = 1.00 × 10−3 J. 2

(h) At t = 0.41 s, the velocity is

v0 x = − ( 20 cm s ) sin ⎡⎣(10 rad s )( 0.40 s ) − 14 π ⎤⎦ = 1.46 cm s

14.15. Model: The block attached to the spring is in simple harmonic motion. Visualize:

Solve:

(a) The conservation of mechanical energy equation K f + U sf = K i + U si is 1 2

mv12 + 12 k ( Δx ) = 12 mv02 + 0 J ⇒ 0 J + 12 kA2 = 12 mv02 + 0 J

⇒ A=

2

m 1.0 kg v0 = ( 0.40 m s ) = 0.10 m = 10.0 cm k 16 N m

(b) We have to find the velocity at a point where x = A/2. The conservation of mechanical energy equation K 2 + U s2 = K i + U si is 2

1 2 1 ⎛ A⎞ 1 2 1 1 1⎛1 1⎛ 1 ⎞ 1 ⎞ mv2 + k ⎜ ⎟ = mv0 + 0 J ⇒ mv22 = mv02 − ⎜ kA2 ⎟ = mv02 − ⎜ mv02 ⎟ = 2 2 ⎝2⎠ 2 2 2 4⎝ 2 2 4 2 ⎠ ⎝ ⎠ ⇒ v2 = The velocity is 35 cm/s.

3 3 v0 = ( 0.40 m s ) = 0.346 m s 4 4

3⎛ 1 2⎞ ⎜ mv0 ⎟ 4⎝ 2 ⎠

14.16. Model: The vertical oscillations constitute simple harmonic motion. Visualize:

Solve:

(a) At equilibrium, Newton’s first law applied to the physics book is

(F )

sp y

− mg = 0 N ⇒ −k Δy − mg = 0 N

⇒ k = − mg Δy = − ( 0.500 kg ) ( 9.8 m s 2 ) ( −0.20 m ) = 24.5 N m (b) To calculate the period:

ω=

k 24.5 N m 2π 2π rad = = 7.0 rad s and T = = = 0.90 s m 0.500 kg ω 7.0 rad s

(c) The maximum speed is

vmax = Aω = ( 0.10 m )( 7.0 rad s ) = 0.70 m s Maximum speed occurs as the book passes through the equilibrium position.

14.17. Model: The vertical oscillations constitute simple harmonic motion. Solve: To find the oscillation frequency using ω = 2π f = k m , we first need to find the spring constant k. In equilibrium, the weight mg of the block and the spring force k ΔL are equal and opposite. That is, mg = k ΔL ⇒ k = mg ΔL . The frequency of oscillation f is thus given as f =

1 2π

k 1 = m 2π

mg ΔL 1 = m 2π

g 1 = ΔL 2π

9.8 m s 2 = 3.5 Hz 0.020 m

14.18. Model: The vertical oscillations constitute simple harmonic motion. Visualize:

Solve:

The period and angular frequency are

T=

20 s 2π 2π = 0.6667 s and ω = = = 9.425 rad s 30 oscillations T 0.6667 s

(a) The mass can be found as follows:

ω=

k k 15 N/m ⇒m= 2 = = 0.169 kg 2 ω m ( 9.425 rad s )

(b) The maximum speed vmax = ω A = ( 9.425 rad s )( 0.060 m ) = 0.57 m/s.

14.19. Model: Assume a small angle of oscillation so there is simple harmonic motion. Solve:

The period of the pendulum is

T0 = 2π

L0 = 4.0 s g

(a) The period is independent of the mass and depends only on the length. Thus T = T0 = 4.0 s. (b) For a new length L = 2 L0 ,

T = 2π

2 L0 = 2T0 = 5.7 s g

(c) For a new length L = L0 /2,

T = 2π

L0 2 1 = T0 = 2.8 s g 2

(d) The period is independent of the amplitude as long as there is simple harmonic motion. Thus T = 4.0 s.

14.20. Model: The pendulum undergoes simple harmonic motion. Solve: (a) The amplitude is 0.10 rad. (b) The frequency of oscillations is f =

ω 5 = Hz = 0.796 Hz 2π 2π

(c) The phase constant φ = π rad. (d) The length can be obtained from the period: 2

2

ω = 2π f =

⎛ ⎞ ⎛ 1 ⎞ g 1 2 ⇒ L=⎜ ⎟⎟ ( 9.8 m s ) = 0.392 m ⎟ g = ⎜⎜ π L 2 0.796 Hz ( ) ⎝ 2π f ⎠ ⎝ ⎠

(e) At t = 0 s, θ 0 = ( 0.10 rad ) cos (π ) = −0.10 rad. To find the initial condition for the angular velocity we take

the derivative of the angular position:

θ ( t ) = ( 0.10 rad ) cos ( 5t + π ) ⇒ At t = 0 s,

( dθ

dt )0 = ( −0.50 rad ) sin (π ) = 0 rad s.

dθ ( t ) = − ( 0.10 rad )( 5 ) sin ( 5t + π ) dt

(f) At t = 2.0 s, θ 2.0 = ( 0.10 rad ) cos ( 5 ( 2.0 s ) + π ) = 0.084 rad.

14.21. Model: Assume the small-angle approximation so there is simple harmonic motion. Solve:

The period is T = 12 s 10 oscillations = 1.20 s and is given by the formula 2

T = 2π

2

L ⎛ T ⎞ ⎛ 1.20 s ⎞ 2 ⇒L=⎜ ⎟ g =⎜ ⎟ ( 9.8 m s ) = 36 cm g ⎝ 2π ⎠ ⎝ 2π ⎠

14.22. Model: Assume a small angle of oscillation so there is simple harmonic motion. Solve:

(a) On the earth the period is

Tearth = 2π

1.0 m L = 2π = 2.0 s 9.80 m s 2 g

(b) On Venus the acceleration due to gravity is

g Venus =

GM Venus ( 6.67 × 10 = 2 RVenus ⇒ TVenus = 2π

−11

N ⋅ m 2 kg 2 )( 4.88 × 1024 kg )

( 6.06 ×106 m )

2

L 1.0 m = 2π = 2.1 s g Venus 8.86 m s 2

= 8.86 m s 2

14.23. Model: Assume the pendulum to have small-angle oscillations. In this case, the pendulum undergoes simple harmonic motion. Solve: Using the formula g = GM R 2 , the periods of the pendulums on the moon and on the earth are

Tearth = 2π

L L R2 L R2 = 2π earth earth and Tmoon = 2π moon moon g GM earth GM moon

Because Tearth = Tmoon , 2

2π

2 ⎛M ⎞⎛ R ⎞ Learth Rearth L R2 = 2π moon moon ⇒ Lmoon = ⎜ moon ⎟⎜ earth ⎟ Learth GM earth GM moon ⎝ M earth ⎠⎝ Rmoon ⎠ 2

⎛ 7.36 × 1022 kg ⎞⎛ 6.37 × 106 m ⎞ =⎜ ⎟⎜ ⎟ ( 2.0 m ) = 33 cm 24 6 ⎝ 5.98 × 10 kg ⎠⎝ 1.74 × 10 m ⎠

14.24. Model: Assume a small angle of oscillation so that the pendulum has simple harmonic motion. Solve:

The time periods of the pendulums on the earth and on Mars are

Tearth = 2π

L g earth

and TMars = 2π

L g Mars

Dividing these two equations, Tearth = TMars

2

2

⎛T ⎞ g Mars ⎛ 1.50 s ⎞ 2 ⇒ g Mars = g earth ⎜ earth ⎟ = ( 9.8 m s 2 ) ⎜ ⎟ = 3.67 m s g earth ⎝ 2.45 s ⎠ ⎝ TMars ⎠

14.25. Visualize: Please refer to Figure Ex14.25. Solve: The mass of the wrench can be obtained from the length that it stretches the spring. From Equation 14.41,

ΔL =

mg k ΔL 360 N/m ( 0.030 m ) ⇒m= = = 1.10 kg k g 9.8 m/s 2

When swinging on a hook the wrench is a physical pendulum. From Equation 14.52, 2π f =

mgl 2π mgl ⇒ = I T I

From the figure, l = 0.14 m. Thus 2

2

⎛ T ⎞ ⎛ 0.90 s ⎞ −2 2 2 I =⎜ ⎟ mgl = ⎜ ⎟ (1.10 kg ) ( 9.8 m/s ) ( 0.14 m ) = 3.1 × 10 kg m ⎝ 2π ⎠ ⎝ 2π ⎠

14.26. Model: The spider is in simple harmonic motion. Solve:

Your tapping is a driving frequency. Largest amplitude at f ext = 1.0 Hz means that this is the resonance

frequency, so f 0 = f ext = 1.0 Hz. That is, the spider’s natural frequency of oscillation f 0 is 1.0 Hz and

ω 0 = 2π f 0 = 2π rad s. We have ω0 =

k 2 ⇒ k = mω 02 = ( 0.0020 kg )( 2π rad s ) = 0.079 N m m

14.27. Model: The motion is a damped oscillation. Solve:

The amplitude of the oscillation at time t is given by Equation 14.58: A ( t ) = A0e− t 2τ , where τ = m b is

the time constant. Using x = 0.368 A and t = 10.0 s, we get

0.368 A = Ae−10.0 s 2τ ⇒ ln ( 0.368 ) =

−10 s 10.0 s ⇒τ = − = 5.00 s 2τ 2ln ( 0.368 )

14.28. Model: The motion is a damped oscillation. The position of a damped oscillator is x ( t ) = Ae− ( t 2τ ) cos (ω t + φ0 ) . The frequency is 1.0 Hz and the

Solve:

damping time constant τ is 4.0 s. Let us assume φ0 = 0 rad and A = 1 with arbitrary units. Thus, x ( t ) = e − t (8.0 s ) cos ⎣⎡ 2π (1.0 Hz ) t ⎦⎤ ⇒ x ( t ) = e −0.125 t cos ( 2π t ) where t is in s. Values of x(t) at selected values of t are displayed in the following table: t(s)

x(t)

t(s)

x(t)

t(s)

x(t)

0

1

2.00

0.779

6.00

0.472

0.25

0

2.50

−0.732

6.50

− 0.444

0.50

−0.939

3.00

0.687

7.00

0.417

0.75

0

3.50

−0.646

7.50

− 0.392

1.00

0.882

4.00

0.607

8.00

0.368

1.25

0

4.50

−0.570

8.50

− 0.346

1.50

−0.829

5.00

0.535

9.00

0.325

1.75

0

5.50

−0.503

9.50

− 0.305

10.00

0.286

14.29. Model: The pendulum is a damped oscillator. Solve:

The period of the pendulum and the number of oscillations in 4 hours are calculated as follows:

T = 2π

4 ( 3600 s ) 15.0 m L = 2π = 7.773 s ⇒ N osc = = 1853 2 9.8 m s 7.773 s g

The amplitude of the pendulum as a function of time is A ( t ) = Ae − bt / 2 m . The exponent of this expression can be calculated to be −

bt ( 0.010 kg s )( 4 × 3600 s ) = −0.6545 =− 2m 2 (110 kg )

We have A ( t ) = (1.50 m ) e −0.6545 = 0.780 m.

14.30. Model: The vertical oscillations are damped and follow simple harmonic motion. Solve: The position of the ball is given by x ( t ) = Ae − ( t 2τ ) cos (ω t + φ0 ) . The amplitude A ( t ) = Ae − ( t 2τ ) is a function of time. The angular frequency is

ω=

k = m

(15.0

N m) 2π = 5.477 rad s ⇒ T = = 1.147 s 0.500 kg ω

Because the ball’s amplitude decreases to 3.0 cm from 6.0 cm after 30 oscillations, that is, after 30 × 1.147 s = 34.41 s, we have 3.0 cm = ( 6.0 cm ) e − (34.414 s 2τ ) ⇒ 0.50 = e − ( 34.41 s 2τ ) ⇒ ln ( 0.50 ) =

−34.41 s ⇒ τ = 25 s 2τ

14.31. Visualize: Please refer to Figure P14.31. Solve:

The position and the velocity of a particle in simple harmonic motion are

x ( t ) = A cos (ω t + φ0 ) and vx ( t ) = − Aω sin (ω t + φ0 ) = −vmax sin (ω t + φ0 ) (a) At t = 0 s, the equation for x yields

( −5.0 cm ) = (10.0 cm ) cos (φ0 ) ⇒ φ0 = cos−1 ( −0.5) = ± 23 π

rad

Because the particle is moving to the left at t = 0 s, it is in the upper half of the circular motion diagram, and the phase constant is between 0 and π radians. Thus, φ0 = 23 π rad. (b) The period is 4.0 s. At t = 0 s,

⎛ 2π ⎞ ⎛ 2π ⎞ v0 x = − Aω sin φ0 = − (10.0 cm ) ⎜ ⎟ sin ⎜ ⎟ = −13.6 cm/s ⎝ T ⎠ ⎝ 3 ⎠ (c) The maximum speed is

⎛ 2π ⎞ vmax = ω A = ⎜ ⎟ (10.0 cm ) = 15.7 cm s ⎝ 4.0 s ⎠ Assess: The negative velocity at t = 0 s is consistent with the position-versus-time graph and the positive sign of the phase constant.

14.32. Visualize: Please refer to Figure P14.32. Solve:

The position and the velocity of a particle in simple harmonic motion are

x ( t ) = A cos (ω t + φ0 ) and vx ( t ) = − Aω sin (ω t + φ0 ) = −vmax sin (ω t + φ0 ) From the graph, T = 12 s and the angular frequency is

ω=

2π 2π π = = rad s T 12 s 6

(a) Because vmax = Aω = 60 cm s, we have

A=

60 cm s

ω

=

60 cm s = 115 cm π 6 rad s

(b) At t = 0 s, v0x = − Aω sin φ0 = −30 cm/s ⇒ − ( 60 cm s ) sin φ0 = −30 cm/s ⇒ φ0 = sin −1 ( 0.5 rad ) = 16 π rad ( 30° ) or 56 π rad (150° )

Because the velocity at t = 0 s is negative and the particle is slowing down, the particle is in the second quadrant of the circular motion diagram. Thus φ0 = 56 π rad.

(c) At t = 0 s, x0 = (115 cm ) cos ( 56 π rad ) = −100 cm.

14.33. Model: The vertical mass/spring systems are in simple harmonic motion. Visualize: Please refer to Figure P14.33. Solve: (a) For system A, the maximum speed while traveling in the upward direction corresponds to the maximum positive slope, which is at t = 3.0 s. The frequency of oscillation is 0.25 Hz. (b) For system B, all the energy is potential energy when the position is at maximum amplitude, which for the first time is at t = 1.5 s. The time period of system B is thus 6.0 s. (c) Spring/mass A undergoes three oscillations in 12 s, giving it a period TA = 4.0 s. Spring/mass B undergoes 2

oscillations in 12 s, giving it a period TB = 6.0 s. We have

TA = 2π

⎛ m ⎞⎛ k ⎞ 4.0 s 2 mA mB T and TB = 2π ⇒ A = ⎜ A ⎟⎜ B ⎟ = = kA kB TB ⎝ mB ⎠⎝ kA ⎠ 6.0 s 3

If mA = mB , then

kB 4 k 9 = ⇒ A = = 2.25 kA 9 kB 4

14.34. Solve: The object’s position as a function of time is x ( t ) = A cos (ω t + φ0 ) . Letting x = 0 m at t = 0 s, gives 0 = A cos φ0 ⇒ φ0 = ± 12 π Since the object is traveling to the right, it is in the lower half of the circular motion diagram, giving a phase constant between −π and 0 radians. Thus, φ0 = − 12 π and x ( t ) = A cos (ω t − 12 π ) ⇒ x ( t ) = A sin ω t = ( 0.10 m ) sin ( 12 π t )

where we have used A = 0.10 m and

ω=

2π 2π rad π = = rad s T 4.0 s 2

Let us now find t where x = 0.60 m:

2 ⎛π ⎞ ⎛ 0.060 m ⎞ 0.060 m = ( 0.10 m ) sin ⎜ t ⎟ ⇒ t = sin −1 ⎜ ⎟ = 0.41 s π ⎝2 ⎠ ⎝ 0.10 m ⎠ Assess:

The answer is reasonable because it is approximately

1 8

of the period.

14.35. Model: The block attached to the spring is in simple harmonic motion. Visualize:

The position and the velocity of the block are given by the equations

x ( t ) = A cos (ω t + φ0 ) and vx ( t ) = − Aω sin (ω t + φ0 ) Solve:

To graph x(t) we need to determine ω , φ0 , and A. These quantities will be found by using the initial

(t = 0 s) conditions on x(t) and vx (t ). The period is

T = 2π

1.0 kg 2π 2π rad m = 2π = 1.405 s ⇒ ω = = = 4.472 rad s 20 N m k T 1.405 s

At t = 0 s, x0 = A cos φ0 and v0 x = − Aω sin φ0 . Dividing these equations, tan φ0 = −

v0 x ( −1.0 m s ) =− = 1.1181 ⇒ φ0 = 0.841 rad ω x0 ( 4.472 rad s )( 0.20 m )

From the initial conditions, 2

⎛v ⎞ A = x + ⎜ 0x ⎟ = ⎝ω ⎠ 2 0

2

⎛ −1.0 m s ⎞ ( 0.20 m ) + ⎜ ⎟ = 0.300 m ⎝ 4.472 rad s ⎠ 2

The position-versus-time graph can now be plotted using the equation x ( t ) = ( 0.300 m ) cos ⎡⎣( 4.472 rad s ) t + 0.841 rad ⎤⎦

14.36. Model: The astronaut attached to the spring is in simple harmonic motion. Visualize: Please refer to Figure P14.36. Solve: (a) From the graph, T = 3.0 s, so we have 2

T = 2π

2

m ⎛ T ⎞ ⎛ 3.0 s ⎞ ⇒m=⎜ ⎟ k =⎜ ⎟ ( 240 N m ) = 55 kg k ⎝ 2π ⎠ ⎝ 2π ⎠

(b) Oscillations occur about an equilibrium position of 1.0 m. From the graph, A =

1 2

( 0.80 m ) = 0.40 m, φ0 =

0 rad, and

ω=

2π 2π = = 2.1 rad s T 3.0 s

The equation for the position of the astronaut is

x ( t ) = A cos ω t + 1.0 m = ( 0.4 m ) cos ⎡⎣( 2.1 rad s ) t ⎤⎦ + 1.0 m ⇒ 1.2 m= ( 0.4 m ) cos ⎡⎣( 2.1 rad s ) t ⎤⎦ + 1.0 m ⇒ cos ⎡⎣( 2.1 rad s ) t ⎤⎦ = 0.5 ⇒ t = 0.50 s The equation for the velocity of the astronaut is vx ( t ) = − Aω sin (ω t ) ⇒ v0.5 s = − ( 0.4 m )( 2.1 rad s ) sin ⎡⎣( 2.1 rad s )( 0.50 s ) ⎤⎦ = −0.73 m s Thus her speed is 0.73 m/s.

14.37. Model: The particle is in simple harmonic motion. Solve:

The equation for the velocity of the particle is

vx ( t ) = − ( 25 cm )(10 rad s ) sin (10 t ) Substituting into K = 2U gives 2 2 1 2 ⎛1 ⎞ 1 mvx ( t ) = 2 ⎜ kx 2 ( t ) ⎟ ⇒ m ⎣⎡ − ( 250 cm s ) sin (10 t ) ⎦⎤ = k ⎣⎡( 25 cm ) cos (10 t ) ⎦⎤ 2 2 2 ⎝ ⎠

sin 2 (10 t ) ⎛ k ⎞ ( 25 cm ) ⎛ 1 ⎞ 2 = 2⎜ ⎟ = 2ω 2 ⎜ ⎟s 2 cos 2 (10 t ) ⎝ m ⎠ ( 250 cm s ) ⎝ 100 ⎠ 2

⇒

1 2⎛ 1 ⎞ 2 −1 ⇒ tan 2 (10 t ) = 2 (10 rad s ) ⎜ 2.0 = 0.096 s ⎟ s = 2.0 ⇒ t = tan 10 ⎝ 100 ⎠

14.38. Model: The spring undergoes simple harmonic motion. Solve:

(a) Total energy is E = 12 kA2 . When the displacement is x = 12 A, the potential energy is

U = 12 kx 2 = 12 k ( 12 A) 2 =

1 4

(

1 2

kA2 ) = 14 E ⇒ K = E − U = 34 E

One quarter of the energy is potential and three-quarters is kinetic. (b) To have U = 12 E requires U = 12 kx 2 = 12 E = 12 ( 12 kA2 ) ⇒ x =

A 2

14.39. Solve: Average speed is vavg = Δx/Δt. During half a period (Δt = 12 T ), the particle moves from x = − A to x = + A( Δx = 2 A). Thus vavg =

Δx 2 A 4 A 4A 2 2 π = = = = (ω A) = vmax ⇒ vmax = vavg Δt T /2 T 2π /ω π π 2

14.40. Model: The ball attached to a spring is in simple harmonic motion. Solve:

(a) Let t = 0 s be the instant when x0 = −5.0 cm and v0 = 20 cm/s. The oscillation frequency is

k 2.5 N m = = 5.0 rad / s m 0.100 kg

ω=

Using Equation 14.27, the amplitude of the oscillation is 2

⎛v ⎞ A = x02 + ⎜ 0 ⎟ = ⎝ω ⎠

2

⎛ 20 cm / s ⎞ 2 ( −5.0 cm ) + ⎜ ⎟ = 6.4 cm ⎝ 5.0 rad / s ⎠

(b) The maximum acceleration is amax = ω 2 A = 160 cm s 2 .

(c) For an oscillator, the acceleration is most positive ( a = amax ) when the displacement is most negative

( x = − xmax = − A) .

So the acceleration is maximum when x = −6.4 cm.

(d) We can use the conservation of energy between x0 = −5.0 cm and x1 = 3.0 cm: 1 2

mv02 + 12 kx02 = 12 mv12 + 12 kx12 ⇒ v1 = v02 +

k 2 ( x0 − x12 ) = 0.283 m s m

The speed is 28 cm/s. Because k is known in SI units of N/m, the energy calculation must be done using SI units of m, m/s, and kg.

14.41. Model: The block on a spring is in simple harmonic motion. Solve:

(a) The position of the block is given by x ( t ) = A cos (ω t + φ0 ) . Because x ( t ) = A at t = 0 s, we have

φ0 = 0 rad, and the position equation becomes x ( t ) = A cos ω t. At t = 0.685 s, 3.00 cm = A cos ( 0.685ω ) and at t = 0.886 s, −3.00 cm = A cos ( 0.886ω ) . These two equations give

cos ( 0.685ω ) = − cos ( 0.886ω ) = cos (π − 0.886ω ) ⇒ 0.685ω = π − 0.886ω ⇒ ω = 2.00 rad s (b) Substituting into the position equation,

3.00 cm = A cos ( ( 2.00 rad s )( 0.685 s ) ) = A cos (1.370 ) = 0.20 A ⇒ A =

3.00 cm = 15.0 cm 0.20

14.42. Model: The oscillator is in simple harmonic motion. Energy is conserved. Solve:

The energy conservation equation E1 = E2 is 1 2

mv12 + 12 kx12 = 12 mv22 + 12 kx22

1 1 1 1 2 2 2 2 ( 0.30 kg )( 0.954 m s ) + k ( 0.030 m ) = ( 0.30 kg )( 0.714 m s ) + k ( 0.060 m ) 2 2 2 2 ⇒ k = 44.48 N m The total energy of the oscillator is 1 1 1 1 2 2 Etotal = mv12 + kx12 = ( 0.30 kg )( 0.954 m s ) + ( 44.48 N m )( 0.030 m ) = 0.1565 J 2 2 2 2 2 , Because Etotal = 12 mvmax

0.1565 J = Assess:

1 2 ⇒ v max = 1.02 m s ( 0.300 kg ) vmax 2

A maximum speed of 1.02 m/s is reasonable.

14.43. Model: The transducer undergoes simple harmonic motion. Solve:

Newton’s second law for the transducer is

Frestoring = ma max ⇒ 40,000 N = ( 0.10 × 10−3 kg ) amax ⇒ amax = 4.0 × 108 m s 2 Because amax = ω 2 A,

A=

amax

ω

2

=

4.0 × 108 m s 2 ⎡ 2π (1.0 × 10 Hz ) ⎤ ⎣ ⎦ 6

2

= 1.01 × 10−5 m = 10.1 μ m

(b) The maximum velocity is

vmax = ω A = 2π (1.0 × 106 Hz )(1.01 × 10−5 m ) = 64 m s

14.44. Model: The block attached to the spring is in simple harmonic motion. Solve:

(a) The frequency is

f =

1 2π

1 k = m 2π

2000 N m = 3.183 Hz 5.0 kg

The frequency is 3.2 Hz. (b) From energy conservation, 2

2

⎛v ⎞ ⎛ 1.0 m/s ⎞ A = x02 + ⎜ 0 ⎟ = (0.050 m) 2 + ⎜ ⎟ = 0.0707 m ω ⎝ ⎠ ⎝ 2π ⋅ 3.183 Hz ⎠ The amplitude is 7.1 cm. (c) The total mechanical energy is 2 E = 12 kA2 = 12 ( 2000 N m )( 0.0707 m ) = 5.0 J

14.45. Model: The tips of the tuning fork are in simple harmonic oscillation. Solve:

(a) The maximum speed is related to the amplitude.

vmax = ω A = 2π fA = 2π ( 440 Hz ) ( 5.0 × 10−4 m ) = 1.38 m/s (b) The acceleration of the flea can not be greater than that allowed by the maximum force with which it can hold on. From Newton’s second law, the maximum acceleration that the flea can withstand is

aflea =

F 1.0 × 10−3 N = = 100 m/s 2 m 10 × 10−6 kg

The maximum acceleration at the tip of the prong is amax = ω 2 A = ( 2π f ) A = ( 2π ( 440 Hz ) ) ( 5.0 × 10−4 m ) = 382 m/s 2 2

The flea will not be able to hold on to the tuning fork.

2

14.46. Model: The block undergoes simple harmonic motion. Visualize:

Solve:

(a) The frequency of oscillation is

f = The frequency is 1.13 Hz. (b) Using conservation of energy,

1 2

1 2π

k 1 = m 2π

10 N/m = 1.125 Hz 0.20 kg

mv12 + 12 kx12 = 12 mv02 + 12 kx02 , we find

m 2 2 0.20 kg (v0 − v1 ) = (−0.20 m) 2 + ((1.00 m/s) 2 − (0.50 m/s) 2 ) k 10 N/m = 0.2345 m or 23 cm

x1 = x02 +

(c) At time t, the displacement is x = A cos(ω t + φ0 ). The angular frequency is ω = 2π f = 7.071 rad/s. The amplitude is 2

2

⎛v ⎞ ⎛ 1.00 m/s ⎞ A = x02 + ⎜ 0 ⎟ = (−0.20 m) 2 + ⎜ ⎟ = 0.245 m ω ⎝ ⎠ ⎝ 7.071 rad/s ⎠ The phase constant is ⎛ x0 ⎞ −1 ⎛ −0.200 m ⎞ ⎟ = cos ⎜ ⎟ = ±2.526 rad or ± 145° ⎝ A⎠ ⎝ 0.245 m ⎠

φ0 = cos −1 ⎜

A negative displacement (below the equilibrium point) and positive velocity (upward motion) indicate that the corresponding circular motion is in the third quadrant, so φ0 = −2.526 rad. Thus at t = 1.0 s,

x = (0.245 m)cos ( (7.071 rad/s)(1.0 s) − 2.526 rad ) = −0.0409 m = −4.09 cm The block is 4.1 cm below the equilibrium point.

14.47. Model: The mass is in simple harmonic motion. Visualize:

The high point of the oscillation is at the point of release. This conclusion is based on energy conservation. Gravitational potential energy is converted to the spring’s elastic potential energy as the mass falls and stretches the spring, then the elastic potential energy is converted 100% back into gravitational potential energy as the mass rises, bringing the mass back to exactly its starting height. The total displacement of the oscillation—high point to low point—is 20 cm. Because the oscillations are symmetrical about the equilibrium point, we can deduce that the equilibrium point of the spring is 10 cm below the point where the mass is released. The mass oscillates about this equilibrium point with an amplitude of 10 cm, that is, the mass oscillates between 10 cm above and 10 cm below the equilibrium point. Solve: The equilibrium point is the point where the mass would hang at rest, with Fsp = FG = mg . At the

equilibrium point, the spring is stretched by Δy = 10 cm = 0.10 m. Hooke’s law is Fsp = k Δy, so the equilibrium condition is k g 9.8 m s 2 ⎡⎣ Fsp = k Δy ⎤⎦ = [ FG = mg ] ⇒ = = = 98 s −2 m Δy 0.10 m

The ratio k m is all we need to find the oscillation frequency:

f =

1 2π

k 1 98 s −2 = 1.58 Hz = m 2π

14.48. Model: The spring is ideal, so the apples undergo SHM. Solve: The spring constant of the scale can be found by considering how far the pan goes down when the apples are added. ΔL =

mg mg 20 N ⇒k= = = 222 N/m k ΔL 0.090 m

The frequency of oscillation is

f =

1 2π

1 k = m 2π

222 N/m = 1.66 Hz ( 20 N 9.8 m/s 2 )

Assess: An oscillation of fewer than twice per second is reasonable.

14.49. Model: The compact car is in simple harmonic motion. Solve:

(a) The mass on each spring is (1200 kg ) 4 = 300 kg. The spring constant can be calculated as follows:

ω2 =

2 k 2 ⇒ k = mω 2 = m ( 2π f ) = ( 300 kg ) ⎡⎣ 2π ( 2.0 Hz ) ⎤⎦ = 4.74 × 104 N m m

The spring constant is

4.7 × 104 N/m. (b) The car carrying four persons means that each spring has, on the average, an additional mass of 70 kg. That is, m = 300 kg + 70 kg = 370 kg. Thus,

f =

ω 1 = 2π 2π

1 k = m 2π

4.74 × 104 N m = 1.80 Hz 370 kg

Assess: A small frequency change from the additional mass is reasonable because frequency is inversely proportional to the square root of the mass.

14.50. Model: Hooke’s law for the spring. The spring’s compression and decompression constitutes simple harmonic motion. Visualize:

Solve: (a) The spring’s compression or decompression is one-half of the oscillation cycle. This means the contact time is Δt = 12 T , where T is the period. The period is calculated as follows:

ω=

k 50 N m 1 2π 2π = = 10 rad s ⇒ T = = = = 0.628 s m 0.500 kg f ω 10 rad s ⇒ Δt =

T = 0.31 s 2

(b) There is no change in contact time, because period of oscillation is independent of the amplitude or the maximum speed.

14.51. Model: The two blocks are in simple harmonic motion, without the upper block slipping. We will also apply the model of static friction between the two blocks. Visualize:

Solve: The net force acting on the upper block m1 is the force of friction due to the lower block m2 . The model of static friction gives the maximum force of static friction as

( fs )max = μs n = μs ( m1g ) = m1amax ⇒ amax = μs g Using μ s = 0.5, amax = μS g = ( 0.5) ( 9.8 m s 2 ) = 4.9 m s 2 . That is, the two blocks will ride together if the maximum acceleration of the system is equal to or less than amax . We can calculate the maximum value of A as follows: amax = ω 2 Amax =

2 a ( m + m2 ) ( 4.9 m s ) (1.0 kg + 5.0 kg ) k Amax ⇒ Amax = max 1 = = 0.59 m m1 + m2 k 50 N m

14.52. Model: Assume simple harmonic motion for the two-block system without the upper block slipping. We will also use the model of static friction between the two blocks. Visualize:

Solve: The net force on the upper block m1 is the force of static friction due to the lower block m2 . The two blocks ride together as long as the static friction doesn’t exceed its maximum possible value. The model of static friction gives the maximum force of static friction as

( fs )max = μs n = μs ( m1g ) = m1amax ⇒ amax = μs g 2

⇒ μs =

2

amax ω 2 Amax ⎛ 2π ⎞ ⎛ Amax ⎞ ⎛ 2π ⎞ ⎛ 0.40 m ⎞ = =⎜ = 0.72 ⎟=⎜ ⎟ ⎜ ⎟ ⎜ 2 ⎟ g g ⎝ T ⎠ ⎝ g ⎠ ⎝ 1.5 s ⎠ ⎝ 9.8 m s ⎠

Assess: Because the period is given, we did not need to use the block masses or the spring constant in our calculation.

14.53. Model: The DNA and cantilever undergo SHM. Visualize: Please refer to figure P14.53. Solve: The cantilever has the same spring constant with and without the DNA molecule. The frequency of oscillation without the DNA is k ω1 = 1 3M With the DNA, the frequency of oscillation is

ω2 =

1 3

k M +m

where m is the mass of the DNA. Divide the two equations, and express ω 2 = ω1 − Δω , where Δω = 2πΔf = 2π ( 50 Hz ) .

ω1 f1 f1 = = = ω 2 f 2 f1 − Δf Thus

k ( 13 M ) = k ( 13 M + m)

f12 ( 13 M ) = ( f1 − Δf )

m=

Since Δf  f1 , (50 Hz  12MHz),

1 3

(f M

2 1

− ( f1 − Δf )

( f1 − Δf )

( f1 − Δf )

−2

2

2

)=

2

1 3

1 3

M +m 1 M 3

( 13 M + m ) ⎛ ⎞ f12 − 1⎟ M⎜ 2 ⎜ ( f − Δf ) ⎟ ⎝ 1 ⎠ −2

Δf ⎛ Δf ⎞ −2 ⎛ = f1−2 ⎜ 1 − ⎟ ≈ f1 ⎜ 1 + 2 f f 1 ⎠ 1 ⎝ ⎝

⎞ ⎟ . Thus ⎠

⎛ ⎞ Δf Δf m = 13 M ⎜1 + 2 − 1⎟ = 23 M f f1 1 ⎝ ⎠ The mass of the cantilever M = (2300 kg/m3 )(4000 × 10−9 m)(100 × 10−9 m) = 3.68 × 10−16 kg Thus the mass of the DNA molecule is 2 ⎛ 50 Hz ⎞ −21 m = ( 3.68 × 10−16 kg ) ⎜ ⎟ = 1.02 × 10 kg 6 3 ⎝ 12 × 10 Hz ⎠ Assess: The mass of the DNA molecule is about 6.2 × 105 atomic mass units, which is reasonable for such a large molecule.

14.54. Model: Assume that the swinging lamp makes a small angle with the vertical so that there is simple harmonic motion. Visualize:

Solve:

(a) Using the formula for the period of a pendulum, 2

T = 2π

2

L ⎛ T ⎞ 2 ⎛ 5.5 s ⎞ ⇒ L = g⎜ ⎟ = ( 9.8 m s ) ⎜ ⎟ = 7.5 m π g 2 ⎝ ⎠ ⎝ 2π ⎠

(b) The conservation of mechanical energy equation K 0 + U g0 = K1 + U g1 for the swinging lamp is 1 2

2 mv02 + mgy0 = 12 mv12 + mgy1 ⇒ 0 J + mgh = 12 mvmax +0 J

⇒ vmax = 2 gh = 2 g ( L − L cos3° ) = 2 ( 9.8 m s 2 ) ( 7.5 m )(1 − cos3° ) = 0.45 m s

14.55. Model: Assume that the angle with the vertical that the pendulum makes is small enough so that there is simple harmonic motion. Solve: The angle θ made by the string with the vertical as a function of time is

θ ( t ) = θ max cos (ω t + φ0 ) The pendulum starts from maximum displacement, thus φ0 = 0. Thus, θ ( t ) = θ max cos ω t. To find the time t when

the pendulum reaches 4.0° on the opposite side:

( −4.0° ) = (8.0° ) cosω t ⇒ ω t = cos −1 ( −0.5) = 2.094 rad Using the formula for the angular frequency,

ω=

g 9.8 m s 2 2.0944 rad 2.094 rad = = 3.130 rad s ⇒ t = = = 0.669 s L 1.0 m 3.130 rad s ω

The time t = 0.67 s. Assess: Because T = 2π ω = 2.0 s, a value of 0.67 s for the pendulum to cover a little less than half the oscillation is reasonable.

14.56. Model: Assume a small angle oscillation of the pendulum so that it has simple harmonic motion. Solve:

(a) At the equator, the period of the pendulum is

Tequator = 2π

1.000 m = 2.009 s 9.78 m s 2

The time for 100 oscillations is 200.9 s. (b) At the north pole, the period is Tpole = 2π

1.000 m = 2.004 s 9.83 m s 2

The time for 100 oscillations is 200.4 s. (c) The difference between the two answers is 0.5 s, and this difference is quite measurable with a hand-operated stopwatch. (d) The period on the top of the mountain is 2.010 s. The acceleration due to gravity can be calculated by rearranging the formula for the period: 2

2

⎛ 2π ⎞ ⎛ 2π ⎞ 2 g mountain = L ⎜ ⎟ = (1.000 m ) ⎜ ⎟ = 9.772 m s T 2.010 s ⎝ ⎠ mountain ⎝ ⎠ Assess:

This last result is reasonable because g decreases with altitude.

14.57. Model: The mass is a particle and the string is massless.

Solve:

Equation 14.52 is ω=

Mgl I

The moment of inertia of the mass on a string is I = Ml 2 , where l is the length of the string. Thus

ω=

Mgl = Ml 2

g l

This is Equation 14.49 with L = l. Assess: Equation 14.49 is really a specific case of the more general physical pendulum described by Equation 14.52.

14.58. Model: The rod is thin and uniform with moment of inertia described in Table 12.2. The clay ball is a particle located at the end of the rod. The ball and rod together form a physical pendulum. The oscillations are small. Visualize:

Solve:

The moment of inertia of the composite pendulum formed by the rod and clay ball is

1 I rod + ball = I rod + I ball = mrod L2 + mball L2 3 1 2 2 = ( 0.200 kg )( 0.15 m ) + ( 0.020 kg )( 0.15 m ) = 1.95 × 10−3 kg m 2 3 The center of mass of the rod and ball is located at a distance from the pivot point of

ycm =

( 0.200 kg ) ⎛⎜

0.15 ⎞ m ⎟ + ( 0.020 kg )( 0.15 m ) ⎝ 2 ⎠ = 8.18 × 10−2 m ( 0.200 kg + 0.020 kg )

The frequency of oscillation of a physical pendulum is 1 2π

Mgl 1 = I 2π

The period of oscillation T =

1 = 0.66 s. f

f =

( 0.220 kg ) ( 9.8 m/s 2 )(8.18 × 10−2 m ) 1.95 × 10−3 kg m 2

= 1.51 Hz

14.59. Model: The circular hoop can be modeled as a cylindrical hoop and its moment of inertia about the point of rotation found with the parallel-axis theorem. Visualize: Please refer to Figure P14.59. Solve: Using the parallel-axis theorem, the moment of inertia of the cylindrical hoop about the rotation point is

I = MR 2 + MR 2 = 2MR 2 The frequency of small oscillations is given by Equation 14.52. f =

1 2π

Mgl I

The center of mass of the hoop is its center, so l = R. Thus f =

1 2π

MgR 1 = 2 2 MR 2π

g 2R

14.60. Model: The motion is a damped oscillation. Solve:

The position of the air-track glider is x ( t ) = Ae − ( t 2τ ) cos (ω t + φ0 ) , where τ = m b and

k b2 − m 4m 2

ω= Using A = 0.20 m, φ0 = 0 rad, and b = 0.015 kg/s,

4.0 N m ( 0.015 kg s ) − = 16 − 9 × 10−4 rad s = 4.0 rad s 0.250 kg 4 ( 0.250 kg )2 2

ω= Thus the period is

T=

2π

ω

=

2π rad = 1.57 s 4.0 rad s

The amplitude at t = 0 s is x0 = A and the amplitude will be equal to e −1 A at a time given by 1 m A = Ae− ( t 2τ ) ⇒ t = 2τ = 2 = 33.3 s e b The number of oscillations in a time of 33.3 s is (33.3 s)/(1.57 s) = 21.

14.61. Model: A completely inelastic collision between the two gliders resulting in simple harmonic motion. Visualize:

Let us denote the 250 g and 500 g masses as m1 and m2 , which have initial velocities vi1 and vi2 . After m1 collides with and sticks to m2 , the two masses move together with velocity vf . Solve:

The momentum conservation equation pf = pi for the completely inelastic collision is ( m1 + m2 ) vf =

m1vi1 + m2vi2 . Substituting the given values,

( 0.750 kg ) vf = ( 0.250 kg )(1.20

m s ) + ( 0.500 kg )( 0 m s ) ⇒ vf = 0.400 m s

We now use the conservation of mechanical energy equation:

( K + U s )compressed = ( K + U s )equilibrium ⇒ 0 J + 12 kA2 = 12 ( m1 + m2 ) vf2 + 0 J ⇒ A=

m1 + m2 0.750 kg vf = ( 0.400 m s ) = 0.110 m k 10 N m

The period is

T = 2π

m1 + m2 0.750 kg = 2π = 1.72 s k 10 N m

14.62. Model: The block attached to the spring is oscillating in simple harmonic motion. Solve: (a) Because the frequency of an object in simple harmonic motion is independent of the amplitude and/or the maximum velocity, the new frequency is equal to the old frequency of 2.0 Hz. (b) The speed v0 of the block just before it is given a blow can be obtained by using the conservation of mechanical energy equation as follows: 1 2

⇒ v0 =

2 = 12 mv02 kA2 = 12 mvmax

k A = ω A = ( 2π f ) A = ( 2π )( 2.0 Hz )( 0.02 m ) = 0.25 m s m

The blow to the block provides an impulse that changes the velocity of the block:

( −20 N ) (1.0 × 10

J x = Fx Δt = Δp = mvf − mv0

−3

s ) = ( 0.200 kg ) vf − ( 0.200 kg )( 0.25 m s ) ⇒ vf = 0.150 m s

Since vf is the new maximum velocity of the block at the equilibrium position, it is equal to Aω . Thus, A=

Assess:

0.150 m s

ω

=

0.150 m s = 0.012 m = 1.19 cm 2π ( 2.0 Hz )

Because vf is positive, the block continues to move to the right even after the blow.

14.63. Model: The pendulum falls, then undergoes small-amplitude oscillations in simple harmonic motion. Visualize:

We placed the origin of the coordinate system at the bottom of the arc. Solve: We need to find the length of the pendulum. The conservation of mechanical energy equation for the pendulum’s fall is ( K + U g ) = ( K + U g ) : top

1 2

bottom

mv + mgy0 = mv12 + mgy1 ⇒ 0 J + mg ( 2 L ) = 12 m ( 5.0 m s ) + 0 J 2 0

2

1 2

1 ( 5.0 m s ) = 0.6377 m 4 g Using L = 0.6377 m, we can find the frequency f as 2

⇒L=

f =

1 2π

g 1 = L 2π

9.8 m s 2 = 0.62 Hz 0.6377 m

14.64. Model: Assume the small-angle approximation. Visualize:

Solve:

The tension in the two strings pulls downward at angle θ . Thus Newton’s second law is

∑ Fy = −2T sinθ = ma y From the geometry of the figure we can see that sin θ =

y L + y2 2

If the oscillation is small, then y  L and we can approximate sin θ ≈ y / L. Since y/L is tanθ , this approximation is equivalent to the small-angle approximation sinθ ≈ tanθ if θ  1 rad. With this approximation, Newton’s second law becomes 2T d2y d2y 2T y = ma y = m 2 ⇒ y −2T sin θ ≈ − =− 2 L dt dt mL This is the equation of motion for simple harmonic motion (see Equations 14.33 and 14.47). The constants 2T/mL are equivalent to k/m in the spring equation or g/L in the pendulum equation. Thus the oscillation frequency is f =

1 2π

2T mL

14.65. Visualize: Please refer to Figure P14.65. Solve:

The potential energy curve of a simple harmonic oscillator is described by U = 12 k ( Δx ) , where 2

Δx = x − x0 is the displacement from equilibrium. From the graph, we see that the equilibrium bond length is x0 = 0.13 nm. We can find the bond’s spring constant by reading the value of the potential energy U at a displacement Δx and using the potential energy formula to calculate k. x (nm)

0.11

Δx (nm)

U (J)

k (N/m)

0.02

0.8 × 10−19 J

400

0.03

0.10

0.04

0.09

1.9 × 10

−19

J

422

3.4 × 10−19 J

425

The three values of k are all very similar, as they should be, with an average value of 416 N/m. Knowing the spring constant, we can now calculate the oscillation frequency of a hydrogen atom on this “spring” to be

f =

1 2π

1 416 N m k = = 7.9 × 1013 Hz m 2π 1.67 × 10−27 kg

14.66. Model: Visualize:

Solve: is

Assume that the size of the ice cube is much less than R and that θ is a small angle.

The ice cube is like an object on an inclined plane. The net force on the ice cube in the tangential direction

− ( FG ) sin θ = ma = mRα = mR

d 2θ d 2θ θ ⇒ − mg sin = mR dt 2 dt 2

where α is the angular acceleration. With the small-angle approximation sinθ ≈ θ , this becomes d 2θ g = − θ = −ω 2θ dt 2 R

This is the equation of motion of an object in simple harmonic motion with a period of

T=

2π

ω

= 2π

R g

14.67.

Visualize:

Solve:

(a) Newton’s second law applied to the penny along the y-axis is

Fnet = n − mg = ma y

G Fnet is upward at the bottom of the cycle (positive a y ), so n > mg . The speed is maximum when passing G through equilibrium, but a y = 0 so n = mg . The critical point is the highest point. Fnet points down and a y is negative. If a y becomes sufficiently negative, n drops to zero and the penny is no longer in contact with the surface. (b) When the penny loses contact (n = 0), the equation for Newton’s law becomes amax = g . For simple harmonic motion,

9.8 m s 2 g = = 15.65 rad 0.040 m A ω 15.65 rad s ⇒ f = = = 2.5 Hz 2π 2π

amax = Aω 2 ⇒ ω =

14.68. Model: The vertical oscillations constitute simple harmonic motion. Visualize:

Solve:

At the equilibrium position, the net force on mass m on Planet X is: Fnet = k ΔL − mg X = 0 N ⇒

k gX = m ΔL

For simple harmonic motion k m = ω 2 , thus 2

ω2 =

2

⎛ 2π ⎞ gX g X 2π ⎛ 2π ⎞ 2 ⇒ω = = ⇒ gX = ⎜ ⎟ ( 0.312 m ) = 5.86 m s ⎟ ΔL = ⎜ 14.5 s 10 ΔL ΔL T ⎝ T ⎠ ⎝ ⎠

14.69. Model: The doll’s head is in simple harmonic motion and is damped. Solve:

(a) The oscillation frequency is

f =

1 2π

k 2 2 2 ⇒ k = m ( 2π f ) = ( 0.015 kg )( 2π ) ( 4.0 Hz ) = 9.475 N m m

The spring constant is 9.5 N/m. (b) The maximum speed is

vmax = ω A =

k 9.475 N m A= ( 0.020 m ) = 0.50 m s m 0.015 kg

(c) Using A ( t ) = A0e− bt / 2 m , we get

( 0.5 cm ) = ( 2.0 cm ) e−b(4.0 s)/(2×0.015 kg) ⇒ 0.25 = e− (133.3 s/kg)b ⇒ − (133.33 s kg ) b = ln 0.25 ⇒ b = 0.0104 kg / s

14.70. Model: The oscillator is in simple harmonic motion. Solve:

(a) The maximum displacement at time t of a damped oscillator is

xmax ( t ) = Ae −t 2τ ⇒ −

⎛ x (t ) ⎞ t = ln ⎜ max ⎟ 2τ ⎝ A ⎠

Using xmax = 0.98 A at t = 0.50 s, we can find the time constant τ to be

τ =−

0.50 s = 12.375 s 2ln ( 0.98 )

25 oscillations will be completed at t = 25T = 12.5 s. At that time, the amplitude will be xmax, 12.5 s = (10 cm ) e−12.5 s ( 2)(12.375 s ) = 6.0 cm (b) The energy of a damped oscillator decays more rapidly than the amplitude: E (t ) = E0e−1/ τ . When the energy

is 60% of its initial value, E (t )/E0 = 0.60. We can find the time this occurs as follows: ⎛ E (t ) ⎞ ⎛ E (t ) ⎞ t − = ln ⎜ ⎟ ⇒ t = −τ ln ⎜ ⎟ = − (12.375 s ) ln ( 0.60 ) = 6.3 s τ ⎝ E0 ⎠ ⎝ E0 ⎠

14.71. Model: The oscillator is in simple harmonic motion. Solve:

The maximum displacement, or amplitude, of a damped oscillator decreases as xmax (t ) = Ae− t 2τ , where

τ is the time constant. We know xmax A = 0.60 at t = 50 s, so we can find τ as follows: −

⎛ x (t ) ⎞ t 50 s = ln ⎜ max ⎟ ⇒ τ = − = 48.9 s 2τ 2ln ( 0.60 ) ⎝ A ⎠

Now we can find the time t30 at which xmax A = 0.30 :

⎛ x (t ) ⎞ t30 = −2τ ln ⎜ max ⎟ = −2 ( 48.9 s ) ln ( 0.30 ) = 118 s ⎝ A ⎠ The undamped oscillator has a frequency f = 2 Hz = 2 oscillations per second. Damping changes the oscillation frequency slightly, but the text notes that the change is negligible for “light damping.” Damping by air, which allows the oscillations to continue for well over 100 s, is certainly light damping, so we will use f = 2.0 Hz. Then the number of oscillations before the spring decays to 30% of its initial amplitude is N = f ⋅ t30 = ( 2 oscillations s ) ⋅ (118 s ) = 236 oscillations

14.72. Solve: The solution of the equation d 2 x b dx k + + x=0 dt 2 m dt m is x ( t ) = Ae − bt 2 m cos (ω t + φ0 ) . The first and second derivatives of x(t) are

dx Ab −bt / 2 m e =− cos (ω t + φ0 ) − Aω e − bt / 2 m sin (ω t + φ0 ) dt 2m ⎤ ⎞ d 2 x ⎡⎛ Ab 2 ω Ab = ⎢⎜ − Aω 2 ⎟ cos (ω t + φ0 ) + sin (ω t + φ0 ) ⎥ e −bt / 2 m dt 2 ⎣⎝ 4m 2 m ⎠ ⎦ Substituting these expressions into the differential equation, the terms involving sin(ω t + φ0 ) cancel and we obtain the simplified result

⎛ −b 2 k⎞ 2 ⎜ 2 − ω + ⎟ cos (ω t + φ0 ) = 0 m⎠ ⎝ 4m Because cos (ω t + φ0 ) is not equal to zero in general, k k b2 −b 2 −ω2 + = 0 ⇒ ω = − 2 4m m m 4m 2

14.73. Model: The two springs obey Hooke’s law. Visualize:

Solve: There are two restoring forces on the block. If the block’s displacement x is positive, both restoring forces—one pushing, the other pulling—are directed to the left and have negative values:

( Fnet ) x = ( Fsp 1 ) x + ( Fsp 2 ) x = −k1x − k2 x = − ( k1 + k2 ) x = −keff x where keff = k1 + k2 is the effective spring constant. This means the oscillatory motion of the block under the influence of the two springs will be the same as if the block were attached to a single spring with spring constant keff . The frequency of the blocks, therefore, is

f =

1 2π

keff 1 = m 2π

k1 + k2 k1 k = + 22 = 2 m 4π m 4π m

f12 + f 22

14.74. Model: The two springs obey Hooke’s law. Assume massless springs. Visualize:

Solve:

Each spring is shown separately. Note that Δx = Δx1 + Δx2 .

Only spring 2 touches the mass, so the net force on the mass is Fm = F2 on m . Newton’s third law tells us

that F2 on m = Fm on 2 and that F2 on 1 = F1 on 2 . From Fnet = ma, the net force on a massless spring is zero. Thus Fw on 1 = F2 on 1 = k1Δx1 and Fm on 2 = F1 on 2 = k2 Δx2 . Combining these pieces of information, Fm = k1Δx1 = k2 Δx2 The net displacement of the mass is Δx = Δx1 + Δx2 , so

Δx = Δx1 + Δx2 =

Fm Fm ⎛ 1 1 ⎞ k +k + = ⎜ + ⎟ Fm = 1 2 Fm k1 k2 ⎝ k1 k2 ⎠ k1k2

Turning this around, the net force on the mass is

Fm =

k1k2 kk Δx = keff Δx where keff = 1 2 k1 + k2 k1 + k2

keff , the proportionality constant between the force on the mass and the mass’s displacement, is the effective spring constant. Thus the mass’s angular frequency of oscillation is

ω=

keff 1 k1k2 = m m k1 + k2

Using ω12 = k1 / m and ω 22 = k2 / m for the angular frequencies of either spring acting alone on m, we have

ω=

( k1 / m) ( k2 / m) ω12ω 22 = ( k1 / m) + ( k2 / m) ω12 + ω 22

Since the actual frequency f is simply a multiple of ω , this same relationship holds for f: f =

f 12 f 22 f 12 + f 22

14.75. Model: The blocks undergo simple harmonic motion. Visualize:

The length of the stretched spring due to a block of mass m is ΔL1. In the case of the two block system, the spring is further stretched by an amount ΔL2 . Solve: The equilibrium equations from Newton’s second law for the single block and double block systems are

( ΔL1 ) k = mg and ( ΔL1 + ΔL2 ) k = ( 2m ) g Using ΔL2 = 5.0 cm, and subtracting these two equations, gives us

( ΔL1 + ΔL2 ) k − ΔL1k = ( 2m ) g − mg ⇒ ( 0.05 m ) k = mg ⇒

k 9.8 m/s 2 = = 196 s –2 m 0.05 m

With both blocks attached, giving total mass 2m, the angular frequency of oscillation is

ω=

k 1k 1 196 s –2 = 9.90 rad/s = = 2m 2m 2

Thus the oscillation frequency is f = ω /2π = 1.58 Hz.

14.76. Model: A completely inelastic collision between the bullet and the block resulting in simple harmonic motion. Visualize:

Solve:

(a) The equation for conservation of energy after the collision is

1 2 1 2500 N m k kA = ( mb + mB ) vf2 ⇒ vf = A= ( 0.10 m ) = 5.0 m s 2 2 1.010 kg mb + mB The momentum conservation equation for the perfectly inelastic collision pafter = pbefore is

(1.010 kg )( 5.0

( mb + mB ) vf = mbvb + mBvB m s ) = ( 0.010 kg ) vb + (1.00 kg )( 0 m s ) ⇒ vb = 5.0 × 102

ms

(b) No. The oscillation frequency k ( mb + mB ) depends on the masses but not on the speeds.

14.77. Model: The block undergoes SHM after sticking to the spring. Energy is conserved throughout the motion. Visualize:

It’s essential to carefully visualize the motion. At the highest point of the oscillation the spring is stretched upward. Solve: We’ve placed the origin of the coordinate system at the equilibrium position, where the block would sit on the spring at rest. The spring is compressed by ΔL at this point. Balancing the forces requires k ΔL = mg . The angular frequency is w2 = k/m = g/ΔL, so we can find the oscillation frequency by finding ΔL. The block hits the spring (1) with kinetic energy. At the lowest point (3), kinetic energy and gravitational potential energy have been transformed into the spring’s elastic energy. Equate the energies at these points: K1 + U1g = U 3s + U 3g ⇒ 12 mv12 + mg ΔL = 12 k (ΔL + A) 2 + mg (− A)

We’ve used y1 = ΔL as the block hits and y3 = − A at the bottom. The spring has been compressed by Δy = ΔL + A. Speed v1 is the speed after falling distance h, which from free-fall kinematics is v12 = 2 gh. Substitute this expression for v12 and mg/ΔL for k, giving

mgh + mg ΔL =

mg (ΔL + A) 2 + mg (− A) 2(ΔL)

The mg term cancels, and the equation can be rearranged into the quadratic equation (ΔL) 2 + 2h(ΔL) − A2 = 0 The positive solution is ΔL = h 2 + A2 − h = (0.030 m)2 + (0.100 m) 2 − 0.030 m = 0.0744 m Now that ΔL is known, we can find

ω=

g 9.80 m/s 2 ω = = 11.48 rad/s ⇒ f = = 1.83 Hz 0.0744 m 2π ΔL

14.78. Model: Model the bungee cord as a spring. The motion is damped SHM. Visualize:

Solve:

(a) For light damping, the oscillation period is 2

2

m ⎛ 2π ⎞ ⎛ 2π ⎞ ⇒ k = m⎜ ⎟ = (75 kg) ⎜ ⎟ = 185 N/m k ⎝ T ⎠ ⎝ 4.0 s ⎠

T = 2π

(b) The maximum speed is vmax = ω A = (2π /4.0 s)(11.0 m) = 17.3 m/s. (c) Jose oscillates about the equilibrium position at which he would hang at rest. Balancing the forces, ΔL = mg / k = (75 kg)(9.80 m/s 2 )(185 N/m) = 3.97 m. Jose’s lowest point is 11.0 m below this point, so the

bungee cord is stretched by Δymax = ΔL + A = 14.97 m. Choose this lowest point as y = 0. Because Jose is instantaneously at rest at this point, his energy is entirely the elastic potential energy of the stretched bungee cord. Initially, his energy was entirely gravitational potential energy. Equating his initial energy to his energy at the lowest point,

U lowest point = U highest point ⇒ 12 k (Δymax ) 2 = mgh h=

k (Δymax ) 2 (185 N/m)(14.97 m)2 = = 28.2 m 2mg 2(75 kg)(9.80 m/s 2)

Jose jumped 28 m above the lowest point. (d) The amplitude decreases due to damping as A(t ) = Ae −bt/2m . At the time when the amplitude has decreased from 11.0 m to 2.0 m, 2.0 m 2m ⎛ 2 ⎞ 2(75 kg) = e − bt / 2 m ⇒ t = − ln ⎜ ⎟ = − (−1.705) = 42.6 s 11.0 m b ⎝ 11 ⎠ 6.0 kg/s

With a period of 4.0 s, the number of oscillations is N osc = (42.6 s)/(4.0 s) = 10.7 oscillations.

14.79. Model: The vertical movement of the car is simple harmonic motion. Visualize:

The fact that the car has a maximum oscillation amplitude at 5 m/s implies a resonance. The bumps in the road provide a periodic external force to the car’s suspension system, and a resonance will occur when the “bump frequency” f ext matches the car’s natural oscillation frequency f 0 . Solve: Now the 5.0 m/s is not a frequency, but we can convert it to a frequency because we know the bumps are spaced every 3.0 meters. The time to drive 3.0 m at 5.0 m/s is the period:

T=

Δx 3.0 m = = 0.60 s v 5.0 m s

The external frequency due to the bumps is thus f ext = 1 T = 1.667 Hz. This matches the car’s natural frequency f 0 , which is the frequency the car oscillates up and down if you push the car down and release it. This is enough information to deduce the spring constant of the car’s suspension: 1.667 Hz =

1 2π

N k 2 ⇒ k = m ( 2π f ext ) = 131,600 m m

where we used m = mtotal = mcar + 2mpassenger = 1200 kg. When at rest, the car is in static equilibrium with Fnet = 0 N. The downward weight mtotal g of the car and passengers is balanced by the upward spring force k Δy of the suspension. Thus the compression Δy of the suspension is Δy =

mtotal g k

Initially mtotal = mcar + 2mpassenger = 1200 kg, causing an initial compression Δyi = 0.0894 m = 8.94 cm. When three

additional passengers get in, the mass increases to mtotal = mcar + 5mpassenger = 1500 kg. The final compression is Δyf = 0.1117 m = 11.17 cm. Thus the three new passengers cause the suspension to “sag” by 11.17 cm − 8.94 cm = 2.23 cm.

14.80. Model: The rod is thin and uniform. Visualize: Please refer to Figure CP14.80.

G Solve: We must derive our own equation for this combination of a pendulum and spring. For small oscillations, Fs remains horizontal. The net torque around the pivot point is ⎛ L⎞ ⎝ ⎠

τ net = Iα = − Fs L cosθ − FG ⎜ ⎟ sinθ 2 With α =

d 2θ 1 , FG = mg , Fs = k Δx = kL sin θ , and I = mL2 , dt 2 3 d 2θ 3k 3g = − sinθ cosθ − sinθ dt 2 m 2L

1 We can use sin θ cosθ = sin 2θ . For small angles, sinθ ≈ θ and sin 2θ ≈ 2θ . So 2

d 2θ ⎛ 3k 3 g ⎞ = −⎜ + ⎟θ dt 2 ⎝ m 2L ⎠ This is the same as Equations 14.33 and 14.47 with

ω=

3k 3 g + m 2L

The frequency of oscillation is thus f =

1 2π

2 3 ( 3.0 N/m ) 3 ( 9.8 m/s ) + = 1.73 Hz ( 0.200 kg ) 2 ( 0.20 m )

1 = 0.58 s. f Assess: Fewer than two oscillations per second is reasonable. The rod’s angle from the vertical must be small enough that sin 2θ ≈ 2θ . This is more restrictive than other examples, which only require that sinθ ≈ θ .

The period T =

14-1

15.1. Solve: The density of the liquid is ρ= Assess:

0.240 kg m 0.240 kg = = = 960 kg m3 V 250 mL 250 × 10−3 × 10−3 m3

The liquid’s density is near that of water (1000 kg/m3 ) and is a reasonable number.

15.2. Solve: The volume of the helium gas in container A is equal to the volume of the liquid in container B. That is, VA = VB . Using the definition of mass density ρ = m V , the above relationship becomes mA

ρA

=

mB

ρB

⇒

mHe

ρ He

=

7000 mHe

ρB

⇒ ρ B = ( 7000 ) ρ He = ( 7000 ) ( 0.18 kg m3 ) = 1260 kg m3

Referring to Table 15.1, we find that the liquid is glycerine.

15.3. Model: The density of water is 1000 kg/m3 . Visualize:

Solve:

Volume of water in the swimming pool is V = 6 m × 12 m × 3 m − 12 ( 6 m × 12 m × 2 m ) = 144 m3

The mass of water in the swimming pool is

m = ρV = (1000 kg m3 )(144 m3 ) = 1.44 × 105 kg

15.4. Model: The densities of gasoline and water are given in Table 15.1. Solve:

(a) The total mass is

mtotal = mgasoline + mwater = 0.050 kg + 0.050 kg = 0.100 kg The total volume is

Vtotal = Vgasoline + Vwater =

mgasoline

ρ gasoline

⇒ ρ avg =

+

mwater

ρ water

=

0.050 kg 0.050 kg + = 1.24 × 10−4 m3 680 kg m3 1000 kg m3

mtotal 0.100 kg = = 8.1× 102 kg m3 Vtotal 1.24 × 10−4 m3

(b) The average density is calculated as follows:

mtotal = mgasoline + mwater = ρ waterVwater + ρ gasolineVgasoline ⇒ ρ avg = Assess:

ρ waterVwater + ρgasolineVgasoline Vwater + Vgasoline

( 50 cm )(1000 kg/m 3

=

3

100 cm

3

+ 680 kg/m3 )

= 8.4 × 102 kg/m3

The above average densities are between those of gasoline and water, and are reasonable.

15.5. Model: The density of sea water is 1030 kg/m3 . Solve:

The pressure below sea level can be found from Equation 15.6 as follows: p = p0 + ρ gd = 1.013 × 105 Pa + (1030 kg m3 )( 9.80 m s 2 )(1.1 × 104 m ) = 1.013 × 105 Pa + 1.1103 × 108 Pa = 1.1113 × 108 Pa = 1.10 × 103 atm

where we have used the conversion 1 atm = 1.013 × 105 Pa. Assess: The pressure deep in the ocean is very large.

15.6. Model: The density of water is 1000 kg/m3 and the density of ethyl alcohol is 790 kg/m3 . Solve:

(a) The volume of water that has the same mass as 8.0 m 3 of ethyl alcohol is

Vwater =

mwater

ρ water

=

malcohol

ρ water

=

ρ alcoholValcohol ⎛ 790 kg/m3 ⎞ 8.0 m3 ) = 6.3 m3 =⎜ 3 ⎟( ρ water ⎝ 1000 kg/m ⎠

(b) The pressure at the bottom of the cubic tank is p = p0 + ρ water gd :

p = 1.013 × 105 Pa + (1000 kg/m3 )( 9.80 m s 2 ) ( 6.3) = 1.19 × 105 Pa 13

where we have used the relation d = (Vwater ) . 13

15.7.

Visualize:

Solve:

The pressure at the bottom of the vat is p = p0 + ρ gd = 1.3 atm. Substituting into this equation gives

1.013 × 105 Pa + ρ ( 9.8 m s 2 ) ( 2.0 m ) = (1.3) (1.013 × 105 ) Pa ⇒ ρ = 1550.5 kg m3 The mass of the liquid in the vat is

m = ρV = ρπ ( 0.50 m ) d = (1550.5 kg m3 )π ( 0.50 m ) ( 2.0 m ) = 2.4 × 103 kg 2

2

15.8. Model: Visualize:

The density of oil ρoil = 900 kg m 3 and the density of water ρ water = 1000 kg m3 .

Solve: The pressure at the bottom of the oil layer is p1 = p0 + ρ oil gd1 , and the pressure at the bottom of the water layer is

p2 = p1 + ρ water gd 2 = p0 + ρoil gd1 + ρ water gd 2

⇒ p2 = (1.013 × 10 Pa ) + ( 900 kg m 3 )( 9.80 m s 2 ) ( 0.50 m ) + (1000 kg m3 )( 9.80 m s 2 ) (1.20 m ) = 1.18 × 105 Pa 5

Assess:

A pressure of 1.18 × 105 Pa = 1.16 atm is reasonable.

15.9. Model: The density of seawater ρseawater = 1030 kg m 2 . Visualize:

The pressure outside the submarine’s window is pout = p0 + ρseawater gd , where d is the maximum safe

Solve:

depth for the window to withstand a force F. This force is F A = pout − pin , where A is the area of the window. With pin = p0 , we simplify the pressure equation to pout − p0 = Assess:

F F = ρseawater gd ⇒ d = A Aρseawater g

d=

1.0 × 106 N

π ( 0.10 m ) (1030 kg m 2 )( 9.8 m s 2 ) 2

A force of 1.0 × 106 N corresponds to a pressure of

ρ= A depth of 3 km is therefore reasonable.

F 1.0 × 106 N = = 314 atm A π ( 0.10 m )2

= 3.2 km

15.10.

Visualize:

We assume that the seal is at a radius of 5 cm. Outside the seal, atmospheric pressure presses on both sides of the cover and the forces cancel. Thus, only the 10 cm diameter opening inside the seal is relevant, not the 20 cm diameter of the cover. Solve: Within the 10 cm diameter area where the pressures differ,

Fto left = patmos A

Fto right = pgas A

where A = π r 2 = 7.85 × 10−3 m 2 is the area of the opening. The difference between the forces is

Fto left − Fto right = ( patmos − pgas ) A = (101,300 Pa − 20,000 Pa ) ( 7.85 × 10−3 m 2 ) = 0.64 kN Normally, the rubber seal exerts a 0.64 kN force to the right to balance the air pressure force. To pull the cover off, an external force must pull to the right with a force ≥ 0.64 kN.

15.11. Model: The density of water is ρ = 1000 kg m3 . Visualize: Please refer to Figure 15.17. Solve: From the figure and the equation for hydrostatic pressure, we have

p0 + ρ gh = patmos

Using p0 = 0 atm, and patmos = 1.013 × 105 Pa, we get 0 Pa + (1000 kg m3 )( 9.8 m s 2 ) h = 1.013 × 105 Pa ⇒ h = 10.3 m Assess:

This large value of h is due to water having a much smaller density than mercury.

15.12. Model: Assume that the oil is incompressible. Its density is 900 kg/m3 . Visualize: Please refer to Figure 15.19. Because the liquid is incompressible, the volume displaced in the left cylinder of the hydraulic lift is equal to the volume displaced in the right cylinder. Solve: Equating the two volumes, 2

2

⎛r ⎞ ⎛ 0.04 m ⎞ A1d1 = A2 d 2 ⇒ (π r12 ) d1 = (π r22 ) d 2 ⇒ d1 = ⎜ 2 ⎟ d 2 = ⎜ ⎟ ( 0.20 m ) = 3.2 m r 0.01 m ⎠ ⎝ ⎝ 1⎠

15.13.

Visualize:

Solve: By sucking on the top and reducing the pressure, the straw is essentially a barometer. Atmospheric pressure pushes the liquid up the straw. The length of the longest straw can be obtained from the formula p = p0 + ρ gd . If one could reduce the mouth pressure to zero, the length of the straw would be

1.013 × 105 Pa = 0 Pa + (1000 kg m3 )( 9.8 m s 2 ) d ⇒ d = 10.3 m

15.14.

Model: Visualize:

The vacuum cleaner can create zero pressure.

Solve: The gravitational force on the dog is balanced by the force resulting from the pressure difference between the atmosphere and the vacuum ( phose = 0) in the hose. The force applied by the hose is

F = ( patmos − phose ) A = patmos A = mg

(10 kg ) ( 9.8 m/s2 )

= 9.7 × 10−4 m 2 1.013 × 105 Pa 2 A ⎛d⎞ = 0.035 m = 3.5 cm. Since A = π ⎜ ⎟ , the diameter of the hose is d = 2 π ⎝2⎠ Assess: The dog will have a terrible skin rash. It’s a good thing the miscreant holding the vacuum does not have a younger sibling. ⇒ A=

15.15. Model: The buoyant force on the sphere is given by Archimedes’ principle. Visualize:

Solve:

The sphere is in static equilibrium because it is neutrally buoyant. That is,

∑F

y

= FB − FG = 0 N ⇒ ρ lVl g − ms g = 0 N

The sphere displaces a volume of liquid equal to its own volume, Vl = Vs , so

ρl =

ms m 0.0893 kg = 4 s3 = = 7.9 × 102 kg m3 3 4 Vs 3 π rs 0.030 m π ( ) 3

A density of 790 kg/m3 in Table 15.1 identifies the liquid as ethyl alcohol. Assess: If the density of the fluid and an object are equal, we have neutral buoyancy.

15.16. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:

Vcyl is the volume of the cylinder and Vw is the volume of the water displaced by the cylinder. Note that the volume displaced is only from the part of the cylinder that is immersed in water. Solve: The cylinder is in static equilibrium, so FB = FG . The buoyant force is the weight ρ wVw g of the displaced water. Thus

FB = ρ wVw g = FG = mg = ρcylVcyl g ⇒ ρ wVw = ρcylVcyl ⇒ ρ cyl = ρ w ⇒ ρ cyl = (1000 kg m3 ) Assess:

A ( 0.040 m ) = 6.7 × 102 kg m3 A ( 0.060 m )

ρcyl < ρ w for a cylinder floating in water is an expected result.

Vw Vcyl

15.17. Model: The buoyant force on the sphere is given by Archimedes’ principle. Visualize:

Solve:

The sphere is in static equilibrium. The free body diagram on the sphere shows that

1 4 = FB − T − FG = 0 N ⇒ FB = T + FG = FG + FG = FG 3 3 4 3 3 3 ⇒ ρ wVsphere g = ρsphereVsphere g ⇒ ρsphere = ρ w = (1000 kg m ) = 750 kg m3 3 4 4

∑F

y

15.18. Model: The buoyant force on the rock is given by Archimedes’ principle. Visualize:

Solve:

Because the rock is in static equilibrium, Newton’s first law is

Fnet = T + FB − ( FG )rock = 0 N ⎞ 1 1 ρ ⎛1 ⎞ ⎛ ⎞ ⎛ ⎞⎛ m g ⎞ ⎛ ⇒ T = ρ rockVrock g − ρ water ⎜ Vrock ⎟ g = ⎜ ρ rock − ρ water ⎟Vrock g = ⎜ ρ rock − ρ water ⎟ ⎜ rock ⎟ = ⎜ 1 − water ⎟ mrock g 2 2 ⎝2 ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ρ rock ⎠ ⎝ 2ρ rock ⎠ Using ρ rock = 4800 kg m3 and mrock = 5.0 kg, we get T = 44 N.

15.19. Model: The buoyant force on the aluminum block is given by Archimedes’ principle. The density of aluminum and ethyl alcohol are ρ Al = 2700 kg m3 and ρ ethyl alcohol = 790 kg m3 . Visualize:

The buoyant force FB and the tension due to the string act vertically up, and the gravitational force on the aluminum block acts vertically down. The block is submerged, so the volume of displaced fluid equals VAl , the volume of the block. Solve: The aluminum block is in static equilibrium, so

∑F

y

= FB + T − FG = 0 N ⇒ ρ fVAl g + T − ρ AlVAl g = 0 N ⇒ T = VAl g ( ρ Al − ρ f )

T = (100 × 10−6 m3 )( 9.80 m s 2 )( 2700 kg m3 − 790 kg m3 ) = 1.87 N

where we have used the conversion 100 cm3 = 100 × (10−2 m ) = 10−4 m3 . 3

Assess: The gravitational force on the aluminum block is ρ AlVAl g = 2.65 N. A similar order of magnitude for T is reasonable.

15.20. Model: The buoyant force on the steel cylinder is given by Archimedes’ principle. Visualize:

The length of the cylinder above the surface of mercury is d. Solve: The cylinder is in static equilibrium with FB = FG . Thus

FB = ρ HgVHg g = FG = mg = ρ cylVcyl g ⇒ ρ HgVHg = ρcylVcyl ⇒ ρ Hg A ( 0.20 m − d ) = ρ cyl A ( 0.20 m ) ⇒ d = 0.20 m −

ρ cyl ⎛ 7900 kg m3 ⎞ = 0.084 m = 8.4 cm ( 0.20 m ) = ( 0.20 m ) ⎜1 − 3 ⎟ ρ Hg ⎝ 13,600 kg m ⎠

That is, the length of the cylinder above the surface of the mercury is 8.4 cm.

15.21.

Model: The buoyant force is determined by Archimedes’ principle. Ignore any compression the air in the beach ball may undergo as a result of submersion. Solve: The mass of the beach ball is negligible, so the force needed to push it below the water is equal to the buoyant force. 3⎞ ⎛4 ⎞ ⎛4 FB = pw ⎜ π R 3 ⎟ g = (1000 kg/m3 ) ⎜ π ( 0.30 m ) ⎟ ( 9.8 m/s 2 ) = 1.11 kN ⎝3 ⎠ ⎝3 ⎠ Assess: It would take a 113 kg (250 lb) person to push the ball below the water. Two people together could do it. This seems about right.

15.22. Model: The buoyant force on the sphere is given by Archimedes’ principle. Visualize:

Solve: For the Styrofoam sphere and the mass not to sink, the sphere must be completely submerged and the buoyant force FB must be equal to the sum of the gravitational force on the Styrofoam sphere and the attached mass. The volume of displaced water equals the volume of the sphere, so

FB = ρ waterVwater g = (1000 kg m3 ) 43π ( 0.25 m ) ( 9.80 m s 2 ) = 641.4 N 3

( FG )Styrofoam = ρStyrofoamVStyrofoam g = (150

3 kg m3 ) ⎡ 34 π ( 0.25 m ) ⎤ ( 9.80 m s 2 ) = 96.2 N ⎣ ⎦

Because ( FG )Styrofoam + mg = FB ,

m= The mass is 56 kg.

FB − ( FG )Styrofoam g

=

641.4 N − 96.2 N = 55.6 kg 9.80 m s 2

15.23. Model: Treat the water as an ideal fluid. The pipe is a flow tube, so the equation of continuity applies. Solve:

The volume flow rate is

Q=

300 L 300 × 10−3 m3 = = 1.0 × 10−3 m3 s 5.0 min 5.0 × 60 s

Using the definition Q = vA, we get

v=

Q 1.0 × 10−3 m3 s = = 3.2 m s A π ( 0.010 m )2

15.24. Model: Treat the water as an ideal fluid. The pipe itself is a flow tube, so the equation of continuity applies.

Visualize:

Note that A1 , A2 , and A3 and v1 , v2 , and v3 are the cross-sectional areas and the speeds in the first, second, and third segments of the pipe. Solve: (a) The equation of continuity is

A1v1 = A2v2 = A3v3 ⇒ π r12v1 = π r22v2 = π r32v3 ⇒ r12v1 = r22v2 = r32v3 ⇒ ( 0.0050 m ) ( 4.0 m s ) = ( 0.010 m ) v2 = ( 0.0025 m ) v3 2

2

2

⎛ 0.0050 m ⎞ ⇒ v2 = ⎜ ⎟ ( 4.0 m s ) = 1.00 m s ⎝ 0.010 m ⎠

2

2

⎛ 0.0050 m ⎞ v3 = ⎜ ⎟ ( 4.0 m s ) = 16.0 m s ⎝ 0.0025 m ⎠

(b) The volume flow rate through the pipe is

Q = A1v1 = π ( 0.0050 m ) ( 4.0 m s ) = 3.1 × 10−4 m3 s 2

15.25. Model: Treat the water as an ideal fluid so that the flow in the tube follows the continuity equation. Visualize:

Solve:

The equation of continuity is v0 A0 = v1 A1 , where A0 = L2 and A1 = π ( 12 L ) . The above equation simplifies 2

to 2

⎛ L⎞ ⎛ 4⎞ v0 L2 = v1π ⎜ ⎟ ⇒ v1 = ⎜ ⎟ v0 = 1.27v0 2 ⎝ ⎠ ⎝π ⎠

15.26. Model: Treat the oil as an ideal fluid obeying Bernoulli’s equation. Consider the path connecting point 1 in the lower pipe with point 2 in the upper pipe a streamline. Visualize: Please refer to Figure EX15.26. Solve: Bernoulli’s equation is

p2 + 12 ρ v22 + ρ gy2 = p1 + 12 ρ v12 + ρ gy1 ⇒ p2 = p1 + 12 ρ ( v12 − v22 ) + ρ g ( y1 − y2 ) Using p1 = 200 kPa = 2.00 × 105 Pa, ρ = 900 kg m3 , y2 − y1 = 10.0 m, v1 = 2.0 m s, and v2 = 3.0 m s, we

get p2 = 1.096 × 105 Pa = 110 kPa.

15.27. Model: Turning the tuning screws on a guitar string creates tensile stress in the string. Solve: The tensile stress in the string is given by T/A, where T is the tension in the string and A is the crosssectional area of the string. From the definition of Young’s modulus,

Y=

T/A T ⎛ L⎞ ⇒ ΔL = ⎜ ⎟ ΔL / L A⎝Y ⎠

Using T = 2000 N, L = 0.80 m, A = π (0.00050 m) 2 , and Y = 20 × 1010 N/m 2 (from Table 15.3), we obtain ΔL = 0.010 m = 1.02 cm. Assess: 1.02 cm is a large stretch for a length of 80 cm, but 2000 N is a large tension.

15.28. Model: The dangling mountain climber creates tensile stress in the rope. Solve:

Young’s modulus for the rope is

Y=

F / A stress = ΔL / L strain

The tensile stress is

( 70 kg ) ( 9.8 m/s 2 ) = 8.734 × 106 Pa 2 π ( 0.0050 m ) and the strain is 0.080 m 50 m = 0.00160. Dividing the two quantities yields Y = 5.5 × 109 N m 2 .

15.29. Model: The hanging mass creates tensile stress in the wire. Solve: The force (F) pulling on the wire, which is simply the gravitational force (mg) on the hanging mass, produces tensile stress given by F/A, where A is the cross-sectional area of the wire. From the definition of Young’s modulus, we have

(π r 2 )Y ΔL = π ( 2.50 × 10−4 m ) ( 20 × 1010 N/m2 )(1.0 × 10−3 m ) = 2.0 kg mg / A ⇒m= ΔL / L gL ( 9.80 m/s 2 ) ( 2.0 m ) 2

Y=

15.30. Model: The load supported by a concrete column creates compressive stress in the concrete column. Solve: The gravitational force on the load produces tensile stress given by F/A, where A is the cross-sectional area of the concrete column and F equals the gravitational force on the load. From the definition of Young’s modulus,

Y= Assess:

2 F/A 3.0 m ⎛ F ⎞⎛ L ⎞ ⎛ 200,000 kg × 9.8 m/s ⎞ ⎛ ⎞ ⎟⎜ ⇒ ΔL = ⎜ ⎟⎜ ⎟ = ⎜ = 1.0 mm 2 10 2 ⎟ ⎜ ⎟ ΔL / L × 3 10 N/m ⎝ A ⎠⎝ Y ⎠ ⎝ ⎝ ⎠ π ( 0.25 m ) ⎠

A compression of 1.0 mm of the concrete column by a load of approximately 200 tons is reasonable.

15.31. Model: Water is almost incompressible and it applies a volume stress. Solve:

(a) The pressure at a depth of 5000 m in the ocean is

p = p0 + ρsea water g ( 5000 m ) = 1.013 × 105 Pa + (1030 kg m3 )( 9.8 m s 2 ) ( 5000 m ) = 5.057 × 107 Pa (b) Using the bulk modulus of water,

ΔV p 5.057 × 107 Pa =− =− = −0.025 V B 0.2 × 1010 Pa (c) The volume of a mass of water decreases from V to 0.975V. Thus the water’s density increases from ρ to ρ /0.975. The new density is

ρ5000m =

1030 kg/m3 = 1056 kg/m3 0.975

15.32. Solve: The pressure p at depth d in a fluid is p = p0 + ρ gd . Using 1.29 kg/m3 for the density of air, pbottom = ptop + ρ air gd ⇒ pbottom − ptop = 202 Pa = 1.99 × 10−3 atm Assuming pbottom = 1 atm, pbottom − ptop pbottom

=

1.99 × 10−3 atm = 0.2% 1 atm

15.33. Model: We assume that there is a perfect vacuum inside the cylinders with p = 0 Pa. We also assume that the atmospheric pressure in the room is 1 atm. Visualize: Please refer to Figure P15.33. 2 Solve: (a) The flat end of each cylinder has an area A = π r 2 = π ( 0.30 m ) = 0.283 m 2 . The force on each end is thus

Fatm = p0 A = (1.013 × 105 Pa )( 0.283 m 2 ) = 2.86 × 104 N The force on each end is 2.9 × 104 N. (b) The net vertical force on the lower cylinder when it is on the verge of being pulled apart is

∑F

y

= Fatm − ( FG )players = 0 N ⇒ ( FG )players = Fatm = 2.86 × 104 N ⇒ number of players =

2.86 × 104 N = 29.2 (100 kg ) ( 9.8 m s 2 )

That is, 30 players are needed to pull the two cylinders apart.

15.34. Model: Assume that the oil is incompressible and its density is 900 kg/m3 . Visualize: Please refer to Figure P15.34. Solve: (a) The pressure at depth d in a fluid is p = p0 + ρ gd . Here, pressure p0 at the top of the fluid is due

both to the atmosphere and to the gravitational force on the piston. That is, p0 = patm + ( FG ) p / A. At point A, pA = patm +

( FG )P A

+ ρ g (1.00 m − 0.30 m )

(10 kg ) ( 9.8 m s 2 ) + ( 900 kg m3 )( 9.8 2 π ( 0.02 m ) 2 ⇒ FA = pA A = (185,460 Pa )π ( 0.10 m ) = 5.8 kN = 1.013 × 105 Pa +

m s 2 ) ( 0.70 m ) = 185,460 Pa

(b) In the same way,

pB = patm + Assess:

( FG )P A

+ ρ g (1.30 m ) = 190,752 Pa ⇒ FB = 6.0 kN

FB is larger than FA , because pB is larger than pA .

15.35. Model: The tire flattens until the pressure force against the ground balances the upward normal force of the ground on the tire. Solve: The area of the tire in contact with the road is A = ( 0.15 m )( 0.13 m ) = 0.0195 m 2 . The normal force on each tire is

n=

2 FG (1500 kg ) ( 9.8 m s ) = = 3675 N 4 4

Thus, the pressure inside each tire is pinside =

3675 N 14.7 psi n = = 188,500 Pa = 1.86 atm × = 27 psi 1 atm A 0.0195 m 2

15.36.

Visualize:

Solve: (a) Because the patient’s blood pressure is 140/100, the minimum fluid pressure needs to be 100 mm of Hg above atmospheric pressure. Since 760 mm of Hg is equivalent to 1 atm and 1 atm is equivalent to 1.013 × 105 Pa, the minimum pressure is 100 mm = 1.333 × 104 Pa. The excess pressure in the fluid is due to force F pushing on the internal 6.0-mm-diameter piston that presses against the liquid. Thus, the minimum force the nurse needs to apply to the syringe is 2 F = fluid pressure × area of plunger = (1.333 × 104 Pa ) ⎡π ( 0.0030 m ) ⎤ = 0.38 N ⎣ ⎦

(b) The flow rate is Q = vA, where v is the flow speed of the medicine and A is the cross-sectional area of the needle. Thus,

v=

Q 2.0 × 10−6 m3 2.0 s = = 20 m s A π ( 0.125 × 10−3 m )2

Assess: Note that the pressure in the fluid is due to F that is not dependent on the size of the plunger pad. Also note that the syringe is not drawn to scale.

15.37. Solve: The fact that atmospheric pressure at sea level is 101.3 kPa = 101,300 N/m 2 means that the weight of the atmosphere over each square meter of surface is 101,300 N. Thus the mass of air over each square meter is m = (101,300 N)/g = (101,300 N)/(9.80 m/s 2 ) = 10,340 kg per m 2 . Multiplying by the earth’s surface

area will give the total mass. Using Re = 6.27 × 106 m for the earth’s radius, the total mass of the atmosphere is

M air = Aearth m = (4π Re2 )m = 4π (6.37 × 106 m) 2 (10,340 kg/m 2 ) = 5.27 × 1018 kg

15.38. Visualize: Let d be the atmosphere’s thickness, p the atmospheric pressure on the earth’s surface, and p0 (= 0 atm) the pressure beyond the earth’s atmosphere. Solve:

The pressure at a depth d in a fluid is p = p0 + ρ gd . This equation becomes

1 atm = 0 atm + ρair gd ⇒ d =

1 atm 1.013 × 105 Pa = = 7.95 km ρair g (1.3 kg/m3 )( 9.8 m/s 2 )

15.39. Solve: (a) We can measure the atmosphere’s pressure by measuring the height of the liquid column in a barometer, because patmos = ρ gh. In the case of the water barometer, the height of the column at a pressure of 1 atm is h=

patmos 1.013 × 105 Pa = = 10.337 m ρ water g (1000 kg/m3 )( 9.8 m/s 2 )

Because the pressure of the atmosphere can vary by 5 percent, the height of the barometer must be at least be 1.05 greater than this amount. That is, hmin = 10.85 m. (b) Using the conversion rate 1 atm = 29.92 inches of Hg, we have 29.55 inches of Hg =

29.55 × 1 atm = 0.9876 atm 29.92

The height of the water in your barometer will be

h=

patmos 0.9876 × 1.013 × 105 Pa = = 10.21 m ρ water g (1000 kg/m3 )( 9.8 m/s 2 )

15.40. Model: Oil is incompressible and has a density 900 kg/m3 . Visualize: Please refer to Figure P15.40. Solve: (a) The pressure at point A, which is 0.50 m below the open oil surface, is

pA = p0 + ρ oil g (1.00 m − 0.50 m ) = 101,300 Pa + ( 900 kg m3 )( 9.8 m s 2 ) ( 0.50 m ) = 106 kPa (b) The pressure difference between A and B is

pB − pA = ( p0 + ρ gd B ) − ( p0 + ρ gd A ) = ρ g (d B − d A ) = (900 kg/m3 )(9.8 m/s 2 )(0.50 m) = 4.4 kPa Pressure depends only on depth, and C is the same depth as B. Thus pC − pA = 4.4 kPa also, even though C isn’t directly under A.

15.41. Model: Assume that oil is incompressible and its density is 900 kg/m3 . Visualize: Please refer to Figure P15.41. Solve: (a) The hydraulic lift is in equilibrium and the pistons on the left and the right are at the same level. Equation 15.11, therefore, simplifies to Fleft piston Fright piston ( FG )student ( FG )elephant = ⇒ = 2 2 Aleft piston Aright piston π ( rstudent ) π ( relephant )

⎛ (F ) ⇒ rstudent = ⎜ G student ⎜ ( FG )elephant ⎝

⎞ ⎟ ( relephant ) = ⎟ ⎠

( 70 kg ) g 1.0 m = 0.2415 m ( ) (1200 kg ) g

The diameter of the piston the student is standing on is therefore 2 × 0.2415 m = 0.48 m. (b) From Equation 15.13, we see that an additional force ΔF is required to increase the elephant’s elevation through a distance d 2 . That is,

ΔF = ρ g ( Aleft piston + Aright piston ) d 2

2 2 ⇒ ( 70 kg ) ( 9.8 m s 2 ) = ( 900 kg m 3 )( 9.8 m s 2 )π ⎡( 0.2415 m ) + (1.0 m ) ⎤ d 2 ⎣ ⎦ ⇒ d 2 = 0.0234 m The elephant moves 2.3 cm.

15.42. Model: Assume that the oil is incompressible and its density is 900 kg/m3 . Visualize:

Solve:

The pressures p1 and p2 are equal. Thus,

p0 +

F1 F F F = p0 + 2 + ρ gh ⇒ 1 = 2 + ρ gh A1 A2 A1 A2

With F1 − m1 g , F2 = 4m2 g , A1 = π r12 , and A2 = π r2 2 , we have 12

⎞ m1 g 4m2 g ⎛ 4m g ⎞ ⎛ m g = + ρ gh ⇒ r2 = ⎜ 2 ⎟ ⎜ 1 2 − ρ gh ⎟ r π r12 π r22 π π ⎝ ⎠ ⎝ 1 ⎠ Using

m1 = 55 kg, m2 = 110 kg, r1 = 0.08 m, ρ = 900 kg/m3 ,

and

−1 2

h = 1.0 m,

the

r2 = 0.276 m. The diameter is 55 cm. Assess: Both pistons are too small to hold the people as shown, but the ideas are correct.

calculation

yields

15.43. Model: Assume that the oil is incompressible. Visualize:

Solve:

When the force F1 balances the gravitational force mg, the pressure p2 is related to the pressure p1 as

p1 = p2 + ρ gh When F1 is increased to F1′, the weight is raised higher through a distance d 2 and the left piston is lowered through a distance d1. The pressures p′2 and p1′ are now related through

p1′ = p′2 + ρ g ( h + d1 + d 2 ) Subtracting these two equations,

p1′ − p1 = ( p′2 − p2 ) + ρ g ( d1 + d 2 )

Because p1 = F1 A1 , p1′ = F1′ A1 , and p2 = p′2 = mg A2 , the above equation simplifies to

( F1′ − F1 )

A1 = ρ g ( d1 + d 2 ) ⇒ ΔF = ρ g ( A1d1 + A1d 2 )

Since the oil is incompressible, A1d1 = A2 d 2 . The equation for ΔF thus becomes ΔF = ρ g ( A2 d 2 + A1d 2 ) = ρ g ( A1 + A2 ) d 2

15.44. Model: Water and mercury are incompressible and immiscible liquids. Visualize:

The water in the left arm floats on top of the mercury and presses the mercury down from its initial level. Because points 1 and 2 are level with each other and the fluid is in static equilibrium, the pressure at these two points must be equal. If the pressures were not equal, the pressure difference would cause the fluid to flow, violating the assumption of static equilibrium. Solve: The pressure at point 1 is due to water of depth d w = 10 cm:

p1 = patmos + ρ w gd w Because mercury is incompressible, the mercury in the left arm goes down a distance h while the mercury in the right arm goes up a distance h. Thus, the pressure at point 2 is due to mercury of depth d Hg = 2h :

p2 = patmos + ρ Hg gd Hg = patmos + 2 ρ Hg gh Equating p1 and p2 gives patmos + ρ w gd w = patmos + 2 ρ Hg gh ⇒ h =

1 ρw 1 1000 kg m3 dw = (10 cm ) = 3.7 mm 2 ρ Hg 2 13,600 kg m3

The mercury in the right arm rises 3.7 mm above its initial level.

15.45. Model: Glycerin and ethyl alcohol are incompressible and do not mix. Visualize:

Solve: The alcohol in the left arm floats on top of the denser glycerin and presses the glycerin down distance h from its initial level. This causes the glycerin to rise distance h in the right arm. Points 1 and 2 are level with each other and the fluids are in static equilibrium, so the pressures at these two points must be equal:

p1 = p2 ⇒ p0 + ρ eth gd eth = p0 + ρ gly gd gly ⇒ ρ eth g (20 cm) = ρ gly g (2h) h=

1 ρ eth 1 790 kg/m3 (20 cm) = (20 cm) = 6.27 cm 2 ρgly 2 1260 kg/m3

You can see from the figure that the difference between the top surfaces of the fluids is Δy = 20 cm − 2h = 20 cm − 2(6.27 cm) = 7.46 cm ≈ 7.5 cm

15.46.

Model: The water is in hydrostatic equilibrium. Visualize: Please refer to figure P15.46. Solve: (a) Can 2 has moved down with respect to Can 1 since the water level in Can 2 has risen. Since the total volume of water stays constant, the water level in Can 1 has fallen by the same amount. The water level is equalized in the two cans at the middle of the height change, so the change in height of the water is half the relative change in height of the cans. Can 2 has moved relative to Can 1 (6.5 cm − 5.0 cm) × 2 = 3.0 cm down. (b) The water level in Can 1 has fallen by the same amount. The new level is 5.0 − 1.5 cm = 3.5 cm Assess: The two cans are an inexpensive method of measuring relative changes in height.

15.47.

Visualize:

The figure shows a dam with water height d. We chose a coordinate system with the origin at the bottom of the dam. The horizontal slice has height dy, width w, and area dA = wdy. The slice is at the depth of d − y. Solve: (a) The water exerts a small force dF = pdA = pwdy on this small piece of the dam, where p = ρ g (d − y ) is the pressure at depth d − y. Altogether, the force on this small horizontal slice at position y is dF = ρ gw(d − y ) dy. Note that a force from atmospheric pressure is not included. This is because atmospheric pressure exerts a force on both sides of the dam. The total force of the water on the dam is found by adding up all the small forces dF for the small slices dy between y = 0 m and y = d . This summation is expressed as the integral d

Ftotal

d

d

1 ⎞ 1 ⎛ = ∫ dF = ∫ dF = ρ gw∫ ( d − y ) dy = ρ gw ⎜ yd − y 2 ⎟ = ρ gwd 2 2 ⎝ ⎠0 2 all slices 0 0

(b) The total force is

Ftotal =

1 2

(1000 kg

m3 )( 9.8 m s 2 ) (100 m )( 60 m ) = 1.76 × 109 N 2

15.48.

Visualize:

The figure shows an aquarium tank with water at a height d. We chose a coordinate system with the origin at the bottom of the tank. The slice has a height dy, length l, and area dA = ldy. It is at depth d − y. Solve: (a) Pressure in excess of atmospheric pressure of the water on the bottom is pbottom = ρ water gd . Atmospheric pressure is ignored because it exerts an equal force on both sides of the bottom. Therefore, the force of the water on the bottom is

Fbottom = pbottom ( lw ) = ( ρ water gd )( lw ) = (1000 kg m3 )( 9.8 m s 2 ) ( 0.40 m )(1.0 m )( 0.35 m ) = 1.37 kN (b) The water exerts a small force dF = pdA = pldy on a small piece of the window, where p = ρ g (d − y ) is the excess pressure at a depth of d − y. The force on this small horizontal slice at position y is

dF = ρ gl ( d − y ) dy. The total force on the front window is d

Ftotal = ∫ dF = ∫ ρ water gL ( d − y ) dy = ρ water gl ( yd − 12 y 2 ) = 0

=

1 2

(1000 kg

d

0

1 ρ water gld 2 2

m3 )( 9.8 m s 2 ) (1.00 m )( 0.40 m ) = 0.78 kN 2

15.49.

Visualize:

The figure shows a small column of air of thickness dz, of cross-sectional area A = 1 m 2 , and of density ρ ( z ). The column is at a height z above the surface of the earth. Solve: (a) The atmospheric pressure at sea level is 1.013 × 105 Pa. That is, the weight of the air column with a

1 m 2 cross section is 1.013 × 105 N. Consider the weight of a 1 m 2 slice of thickness dz at a height z. This slice has volume dV = Adz = (1 m 2 ) dz, so its weight is dw = ( ρ dV ) g = ρ g (1 m 2 )dz = ρ 0e− z/z0 g (1 m 2 )dz. The total weight of the 1 m 2 column is found by adding all the dw. Integrating from z = 0 to z = ∞, ∞

w = ∫ ρ 0 g (1 m 2 )e− z z0 dz 0

= ( − ρ 0 g (1 m 2 ) z0 ) ⎡⎣e− z z0 ⎤⎦

∞ 0

= ρ 0 g (1 m 2 ) z0 Because w = 101,300 N = ρ 0 g (1 m 2 ) z0 ,

z0 =

101,300 N = 8.08 × 103 m (1.28 kg/m3 )( 9.8 m/s2 ) (1.0 m2 )

(b) Using the density at sea level from Table 15.1,

ρ = (1.28 kg m3 ) e − z /(8.08×10

3

This is 82% of ρ 0 .

m)

= (1.28 kg m3 ) e −1600 m/(8.08×10

3

m)

= 1.05 kg m3

15.50. Model: The buoyant force on the ceramic statue is given by Archimedes’ principle. Visualize:

Solve: The gravitational force on the statue is the 28.4 N registered on the scale in air. In water, the gravitational force on the statue is balanced by the sum of the buoyant force FB and the spring’s force on the statue. That is,

( FG )statue = FB + Fspring on statue ⇒ 28.4 N = ρ wVstatue g + 17.0 N ⇒ Vstatue = ⇒ ρstatue =

( mstatue g ) ρ w = ( 28.4 N ) (1000 kg (11.4 N ) (11.4 N )

m3 )

11.4 N mstatue = g ρw ρstatue

= 2.49 × 103 kg m3

15.51. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:

Solve: (a) Initially, as it floats, the cylinder is in static equilibrium, with the buoyant force balancing the gravitational force on the cylinder. The volume of displaced liquid is Ah, so

FB = ρ liq ( Ah) g = FG Force F pushes the cylinder down distance x, so the submerged length is h + x and the volume of displaced liquid is A(h + x ). The cylinder is again in equilibrium, but now the buoyant force balances both the gravitational force and force F. Thus

FB = ρ liq ( A(h + x)) g = FG + F Since ρ liq ( Ah) g = FG , we’re left with

F = ρ liq Agx (b) The amount of work dW done by force F to push the cylinder from x to x + dx is dW = Fdx = ( ρ liq Agx)dx.

To push the cylinder from xi = 0 m to xf = 10 cm = 0.10 m requires work xf

xf

xi

xi

W = ∫ F dx = ρliq Ag ∫ x dx = 12 ρ liq Ag ( xi2 − xf2 ) = 12 (1000 kg/m3 )π (0.020 m) 2 (9.8 m/s 2 )(0.10 m) 2 = 0.62 J

15.52. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:

Let d1 be the length of the cylinder in the less-dense liquid with density ρ1 , and d 2 be the length of the cylinder in the more-dense liquid with density p2 . Solve: The cylinder is in static equilibrium, so

∑F

y

= FB1 + FB2 − FG = 0 N ⇒ ρ1 ( Ad1 ) g + ρ 2 ( Ad 2 ) g = ρ A ( d1 + d 2 ) g ⇒

ρ1 ρ d1 + d 2 = ( d1 + d 2 ) ρ2 ρ2

Since l = d1 + d 2 ⇒ d1 = l − d 2 , we can simplify the above equation to obtain

⎛ ρ − ρ1 ⎞ d2 = ⎜ ⎟l ⎝ ρ 2 − ρ1 ⎠ That is, the fraction of the cylinder in the more dense liquid is f = ( ρ − ρ1 ) ( ρ 2 − ρ1 ) . Assess:

As expected f = 0 when ρ is equal to ρ1 , and f = 1 when ρ = ρ 2 .

15.53. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:

Solve:

The tube is in static equilibrium, so

∑F

y

= FB − ( FG ) tube − ( FG )Pb = 0 N ⇒ ρ liquid A ( 0.25 m ) g = ( 0.030 kg ) g + ( 0.250 kg ) g ⇒ ρ liquid =

Assess:

( 0.280 kg ) ( 0.280 kg ) = = 8.9 × 102 A ( 0.25 m ) π ( 0.020 m )2 ( 0.25 m )

This is a reasonable value for a liquid.

kg m 3

15.54. Model: Archimedes’ Principle determines the buoyant force. Visualize:

Solve: The plastic hemisphere will hold the most weight when its rim is at the surface of the water. The buoyant force balances the gravitational force on the bowl and rock.

∑F

y

= FB − ( FG )rock − ( FG )bowl = 0 N

Thus 3⎞ ⎛ 1 ⎞⎛ 4 mg = pwVbowl g − m bowl g = (1000 kg/m3 ) ⎜ ⎟⎜ π ( 0.040 m ) ⎟ ( 9.8 m/s 2 ) − ( 0.021 kg ) ( 9.8 m/s 2 ) ⇒ m = 0.155 kg 2 3 ⎝ ⎠⎝ ⎠

Assess:

Putting a rock as big as 155 g in an 8 cm diameter bowl before it sinks is reasonable.

15.55. Model: The buoyant force is determined by Archimedes’ principle. The spring is ideal. Visualize:

Solve: The spring is stretched by the same amount that the cylinder is submerged. The buoyant force and spring force balance the gravitational force on the cylinder.

∑F

y

= FB + FS − mg = 0 N ⇒ pw Ayg + ky = mg

y=

(1.0 kg ) ( 9.8 m/s 2 ) mg = pw Ag + k (1000 kg/m3 )π ( 0.025 m )2 ( 9.8 m/s 2 ) + 35 N/m

= 0.181 m = 18.1 cm Assess: This is difficult to assess because we don’t know the height h of the cylinder and can’t calculate it without the density of the metal material.

15.56. Model: The balloon displaces air, so the air exerts an upward buoyant force on the balloon. The buoyant force on the balloon is given by Archimedes’ principle.

Visualize:

FB is the buoyant force on the balloon, ( FG ) b is the gravitational force on the balloon, ( FG ) He is the gravitational force on the helium gas, and FG is the gravitational force on the mass tied to the balloon. Solve:

The volume of the balloon Vballoon = 34 π ( 0.10 m ) = 4.1888 × 10−3 m3 . We have 3

FB = ρ air gVballoon = (1.28 kg m3 )( 9.8 m s 2 )( 4.188 × 10−3 m3 ) = 0.05254 N

( FG )He = ρ He gVballoon = ( 0.18 kg m3 )( 9.8 m s 2 )( 4.188 × 10−3 m3 ) = 0.007387 N ( FG )b = ( 0.0010 kg ) g = 0.0098 N For the balloon to be in static equilibrium,

FB = ( FG )He + ( FG )b + FG = 0.05254 N=0.01719 N + mg ⇒ m = 0.0036 kg = 3.6 g

15.57. Model: The buoyant force on the can is given by Archimedes’ principle. Visualize:

The length of the can above the water level is d, the length of the can is L, and the cross-sectional area of the can is A. Solve: The can is in static equilibrium, so

∑F

y

= FB − ( FG )can − ( FG ) water = 0 N ⇒ ρ water A ( L − d ) g = ( 0.020 kg ) g + mwater g

The mass of the water in the can is

355 × 10 ⎛V ⎞ mwater = ρ water ⎜ can ⎟ = (1000 kg m3 ) 2 2 ⎝ ⎠

−6

m3

= 0.1775 kg

⇒ ρ water A ( L − d ) = 0.020 kg + 0.1775 kg = 0.1975 kg ⇒ d − L = − Because

Vcan = π ( 0.031 m ) L = 355 × 10−6 m3 , 2

L = 0.1176 m.

Using

0.1975 kg = 0.0654 m ρ water A this

value

of

L,

we

get

d = 0.0522 m ≈ 5.2 cm. Assess: d L = 5.22 cm 11.76 cm = 0.444, thus 44.4% of the length of the can is above the water surface. This is reasonable.

15.58. Model: The buoyant force on the boat is given by Archimedes’ principle. Visualize:

The minimum height of the boat that will enable the boat to float in perfectly calm water is h. Solve: The boat barely floats if the water comes completely to the top of the sides. In this case, the volume of displaced water is the volume of the boat. Archimedes’ principle in equation form is FB = ρ wVboat g . For the boat

to float FB = ( FG )boat . Let us first calculate the gravitational force on the boat:

( FG )boat = ( FG )bottom + 2 ( FG )side 1 + 2 ( FG )side 2 , where

( FG )bottom = ρsteelVbottom g = ( 7900 kg m3 )( 5 × 10 × 0.02 m3 ) g = 7900g N ( FG )side 1 = ρsteelVside 1g = ( 7900 kg m3 )( 5 × h × 0.005 m3 ) g = 197.5gh N ( FG )side 2 = ρsteelVside 2 g = ( 7900 kg m3 )(10 × h × 0.005 m3 ) g = 395gh ⇒ ( FG )boat = ⎡⎣7900g + 2 (197.5gh ) + 2 ( 395gh ) ⎤⎦ N = ( 7900+1185h ) g N Going back to the Archimedes’ equation and remembering that h is in meters, we obtain

ρ wVboat g = ( 7900 + 1185h ) g N ⇒ (1000 ) (10 × 5 × ( h + 0.020 ) ) = 7900 + 1185h ⇒ h = 14.1 cm

15.59. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. There is a streamline connecting point 1 in the wider pipe with point 2 in the narrower pipe.

Visualize:

Solve:

Bernoulli’s equation, Equation 15.28, relates the pressure, water speed, and heights at points 1 and 2:

p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 For no height change y1 = y2 . The flow rate is given as 5.0 L/s, which means

Q = v1 A1 = v2 A2 = 5.0 × 10−3 m3 s ⇒ v1 =

5.0 × 10−3 m3 s

π ( 0.05 m )

2

= 0.6366 m s

2

⇒ v2 = v1

A1 ⎛ 0.050 m ⎞ = ( 0.6366 m s ) ⎜ ⎟ = 2.546 m s A2 ⎝ 0.025 m ⎠

Bernoulli’s equation now simplifies to

p1 + 12 ρ v12 = p2 + 12 ρ v22 2 2 ⇒ p1 = p2 + 12 ρ ( v22 − v12 ) = 50 × 103 Pa + 12 (1000 kg m3 ) ⎡( 2.546 m s ) − ( 0.6366 m s ) ⎤ = 53 kPa ⎣ ⎦

Assess:

Reducing the pipe size reduces the pressure because it makes v2 > v1.

15.60. Model: The two pipes are identical. Visualize:

Solve:

The water speed is the same in both pipes. The flow rate Q = 3.0 × 106 L/min = 2 ( vA )

( 3.0 ×10 )(10 ) ⎛⎜⎝ 601 ⎞⎟⎠ m /s 6

Q ⇒v= = 2A

−3

2π (1.5 m )

3

2

= 3.5 m/s

15.61. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. A streamline begins in the bigger size pipe and ends at the exit of the narrower pipe.

Visualize: Please see Figure P15.61. Let point 1 be beneath the standing column and point 2 be where the water exits the pipe. Solve: (a) The pressure of the water as it exits into the air is p2 = patmos . (b) Bernoulli’s equation, Equation 15.28, relates the pressure, water speed, and heights at points 1 and 2:

p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 ⇒ p1 − p2 = 12 ρ ( v22 − v12 ) + ρ g ( y2 − y1 ) From the continuity equation,

v1 A1 = v2 A2 = ( 4 m s ) ( 5 × 10−4 m 2 ) ⇒ v1 (10 × 10−4 m 2 ) = 20 × 10−4 m3 s ⇒ v1 = 2.0 m s Substituting into Bernoulli’s equation, 2 2 m3 ) ⎡( 4.0 m s ) − ( 2.0 m s ) ⎤ + (1000 kg m 3 ) ( 9.8 m s )( 4.0 m ) ⎣ ⎦ = 6000 Pa + 39,200 Pa = 45 kPa

p1 − p2 = p1 − patmos =

1 2

(1000 kg

But p1 − p2 = ρ gh, where h is the height of the standing water column. Thus h=

45 × 103 Pa = 4.6 m (1000 kg/m3 )( 9.8 m/s2 )

15.62. Model: Treat the water as an ideal fluid obeying Bernoulli’s equation. A streamline begins at the faucet and continues down the stream.

Visualize:

The pressure at point 1 is p1 and the pressure at point 2 is p2 . Both p1 and p2 are atmospheric pressure. The velocity and the area at point 1 are v1 and A1 and they are v2 and A2 at point 2. Let the distance of point 2 below point 1 be d. Solve: The flow rate 3

Q = v1 A =

2.0 × 10−4 m 3 2.0 × 1000 × 10−6 m3 2 = 1.0 m/s = 2.0 × 10−4 m ⇒ v1 = 2 s 10 s π ( 0.0080 m )

Bernoulli’s equation at points 1 and 2 is

p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 ⇒ ρ gd = 12 ρ ( v22 − v12 )

From the continuity equation,

v1 A1 = v2 A2 ⇒ (1.0 m s )π ( 8.0 × 10−3 m ) = v2π ( 5.0 × 10−3 m ) ⇒ v2 = 2.56 m s 2

2

Going back to Bernoulli’s equation, we have 2 2 gd = 12 ⎡( 2.56 m s ) − (1.0 m s ) ⎤ ⇒ d = 0.283 m ≈ 28 cm ⎣ ⎦

15.63. Model: Treat the air as an ideal fluid obeying Bernoulli’s equation. Solve: (a) The pressure above the roof is lower due to the higher velocity of the air. (b) Bernoulli’s equation, with yinside ≈ youtside , is 2

pinside = poutside + 12 ρair v 2 ⇒ Δp =

1 1 ⎛ 130 × 1000 m ⎞ ρ air v 2 = (1.28 kg m3 ) ⎜ ⎟ = 835 Pa 2 2 ⎝ 3600 s ⎠

The pressure difference is 0.83 kPa (c) The force on the roof is ( Δp ) A = ( 835 Pa )( 6.0 m × 15.0 m ) = 7.5 × 104 N. The roof will blow up, because pressure inside the house is greater than pressure on the top of the roof.

15.64. Model: The ideal fluid obeys Bernoulli’s equation. Visualize: Please refer to Figure P15.64. There is a streamline connecting point 1 in the wider pipe on the left with point 2 in the narrower pipe on the right. The air speeds at points 1 and 2 are v1 and v2 and the crosssectional area of the pipes at these points are A1 and A2 . Points 1 and 2 are at the same height, so y1 = y2 . Solve:

The volume flow rate is Q = A1v1 = A2v2 = 1200 × 10−6 m3 s. Thus v2 =

1200 × 10−6 m3 s = 95.49 m s π (0.0020 m) 2

v1 =

3 1200 × 10−6 m

π ( 0.010 m )

2

s = 3.82 m s

Now we can use Bernoulli’s equation to connect points 1 and 2:

p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 ⇒ p1 − p2 = 12 ρ ( v22 − v12 ) + ρ g ( y2 − y1 ) =

1 2

(1.28 kg

2 2 m3 ) ⎡( 95.49 m s ) − ( 3.82 m s ) ⎤ + 0 Pa = 5.83 kPa ⎣ ⎦

Because the pressure above the mercury surface in the right tube is p2 and in the left tube is p1 , the difference in the pressures p1 and p2 is ρ Hg gh. That is, p1 − p2 = 5.83 kPa = ρ Hg gh ⇒ h =

5.83 × 103 Pa = 4.4 cm (13,600 kg m3 )(9.8 m s 2 )

15.65. Model: The ideal fluid (that is, air) obeys Bernoulli’s equation. Visualize: Please refer to Figure P15.65. There is a streamline connecting points 1 and 2. The air speeds at points 1 and 2 are v1 and v2 , and the cross-sectional areas of the pipes at these points are A1 and A2 . Points 1 and 2 are at the same height, so y1 = y2 . Solve: (a) The height of the mercury is 10 cm. So, the pressure at point 2 is larger than at point 1 by

ρ Hg g ( 0.10 m ) = (13,600 kg m3 )( 9.8 m s 2 ) ( 0.10 m ) = 13,328 Pa ⇒ p2 = p1 + 13,328 Pa

Using Bernoulli’s equation,

p1 + 12 ρair v12 + ρ air gy1 = p2 + 12 ρ air v22 + ρ air gy2 ⇒ p2 − p1 = 12 ρair ( v12 − v22 ) ⇒ v12 − v22 =

2 ( p2 − p1 )

ρ air

=

2 (13,328 Pa )

(1.28 kg/m ) 3

= 20,825 m 2 s 2

From the continuity equation, we can obtain another equation connecting v1 and v2 :

π ( 0.005 m ) A A1v1 = A2v2 ⇒ v1 = 2 v2 = v = 25 v2 2 2 A1 π ( 0.001 m ) 2

Substituting v1 = 25v2 in the Bernoulli equation, we get

( 25 v2 )

2

− v22 = 20,825 m 2 s 2 ⇒ v2 = 5.78 m s

Thus v2 = 5.8 m/s and v1 = 25v2 = 144 m/s. (b) The volume flow rate A2v2 = π ( 0.0050 m ) ( 5.78 m s ) = 4.5 × 10−3 m3 s. 2

15.66. Model: Treat the water as an ideal fluid that obeys Bernoulli’s equation. There is a streamline from the top of the water to the hole. Visualize: Please refer to Figure P15.66. The top of the water (at y1 = h) and the hole (at y2 = y ) are at atmospheric pressure. The speed of the water at the top is zero because the tank is kept filled. Solve: (a) Bernoulli’s equation connecting the two points is

0 + ρ gh = 12 ρ v 2 + ρ gy ⇒ v = 2 g ( h − y ) (b) For a particle shot horizontally from a height y with speed v, the range can be found using kinematic equations. For the y-motion, using t0 = 0 s, we have y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ 0 m = y + ( 0 m s ) t1 + 12 ( − g ) t12 ⇒ t1 = 2 y g 2

For the x-motion, x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 ) ⇒ x = 0 m + vt1 + 0 m ⇒ x = v 2 y g 2

(c) Combining the results of (a) and (b), we obtain x = 2g ( h − y ) 2 y g = 4 y ( h − y )

To find the maximum range relative to the vertical height, 1 dx h =0⇒ ⎡⎣ 4 ( h − y ) − 4 y ⎤⎦ = 0 ⇒ y = 2 dy 4y(h − y)

With y = 12 h, the maximum range is h⎞ ⎛ h ⎞⎛ xmax = 4 ⎜ ⎟⎜ h − ⎟ = h 2⎠ ⎝ 2 ⎠⎝

15.67. Model: Treat water as an ideal fluid that obeys Bernoulli’s equation. There is a streamline connecting the top of the tank with the hole. Visualize: Please refer to Figure P15.67. We placed the origin of the coordinate system at the bottom of the tank so that the top of the tank (point 1) is at a height of h + 1.0 m and the hole (point 2) is at a height h. Both points 1 and 2 are at atmospheric pressure. Solve: (a) Bernoulli’s equation connecting points 1 and 2 is

p1 + 12 ρ v12 + ρ gy1 = p2 + 12 ρ v22 + ρ gy2 ⇒ patmos + 12 ρ v12 + ρ g ( h + 1.0 m ) = patmos + 12 ρ v22 + ρ gh ⇒ v22 − v12 = 2g (1.0 m ) = 19.6 m 2 s 2 Using the continuity equation A1v1 = A2v2 ,

π ( 2.0 × 10 m ) ⎛A ⎞ v2 v1 = ⎜ 2 ⎟ v2 = v2 = 2 A 250,000 π (1.0 m ) ⎝ 1⎠ −3

2

Because v1  v2 , we can simply put v1 ≈ 0 m/s. Bernoulli’s equation thus simplifies to

v22 = 19.6 m 2 s 2 ⇒ v2 = 4.43 m s Therefore, the volume flow rate through the hole is Q = A2v2 = π ( 2.0 × 10−3 m ) ( 4.43 m s ) = 5.56 × 10−5 m3 s = 3.3 L min 2

(b) The rate at which the water level will drop is v2 4.43 m s = = 1.77 × 10−2 mm s = 1.06 mm min v1 = 250,000 250,000 Assess: Because the hole through which water flows out of the tank has a diameter of only 4.0 mm, a drop in the water level at the rate of 1.06 mm/min is reasonable.

15.68. Model: The aquarium creates tensile stress. Solve:

Weight of the aquarium is

FG = mg = ρ waterVg = (1000 kg m3 )(10 m3 )( 9.8 m s 2 ) = 9.8 × 104 N

where we have used the conversion 1 L = 10−3 m 3 . The weight supported by each wood post is 14 (9.8 × 104 N) = 2.45 × 104 N. The cross-sectional area of each post is A = ( 0.040 m ) = 1.6 × 10−3 m 2 . Young’s modulus for the 2

wood is F/A FL = ΔL / L AΔL 4 2.45 10 N 0.80 m × ( ) ( ) FL ⇒ ΔL = = = 1.23 × 10−3 m = 1.23 mm 2 10 −3 AY (1.6 × 10 m )(1× 10 N/m 2 ) Y = 1 × 1010 N m 2 =

Assess:

A compression of 1.23 mm due to a weight of 2.45 × 104 N is reasonable.

15.69. Model: The water pressure applies a volume stress to the sphere. Solve:

The volume change is ΔV = −1.0 × 10−3 V . The volume stress is

F ΔV −1.0 × 10−3V = p = −B = − ( 7 × 1010 N m 2 ) = 7.0 × 107 Pa A V V Using p = p0 + ρ gh, we get

7.0 × 107 Pa = 1.013 × 105 Pa + (1030 kg m3 )( 9.8 m s 2 ) d ⇒ d = 6.9 km Assess: A pressure of 7.0 × 107 Pa = 690 atm causes only a volume change of 0.1%. This is reasonable because liquids and solids are nearly incompressible.

15.70. Model: Pressure applies a volume stress to water in the cylinder. Solve:

The volume strain of water due to the pressure applied is

ΔV p 2 × 106 Pa =− =− = −1.0 × 10−3 V B 0.20 × 1010 Pa ⇒ ΔV = V ′ − V = − (1.0 × 10−3 )(1.30 m 3 ) = −1.30 × 10−3 m3 = −1.3 L As the safety plug on the top of the cylinder bursts, the water comes back to atmospheric pressure. The volume of water that comes out is 1.30 L.

15.71. Model: Air is an ideal gas and obeys Boyle’s law. Visualize: Please refer to Figure CP15.71. h is the length of the air column when the mercury fills the cylinder to the top. A is the cross-sectional area of the cylinder. Solve: For the column of air, Boyle’s law is p0V0 = p1V1 , where p0 and V0 are the pressure and volume before any mercury is poured, and p1 and V1 are the pressure and volume when mercury fills the cylinder above the air. Using p1 = p0 + ρ Hg g (1.0 m − h ) , Boyle’s law becomes

p0V0 = ⎡⎣ p0 + ρ Hg g (1.0m − h ) ⎤⎦ V1 ⇒ p0 A (1.0 m ) = ⎡⎣ p0 + ρ Hg g (1.0m − h ) ⎤⎦ Ah p 1.013 × 105 Pa p0 (1.0m − h ) = ρ Hg g (1.0m − h ) h ⇒ h = 0 = = 0.76 m = 76 cm ρ Hg g (13,600 kg/m3 )( 9.8 m/s 2 )

15.72. Model: The buoyant force on the cone is given by Archimedes’ principle. Visualize:

Solve: It may seem like we need the formula for the volume of a cone. You can use that formula if you know it, but it isn’t essential. The volume is clearly the area of the base multiplied by the height multiplied by some constant. That is, the cone shown above has V = aAd , where a is some constant. But the radius of the base is r = d tan α , where α is the angle of the apex of the cone, and A = π r 2 , making A proportional to d 2 . Thus the volume of a cone of height d is V = cd 3 , where c is a constant. Because the cone is floating in static equilibrium, we must have FB = FG . The cone’s density is ρ 0 , so the gravitational force on it is FG = ρ0Vg = ρ0cl 3 g. The buoyant force is the gravitational force on the displaced fluid. The volume of displaced fluid is the full volume of the cone minus the volume of the cone of height h above the water, or Vdisp = cl 3 = ch3 . Thus

FB = ρf Vdisp g = ρ f c(l 3 − h3 ) g , and the equilibrium condition is 1/ 3

⎛ h3 ⎞ h ⎛ ρ ⎞ FB = FG ⇒ ρ f c(l 3 − h3 ) g = ρ 0cl 3 g ⇒ ρ f ⎜1 − 3 ⎟ = ρ 0 ⇒ = ⎜1 − 0 ⎟ l ⎠ l ⎝ ρf ⎠ ⎝

15.73. Model: The grinding wheel is a uniform disk. We will use the model of kinetic friction and hydrostatics. Visualize: Please refer to figure CP15.73. Solve: This is a three-part problem. First find the desired angular acceleration, then use that to find the force applied by each brake pad, then finally the needed oil pressure. The angular acceleration required to stop the wheel is found using rotational kinematics. 1 Δω ωf − ωi 0 rad/s − 900 × 2π × 60 rad/s α= = = = −18.85 rad/s 2 5.0 s Δt Δt Each brake pad applies a frictional force f k = μ k n to the wheel. The normal force is equal to the force applied by the piston by Newton’s third law. Rotational dynamics can be used to find the magnitude of the force. f k is applied 12 cm from the rotation axis on both sides of the disk. τ net = Iα

⎛1 ⎞ −2 f k ( 0.12 m ) = ⎜ MR 2 ⎟α ⎝2 ⎠

(15 kg )( 0.13 m ) ( −18.85 rad/s2 ) MR 2α ⇒n= = = 16.6 N −4μ k ( 0.12 m ) −4 ( 0.60 )( 0.12 m ) 2

The oil pressure required to generate this much force at each brake pad is F 16.6 N = 53 kPa p= = A π ( 0.010 m )2 relative to atmospheric pressure. Assess: The required oil pressure is about half an atmosphere, which is quite reasonable.

15.74. Model: The buoyant force on the cylinder is given by Archimedes’ principle. Visualize:

A is cross-sectional area of the cylinder. Solve: (a) The gravitational force on the cylinder is FG = ρ0 ( Al ) g and the weight of the displaced fluid is

FB = ρf ( Ah ) g. In static equilibrium, FB = FG and we can write

ρf Ahg = ρ0 Alg ⇒ h = ( ρ0 ρf ) l (b) Now the volume of the displaced liquid is A(h − y ). Applying Newton’s second law in the y-direction,

∑F

y

= − FG + FB = − ρ0 Alg + ρf A ( h − y ) g

Using ρf Ahg = ρ0 Alg from part (a), we find

( Fnet ) y = − ρf Agy

(c) The result in part (b) is F = −ky, where k = ρ f Ag . This is Hooke’s law. (d) Since the cylinder’s equation of motion is determined by Hooke’s law, the angular frequency for the resulting simple harmonic motion is ω = k m , and the period is

T=

2π

ω

= T = 2π

m m ρ0 Al h = 2π = 2π = 2π k ρf Ag ρf Ag g

where we have used the expression for h from part (a). (e) The oscillation period for the 100-m-tall iceberg ( pice = 917 kg/m3 ) in sea water is T = 2π

ρ 0l h = 2π = 2π ρf g g

( 917 kg/m ) (100 m ) = 18.9 s (1030 kg/m )( 9.8 m/s ) 3

3

2

15.75. Model: A streamline connects every point on the surface of the liquid to a point in the drain. The drain diameter is much smaller than the tank diameter (r  R ). Visualize:

Solve: The pressures at the surface and drain (points 1 and 2) are equal to one atmosphere. When the liquid is a depth y, Bernoulli’s principle comparing points 1 and 2 is 1 1 p1 + ρ v12 + ρ gy = p2 + ρ v22 + ρ g ( 0 ) 2 2 ⇒ v22 = v12 + 2 gy The flow rate through the drain is the same as through a horizontal layer in the tank: A Q = v1 A1 = v2 A2 ⇒ v1 = v2 2 A1 Thus the velocity of the liquid through the drain is 2

⎛ A ⎞ 2 gy v22 = ⎜ v2 2 ⎟ + 2 gy ⇒ v2 = π r2 ⎝ A1 ⎠ 1− π R2 Since r  R, v2 ≈ 2 gy , and the flow rate through the drain is

Q ≈ π r 2 2 gy The flow rate gives the volume of liquid flowing out per unit time. The inverse gives the time needed for a unit volume of liquid to flow out. For the volume of liquid to decrease by dV requires a time 2 − R 2 dy dV − (π R dy ) = = dt = Q π r 2 2 gy 2g r 2 y Note that dV < 0 implies the volume of liquid is decreasing. Integrating both sides from the initial condition (t = 0, y = d ) to the final condition (t , y = 0) yields 0

R2 2 R 2 1 2 0 R 2 2d − 12 = − = 2 y dy y d r g 2 gr 2 ∫d 2 gr 2 Assess: The time for the tank to drain depends on the ratio of the cross-sectional areas of the tank to drain, which makes sense, as well as on the strength of the free-fall acceleration. Note that if g were larger (say we were on Jupiter) the time to drain would be shorter, while if the tank were taller (larger d) the time would be longer. t=−

15.76.

Visualize:

The column of air has a cross-sectional area A. We chose the origin of the coordinate system at z = 0 m and denote the vertical axis as the z-axis. Solve: (a) The pressure at height z is pz . The pressure at a height z + dz is

pz + dz = pz −

weight of air column of height dz ρ Adzg = pz − = pz − ρ gdz A A

(b) The change in pressure is dp = pz + dz − pz = − ρ gdz. The sign is minus because pressure decreases as z increases. (c) From the ideal gas law,

p

ρ

=

p0

ρ0

⇒ρ=

p ρ0 p0

The result for part (b) thus becomes dp = − ( p ρ 0 p0 ) gdz. (d) To find the pressure p at a height z, we rewrite the result of part (c) as

⎛ρ ⎞ dp = − ⎜ 0 g ⎟ dz p ⎝ p0 ⎠ Integrating both sides p

⎛ p⎞ dp ρ0 g z ρg = − dz ⇒ ln ⎜ ⎟ = − 0 z ⇒ p = p0e − ρ0 gz ∫p p ∫ p p p0 0 0 ⎝ 0⎠ 0 where z0 = p0 ρ 0 g . (e) The scale height is z0 =

p0

= p0e− z z0

(1.013 × 105 Pa ) = 8076 m. p0 = ρ0 g (1.28 kg m3 )( 9.8 m s 2 )

(f) A table showing values of p = p0e − z z0 = e − z (8076 m ) atm at selected values of z follows.

z (m) 0 1000 2000 4000 6000 8000 10,000 12,000 14,000 15,000

p (atm) 1 0.884 0.781 0.609 0.476 0.371 0.290 0.226 0.177 0.156

15-1

Fluids and Elasticity

15-2

16.1. Model: Recall the density of water is 1000 kg/m3. Solve: The mass of lead mPb = ρ PbVPb = (11,300 kg m3 )( 2.0 m3 ) = 22,600 kg . For water to have the same mass its volume must be

Vwater = Assess:

mwater

ρ water

=

22,600 kg = 22.6 m3 1000 kg m3

Since the lead is 11.3 times as dense we expect the water to take 11.3 times the volume.

16.2. Model: Assume the nucleus is spherical. Solve:

The volume of the uranium nucleus is

V = 34 π R 3 = 43 π ( 7.5 × 10−15 m ) = 1.767 × 10−42 m3 3

The density of the uranium nucleus is

ρ nucleus = Assess:

mnucleus 4.0 × 10−25 kg = = 2.3 × 1017 kg m3 Vnucleus 1.767 × 10−42 m3

This density is extremely large compared to the typical density of materials.

16.3. Solve: The volume of the aluminum cube is 10−3 m3 and its mass is mAl = ρ AlVAl = ( 2700 kg m3 )(1.0 × 10−3 m3 ) = 2.7 kg

The volume of the copper sphere with this mass is

VCu =

m 4π 2.7 kg 3 = 3.027 × 10−4 m3 ( rCu ) = Cu = 3 ρ Cu 8920 kg m3 ⎡ 3 ( 3.027 × 10−4 m3 ) ⎤ ⎥ ⇒ rCu = ⎢ 4π ⎢⎣ ⎥⎦

1

3

= 0.042 m

The diameter of the copper sphere is 0.0833 m = 8.33 cm. Assess: The diameter of the sphere is a little less than the length of the cube, and this is reasonable considering the density of copper is greater than the density of aluminum.

16.4. Model: The volume of a hollow sphere is V= Solve:

4p 3 ( rout − rin3 ) 3

We are given m = 0.690 kg, rout = 0.050 m, and we know that for aluminum r = 2700 kg/m3 .

Solve the above equation for rin . 3 rin = 3 rout −

3 = 3 rout −

V p

4 3

m/r 4 p 3

= 3 0.050 m3 −

0.690 kg/2700 kg/m3 4 p 3

= 0.040 m So the inner diameter is 8.0 cm. Assess: We are happy that the inner diameter is less than the outer diameter, and in a reasonable range.

16.5. Solve: The volume of the aluminum cube V = 8.0 × 10−6 m3 and its mass is M = ρV = (2700 kg/m3 )(8.0 × 10−6 m3 ) = 0.0216 kg = 21.6 g One mole of aluminum (27Al) has a mass of 27 g. The number of atoms is ⎛ 6.02 × 1023 atoms ⎞⎛ 1 mol ⎞ 23 N =⎜ ⎟⎜ ⎟ ( 21.6 g ) = 4.8 × 10 atoms 1 mol 27 g ⎠ ⎝ ⎠⎝ Assess:

This is almost one mole of atoms, which is a reasonable value.

16.6. Solve: The volume of the copper cube is 8.0 × 10−6 m3 and its mass is M = ρV = (8920 kg/m3 )(8.0 × 10−6 m3 ) = 0.07136 kg = 71.36 g Because the atomic mass number of Cu is 64, one mole of Cu has a mass of 64 g. The number of moles in the cube is ⎛ 1 mol ⎞ n=⎜ ⎟ ( 71.36 g ) = 1.1 mol ⎝ 64 g ⎠ Assess:

This answer is in the same ballpark as the previous exercise.

16.7. Solve: (a) The number density is defined as N V , where N is the number of particles occupying a volume V. Because Al has a mass density of 2700 kg/m3, a volume of 1 m3 has a mass of 2700 kg. We also know that the molar mass of Al is 27 g/mol or 0.027 kg/mol. So, the number of moles in a mass of 2700 kg is

⎛ 1 mol ⎞ 5 n = ( 2700 kg ) ⎜ ⎟ = 1.00 × 10 mol ⎝ 0.027 kg ⎠ The number of Al atoms in 1.00 × 105 mols is

N = nN A = (1.00 × 105 mol )( 6.02 × 1023 atoms mol ) = 6.02 × 1028 atoms Thus, the number density is N 6.02 × 1028 atoms = = 6.02 × 1028 atoms m3 V 1 m3 (b) Pb has a mass of 11,300 kg in a volume of 1 m3. Since the atomic mass number of Pb is 207, the number of moles in 11,300 kg is

⎛ 1 mole ⎞ n = (11,300 kg ) ⎜ ⎟ ⎝ 0.207 kg ⎠ The number of Pb atoms is thus N = nNA, and hence the number density is atoms N nN A ⎛ 11,300 kg ⎞ ⎛ 23 atoms ⎞ (1 mol ) = =⎜ = 3.28 ×1028 ⎟⎜ 6.02 ×10 ⎟ 3 V V mol ⎠ 1 m m3 ⎝ 0.207 kg ⎠ ⎝ Assess:

We expected to get very large numbers like this.

16.8. Solve: The mass density is ρ = M/V . The mass M of the sample is related to the number of atoms N and the mass m of each atom by M = Nm. Combining these, the atomic mass is

m=

M ρV ρ 1750 kg/m3 = = = = 3.986 × 10−26 kg/atom N N N / V 4.39 × 1028 atoms/m3

the atomic mass in m = A u, where A is the atomic mass number. Thus A=

m 3.986 × 10−26 kg = = 24 1 u 1.661 × 10−27 kg

The element’s atomic mass number is 24. Assess: This is a reasonable answer for an isotope of neon (although neon is a gas at normal temperatures), sodium, or magnesium.

4 16.9 Model: Assume the gold is shaped into a solid sphere of volume V = π r 3 . 3 Visualize: We want to know D = 2r. Because the atomic mass number of gold is 197, 1.0 mol of gold has a mass of 197 g or 0.197 kg. Table 16.1 gives ρ = 19,300 kg/m3 . Solve:

4 3 M 3 0.197 kg =3 = 0.0135 m = 1.35 cm V = π r3 = M / ρ ⇒ r = 3 3 4π ρ 4π 19,300 kg/m3 D = 2r = 2(1.35 cm) = 2.70 cm. Assess: This is about the right size for a chunk that contains one mole of material.

16.10. Solve: The mass of mercury is ⎛ 10−6 m3 ⎞ M = ρV = (13,600 kg m3 )(10 cm3 ) ⎜ = 0.136 kg = 136 g 3 ⎟ ⎝ 1 cm ⎠ and the number of moles is

n=

M 0.136 g = = 0.6766 mol M mol 201 g mol

The mass of aluminum with 0.6766 mol of Al is ⎛ 27 g ⎞ M = ( 0.6766 mol ) M mol = ( 0.6766 mol ) ⎜ ⎟ = 18.27 g = 0.01827 kg ⎝ mol ⎠ This mass M of aluminum corresponds to a volume of V=

M

ρ

=

0.01827 kg = 6.8 × 10−6 m3 = 6.8 cm3 2700 kg m3

Assess: We expected an answer in the same order of magnitude. The size of atoms doesn’t vary as much as the density of atoms from element to element.

16.11. Solve: The lowest temperature is TF = 95 TC + 32° ⇒ −127°F = 95 TC + 32° ⇒ TC = −88°C ⇒ Tk = ( −88.3 + 273) K = 185 K In the same way, the highest temperature is 136°F = 95 TC + 32° ⇒ TC = 58°C = 331 K Assess:

On the absolute scale the highest recorded temperature is not quite twice the lowest.

16.12. Solve: Let TF = TC = T : TF = 95 TC + 32° ⇒ T = 95 T + 32° ⇒ T = −40° That is, the Fahrenheit and the Celsius scales give the same numerical value at −40° . Assess: It is usually unnecessary to specify the scale when the temperature is reported as −40° .

16.13. Model: A temperature scale is a linear scale.

Solve: (a) We need a conversion formula for °C to °Z, analogous to the conversion of °C to °F. Since temperature scales are linear, TC = aTZ + b, where a and b are constants to be determined. We know the boiling

point of liquid nitrogen is 0°Z and –196°C. Similarly, the melting point of iron is 1000°Z and 1538°C. Thus −196 = 0a + b 1538 = 1000a + b

From the first, b = −196°. Then from the second, a = (1538 + 196)/1000 = 1734/1000. Thus the conversion is TC = (1734/1000)TZ − 196°. Since the boiling point of water is TC = 100°C, its temperature in °Z is ⎛ 1000 ⎞ TZ = ⎜ ⎟ (100° + 196°) = 171°Z ⎝ 1734 ⎠

(b) A temperature TZ = 500°Z is

⎛ 1734 ⎞ TC = ⎜ ⎟ 500° − 196° = 671°C = 944 K ⎝ 1000 ⎠

16.14. Solve: (a) The triple point of water is T = 0.01°C and p = 0.006 atm. Thus TF = 95 TC + 32° = 95 ( 0.01) + 32 = 32.02°F p = 0.006 atm ×

1.013 × 105 Pa = 608 Pa 1 atm

(b) The triple point of carbon dioxide is T = −56°C and p = 5 atm. Thus

TF = 95 TC + 32° = 95 ( −56° ) + 32° = −68.8°F

⎛ 1.013 × 105 Pa ⎞ 5 p = 5.0 atm = ( 5.0 atm ) ⎜ ⎟ = 5.06 × 10 Pa 1 atm ⎝ ⎠

16.15. Model: Treat the gas in the container as an ideal gas. Solve:

From the ideal-gas law pV = nRT , the pressure of the gas is

p= Assess:

nRT ( 3.0 mol )( 8.31 J mol K ) ⎡⎣( 273 − 120 ) K ⎤⎦ = = 1.9 × 106 Pa = 19 atm V ( 2.0 × 10−3 m3 )

19 atm is a high pressure, but not unreasonable.

16.16. Model: Treat the nitrogen gas in the closed cylinder as an ideal gas. Solve: (a) The density before and after the compression are ρ before = m1 V1 and ρ after = m2 V2 . Noting that m1 = m2 and V2 = 12 V1 ,

ρ after m V1 = = 2 ⇒ ρ after = 2 ρ before ρ before V2 m The mass density has changed by a factor of 2. (b) The number of atoms in the gas is unchanged, implying that the number of moles in the gas remains the same; hence the number density is unchanged.

16.17. Model: Treat the gas in the sealed container as an ideal gas. Solve:

(a) From the ideal gas law equation pV = nRT , the volume V of the container is

V=

nRT ( 2.0 mol )( 8.31 J mol K ) ⎡⎣( 273 + 30 ) K ⎤⎦ = = 0.050 m3 1.013 × 105 Pa p

Note that pressure must be in Pa in the ideal-gas law. (b) The before-and-after relationship of an ideal gas in a sealed container (constant volume) is p1V p2V T ( 273 + 130 ) K = 1.3 atm = ⇒ p2 = p1 2 = (1.0 atm ) T1 T2 T1 ( 273 + 30 ) K

Note that gas-law calculations must use T in kelvins.

16.18. Model: Treat the gas as an ideal gas in the sealed container. Solve:

(a) For p1 = p0 , the before-and-after relationship of an ideal gas in a sealed container is

V1 V0 T 473 K = ⇒ V1 = V0 1 = V0 = 1.27V0 T1 T0 T0 373 K where T0 = (273 + 100) K and T1 = (273 + 200) K. (b) When the Kelvin temperature is doubled, T1 = 2T0 = 2(373 K) = 746 K and the above equation becomes

V1 = V0

T1 746 K = V0 = 2V0 T0 373 K

Assess: When we use the Kelvin scale we expect the volume to double when the temperature doubles (if the pressure is kept constant).

16.19. Model: Treat the air in the compressed-air tank as an ideal gas. Solve:

(a) From the ideal-gas law pV = nRT , the number of moles n is

pV p (π r h ) = = RT RT 2

n=

⎛ 1.013 × 105 Pa ⎞ ⎡ 2 ⎟ ⎣π ( 0.075 m ) ( 0.50 m ) ⎤⎦ 1 atm ⎝ ⎠ = 55.1 mol ≈ 55 mol ( 8.31 J mol K ) ⎣⎡( 273 + 20 ) K ⎦⎤

(150 atm ) ⎜

(b) At STP, the ideal-gas law yields V=

nRT ( 55.1 mol )( 8.31 J mol K )( 273 K ) = = 1.234 m3 ≈ 1.2 m3 p 1.013 × 105 Pa

Assess: Because the volume of the compressed air tank is (π r 2 )h = 8.8357 × 10−3 m3 , the volume at STP is 140 times the volume of the tank.

16.20. Model: Treat the oxygen gas in the cylinder as an ideal gas. Solve:

(a) The number of moles of oxygen is

n=

M 50 g = = 1.563 mol ≈ 1.6 mol M mol 32 g mol

(b) The number of molecules is

N = nN A = (1.563 mol ) ( 6.02 × 1023 mol−1 ) = 9.41× 1023 ≈ 9.4 ×1023 (c) The volume of the cylinder V = π r 2 L = π ( 0.10 m ) ( 0.40 m ) = 1.257 × 10−2 m3 . Thus, 2

N 9.41 × 1023 = = 7.5 × 1025 m −3 V 1.257 × 10−2 m3 (d) From the ideal-gas law pV = nRT we can calculate the absolute pressure to be

p=

nRT (1.563 mol )( 8.31 J mol K )( 293 K ) = = 303 kPa V 1.257 × 10−2 m3

where we used T = 20°C = 293 K. But a pressure gauge reads gauge pressure:

pg = p − 1 atm = 303 kPa − 101 kPa = 202 kPa ≈ 200 kPa

16.21. Model: Treat the helium gas in the sealed cylinder as an ideal gas. 2 Solve: The volume of the cylinder is V = π r 2 h = π ( 0.05 m ) ( 0.30 m ) = 2.356 × 10−3 m3. The gauge pressure of the gas is 120 psi ×

1 atm 1.013 × 105 Pa × = 8.269 ×105 Pa, so the absolute pressure of the gas is 14.7 psi 1 atm

8.269 ×105 Pa + 1.013 × 105 Pa = 9.282 × 105 Pa. The temperature of the gas is T = (273 + 20) K = 293 K. The number of moles of the gas in the cylinder is n=

5 −3 3 pV ( 9.282 × 10 Pa )( 2.356 × 10 m ) = = 0.898 mol RT ( 8.31 J mol K )( 293 K )

(a) The number of atoms is

N = nN A = ( 0.898 mol ) ( 6.02 × 1023 mol−1 ) = 5.41× 1023 atoms ≈ 5.4 × 1023 atoms (b) The mass of the helium is

M = nM mol = ( 0.898 mol )( 4 g mol ) = 3.59 g = 3.59 × 10−3 kg ≈ 3.6 × 10−3 kg (c) The number density is

N 5.41 × 1023 atoms = = 2.3 × 1026 atoms m3 V 2.356 × 10−3 m 3 (d) The mass density is

ρ=

M 3.59 × 10−3 kg = = 1.5 kg m3 V 2.356 × 10−3 m3

16.22. Model: The gas is assumed to be ideal and it expands isothermally. Solve: (a) Isothermal expansion means the temperature stays unchanged. That is T2 = T1. (b) The before-and-after relationship of an ideal gas under isothermal conditions is

⎛V ⎞ p p1V1 p2V2 V = ⇒ p2 = p1 1 = p1 ⎜ 1 ⎟ = 1 T1 T1 V2 ⎝ 2V1 ⎠ 2

16.23. Model: The gas is assumed to be ideal. Solve: (a) Isochoric means the volume stays unchanged. That is V2 = V1. (b) The before-and-after relationship of an ideal gas under isochoric conditions is

⎛1 ⎞ ⎜ p1 ⎟ T p1V1 p2V2 p2 = ⇒ T2 = T1 = T1 ⎜ 3 ⎟ = 1 T1 T1 p1 ⎜⎜ p1 ⎟⎟ 3 ⎝ ⎠

16.24. Model: The gas is assumed to be ideal, and since the container is rigid V2 = V1. Solve:

Convert both temperatures to the Kelvin scale. p1V1 p2V2 T ⎛ 283 K ⎞ = ⇒ p2 = p1 2 = 3 atm ⎜ ⎟ = 3.1 atm T1 T1 T1 ⎝ 275 K ⎠

Assess: bit.

On the absolute scale the temperature only went up a little bit, so we expect the pressure to rise a little

16.25. Model: The rigid sphere’s volume does not change, so this is an isochoric process. The air is assumed to be an ideal gas. Solve: (a) When the valve is closed, the air inside is at p1 = 1 atm and T1 = 100°C. The before-and-after relationship of an ideal gas in the closed sphere (constant volume) is

⎛T ⎞ p1V p2V ( 273 + 0 ) K = 0.73 atm = ⇒ p2 = p1 ⎜ 2 ⎟ = (1.0 atm ) T1 T2 ( 273 + 100 ) K ⎝ T1 ⎠ (b) Dry ice is CO2. From Figure 16.4, we can see that the solid-gas transition line gives a temperature of −78°C when p = 1 atm. Cooling the sphere to –78°C gives

⎛T ⎞ ( 273 − 78) K = 0.52 atm p2 = p1 ⎜ 2 ⎟ = (1.0 atm ) 373 K ⎝ T1 ⎠

16.26. Model: Assume the gas is an ideal gas. Visualize: The pressure in the gas must exert exactly enough upward force to counteract the gravitational force on the piston; this is true at both temperatures ( p2 = p1 = p ).

The initial volume of the cylinder is V1 = pr 2 h1 = p ( 0.12m ) ( 0.84m ) = 0.038m3 . The initial temperature is 2

T1 = 303°C + 273 = 576K. Solve: (a)

p=

F mg ( 20kg ) ( 9.80m/s = = 2 A pr 2 p ( 0.12m )

2

) = 4333Pa < 4300Pa

(b) Knowing the initial temperature, pressure, and volume allows us to compute the number of moles of gas (which will stay constant).

n=

( 4333Pa ) ( 0.038m3 ) pV1 = = 0.0344mol < 0.034mol RT1 ( 8.31J/mol ⋅ K )( 576K )

Now apply the n just found and the new temperature (T2 = 15°C + 273 = 288K) to find V2 . V2 =

nRT2 ( 0.0344 mol )( 8.31 J/mol ⋅ K )( 288 K ) = = 0.019 m3 p 4333 Pa

Solve V2 = pr 2 h2 for h2 .

h2 =

V2 0.019m3 = = 0.42m pr 2 p ( 0.12m )2

The new height of the piston is 42 cm. Assess:

Due to a coincidence in the data we could have used the shortcut that T2 = 12 T1 on the absolute scale to

deduce that V2 = 12 V1 1 h2 = 12 h1 .

16.27. Model: In an isochoric process, the volume of the container stays unchanged. Argon gas in the container is assumed to be an ideal gas. Solve: (a) The container has only argon inside with n = 0.1 mol, V1 = 50 cm3 = 50 × 10−6 m3 , and

T1 = 20°C = 293 K. This produces a pressure p1 =

nRT ( 0.1 mol )( 8.31 J mol K )( 293 K ) = = 4.87 × 106 Pa = 4870 kPa ≈ 4900 kPa V1 50 × 10−6 m3

An ideal gas process has p2V2 /T2 = p1V1 / T1. Isochoric heating to a final temperature T2 = 300°C = 573 K has V2 = V1 , so the final pressure is

p2 =

V1 T2 573 p1 = 1 × × 4870 kPa = 9520 kPa ≈ 9500 kPa V2 T1 293

Note that it is essential to express temperatures in kelvins. (b)

Assess:

All isochoric processes will be a straight vertical line on a pV diagram.

16.28. Model: The isobaric heating means that the pressure of the argon gas stays unchanged. Solve:

(a) The container has only argon inside with n = 0.1 mol, V1 = 50 cm3 = 50 × 10−6 m3 , and

T1 = 20°C = 293 K. This produces a pressure p1 =

nRT1 ( 0.1 mol )( 8.31 J mol K )( 293 K ) = = 4.87 × 106 Pa = 4870 kPa ≈ 4900 kPa V1 50 × 10−6 m3

An ideal gas process has p2V2 /T2 = p1V1/T1. Isobaric heating to a final temperature T2 = 300°C = 573 K has p2 = p1 , so the final volume is

V2 =

p1 T2 573 V1 = 1 × × 50 cm3 = 97.8 cm3 ≈ 98 cm3 p2 T1 293

(b)

Assess:

All isobaric processes will be a straight horizontal line on a pV diagram.

16.29. Model: In an isothermal expansion, the temperature stays the same. The argon gas in the container is assumed to be an ideal gas. Solve: (a) The container has only argon inside with n = 0.1 mol,

V1 = 50 cm3 = 50 × 10−6 m3 , and

T1 = 20°C = 293 K. This produces a pressure

p1 =

nRT1 (0.1 mol)(8.31 J/mol K) (293 K) = = 4.87 ×106 Pa = 12.02 atm ≈ 12 atm V1 50 ×10−6 Pa

An ideal-gas process obeys p2V2 /T2 = p1V1/T1. Isothermal expansion to V2 = 200 cm3 gives a final pressure p2 = (b)

T2 V1 200 p1 = 1 × × 12.02 atm = 48 atm T1 V2 50

16.30. Model: Assume the gas to be an ideal gas. Solve: (a) Because the volume stays unchanged, the process is isochoric. (b) The ideal-gas law p1V1 = nRT1 gives

T1 =

5 −6 3 p1V1 ( 3 × 1.013 × 10 Pa )(100 × 10 m ) = = 914 K = 641°C nR ( 0.0040 mol )( 8.31 J mol K )

The final temperature T2 is calculated as follows for an isochoric process:

p1 p2 p ⎛ 1 atm ⎞ ⇒ T2 = T1 2 = ( 914 K ) ⎜ = ⎟ = 305 K = 32°C p1 T1 T2 ⎝ 3 atm ⎠ Assess:

All straight vertical lines on a pV diagram represent isochoric processes.

16.31. Model: Assume that the gas is an ideal gas. Solve: (a) The graph shows that the pressure is inversely proportional to the volume. The process is isothermal. (b) From the ideal-gas law,

T1 =

5 −6 3 p1V1 ( 3 × 1.013 × 10 Pa )(100 × 10 m ) = = 914 K = 641°C nR ( 0.0040 mol )( 8.31 J mol K )

T2 is also 914 K, because the process is isothermal. (c) The before-and-after relationship of an ideal gas under isothermal conditions is p1V1 = p2V2 ⇒ V2 = V1

p1 ⎛ 3 atm ⎞ 3 = (100 cm3 ) ⎜ ⎟ = 300 cm p2 ⎝ 1 atm ⎠

16.32. Model: Assume that the gas is ideal. Solve: (a) Because the process is at a constant pressure, it is isobaric. (b) For an ideal gas at constant pressure,

V2 V1 V 100 cm3 ⇒ T2 = T1 2 = ⎡⎣( 273 + 900 ) K ⎤⎦ = 391 K = 118°C = V1 300 cm3 T2 T1 (c) Using the ideal-gas law p2V2 = nRT2 ,

n=

5 −6 3 p2V2 ( 3 × 1.013 × 10 Pa )(100 × 10 m ) = = 9.35 × 10−3 mol RT2 ( 8.31 J mol K )( 391 K )

Assess: All straight horizontal lines on a pV diagram represent isobaric processes.

16.33. Visualize:

Solve: Suppose we have a 1 m × 1 m × 1 m block of copper of mass M containing N atoms. The atoms are spaced a distance L apart along all three axes of the cube. There are Nx atoms along the x-edge of the cube, Ny atoms along the y-edge, and Nz atoms along the z-edge. The total number of atoms is N = Nx Ny Nz . If L is

expressed in meters, then the number of atoms along the x-edge is N x = (1 m)/L. Thus, 13

N=

⎛ 1 m3 ⎞ 1 m 1 m 1 m 1 m3 × × = 3 ⇒L=⎜ ⎟ L L L L ⎝ N ⎠

This relates the spacing between atoms to the number of atoms in a 1-meter cube. The mass of the large cube of copper is

M = ρ CuV = ( 8920 kg m 3 )(1 m3 ) = 8920 kg But M = mN , where m = 64 u = 64 × (1.661 × 10−27 kg ) is the mass of an individual copper atom. Thus, N=

M 8920 kg = = 8.39 × 1028 atoms m 64 × (1.661 × 10−27 kg ) 13

⎛ 1 m3 ⎞ ⇒ L=⎜ = 2.28 × 10−10 m = 0.228 nm 28 ⎟ 8.39 10 × ⎝ ⎠ Assess:

This is a reasonable interatomic spacing in a crystal lattice.

16.34. Solve: The volume of the cube associated with each atom is Vatom = ( 0.227 × 10−9 m ) = 1.1697 × 10−29 m3 3

The volume of a mole of atoms is

V = Vatom N A = (1.1697 × 10−29 m3 )( 6.02 × 1023 mol−1 ) = 7.0416 × 10−6 m3 mol Thus, the mass of a mole of atoms is

M mol = V ρ = ( 7.0416 m3 mol )( 7950 kg m3 ) = 0.056 kg/mol = 56 g/mol The atomic mass number of the element is 56. Assess: This is likely the most common isotope of iron.

1.0 g . 1.0 cm3 There are 10 protons in each water molecule. 1 mol of water molecules has a mass of 18 g.

16.35. Model: Assume the density of the liquid water is Visualize: Solve:

23 ⎛ 1.0 g ⎞ ⎛ 1 mol ⎞ ⎛ 6.02 × 10 molecules ⎞ ⎛ 10 protons ⎞ 26 1.0L = 1000cm3 ⎜ ⎜ ⎟⎜ ⎜ ⎟ ⎟ ⎟ = 3.3 × 10 protons 3 1 mol ⎝ 1.0 cm ⎠ ⎝ 18 g ⎠⎝ ⎠ ⎝ 1 molecule ⎠

Assess: This is an unimaginably large number, but reasonable considering the data.

16.36. Model: Air is an ideal gas. The pressure at sea level is 1 atm. Solve:

From the ideal gas law pV = NkBT , the number density is

N p 1.013 ×105 Pa = = = 2.5 ×1025 molecules / m3 V kBT (1.38 ×10−23 J/K)(293 K) Assess:

Each cubic meter of air holds an unimaginably large number of molecules.

16.37. Model: Assume the gas in the solar corona is an ideal gas. Solve:

The number density of particles in the solar corona is N V . Using the ideal-gas equation,

pV = NkBT ⇒

N p = V kBT

N ( 0.03 Pa ) = = 1.1 × 1015 particles m3 −23 V (1.38 × 10 J K )( 2 × 106 K ) Assess:

This answer is a lot smaller than the one in the previous problem.

16.38. Model: Assume the gas in the evacuated volume is an ideal gas. Solve:

The number density of particles is N V . Using the ideal-gas equation,

pV = NkBT ⇒

N p = V kBT

The pressure is

p = (1 × 10−10 mm of Hg ) × ⇒

1 atm 1.013 × 105 Pa × = 1.33 × 10−8 Pa 760 mm of Hg 1 atm

N 1.33 × 10−8 Pa = = 3.3 × 1012 m −3 =3.3 × 106 cm −3 V (1.38 × 10−23 J/K)(293 K)

Assess: Even the best vacuum we can achieve in the laboratory contains 3 million molecules per cubic centimeter.

16.39. Model: Assume that the gas in the vacuum chamber is an ideal gas. Solve:

(a) The fraction is

pvacuum chamber 1.0 × 10−10 mm of Hg = = 1.3 × 10−13 patmosphere 760 mm of Hg (b) The volume of the chamber V = π ( 0.20 m ) ( 0.30 m ) = 0.03770 m3 . From the ideal-gas equation 2

pV = NkBT , the number of molecules of gas in the chamber is N=

−13 5 3 pV (1.32 × 10 )(1.013 × 10 Pa )( 0.03770 m ) = = 1.2 × 1011 molecules −23 kBT 1.38 × 10 J K 293 K ) ( )(

16.40 Model: Assume the nebula gas is ideal. Visualize: Use pV = NkBT . We are given N / V = 100 atoms/cm3 = 1× 108 atoms/m3 . Solve:

p=

NkBT = (1.0 × 108 atoms/m3 )(1.38 × 10−23 J/K)(7500 K) = 1.0 ×10−11 Pa = 1.0 × 10−16 atm V

Assess: This is much lower pressure than the best vacuum we can achieve in a laboratory, but it is a higher pressure than in non-nebula space.

16.41. Model: Assume the air is pure N 2 , with a molar mass of 28g/mol. Visualize:

We will use pV = nRT both before and after. Our intermediate goal is n1 − n2 . We are given

T2 = T1 = T = 20°C + 273 = 293 K and V2 = V1 = V = pr 2C = p(0.011 m 2 )(2.0 m) = 7.60 × 1024 m3 . Solve:

We convert the gauge pressures to absolute pressures with

p = pg + 1 atm = pg + 14.7 psi.

p1 = 110 psi + 14.7 psi = 124.7 psi = 860 kPa. p2 = 80 psi + 14.7 psi = 94.7 psi = 653 kPa. n1 − n2 =

p1V1 p2V2 − RT1 RT2

V ( p1 − p2 ) RT 7.60 × 10−4 m3 = (860 kPa − 653 kPa) (8.31 J/mol ⋅ K)(293 K) = 0.0646 mol =

Thus 0.0646 mol of N2 was lost; this is

⎛ 28g ⎞ 0.0646mol ⎜ ⎟ = 1.8g ⎝ 1mol ⎠ Assess:

The result seems to be a reasonable number.

16.42. Model: The carbon dioxide in the cube is an ideal gas. Solve:

Using the ideal gas equation and n = M/M mol ,

pV = nRT ⇒ V =

nRT MRT = p pM mol

The molar mass of CO2 is 44 g/mol or 0.044 kg/mol. Thus, V=

(10,000kg )(8.31 J mol K )( 273 K ) = 5090 m3 (1.013 × 105 Pa ) ( 0.044 kg mol )

The length of the cube is L = (V )1/3 = 17.2 m. Assess: This is sobering when you multiply it by the number of people in the industrialized world. Good thing plants take up CO2 in large quantities.

16.43. Model: Assume the gas is an ideal gas. Solve:

The before-and-after relationship of an ideal gas is 1 p 3V p1V1 p2V2 p V = ⇒ T2 = T1 2 2 = ( 298 K ) 2 1 1 = 447 K = 174°C T1 T2 p1 V1 p1 V1

16.44. Model: Assume the evaporated water is an ideal gas with a molar mass of 18 g/mol. Assume the pressure is 1 atm = 101.3 kPa. Visualize: We are given T = 35°C + 273 = 308 K. n = 1000 g (1 mol/ 18 g ) = 55.6 mol. Solve:

(a) pV = nRT 1V =

nRT ( 55.6 mol )( 8.31 J/mol ⋅ K )( 308 K ) = = 1.4 m3 p 101.3 kPa

(b) In the liquid state r = 1000 kg/m3 .

V=

m 1.0 kg = = 0.0010 m3 r 1000 kg/m3

The factor by which the volume of the evaporated water is larger than the liquid water is 1.4 m3 = 1400 0.0010 m3 Assess:

Gases really do take up a lot more volume than the equivalent mass of a liquid.

16.45. Model: Assume that the steam (as water vapor) is an ideal gas. Solve:

The volume of the liquid water is

V=

m

ρ

=

nM mol

ρ

⎛ pV =⎜ ⎝ RT

5 −6 3 ⎞ M mol 20 (1.013 × 10 Pa )(10,000 × 10 m ) ( 0.018 kg mol ) = ⎟ ⎠ ρ (8.31 J mol K )( 473 K ) (1000 kg m3 )

= 9.28 × 10−5 m 3 = 92.8 cm3 ≈ 93 cm3 Assess:

The liquid takes a lot smaller volume than the same number of atoms as a gas.

16.46. Model: Assume that the steam is an ideal gas. Solve:

V=

(a) The volume of water is

M

ρ

=

nM mol

ρ

=

5 3 pV M mol 50 (1.013 × 10 Pa )( 5.0 m ) ( 0.018 kg mol ) = 0.0815 m3 = 81.5 L ≈ 82 L = RT ρ ( 8.31 J mol K )( 673 K ) (1000 kg m3 )

(b) Using the before-and-after relationship of an ideal gas,

⎛ ( 273 + 150 ) K ⎞ ⎛ 50 atm ⎞ p2V2 p1V1 T p 3 3 3 = ⇒ V2 = 2 1 V1 = ⎜ ⎟⎜ ⎟ ( 5.0 m ) = 78.6 m ≈ 79 m T2 T1 T1 p2 673 K 2.0 atm ⎠ ⎝ ⎠⎝

16.47. Model: We assume that the volume of the tire and that of the air in the tire is constant.

Solve: A gauge pressure of 30 psi corresponds to an absolute pressure of (30 psi) + (14.7 psi) = 44.7 psi. Using the before-and-after relationship of an ideal gas for an isochoric (constant volume) process,

p1 p2 T ⎛ 273 + 45 ⎞ ⇒ p2 = 2 p1 = ⎜ = ⎟ ( 44.7 psi ) = 49.4 psi T1 T1 T2 ⎝ 273 + 15 ⎠ Your tire gauge will read a gauge pressure pg = 49.4 psi − 14.7 psi = 34.7 psi. ≈ 35 psi.

16.48. Model: The air is assumed to be an ideal gas. Solve:

At 20°C and 1 atm pressure, the number of moles in the container is

n1 =

5 −3 3 p1V1 (1.013 × 10 Pa )(10 m ) = = 0.0416 mol RT1 ( 8.31 J mol K )( 293 K )

At 100°C and 1 atm pressure, the number of moles is n2 =

5 −3 3 p2V2 (1.013 × 10 Pa )(10 m ) = = 0.0327 mol RT2 ( 8.31 J mol K )( 373 K )

When heated, the pressure will rise as the number of moles remains n1. When opened, the pressure drops to 1 atm as gas escapes. Thus, the number of moles of air that escape as the container is opened is n1 − n2 = 0.0416 mol − 0.0327 mol = 0.0089 mol.

16.49. Model: The gas’s pressure does not change, so this is an isobaric process. Solve:

The triple point of water is 0.01°C or 273.16 K, so T1 = 273.16 K. Because the pressure is a constant,

V V1 V2 ⎛ 1638 mL ⎞ ⇒ T2 = T1 2 = ( 273.16 K ) ⎜ = ⎟ = 447.44 K = 174.3°C V1 T1 T2 ⎝ 1000 mL ⎠

16.50. Model: Assume the gas in the manometer is an ideal gas. Solve:

In the ice-water mixture the pressure is

p1 = patoms + ρ Hg g (0.120 m) = 1.013 × 105 Pa + (13,600 kg/m3 )(9.8 m/s 2 )(0.120 m) = 1.173 × 105 Pa In the freezer the pressure is

p2 = patm + ρ Hg g (0.030 m) = 1.013 × 105 Pa + (13,600 kg/m3 )(9.8 m/s 2 )(0.030 m) = 1.053 × 105 Pa Assume that a drop in length of 90 mm produces a very small change in gas volume compared with the total volume of the gas cell. This means the volume of the chamber can be considered constant. Hence, ⎛p ⎞ p1 p2 1.053 × 105 Pa = ⇒ T2 = T1 ⎜ 2 ⎟ = ( 273 K ) = 245 K= − 28°C T1 T2 1.173 × 105 Pa ⎝ p1 ⎠ Assess:

This is a reasonable temperature for an industrial freezer.

16.51. Model: The air in the closed section of the U-tube is an ideal gas.

Visualize: The length of the tube is l = 1.0 m and its cross-sectional area is A. Solve: Initially, the pressure of the air in the tube is p1 = patmos and its volume is V1 = Al. After the mercury is poured in, compressing the air, the air-pressure force supports the weight of the mercury. Thus the compressed pressure equals the pressure at the bottom of the column: p2 = patmos + ρ gL. The volume of the compressed air is V2 = A(l − L). Because the mercury is poured in slowly, we will assume that the gas remains in thermal

equilibrium with the surrounding air, so T2 = T1. In an isothermal process, pressure and volume are related by p1V1 = patmos Al = p2V2 = ( patmos + ρ gL) A(l − L)

Canceling the A, multiplying through, and solving for L gives

L=l−

patmos 101,300 Pa = 1.00 m − = 0.240 m = 24.0 cm (13,600 kg/m3 )(9.8 m/s 2 ) ρg

16.52. Model: Assume that the air bubble is always in thermal equilibrium with the surrounding water, and the air in the bubble is an ideal gas. Solve: The pressure inside the bubble matches the pressure of the surrounding water. At 50 m deep, the pressure is

p1 = p0 + ρ water gd = 1.013 × 105 Pa + (1000 kg/m3 )(9.8 m/s 2 )(50 m) = 5.913 × 105 Pa At the lake’s surface, p2 = p0 = 1.013 × 105 Pa. Using the before-and-after relationship of an ideal gas, p2V2 p1V1 p T = ⇒ V2 = V1 1 2 T2 T1 p2 T1 ⇒

5 4π 3 4π 3 ⎛ 5.913 × 10 Pa ⎞ ⎛ 293 K ⎞ r2 = ( 0.005 m ) ⎜ ⎟⎜ ⎟ ⇒ r2 = 0.0091 m 5 3 3 ⎝ 1.013 × 10 Pa ⎠ ⎝ 283 K ⎠

The diameter of the bubble is 2r2 = 0.0182 m = 1.82 cm. Assess: The bubble’s diameter just about doubled; this seems reasonable.

16.53. Model: Assume that the compressed air in the cylinder is an ideal gas. The volume of the air in the cylinder is a constant. Solve: Using the before-and-after relationship of an ideal gas, p2V2 p1V1 T V ⎛ 1223 K ⎞ V1 = ⇒ p2 = p1 2 1 = ( 25 atm ) ⎜ ⎟ = 104 atm T2 T1 T1 V2 ⎝ 293 K ⎠ V1

where we have converted to the Kelvin temperature scale. Because the pressure does not exceed 110 atm, the compressed air cylinder does not blow.

16.54. Model: Assume that the gas is an ideal gas. Solve:

Assess: For the isothermal process, the pressure must be halved as the volume doubles. This is because p1 is proportional to 1/V1 for isothermal processes.

16.55. Model: Assume that the gas is an ideal gas. Solve:

Assess: For the isothermal process, the pressure must double as the volume is halved. This is because p is proportional to 1/V for isothermal processes.

16.56. Model: Assume that the helium gas is an ideal gas. Visualize: Process 1 → 2 is isochoric, process 2 → 3 is isothermal, and process 3 → 1 is isobaric. Solve: The number of moles of helium is

n=

M 8.0 g = = 2.0 mol M mol 4 g mol

Using the ideal-gas equation, V1 =

nRT1 ( 2.0 mol )( 8.31 J mol K ) ⎡⎣( 273 + 37 ) K ⎤⎦ = = 0.0254 m3 ≈ 0.025 m3 p1 2 (1.013 × 105 Pa )

For the isochoric process V2 = V1 , and

p1 p2 T ⎛ 657 + 273 ⎞ = ⇒ p2 = p1 2 = ( 2 atm ) ⎜ ⎟ = 6.0 atm T1 T1 T2 ⎝ 37 + 273 ⎠ For the isothermal process, the equation p3V3 = p2V2 is V3 = V2

p2 ⎛ 6 atm ⎞ 3 3 = ( 0.0254 m3 ) ⎜ ⎟ = 0.0762 m ≈ 0.076 m p3 ⎝ 2 atm ⎠

For the isothermal process, T3 = T2 = 657°C.

16.57. Model: Assume the nitrogen gas is an ideal gas. Solve:

(a) The number of moles of nitrogen is

n=

M 1g ⎛ 1 ⎞ = = ⎜ ⎟ mol M mol 28 g mol ⎝ 28 ⎠

Using the ideal-gas equation,

p1 =

nRT1 (1 28 mol )( 8.31 J mol K )( 298 K ) = = 8.84 × 105 Pa = 884 kPa ≈ 880 kPa V1 (100 × 10−6 m3 )

(b) For the process from state 1 to state 3:

⎛ 1.5 p1 ⎞⎛ 50 cm3 ⎞ p1V1 p3V3 p V = ⇒ T3 = T1 3 3 = ( 298 K ) ⎜ = 223.5 K ≈ −49°C ⎟⎜ 3 ⎟ T1 T3 p1 V1 ⎝ p1 ⎠ ⎝ 100 cm ⎠ For the process from state 3 to state 2: ⎛ p ⎞⎛ V ⎞ ⎛ 2.0 p1 ⎞⎛ 100 cm3 ⎞ p2V2 p3V3 = ⇒ T2 = T3 ⎜ 2 ⎟⎜ 2 ⎟ = ( 223.5 K ) ⎜ = 596 K = 323°C ⎟⎜ 3 ⎟ T2 T3 ⎝ 1.5 p1 ⎠ ⎝ 50 cm ⎠ ⎝ p3 ⎠⎝ V3 ⎠ For the process from state 1 to state 4: ⎛ 1.5 p1 ⎞⎛ 150 cm3 ⎞ p4V4 p1V1 p V = ⇒ T4 = T1 4 4 = ( 298 K ) ⎜ = 670.5 K ≈ 398°C ⎟⎜ 3 ⎟ T4 T1 p1 V1 ⎝ p1 ⎠ ⎝ 100 cm ⎠

16.58. Model: The gas is an ideal gas. Solve:

(a) Using the ideal-gas equation,

T1 =

(1.0 × 105 Pa )( 2.0 m3 ) = 301 K p1V1 = nR ( 80 mol )( 8.31 J mol K )

Because points 1 and 2 lie on the isotherm, T2 = T1 = 301 K. The temperature of the isothermal process is 301 K. (b) The straight-line process 1 → 2 can be represented by the equation

p = (3 − V ) × 105 where V is in m3 and p is in Pa. We can use the ideal gas law to find that the temperature along the line varies as T=

pV 105 = (3V − V 2 ) × nR nR

We can maximize T by setting the derivative dT/dV to zero: dT 105 3 2 = (3 − 2Vmax = 0 ⇒ Vmax = m3 = 1.50 m3 )× nR dV 2 At this volume, the pressure is pmax = 1.5 × 105 Pa and the temperature is Tmax =

pmaxVmax (1.50 × 105 Pa)(1.5 m3 ) = = 338 K nR (80 mol)(8.31 J/mol K)

16.59. Model: Assume the gas is an ideal gas. Solve: units:

(a) We can find the temperatures directly from the ideal-gas law after we convert all quantities to SI

T1 =

3 −6 3 3 p1V1 ( 3.0 atm × 101,300 Pa atm ) (1000 cm × 10 m cm ) = = 366 K = 93°C nR ( 0.10 mol )(8.31 J mol K )

T2 =

3 −6 3 3 p2V2 (1.0 atm × 101,300 Pa atm ) ( 3000 cm × 10 m cm ) = = 366 K = 93°C nR ( 0.10 mol )( 8.31 J mol K )

(b) T2 = T1 , so this is an isothermal process. (c) A constant volume process has V3 = V2 . Because p1 = 3 p2 , restoring the pressure to its original value means

that p3 = 3p2. From the ideal-gas law,

⎛ p ⎞⎛ V ⎞ p3V3 p2V2 = ⇒ T3 = ⎜ 3 ⎟⎜ 3 ⎟ T2 = 3 × 1 × T2 = 3 × 366 K = 1098 K = 825°C T3 T2 ⎝ p2 ⎠⎝ V2 ⎠

16.60. Model: Assume the gas is an ideal gas. Solve:

(a) Using the ideal-gas law,

T1 =

5 −6 3 p1V1 (1.013 × 10 Pa )(100 × 10 m ) = = 244 K = −29°C nR ( 5.0 × 10−3 mol ) ( 8.31 J mol K )

(b) Using the before and after relationship of an ideal gas, 3 p1V1 p2V2 T V ⎛ 2926 K ⎞ ⎛ 100 cm ⎞ ⇒ p2 = p1 2 1 = (1 atm ) ⎜ = = 4.0 atm ⎟⎜ 3 ⎟ T1 V2 T1 T2 ⎝ 244 K ⎠ ⎝ 300 cm ⎠

(c) Using the before and after relationship of an ideal gas,

p3V3 p2V2 p T ⎛ 4 atm ⎞⎛ 2438 K ⎞ 3 3 = ⇒ V3 = 2 3 V2 = ⎜ ⎟⎜ ⎟ ( 300 cm ) = 500 cm T3 T2 p3 T2 ⎝ 2 atm ⎠⎝ 2926 K ⎠

16.61. Model: We assume the oxygen gas is ideal. Visualize: From the figure we glean p2 = 3 p1 and V2 = 3V1 . We are given T1 = 20°C + 273 = 293K. Solve: Use the before-and-after version of the ideal-gas law.

p1V1 p2V2 = T1 T2 T2 =

Assess:

( 3 p1 )( 3V1 ) T = 9T = 9 293K = 2637 K = 2364°C p2V2 T1 = ( ) 1 1 p1V1 p1V1

This is a hot temperature—higher than the melting point of many elements.

16.62. Model: Assume CO2 gas is an ideal gas. Solve:

(a) The molar mass for CO2 is M mol = 44 g/mol, so a 10 g piece of dry ice is 0.2273 mol. This becomes

0.227 mol of gas at 0°C. With V1 = 10,000 cm3 = 0.010 m3 and T1 = 0°C = 273 K, the pressure is p1 =

nRT1 ( 0.2273 mol )( 8.31 J mol K )( 273 K ) = = 5.156 × 104 Pa = 0.509 atm ≈ 0.50 atm V1 0.010 m3

(b) From the isothermal compression,

p2V2 = p1V1 ⇒ V2 = V1

p1 ⎛ 0.509 atm ⎞ −3 3 3 = ( 0.010 m3 ) ⎜ ⎟ = 1.70 × 10 m = 1700 cm p2 ⎝ 3.0 atm ⎠

From the isobaric compression,

T3 = T2 (c)

⎛ 1000 cm 3 ⎞ V3 = ( 273 K ) ⎜ = 161 K = −112°C 3 ⎟ V2 ⎝ 1700 cm ⎠

16.63. Model: The gas in the container is assumed to be an ideal gas. Solve:

(a) The gas starts at pressure p1 = 2.0 atm, temperature T1 = 127°C = (127 + 273) K = 400 K and

volume V1. It is first compressed at a constant temperature T2 = T1 until V2 = 12 V1 and the pressure is p2. It is then further compressed at constant pressure p3 = p2 until V3 = 12 V2 . From the ideal-gas law,

p2V2 p1V1 V T V = ⇒ p2 = p1 1 2 = ( 2.0 atm ) 1 1 × 1 = 4.0 atm V2 T1 V T2 T1 2 1 Note that T2 = T1 = 400 K. Using the ideal-gas law once again, 1 V V p p3V3 p2V2 ⇒ T3 = T2 3 3 = ( 400 K ) 2 2 × 1 = 200 K = −73°C = V2 p2 V2 T3 T2

The final pressure and temperature are 4.0 atm and –73°C. (b)

16.64. Model: Assume that the nitrogen gas is an ideal gas. Solve:

(a) The molar mass of N2 gas is 28 g/mol. The number of moles is n = (5 g)/(28 g / mol) = 0.1786 mol.

The initial conditions are p1 = 3.0 atm and T1 = 293 K. We use the ideal gas law to find the initial volume as follows: V1 =

nRT1 ( 0.1786 mol )( 8.31 J mol K )( 293 K ) = = 1.430 × 10−3 m3 = 1430 cm 3 ≈ 1400 cm3 p1 3.0 atm × 101,300 Pa atm

An isobaric expansion until the volume triples results in V2 = 3V1 = 4290 cm3 . (b) After the expansion,

p2V2 p1V1 p V = ⇒ T2 = 2 2 T1 = 1 × 3 × T1 = 3T1 = 879 K = 606°C T2 T1 p1 V1 (c) A constant volume decrease at V3 = V2 = 4290 cm3 back to T3 = T1 = 13 T2 results in the following:

p3V3 p2V2 T V 1 1 = ⇒ p3 = 3 2 p2 = × 1 × p2 = × 3.0 atm = 1.0 atm T3 T2 T2 V3 3 3 (d) An isothermal compression at T4 = T3 back to the initial volume V4 = V1 = 13 V3 results in the following:

p4V4 p3V3 T V 1 = ⇒ p4 = 4 3 p3 = 1 × 1 × p3 = 3 × 1.0 atm = 3.0 atm T4 T3 T3 V4 3 (e)

16.65. Solve: (a) A gas is compressed isothermally from a volume 300 cm3 at 2 atm to a volume of 100 cm3. What is the final pressure? (b)

(c) The final pressure is p2 = 6 atm.

16.66. Solve: (a) A gas at 400°C and 500 kPa is cooled at a constant volume to a pressure of 200 kPa. What is the final temperature in C°? (b)

(c) The final temperature is T2 = −3.8°C.

16.67. Solve: (a) A gas expands at constant pressure from 200 cm3 at 50°C until the temperature is 400°C. What is the final volume? (b)

(c) The final volume is V2 = 417 cm3 .

16.68. Solve: (a) 0.12 g of neon gas at 2.0 atm and 100 cm3 expands isobarically to twice its initial volume. What is the final temperature of the gas? (b)

(c) The number of moles of neon is

n=

M 0.12 g = = 0.006 mol M mol 20 g

The initial temperature is T1 =

5 −4 3 p1V1 2 (1.013 × 10 Pa )(1.0 × 10 m ) = = 406 K nR ( 0.006 mol )(8.31 J mol K )

The final temperature is T2 =

p2 V2 p 2V1 T1 = ( 406 K ) = 812 K p1 V1 p V1

16.69. Model: Assume that the compressed air is an ideal gas. Solve:

(a) Because the piston is floating in equilibrium,

Fnet = ( p1 − patoms ) A − w = 0 N where the piston’s cross-sectional area A = π (r 2 ) = π (0.050 m) 2 = 7.854 ×10−3 m 2 and the piston’s weight w = (50 kg)(9.8) = 490 N. Thus, p1 =

w 490 N + patmos = + 1.013 × 105 Pa=1.637 × 105 Pa A 7.854 × 10−3 m 2

Using the ideal-gas equation p1V1 = nRT1 ,

(1.637 × 105 Pa)Ah1 = (0.12 mol)(8.31 J/mol K)[(273 + 30) K] With the value of A given above, this equation yields h1 = 0.235 m = 23.5 cm. (b) When the temperature is increased from T1 = 303 K to T2 = (303 + 100) K = 403 K, the volume changes

from V1 = Ah1 to V2 = Ah2 at a constant pressure p2 = p1. From the before-and-after relationship of the ideal gas:

p1 ( Ah1 ) p2 ( Ah2 ) = T1 T2 ⇒ h2 =

p1 T2 ⎛ 403 K ⎞ h1 = (1) ⎜ ⎟ ( 0.235 m ) = 0.313 m = 31.3 cm p2 T1 ⎝ 303 K ⎠

Thus, the piston moves h2 − h1 = 7.8 cm.

16.70. Model: The air in the diving bell is an ideal gas. Visualize:

Solve:

(a) Initially p1 = p0 (atmospheric pressure), V1 = AL, and T1 = 293 K. When the diving bell is

submerged to d = 100 m at the bottom edge, the water comes up height h inside. The volume is V2 = A( L − h) and the temperature is T2 = 283 K. Like a barometer, the pressure at points a and b must be the same. Thus p2 + ρ gh = p0 + ρ gd , or p2 = p0 + ρ g (d − h). Using the before and after relationship of an ideal gas,

p1V1 p2V2 p AL [ p0 + ρ g (d − h)] A( L − h) = ⇒ 0 = T1 T2 293 K 283 K Multiplying this out gives the following quadratic equation for h: ⎛ ⎝

ρ gh 2 − [ p0 + ρ g (d + L)]h + ⎜1 −

283 ⎞ ⎟ p0 L + ρ gLd = 0 293 ⎠

Inserting the known values (using ρ = 1030 kg/m3 for seawater) and dividing by ρ g gives h 2 − 113.04h + 301.03 = 0 ⇒ h = 110 m or 2.7 m

The first solution is not physically meaningful, so the water rises to height h = 2.7 m. (b) To expel all the water, the air pressure inside the bell must be increased to match the water pressure at the bottom edge of the bell, d = 100 m. The necessary pressure is

p = p0 + ρ gd = 101,300 Pa + (1030 kg/m3 )(9.8 m/s 2 )(100 m) = 1111 kPa = 10.96 atm ≈ 11 atm

16.71. Model: Assume the trapped air to be an ideal gas. Visualize:

Initially, as the pipe is touched to the water surface and the gas inside is thus closed off from the air, the pressure p1 = patmos = 1 atm and the volume is V1 = L1 A, where A is the cross-sectional area of the pipe. By pushing the pipe in slowly, the gas temperature in the pipe remains the same as the water temperature. Thus, this is an isothermal compression of the gas with T2 = T1. Solve: From the ideal-gas law,

p2V2 = p1V1 ⇒ p2 L2 A = p1L1 A ⇒ p2 L2 = patmos L1 ⇒ p2 = patmos ( L1 L2 ) As the pipe is pushed down, the increasing water pressure pushes water up into the pipe, compressing the air. In equilibrium, the pressure at points a and b, along a horizontal line, must be equal. (This is like the barometer. If the pressures at a and b weren’t equal, the pressure difference would cause the liquid level in the pipe to move up or down.) The pressure at point a is just the gas pressure inside the pipe: pa = p2 . The pressure at point b is the pressure at depth L2 in water: pb = patmos + ρ gL2 . Equating these gives

p2 = patmos = ρ gL2 Substituting the expression for p2 from the ideal-gas equation above, the pressure equation becomes

patmos L1 = patmos + ρ gL2 ⇒ ρ gL22 + patmos L2 − patmos L1 = 0 L2 This is a quadratic equation for L2 with solutions

L2 =

− patmos ±

( patmos )

2

+ 4 ρ gpatmos L1

2ρ g

Length has to be a positive quantity, so the one physically acceptable solution is L2 =

−101,300 Pa +

(101,300 Pa )

2

+ 4 (1000 kg m3 )( 9.8 m s 2 ) (101,300 Pa )( 3.0 m )

2 (1000 kg m3 )( 9.8 m s 2 )

= 2.4 m

16.72. Model: The gas in the cylinder is assumed to be an ideal gas. Solve:

The gas at T1 = 20°C = 293 K has a pressure p1 = 1 atm and a volume V1 = AL0 . (The pressure has to be

1 atm to balance the external force of the air on the piston.) At a temperature T2 = 100°C = 373 K, its volume becomes V2 = A( L0 + Δx) and its pressure increases to p2 = p1 +

F k Δx = p1 + A A

where F = k Δx is the spring force on the piston. Using the before-and-after relationship of an ideal-gas equation,

p1V1 p2V2 p L A ( p + k Δx A )( L0 + Δx ) A = ⇒ 1 0 = 1 T1 T2 T1 T2 ⇒

T2 ( p1 + k Δx A )( L0 + Δx ) ⎛ k Δx ⎞ ⎛ Δx ⎞ = = ⎜1 + ⎟ ⎟ ⎜1 + T1 p1L0 p1 A ⎠⎝ L0 ⎠ ⎝

L0 can be obtained from the ideal-gas law as follows: L0 =

V1 ⎛ nRT1 ⎞ 1 ( 0.004 mol )( 8.31 J mol K )( 293 K ) =⎜ = 0.0961 m = 9.61 cm ⎟ = A ⎝ p1 ⎠ A (1.013 ×105 Pa )(10 ×10−4 m2 )

Substituting this value of L0 and the values for p1, T1, T2, and k, the above equation can be simplified to 154.02Δx 2 + 25.210Δx − 0.2730 = 0 ⇒ Δx = 0.0102 m = 1.02 cm The spring is compressed by 1.0 cm.

16.73. Model: The gas in containers A and B is assumed to be an ideal gas. Solve:

(a) The number of moles of gas in containers A and B can be expressed as follows:

nA =

5 ( 5.0 ×105 Pa ) ( 4VA ) = 5000 VA pAVA (1.0 × 10 Pa )VA V pV = = 333.3 A nB = B B = RTA R ( 300 K ) R RTB R ( 400 K ) R

where nA and nB are expressed in moles. Let n′A and n′B be the number of moles after the valve is opened. Since the total number of moles is the same, nA + nB = 5333.3

VA = nA′ + nB′ R

The pressure is now the same in both containers:

p′A = ⇒ n′A = nB′

n′A RTA n′ RT = p′B = B B VA VB

TB VA ⎛ 400 K ⎞ VA = n′B ⎜ ⇒ 3n′A = n′B ⎟ TA VB ⎝ 300 K ⎠ 4VA

Solving the above equations, 5333.3

VA V V = nA′ + 3nA′ = 4nA′ ⇒ n′A = 1333.3 A ⇒ n′B = 4000 A R R R

The new pressure is p′A =

n′A RTA 1333.3 ⎛ VA ⎞ 5 = ⎜ ⎟ RTA = 4.0 × 10 Pa VA VA ⎝ R ⎠

p′B =

n′B RTB 4000 ⎛ VA ⎞ 5 = ⎜ ⎟ RTB = 4.0 × 10 Pa VB VB ⎝ R ⎠

Gas flows from B to A until the pressure is 4.0 × 105 Pa. (b) The change is irreversible because opening the valve is like breaking a membrane. It is not a quasi-static process.

16.74 Model: Assume the gas in the left chamber is ideal. Visualize: For the piston to be in equilibrium ( Fx ) net = 0. The initial volume is V0 = L0 A. Solve: (a) The force to the right on the piston by the gas in the left chamber is simply p0 A. . The magnitude of the force to the left on the piston by the spring is k (ΔL). p A ( Fx ) net = p0 A − k (ΔL) ⇒ ΔL = 0 k (b) When the piston is moved x to the right the spring is compressed a total of ΔL + x and the pressure in the left chamber changes to p1 according to the ideal gas law. When the temperature doesn’t change the ideal gas equation reduces to p0V0 = p1V1 , where V0 = L0 A and V1 = ( L0 + x) A . Therefore pL p1 = 0 0 ( L0 + x) Now, similar to part (a) above, ( Fx ) net = p1 A − k (ΔL + x) Substitute in ΔL from part (a) and p1 from above. ( Fx ) net =

p0 L0 ⎛p A ⎞ A− k ⎜ 0 + x⎟ ( L0 + x) ⎝ k ⎠

Now it is time to apply the binomial approximation: ( L0 + x) −1 ≈ 1/ L0 (1 − x / L0 ) for x << L0 .

( Fx ) net = ( p0 L0 ) (1/ L0 (1 − x / L0 ) ) A − p0 A − kx = p0 A − p0 Ax / L0 − p0 A − kx

= − ( p0 A / L0 + k ) x This is clearly a linear restoring force of the form of Hooke’s law where the usual k is replaced with ( p0 A / L0 + k ) . That is, the gas in the left chamber is increasing the k of the system (by p0 A / L0 ) and making it act like a stiffer spring. (c) Since we have a Hooke’s law for which we know the modified k, we simply replace k in the period equation for harmonic oscillators with the modified constant. M T = 2π p0 A / L0 + k Assess: This is simple harmonic motion when the approximation x << L0 is valid.

16-1

17.1. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve. Visualize: The gas is compressing, so we expect the work to be positive. Solve: The work done on the gas is

W = − ∫ p dV = − ( area under the pV curve )

(

)

= − − ( 200 cm3 ) ( 200 kPa ) = ( 200 × 10−6 m3 )( 2.0 × 105 Pa ) = 40 J

Assess: The area under the curve is negative because the integration direction is to the left. Thus, the environment does positive work on the gas to compress it.

17.2. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve. Visualize: The gas is expanding, so we expect the work to be negative. Solve: The area under the pV curve is the area of the rectangle and triangle. We have

( 200 × 10

−6

m3 )( 200 × 103 Pa ) + 12 ( 200 × 10−6 m3 )( 200 × 103 Pa ) = 60 J

Thus, the work done on the gas is W = −60 J. Assess: The environment does negative work on the gas as it expands.

17.3. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve. Solve:

The work done on gas in an isobaric process is

W = − pΔV = − p (Vf − Vi ) Substituting into this equation,

80 J = − ( 200 × 103 Pa ) (V1 − 3V1 ) ⇒ Vi = 2.0 × 10−4 m3 = 200 cm3 Assess:

The work done to compress a gas is positive.

17.4. Model: Helium is an ideal gas that undergoes isobaric and isothermal processes. Solve:

(a) Since the pressure ( pi = pf = p ) is constant the work done is

Won gas = − pΔV = − p (Vf − Vi ) = −

nRTi (Vf − V i ) Vi

= −( 0.10 mol )( 8.31 J mol K )( 573 K )

(1000 cm

3

− 2000 cm3 )

2000 cm3

= 240 J

(b) For compression at a constant temperature,

Won gas = −nRT ln (Vf Vi ) ⎛ 1000 × 10−6 m3 ⎞ = − ( 0.10 mol )( 8.31 J mol K )( 573 K ) ln ⎜ = 330 J −6 3 ⎟ ⎝ 2000 × 10 m ⎠ (c) For the isobaric case,

p=

nRTi = 2.38 × 105 Pa Vi

For the isothermal case, pi = 2.38 × 105 Pa and the final pressure is

pf =

nRTf = 4.76 × 105 Pa Vf

17.5.

Visualize:

Solve:

Because W = − ∫ p dV and this is an isochoric process, W = 0 J. The final point is on a higher isotherm

than the initial point, so Tf > Ti . Heat energy is thus transferred into the gas (Q > 0) and the thermal energy of the gas increases ( Eth f > Eth i ) as the temperature increases.

17.6.

Visualize:

Solve:

Because this is an isobaric process W = − ∫ pdV = − p (Vf − Vi ) . Since Vf is smaller than Vi, W is positive.

That is, the gas is compressed. Since the final point is on a lower isotherm than the initial point, Tf < Ti . In other words, the thermal energy decreases. For this to happen, the heat energy transferred out of the gas must be larger than the work done.

17.7.

Visualize:

Solve:

Because the process is isothermal, ΔEth = Eth f − Eth i = 0 J. According to the first law of

thermodynamics, ΔEth = W + Q. This can only be satisfied if W = −Q. W is positive because the gas is compressing, hence Q is negative. That is, heat energy is removed from the gas.

17.8.

Visualize:

Solve: This is an adiabatic process of gas compression so no heat energy is transferred between the gas and the environment. That is, Q = 0 J. According to the first law of thermodynamics, the work done on a gas in an adiabatic process goes entirely to changing the thermal energy of the gas. The work W is positive because the gas is compressed.

17.9. Solve: The first law of thermodynamics is ΔEth = W + Q ⇒ −200 J = 500 J + Q = Q ⇒ Q = −700 J The negative sign means a transfer of energy from the system to the environment. Assess: Because W > 0 means a transfer of energy into the system, Q must be less than zero and larger in magnitude than W so that Eth f < Eth i .

17.10. Solve: This is an isobaric process. W > 0 because the gas is compressed. This transfers energy into the system. Also, 100 J of heat energy is transferred out of the gas. The first law of thermodynamics is

ΔEth = W + Q = − pΔV + Q = −(4.0 × 105 Pa)(200 − 600) × 10−6 m3 − 100 J = 60 J Thermal energy increases by 60 J.

17.11. Model: The removal of heat from the ice reduces its thermal energy and its temperature. Solve:

The heat needed to change an object’s temperature is Q = McΔT. The mass of the ice cube is

M = ρiceV = (920 kg/m3 )(0.06 × 0.06 × 0.06)m3 = 0.1987 kg The specific heat of ice from Table 17.2 is cice = 2090 J/kg K, so Q = (0.199 kg)(2090 J/kg K)(243 K − 273 K) = −12,460 J ≈ 12,000 J

Thus, the energy removed from the ice block is 12,000 J. Assess: The negative sign with Q means loss of energy.

17.12. Model: The spinning paddle wheel does work and changes the water’s thermal energy and its temperature. Solve: (a) The temperature change is ΔT = Tf − Ti = 25°C − 21°C = 4 K. The mass of the water is

M = (200 × 10−6 m3 )(1000 kg/m3 ) = 0.20 kg The work done is W = ΔEth = Mcwater ΔT = (0.20 kg)(4190 J/kg K)(4 K) = 3350 J ≈ 3400 J (b) Q = 0. No energy is transferred between the system and the environment because of a difference in temperature.

17.13. Model: Heating the mercury at its boiling point changes its thermal energy without a change in temperature. Solve: The mass of the mercury is M = 20 g = 2.0 × 10−2 kg, the specific heat cmercury = 140 J/kg K, the boiling

point Tb = 357°C, and the heat of vaporization Lv = 2.96 × 105 J/kg. The heat required for the mercury to change to the vapor phase is the sum of two steps. The first step is Q1 = Mcmercury ΔT = (2.0 × 10−2 kg)(140 J/kg K)(357°C − 20°C) = 940 J

The second step is Q2 = MLV = (2.0 ×10−2 kg)(2.96 × 105 J/kg) = 5920 J

The total heat needed is 6860 J.

17.14. Model: Heating the mercury changes its thermal energy and its temperature. Solve:

(a) The heat needed to change the mercury’s temperature is

Q = McHg ΔT ⇒ ΔT =

Q 100 J = = 35.7 K ≈ 35°C McHg ( 0.020 kg )(140 J kg K )

(b) The amount of heat required to raise the temperature of the same amount of water by the same number of degrees is

Q = Mcwater ΔT = (0.020 kg)(4190 J/kg K)(35.7 K) = 3000 J Assess: Q is directly proportional to cwater and the specific heat for water is much higher than the specific heat for mercury. This explains why Qwater > Qmercury.

17.15. Model: Changing ethyl alcohol at 20°C to solid ethyl alcohol at its melting point requires two steps:

lowering its temperature from 20°C to −114°C, then changing the ethyl alcohol to its solid phase at −114°C. Solve: The change in temperature is −114°C − 20°C = −134°C = −134 K. The mass is

M = ρV = ( 789 kg/m3 )( 200 × 10−6 m3 ) = 0.1578 kg The heat needed for the two steps is

Q1 = Mcalcohol ΔT = (0.1578 kg)(2400 J/kg K)(−134 K) = −5.07 × 104 J Q2 = − MLf = −(0.1578 kg)(1.09 × 105 J/kg) = −1.72 × 104 J The total heat required is

Q = Q1 + Q2 = −6.79 × 104 J ≈ −6.8 × 104 J Thus, the minimum amount of energy that must be removed is 6.8 × 104 J. Assess: The negative sign with Q indicates that 6.8 × 104 J will be removed from the system.

17.16. Model: Changing solid lead at 20°C to liquid lead at its melting point (Tm = 328°C) requires two steps: raising the temperature to Tm and then melting the solid at Tm to a liquid at Tm. Solve: The equation for the total heat is Q = Q1 + Q2 ⇒ 1000 J = Mclead (Tf − Ti ) + MLf

⇒ 1000 J = M (128 J/kg K)(328 − 20) K + M (0.25 × 105 J/kg) ⇒ M = The maximum mass of lead you can melt with 1000 J of heat is 15.5 g.

1000 J = 15.5 g ( 64,424 J/kg)

17.17. Model: We have a thermal interaction between the copper pellets and the water. Solve:

The conservation of energy equation Qc + Ww = 0 is

Mc cc (Tf − 300°C) + Mw cw (Tf − 20°C) = 0 J Solving this equation for the final temperature Tf gives Tf = =

Mc cc (300°C) + Mw cw (20°C) Mc cc + Mw cw (0.030 kg)(385 J/kg K) (300°C) + (0.10 kg)(4190 J/kg K)(20°C) = 28°C (0.030 kg)(385 J/kg K) + (0.10 kg)(4190 J/kg K)

The final temperature of the water and the copper is 28°C.

17.18. Model: We have a thermal interaction between the copper block and water. Solve:

The conservation of energy equation Qcopper + Qwater = 0 J is

M copper ccopper (Tf − Ti copper ) + M water cwater (Tf − Ti water ) = 0 J Both the copper and the water reach the common final temperature Tf = 25.5°C. Thus M copper (385 J/kg K)(25.5°C − 300°C) + (1.00 × 10−3 m3 )(1000 kg/m3 )(4190 J/kg K)(25.5°C − 20°C) = 0 J ⇒ M copper = 0.218 kg

17.19. Model: We have a thermal interaction between the thermometer and the water. Solve:

The conservation of energy equation Qthermo + Qwater = 0 J is

M thermocthermo (Tf − (Ti ) thermo ) + M water cwater (Tf − (Ti ) water ) = 0 J The thermometer slightly cools the water until both have the same final temperature Tf = 71.2°C. Thus (0.050 kg)(750 J/kg K)(71.2°C − 20.0°C) + (200 × 10−6 m3 )(1000 kg/m3 )(4190 J/kg K)(71.2° C − Ti water ) = 1920 J + 838 (J/K)(71.2°C − Ti water ) = 0 J ⇒ Ti water = 73.5°C Assess:

The thermometer reads 71.2°C for a real temperature of 73.5°C. This is reasonable.

17.20. Model: We have a thermal interaction between the aluminum pan and the water. Solve:

The conservation of energy equation QAl + Qwater = 0 J is

M AlcAl (Tf − Ti A1 ) + M water cwater (Tf − Ti water ) The pan and water reach a common final temperature Tf = 24.0°C

(0.750 kg)(900 J/kg K)(24.0°C − Ti Al ) + (10.0 × 10−3 m3 )(1000 kg/m3 )(4190 J/kg K)(24.0°C − 20.0°C ) = (675.0 J/K)(24.0°C − Ti Al ) + 167,600 J = 0 J ⇒ Ti Al = 272°C = [(272)(9/5) + 32]°F = 522°F

17.21. Model: We have a thermal interaction between the metal sphere and the mercury. Solve:

The conservation of energy equation Qmetal + QHg = 0 J is

M metalcmetal (Tf − Ti metal ) + M Hg cHg (Tf − Ti Hg ) = 0 J The metal and mercury reach a common final temperature Tf = 99.0°C. Thus (0.500 kg)cmetal (99°C − 300°C) + (300 × 10−6 m3 )(13,600 kg/m3 )(140 J/kg K)(99°C − 20°C) = 0 J We find that cmetal = 449 J kg K. The metal is iron.

17.22. Model: Use the models of isochoric and isobaric heating. Note that the change in temperature on the Kelvin scale is the same as the change in temperature on the Celsius scale. Solve: (a) The atomic mass number of argon is 40. That is, M mol = 40 g/mol. The number of moles of argon gas in the container is

n=

M 1.0 g = = 0.025 mol M mol 40 g mol

The amount of heat is Q = nCV ΔT = (0.025 mol)(12.5 J/mol K)(100°C) = 31.25 J ≈ 31 J (b) For the isobaric process Q = nCP ΔT becomes

31.25 J = (0.025 mol)(20.8 J/mol K) ΔT ⇒ ΔT = 60°C

17.23. Model: The heating processes are isobaric and isochoric. O2 is a diatomic ideal gas. Solve:

(a) The number of moles of oxygen is

n=

M 1.0 g = = 0.03125 mol M mol 32 g mol

For the isobaric process, Q = nCp ΔT = (0.03125 mol)(29.2 J/mol K)(100°C) = 91.2 J ≈ 91 J (b) For the isochoric process,

Q = nCV ΔT = 91.2 J = (0.03125 mol)(20.9 J/mol K) ΔT ⇒ ΔT = 140°C

17.24. Model: The heating is an isochoric process. Solve:

The number of moles of helium is

n=

M 2.0 g = = 0.50 mol M mol 4 g mol

For the isochoric processes,

QHe = nCV ΔT = ( 0.50 mol )(12.5 J mol K ) ΔT

⎛ ⎞ M QO2 = nCV ΔT = ⎜ ⎟ ( 20.9 J mol K ) ΔT 32 g mol ⎝ ⎠

Because QHe = QO2 ,

( 0.50 mol )(12.5

⎛ ⎞ M J mol K ) = ⎜ ⎟ ( 20.9 J mol K ) ⇒ M = 9.6 g 32 g mol ⎝ ⎠

17.25. Model: The gas is an ideal gas that is subjected to an adiabatic process. Solve:

(a) For an adiabatic process, γ

γ pf Vfγ = pV i i ⇒

pf ⎛ Vi ⎞ ln(2.5) γ = 1.32 = ⎜ ⎟ ⇒ 2.5 = ( 2.0 ) ⇒ γ = pi ⎝ Vf ⎠ ln(2.0)

(b) Equation 17.39 for an adiabatic process is γ −1 Tf Vfγ −1 = TV ⇒ i i

Tf ⎛ Vi ⎞ =⎜ ⎟ Ti ⎝ Vf ⎠

γ −1

= ( 2.0 )

1.32 −1

= ( 2.0 )

0.32

= 1.25

17.26. Model: We assume the gas is an ideal gas and γ = 1.40 for a diatomic gas. Solve:

Using the ideal-gas law, Vi =

nRTi ( 0.10 mol )( 8.31 J mol K )( 423 K ) = = 1.157 × 10−3 m3 pi ( 3 × 1.013 × 105 Pa )

For an adiabatic process, γ γ pV i i = pf Vf 1γ

⎛ p ⎞ ⇒ Vf = Vi ⎜ i ⎟ ⎝ pf ⎠

1 1.40

⎛ pi ⎞ = (1.157 × 10 m ) ⎜ ⎟ ⎝ 0.5 pi ⎠ −3

3

= 1.9 × 10−3 m3

To find the final temperature, we use the ideal-gas law once again as follows:

Tf = Ti

⎛ 0.5 pi ⎞⎛ 1.90 × 10−3 m3 ⎞ pf Vf = ( 423 K ) ⎜ = 346.9 K ≈ 74°C ⎟⎜ −3 3 ⎟ pi Vi ⎝ pi ⎠ ⎝ 1.157 × 10 m ⎠

17.27. Model: The O2 gas has γ = 1.40 and is an ideal gas. Solve:

(a) For an adiabatic process, pV γ remains a constant. That is, γ

1.40

⎛ Vi ⎞ ⎛ Vi ⎞ pV ⎟ = ( 3.0 atm ) ⎜ ⎟ i i = pf Vf ⇒ pf = pi ⎜ V ⎝ f⎠ ⎝ 2Vi ⎠ γ

γ

1.40

⎛1⎞ = ( 3.0 atm ) ⎜ ⎟ ⎝2⎠

= 1.14 atm ≈ 1.1 atm

(b) Using the ideal-gas law, the final temperature of the gas is calculated as follows:

pV pV p V ⎛ 1.14 atm ⎞ ⎛ 2Vi ⎞ i i = f f ⇒ Tf = Ti f f = ( 423 K ) ⎜ ⎟ = 321.5 K ≈ 48°C ⎟⎜ Ti Tf pi Vi ⎝ 3.0 atm ⎠ ⎝ Vi ⎠

17.28. Visualize: We are asked for the heat-loss rate which is given by Equation 17.48: Q A = k DT Dt L We are given A = 10 m ×14 m = 140 m 2 , L = 0.12 m, and DT = 22°C − 5°C = 17°C = 17 K. We look up the thermal conductivity of concrete in Table 17.5: k = 0.8W/m ⋅ K. Solve:

⎛ 140 m 2 ⎞ Q A = k DT = ( 0.8W/m ⋅ K ) ⎜ ⎟ (17 K ) = 16 kW L Dt ⎝ 0.12 m ⎠ Assess: The answer is in a reasonable range. The heat loss could be reduced with thicker concrete or adding a layer of a different material.

17.29. Visualize: To determine the material we will solve for k in Equation 17.48: Q A = k DT Dt L We are given L = 0.20m and DT = 100 K. We compute A = pr 2 = p ( 0.010m ) = 3.14 × 1024 m 2 . 2

We also convert the heat conduction to watts.

⎛ 1h ⎞ 4.5 ×104 J/h ⎜ ⎟ = 12.5 W ⎝ 3600 s ⎠ Solve:

Solve the equation for k . ⎛ ⎞⎛ 1 ⎞ 0.20m ⎛ Q ⎞L 1 k =⎜ ⎟ = (12.5 W ) ⎜ ⎟ = 80 W/m ⋅ K 24 2 ⎟⎜ 3 14 10 m . × ⎝ Dt ⎠ A DT ⎝ ⎠⎝ 100K ⎠

Look this up in Table 17.5; the value corresponds to iron. Assess: We are grateful that our answer was one of the entries in the table.

17.30. Model: Assume the lead sphere is an ideal radiator with e = 1 . Also assume that the highest temperature the solid lead sphere can have is the melting temperature of lead. Visualize: Use Equation 17.49. First look up the melting temperature of lead in Table 17.3: Tm = 328°C = 601 K. Then compute the surface area of the sphere: A = 4pR 2 = 4p(0.050 m) 2 = 0.0314 m 2 * Solve:

Q = esAT 4 = (1)(5.6731028 W/m 2?K 4 )(0.0314 m 2 )(601 K) 4 = 230 W Dt Assess:

If the sphere were larger it could radiate more power without melting.

17.31. Model: We will ignore the bottom of the head and model it with just the cylindrical sides and top, all

covered in skin. As instructed, assume an emissivity of e = 0.95 . Visualize: The area of the 2 2 A = Aside + Atop = 2prh + pr = 2p(0.10 m)(0.20 m) + p(0.10 m) = 0.157 m 2 .

cylinder

Solve: Use Equation 17.50.

Qnet = esA(T 4 − T04 ) = (0.95)((5.6731028 W/m 2?K 4 )(0.157 m 2 ))((308 K) 4 − (278 K) 4 ) = 26 W Dt Assess:

This is a significant amount of the heat lost by the body. Wearing a hat can help a lot.

is

17.32. Model: There are various steps to the problem. We must (1) raise the the temperature of the ice to

0°C, (2) melt the ice to liquid water at 0°C, (3) raise the water temperature to 100°C, (4) boil the water to produce steam at 100°C , and (5) raise the temperature of the steam to 200°C. Solve: The heat needed for each step is Q1 = MciceDTice = ( 0.0050 kg )( 2090 J/kg?K )( 20 K ) = 209 J < 210 J Q2 = MLf = ( 0.0050 kg ) ( 3.333105 J/kg ) = 1665 J < 1670 J

Q3 = Mcwater DTwater = ( 0.0050 kg )( 4190 J/kg?K )(100 K ) = 2095 J < 2100 J Q4 = MLv = ( 0.0050 kg ) ( 22.63105 J/kg ) = 11,300 J Q5 = Mcsteam DTsteam = ( 0.0050 kg )( 2009 J/kg?K )(100 K ) = 1005 J < 1000 J The total heat is Q = Q1 + Q2 + Q3 + Q4 + Q5 = 16,300 J Assess: It is interesting to note which step required the most heat: The boiling to change the liquid to the gas is by far the largest contributor to the total.

17.33. Solve: The area of the garden pond is A = π ( 2.5 m ) = 19.635 m 2 and its volume is V = A ( 0.30 m ) = 2

5.891 m 3 . The mass of water in the pond is

M = ρV = (1000 kg m3 )(19.635 m3 ) = 5891 kg The water absorbs all the solar power which is

( 400 W

m 2 )(19.635 m 2 ) = 7854 W

This power is used to raise the temperature of the water. That is,

Q = ( 7854 W ) Δt = Mcwater ΔT = ( 5891 kg )( 4190 J kg K )(10 K ) ⇒ Δt = 31,425 s ≈ 8.7 h

17.34. Model: The potential energy of the bowling ball is transferred into the thermal energy of the mixture. We assume the starting temperature of the bowling ball to be 0°C. Solve: The potential energy of the bowling ball is

U g = M ball gh = (11 kg ) ( 9.8 m s 2 ) h = (107.8 kg m s 2 ) h This energy is transferred into the mixture of ice and water and melts 5 g of ice. That is,

(107.8 kg m s2 ) h = ΔEth = M w Lf ⇒ h =

( 0.005 kg ) ( 3.33 × 105

(107.8 kg m s ) 2

J kg )

= 15.4 m

17.35. Model: Heating the water and the kettle raises the temperature of the water to the boiling point and raises the temperature of the kettle to 100°C. Solve: The amount of heat energy from the electric stove’s output in 3 minutes is

Q = ( 2000 J s )( 3 × 60 s ) = 3.6 × 105 J This heat energy heats the kettle and brings the water to a boil. Thus,

Q = M water cwater ΔT + M kettle ckettle ΔT Substituting the given values into this equation, 3.6 × 105 J = M water ( 4190 J kg K )(100°C − 20°C ) + ( 0.750 kg )( 449 J kg K )(100°C − 20°C )

⇒ M water = 0.994 kg The volume of water in the kettle is

V= Assess:

M water

ρ water

=

0.994 kg = 0.994 × 10−3 m3 = 994 cm3 ≈ 990 cm3 1000 kg m3

1 L = 103 cm3, so V ≈ 1 L. This is a reasonable volume of water.

17.36. Model: Each car’s kinetic energy is transformed into thermal energy. Solve:

For each car,

K = 12 Mv 2 = ΔEth = Mccar ΔT ⇒ ΔT =

v2 2ccar

Assume ccar = ciron. The speed of the car is

80 × 1000 m ( 22.22 m s ) = 0.55°C = 22.22 m s ⇒ ΔT = 2 ( 449 J kg K ) 3600 s 2

v = 80 km hr = Assess:

Notice the answer is independent of the car’s mass.

17.37. Model: There are three interacting systems: aluminum, copper, and ethyl alcohol.

Solve: The aluminum, copper, and alcohol form a closed system, so Q = QAl + QCu + Qeth = 0 J. The mass of the alcohol is

M eth = ρV = ( 790 kg m3 )( 50 × 10−6 m3 ) = 0.0395 kg Expressed in terms of specific heats and using the fact that ΔT = Tf – Ti, the Q = 0 J condition is M AlcAl ΔTAl + M Cu cCu ΔTCu + M eth ceth ΔTeth = 0 J Substituting into this expression,

( 0.010 kg )( 900 J kg K )( 298 K − 473 K ) + ( 0.020 kg )( 385 J kg K )( 298 K − T ) + ( 0.0395 kg )( 2400 J kg K )( 298 K − 288 K ) = −1575 J + ( 7.7 J K )( 298 − T ) + 948 J = 0 J ⇒ T = 216.6 K = −56.4°C ≈ −56°C

17.38. Model: There are two interacting systems: aluminum and ice. The system comes to thermal

equilibrium in four steps: (1) the ice temperature increases from −10°C to 0°C, (2) the ice becomes water at 0°C, (3) the water temperature increases from 0°C to 20°C, and (4) the cup temperature decreases from 70°C to 20°C. Solve: The aluminum and ice form a closed system, so Q = Q1 + Q2 + Q3 + Q4 = 0 J. These quantities are

Q1 = M icecice ΔT = ( 0.100 kg )( 2090 J kg K )(10 K ) = 2090 J

Q2 = M ice Lf = ( 0.100 kg ) ( 3.33 × 105 J kg ) = 33,300 J

Q3 = M icecwater ΔT = ( 0.100 kg )( 4190 J kg K )( 20 K ) = 8380 J Q4 = MAl cAl ΔT = MAl ( 900 J kg K )( −50 K ) = − ( 45,000 J kg ) MAl The Q = 0 J equation now becomes 43,770 J – (45,000 J/kg)MAl = 0 J The solution to this is MAl = 0.973 kg.

17.39. Model: There are three interacting systems: metal, aluminum, and water.

Solve: The metal, aluminum container, and water form a closed system, so Qm + QAl + Qw = 0 J, where Qm is the heat transferred to the metal sample. This equation can be written: MmcmΔTm + MAlcAlΔTAl + MwcwΔT = 0 J Substituting in the given values,

( 0.512 kg ) cm ( 351 K − 288 K ) + ( 0.100 kg )( 900 J kg K )( 351 K − 371 K ) + ( 0.325 kg )( 4190 J kg K )( 351 K − 371 K ) = 0 J ⇒ cm = 900 J kg K From Table 17.2, we see that this is the specific heat of aluminum.

17.40. Solve: For a monatomic gas, the molar specific heat at constant volume is CV = 12.5 J mol K . From Equation 17.22,

12.5 J mol K = The gas is therefore neon.

M mol 625 J kg K ⇒ M mol = 20 g mol 1000 g kg

17.41. Model: Heating the water raises its thermal energy and its temperature. Solve: A 5.0 kW heater has power P = 5000 W. That is, it supplies heat energy at the rate 5000 J/s. The heat supplied in time ∆t is Q = 5000∆t J. The temperature increase is ∆TC = (5/9)∆TF = (5/9)(75°) = 41.67°C. Thus

Q = 5000Δt J = Mcw ΔT = (150 kg)(4190 J/kg K)(41.67°C) ⇒ Δt = 5283 s ≈ 87 min Assess:

A time of ≈1.5 hours to heat 40 gallons of water is reasonable.

17.42. Model: Heating the material increases its thermal energy. Visualize: Heat raises the temperature of the substance from –40°C to –20°C, at which temperature a solid to liquid phase change occurs. From –20°C, heat raises the liquid’s temperature up to 40°C. Boiling occurs at 40°C where all of the liquid is converted into the vapor phase. Solve: (a) In the solid phase,

⎛ ΔQ ⎞ 1 ⎛ 20 kJ ⎞ ⎛ 1 ⎞ ΔQ = McΔT ⇒ c = ⎜ ⎟ = 2000 J kg K ⎟ =⎜ ⎟⎜ ⎝ ΔT ⎠ M ⎝ 20 K ⎠ ⎝ 0.50 kg ⎠ (b) In the liquid phase,

c=

1 M

⎛ ΔQ ⎞ ⎛ 1 ⎞ ⎛ 80 kJ ⎞ ⎟⎜ ⎜ ⎟=⎜ ⎟ = 2667 J kg K ⎝ ΔT ⎠ ⎝ 0.50 kg ⎠ ⎝ 60 K ⎠

(c) The melting point Tm = −20°C and the boiling point Tb = +40°C . (d) The heat of fusion is

Lf =

Q 20,000 J = = 4.0 × 104 J kg M 0.50 kg

Lv =

Q 60,000 J = = 1.2 × 105 J kg M 0.50 kg

The heat of vaporization is

17.43. Model: The liquefaction of the nitrogen occurs in two steps: lowering nitrogen’s temperature from

20°C to −196°C, and then liquefying it at −196°C. Assume the cooling occurs at a constant pressure of 1 atm. Solve: The mass of 1.0 L of liquid nitrogen is M = ρV = ( 810 kg m3 )(10−3 m3 ) = 0.810 kg . This mass corres-

ponds to

n=

M 810 g = = 28.9 mols M mol 28 g mol

At constant atmospheric pressure, the heat to be removed from 28.93 mols of nitrogen is Q = MLv + nCP ΔT

= − ( 0.810 kg ) (1.99 × 105 J kg ) + ( 28.9 mols )( 29.1 J mol K )( 77 K − 293 K ) = −3.4 × 105 J

17.44. Model: There are two interacting systems: coffee (i.e., water) and ice. Changing the coffee

temperature from 90°C to 60°C requires four steps: (1) raise the temperature of ice from −20°C to 0°C, (2) change ice at 0°C to water at 0°C, (c) raise the water temperature from 0°C to 60°C, and (4) lower the coffee temperature from 90°C to 60°C. Solve: For the closed coffee-ice system,

Q = Qice + Qcoffee = ( Q1 + Q2 + Q3 ) + ( Q4 ) = 0 J Q1 = M icecice ΔT = M ice ( 2090 J kg K )( 20 K ) = M ice ( 41,800 J kg )

Q2 = M ice Lf = M ice ( 330,000 J kg ) Q3 = M icecwater ΔT = M ice ( 4190 J kg K )( 60 K ) = M ice ( 251,400 J kg ) Q4 = M coffeeccoffee ΔT = ( 300 × 10−6 m3 )(1000 kg m3 ) ( 4190 J kg K )( −30 K ) = −37,000 J The Q = 0 J equation thus becomes M ice ( 41,800 + 330,000 + 251,400 ) J kg − 37,710 J = 0 J ⇒ M ice = 0.061 kg = 61 g Assess:

61 g is the mass of approximately 1 ice cube.

17.45. Model: We have three interacting systems: the aluminum, the air, and the firecracker. The energy released by the firecracker raises the temperature of the aluminum and the air. We will assume that air is predominantly N2. Assume that the surrounding insulation is excellent. Solve: For the closed firecracker + air + aluminum system, energy conservation requires that Q = Qfirecracker + QAl + Qair = 0 J QAl = mAlcAlΔT = (2.0 kg)(900 J/kg K)(3 K) = 5400 J

⎛ pV Qair = nCV ΔT = ⎜ ⎝ RT

=

(1.01×10

5

Pa )( 20 × 10−3 m3 )

(8.31 J/mol K )( 298 K )

⎞ ⎟ CV ΔT ⎠

( 20.8 J/mol K )( 3 K )

= 51 J Thus,

Qfirecracker = −QAl − Qair

= –5450 J That is, ≈ 5500 J of energy are released on explosion of the firecracker. Assess: The negative sign with Qfirecracker means that the firecracker has lost energy.

17.46. Model: There are two interacting systems: the nuclear reactor and the water. The heat generated by the nuclear reactor is used to raise the water temperature. Solve: For the closed reactor-water system, energy conservation per second requires Q = Qreactor + Qwater = 0 J The heat from the reactor in Δt = 1 s is Qreactor = −2000 MJ = −2.0 × 109 J and the heat absorbed by the water is

Qwater = mwater cwater ΔT = mwater ( 4190 J kg K )(12 K ) ⇒ −2.0 × 109 J + mwater ( 4190 J kg K )(12 K ) = 0 J ⇒ mwater = 3.98 × 104 kg Each second, 3.98 × 104 kg of water is needed to remove heat from the nuclear reactor. Thus, the water flow per minute is

3.98 × 104

kg 60 s 1m3 1L × × × = 2.4 × 106 L/min s min 1000 kg 10−3 m3

17.47. Model: We have two interacting systems: the water and the gas. For the closed system comprised of water and gas to come to equilibrium, heat is transferred from one interacting system to the other. Solve: Energy conservation requires that Qair + Qwater = 0 J ⇒ ngasCV (Tf – Ti gas) + mwaterc(Tf – Ti water) = 0 J Using the ideal-gas law,

Ti gas =

pgasVgas ngas R

(10 ×1.013 × 10 Pa )( 4000 × 10 5

=

( 0.40 mol )(8.31 J

−6

mol K )

m3 )

=1219 K

The energy conservation equation with Ti water = 293 K becomes

( 0.40 mol )(12.5

J mol K )(T − 1219 K ) + ( 20 × 10−3 kg ) ( 4190 J kg K )(T − 293 K ) = 0 J ⇒ Tf = 345 K

We can now use the ideal-gas equation to find the final gas pressure.

pV pV T ⎛ 345 K ⎞ i i = f f ⇒ pf = pi f = ⎜ ⎟ (10 atm ) = 2.8 atm Ti Tf Ti ⎝ 1219 K ⎠

17.48. Model: These are isothermal and isobaric ideal-gas processes. Solve:

(a) The work done at constant temperature is

nRT dV = − nRT ( ln Vf − ln Vi ) = − nRT ln (Vf Vi ) V = − ( 2.0 mol )( 8.31 J mol K )( 303 K ) ln ( 13 ) = 5500 J

W = −∫

Vf

Vi

pdV = − ∫

Vf

Vi

(b) The work done at constant pressure is Vf ⎛V ⎞ 2 W = − ∫ p dV = − p (Vf − Vi ) = − p ⎜ i − Vi ⎟ = pVi Vi ⎝3 ⎠ 3 2 2 = nRT = ( 2.0 mol )( 8.31 J mol K )( 303 K ) = 3400 J 3 3

(c) For an isothermal process in which Vf = 13 Vi , the pressure changes to pf = 3 pi = 4.5 atm.

17.49. Model: This is an isothermal process. The work done is positive for a compression. Solve:

For an isothermal process,

W = − nRT ln (Vf Vi ) For the first process,

W = 500 J = −nRT ln ( 12 ) ⇒ nRT = 721.35 J

For the second process, W = − nRT ln ( 101 ) ⇒ W = − ( 721.35 J ) ln ( 101 ) = 1660 J

17.50.

Solve:

Visualize:

(a) The gas exerts a force on the piston of magnitude 2 Fgas on piston = pgas A = ( 3 atm × 101,300 Pa atm ) ⎡π ( 0.080 m ) ⎤ = 6100 N ⎣ ⎦

This force is directed toward the right. (b) The piston is in static equilibrium, so the environment must exert a force on the piston of equal magnitude Fenviron on piston = 6100 N but in the opposite direction, toward the left. (c) The work done by the environment is G G G Wenviron = Fenviron on piston ⋅ Δr = − Fenviron on piston Δx = − ( 6110 N )( 0.10 m ) = −610 J

The work is negative because the force and the displacement are in opposite directions. (d) The work done by the gas is G G Wgas = Fgas on piston ⋅ Δr = + Fgas on piston Δx = ( 6110 N )( 0.10 m ) = 610 J This work is positive because the force and the displacement are in the same direction. (e) The first law of thermodynamics is W + Q = ΔEth where W is Wenviron. So here W = −610 J and we find Q = ΔEth − W = (196 J ) − ( −610 J ) = 806 J

Thus, 806 J of heat is added to the gas.

17.51. Model: This is an isobaric process. Visualize:

Solve: (a) The initial conditions are p1 = 10 atm = 1.013 × 106 Pa, T1 = 50°C = 323 K, V1 = πr2L1 = π(0.050 m)2 (0.20 m) = 1.57 × 10−3 m3. The gas is heated at a constant pressure, so heat and temperature change are related by Q = nCPΔT. From the ideal gas law, the number of moles of gas is

n=

6 −3 3 p1V1 (1.013 × 10 Pa )(1.57 × 10 m ) = = 0.593 mol RT1 ( 8.31 J mol K )( 323 K )

The temperature change due to the addition of Q = 2500 J of heat is thus ΔT =

Q 2500 J = = 203 K nCP ( 0.593 mol )( 20.8 J mol K )

The final temperature is T2 = T1 + ΔT = 526 K = 253°C. (b) Noting that the volume of a cylinder is V = πr2L and that r doesn’t change, the ideal gas relationship for an isobaric process is

V2 V1 L L T 526 K = ⇒ 2 = 1 ⇒ L2 = 2 L1 = ( 20 cm ) = 33 cm T2 T1 T2 T1 T1 323 K

17.52. Model: The process in part (a) is isochoric and the process in part (b) is isobaric. Solve: (a) Initially V1 = (0.20 m)3 = 0.0080 m3 = 8.0 L and T1 = 293 K. Helium has an atomic mass number A = 4, so 3 g of helium is n = M/Mmol = 0.75 mole of helium. We can find the initial pressure from the ideal-gas law: p1 =

nRT1 ( 0.75 mol )( 8.31 J mol K )( 293 K ) = = 228 kPa = 2.25 atm V1 0.0080 m3

Heating the gas will raise its temperature. A constant volume process has Q = nCVΔT, so

ΔT =

Q 1000 J = = 107 K nCV ( 0.75 mol )(12.5 J mol K )

This raises the final temperature to T2 = T1 + ΔT = 400 K. Because the process is isochoric,

p2 p1 T 400 K = ⇒ p2 = 2 p1 = ( 2.25 atm ) = 3.1 atm T2 T1 T1 293 K (b) The initial conditions are the same as part a, but now Q = nCPΔT. Thus, ΔT =

1000 J Q = = 64.1 K nCP ( 0.75 mol )( 20.8 J mol K )

Now the final temperature is T2 = T1 + ΔT = 357 K. Because the process is isobaric,

V2 V1 T 357 K = ⇒ V2 = 2 V1 = ( 0.0080 m3 ) = 0.0097 m3 = 9.7 L T2 T1 T1 293 K

(c)

17.53. Model: Assume the air and the nitrogen are ideal gases. For the piston to be in equilibrium the forces on it must sum to zero. Also assume that the temperature doesn’t change in the second part. 2 Solve: The area is A = pr 2 = p ( 0.040 m ) = 5.031023 m 2 . V1 = Ah1 = ( 5.031023 m 2 ) ( 0.26 m ) = 0.0013 m3 .

Fnet = pin A − mg − pabove A = 0 N

(a)

pin =

2 23 2 mg + pabove A ( 5.1 kg ) ( 9.8 m/s ) + (100 kPa ) ( 5.0310 m ) = = 110 kPa A 5.031023 m 2

(b) Since T1 = T2 , then p1V1 = p2V2 where p1 is pin from above, and p2 is the inside pressure after the new

weight is added. Compute p2 exactly as above replacing 5.1 kg with 5.1 kg + 3.5kg = 8.6kg. p2 =

2 23 2 mg + pabove A ( 8.6 kg ) ( 9.8 m/s ) + (100 kPa ) ( 5.0310 m ) = = 117 kPa A 5.031023 m 2

V2 =

3 p1V1 (110 kPa ) ( 0.0013 m ) = = 1.2231023 m3 p2 117 kPa

h2 =

V2 1.2231023 m 3 = = 0.2447 m < 24 cm A 5.031023 m 2

Assess: The result seems to be a reasonable number; we expected the piston to be a little lower than before due to the increased mass on it.

17.54. Model: This is an isothermal process.

Solve: (a) The final temperature is T2 = T1 because the process is isothermal. (b) The work done on the gas is V2

V2

V1

V1

W = − ∫ pdV = − ∫

nRT1 V dV = − nRT1 ln 2 = −nRT1ln 2 V V1

(c) From the first law of thermodynamics ΔEth = W + Q = 0 J because ΔT = 0 K. Thus, the heat energy transferred to the gas is Q = −W = nRT1 ln 2 .

17.55. Model: The gas is an ideal gas and it goes through an isobaric and an isochoric process.

Solve: (a) The initial conditions are p1 = 3.0 atm = 304,000 Pa and T1 = 293 K. Nitrogen has a molar mass Mmol = 28 g/mol, so 5 g of nitrogen gas has n = M/Mmol = 0.1786 mol. From this, we can find the initial volume:

V1 =

nRT1 ( 0.1786 mol )( 8.31 J mol K )( 293 K ) = = 1.430 × 10−3 m3 ≈ 1400 cm 3 p1 304,000 Pa

The volume triples, so V2 = 3V1 = 4300 cm3. The expansion is isobaric (p 2 = p1 = 3.0 atm), so

V2 V1 V = ⇒ T2 = 2 T1 = ( 3) 293 K = 879 K = 606°C T2 T1 V1 (b) The process is isobaric, so

Q = nCP ΔT = ( 0.1786 mol )( 29.1 J mol K )( 879 K − 293 K ) = 3000 J (c) The pressure is decreased at constant volume (V3 = V2 = 4290 cm3) until the original temperature is reached (T3 = T1 = 293 K). For an isochoric process,

p3 p2 T 293 K = ⇒ p3 = 3 p2 = ( 3.0 atm ) = 1.0 atm T3 T2 T2 879 K (d) The process is isochoric, so

Q = nCV ΔT = ( 0.1786 mol )( 20.8 J mol K )( 293 K − 879 K ) = −2200 J So, 2200 J of heat was removed to decrease the pressure. (e)

17.56. Model: The gas is an ideal gas. Visualize: In the figure, call the upper left corner (on process A) 2 and the lower right corner (on process B) 3. Solve: The change in thermal energy is the same for any gas process that has the same ∆T. Processes A and B have the same ∆T, since they start and end at the same points, so (∆Eth)A = (∆Eth)B. The first law is then (∆Eth)A = QA + WA = (∆Eth)B = QB + WB ⇒ QA – QB = WB – WA In process B, work W = –p∆V = –pi(2Vi – Vi) = –piVi is done during the isobaric process i → 3. No work is done during the isochoric process 3 → f. Thus WB = –piVi. Similarly, no work is done during the isochoric process i → 2 of process A, but W = –p∆V = –2pi(2Vi – Vi) = –2piVi is done during the isobaric process 2→ f. Thus WA = – 2piVi. Combining these, QA – QB = WB – WA = –piVi – (–2piVi) = piVi

17.57. Model: The two processes are isochoric and isobaric. Solve:

Process A is isochoric which means

Tf Ti = pf pi ⇒ Tf = Ti ( pf pi ) = Ti (1 atm 3 atm ) = 13 Ti From the ideal-gas equation,

Ti =

( 3 ×1.013 ×105 Pa )( 2000 × 10−6 m3 ) = 731.4 K ⇒ T = 1 T = 243.8 K pV i i = f 3 i nR ( 0.10 mol )(8.31 J mol K ) ⇒ Tf − Ti = −487.6 K

Thus, the heat required for process A is

QA = nCV ΔT = ( 0.10 mol )( 20.8 J mol K )( −487.6 K ) = −1000 J Process B is isobaric which means

Tf Vf = Ti Vi ⇒ Tf = Ti (Vf Vi ) = Ti ( 3000 cm3 1000 cm3 ) = 3Ti From the ideal-gas equation, Ti =

( 2 × 1.013 × 105 Pa )(1000 × 10−6 m3 ) = 243.8 K pV i i = nR ( 0.10 mol )( 8.31 J mol K ) ⇒ Tf = 3Ti = 731.4 K ⇒ Tf − Ti = 487.6 K

Thus, heat required for process B is

QB = nCP ΔT = ( 0.10 mol )( 29.1 J mol K )( 487.6 K ) = 1400 J Assess:

Heat is transferred out of the gas in process A, but transferred into the gas in process B.

17.58. Model: We have an adiabatic and an isothermal process.

Solve: For the adiabatic process, no heat is added or removed. That is Q = 0 J. Isothermal processes occur at a fixed temperature, so ∆T = 0 K. Thus ∆Eth = 0 J, and the first law of thermodynamics gives

Q = −W = nRT ln (Vf Vi )

The temperature T can be obtained from the ideal-gas equation as follows: pV i i = nRT ⇒ T =

(1.013 ×105 Pa )( 3000 × 10−6 m3 ) = 366 K pV i i = nR ( 0.10 mol )(8.31 J mol K )

Substituting into the equation for Q we get

⎛ 1000 × 10−6 m3 ⎞ Q = ( 0.10 mol )( 8.31 J mol K )( 366 K ) ln ⎜ = −330 J −6 3 ⎟ ⎝ 3000 × 10 m ⎠ That is, 330 J of heat energy is removed from the gas.

17.59. Model: The monatomic gas is an ideal gas which is subject to isobaric and isochoric processes.

Solve: (a) For the isochoric process, V2 = V1 = 800 × 10−6 m3, p1 = 4.0 atm, p2 = 2.0 atm. The temperature T1 of the gas is obtained from the ideal-gas equation as:

T1 =

p1V1 = 390 K nR

where n = 0.10 mol. T2 can be obtained from the ideal-gas equation as follows: p1V1 p2V2 ⎛ 2.0 atm ⎞ = ⇒ T2 = T1 ( p2 p1 ) = ( 390 K ) ⎜ ⎟ = 195 K T1 T2 ⎝ 4.0 atm ⎠

The heat required for the process 1 → 2 is

Q = nCV (T2 − T1 ) = ( 0.10 mol )( 20.8 J/mol K )(195 K − 390 K ) = −406 J ≈ −410 J Because of the negative sign, this is the amount of heat removed from the gas. (b) For this isobaric process, p2 = p3 = 2.0 atm, V2 = 800 × 10−6 m3, and V3 = 1600 × 10−6 m3.

T3 = T2

⎛ 1600 × 10−6 m3 ⎞ V3 = T2 ⎜ = 2T2 = 390 K −6 3 ⎟ V2 ⎝ 800 × 10 m ⎠

Thus, the heat required for the process 2 → 3 is Q = nCP (T3 − T2 ) = ( 0.10 mol )( 29.1 J mol K )(195 K ) = 567 J ≈ 570 J This is heat transferred to the gas. (c) The change in the thermal energy of the gas is ΔEth = ( Q1→ 2 + Q2 →3 ) + (W1→ 2 + W2 →3 ) = −406 J + 567 J + 0 J +W2 → 3 = 162 J − pΔV = 162 J – (2.0 × 1.013 × 105 Pa)(1600 × 10−6 m3 – 800 × 10−6 m3) = 0 J Assess:

This result was expected since T3 = T1.

17.60. Model: Assume that the gas is an ideal gas. Visualize:

The volume of container A is a constant. On the other hand, heating container B causes the volume to change, but the pressure remains the same. Solve: (a) For the heating of the gas in container A, ΔTA = QA / nCV . Similarly, for the gas in container B,

ΔTB = QB / nCP . Because QA = QB and CP > CV, we see that ΔTA > ΔTB . The gases started at the same temperature, so TA > TB.

(b)

(c) The pressure in container B exerted by the gas is equal to the pressure on the gas by the piston. That is,

pB = patmos +

wpiston Apiston

= 1.013 × 105 Pa +

(10 kg ) ( 9.8 m 1.0 × 10

−4

m

s2 ) 2

= 1.08 × 106 Pa

Container A has the same volume, temperature, and number of moles of gas as container B, so PA = PB = 1.08 × 106 Pa. (d) The heating of container B is isobaric, so

Vf Vi T = ⇒ Vf = Vi f Tf Ti Ti We have Ti = 293 K, and Tf can be obtained from

Q = PΔt = nCP (Tf − Ti ) The number of moles of gas is n = PV i i / RTi = 0.355 mol. Thus

( 25 W )(15 s ) = ( 0.355 mol )( 20.8 J

mol K )(Tf − 293 K )

⇒ Tf = 344 K ⇒ Vf = ( 8.0 × 10−4 m3 ) ( 344 K 293 K ) = 9.39 × 10−4 m3 = 939 cm3 ≈ 940 cm3

17.61. Model: Assume that the gas is an ideal gas. A diatomic gas has γ = 1.40. Solve:

(a) For container A,

ViA =

nRTAi ( 0.10 mol )( 8.31 J mol K )( 300 K ) = = 8.20 × 10−4 m3 pAi 3.0 × 1.013 × 105 Pa

For an isothermal process pAfVAf = pAiVAi. This means TAf = TAi = 300 K and

VAf = VAi ( pAi pAf ) = ( 8.20 × 10−4 m3 ) ( 3.0 atm 1.0 atm ) = 2.5 × 10−3 m3 The gas in container B starts with the same initial volume. For an adiabatic process,

⎛p ⎞ pBf VBf = pBiVBi ⇒ VBf = VBi ⎜ Bi ⎟ ⎝ pBf ⎠ γ

γ

1

γ

⎛ 3.0 atm ⎞ = ( 8.20 × 10 m ) ⎜ ⎟ ⎝ 1.0 atm ⎠ −4

3

1 1.40

= 1.8 × 10−3 m3

The final temperature TBf can now be obtained by using the ideal-gas equation: TBf = TiB

(b)

−3 3 pBf VBf ⎛ 1.0 atm ⎞ ⎛ 1.80 × 10 m ⎞ = ( 300 K ) ⎜ = 220 K ⎟⎜ 3 ⎟ −4 pBi VBi ⎝ 3.0 atm ⎠ ⎝ 8.20 × 10 m ⎠

17.62. Model: The gas is assumed to be an ideal gas that is subjected to isobaric and isochoric processes.

Solve: (a) The initial conditions are p1 = 3.0 atm = 304,000 Pa, V1 = 100 cm3 = 1.0 × 10−4 m3, and T1 = 100°C = 373 K. The number of moles of gas is

n=

−4 3 p1V1 ( 304,000 Pa ) (1.0 × 10 m ) = = 9.81 × 10−3 mol RT1 ( 8.31 J mol K ) (373 K)

At point 2 we have p2 = p1 = 3.0 atm and V2 = 300 cm3 = 3V1. This is an isobaric process, so

V2 V1 V = ⇒ T2 = 2 T1 = 3(373 K) = 1119 K T2 T1 V1 The gas is heated to raise the temperature from T1 to T2. The amount of heat required is

Q = nCP ΔT = ( 9.81 × 10−3 mol ) ( 20.8 J mol K )(1119 K − 373 K ) = 152 J This amount of heat is added during process 1 → 2. (b) Point 3 returns to T3 = 100°C = 373 K. This is an isochoric process, so

Q = nCV ΔT = ( 9.81 × 10−3 mol ) (12.5 J mol K )( 373 K − 1119 K ) = −91.5 J This amount of heat is removed during process 2 → 3.

17.63. Assume the gas to be an ideal gas. Solve:

(a) The work done on the gas is the negative of the area under the p-versus-V graph, that is

W = −area under curve = −50.7 J (b) The change in thermal energy is

ΔEth = nCV ΔT = nCV (Tf − Ti )

Using the ideal-gas law to calculate the initial and final temperatures, Ti =

( 4.0 ×1.013 × 105 Pa )(100 × 10−6 m3 ) = 325 K pV i i = nR ( 0.015 mol )(8.31 J mol K )

Tf =

5 −6 3 pfVf (1.013 × 10 Pa )( 300 × 10 m ) = = 244 K nR ( 0.015 mol )( 8.31 J mol K )

⇒ ΔEth = ( 0.015 mol )(12.5 J mol K )( 244 K − 325 K ) = −15.2 J ≈ −15 J (c) From the first law of thermodynamics, ΔEth = Q + W ⇒ Q = ΔEth − W = −15.2 J − ( −50.7 J ) = 35.5 J ≈ 36 J

That is, 36 J of heat energy is transferred to the gas.

17.64. Model: Assume that the gas is an ideal gas and that the work, heat, and thermal energy are connected by the first law of thermodynamics. Solve: (a) For point 1, V1 = 1000 cm3 = 1.0 × 10−3 m3, T1 = 133°C = 406 K, and the number of moles is

n=

⎛ 120 × 10−3 g ⎞ M =⎜ ⎟ = 0.030 mol M mol ⎝ 4 g/mol ⎠

Thus, the pressure p1 is

p1 =

nRT1 = 1.012 × 105 Pa = 1.0 atm V1

The process 1 → 2 is isochoric (V2 = V1) and p2 = 5p1 = 5.0 atm. Thus,

T2 = T1 ( p2 p1 ) = ( 406 K )( 5 ) = 2030 K = 1757°C The process 2 → 3 is isothermal (T2 = T3), so V3 = V2(p2/p3) = V2(p2/p1) = 5V2 = 5000 cm3 p (atm)

T (°C)

V (cm3)

1.0 5.0 1.0

133 1757 1757

1000 1000 5000

Point 1 Point 2 Point 3

(b) The work W1→ 2 = 0 J because it is an isochoric process. The work in process 2 → 3 can be found using Equation 17.16 as follows:

W2 →3 = − nRT2 ln (V3 V2 ) = − ( 0.030 mol )( 8.31 J mol K )( 2030 K ) ln ( 5 ) = −815 J The work in the isobaric process 3 → 1 is

W3→1 = − p (Vf − Vi ) = − (1.012 × 105 Pa )(1.0 × 10−3 m3 − 5.0 × 10−3 m3 ) = 405 J (c) The heat transferred in process 1 → 2 is

Q1→ 2 = nCV ΔT = ( 0.030 mol )(12.5 J mol K )( 2030 K − 406 K ) = 609 J

The heat transferred in the isothermal process 2 → 3 is Q2 →3 = −W2 →3 = 815 J . The heat transferred in the isobaric process 3 → 1 is Q3→1 = nCP ΔT = ( 0.030 mol )( 20.8 J mol K )( 406 K − 2030 K ) = −1013 J

17.65. Model: The air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process.

Solve: The air admitted into the cylinder at T0 = 30°C = 303 K and p0 = 1 atm = 1.013 × 105 Pa has a volume V0 = 600 × 10−6 m3 and contains pV n = 0 0 = 0.024 mol RT0

Using Equation 17.36 and the fact that Q = 0 J for an adiabatic process, ΔEth = Q + W = nCV ΔT ⇒ W = nCVT

⇒ 400 J = (0.024 mol)(20.8 J/mol K)(Tf – 303 K) ⇒ Tf = 1100 K For an adiabatic process Equation 17.40 is 1

TfVf Assess:

γ −1

γ −1

= T0V0

1

⎛ T ⎞ γ −1 ⎛ 303 K ⎞1.4 −1 ⇒ Vf = V0 ⎜ 0 ⎟ = ( 6.0 × 10−4 m3 ) ⎜ = 2.39 × 10−5 m3 ≈ 24 cm3 ⎟ T ⎝ 1100 K ⎠ ⎝ f⎠

Note that W is positive because the environment does work on the gas.

17.66. Model: γ is 1.40 for a diatomic gas and 1.67 for a monoatomic gas. Solve:

(a) We will assume that air is a diatomic gas. For an adiabatic process, γ −1 Tf Vfγ −1 = TV i i

Thus 1

1

⎛ Vi ⎞ ⎛ Tf ⎞ γ −1 ⎛ 1123 K ⎞1.40 −1 = 26.4 ⎜ ⎟=⎜ ⎟ =⎜ ⎟ ⎝ 303 K ⎠ ⎝ Vf ⎠ ⎝ Ti ⎠ (b) For argon, a monatomic gas, 1

⎛ Vi ⎞ ⎛ 1123 K ⎞1.67 −1 = 7.07 ⎜ ⎟=⎜ ⎟ ⎝ Vf ⎠ ⎝ 303 K ⎠

17.67. Model: Air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process. Solve:

(a) Equation 17.40 for an adiabatic process is 1

γ −1

TfVf

= TV i i

γ −1

V ⎛ T ⎞ γ −1 ⇒ f =⎜ i ⎟ Vi ⎝ Tf ⎠

For the temperature to increase from Ti = 20°C = 293 K to Tf = 1000°C = 1273 K, the compression ratio will be 1

V Vf ⎛ 293 K ⎞1.4 −1 1 = 39.3 =⎜ = 0.02542 ⇒ max = ⎟ Vmin 0.02542 Vi ⎝ 1273 K ⎠ (b) From the Equation 17.39, γ

pf ⎛ Vi ⎞ 1.4 pfVf = pV = ⎜ ⎟ = ( 39.3) = 171 i i ⇒ pi ⎝ Vf ⎠ γ

γ

17.68. Model: The helium gas is assumed to be an ideal gas that is subjected to an isobaric process. Solve:

(a) The number of moles in 2.0 g of helium is

n=

M 2.0 g = = 0.50 mol M mol 4.0 g mol

At Ti = 100°C = 373 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume Vi =

nRTi = 0.0153 m3 = 15.3 L pi

For an isobaric process (pf = pi) that doubles the volume Vf = 2Vi,

(Tf

Ti ) = (Vf Vi ) = 2 ⇒ Tf = 2Ti = 2 ( 373 K ) = 746 K = 473°C

(b) The work done by the environment on the gas is 5 3 W = − pi (Vf − Vi ) = − pV i i ( 2 − 1) = − (1.013 × 10 Pa )( 0.0153 m ) = −1550 J

(c) The heat input to the gas is

Q = nCP (Tf − Ti ) = ( 0.50 mol )( 20.8 J mol K )( 746 K − 373 K ) = 3880 J ≈ 3900 J (d) The change in the thermal energy of the gas is

ΔEth = Q + W = 3880 J − 1550 J = 2330 J ≈ 2300 J

(e)

Assess:

The internal energy can also be calculated as follows:

ΔEth = nCV ΔT = ( 0.5 mol )(12.5 J mol K )( 746 K − 373 K ) = 2330 J This is the same result as we got in part (d).

17.69. Model: The helium gas is assumed to be an ideal gas that is subjected to an isothermal process. Solve:

(a) The number of moles in 2.0 g of helium gas is

n=

M 2.0 g = = 0.50 mol M mol 4.0 g mol

At Ti = 100°C = 373 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume Vi =

nRTi = 0.0153 m3 = 15.3 L pi

For an isothermal process (Tf = Ti) that doubles the volume Vf = 2Vi, 1 pf Vf = pV i i ⇒ pf = pi (Vi Vf ) = (1.0 atm ) ( 2 ) = 0.50 atm

(b) The work done by the environment on the gas is

W = − nRTi ln (Vf Vi ) = − ( 0.50 mol )( 8.31 J mol K )( 373 K ) ln ( 2 ) = −1074 J ≈ −1070 J (c) Because ΔEth = Q + W = 0 J for an isothermal process, the heat input to the gas is Q = −W = 1074 J ≈ 1070 J . (d) The change in internal energy ΔEth = 0 J.

(e)

17.70. Model: The nitrogen gas is assumed to be an ideal gas that is subjected to an adiabatic process. Solve:

(a) The number of moles in 14.0 g of N2 gas is

n=

M 14.0 g = = 0.50 mol M mol 28 g mol

At Ti = 273 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume Vi =

nRTi = 0.0112 m3 = 11.2 L pi

For an adiabatic process that compresses to a pressure pf = 20 atm, we can use Equation 17.39 and Equation 17.40 as follows: TfVf

γ −1

= TV i i

γ −1

⎛T ⎞ ⎛V ⎞ ⇒⎜ f ⎟=⎜ i ⎟ ⎝ Ti ⎠ ⎝ Vf ⎠

γ −1

1

⎛ p ⎞γ V pf Vf = piVi ⇒ ⎜ f ⎟ = i Vf ⎝ pi ⎠ γ

γ

Combining the above two equations yields

(Tf

Ti ) = ( pf pi )

γ −1 γ

⇒ Tf = Ti ( 20 )

1.4 −1.0 1.4

= Ti ( 2.3535 ) = 643 K

(b) The work done on the gas is

W = ΔEth = nCV (Tf − Ti ) = ( 0.50 mol )( 20.8 J mol K )( 643 K − 273 K ) = 3850 J ≈ 3800 J (c) The heat input to the gas is Q = 0 J. (d) From the above equation, 1

1 Vi ⎛ pf ⎞ γ V = ⎜ ⎟ = ( 20 )1.4 = 8.5 = max Vf ⎝ pi ⎠ Vmin

(e)

17.71. Model: The gas is assumed to be an ideal gas that is subjected to an isochoric process. Solve:

(a) The number of moles in 14.0 g of N2 gas is

n=

M 14.0 g = = 0.50 mol M mol 28 g mol

At Ti = 273 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume Vi =

nRTi = 0.0112 m3 = 11.2 L pi

For an isochoric process (Vi = Vf), Tf pf 20 atm = = = 20 ⇒ Tf = 20 ( 273 K ) = 5460 K ≈ 5500 K Ti pi 1 atm (b) The work done on the gas is W = − pΔV = 0 J. (c) The heat input to the gas is Q = nCV (Tf − Ti ) = ( 0.50 mol )( 20.8 J mol K )( 5460 K − 273 K ) = 5.4 × 10 4 J

(d) The pressure ratio is

pmax pf 20 atm = = = 20 pmin pi 1 atm

(e)

17.72. Model: The air is assumed to be an ideal gas. Because the air is compressed without time to exchange heat with its surroundings, the compression is an adiabatic process. Solve: The initial pressure of air in the mountains behind Los Angeles is pi = 60 × 103 Pa at Ti = 273 K. The pressure of this air when it is carried down to the elevation near sea level is pf = 100 × 103 Pa. The adiabatic compression of a gas leads to an increase in temperature according to Equation 17.39 and Equation 17.40, which are γ −1

TfVf

= TV i i

γ −1

⎛T ⇒⎜ f ⎝ Ti

γ −1

⎞ ⎛ Vi ⎞ ⎟=⎜ ⎟ ⎠ ⎝ Vf ⎠

1

⎛ p ⎞γ V pfVf = piVi ⇒ ⎜ f ⎟ = i Vf ⎝ pi ⎠ γ

γ

Combining these two equations,

(Tf

Ti ) = ( pf pi )

γ −1 γ

⎛ 100 × 103 Pa ⎞ ⇒ Tf = Ti ⎜ ⎟ 3 ⎝ 60 × 10 Pa ⎠

1.4 −1 1.4

⎛5⎞ = ( 273 K ) ⎜ ⎟ ⎝3⎠

0.286

= 316 K = 43ºC = 109ºF

17.73. Model: Assume the collector has emissivity e = 1.0 . We will use Equation 17.50 and divide both sides by A to get intensity in W/m 2 . At the equilibrium temperature the collector will absorb 800 W/m 2 on the sun side and must radiate the same amount from the same side. T0 = 20°C = 293 K . Solve: Q/ Dt = es (T 4 − T04 ) A Solve this for T . Q/ Dt 1 = T 4 − T04 A es

T=

4

Q/ Dt 1 + T04 = A es

4

800 W/m 2 4 + ( 293K ) = 383K = 110°C 28 2 4 1 0 5 67 10 W/m K . . × ⋅ ( )( )

Assess: 110°C is hot enough to boil water, so this seems feasible as a hot water heater, at least during the 5 hours a day the collector gets 800 W/m 2 . You can store the hot water in an insulated tank for use at other times of the day.

17.74. Model: This is a problem about conduction of the heat from inside the box to outside. 100 W of heat is generated by the light bulb inside the box, so in equilibrium that is how much must be conducted away through the sides of the box. Visualize: We are given L = 0.012 m , Q/ Dt = 100 W and the thermal conductivity of concrete k = 0.8 W/m ⋅ K . We compute A = 6 ( 0.20m × 0.20 m ) = 0.24m 2 . TC = 20°C = 293K .

Solve:

Use Equation 17.48 for the rate of heat transfer by conduction. Q A = k (TH − TC ) Dt L

Solve for TH . ⎛Q⎞ L = (TH − TC ) ⎜ ⎟ ⎝ Dt ⎠ kA

0.012 m ⎛Q⎞ L TH = ⎜ ⎟ + TC = (100 W ) + 293 K = 299 K = 26°C D t kA 0 8 W/m . ?K ) ( 0.24 m 2 ) ⎝ ⎠ ( Assess:

We expected TH to be a few degrees hotter than TC , and so it is.

17.75. Model: Refer to Example 17.11. Assume the surface of the earth is an ideal radiator with e = 1 . Visualize: Only half the earth faces the sun at a time, so the earth intercepts 1370 W/m 2 over a cross section of pRe2 = p ( 6.37 × 106 m ) = 1.275 × 1014 m 2 . If the earth’s surface absorbs 70% of the incident power then the total 2

power absorbed is

( 0.70 ) (1370 W/m 2 )(1.275 × 1014 m 2 ) = 1.222 × 1017 W

. This much power must also be

radiated away from the earth in equilibrium. Solve: The power is radiated from the whole spherical surface of the earth: A = 4pRe2 . Use Equation 17.49 to find the temperature of the earth.

⎡ Q/ Dt ⎤ ⎥ T =⎢ 2 ⎢⎣ es ( 4pRe ) ⎥⎦

1/ 4

⎡ ⎤ 1.22231017 W ⎥ =⎢ 2 ⎢ (1) ( 5.6731028 W/m 2?K ) 4p ( 6.373106 m ) ⎥ ⎣ ⎦

1/ 4

= 255 K = 218°C

Assess: Without a moderate greenhouse effect the earth would be too cold for mammal life. On the other hand, when the greenhouse effect gets out of hand, such as on Venus, it can be too hot for life.

17.76. Solve: (a) 50 J of work are done on a gas to compress it to one-third of its original volume at a constant temperature of 77°C. How many moles of the gas are in the sample? (b) The number of moles is

n=

50 J = 0.0156 mol − ( 8.31 J mol K )( 350 K ) ( ln 13 )

17.77. Solve: (a) A heated 500 g iron slug is dropped into a 200 cm3 pool of mercury at 15°C. If the mercury temperature rises to 90°C, what was the initial temperature of the iron slug? (b) The initial temperature was 217°C .

17.78. Solve: (a) A diatomic gas is adiabatically compressed from 1 atm pressure to 10 atm pressure. What is the compression ratio Vmax/Vmin? 1

(b) The ratio is Vmax Vmin = 101.4 = 5.18 .

17.79. Model: Assume the helium gas to be an ideal gas. The gas is subjected to isothermal, isochoric, and adiabatic processes. Visualize: Please refer to Figure CP17.79. The gas at point 1 has volume V1 = 1000 cm3 = 1.0 × 10−3 m3 and pressure p1 = 3.0 atm. At point 2, V2 = 3000 cm3 = 3.0 × 10−3 m3 and p2 = 1.0 atm. These values mean that T2 = T1, so process 1 → 2 is an isothermal process. The process 2 → 3 occurs at constant volume and is thus an isochoric process. Finally, because temperature T3 is lower than T1 or T2, the process 3 → 1 is an adiabatic process. Solve: (a) The number of moles of gas is n=

0.120 g = 0.030 mols 4 g mol

The temperature T1 can be calculated to be T1 =

5 −3 3 p1V1 ( 3.0 × 1.013 × 10 Pa )(1.0 × 10 m ) = = 1219 K = 946°C nR ( 0.030 mol )(8.31 J mol K )

For the isothermal process 1 → 2, T2 = 1219 K. For the adiabatic process 3 → 1, p3 = p1 (V1 V3 )

γ

1.67

⎛ 1000 cm3 ⎞ = ( 3.0 atm ) ⎜ 3 ⎟ ⎝ 3000 cm ⎠ Point

= 0.48 atm

p (atm)

T3 = T1 (V1 V3 )

γ −1

T (°C)

= (1219 K ) ( 13 )

0.67

= 583 K=310°C

V (cm3)

1 3.0 946 1000 2 1.0 946 3000 3 0.48 310 3000 Note that the values obtained above are consistent with the isochoric process 2 → 3, for which p2 p3 = ⇒ p2 = (T2 T3 ) p3 = (1219 K 583 K )( 0.48 atm ) = 1 atm T2 T3 (b) From Equation 17.15,

⎛V ⎞ W1→ 2 = − nRT1 ln ⎜ 2 ⎟ = − ( 0.030 mol )( 8.31 J mol K )(1219 K ) ln ( 3) = −334 J ⎝ V1 ⎠

The work done in the ischochoric process is W2 →3 = 0 J . The work done in the adiabatic process is

W3→1 = nCV (T1 − T3 ) = ( 0.030 mol )(12.5 J mol K )(1219 K − 583 K ) = 239 J (c) For the process 1 → 2, ΔT = 0 K ⇒ ΔEth = 0 J ⇒ Q = −W = 334 J

For the process 2 → 3, W = 0 J,

ΔEth = Q = nCV (T3 − T2 ) = −239 J For the process 3 → 1, Q = 0 J.

17.80. Model: The gas is an ideal gas, and its thermal energy is the total kinetic energy of the moving molecules. Visualize: Please refer to Figure P17.80. Solve: (a) The piston is floating in static equilibrium, so the downward force of gravity on the piston’s mass must exactly balance the upward force of the gas, Fgas = pA where A = πr2 is the area of the face of the piston. Since the upper part of the cylinder is evacuated, there is no gas pressure force pushing downward. Thus,

M piston g = pA ⇒ p =

M piston g A

=

ρCuVpiston g A

= ρ Cu gh = ( 8920 kg m3 )( 9.80 m s 2 ) ( 0.040 m ) = 3500 Pa

(b) The gas volume is V1 = π r 2 L = π (0.030) 2 (0.20 m) = 5.65 × 10−4 m3 . The number of moles is

n=

−4 3 p1V1 ( 3500 Pa ) ( 5.65 × 10 m ) = = 8.12 × 10−4 mol RT1 ( 8.31 J mol K )( 293 K )

The number of molecules is N = nNA = (8.12 × 10−4 mol)(6.02 × 1023 mol−1) = 4.9 × 1020 (c) The pressure in the gas is determined simply by the weight of the piston. That will not change as heat is added, so the heating takes place at constant pressure with Q = nCPΔT. The temperature increase is

ΔT =

2.0 J Q = = 85 K nCP ( 8.12 × 10−4 mol ) ( 29.1 J mol K )

This raises the gas temperature to T2 = T1 + ΔT = 378 K = 105°C. (d) Noting that the volume of a cylinder is V = πr2L and that r doesn’t change, the ideal-gas relationship for an isobaric process is V2 V1 L L T ⎛ 378 K ⎞ = ⇒ 2 = 1 ⇒ L2 = 2 L1 = ⎜ ⎟ 20 cm = 25.8 cm T2 T1 T2 T1 T1 ⎝ 293 K ⎠ (e) The work done by the gas is Wgas = Fgas Δy . The force exerted on the piston by the gas is

Fgas = pA = pπ r 2 = 9.90 N This force is applied through Δy = 5.8 cm = 0.058 m, so the work done is Wgas = (9.90 N)(0.058 m) = 0.574 J ≈ 0.57 J

Thus, 0.57 J is the work done by the gas on the piston. The work done on the gas is 0.57 J.

17.81. Model: There is a thermal interaction between the iron, assumed to be initially at room temperature (20°C), and the liquid nitrogen. The boiling point of liquid nitrogen is –196°C = 77 K. Solve: The piece of iron has mass Miron = 197 g = 0.197 kg and volume Viron = Miron/ρiron = (0.197 kg)/(7870 kg/m3) = 25 × 10−6 m3 = 25 cm3 = 25 mL. The heat lost by the iron is Qiron = Mironciron∆T = (0.197 kg)(449 J/kg K)(77 K – 293 K) = –1.911 × 104 J This heat causes mass M of the liquid nitrogen to boil. Energy conservation requires Qiron + QN2 = Qiron + MLf = 0 ⇒ M = −

Qiron 1.911 × 104 J = = 0.0960 kg Lf 1.99 × 105 J/kg

The volume of liquid nitrogen boiled away is thus

Vboil =

M

ρ N2

=

0.0960 kg = 1.19 × 10−4 m3 = 119 mL 810 kg/m3

Now the volume of nitrogen gas (at 77 K) is 1500 mL before the iron is dropped in. The volume of the piece of iron excludes 25 mL of gas, so the initial gas volume, when the lid is sealed and the liquid starts to boil, is V1 = 1475 mL. The pressure is p1 = 1.0 atm and the temperature is T1 = 77 K. Thus the number of moles of nitrogen gas is n1 =

p1V1 (101,300 Pa)(1.475 × 10−3 m3 ) = = 0.234 mol RT1 (8.31 J/mol K)(77 K)

119 mL of liquid boils away, so the gas volume increases to V2 = V1 + Vboil = 1475 mL + 119 mL = 1594 mL. The temperature is still T2 = 77 K, but the number of moles of gas has been increased by the liquid that boiled. The number of moles that boiled away is 96.0 g nboil = = 3.429 mol 28 mol/g

Thus the number of moles of nitrogen gas increases to n2 = n1 + nboil = 0.234 mol + 3.429 mol = 3.663 mol. Consequently, the gas pressure increases to p2 =

n2 RT2 (3.663 mol)(8.31 J/mol K)(77 K) = = 1.470 × 106 Pa = 14.5 atm ≈ 15 atm V2 1.594 × 10−3 m3

Assess: Don’t try this! The large pressure increase could cause a flask of liquid nitrogen to explode, leading to serious injuries.

17.82.

Model: Ignore any radiation and assume all of the heat is transferred by conduction through the compound rod from the hot end to the cold end. Visualize: Look up the thermal conductivities of the two metals in Table 17.5: kCu = 400 W/m? k and

kFe = 80 W/m?K . TH = 373 K and TC = 273 K. Equation 17.48 will be applied to each rod and all of the heat must go through both rods at the same rate. ⎛Q⎞ ⎛Q⎞ ⎜ ⎟ =⎜ ⎟ D t ⎝ ⎠Cu ⎝ Dt ⎠ Fe

Solve:

Since A and L are the same for the two rods, then kCu DTCu = kFeDTFe

Label the temperature at the joining point in the middle with a subscript M.

kCu (TH − TM ) = kFe (TM − TC ) Solve for TM . kCuTH − kCuTM = kFeTM − k FeTC kCuTH + kFeTC = kFeTM + kCuTM

( kFe + kCu )TM = kCuTH + kFeTC TM = TM =

kCuTH + kFeTC kCu + kFe

(400 W/m ⋅ K)(373 K) + (80 W/m ⋅ K)(273 K) = 356 K = 83°C 400 W/m ⋅ K + 80 W/m ⋅ K

Assess: Had we not arrived at an answer between 100°C and 0°C we would have been very worried. Furthermore, because the conductivity of copper is greater than the conductivity of iron we expected the answer to be above 50°C.

17.83. Model: Assume the gas is ideal. The process is not adiabatic, despite Equation 17.38, because the exponent

given in the problem is not equal to g . The gas is identified as diatomic which means that g = 1.40 , but the exponent is

explicitly given as 2, so while we are given pV 2 = constant, it is not true that pV g = constant. We will use the fact that the gas is diatomic to deduce CV = 52 R = 20.8J/mol ⋅ k , although we won’t specifically need g = 1.40 . We are given n = 0.020mol , Ti = 293K , Vi = 1500cm3 , and Vf = 500cm3 .

Visualize: Solve:

(a) If we use the ideal-gas-law expression p = nRT/V in pV 2 = constant we get TV = constant .

Tf =

TV (293 K)(1500 cm3 ) i i = = 879 K = 606°C Vf 500 cm3

(b) The strategy to find Q will be to use the first law Q = DEth − W . We will use Eth = nCV DT and W = 2∫ pdV.

DEth = nCV DT = ( 0.020 mol )( 20.8 J/mol ⋅ k )( 879 K − 293 K ) = 243.8 J To do the W = 2∫ pdV integral we need to know what the constant is in pV 2 = constant . Use the ideal gas law to compute pi . pi =

nRTi ( 0.020 mol )( 8.31 J/mol ⋅ K )( 293 K ) = = 32,460 Pa Vi 1.5 ×1023 m3

23 2 constant = pV m 3 ) = 0.0730 Pa ⋅ m 6 i i = ( 32,460Pa ) (1.5 × 10 2

Vf

W = − ∫ pdV = 2Ñ

Vi

constant V2

dV = 2( 0.0730 Pa ⋅ m

6

)Ñ

0.00050 m3

0.0015 m3

0.00050 m3

dv ⎡1⎤ = ( 0.0730 Pa ⋅ m 6 ) ⎢ ⎥ = 97.4 J 2 V ⎣V ⎦ 0.0015 m3

Q = DEth − W = 243.8 J − 97.4 J = 146.4 J < 150 J (c) We also need pf to complete the pV diagram. We could use the ideal-gas law again to compute pf , but let’s do it another way. 2 2 pV i i = pf Vf 2

2

⎛V ⎞ ⎛ 1500 cm3 ⎞ pf = pi ⎜ i ⎟ = pi ⎜ = 9 pi = 292,200 Pa 3 ⎟ ⎝ 500 cm ⎠ ⎝ Vf ⎠

Assess: Both Q and W are positive because heat was added to the system and work was done on the system.

17-1

18.1. Solve: We can use the ideal-gas law in the form pV = NkBT to determine the Loschmidt number (N/V):

(1.013 × 105 Pa ) = 2.69 × 1025 m−3 N p = = V kBT (1.38 × 10−23 J K ) ( 273 K )

18.2. Solve: The volume of the nitrogen gas is 1.0 m3 and its temperature is 20°C or 293 K. The number of gas molecules can be found as N = nN A =

(1.013 × 105 Pa )(1.0 m3 ) 6.02 × 1023 mol−1 = 2.5 × 1025 pV NA = ( ) RT ( 8.31 J mol K )( 293 K )

According to Figure 18.2, 12% of the molecules have a speed between 700 and 800 m/s, 7% between 800 and 900 m/s, and 3% between 900 and 1000 m/s. Thus, the number of molecules in the cube with a speed between 700 m/s and 1000 m/s is (0.22)(2.51 × 1025) = 5.5 × 1024.

18.3. Solve: Nitrogen is a diatomic molecule, so r ≈ 1.0 × 10−10 m. We can use the ideal-gas law in the form pV = NkBT and Equation 18.3 for the mean free path to obtain p:

λ=

(1.38 ×10−23 J k ) ( 293 K ) = 0.023 Pa kBT 1 kBT = ⇒ p = = 2 4 2π ( N V ) r 2 4 2π pr 2 4 2πλ r 2 4 2π (1.0 m ) (1.0 × 10−10 m )

Assess: In Example 18.1 λ = 225 nm at STP for nitrogen. λ = 1.0 m must therefore require a very small pressure.

18.4. Solve: (a) Air is primarily comprised of diatomic molecules, so r ≈ 1.0 × 10−10 m. Using the ideal-gas

law in the form pV = NkBT, we get

N p = = V kBT

1.013 × 105 Pa 760 mm of Hg = 3.30 × 1012 m −3 J K ) ( 293 K )

1.0 × 10−10 mm of Hg ×

(1.38 ×10

−23

(b) The mean free path is

λ=

1 1 = = 1.71 × 106 m 2 4 2π ( N V ) ( r 2 ) 4 2π ( 3.30 × 1012 m −3 )(1.0 × 10−10 m )

Assess: The pressure p in the vacuum chamber is 1.33 × 10−8 Pa = 1.32 × 10−13 atm. A mean free path of 1.71 × 106 m is large but not unreasonable.

18.5. Solve: (a) The mean free path of a molecule in a gas at temperature T1, volume V1, and pressure p1 is λ1 = 300 nm. We also know that

λ=

1 ⇒ λ ∝V 4 2π ( N / V )r 2

Although T2 = 2T1, constant volume (V2 = V1) means that λ2 = λ1 = 300 nm. (b) For T2 = 2 T1 and p2 = p1, the ideal gas equation gives

p1V1 pV p1V2 ⇒ V2 = 2V1 = 2 2 = NkBT1 NkBT2 NkB ( 2T1 )

Because λ ∝ V , λ2 = 2λ1 = 2(300 nm) = 600 nm.

18.6. Solve: Neon is a monatomic gas and has a radius r ≈ 5.0 × 10−11 m. Using the ideal-gas equation,

(150 ) (1.013 × 105 Pa ) N p = = = 3.695 × 1027 m −3 V kBT (1.38 × 10−23 J/K ) ( 298 K ) Thus, the mean free path of a neon atom is

λ=

1 1 = = 6.09 × 10−9 m 2 2 27 4 2π ( N V ) r 4 2π ( 3.695 × 10 m −3 )( 5.0 × 10−11 m )

Since the atomic diameter of neon is 2 × (5.0 × 10−11 m) = 1.0 × 10–10 m,

λ=

6.09 × 10−9 m = 61 atomic diameters 1.0 × 10−10 m

18.7. Solve: The number density of the Ping-Pong balls inside the box is 2000 N = = 2000 m −3 V 1.0 m3 With r = (3.0 cm)/2 = 1.5 cm, the mean free path of the balls is

λ=

1 = 0.125 m = 12.5 cm 4 2π ( N V ) ( r 2 )

18.8. Solve: (a) The average speed is vavg =

1 ⎛ n = 25 ⎞ 220 m/s = 20.0 m/s ⎜ ∑ n ⎟ m/s = 11 ⎝ n =15 ⎠ 11

(b) The root-mean-square speed is vrms =

(v2 )

1

avg

1

⎛ 1 n = 25 ⎞ 2 ⎛ 4510 ⎞ 2 = ⎜ ∑ n 2 ⎟ m/s = ⎜ ⎟ = 20.2 m/s ⎝ 11 ⎠ ⎝ 11 n =15 ⎠

18.9. Solve: (a) In tabular form we have Particle

vx (m/s)

vy (m/s)

1 20 30 2 70 −40 3 −80 −10 4 60 −20 5 0 −50 6 40 −20 Average 0 0 G G G ˆ The average velocity is vavg = 0 i + 0 ˆj .

vx2 (m/s)2

v y2 (m/s)2

v2 (m/s)2

v (m/s)

400 1600 6400 3600 0 1600

900 4900 100 400 2500 400

1300 6500 6500 4000 2500 2000 3800

36.06 80.62 80.62 63.25 50.00 44.72 59.20

(b) The average speed is vavg = 59 m/s . (c) The root-mean-square speed is vrms =

(v ) 2

avg

= 3800 m 2 / s 2 = 62 m/s .

18.10. Solve: (a) The most probable speed is 4.0 m/s. (b) The average speed is vavg =

2 × 2 m/s + 4 × 4 m/s + 3 × 6 m/s + 1 × 8 m/s = 4.6 m/s 2 + 4 + 3 +1

(c) The root-mean-square speed is

2 × ( 2 m/s ) + 4 × ( 4 m/s ) + 3 × ( 6 m/s ) + 1 × ( 8 m/s ) = 4.9 m/s 2 + 4 + 3 +1 2

vrms =

2

2

2

18.11. Solve: (a) The atomic mass number of argon is 40. This means the mass of an argon atom is m = 40 u = 40(1.661 × 10−27 kg) = 6.64 × 10−26 kg The pressure of the gas is 2 ⎛N⎞ 2 p = 13 ⎜ ⎟ mvrms = 13 ( 2.00 × 1025 m −3 )( 6.64 × 10−26 kg ) ( 455 m/s ) = 9.16 × 104 Pa ⎝V ⎠

(b) The temperature of the gas in the container can be obtained from the ideal-gas equation in the form pV = NkBT:

T=

pV 9.16 × 104 Pa = = 332 K 25 NkB ( 2.00 × 10 m −3 )(1.38 × 10−23 J/K )

18.12. Model: Pressure is due to random collisions of gas molecules with the walls. Solve:

According to Equation 18.8, the collision rate with one wall is

rate of collisions =

N coll F pA = net = Δtcoll 2mvx 2mvx

where Fnet = pA is the force exerted on area A by the gas pressure. However, this equation assumed that all molecules are moving in the x-direction with constant speed. The rms speed vrms is for motion in three dimensions at varying speeds. Consequently, we need to replace vx not with (vx)avg, which is zero, but with vx → (vx2)avg =

2 vrms v = rms 3 3

With this change, rate of collisions =

3 pA 3(2 × 101,300 Pa)(0.10 m × 0.10 m) = = 6.5 × 1025 s −1 2mvrms 2(28 × 1.661× 10−27 kg)(576 m/s)

This collision rate can also be found by using the expression in Eq. 18.10, making the same change in vx, and using the ideal-gas law to determine N/V.

18.13. Visualize: We will use Equation 18.18 to find m of the atoms. p=

1N 2 mvrms 3V

We are given p = 2.0 atm = 202,650 Pa , N /V = 4.2 × 1025 m −3 , and vrms = 660 m/s. Solve: Solve for m.

m=

3 p V 3(202,650 Pa) 1 = = 3.32 × 10−26 kg = 20 u 2 vrms N (660 m/s) 2 4.2 ×1025 m −3

With an atomic mass of 20 the gas is likely neon. Assess: Neon is gaseous at room temperature, so is a likely choice. As a noble gas, it also doesn’t form many molecules.

18.14. Visualize: Equation 18.26 will give us vrms from the temperature and mass of the particles. vrms =

3kBT m

We are given T = 1100°C = 1373 K . The atomic mass of neon atoms is 20 u = 3.321 × 10−26 kg; the molecular

mass of oxygen molecules is 32 u = 5.31 × 10−26 kg. Solve: (a) vrms =

3(1.38 × 10−23 J/K)(1373 K) = 1310 m/s 3.321 × 10−26 kg

vrms =

3(1.38 × 10−23 J/K)(1373 K) = 1030 m/s 5.31 × 10−26 kg

(b)

Assess: These are fast speeds, but the temperature is high.

18.15. Visualize: Use Equation 18.26. vrms =

3kBT m

We are given vrms = 1.5 m/s and m = 28 u = 4.650 × 10−26 kg. Solve: Solve the equation for T . T=

Assess:

2 vrms m (1.5 m/s) 2 (4.650 × 10−26 kg) = = 2.5 mK 3kB 3(1.38 × 10−23 J/K)

2.5 mK is close to absolute zero, but that’s how cold it would have to be for vrms to be 1.5 m/s.

18.16.

Solve: Because the neon and argon atoms in the mixture are in thermal equilibrium, the temperature of each gas in the mixture must be the same. That is, using Equation 18.26, 2 2 mAr vrms Ar = mNe vrms Ne

vrms Ar = vrms Ne

20 u mNe = ( 400 m/s ) = 283 m/s 40 u mAr

18.17. Solve: The average translational kinetic energy per molecule is 3kBT 1 2 3 Pavg = mvrms = kBT ⇒ vrms = 2 2 m Since we want the vrms for H2 and N2 to be equal,

3kBTH2 mH 2

=

mH 3kBT ⎛ 2u ⎞ ⇒ TH2 = 2 TN2 = ⎜ ⎟ ( 373 K ) = 27 K = −246°C mN2 mN 2 ⎝ 28 u ⎠

18.18. Solve: The formula for the root-means-square speed as a function of temperature is

( vrms )T =

3kBT m

(a) For ( vrms )T = 12 ( vrms )STP ,

3kBT 1 3kB ( 273 K ) 1 =2 ⇒ T = (273 K) = 68 K= − 205°C m m 4 (b) For ( vrms )T = 2 ( vrms )STP ,

3kB ( 273 K ) 3kBT =2 ⇒ T = 4 ( 273 K ) = 1090 K = 817°C m m

18.19.

Solve:

Solve Equation 18.18 for vrms to see that if both p and V are doubled then vrms is also doubled.

vrms =

3 pV Nm

So the new rms speed will be 800 m/s. Assess: Think microscopically; for the pressure to double when the volume is doubled the particles will have to be going a lot faster and hit the walls more often.

18.20. Solve: The formula for the root-means-square speed as a function of temperature is

( vrms )T =

3kBT m

The ratio at 20°C and at 100°C is

( vrms )100 ( vrms )20

=

373 K = 1.13 293 K

18.21. Solve: molecule is

Assuming ideal-gas behavior and ignoring relativistic effects, the root-mean-square speed of a

vrms =

3kBT m

The temperature where vrms is the speed of light (c) for a hydrogen molecule

mv 2 ( 2 u ) c 2 = 2 (1.66 ×10−27 kg )( 3.0 ×108 m/s ) = 7.22 ×1012 K T = rms = 3kB 3kB 3 (1.38 × 10−23 J/K ) 2

18.22. Solve: (a) The average translational kinetic energy per molecule is 1 2 3 Pavg = mvrms = kBT 2 2

This means Pavg doubles if the temperature T doubles. (b) The root-mean-square speed vrms increases by a factor of (c) The mean free path is

λ=

2 as the temperature doubles.

1 4 2π ( N V ) r 2

Because N/V and r do not depend on T, doubling temperature has no effect on λ.

18.23. Solve: (a) The total translational kinetic energy of a gas is K micro = 32 N A kBT = 32 nRT . For H2 gas at STP, K micro =

3 (1.0 mol )(8.31 J/mol K )( 273 K ) = 3400 J 2

K micro =

3 (1.0 mol )(8.31 J/mol K )( 273 K ) = 3400 J 2

(b) For He gas at STP,

(c) For O2 gas at STP, K micro = 3400 J. Assess: The translational kinetic energy of a gas depends on the temperature and the number of molecules but not on the molecule’s mass.

18.24. Solve: (a) The mean free path is λ=

1 4 2π ( N V ) r 2

where r ≈ 0.5 × 10−10 m is the atomic radius for helium and N/V is the gas number density. From the ideal-gas law,

N p 0.10 atm × 101,300 Pa/atm = = = 7.34 × 1025 m −3 V kT (1.38 ×10−23 J/K ) (10 K ) ⇒λ =

1

4 2π ( 7.34 × 10 m 25

−3

)( 0.5 × 10

−10

m)

2

= 3.1 × 10−7 m = 310 nm

(b) The root-mean-square speed is

vrms =

3 (1.38 × 10−23 J/K ) (10 K ) 3kBT = = 250 m/s m 4 × (1.661 × 10−27 kg )

where we used A = 4 u as the atomic mass of helium. (c) The average energy per atom is ε avg = 32 kBT = 32 (1.38 × 10−23 J/K ) (10 K ) = 2.1 × 10−22 J .

18.25. Solve: (a) The average kinetic energy of a proton at the center of the sun is 3 Pavg = kBT ≈ 2

3 2

(1.38 × 10

−23

J/K )( 2.0 × 107 K ) = 4.1× 10−16 J

(b) The root-mean-square speed of the proton is

vrms =

3 (1.38 × 10−23 J/K )( 2.0 × 107 K ) 3kBT ≈ = 7.0 × 105 m/s m 1.67 × 10−27 kg

18.26. Solve: (a) Since the hydrogen in the sun’s atmosphere is monatomic, the average translational kinetic energy per atom is 3 Pavg = kBT = 32 (1.38 × 10−23 J/K ) ( 6000 K ) = 1.24 × 10−19 J 2 (b) The root-mean-square speed is

vrms =

3 (1.38 × 10−23 J/K ) ( 6000 K ) 3kBT = = 1.22 × 104 m/s mH 1.67 × 10−27 kg

18.27. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth = Kmicro. Solve: The number of atoms is

N=

M 0.0020 kg = = 3.01 × 1023 m 6.64 × 10−27 kg

Because helium atoms have an atomic mass number A = 4, the mass of each helium atom is m = 4 u = 4 (1.661 × 10−27 kg ) = 6.64 × 10−27 kg The average kinetic energy of each atom is 2 K avg = 12 mvavg = 12 ( 6.64 × 10−27 kg ) ( 700 m s ) = 1.63 × 10−21 J 2

Thus the thermal energy of the gas is Eth = K micro = NK avg = ( 3.01× 1023 )(1.63 × 10−21 J ) = 490 J

18.28. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. Solve:

Neon atoms have an atomic mass number A = 14, so the mass of each molecule is m = 28 u = 28(1.661 × 10−27 kg) = 4.51 × 10−26 kg

The number of molecules in the gas is

N=

M 0.010 kg = = 2.218 × 1023 m 4.51 × 10−26 kg

The thermal energy is 2 Eth = NK avg = N ( 12 mvavg ) ⇒ vavg =

2 Eth = Nm

2 (1700 J )

( 2.218 ×10 )( 4.51×10 23

−26

kg )

= 580 m/s

18.29. Solve: The volume of the air is V = 6.0 m × 8.0 m × 3.0 m = 144.0 m3, the pressure p = 1 atm = 1.013

× 105 Pa, and the temperature T = 20°C = 293 K. The number of moles of the gas is pV n= = 5991 mol RT This means the number of molecules is

N = nN A = ( 5991 mols ) ( 6.022 × 1023 mol−1 ) = 3.61 × 1027 molecules

Since air is a diatomic gas, the room’s thermal energy is Eth = N Pavg = N ( 52 kBT ) = 3.6 × 107 J Assess:

The room’s thermal energy can also be obtained as follows:

Eth = nCVT = (5991 mol)(20.8 J/mol K)(293 K) = 3.6 × 107 J

18.30. Solve: The thermal energy of a solid is Eth = 3 NkBT = 3nRT The volume of lead V = 100 cm3 = 10−4 m3, which means the mass is M = ρV = (11,300 kg/m3)(10−4 m3) = 1.13 kg

Because the atomic mass number of Pb is 207, the number of moles is n=

M 1.13 kg = = 5.459 mol M mol 0.207 kg/mol

⇒ Eth = 3( 5.459 mols )( 8.31 J/mol K )( 293 K ) = 3.99 × 104 J

18.31. Solve: (a) For a monatomic gas, ΔEth = nCV ΔT = 1.0 J = (1.0 mol )(12.5 J/mol K ) ΔT ⇒ ΔT = 0.080°C or 0.080 K (b) For a diatomic gas,

1.0 J = (1.0 mol)(20.8 J/mol K)ΔT ⇒ ΔT = 0.048°C or K (c) For a solid,

1.0 J = (1.0 mol)(25.0 J/mol K) ΔT ⇒ ΔT = 0.040°C or K

18.32. Solve: The conservation of energy equation (ΔEth)gas + (ΔEth)solid = 0 J is ngas ( CV )gas (Tf − Ti )gas + nsolid ( CV )solid (Tf − Ti )solid = 0 J

⇒ (1.0 mol)(12.5 J/mol K)(−50 K) + (1.0 mol)(25.0 J/mol K)(ΔT)solid = 0 ⇒ (ΔT)solid = 25°C The temperature of the solid increases by 25°C.

Refer to Figure 18.13. At low temperatures, CV = 32 R = 12.5 J/mol K. At room temperature and modestly hot temperatures, CV = 52 R = 20.8 J/mol K. At very hot temperatures, CV = 72 R = 29.1 J/mol K. Solve: (a) The number of moles of diatomic hydrogen gas in the rigid container is

18.33. Visualize:

0.20 g = 0.10 mol 2 g/mol

The heat needed to change the temperature of the gas from 50 K to 100 K at constant volume is Q = ΔEth = nCV ΔT = (0.10 mol)(12.5 J/mol K)(100 K – 50 K) = 62 J (b) To raise the temperature from 250 K to 300 K, Q = ΔEth = (0.10 mol)(20.8 J/mol K)(300 K – 250 K) = 100 J (c) To raise the temperature from 550 K to 600 K, Q = 100 J. (d) To raise the temperature from 2250 K to 2300 K, Q = ΔEth = nCVΔT = (0.10 mol)(29.1 J/mol K)(50 K) = 150 J.

18.34. Model: The potential energy of the bowling ball is transferred into the thermal energy of the mixture. We assume the starting temperature of the bowling ball to be 0°C. Solve: The potential energy of the bowling ball is

U g = M ball gh = (11 kg)(9.8 m/s 2 ) h = (107.8 kg m/s 2 )h This energy is transferred into the mixture of ice and water and melts 5 g of ice. That is, (107.8 kg m/s 2 )h = ΔEth = M w Lf ⇒ h =

(0.005 kg)(3.33 × 105 J/kg) = 15.4 m (107.8 kg m/s 2 )

18.35. Model: Heating the water and the kettle raises the temperature of the water to the boiling point and raises the temperature of the kettle to 100°C. Solve: The amount of heat energy from the electric stove’s output in 3 minutes is

Q = ( 2000 J s )( 3 × 60 s ) = 3.6 × 105 J This heat energy heats the kettle and brings the water to a boil. Thus,

Q = M water cwater ΔT + M kettle ckettle ΔT Substituting the given values into this equation, 3.6 × 105 J = M water ( 4190 J kg K )(100°C − 20°C ) + ( 0.750 kg )( 449 J kg K )(100°C − 20°C )

⇒ M water = 0.994 kg The volume of water in the kettle is

V= Assess:

3

3

M water

ρ water

=

0.994 kg = 0.994 × 10−3 m3 = 994 cm3 ≈ 990 cm3 1000 kg m3

1 L = 10 cm , so V ≈ 1 L. This is a reasonable volume of water.

18.36. Solve: The equilibrium condition is (εA)avg = (εB)avg = (εtot)avg ⇒ EAf = EBf = Etot nA nB nA + nB Thus the final thermal energies are

⎛ nA ⎞ 4.0 mols ⎛ ⎞ EAf = ⎜ ⎟ Etot = ⎜ ⎟ ( 9000 J + 5000 J ) = 8000 J 4.0 mols 3.0 mols n n + + ⎝ ⎠ B ⎠ ⎝ A ⎛ nB ⎞ ⎛ 3.0 mols ⎞ EBf = ⎜ ⎟ Etot = ⎜ ⎟ (14,000 J ) = 6000 J n n + ⎝ 7.0 mols ⎠ B ⎠ ⎝ A Because EAi = 9000 J and EAf = 8000 J, 1000 J of heat energy is transferred from gas A to gas B.

18.37. Solve: The mean free path for a monatomic gas is λ=

1 V ⇒ = 4π 2λ r 2 N 4π 2 ( N V ) r 2

For λ = 20r, meaning that the mean free path equals the atomic diameter, V V /N ⎛ 4π 3 ⎞ = 4π 2 ( 20r ) ( r 2 ) = ⎜ r ⎟ 60 2 ⇒ 4π 3 = 60 2 = 84.8 N r ⎝ 3 ⎠ 3

18.38. Visualize:

Solve:

The average energy of an oxygen molecule at 300 K is

ε avg =

Eth 5 = kBT = N 2

5 2

(1.38 × 10

−23

J/K ) ( 300 K ) = 1.035 × 10−20 J

The energy conservation equation Ugf + Kf = Ugi + Ki with K f = ε avg is

mgyf + ε avg = mgyi + 0 J With yi = h and yf = 0 m, we have

mgh = 1.035 × 10−20 J ⇒ h =

1.035 × 10−20 J = 1.99 × 104 m ( 32 ×1.66 ×10−27 kg )(9.8 m/s 2 )

18.39. Solve: (a) To identify the gas, we need to determine its atomic mass number A or, equivalently, the mass m of each atom or molecule. The mass density ρ and the number density (N/V ) are related by ρ = m(N/V ) , so the mass is m = ρ (V/N ) . From the ideal-gas law, the number density is 50,000 Pa N p = = = 1.208 × 1025 m −3 V kT (1.38 × 10−23 J/K ) ( 300 K ) Thus, the mass of an atom is m=ρ

V 8.02 × 10−2 kg/m3 = = 6.64 × 10−27 kg N 1.208 × 1025 m −3

Converting to atomic mass units,

A = 6.64 × 10−27 kg ×

1u = 4.00 u 1.661 × 10−27 kg

This is the atomic mass of helium. (b) Knowing the mass, we find vrms to be vrms =

3 (1.38 × 10−23 J/K ) ( 300 K ) 3kBT = = 1370 m/s m 6.64 × 10−27 kg

(c) A typical atomic radius is r ≈ 0.5 × 10−10 m. The mean free path is thus

λ=

1 1 = = 1.86 × 10−6 m = 1.86 μ m 2 4 2π ( N V ) r 2 4 2π (1.208 × 1025 m −3 )( 0.5 × 10−10 m )

18.40. Solve: (a) The number density is N / V = 1 cm −3 = 106 m −3 . Using the ideal-gas equation, N kBT ≈ (1 × 106 m −3 )(1.38 × 10−23 J/K ) ( 3 K ) V 1 atm = 4 × 10−17 Pa × = 4 × 10−22 atm 1.013 × 105 Pa

p=

(b) For a monatomic gas, 3 (1.38 × 10−23 J/K ) ( 3 K ) 3kBT = 270 m/s = m 1.67 × 10−27 kg

vrms =

(c) The thermal energy is Eth = 32 NkBT , where N = (106 m −3 )V . Thus

Eth = 1.0 J =

3 2

(10

6

m −3 )V (1.38 × 10−23 J/K ) ( 3 K ) ⇒ V = 1.6 × 1016 m3 = L3 ⇒ L = 2.5 × 105 m

18.41. Solve: Mass m of a dust particle is 3 m = ρV = ρ ( 43 π r 3 ) = ( 2500 kg/m3 ) ⎡ 43 π ( 5 × 10−6 m ) ⎤ = 1.3 × 10−12 kg ⎥⎦ ⎣⎢

The root-mean-square speed of the dust particles at 20°C is vrms =

3 (1.38 × 10−23 J/K ) ( 293 K ) 3kBT = = 9.6 × 10−5 m/s m 1.3 × 10−12 kg

18.42. Solve: Fluorine has atomic mass number A = 19. Thus the root-mean-square speed of 238UF6 is vrms ( 238 UF6 ) =

3kBT 3kBT = m 238 u + 6 × 19 u

The ratio of the root-mean-square speed for the molecules of this isotope and the 235UF6 molecules is

vrms ( 235 UF6 ) vrms (

238

UF6 )

=

( 238 + 6 × 19 ) u ( 235 + 6 × 19 ) u

=

352 = 1.0043 349

18.43. Solve: (a) If the electron can be thought of as a point particle with zero radius, then it will collide with any gas particle that is within r of its path. Hence, the number of collisions Ncoll is equal to the number of gas particles in a cylindrical volume of length L. The volume of a cylinder is Vcyl = AL = (π r 2 ) L . If the number density of the gas is (N/V) particles per m3, then the number of collisions along a trajectory of length L is

N coll = Introducing a factor of

N N L 1 Vcyl = (π r 2 L ) ⇒ λelectron = = V V N coll π ( N / V ) r 2

2 to account for the motion of all particles,

λelectron =

L 1 = N coll 2π ( N / V ) r 2

(b) Assuming that most of the molecules in the accelerator are diatomic,

5.0 × 104 m =

1

2π ( N / V ) (1.0 × 10

−10

m)

2

⇒ N / V = 4.50 × 1014 m −3

From the ideal-gas equation, p=

N kBT = ( 4.50 × 1014 m −3 )(1.38 × 10−23 J/K ) ( 293 K ) = 1.82 × 10−6 Pa = 1.80 × 10−11 atm V

18.44. Solve: The pressure on the wall with area A = 10 cm2 = 10 × 10−4 m2 is p=

F Δ ( mv ) N = A AΔt

where N Δt is the number of N2 molecules colliding with the wall every second and Δ(mv) is the change in momentum for one collision. The mass of the nitrogen molecule is m = 28 u = 28 (1.66 × 10−27 kg) = 4.648 × 10−26 kg and Δv = 400 m/s − (−400 m/s) = 800 m/s . Thus,

( 4.648 ×10 p=

−26

kg ) ( 800 m/s ) ( 5.0 × 1023 s −1 ) 1.0 × 10−3 m 2

= 1.9 × 104 Pa

18.45. Solve: (a) The cylinder volume is V = πr2L = 1.571 × 10–3 m3. Thus the number density is N 2.0 × 1022 = = 1.273 × 1025 m −3 ≈ 1.3 × 1025 m −3 V 1.571 × 10−3 m3 (b) The mass of an argon atom is m = 40 u = 40(1.661 × 10−27 kg) = 6.64 × 10−26 kg ⇒ vrms =

3 (1.38 × 10−23 J/K ) ( 323 K ) 3kBT = = 449 m/s ≈ 450 m/s m 6.64 × 10−26 kg

(c) vrms is the square root of the average of v2. That is, 2 vrms = ( v2 )

avg

= ( vx2 )

avg

+ ( v y2 )

avg

+ ( vz2 )

avg

An atom is equally likely to move in the x, y, or z direction, so on average ( vx2 ) 2 vrms = 3 ( vx2 )

avg

⇒ ( vx )rms =

(v )

2 x avg

=

avg

= ( v y2 )

avg

= ( vz2 )

avg

. Hence,

vrms = 259 m/s ≈ 260 m/s 3

(d) When we considered all the atoms to have the same velocity, we found the collision rate to be 12 ( N/V ) Avx (see Equation 18.10). Because the atoms move with different speeds, we need to replace vx with (vx)rms. The end of the cylinder has area A = πr2 = 7.85 × 10–3 m2. Therefore, the number of collisions per second is 1 2

( N / V ) A ( vx )rms = 12 (1.273 × 1025 m−3 )( 7.85 × 10−3 m2 ) ( 259 m/s ) = 1.3 × 1025 s−1

(e) From kinetic theory, the pressure is 2 ⎛N⎞ ⎛N⎞ 2 p = 13 ⎜ ⎟ m ( v 2 ) = 13 ⎜ ⎟ mvrms = 13 (1.273 × 1025 m −3 )( 6.64 × 10−26 kg ) ( 449 m/s ) = 56,800 Pa ≈ 57,000 Pa avg V V ⎝ ⎠ ⎝ ⎠ (f) From the ideal-gas law, the pressure is

22 −23 NkBT ( 2.0 × 10 )(1.38 × 10 J/K ) ( 323 K ) = = 56,700 Pa ≈ 57,000 Pa V 1.571 × 10−3 m3 Assess: The very slight difference between parts (e) and (f) is due to rounding errors; to two significant figures they are the same.

p=

18.46. Model: Pressure is due to random collisions of gas molecules with the walls. Solve:

According to Equation 18.9, the collision rate with one wall is N 1N rate of collisions = coll = Avx Δtcoll 2 V

However, this equation assumed that all molecules are moving in the x-direction with constant speed. The rms speed vrms is for motion in three dimensions at varying speeds. Consequently, we need to replace vx not with (vx)avg, which is zero, but with vx → (vx2)avg =

2 vrms v = rms 3 3

With this change,

rate of collisions =

1 N Avrms 2 3V

The molecular mass of nitrogen is A = 28 u, thus the rms speed of the molecules at 20°C is vrms =

3kBT 3(1.38 × 10 −23 J/K)(293 K) = = 510 m/s m 28(1.661 × 10 –27 kg)

With N = 0.10NA = 6.02 × 1022 molecules, the number density is N 6.02 × 1022 = = 6.02 × 1025 m −3 V 0.10 m × 0.10 m × 0.10 m Thus 1 rate of collisions = (6.02 × 1025 m −3 )(0.10 m × 0.10 m)(510 m/s)=8.9 × 1025 s −1 2 3

18.47.

Model:

Assume the gas is ideal so that Equation 18.30 will apply.

ΔEth = nCV ΔT Visualize: Solve:

We see from the graph that ΔT = 200 K and ΔEth = 800 J. We are also given n = 0.14 mol.

Solve for CV .

CV = Assess:

This is about

7 2

R.

ΔEth 800 J = = 29 J/mol ⋅ K nΔT (0.14 mol)(200 K)

18.48. Solve: (a) The number of molecules of helium is 5 −6 3 pV ( 2.0 × 1.013 × 10 Pa )(100 × 10 m ) = = 3.936 × 1021 kBT (1.38 ×10−23 J/K ) ( 373 K )

N helium =

⇒ nhelium =

3.936 × 1021 = 6.536 × 10−3 mol 6.022 × 1023 mol−1

The initial internal energy of helium is

Ehelium i = 32 N helium kBT = 30.4 J ≈ 30 J The number of molecules of argon is N argon =

5 −6 3 pV ( 4.0 × 1.013 × 10 Pa )( 200 × 10 m ) = = 8.726 × 1021 kBT (1.38 ×10−23 J/K ) ( 673 K )

⇒ nargon =

8.726 × 1021 = 1.449 × 10−2 mol 6.022 × 1023 mol−1

The initial thermal energy of argon is Eargon i = 32 N argon kBT = 121.6 J ≈ 122 J (b) The equilibrium condition for monatomic gases is

(εhelium f)avg = (εargon f)avg = (εtotal)avg ⇒

Ehelium f nhelium

=

Eargon f nargon

=

Etot ( 30.4 + 121.6 ) J = = 7228 J/mol ntot ( 6.54 × 10−3 + 1.449 × 10−2 ) mol

⇒ Ehelium f = ( 7228 J/mol ) nhelium = ( 7228 J/mol ) ( 6.54 × 10−3 mol ) = 47.3 J ≈ 47 J Eargon f = ( 7228 J/mol ) nargon = ( 7228 J/mol ) (1.449 × 10−2 mol ) = 104.7 J ≈ 105 J (c) The amount of heat transferred is

Ehelium f − Ehelium i = 47.3 J − 30.4 J = 16.9 J

Eargon f − Eargon i = 104.7 J − 121.6 J = −16.9 J

The helium gains 16.9 J of heat energy and the argon loses 16.9 J. Thus approximately 17 J are transferred from the argon to the helium. (d) The equilibrium condition for monatomic gases is

( ε helium )avg = (ε argon )avg

⇒

Ehelium f N helium

=

Eargon f N argon

= 32 kBTf

Substituting the above values, 47.3 J 104.7 J = = 3.936 × 1021 8.726 × 1021

3 2

(1.38 × 10

−23

J/K ) Tf ⇒ TF = 580 K = 307°C

(e) The final pressure of the helium and argon are

phelium f =

pargon f =

21 −23 N helium kBT ( 3.936 × 10 )(1.38 × 10 J/K ) ( 580 K ) = = 3.15 × 105 Pa ≈ 3.1 atm Vhelium 100 × 10−6 m 3

N argon kBT Vargon

(8.726 ×10 )(1.38 ×10 21

=

200 × 10

−6

J/K ) ( 580 K )

−23

m

3

= 3.49 × 105 Pa ≈ 3.4 atm

18.49. Solve: (a) The number of moles of helium and oxygen are nhelium =

2.0 g = 0.50 mol 4.0 g/mol

noxygen =

8.0 g = 0.25 mol 32.0 g/mol

Since helium is a monoatomic gas, the initial thermal energy is Ehelium i = nhelium ( 32 RThelium ) = ( 0.50 mol ) ( 32 ) ( 8.31 J/mol K )( 300 K ) = 1870 J ≈ 1900 J Since oxygen is a diatomic gas, the initial thermal energy is

Eoxygen i = noxygen ( 52 RToxygen ) = ( 0.25 mol ) ( 52 ) ( 8.31 J/mol K )( 600 K ) = 3116 J ≈ 3100 J (b) The total initial thermal energy is Etot = Ehelium i + Eoxygen i = 4986 J

As the gases interact, they come to equilibrium at a common temperature Tf. This means 4986 J = nhelium ( 32 RTf ) + noxygen ( 52 RTf )

⇒ Tf =

(

1 2

4986 J = R ) ( 3nhelium + 5noxygen )

1 2

4986 J = 436.4 K = 436 K ( 8.31 J/mol K )( 3 × 0.50 mol + 5 × 0.25 mol )

The thermal energies at the final temperature Tf are Ehelium f = nhelium ( 32 RTf ) = ( 32 ) ( 0.50 mol )( 8.31 J/mol K )( 436.4 K ) = 2700 J Eoxygen f = noxygen ( 52 RTf ) = ( 52 ) ( 0.25 mol )( 8.31 J/mol K )( 436.4 K ) = 2300 J (c) The change in the thermal energies are

Ehelium f − Ehelium i = 2720 J − 1870 J = 850 J

Eoxygen f − Eoxygen i = 2266 J − 3116 J = −850 J

The helium gains energy and the oxygen loses energy. (d) The final temperature can also be calculated as follows:

Ehelium f = ( nhelium ) 32 RTf ⇒ 2720 J = (0.50 mol)(1.5)(8.31 J/mol K)Tf ⇒ Tf = 436.4 K ≈ 436 K

18.50. Solve: The thermal energy of any ideal gas is related to the molar specific heat at constant volume by Eth = nCVT Since CP = CV + R ,

20.8 J/mol K = CV + R ⇒ CV = 12.5 J/mol K The number of moles of the gas is n=

1.0 × 1020 = 1.66 × 10−4 mol 6.02 × 1023 mol −1

Thus T=

(1.66 ×10

−4

(1.0 J ) = 482 K mols ) (12.5 J/mol K )

18.51. Solve: For a system with n degrees of freedom, the molar specific heat is CV = nR/2. The specific heat

ratio is

γ=

CP CV + R R R R ⇒ = γ − 1 = 1.29 − 1 = 0.29 ⇒ CV = = =1+ = 3.5 R = 72 R 0.29 CV CV CV CV

Thus, the system has 7 degrees of freedom.

18.52. Solve: As the volume V1 of a gas increases to V2 = 2V1 at a constant pressure p1 = p2, the temperature of the gas changes from T1 to T2 as follows: V p p2V2 p1V1 ⇒ T2 = T1 2 2 = 2T1 = V1 p1 T2 T1

Since the process occurs at constant pressure the heat transferred is

Q = nCP ΔT = nCP (T2 − T1 ) = nCP ( 2T1 − T1 ) = nCPT1 For a monatomic gas, CP = CV + R = 32 R + R = 52 R

For the diatomic gas, CP = CV + R = 52 R + R = 72 R

Thus

n ( 7 R ) T1 Qdiatomic = 52 = 1.40 Qmonatomic n ( 2 R ) T1

18.53. Solve: Assuming that the systems are thermally isolated except from each other, the total energy for the two thermally interacting systems must remain the same. That is, E1i + E2i = E1f + E2f ⇒ E1f − E1i = E2i − E2f = − ( E2f − E1f ) ⇒ ΔE1 = −ΔE2 No work is done on either system, so from the first law ΔE = Q. Thus

Q1 = −Q2 That is, the heat lost by one system is gained by the other system.

18.54. Solve: (a) From Equation 18.26 vrms = 3kBT m . For an adiabatic process γ −1

γ −1

TfVf

= TV i i

γ −1

⎛V ⎞ ⇒ Tf = Ti ⎜ i ⎟ ⎝ Vf ⎠

⇒ Tf = Ti ( 8 ) 3 = 4Ti 5 −1

The root-mean-square speed increases by a factor of 2 with an increase in temperature. (b) From Equation 18.3 λ = [4 2π ( N / V )r 2 ]−1. A decrease in volume decreases the mean free path by a factor of 1/8. (c) For an adiabatic process, γ −1

γ −1

Tf Vf

= TV i i

γ −1

⎛V ⎞ ⇒ Tf = Ti ⎜ i ⎟ ⎝ Vf ⎠

= Ti ( 8 ) 3 = 4Ti 5 −1

Because the decrease in volume increases Tf , the thermal energy increases by a factor of 4. (d) The molar specific heat at constant volume is CV = 32 R, a constant. It does not change.

18.55. Model: Assume the gas is monatomic. Visualize: From the equipartition theorem there is 12 nRT of energy for each degree of freedom. For a twodimensional monatomic gas there are only two degrees of freedom. 2 Solve: (a) CV = R = R 2 (b) Equation 17.34 gives Cp = CV + R , so

Cp = R + R = 2 R Assess:

It would be nice to measure the values and compare with these predictions.

18.56. Solve: (a) The thermal energy of a monatomic gas of N molecules is Eth = N Pavg , where Pavg = 32 kBT . A monatomic gas molecule has 3 degrees of freedom. However, a two-dimensional monatomic gas molecule has only 2 degrees of freedom. Thus, ⎛2 ⎞ Eth = N ⎜ kBT ⎟ = NkBT = nRT 2 ⎝ ⎠ If the temperature changes by ΔT, then the thermal energy changes by ΔEth = nRΔT . Comparing this form with ΔEth = nCV ΔT , we have CV = R = 8.31 J/mol K. (b) A two-dimensional solid has 2 degrees of freedom associated with the kinetic energy and 2 degrees of freedom associated with the potential energy (or 2 spring directions), giving a total of 4 degrees of freedom. Thus, Eth = N Pavg and Pavg = 42 kBT . Or Eth = 2NkBT = 2nRT. For a temperature change of ΔT, ΔEth = 2nRΔT = nCVΔT ⇒ CV = 2 R = 16.6 J/mol K.

18.57. Solve: (a) The rms speed is vrms =

v 3kBT 32 u ⇒ rms hydrogen = =4 vrms oxygen 2u m

(b) The average translational energy is P = 32 kBT . Thus

Pavg hydrogen Pavg oxygen

=

Thydrogen Toxygen

=1

(c) The thermal energy is

Eth = 52 nRT

⇒

Eth hydrogen Eth oxygen

=

nhydrogen noxygen

=

mhydrogen 32.0 g/mol = 16 2.0 g/mol moxygen

18.58. Visualize: We are given vrms = 1800 m/s. For 1.0 g of molecular hydrogen gas ⎛ 1 mol ⎞ ⎛ 23 molecules ⎞ 23 N = 1.0 g ⎜ ⎟ = 0.5 mol ⎜ 6.02 × 10 ⎟ = 3.01× 10 molecules mol ⎠ ⎝ ⎝ 2.0 g ⎠ −27 The mass of one molecule is 2(1.661× 10 kg) = 3.322 ×10−27 kg. Solve: (a) 2 ⎛1⎞ 2 ⎛1⎞ ∈total = N ⎜ ⎟ mvrms = ( 3.01× 1023 ) ⎜ ⎟ ( 3.322 × 10−27 kg ) (1800 m/s ) = 1.6 kJ ⎝2⎠ ⎝2⎠ (b) The temperature is given by Equation 18.25. 2 2 ⎛1⎞ 2 3.322 ×10−27 kg 2 T= ∈ave = (1800 m/s ) = 260 K ⎜ ⎟ mvrms = −23 3kB 3kB ⎝ 2 ⎠ 3 (1.38 × 10 J/K )

For diatomic gases Equation 18.37 gives the thermal energy. 5 5 Eth = nRT = ( 0.50 mol )( 8.31 J/mol ⋅ K )( 260 K ) = 2.7 kJ 2 2 (c) There is a net loss of 700 J of energy from the system, so the new Eth = 2000 J. Solve Equation 18.37 for T. 2 2 T= Eth = ( 2000 J ) = 192.5 K 5nR 5 ( 0.50 mol )( 8.31 J/mol ⋅ K ) Now Equation 18.26 gives the new rms speed.

3 (1.38 ×10−23 J/K ) (192.5 K ) 3kBT = = 1549 m/s ≈ 1500 m/s vrms = m 3.322 × 10−27 kg Assess: With the net loss of energy we expected vrms to decrease as the temperature decreased.

18.59. Solve: (a) The escape speed is the speed with which a mass m can leave the earth’s surface and

escape to infinity (rf = ∞) with no left over speed (vf = 0). The conservation of energy equation Kf + Uf = Ki + Ui is 2 0 + 0 = 12 mvesc −

GM e m 2GM e ⇒ vesc = Re Re

The rms speed of a gas molecule is vrms = (3kBT/m)1/2. Equating vesc and vrms, and squaring both sides, the temperature at which the rms speed equals the escape speed is

⎛ 2GM e ⎞ 3kBT 2GM e = ⇒ T = m⎜ ⎟ m Re ⎝ 3kB Re ⎠ For a nitrogen molecule, with m = 28 u, the temperature is

⎛ 2(6.67 × 10−11 N m 2 / kg 2 )(5.98 × 1024 kg) ⎞ T = (28 × 1.661 × 10−27 kg) ⎜ ⎟ = 141,000 K 3(1.38 × 10−23 J/K)(6.37 × 106 m) ⎝ ⎠ (b) For a hydrogen molecule, with m = 2 u, the temperature is less by a factor of 14, or T = 10,100 K. (c) The average translational kinetic energy of a molecule is ε avg = 32 kBT = 6.1 × 10–21 J at a typical atmosphere 2 temperature of 20°C. The kinetic energy needed to escape is K esc = 12 mvesc . For nitrogen molecules, Kesc =

2.9 × 10–18 J. Thus εavg/Kesc = 0.002 = 0.2%. Earth will retain nitrogen in its atmosphere because the molecules are moving too slowly to escape. But for hydrogen molecules, with Kesc = 2.1 × 10–19 J, the ratio is εavg/Kesc = 0.03 = 3%. Thus a large enough fraction of hydrogen molecules are moving at escape speed, or faster, to allow hydrogen to leak out of the atmosphere into space. Consequently, earth’s atmosphere does not contain hydrogen.

18.60. Solve: (a) The thermal energy of a monatomic gas of n1 moles is E1 = 32 n1RT . The thermal energy of a diatomic gas of n2 moles is E2 = 52 n2 RT . The total thermal energy of the mixture is Eth = 12 ( 3n1 + 5n2 ) RT ⇒ ΔEth = 12 ( 3n1 + 5n2 ) RΔT Comparing this expression with

ΔEth = ntotal CV ΔT = ( n1 + n2 ) CV ΔT

we get CV =

( 3n1 + 5n2 ) R 2 ( n1 + n2 )

(b) For a diatomic gas, n1 → 0, and CV = 52 R . For a monotomic gas, n2 → 0, and CV = 32 R .

18.61. Solve: (a) The thermal energy is

Eth = ( Eth ) N + ( Eth )O = 52 N N2 kBT + 52 N O2 kBT = 52 N total kBT 2

2

where Ntotal is the total number of molecules. The identity of the molecules makes no difference since both are diatomic. The number of molecules in the room is pV (101,300 Pa )( 2 m × 2 m × 2 m ) N total = = = 2.15 × 10 26 −23 kBT 1.38 × 10 J/K 273 K ) ( )( The thermal energy is Eth =

5 2

( 2.15 × 10 )(1.38 × 10 26

−23

J/K ) ( 273 K ) = 2.03 × 106 J ≈ 2.0 × 106 J

(b) A 1 kg ball at height y = 1 m has a potential energy U = mgy = 9.8 J. The ball would need 9.8 J of initial kinetic energy to reach this height. The fraction of thermal energy that would have to be conveyed to the ball is 9.8 J = 4.8 × 10−6 2.03 × 106 J (c) A temperature change ΔT corresponds to a thermal energy change ΔEth = 52 N total kBΔT . But Using this, we can write E ΔEth −9.8 J ΔEth = th ΔT ⇒ ΔT = T= 273 K = −0.0013 K T Eth 2.03 × 106 J

5 2

N total kB = Eth T .

The room temperature would decrease by 0.0013 K or 0.0013°C. (d) The situation with the ball at rest on the floor and in thermal equilibrium with the air is a very probable distribution of energy and thus a state with high entropy. Although energy would be conserved by removing energy from the air and transferring it to the ball, this would be a very improbable distribution of energy and thus a state of low entropy. The ball will not be spontaneously launched from the ground because this would require a decrease in entropy, in violation of the second law of thermodynamics. As another way of thinking about the situation, the ball and the air are initially at the same temperature. Once even the slightest amount of energy is transferred from the air to the ball, the air’s temperature will be less than that of the ball. Any further flow of energy from the air to the ball would be a situation in which heat energy is flowing from a colder object to a hotter object. This cannot happen because it would violate the second law of thermodynamics.

18.62. Solve: This is not a wise investment. Although an invention to move energy from the hot brakes to the forward motion of the car would not violate energy conservation, it would violate the second law of thermodynamics. The forward motion of the car is a very improbable distribution of energy. It happens only when a force accelerates the car and then sustains the motion against the retarding forces of friction and air resistance. The moving car is a state of low entropy. By contrast, the random motion of many atoms in the hot brakes is a state of high probability and high entropy. To convert the random motion of the atoms in the brakes back into the forward motion of the car would require a decrease of entropy and thus would violate the second law of thermodynamics. In other words, the increasing temperature of the brakes as the car stops is an irreversible process that cannot be undone.

18.63. Solve: (a) We are given that

( vrms i ) =

3kBTi m

( vrms f ) =

3kBTf = 2 ( vrms i ) m

This means that Tf = 4Ti. Using the ideal-gas law, it also means that pfVf = 4piVi. Since the pressure is directly proportional to the volume during the process, we have pi / V i = pf /Vf. Combining these two equations gives pf = 2pi and Vf = 2Vi.

(b) The change in thermal energy for any ideal gas process is related to the molar specific heat at constant volume by ΔEth = nCV (Tf − Ti ) . The work done on the gas is

W = − ∫ pdV = − (area under the p-versus-V graph) = − 32 pV i i The first law of thermodynamics ΔEth = Q + W can be written 3 Q = ΔEth − W = nCV (Tf − Ti ) + 32 pV i i = 3nCVTi + 2 pV i i 15 3 = 3n ( 52 R ) Ti + 32 pV i i = 2 piVi + 2 pV i i = 9 pV i i

18.64. Solve: The thermal energy of a monatomic gas of n1 moles is E1 = 32 n1RT . The thermal energy of a diatomic gas of n2 moles is E2 = 52 n2 RT . The total thermal energy of the mixture is

Eth = E1 + E2 = Comparing this expression with

1 2

( 3n1 + 5n2 ) RT

⇒ ΔEth = 12 ( 3n1 + 5n2 ) RΔT

ΔEth = ntotal CV ΔT = ( n1 + n2 ) CV ΔT

we get

( n1 + n2 ) CV =

( 3n1 + 5n2 ) R 2

The requirement that the ratio of specific heats is 1.50 means

γ = 1.50 = The above equation is then

( n1 + n2 )( 2 R ) =

CP CV + R R = =1+ ⇒ CV = 2 R CV CV CV

( 3n1 + 5n2 ) R ⇒ 4n 2

1

+ 4n2 = 3n1 + 5n2 ⇒ n1 = n2

Thus, monatomic and diatomic molecules need to be mixed in the ratio 1:1. Or the fraction of the molecules that are monatomic needs to be 12 .

18.65.

Solve: (a) The thermal energy of a monatomic gas of n1 moles at an initial temperature T1i is E1i = 32 n1RT1i . The thermal energy of a diatomic gas of n2 moles at an initial temperature T2i is E2i = 52 n2 RT2i . Consequently, the total initial energy is

Etot = E1i + E2i =

3n1RT1i + 5n2 RT2i ( 3n1T1i + 5n2T2i ) R = 2 2

After the gases interact, they come to equilibrium with T1f = T2f = Tf. Then their total energy is Etot =

( 3n1 + 5n2 ) RTf 2

No energy is lost, so these two expressions for Etot must be equal. Thus,

( 3n1T1i + 5n2T2i ) R = ( 3n1 + 5n2 ) RTf 2

⇒ Tf =

2

3n1T1i + 5n2T2i 2 Etot 2 ( E1i + E2i ) = = 3n1 + 5n2 R ( 3n1 + 5n2 ) R ( 3n1 + 5n2 )

The thermal energies at the final temperature Tf are E1f = 32 n1RTf = 32 n1R

⎛ 3n1 ⎞ 2 E1i + E2i =⎜ ⎟ ( E1i + E2i ) R ( 3n1 + 5n2 ) ⎝ 3n1 + 5n2 ⎠

E2f = 52 n2 RTf = 52 n2 R

2 E1i + E2i ⎛ 5n2 ⎞ =⎜ ⎟ ( E1i + E2i ) R 3n1 + 5n2 ⎝ 3n1 + 5n2 ⎠

(b) In part (a) we found that

Tf =

3n1T1i + 5n2T2i 3n1 + 5n2

(c) 2 g of He at T1i = 300 K are n1 = (2 g)/(4 g/mol) = 0.50 mol. Oxygen has an atomic mass of 16, so the molecular mass of oxygen gas (O2) is A = 32 g/mol. 8 g of O2 at T2i = 600 K are n2 = (8 g)/(32 g/mol) = 0.25 mol. The final temperature is

Tf =

3 ( 0.50 mol )( 300 K ) + 5 ( 0.25 mol )( 600 K ) = 436 K 3 ( 0.50 mol ) + 5 ( 0.25 mol )

The heat flows to the two gases are found from Q = nCVΔT. For helium,

Q = n1CV ΔT = ( 0.50 mol )(12.5 J/mol K )( 436 K − 300 K ) = 850 J For O2,

Q = n2CV ΔT = ( 0.25 mol )( 20.8 J/mol K )( 436 K − 600 K ) = −850 J

So 850 J of heat is transferred from the oxygen to the helium.

Solve: (a) The engine has a thermal efficiency of η = 40% = 0.40 and a work output of 100 J per cycle. The heat input is calculated as follows:

19.1.

η=

Wout 100 J ⇒ 0.40 = ⇒ QH = 250 J QH QH

(b) Because Wout = QH − QC , the heat exhausted is

QC = QH − Wout = 250 J − 100 J = 150 J

19.2.

Solve: During each cycle, the work done by the engine is Wout = 20 J and the engine exhausts

QC = 30 J of heat energy. Because Wout = QH − QC , QH = Wout + QC = 20 J + 30 J = 50 J Thus, the efficiency of the engine is

η =1−

QC 30 J =1− = 0.40 QH 50 J

19.3. Solve: (a) During each cycle, the heat transferred into the engine is QH = 55 kJ , and the heat exhausted is QC = 40 kJ . The thermal efficiency of the heat engine is

η = 1−

QC 40 kJ =1− = 0.27 = 27% QH 55 kJ

(b) The work done by the engine per cycle is

Wout = QH − QC = 55 kJ − 40 kJ = 15 kJ

19.4. Solve: The coefficient of performance of the refrigerator is K=

QC QH − Win 50 J − 20 J = = = 1.5 Win Win 20 J

19.5. Solve: (a) The heat extracted from the cold reservoir is calculated as follows: K=

QC Q ⇒ 4.0 = C ⇒ QC = 200 J Win 50 J

(b) The heat exhausted to the hot reservoir is

QH = QC + Win = 200 J + 50 J = 250 J

19.6. Solve:

Model: Assume that the car engine follows a closed cycle. (a) Since 2400 rpm is 40 cycles per second, the work output of the car engine per cycle is

Wout = 500

kJ 1s kJ × = 12.5 s 40 cycles cycle

(b) The heat input per cycle is calculated as follows:

η=

Wout 12.5 kJ ⇒ QH = = 62.5 kJ QH 0.20

The heat exhausted per cycle is QC = QH − Win = 62.5 kJ − 12.5 kJ = 50 kJ

19.7.

Solve: The amount of heat discharged per second is calculated as follows:

η=

⎛1 ⎞ Wout Wout ⎛ 1 ⎞ = ⇒ QC = Wout ⎜ − 1⎟ = ( 900 MW ) ⎜ − 1⎟ = 1.913 × 109 W η 0.32 QH QC + Wout ⎝ ⎠ ⎝ ⎠

That is, each second the electric power plant discharges 1.913 × 109 J of energy into the ocean. Since a typical American house needs 2.0 × 104 J of energy per second for heating, the number of houses that could be heated with the waste heat is (1.913 × 109 J ) ( 2.0 × 104 J ) = 96,000 .

19.8. Solve: The amount of heat removed from the water in cooling it down in 1 hour is QC = mwater cwater ΔT . The mass of the water is

mwater = ρ waterVwater = (1000 kg/m3 ) (1 L ) = (100 kg/m3 )(10−3 m3 ) = 1.0 kg ⇒ QC = (1.0 kg )( 4190 J/kg K )( 20°C − 5°C ) = 6.285 × 10 4 J The rate of heat removal from the refrigerator is 6.285 × 104 J = 17.46 J/s 3600 s The refrigerator does work W = 8.0. W = 8.0 J/s to remove this heat. Thus the performance coefficient of the refrigerator is QC =

K=

17.46 J/s = 2.2 8.0 J/s

19.9.

Model: Process A is isochoric, process B is isothermal, process C is adiabatic, and process D is isobaric. Solve: Process A is isochoric, so the increase in pressure increases the temperature and hence the thermal energy. Because ΔEth = Q − Ws and Ws = 0 J , Q increases for process A. Process B is isothermal, so T is constant

and hence ΔEth = 0 J. The work done Ws is positive because the gas expands. Because Q = Ws + ΔEth , Q is positive for process B. Process C is adiabatic, so Q = 0 J. Ws is positive because of the increase in volume. Since Q = 0 J = Ws + ΔEth , ΔEth is negative for process C. Process D is isobaric, so the decrease in volume leads to a decrease in temperature and hence a decrease in the thermal energy. Due to the decrease in volume, Ws is negative. Because Q = Ws + ΔEth , Q also decreases for process D. A B C D

ΔEth + 0 − −

Ws 0 + + −

Q + + 0 −

19.10. Model: Process A is adiabatic, process B is isothermal, and process C is isochoric.

Solve: Process A is adiabatic, so Q = 0 J. Work Ws is positive as the gas expands. Since Q = Ws + ∆Eth = 0 J, ∆Eth must be negative. The temperature falls during an adiabatic expansion. Process B is isothermal so ∆T = 0 and ∆Eth = 0 J. The gas is compressed, so Ws is negative. Q = Ws for an isothermal process, so Q is negative. Heat energy is withdrawn during the compression to keep the temperature constant. Process C is isochoric. No work is done (Ws = 0 J), and Q is positive as heat energy is added to raise the temperature (∆Eth positive). ∆Eth Ws Q A – 0 + B 0 – – C 0 + +

19.11. Solve: The work done by the gas per cycle is the area inside the closed p-versus-V curve. The area inside the triangle is Wout =

1 2

( 3 atm − 1 atm ) ( 600 × 10−6 m3 − 200 × 10−6 m3 )

⎛ 1.013 × 105 Pa ⎞ −6 3 = 12 ⎜ 2 atm × ⎟ ( 400 × 10 m ) = 40.5 J 1 atm ⎝ ⎠

19.12.

Solve:

The work done by the gas per cycle is the area enclosed within the pV curve. We have

60 J = 12 ( pmax − 100 kPa ) ( 800 cm3 − 200 cm3 ) ⇒

2 ( 60 J ) = pmax − 1.0 × 105 Pa 600 × 10−6 m3

⇒ pmax = 3.0 × 105 Pa = 300 kPa

19.13.

Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycle consists of three individual processes. Solve: (a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get

Wout =

1 2

( 200 kPa ) (100 × 10−6 m3 ) = 10 J

The heat energy transferred into the engine is QH = 120 J. Because Wout = QH − QC , the heat energy exhausted is QC = QH − Wout = 120 J − 10 J = 110 J (b) The thermal efficiency of the engine is

η= Assess:

Wout 10 J = = 0.0833 QH 120 J

Practical engines have thermal efficiencies in the range η ≈ 0.1 − 0.4 .

19.14. Solve:

Model: The heat engine follows a closed cycle, which consists of four individual processes. (a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get

Wout = ( 400 kPa − 100 kPa ) (100 × 10−6 m3 ) = 30 J The heat energy leaving the engine is QC = 90 J + 25 J = 115 J . The heat input is calculated as follows:

Wout = QH − QC ⇒ QH = QC + Wout = 115 J + 30 J = 145 J (b) The thermal efficiency of the engine is

η= Assess:

Wout 30 J = = 0.207 QH 145 J

Practical engines have thermal efficiencies in the range η ≈ 0.1 − 0.4 .

19.15. Solve:

Model: The heat engine follows a closed cycle. The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get

Wout =

1 2

( 300 kPa − 100 kPa ) ( 600 cm3 − 300 cm3 ) = 12 ( 200 × 103 Pa )( 300 × 10−6 m3 )

Because Wout = QH − QC , the heat exhausted is QC = QH − Wout = (225 J + 90 J) − 30 J = 315 J − 30 J = 285 J

= 30 J

19.16. Solve:

Model: The heat engine follows a closed cycle. (a) The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get

Wout =

1 2

( 300 kPa − 100 kPa ) ( 600 cm3 − 200 cm3 ) = 12 ( 200 × 103 Pa )( 400 × 10−6 m3 )

The heat exhausted is QC = 180 J + 100 J = 280 J . The thermal efficiency of the engine is

η=

Wout Wout 40 J = = = 0.125 QH QC + Wout 280 J + 40 J

(b) The heat extracted from the hot reservoir is QH = QC + Wout = 320 J.

= 40 J

19.17. Solve:

Model: The Brayton cycle involves two adiabatic processes and two isobaric processes. From Equation 19.21, the efficiency of a Brayton cycle is ηB = 1 − rp(1−γ ) γ , where rp is the pressure ratio

pmax/pmin. The specific heat ratio for a diatomic gas is

γ=

CP = CV

7 2 5 2

R = 1.4 R

Solving the above equation for rp,

(1 − ηB ) = rp(1−γ ) / γ

γ

−1.4

⇒ rp = (1 − ηB )1−γ = (1 − 0.60 ) 0.4 = 25

19.18.

Model: The Brayton cycle involves two adiabatic processes and two isobaric processes. The adiabatic processes involve compression and expansion through the turbine. Solve: The thermal efficiency for the Brayton cycle is ηB = 1 − rp(1−γ ) / γ , where γ = CP / CV and rp is the pressure

ratio. For a diatomic gas γ = 1.4. For an adiabatic process, p1V1γ = p2V2γ ⇒ p2 / p1 = (V1 / V2 )

γ

Because the volume is halved, V2 = 12 V1 and hence rp = p2 / p1 = ( 2 ) = 21.4 = 2.639 γ

The efficiency is −0.4

ηB = 1 − ( 2.639 ) 1.4 = 0.242

19.19. Model: The efficiency of a Carnot engine (ηCarnot) depends only on the temperatures of the hot and cold reservoirs. On the other hand, the thermal efficiency (η) of a heat engine depends on the heats QH and QC. Solve:

(a) According to the first law of thermodynamics, QH = Wout + QC . For engine (a), QH = 50 J, QC = 20 J

and Wout = 30 J, so the first law of thermodynamics is obeyed. For engine (b), QH = 10 J, QC = 7 J and Wout = 4 J, so the first law is violated. For engine (c) the first law of thermodynamics is obeyed. (b) For the three heat engines, the maximum or Carnot efficiency is

ηCarnot = 1 −

TC 300 K =1− = 0.50 TH 600 K

Engine (a) has

η = 1−

QC Wout 30 J = = = 0.60 QH QH 50 J

This is larger than ηCarnot, thus violating the second law of thermodynamics. For engine (b),

η=

Wout 4 J = = 0.40 < ηCarnot QH 10 J

so the second law is obeyed. Engine (c) has a thermal efficiency that is 10 J η= = 0.333 < ηCarnot 30 J so the second law of thermodynamics is obeyed.

19.20. Model: For a refrigerator QH = QC + Win, and the coefficient of performance and the Carnot coefficient of performance are

K=

QC Win

K Carnot =

TC TH − TC

Solve: (a) For refrigerator (a) QH = QC + Win (60 J = 40 J + 20 J) , so the first law of thermodynamics is obeyed. For refrigerator (b) 50 J = 40 J + 10 J , so the first law of thermodynamics is obeyed. For the refrigerator (c) 40 J ≠ 30 J + 20 J , so the first law of thermodynamics is violated. (b) For the three refrigerators, the maximum coefficient of performance is

K Carnot =

TC 300 K = =3 TH − TC 400 K − 300 K

For refrigerator (a), K=

QC 40 J = = 2 < K Carnot Win 20 J

so the second law of thermodynamics is obeyed. For refrigerator (b), K=

QC 40 J = = 4 > K Carnot Win 10 J

so the second law of thermodynamics is violated. For refrigerator (c), K= so the second law is obeyed.

30 J = 1.5 < K Carnot 20 J

19.21. Model: The efficiency of a Carnot engine depends only on the absolute temperatures of the hot and cold reservoirs. Solve: The efficiency of a Carnot engine is

ηCarnot = 1 −

TC TC ⇒ TC = 280 K = 7°C ⇒ 0.60 = 1 − TH ( 427 + 273) K

Assess: A “real” engine would need a lower temperature than 7oC to provide 60% efficiency because no real engine can match the Carnot efficiency.

19.22. Model: Assume that the heat engine follows a closed cycle. Solve:

(a) The engine’s efficiency is

η=

Wout Wout 200 J = = = 0.25 = 25% QH QC + Wout 600 J + 200 J

(b) The thermal efficiency of a Carnot engine is ηCarnot = 1 − TC TH . For this to be 25%, 0.25 = 1 −

TC

( 400 + 273) K

⇒ TC = 504.8 K = 232°C

19.23. Model: The efficiency of an ideal engine (or Carnot engine) depends only on the temperatures of the hot and cold reservoirs. Solve: (a) The engine’s thermal efficiency is

η=

Wout Wout 10 J = = = 0.40 = 40% QH QC + Wout 15 J + 10 J

(b) The efficiency of a Carnot engine is ηCarnot = 1 − TC TH . The minimum temperature in the hot reservoir is found as follows:

0.40 = 1 −

293 K ⇒ TH = 488 K = 215°C TH

This is the minimum possible temperature. In a real engine, the hot-reservoir temperature would be higher than 215°C because no real engine can match the Carnot efficiency.

19.24. Solve: (a) The efficiency of the Carnot engine is ηCarnot = 1 −

TC 300 K =1− = 0.40 = 40% TH 500 K

(b) An engine with power output of 1000 W does Wout = 1000 J of work during each Δt = 1 s. A Carnot engine has a heat input that is Qin =

Wout

ηCarnot

=

1000 J = 2500 J 0.40

during each Δt = 1 s. The rate of heat input is 2500 J/s = 2500 W. (c) Wout = Qin − Qout , so the heat output during Δt = 1 s is Qout = Qin − Wout = 1500 J . The rate of heat output is

thus 1500 J/s = 1500 W.

19.25.

Visualize:

We will use Equation 19.29 for the efficiency of a Carnot engine.

ηCarnot = 1 −

TC TH

We are given TH = 673K and the original efficiency ηCarnot = 0.40 . Solve:

First solve for TC . TC = TH (1 − ηCarnot ) = (673K)(1 − 0.40) = 404K

′ Solve for TC′ again with ηCarnot = 0.60. ′ ) = (673K)(1 − 0.60) = 269K TC′ = TH (1 − ηCarnot The difference of these TC ’s is 135 K, so the temperature of the cold reservoir should be decreased by 135 degrees to raise the efficiency from 40% to 60%. Assess: We expected to have to lower TC by quite a bit to get the better efficiency.

19.26. Model: The maximum possible efficiency for a heat engine is provided by the Carnot engine. Solve:

The maximum efficiency is

ηmax = ηCarnot = 1 −

TC ( 273 + 20 ) K = 0.6644 =1− TH ( 273 + 600 ) K

Because the heat engine is running at only 30% of the maximum efficiency, η = ( 0.30 )ηmax = 0.1993 . The

amount of heat that must be extracted is QH =

Wout

η

=

1000 J = 5.0 kJ 0.1993

19.27. Model: The coefficient of performance of a Carnot refrigerator depends only on the temperatures of the cold and hot reservoirs. Solve: (a) The Carnot performance coefficient of a refrigerator is

K Carnot =

TC ( −20 + 273) K = = 6.33 TH − TC ( 20 + 273) K − ( −20 + 273) K

(b) The rate at which work is done on the refrigerator is found as follows:

K=

QC Q 200 J/s ⇒ Win = C = = 32 J/s = 32 W Win K 6.325

(c) The heat exhausted to the hot side per second is

QH = QC + Win = 200 J/s + 32 J/s = 232 J/s = 232 W

19.28. Visualize: We are given TH = 773K and TC = 273K , therefore (by Equation 19.29) the Carnot 273 K efficiency is nCarnot = 1 − 773 K = 0.647

We are also given η = 0.60(ηCarnot ) . Solve:

Rearrange Equation 19.7: QH = Wout (1 − η ) . Wout is the same for both engines, so it cancels.

QH Wout (1 − η ) 1 − 0.60(ηCarnot ) 1 − 0.60(0.647) = = = = 1.73 (QH )Carnot Wout (1 − ηCarnot ) 1 − ηCarnot 1 − 0.647 Assess: This engine requires 1.73 times as much heat energy in during each cycle as a Carnot engine to do the same amount of work.

19.29. Model: The minimum possible value of TC occurs with a Carnot refrigerator.

Solve:

(a) For the refrigerator, the coefficient of performance is

K=

QC ⇒ QC = KWin = ( 5.0 )(10 J ) = 50 J Win

The heat energy exhausted per cycle is QH = QC + Win = 50 J + 10 J = 60 J (b) If the hot-reservoir temperature is 27°C = 300 K, the lowest possible temperature of the cold reservoir can be obtained as follows:

K Carnot =

TC TC ⇒ 5.0 = ⇒ TC = 250 K = −23°C TH − TC 300 K − TC

19.30. Model: The coefficient of performance of a Carnot refrigerator depends only on the temperatures of the cold and hot reservoirs. Solve: Using the definition of the coefficient of performance for a Carnot refrigerator,

K Carnot =

TC ( −13 + 273) K = 260 K ⇒ T = 312 K ⇒ 5.0 = H TH − TC TH − ( −13 + 273) K TH − 260 K

To increase the coefficient of performance to 10.0, the hot-reservoir temperature changes to TH′ . Thus, K Carnot = 10.0 =

260 K ⇒ TH′ = 286 K TH′ − 260 K

Since TH′ is less than TH, the hot-reservoir temperature must be decreased by 26 K or 26 ° C.

19.31.

Solve:

The work done by the engine is equal to the change in the gravitational potential energy.

Thus, Wout = ΔUgrav = mgh = (2000 kg)(9.8 m/s2)(30 m) = 588,000 J The efficiency of this engine is

⎛

η = 0.40 ηCarnot = 0.40 ⎜1 − ⎝

⎛ ( 273 + 20 ) K ⎞ = 0.3484 TC ⎞ ⎟⎟ ⎟ = 0.40 ⎜⎜1 − TH ⎠ ⎝ ( 273 + 2000 ) K ⎠

The amount of heat energy transferred is calculated as follows:

η=

Wout W 588,000 J ⇒ QH = out = = 1.7 × 106 J η QH 0.3484

19.32.

Solve:

The mass of the water is

100 × 10−3 L ×

10−3 m3 1000 kg × = 0.100 kg 1L m3

The heat energy is removed from the water in three steps: (1) cooling from +15°C to 0°C, (2) freezing at 0°C, and (3) cooling from 0°C to −15°C . The three heat energies are Q1 = mcΔT = ( 0.100 kg )( 4186 J/kg K )(15 K ) = 6279 J

Q2 = mLf = ( 0.100 kg ) ( 3.33 × 105 J/kg ) = 33,300 J

Q3 = mcΔT = ( 0.100 kg )( 2090 J/kg K )(15 K ) = 3135 J ⇒ QC = Q1 + Q2 + Q3 = 42,714 J

Using the performance coefficient, Q 42,714 J 42,714 J K = C ⇒ 4.0 = ⇒ Win = = 10,679 J Win Win 4.0

The heat exhausted into the room is thus QH = QC + Win = 42,714 J + 10,679 J = 5.3 × 104 J

19.33. Solve: An adiabatic process has Q = 0 and thus, from the first law, Ws = –∆Eth. For any ideal-gas process, ∆Eth = nCV∆T, so Ws = –nCV∆T. We can use the ideal-gas law to find T=

pV Δ ( pV ) ( pV )f − ( pV )i pfVf − pV i i ⇒ ΔT = = = nR nR nR nR

Consequently, the work is

CV ⎛ p V − pV i i ⎞ Ws = −nCV ΔT = − nCV ⎜ f f ( pfVf − pV i i) ⎟=− nR R ⎝ ⎠ Because CP = CV + R, we can use the specific heat ratio γ to find

γ=

C P CV + R C V / R + 1 = = ⇒ CV CV CV / R

CV 1 = R γ −1

With this, the work done in an adiabatic process is Ws = −

CV 1 ( pfVf − pV ( pf Vf − pV i i) = − i i ) = ( pf Vf − pV i i )/(1 − γ ) R γ −1

19.34. Visualize: We

are

QH = msteam Lv = (10kg)(22.6 × 105 J/kg) = 22.6 × 106 J

given

QC = mice Lf = (55kg) (3.33 × 10 J/kg) = 18.3 × 10 J . 5

Solve:

Equation 19.6 gives Wout = QH − QC .

P= Assess:

6

Wout QH − QC 22.6 × 106 J − 18.3 × 106 J ⎛ 1h ⎞ = = ⎜ ⎟ = 1200 W 1h Δt Δt ⎝ 3600s ⎠

This is a reasonable answer.

and

Solve: For any heat engine, η = 1 – QC/QH. For a Carnot heat engine, ηCarnot = 1 – TC/TH. Thus a property of the Carnot cycle is that QC/QH = TC/TH. Consequently, the coefficient of performance of a Carnot refrigerator is

19.35.

K Carnot =

QC QC QC /QH T C / TH TC = = = = Win QH − QC 1 − QC /QH 1 − T C / TH TH − TC

19.36. Visualize: Equation 19.29 gives ηCarnot = 1 − TT . We are given ηCarnot = 1/ 3. C

H

Solve:

TC 1 T 2 3 = ⇒ C = ⇒ TH = TC TH 3 TH 3 2 Equation 19.30 gives the coefficient of performance for the Carnot refrigerator.

ηCarnot = 1 −

K Carnot =

Assess:

1 TC T =3 C =3 =2 TH − TC 2 TC − TC 2 − 1

This result is in the ballpark for coefficients of performance.

19.37. Visualize: We are given TH = 298 K and TC = 273 K . See Figure 19.11. Solve:

QC = mLf = (10kg)(3.33 × 105 J/kg) = 3.33 × 106 J .

(a) For a Carnot cycle ηCarnot = 1 − THC but that must also equal η = 1 − QHC , so T

QH = QC

Q

QC QH

T

= THC .

⎛ 298K ⎞ TH 6 = (3.33 × 106 J) ⎜ ⎟ = 3.63 × 10 J TC ⎝ 273K ⎠

(b)

Win = QH − QC = 3.63 × 106 J − 3.33 × 106 J = 0.30 × 106 J = 3.0 × 105 J Assess:

This is a reasonable amount of work to freeze 10 kg of water.

19.38. Model: We will use the Carnot engine to find the maximum possible efficiency of a floating power plant. Solve:

The efficiency of a Carnot engine is

ηmax = ηCarnot = 1 −

TC ( 273 + 5) K = 0.0825 ≈ 8% =1− TH ( 273 + 30 ) K

19.39.

Model: The ideal gas in the Carnot engine follows a closed cycle in four steps. During the isothermal expansion at temperature TH, heat QH is transferred from the hot reservoir into the gas. During the isothermal compression at TC, heat QC is removed from the gas. No heat is transferred during the remaining two adiabatic steps. Solve: The thermal efficiency of the Carnot engine is

ηCarnot = 1 −

TC Wout 323 K Wout = ⇒ Wout = 436 J ⇒1− = TH QH 573 K 1000 J

Using QH = QC + Wout , we obtain Qisothermal = QC = QH − Wout = 1000 J − 436 J ≈ 560 J

19.40.

Solve:

Substituting into the formula for the efficiency of a Carnot engine,

ηCarnot = 1 −

TC TC ⇒ 0.25 = 1 − ⇒ TC = 240 K = −33°C TH TC + 80 K

The hot-reservoir temperature is TH = TC + 80 K = 320 K = 47 ° C.

19.41. Model: Assume the gas is monatomic. Visualize:

We are given TC = 273 K and QC = 15 J per cycle.

For 98 cycles Wout = mgh = (10kg)(9.8kg/m 2 )(10m) = 9800 J ; for one cycle this would be Wout = 100 J. Solve: Do the computation for one cycle. QH = Wout + QC = 100J + 15J = 115J For a Carnot cycle ηCarnot = 1 − THC but that must also equal η = 1 − QHC , so T

Q

TH = TC Assess:

QC QH

T

= THC .

⎛ 115J ⎞ QH = (273K) ⎜ ⎟ = 2093K ≈ 2100K QC ⎝ 15J ⎠

TH must be this high to give the 87% efficiency.

19.42.

Solve:

If QC =

2 3

QH, then Wout = QH – QC =

η=

1 3

QC. Thus the efficiency is

Wout 13 QH 1 = = QH QH 3

The efficiency of a Carnot engine is η = 1 – TC/TH. Thus T 1 TC 2 ⇒ = 1− C = TH 3 TH 3

19.43.

Solve:

(a) Q1 is given as 1000 J. Using the energy transfer equation for the heat engine,

QH = QC + Wout ⇒ Q1 = Q2 + Wout ⇒ Q2 = Q1 − Wout The thermal efficiency of a Carnot engine is

η = 1−

TC W 300 K =1− = 0.50 = out TH 600 K Q1

⇒ Q2 = Q1 − ηQ1 = Q1 (1 − η ) = (1000 J )(1 − 0.50 ) = 500 J To determine Q3 and Q4, we turn our attention to the Carnot refrigerator, which is driven by the output of the heat engine with Win = Wout. The coefficient of performance is K=

TC 400 K Q Q Q = = 4.0 = C = 4 = 4 TH − TC 500 K − 400 K Win Wout ηQ1

⇒ Q4 = KηQ1 = ( 4.0 )( 0.50 )(1000 J ) = 2000 J Using now the energy transfer equation Win + Q4 = Q3 , we have

Q3 = Wout + Q4 = ηQ1 + Q4 = ( 0.50 )(1000 J ) + 2000 J = 2500 J (b) From part (a) Q3 = 2500 J and Q1 = 1000 J , so Q3 > Q1 . (c) Although Q1 = 1000 J and Q3 = 2500 J, the two devices together do not violate the second law of thermodynamics. This is because the hot and cold reservoirs are different for the heat engine and the refrigerator.

19.44.

Model: A heat pump is a refrigerator that is cooling the already cold outdoors and warming the indoors with its exhaust heat. Solve: (a) The coefficient of performance for this heat pump is K = 5.0 = QC Win , where QC is the amount of

heat removed from the cold reservoir. QH is the amount of heat exhausted into the hot reservoir. QH = QC + Win, where Win is the amount of work done on the heat pump. We have QC = 5.0Win ⇒ QH = 5.0Win + Win = 6.0Win

If the heat pump is to deliver 15 kJ of heat per second to the house, then 15 kJ = 2.5 kJ 6.0 In other words, 2.5 kW of electric power is used by the heat pump to deliver 15 kJ/s of heat energy to the house. (b) The monthly heating cost in the house using an electric heater is QH = 15 kJ = 6.0Win ⇒ Win =

15 kJ 3600 s 1$ × 200 h × × = $270 s 1h 40 MJ

The monthly heating cost in the house using a heat pump is 2.5 kJ 3600 s 1$ × 200 h × × = $45 s 1h 40 MJ

19.45. Visualize: We are given TC = 275 K , TH = 295 K. We are also given that in one second Win = 100 J and QC = (1 s)(100 kJ/min)(1 min/60 s) = 1667 J. Solve: The coefficient of performance of a refrigerator is given in Equation 19.10.

K=

QC 1667 J = = 16.67 Win 100 J

However the coefficient of performance of a Carnot refrigerator is given in Equation 19.30. K Carnot =

TC 275 K = = 13.75 TH − TC 20 K

However, informal statement #8 of the second law says that the coefficient of performance cannot exceed the Carnot coefficient of performance, so the salesman is making false claims. You should not buy the DreamFridge. Assess: The second law imposes real-world restrictions.

19.46. Solve: The maximum possible efficiency of the heat engine is ηmax = 1 −

TC 300 K =1− = 0.40 TH 500 K

The efficiency of the engine designed by the first student is Wout 110 J = = 0.44 QH 250 J Because η1 > ηmax, the first student has proposed an engine that would violate the second law of thermodynamics. His or her design will not work. The efficiency of the engine designed by the second student is η2 = 90 J 250 J = 0.36 < ηmax in agreement with the second law of thermodynamics. Applying the first law of thermodynamics,

η1 =

QH = QC + Wout ⇒ 250 J = 170 J + 90 J we see the first law is violated. This design will not work as claimed. The design by the third student satisfies the first law of thermodynamics because QH = QC + Wout = 250 J . The thermal efficiency of this engine is η3 = 0.36 < ηmax , which satisfies the second law of thermodynamics. The data presented by students 1 and 2 are faulty. Only student 3 has an acceptable design.

19.47. Model: The power plant is to be treated as a heat engine. Solve:

(a) Every hour 300 metric tons or 3 × 105 kg of coal is burnt. The volume of coal is 3 ×105 kg m3 × × 24 h = 4800 m3 1h 1500 kg

The height of the room will be 48 m. (b) The thermal efficiency of the power plant is

η=

Assess:

7.50 ×108 J/s 7.50 ×108 J Wout = = = 0.32 = 32% 5 6 1h QH 3 ×10 kg 28 ×10 J 2.333 ×109 J × × 1h kg 3600 s

An efficiency of 32% is typical of power plants.

19.48. Model: The power plant is treated as a heat engine. Solve:

(a) The maximum possible thermal efficiency of the power plant is

ηmax = 1 −

TC 303 K =1− = 0.47 = 47% TH 573 K

(b) The plant’s actual efficiency is

η=

Wout 700 × 106 J/s = 0.35 = 35% QH 2000 × 106 J/s

(c) Because QH = QC + Wout,

QC = QH − Wout ⇒ QC = 2.0 × 109 J/s − 0.7 × 109 J/s = 1.3 × 109 J/s The mass of water that flows per second through the condenser is m = 1.2 ×108

L 1 hr 10−3 m3 1000 kg × × × = 3.333 × 104 kg h 3600 s 1L m3

The change in the temperature as QC = 1.3 × 109 J of heat is transferred to m = 3.333 × 104 kg of water is calculated as follows: QC = mcΔT ⇒ 1.3 × 109 J = ( 3.333 × 104 kg ) ( 4186 J/kg K ) ΔT ⇒ ΔT = 9°C The exit temperature is18°C + 9°C = 27°C .

19.49. Model: The power plant is treated as a heat engine. Solve:

The mass of water per second that flows through the plant every second is

m = 1.0 ×108

L 1 hr 10−3 m3 1000 kg × × × = 2.778 × 104 kg/s h 3600 s 1L m3

The amount of heat transferred per second to the cooling water is thus

QC = mcΔT = ( 2.778 × 104 kg/s ) ( 4186 J/kg K )( 27°C − 16°C ) = 1.279 × 109 J/s The amount of heat per second into the power plant is

QH = Wout + QC = 0.750 × 109 J/s + 1.279 × 109 J/s = 2.029 × 109 J/s Finally, the power plant’s thermal efficiency is

η=

Wout 0.750 × 109 J/s = = 0.37 = 37% QH 2.029 × 109 J/s

19.50. Solve: (a) The energy supplied in one day is J 3600 s 24 h Wout = 1.0 ×109 × × = 8.64 × 1013 J s 1h 1d (b) The volume of water is V = 1 km3 = 109 m3 . The amount of energy is

⎛ 1000 kg ⎞ 15 QH = mcΔT = (109 m3 ) ⎜ ⎟ ( 4190 J/kg K )(1.0 K ) = 4.19 × 10 J 3 ⎝ m ⎠ (c) While it’s true that the ocean contains vast amounts of thermal energy, that energy can be extracted to do useful work only if there is a cold reservoir at a lower temperature. That is, the ocean has to be the hot reservoir of a heat engine. But there’s no readily available cold reservoir, so the ocean’s energy cannot readily be tapped. There have been proposals for using the colder water near the bottom of the ocean as a cold reservoir, pumping it up to the surface where the heat engine is. Although possible, the very small temperature difference between the surface and the ocean depths implies that the maximum possible efficiency (the Carnot efficiency) is only a few percent, and the efficiency of any real ocean-driven heat engine would likely be less than 1%—perhaps much less. Thus the second law of thermodynamics prevents us from using the thermal energy of the ocean. Save your money. Don’t invest.

19.51. Visualize: If we do this problem on a “per second” basis then in one second QC = (1 s)(5.0 ×105 J/min) (1 min/60 s) = 8.33 ×103 J. QH = (1 s)(8.0 ×105 J/min)(1 min / 60 s) = 13.33 × 103 J. Solve: (a) Again, in one second Win = QH − QC = 13.33 × 103 J − 8.33 × 103 J = 5.0 × 103 J Since this is per second then the power required by the compressor is P = 5.0 kW. (b) Assess:

K= The result is typical for air conditioners.

QC 8.33 × 103 J = = 1.7 Win 5.0 × 103 J

19.52. Model: The heat engine follows a closed cycle with process 1 → 2 and process 3 → 4 being isochoric and process 2 → 3 and process 4 → 1 being isobaric. For a monatomic gas, CV = 32 R and CP = 52 R. Visualize: Please refer to Figure P19.52. Solve: (a) The first law of thermodynamics is Q = ΔEth + WS . For the isochoric process 1 → 2, WS 1 → 2 = 0 J. Thus, Q1→ 2 = 3750 J = ΔEth = nCV ΔT ⇒ ΔT =

3750 J 3750 J 3750 J = = = 301 K nCV (1.0 mol ) ( 32 R ) (1.0 mol ) ( 32 ) (8.31 J/mol K ) ⇒ T2 − T1 = 300.8 K ⇒ T2 = 300.8 K + 300 K = 601 K

To find volume V2, V2 = V1 =

nRT1 (1.0 mol )( 8.31 J/mol K )( 300 K ) = = 8.31× 10−3 m3 p1 3.0 × 105 Pa

The pressure p2 can be obtained from the isochoric condition as follows: p2 p1 T ⎛ 601 K ⎞ 5 5 = ⇒ p2 = 2 p1 = ⎜ ⎟ ( 3.00 × 10 Pa ) = 6.01 × 10 Pa T2 T1 T1 300 K ⎝ ⎠ With the above values of p2, V2 and T2, we can now obtain p3, V3 and T3. We have V3 = 2V2 = 1.662 × 10−2 m3

p3 = p2 = 6.01× 105 Pa

T3 T2 V = ⇒ T3 = 3 T2 = 2T2 = 1202 K V3 V2 V2 For the isobaric process 2 → 3,

Q2 →3 = nCP ΔT = (1.0 mol ) ( 52 R ) (T3 − T2 ) = (1.0 mol ) ( 52 ) ( 8.31 J/mol K )( 601 K ) = 12,480 J WS 2 →3 = p3 (V3 − V2 ) = ( 6.01× 105 Pa )( 8.31× 10−3 m3 ) = 4990 J ΔEth = Q2 →3 − WS 2 →3 = 12,480 J − 4990 J = 7490 J We are now able to obtain p4, V4 and T4. We have V4 = V3 = 1.662 × 10−2 m3

p4 = p1 = 3.00 × 105 Pa

⎛ 3.00 × 105 Pa ⎞ T4 T3 p = ⇒ T4 = 4 T3 = ⎜ ⎟ (1202 K ) = 600 K 5 p4 p3 p3 ⎝ 6.01 × 10 Pa ⎠ For isochoric process 3 → 4,

Q3→ 4 = nCV ΔT = (1.0 mol ) ( 32 R ) (T4 − T3 ) = (1.0 mol ) ( 32 ) ( 8.31 J/mol K )( −602 ) = −7500 J WS 3→ 4 = 0 J ⇒ ΔEth = Q3→ 4 − WS 3→ 4 = −7500 J For isobaric process 4 → 1, Q4 →1 = nCP ΔT = (1.0 mol ) 52 ( 8.31 J/mol K )( 300 K − 600 K ) = −6230 J

WS 4 →1 = p4 (V1 − V4 ) = ( 3.00 × 105 Pa ) × (8.31 × 10−3 m3 − 1.662 × 10−2 m3 ) = −2490 J ΔEth = Q4 →1 − WS 4 →1 = −6230 J − ( −2490 J ) = −3740 J WS (J)

Q (J)

ΔEth (J)

1→2

0

3750

3750

2→3

4990

12,480

7490

3→4

0

–7500

–7500

4→1 Net

–2490

–6230

–3740

2500

2500

0

(b) The thermal efficiency of this heat engine is

η= Assess:

Wout Wout 2500 J = = = 0.15 = 15% QH Q1→ 2 + Q2 →3 3750 J + 12,480 J

For a closed cycle, as expected, (Ws)net = Qnet and (ΔEth)net = 0 J

19.53. Model: The heat engine follows a closed cycle. For a diatomic gas, CV = 52 R and CP = 72 R. Visualize: Please refer to Figure P19.53. Solve: (a) Since T1 = 293 K, the number of moles of the gas is n=

5 −6 3 p1V1 ( 0.5 × 1.013 × 10 Pa )(10 × 10 m ) = = 2.08 × 10−4 mol RT1 ( 8.31 J/mol K )( 293 K )

At point 2, V2 = 4V1 and p2 = 3 p1 . The temperature is calculated as follows:

p1V1 p2V2 p V = ⇒ T2 = 2 2 T1 = ( 3)( 4 )( 293 K ) = 3516 K T1 T2 p1 V1 At point 3, V3 = V2 = 4V1 and p3 = p1 . The temperature is calculated as before: T3 =

p3 V3 T1 = (1)( 4 )( 293 K ) = 1172 K p1 V1

For process 1 → 2, the work done is the area under the p-versus-V curve. That is, Ws = ( 0.5 atm ) ( 40 cm3 − 10 cm3 ) + 12 (1.5 atm − 0.5 atm ) ( 40 cm3 − 10 cm3 ) ⎛ 1.013 × 105 Pa ⎞ = ( 30 × 10−6 m3 ) (1 atm ) ⎜ ⎟ = 3.04 J 1 atm ⎝ ⎠ The change in the thermal energy is

ΔEth = nCV ΔT = ( 2.08 × 10−4 mol ) 52 ( 8.31 J/mol K )( 3516 K − 293 K ) = 13.93 J The heat is Q = Ws + ΔEth = 16.97 J . For process 2 → 3, the work done is Ws = 0 J and Q = ΔEth = nCV ΔT = n ( 52 R ) (T3 − T2 )

= ( 2.08 × 10−4 mol ) 52 ( 8.31 J/mol K )(1172 K − 3516 K ) = −10.13 J

For process 3 → 1,

Ws = ( 0.5 atm ) (10 cm3 − 40 cm3 ) = ( 0.5 × 1.013 × 105 Pa )( −30 × 10−6 m3 ) = −1.52 J ΔEth = nCV ΔT = ( 2.08 × 10−4 mol ) 52 ( 8.31 J/mol K )( 293 K − 1172 K ) = −3.80 J The heat is Q = ΔEth + Ws = −5.32 J . 1→2 2→3 3→1 Net

Ws (J)

Q (J)

ΔEth

3.04 0 −1.52 1.52

16.97 −10.13 −5.32 1.52

13.93 −10.13 −3.80 0

(b) The efficiency of the engine is

η=

Wnet 1.52 J = = 0.090 = 9.0% QH 16.97 J

(c) The power output of the engine is 500

Assess:

revolutions 1 min Wnet 500 × × = × 1.52 J/s = 13 W min 60 s revolution 60

For a closed cycle, as expected, (Ws)net = Qnet and (ΔEth)net = 0 J.

19.54. Model: For the closed cycle, process 1 → 2 is isothermal, process 2 → 3 is isobaric, and process 3 → 1 is isochoric. Solve: (a) We first need to find the conditions at points 1, 2, and 3. We can then use that information to find WS and Q for each of the three processes that make up this cycle. Using the ideal-gas equation the number of moles of the gas is n=

5 −6 3 p1V1 (1.013 × 10 Pa )( 600 × 10 m ) = = 0.0244 mol RT1 (8.31 J/mol K )( 300 K )

We are given that γ = 1.25, which means this is not a monatomic or a diatomic gas. The specific heats are

CV =

R = 4R 1−γ

C P = CV + R = 5 R

At point 2, process 1 → 2 is isothermal, so we can find the pressure p2 as follows: p1V1 = p2V2 ⇒ p2 =

V1 6.00 × 10−4 m3 p1 = p1 = 3 p1 = 3 atm = 3.039 × 105 Pa V2 2.00 × 10−4 m3

At point 3, process 2 → 3 is isobaric, so we can find the temperature T3 as follows: V2 V3 V 6.00 × 10−4 m 3 T2 = 3T2 = 900 K = ⇒ T3 = 3 T2 = T2 T3 V2 2.00 × 10−4 m3 Point

V (m3 )

P (Pa)

6.00 × 10 2.00 × 10−4 6.00 × 10−4

1.0 atm = 1.013 × 10 3.0 atm = 3.039 × 105 3.0 atm = 3.039 × 105

1 2 3

T (K) −4

5

300 300 900

Process 1 → 2 is isothermal:

(WS )12 = p1V1 ln (V2

V1 ) = −66.8 J

Q12 = (WS )12 = −66.8 J

Process 2 → 3 is isobaric:

(WS )23 = p2ΔV = p2 (V3 − V2 ) = 121.6 J

Q23 = nCP ΔT = nCP (T3 − T2 ) = 608.3 J

Process 3 → 1 is isochoric:

(WS )31 = 0 J

Q31 = nCV ΔT = nCV (T1 − T3 ) = −486.7 J

We find that

(WS )cycle = −66.8 J + 121.6 J + 0 J = 54.8 J

Qcycle = −66.8 J + 608.3 J − 486.7 J = 54.8 J

These are equal, as they should be. Knowing that the work done is Wout = (WS)cycle = 54.8 J/cycle, an engine operating at 20 cycles/s has power output Pout =

54.8 J 20 cycle J × = 1096 = 1096 W ≈ 1.10 kW cycle s s

(b) Only Q23 is positive, so Qin = Q23 = 608 J. Thus, the thermal efficiency is

η=

Wout 54.8 J = = 0.090 = 9.0% Qin 608.3 J

19.55. Model: For the closed cycle of the heat engine, process 1 → 2 is isochoric, process 2 → 3 is adiabatic, and process 3 → 1 is isobaric. CV = 32 R and CP = 52 R for a monatomic gas. Visualize: Please refer to Figure P19.55. Solve: (a) Using the ideal-gas equation, the number of moles is n=

5 −6 3 p1V1 (1.0 × 10 Pa ) (100 × 10 m ) = = 0.00401 mol RT1 (8.31 J/mol K)(300 K)

We can use the adiabat 2 → 3 to calculate p2 as follows: γ

γ

⎛V ⎞ ⎛V ⎞ ⎛ 600 cm3 ⎞ p3V3 = p2V2 ⇒ p2 = p3 ⎜ 3 ⎟ = p1 ⎜ 3 ⎟ = (1.0 × 105 Pa ) ⎜ 3 ⎟ ⎝ 100 cm ⎠ ⎝ V2 ⎠ ⎝ V2 ⎠ γ

5/3

γ

= 1.981 × 106 Pa

T2 can be determined from the ideal-gas equation and is

T2 =

6 −6 3 p2V2 (1.981 × 10 Pa )(100 × 10 m ) = = 5943 K nR (8.31 J/mol K)(0.00401 mol)

At point 3, p3 = p1 and T3 T1 V 600 × 10−6 m3 = ⇒ T3 = 3 T1 = ( 300 K ) = 1800 K V3 V1 V1 100 × 10−6 m3 Now we can calculate Ws, Q, and ΔEth for the three processes involved in the cycle. For process 1 → 2, Ws 1→2 = 0 J and ΔEth = Q1→ 2 = nCV (T2 − T1 ) = n ( 32 R ) (T2 − T1 ) = 282.2 J

For process 2 → 3, Q2→3 = 0 J and ΔEth = nCV (T3 − T2 ) = n ( 32 R ) (T3 − T2 ) = –207.2 J

Because, ΔEth = W + Q, W = −Ws, so Ws 2→3 = −ΔEth = +207.2 J for process 2 → 3. For process 3 → 1, ΔEth = nCV (T1 − T3 ) = n ( 32 R ) (T1 − T3 ) = −75.0 J Q3→1 = nCP (T1 − T3 ) = n ( 52 R ) (T1 − T3 ) = −125.0 J

The work done Ws 3→1 is the area under the p-versus-V graph. We have

Ws 3→1 = (100 × 103 Pa )(100 × 10−6 m3 − 600 × 10−6 m3 ) = −50.0 J

1→2 2→3 3→1 Net

Ws (J)

Q (J)

ΔEth (J)

0 207.2 −50.0 157.2

282.2 0 −125.0 157.2

282.2 −207.2 −75.0 0

(b) The thermal efficiency of the engine is

Wout 157.2 J = = 0.52 = 52% QH 282.2 J As expected for a closed cycle, (Ws)net = Qnet and (ΔEth)net = 0 J.

η=

Assess:

19.56. Model: For the closed cycle of the heat engine, process 1 → 2 is isochoric, process 2 → 3 is adiabatic, and process 3 → 1 is isothermal. For a diatomic gas CV = 52 R and γ = 75 . Solve:

(a) From the graph V2 = 1000 cm3 .

The pressure p2 lies on the adiabat from 2 → 3. We can find the pressure as follows: γ

⎛V ⎞ ⎛ 4000 cm3 ⎞ p2V2γ = p3V3γ ⇒ p2 = p3 ⎜ 3 ⎟ = (1.0 ×105 Pa ) ⎜ 3 ⎟ ⎝ 1000 cm ⎠ ⎝ V2 ⎠

7/5

= 6.964 ×105 Pa ≈ 700 kPa

The temperature T2 can be obtained from the ideal-gas equation relating points 1 and 2:

⎛ 6.964 × 105 Pa ⎞ p1V1 p2V2 p V = ⇒ T2 = T1 2 2 = ( 300 K ) ⎜ ⎟ (1) = 522.3 K ≈ 522 K 5 T1 T2 p1 V1 ⎝ 4.0 × 10 Pa ⎠ (b) The number of moles of the gas is

R=

5 −3 3 p1V1 ( 4.0 × 10 Pa )(1.0 × 10 m ) = = 0.1604 mol RT1 (8.31 J/mol K)(300 K)

For isochoric process 1 → 2 , Ws = 0 J and Q = ΔEth = nCV ΔT = n ( 52 R ) ΔT = 741.1 J For adiabatic process 2 → 3, Q = 0 J and ΔEth = nCV ΔT = n ( 52 R ) (T3 − T2 ) = −741.1 J Using the first law of thermodynamics, ΔEth = ΔWs + Q, which means Ws = −ΔEth = +741.1 J. Ws can also be determined from

Ws =

p3V3 − p2V2 nR (T3 − T2 ) ( 34 J/K ) ( 300 K − 522.3 K ) = = = 741.1 J 1−γ 1−γ ( − 52 )

For isothermal process 3 → 1, ΔEth = 0 J and

Ws = nRT1 ln

V1 = −554.5 J V3

Using the first law of thermodynamics, ΔEth = −Ws + Q , Q = Ws = −554.5 J . 1→2 2→3 3→1 Net

ΔEth (J)

Ws (J)

Q (J)

741.1 −741.1 0 0

0 741.1 −554.5 186.6

741.1 0 −554.5 186.6

(c) The work per cycle is 186.6 J and the thermal efficiency is

η=

Ws 186.6 J = = 0.25 = 25% QH 741.1 J

19.57. Model: For the closed cycle of the heat engine, process 1 → 2 is adiabatic, process 2 → 3 is isothermal, and process 3 → 1 is isochoric. For a diatomic gas CV = 52 R and γ = 75 . Solve: (a) From the graph V1 = 4000 cm3 . The number of moles of gas is n=

5 −6 3 p2V2 ( 4.0 × 10 Pa )(1000 × 10 m ) = = 0.1203 mol RT2 (8.31 J/mol K)(400 K)

Using p1V1γ = p2V2γ , and reading V1 from the graph, γ

⎛V ⎞ 7/5 p1 = p2 ⎜ 2 ⎟ = ( 4.0 × 105 Pa ) ( 14 ) = 5.743 × 104 Pa ≈ 5.7 kPa ⎝ V1 ⎠ With p1, V1, and nR having been determined, we can find T1 using the ideal-gas equation: T1 =

4 −6 3 p1V1 ( 5.743 × 10 )( 4000 × 10 m ) = = 229.7 K ≈ 230 K nR (8.31 J/mol K)(0.1203 mol)

(b) For adiabatic process 1 → 2, Q = 0 J and

Ws =

5 −3 3 4 −3 3 p2V2 − p1V1 ( 4.0 × 10 Pa )(1.00 × 10 m ) − ( 5.743 × 10 Pa )( 4.00 × 10 m ) = = −425.7 K 1−γ 1 − 1.4

Because ΔEth = −Ws + Q , ΔEth = −Ws = 425.7 K . For isothermal process 2 → 3, ΔEth = 0 J . From Equation 17.15,

Ws = nRT2 ln

V3 = 554.5 J V2

From the first law of thermodynamics, Q = Ws = 554.5 J for process 2 → 3. For isochoric process 3 → 1, Ws = 0 J and Q = ΔEth = nCV ΔT = n ( 52 R ) (T1 − T3 ) = −425.7 J

1→2 2→3 3→1 Net

ΔEth (J)

Ws (J)

Q (J)

425.7 0 −425.7 0

−425.7 554.5 0 128.8

0 554.5 −425.7 128.8

(c) The work per cycle is 128.8 J and the thermal efficiency of the engine is

η= Assess:

Ws 128.8 J = = 0.23 = 23% QH 554.5 J

As expected, for a closed cycle (Ws )net = Qnet and ( ΔEth )net = 0 J.

19.58. Model: In the Brayton cycle, process 1 → 2 and process 3 → 4 are adiabatic, and process 2 → 3 and process 4 → 2 are isobaric. We will assume the gas to be diatomic, with CP = 72 R and γ = 1.40. Visualize: Please refer to Figure P19.58. Solve: The number of moles of gas is n=

5 3 p1V1 (1.013 × 10 Pa )( 2.0 m ) = = 81.27 mol RT1 (8.31 J/mol K)(300 K)

We can find the volume at 2 from the adiabatic equation p1V1γ = p2V2γ , 1

1

⎛ p ⎞γ ⎛ 1 ⎞1.4 V2 = V1 ⎜ 1 ⎟ = ( 2.0 m3 ) ⎜ ⎟ = 0.3861 m3 ⎝ 10 ⎠ ⎝ p2 ⎠ The temperature T2 is determined from the ideal-gas equation as follows: 2 p1V1 p2V2 p V ⎛ 10 atm ⎞ ⎛ 0.3861 m ⎞ = ⇒ T2 = T1 2 2 = ( 300 K ) ⎜ ⎟ = 579.2 K ⎟⎜ 3 T1 T2 p1 V1 ⎝ 1 atm ⎠ ⎝ 2.0 m ⎠

To find T3 we use the heat input: QH = 2.0 × 106 J = nCP ΔT = 72 n RΔT =

7 2

( 675.33 J/K ) ΔT

⇒ ΔT = 846.1 K ⇒ T3 = T2 + ΔT = 579.2 K + 846.1 K = 1425.3 K We are now able to find V3 using the ideal-gas law. We have V3 V2 T ⎛ 1425.3 K ⎞ 3 = ⇒ V3 = V2 3 = ( 0.3861 m3 ) ⎜ ⎟ = 0.9501 m T3 T2 T2 ⎝ 579.2 K ⎠

The volume at 4 is found by again using the adiabatic equation p3V3γ = p4V4γ , 1

1 ⎛ p ⎞γ V4 = V3 ⎜ 3 ⎟ = ( 0.9501 m 3 ) (10 )1.4 = 4.921 m3 ⎝ p4 ⎠

T4 is now obtained from T4 =

5 3 p4V4 (1.013 × 10 Pa )( 4.921 m ) = = 738.2 K nR (81.27 mol)(8.31 J/mol K)

To find the engine’s efficiency, we must first find the total work done per cycle which is Wnet = W1→ 2 + W2 →3 + W3→ 4 + W4 →1 The four contributions are W1→ 2 =

p2V2 − p1V1 nR (T2 − T1 ) ( 81.27 mol )( 8.31 J/mol K )( 579.2 K − 300 K ) = = = −4.714 × 105 J 1−γ 1−γ 1 − 1.4

W2 →3 = p2 ΔV = (1.013 × 106 Pa )( 0.9501 m3 − 0.3861 m3 ) = 5.713 × 105 J W3→ 4 =

nR (T4 − T3 ) ( 81.27 mol )( 8.31 J/mol K )( 738.2 K − 1425.3 K ) = = +1.160 × 106 J 1−γ 1 − 1.4

W4 →1 = p4 ΔV = (1.013 × 105 Pa )( 2.0 m3 − 4.921 m3 ) = −2.959 × 105 J Thus, Wnet = 9.64 × 105 J . The thermal efficiency is

η=

Wnet 9.64 × 105 J ⇒η = = 0.482 = 48.2% QH 2.0 × 106 J

As a comparison, we can calculate η from Equation 19.21 which is

η =1−

1

(r ) p

where rp = pmax pmin = 10 .

( ) γ −1 γ

=1−

1

(10 )

0.4 1.4

= 48.2%

19.59. Model: Process 1 → 2 of the cycle is isochoric, process 2 → 3 is isothermal, and process 3 → 1 is isobaric. For a monatomic gas, CV = 32 R and CP = 52 R. Visualize: Please refer to Figure P19.59. Solve: (a) At point 1: The pressure p1 = 1.0 atm = 1.013 × 105 Pa and the volume V1 = 1000 ×10−6 m 3 = 1.0 × 10−3 m3 . The number of moles is

n=

0.120 g = 0.030 mol 4 g/mol

Using the ideal-gas law, T1 =

5 −3 3 p1V1 (1.013 × 10 Pa )(1.0 × 10 m ) = = 406 K nR ( 0.030 mol )(8.31 J/mol K )

At point 2: The pressure p2 = 5.0 atm = 5.06 × 105 Pa and V2 = 1.0 × 10−3 m3 . The temperature is T2 =

5 −3 3 p2V2 ( 5.06 × 10 Pa )(1.0 × 10 m ) = = 2030 K nR ( 0.030 mol )(8.31 J/mol K )

At point 3: The pressure is p3 = 1.0 atm = 1.013 × 105 Pa and the temperature is T3 = T2 = 2030 K . The volume is V3 = V2

p2 ⎛ 5 atm ⎞ −3 3 = (1.0 × 10−3 m3 ) ⎜ ⎟ = 5.0 × 10 m p3 1 atm ⎝ ⎠

(b) For isochoric process 1 → 2, W1→2 = 0 J and Q1→ 2 = nCV ΔT = ( 0.030 mol ) ( 32 R ) ( 2030 K − 406 K ) = 607 J

For isothermal process 2 → 3, ΔEth 2 →3 = 0 J and

Q2 →3 = W2 →3 = nRT2 ln

⎛ 5.0 × 10−3 m3 ⎞ V3 = ( 0.030 mol )(8.31 J/mol K )( 2030 K ) ln ⎜ = 815 J −3 3 ⎟ V2 ⎝ 1.0 × 10 m ⎠

For isobaric process 3 → 1,

W3→1 = p3ΔV = (1.013 × 105 Pa )(1.0 × 10−3 m3 − 5.0 × 10−3 m3 ) = −405 J Q3→1 = nCP ΔT = ( 0.030 mol ) ( 52 ) ( 8.31 J/mol K )( 406 K − 2030 K ) = −1012 J The total work done is Wnet = W1→ 2 + W2 →3 + W3→1 = 410 J. The total heat input is QH = Q1→ 2 + Q2 →3 = 1422 J . The efficiency of the engine is

η=

Wnet 410 J = = 20% QH 1422 J

(c) The maximum possible efficiency of a heat engine that operates between Tmax and Tmin is

ηmax = 1 − Assess:

Tmin 406 K =1− = 80% Tmax 2030 K

The actual efficiency of an engine is less than the maximum possible efficiency.

19.60. Model: The process 2 → 3 of the heat engine cycle is isochoric and the process 3 → 1 is isobaric. For a monatomic gas CV = 32 R and CP = 52 R . Solve: (a) The three temperatures are T1 =

5 3 p1V1 ( 4.0 × 10 Pa )( 0.025 m ) = = 601.7 K ≈ 602 K nR ( 2.0 mol )( 8.31 J/mol K )

T2 =

5 3 p2V2 ( 6.0 × 10 Pa )( 0.050 m ) = = 1805.1 K ≈ 1805 K nR ( 2.0 mol )( 8.31 J/mol K )

T3 =

5 3 p3V3 ( 4.0 × 10 Pa )( 0.050 m ) = = 1203.4 K ≈ 1203 K nR ( 2.0 mol )( 8.31 J/mol K )

(b) For process 1 → 2, the work done is the area under the p-versus-V graph. The work and the change in internal energy are

Ws = 12 ( 6.0 × 105 Pa − 4.0 × 105 Pa )( 0.050 m3 − 0.025 m3 ) + ( 4.0 × 105 Pa )( 0.050 m3 − 0.025 m3 ) = 1.25 × 104 J ΔEth = nCV ΔT = ( 2.0 mol ) ( 32 R ) (T2 − T1 ) = ( 2.0 mol ) ( 32 ) ( 8.31 J/mol K )(1805.1 K − 601.7 K ) = 3.00 × 104 J The heat input is Q = Ws + ΔEth = 4.25 × 104 J . For isochoric process 2 → 3, Ws = 0 J and Q = ΔEth = nCV ΔT = ( 2.0 mol ) 32 ( 8.31 J/mol K )(1203.4 K − 1805.1 K ) = −1.50 × 104 J For isobaric process 3 → 1, the work done is the area under the p-versus-V curve. Hence,

Ws = ( 4.0 × 105 Pa )( 0.025 m3 − 0.050 m3 ) = −1.0 × 104 J ΔEth = nCV ΔT = n ( 32 R ) (T1 − T3 ) = ( 2.0 mol ) 32 ( 8.31 J/mol K )( 601.7 K − 1203.4 K ) = −1.5 × 104 J The heat input is Q = WS + ΔEth = −2.50 × 104 J .

ΔEth (J) 1→2 2→3 3→1 Net

Ws (J) 4

3.0 × 10 −1.5 × 104 −1.5 × 104 0

1.25 × 10 0 −1.0 ×104 2.5 × 103

(c) The thermal efficiency is

η=

Q (J) 4

Wnet 2.5 × 103 J = = 5.9% QH 4.25 × 104 J

4.25 × 104 −1.50 × 104 −2.50 × 104 2.5 × 103

19.61. Model: The closed cycle in this heat engine includes adiabatic process 1 → 2, isobaric process 2 → 3, and isochoric process 3 → 1. For a diatomic gas, CV = 52 R, CP = 72 R, and γ = 75 =1.4. Visualize: Please refer to Figure P19.61. Solve: (a) We can find the temperature T2 from the ideal-gas equation as follows: T2 =

5 −3 3 p2V2 ( 4.0 × 10 Pa )(1.0 × 10 m ) = = 2407 K nR ( 0.020 mol )( 8.31 J/mol K )

We can use the equation p2V2γ = p1V1γ to find V1, 1/ γ

⎛p ⎞ V1 = V2 ⎜ 2 ⎟ ⎝ p1 ⎠

1/1.4

⎛ 4.0 × 105 Pa ⎞ = (1.0 × 10−3 m3 ) ⎜ ⎟ 5 ⎝ 1.0 × 10 Pa ⎠

= 2.692 × 10−3 m3

The ideal-gas equation can now be used to find T1,

T1 =

5 −3 3 p1V1 (1.0 × 10 Pa )( 2.692 × 10 m ) = = 1620 K nR ( 0.020 mol )(8.31 J/mol K )

At point 3, V3 = V1 so we have

T3 =

5 −3 3 p3V3 ( 4 × 10 Pa )( 2.692 × 10 m ) = = 6479 K nR ( 0.020 mol )( 8.31 J/mol K )

(b) For adiabatic process 1 → 2, Q = 0 J, ΔEth = −Ws , and

Ws =

p2V2 − p1V1 nR (T2 − T1 ) ( 0.020 mol )( 8.31 J/mol K )( 2407 K − 1620 K ) = = = −327.0 J 1−γ 1−γ (1 − 1.4 )

For isobaric process 2 → 3, Q = nCP ΔT = n ( 72 R ) ( ΔT ) = ( 0.020 mol ) 72 ( 8.31 J/mol K )( 6479 K − 2407 K ) = 2369 J ΔEth = nCV ΔT = n ( 52 R ) ΔT = 1692 J The work done is the area under the p-versus-V graph. Hence,

Ws = ( 4.0 × 105 Pa )( 2.692 × 10−3 m3 − 1.0 × 10−3 m3 ) = 677 J For isochoric process 3 → 1, Ws = 0 J and ΔEth = Q = nCV ΔT = ( 0.020 mol ) ( 52 ) ( 8.31 J/mol K )(1620 K − 6479 K ) = −2019 J 1→2 2→3 3→1 Net

ΔEth (J)

Ws (J)

Q (J)

327 1692 −2019 0

−327 677 0 350

0 2369 −2019 350

(c) The engine’s thermal efficiency is

η=

350 J = 0.15 = 15% 2369 J

19.62. Model: The closed cycle of the heat engine involves the following processes: one isothermal compression, one isobaric expansion, and one isochoric cooling.

Visualize:

Solve:

The number of moles of helium is

n=

2.0 g = 0.50 mol 4.0 g/mol

The total work done by the engine per cycle is Wnet = W1→ 2 + W2 →3 + W3→1 . For the isothermal process, W1→ 2 = nRT1 ln

V2 = ( 0.50 mol )( 8.31 J/mol K )( 273 K ) ln ( 0.5) = −786.2 J V1

For the isobaric process, ⎛V ⎞ W2 →3 = p2 ( ΔV ) = ( 2 p1 ) ⎜ 1 ⎟ = p1V1 = nRT1 = ( 0.50 mol )( 8.31 J/mol K )( 273 K ) = 1134.3 J ⎝2⎠ For the isochoric process, W3→1 = 0 J. Thus, Wnet = 348.1 J ≈ 350 J . For calculating heat transfers, note that T2 = T1 = 273 K and, since process 3 → 1 is isochoric, T3 = (2 atm/1 atm) T1 = 546 K. For the isothermal process, Q1→ 2 = W1→ 2 = −786.2 J because ΔEth 1→ 2 = 0 J. For the isobaric process, Q2 →3 = nCP ΔT = n ( 52 R ) (T3 − T2 ) =

5 2

( 0.50 mol )(8.31 J/mol K )( 273 K ) = 2835.8 J

For the isochoric process, Q3→1 = nCv ΔT = n ( 32 R ) (T1 − T3 ) = ( 32 ) ( 0.50 mol )( 8.31 J/mol K )( –273 K ) = −1701.5 J From Q1→2, Q2→3, and Q3→1, we see that QH = 2835.8 J. Thus, the thermal efficiency is

η=

348.1 J = 0.12 = 12% 2835.8 J

19.63. Model: The closed cycle of the heat engine involves the following four processes: isothermal expansion, isochoric cooling, isothermal compression, and isochoric heating. For a monatomic gas CV = 32 R .

Visualize:

Solve:

Using the ideal-gas law,

p1 =

nRT1 ( 0.20 mol )( 8.31 J/mol K )( 600 K ) = = 4.986 × 105 Pa V1 2.0 × 10−3 m3

At point 2, because of the isothermal conditions, T2 = T1 = 600 K and

p2 = p1

⎛ 2.0 × 10−3 m3 ⎞ V1 = ( 4.986 × 105 Pa ) ⎜ = 2.493 × 105 Pa −3 3 ⎟ V2 ⎝ 4.0 × 10 m ⎠

At point 3, because it is an isochoric process, V3 = V2 = 4000 cm3 and p3 = p2

T3 ⎛ 300 K ⎞ 5 = ( 2.493 × 105 Pa ) ⎜ ⎟ = 1.247 × 10 Pa T2 ⎝ 600 K ⎠

Likewise at point 4, T4 = T3 = 300 K and p4 = p3

⎛ 4.0 × 10−3 m3 ⎞ V3 = (1.247 × 105 Pa ) ⎜ = 2.493 × 105 Pa 3 ⎟ −3 V4 2.0 10 m × ⎝ ⎠

Let us now calculate Wnet = W1→2 + W2→3 + W3→4 + W4→1. For the isothermal processes,

W1→ 2 = nRT1 ln

W3→ 4 = nRT3 ln

V2 = ( 0.20 mol )( 8.31 J/mol K )( 600 K ) ln ( 2 ) = 691.2 J V1

V4 = ( 0.20 mol )( 8.31 J/mol K )( 300 K ) ln ( 12 ) = −345.6 J V3

For the isochoric processes, W2→3 = W4→1 = 0 J. Thus, Wnet = 345.6 J ≈ 350 J . Because Q = Ws + ΔEth, Q1→ 2 = W1→ 2 + ( ΔEth )1→ 2 = 691.2 J + 0 J = 691.2 J For the first isochoric process, Q2 →3 = nCV ΔT = ( 0.20 mol ) ( 32 R ) (T3 − T2 ) = ( 0.20 mol ) 32 ( 8.31 J/mol K )( 300 K − 600 K ) = −747.9 K For the second isothermal process Q3→ 4 = W3→ 4 + ( ΔEth )3→ 4 = −345.6 J + 0 J = −345.6 J

For the second isochoric process, Q4 →1 = nCV ΔT = n ( 32 R ) (T1 − T4 ) = ( 0.20 mol ) ( 32 ) ( 8.31 J/mol K )( 600 K − 300 K ) = 747.9 K Thus, QH = Q1→ 2 + Q4 →1 = 1439.1 J . The thermal efficiency of the engine is

η=

Wnet 345.6 J = = 0.24 = 24% QH 1439.1 J

19.64. Model: Processes 2 → 1 and 4 → 3 are isobaric. Processes 3 → 2 and 1 → 4 are isochoric. Visualize:

Solve: (a) Except in an adiabatic process, heat must be transferred into the gas to raise its temperature. Thus heat is transferred in during processes 4 → 3 and 3 → 2. This is the reverse of the heat engine in Example 19.2. (b) Heat flows from hot to cold. Since heat energy is transferred into the gas during processes 4 → 3 and 3 → 2, which end with the gas at temperature 2700 K, the reservoir temperature must be T > 2700 K. This is the hot reservoir, so the heat transferred is QH. Similarly, heat energy is transferred out of the gas during processes 2 → 1 and 1 → 4. This requires that the reservoir temperature be T < 300 K. This is the cold reservoir, and the energy transferred during these two processes is QC. (c) The heat energies were calculated in Example 19.2, but now they have the opposite signs. QH = Q43 + Q32 = 7.09 × 105 J + 15.19 × 105 J = 22.28 × 105 J QC = Q21 + Q14 = 21.27 × 105 J + 5.06 × 105 J = 26.33 × 105 J (d) For a counterclockwise cycle in the pV diagram, the work is Win. Its value is the area inside the curve, which is Win = (∆p)(∆V) = (2 × 101,300 Pa)(2 m3) = 4.05 × 105 J. Note that Win = QC – QH, as expected from energy conservation. (e) Since QC > QH, more heat is exhausted to the cold reservoir than is extracted from the hot reservoir. In this device, work is used to transfer energy “downhill,” from hot to cold. The exhaust energy is QC = QH + Win > QH. This is the energy-transfer diagram of Figure 19.19. (f) No. A refrigerator uses work input to transfer heat energy from the cold reservoir to the hot reservoir. This device uses work input to transfer heat energy from the hot reservoir to the cold reservoir.

19.65. Solve: (a) If you wish to build a Carnot engine that is 80% efficient and exhausts heat into a cold reservoir at 0°C, what temperature (in °C) must the hot reservoir be? (b) 0.80 = 1 −

( 0°C + 273) ⇒ 273 = 0.20 ⇒ T = 1092°C H (TH + 273) TH + 273

19.66. Solve: (a) A refrigerator with a coefficient of performance of 4.0 exhausts 100 J of heat in each cycle. What work is required each cycle and how much heat is removed each cycle from the cold reservoir? (b) We have 4.0 = QC Win ⇒ QC = 4Win . This means QH = QC + Win = 4Win + Win = 5Win ⇒ Win = QH = 100 J = 20 J 5 5 Hence, QC = QH − Win = 100 J − 20 J = 80 J.

19.67. Solve: (a) A heat engine operates at 20% efficiency and produces 20 J of work in each cycle. What is the net heat extracted from the hot reservoir and the net heat exhausted in each cycle? (b) We have 0.20 = 1 − QC QH . Using the first law of thermodynamics, Wout = QH − QC = 20 J ⇒ QC = QH − 20 J Substituting into the definition of efficiency,

0.20 = 1 −

QH − 20 J 20 J 20 J 20 J = 1−1+ = ⇒ QH = = 100 J 0.20 QH QH QH

The heat exhausted is QC = QH − 20 J = 100 J − 20 J = 80 J .

19.68.

Solve: (a)

In this heat engine, 400 kJ of work is done each cycle. What is the maximum pressure? (b)

1 ( pmax − 1.0 ×105 Pa )( 2.0 m3 ) = 4.0 × 105 J ⇒ pmax = 5.0 ×105 Pa = 500 kPa 2

19.69. Model: The heat engine follows a closed cycle, starting and ending in the original state. Visualize: The figure indicates the following seven steps. First, the pin is inserted when the heat engine has the initial conditions. Second, heat is turned on and the pressure increases at constant volume from 1 atm to 3 atm. Third, the pin is removed. The flame continues to heat the gas and the volume increases at constant pressure from 50 cm3 to 100 cm3. Fourth, the pin is inserted and some of the weights are removed. Fifth, the container is placed on ice and the gas cools at constant volume to a pressure of 1 atm. Sixth, with the container still on ice, the pin is removed. The gas continues to cool at constant pressure to a volume of 50 cm3. Seventh, with no ice or flame, the pin is inserted back in and the weights returned bringing the engine back to the initial conditions and ready to start over.

Solve:

(a)

(b) The work done per cycle is the area inside the curve: Wout = (∆p)(∆V) = (2 × 101,300 Pa)(50 × 10–6 m3) = 10.13 J (c) Heat energy is input during processes 1 → 2 and 2 → 3, so QH = Q12 + Q23. This is a diatomic gas, with CV = 3 R and CP = 52 R. The number of moles of gas is 2

n=

p1V1 (101,300 Pa)(50 × 10−6 m3 ) = = 0.00208 mol RT1 (8.31 J/mol K) (293 K)

Process 1 → 2 is isochoric, so T2 = (p2/p1)T1 = 3T1 = 879 K. Process 2 → 3 is isobaric, so T3 = (V3/V2)T2 = 2T2 = 1758 K. Thus

Q12 = nCV ΔT = 52 nR(T2 − T1 ) = 52 (0.00208 mol)(8.31 J/mol K)(586 K) = 25.32 J Similarly, Q23 = nCP ΔT = 72 nR (T3 − T2 ) = 72 (0.00208 mol)(8.31 J/mol K)(879 K) = 53.18 J Thus QH = 25.32 J + 53.18 J = 78.50 J and the engine’s efficiency is

η=

Wout 10.13 J = = 0.13 = 13% QH 78.50 J

19.70. Model: System 1 undergoes an isochoric process and system 2 undergoes an isobaric process. Solve: (a) Heat will flow from system 1 to system 2 because system 1 is hotter. Because there is no heat input from (or loss to) the outside world, we have Q1 + Q2 = 0 J. Heat Q1, which is negative, will change the temperature of system 1. Heat Q2 will both change the temperature of system 2 and do work by lifting the piston. But these consequences of heat flow don’t change the fact that Q1 + Q2 = 0 J. System 1 undergoes constant volume cooling from T1i = 600 K to Tf. System 2, whose pressure is controlled by the weight of the piston, undergoes constant pressure heating from T2i = 300 K to Tf. Thus, Q1 + Q2 = 0 J = n1CV (Tf − T1i ) + n2CP (Tf − T2i ) = n1 ( 32 R ) (Tf − T1i ) + n2 ( 52 R ) (Tf − T2i ) Solving this equation for Tf gives

Tf =

3n1T1i + 5n2T2i 3 ( 0.060 mol )( 600 K ) + 5 ( 0.030 mol )( 300 K ) = = 464 K 3n1 + 5n2 3 ( 0.060 mol ) + 5 ( 0.030 mol )

(b) Knowing Tf , we can compute the heat transferred from system 1 to system 2:

Q2 = n2CP (Tf − T2i ) = n2 ( 52 R ) (Tf − T2i ) = 102 J (c) The change of thermal energy in system 2 is

ΔEth = n2CV ΔT = n2 ( 32 R ) (Tf − T2i ) = 53 Q2 = 61.2 J According to the first law of thermodynamics, Q2 = Ws + ΔEth. Thus, the work done by system 2 is Ws = Q − ΔEth = 102.0 J − 61.2 J = 40.8 J. The work is done to lift the weight of the cylinder and the air above it by a height Δy. The weight of the air is wair = pA = pπ r 2 = (101.3 × 10 N/m 2 )π (0.050 m) 2 = 795.6 N. So,

Ws = ( wcyl + wair )Δy ⇒ Δy =

Ws 40.8 J = = 0.050 m ( wcyl + wair ) ( 2.0 kg ) ( 9.8 m/s 2 ) + 795.6 N

(d) The fraction of heat converted to work is

Ws 40.8 J = = 0.40 = 40% Q2 102.0 J

19.71. Model: Process 1 → 2 and process 3 → 4 are adiabatic, and process 2 → 3 and process 4 → 1 are isochoric. Visualize: Please refer to Figure CP19.71. Solve: (a) For adiabatic process 1 → 2, Q12 = 0 J and W12 =

p2V2 − p1V1 nR (T2 − T1 ) = 1−γ 1−γ

For isochoric process 2 → 3, W23 = 0 J and Q23 = nCV (T3 − T2 ) . For adiabatic process 3 → 4, Q34 = 0 J and

W34 =

p4V4 − p3V3 nR (T4 − T3 ) = 1−γ 1−γ

For isochoric process 4 → 1, W41 = 0 J and Q41 = nCV (T1 − T4 ) . The work done per cycle is

Wnet = W12 + W23 + W34 + W41 =

nR (T2 − T1 ) nR (T4 − T3 ) nR +0 J+ +0 J = (T2 − T1 + T4 − T3 ) 1− γ 1−γ 1−γ

(b) The thermal efficiency of the heat engine is

η=

Q nC (T − T ) Wout Q = 1 − C = 1 − 41 = 1 − V 4 1 QH QH Q23 nCV (T3 − T2 )

The last step follows from the fact that T3 > T2 and T4 > T1. We will now simplify this expression further as follows: γ −1

⎛V ⎞ γ −1 pV γ = pVV γ −1 = nRTV γ −1 ⇒ nRTV = nRT2V2γ −1 ⇒ T2 = T1 ⎜ 1 ⎟ 1 1 ⎝ V2 ⎠ Similarly, T3 = T4 r γ −1 . The equation for thermal efficiency now becomes

η =1−

(c)

1 T4 − T1 = 1 − γ −1 γ −1 r − T4 r T1r γ −1

= T1r γ −1

19.72. Model: For the Diesel cycle, process 1 → 2 is an adiabatic compression, process 2 → 3 is an isobaric expansion, process 3 → 4 is an adiabatic expansion, and process 4 → 1 is isochoric. Visualize: Please refer to CP19.72. Solve: (a) It will be useful to do some calculations using the compression ratio, which is r=

Vmax V1 1050 cm3 = = = 21 Vmin V2 50 cm3

The number of moles of gas is

n=

5 −6 3 p1V1 (1.013 × 10 Pa )(1050 × 10 m ) = =0.0430 mol RT1 ( 8.31 J/mol K ) ⎡⎣( 25+273) K ⎤⎦

For an adiabatic process, γ

⎛V ⎞ p1V1 = p2V2 ⇒ p2 = ⎜ 1 ⎟ p1 = r γ p1 = 211.40 × 1 atm = 71.0 atm = 7.19 × 106 Pa ⎝ V2 ⎠ γ

γ

γ −1

⎛V ⎞ γ −1 TV = T2V2γ −1 ⇒ T2 = ⎜ 1 ⎟ 1 1 ⎝ V2 ⎠

T1 = r γ −1T1 = 210.41 × 298 K = 1007 K

Process 2 → 3 is an isobaric heating with Q = 1000 J. Constant pressure heating obeys

Q = nCP ΔT ⇒ ΔT =

Q nCP

The gas has a specific heat ratio γ = 1.40 = 7 / 5, thus CV = 52 R and CP = 72 R. Knowing CP, we can calculate first

ΔT = 800 K and then T3 = T2 + ΔT = 1807 K. Finally, for an isobaric process we have V2 V3 T 1807 K = ⇒ V3 = 3 V2 = ( 50 × 10−6 m3 ) = 89.7 × 10−6 m3 T2 T3 T2 1007 K Process 3 → 4 is an adiabatic expansion to V4 = V1. Thus, γ

1.4

⎛V ⎞ ⎛ 89.7 ×10−6 m 3 ⎞ p3V3γ = p4V4γ ⇒ p4 = ⎜ 3 ⎟ p3 = ⎜ −6 3 ⎟ ⎝ 1050 ×10 m ⎠ ⎝ V4 ⎠ γ −1

γ −1

T3V3

= T4V4

Point 1 2 3 4

γ −1

⎛V ⎞ ⇒ T4 = ⎜ 3 ⎟ ⎝ V4 ⎠ p

( 7.19 ×10

6

⎛ 89.7 ×10−6 m3 ⎞ T3 = ⎜ −6 3 ⎟ ⎝ 1050 ×10 m ⎠

Pa ) = 2.30 ×105 Pa = 2.27 atm 0.4

V 5

1.00 atm = 1.013 × 10 Pa 6

71.0 atm = 7.19 × 10 Pa 6

71.0 atm = 7.19 × 10 Pa 5

2.27 atm = 2.30 × 10 Pa

T −6

3

298 K = 25°C

−6

3

1007 K = 734°C

−6

3

1807 K = 1534°C

−6

3

675 K = 402°C

1050 × 10 m 50.0 × 10 m 89.7 × 10 m

1050 × 10 m

(b) For adiabatic process 1 → 2,

(Ws )12 =

p2V2 − p1V1 = −633 J 1−γ

For isobaric process 2 → 3,

(Ws )23 = p2ΔV = p2 (V3 − V2 ) = 285 J For adiabatic process 3 → 4,

(1807 K ) = 675 K

(Ws )34 =

p4V4 − p3V3 = 1009 J 1−γ

For isochoric process 4 → 1, (Ws )41 = 0 J. Thus,

(Ws )cycle = Wout = (Ws )12 + (Ws )23 + (Ws )34 + (Ws )41 = 661 J (c) The efficiency is

η=

Wout 661 J = = 66.1% ≈ 66% QH 1000 J

(d) The power output of one cylinder is

661 J 2400 cycle 1 min J × × = 26,440 = 26.4 kW cycle min 60 sec s For an 8-cylinder engine the power will be 211 kW or 283 horsepower.

20.1.

Model: The wave is a traveling wave on a stretched string. Solve: The wave speed on a stretched string with linear density μ is vstring = TS / μ . The wave speed if the tension is doubled will be ′ = vstring

2TS

μ

= 2vstring = 2 ( 200 m/s ) = 283 m/s

20.2. Solve:

Model: The wave is a traveling wave on a stretched string. The wave speed on a stretched string with linear density μ is T 75 N ⇒ μ = 3.333 × 10−3 kg/m vstring = S ⇒ 150 m/s =

μ

μ

For a wave speed of 180 m/s, the required tension will be 2 2 TS = μ vstring = ( 3.333 × 10−3 kg/m ) (180 m/s ) = 110 N

20.3.

Model: The wave pulse is a traveling wave on a stretched string. The wave speed on a stretched string with linear density μ is T TS LTS 2.0 m ( 2.0 m )( 20 N ) ⇒ m = 0.025 kg = 25 g = S = = ⇒ = m/L m 50 × 10−3 s m μ

Solve:

vstring

20.4.

Model: This is a wave traveling at constant speed. The pulse moves 1 m to the right every second. Visualize: The snapshot graph shows the wave at all points on the x-axis at t = 0 s. The wave is just reaching x = 5.0. The first part of the wave causes an upward displacement of the medium. The rising portion of the wave is 2 m wide, so it will take 2 s to pass the x = 5.0 m point. The constant part of the wave, whose width is 2 m, will take 2 seconds to pass x = 5.0 m and during this time the displacement of the medium will be a constant (Δy = 1 cm). The trailing edge of the pulse arrives at t = 4 s at x = 5.0 m. The displacement now becomes zero and stays zero for all later times.

20.5.

Model: This is a wave traveling at constant speed. The pulse moves 1 m to the left every second. Visualize: This snapshot graph shows the wave at all points on the x-axis at t = 2 s. You can see that the leading edge of the wave at t = 2 s is precisely at x = 0 m. That is, in the first 2 seconds, the displacement is zero at x = 0 m. The first part of the wave causes a downward displacement of the medium, so immediately after t = 2 s the displacement at x = 0 m will be negative. The negative portion of the wave pulse is 3 m wide and takes 3 s to pass x = 0 m. The positive portion begins to pass through x = 0 m at t = 5 s and until t = 8 s the displacement of the medium is positive. The displacement at x = 0 m returns to zero at t = 8 s and remains zero for all later times.

20.6.

Model: This is a wave traveling at constant speed to the right at 1 m/s. Visualize: This is the history graph of a wave at x = 0 m. The graph shows that the x = 0 m point of the medium first sees the negative portion of the pulse wave at t = 1.0 s. Thus, the snapshot graph of this wave at t = 1.0 s must have the leading negative portion of the wave at x = 0 m.

20.7.

Model: This is a wave traveling at constant speed to the left at 1 m/s. Visualize: This is the history graph of a wave at x = 2 m. Because the wave is moving to the left at 1 m/s, the wave passes the x = 2 m position a distance of 1 m in 1 s. Because the flat part of the history graph takes 2 s to pass the x = 2 m position, its width is 2 m. Similarly, the width of the linearly increasing part of the history graph is 2 m. The center of the flat part of the history graph corresponds to both t = 0 s and x = 2 m.

20.8.

Visualize:

Figure EX20.8 shows a snapshot graph at t = 0 s of a longitudinal wave. This diagram shows a row of particles with an inter-particle separation of 1.0 cm at equilibrium. Because the longitudinal wave has a positive amplitude of 0.5 cm between x = 3 cm and x = 8 cm, the particles at x = 3, 4, 5, 6, 7 and 8 cm are displaced to the right by 0.5 cm.

20.9.

Visualize:

We first draw the particles of the medium in the equilibrium positions, with an inter-particle spacing of 1.0 cm. Just underneath, the positions of the particles as a longitudinal wave is passing through are shown at time t = 0 s. It is clear that relative to the equilibrium the particle positions are displaced negatively on the left side and positively on the right side. For example, the particles at x = 0 cm and x = 1 cm are at equilibrium, the particle at x = 2 cm is displaced left by 0.5 cm, the particle at x = 3 cm is displaced left by 1.0 cm, the particle at x = 4 cm is displaced left by 0.5 cm, and the particle at x = 5 cm is undisplaced. The behavior of particles for x > 5 cm is opposite of that for x < 5 cm.

20.10. Solve: (a) The wave number is k=

2π

λ

=

2π = 3.1 rad/m 2.0 m

(b) The wave speed is

⎛ω v = λ f = λ⎜ ⎝ 2π

⎞ ⎛ 30 rad/s ⎞ ⎟ = (2.0 m) ⎜ ⎟ = 9.5 m/s ⎠ ⎝ 2π ⎠

20.11. Solve: (a) The wavelength is λ=

2π 2π = = 4.2 m k 1.5 rad/m

(b) The frequency is

f =

v

λ

=

200 m/s = 48 Hz 4.19 m

20.12. Model: The wave is a traveling wave.

Solve: (a) A comparison of the wave equation with Equation 20.14 yields: A = 3.5 cm, k = 2.7 rad/m, ω = 124 rad/s, and φ0 = 0 rad. The frequency is

f =

ω 2π

=

124 rad/s = 19.7 Hz ≈ 20 Hz 2π

(b) The wavelength is

λ= (c) The wave speed v = λ f = 46 m/s .

2π 2π = = 2.33 m ≈ 2.3 m k 2.7 rad/m

20.13. Model: The wave is a traveling wave.

Solve: (a) A comparison of the wave equation with Equation 20.14 yields: A = 5.2 cm, k = 5.5 rad/m, ω = 72 rad/s, and φ0 = 0 rad. The frequency is

f =

ω 2π

=

72 rad/s = 11.5 Hz ≈ 11 Hz 2π

(b) The wavelength is

λ= (c) The wave speed v = λ f = 13 m/s.

2π 2π = = 1.14 m ≈ 1.1 m k 5.5 rad/m

20.14.

Solve: The amplitude of the wave is the maximum displacement, which is 6.0 cm. The period of the wave is 0.60 s, so the frequency f = 1 T = 1 0.60 s = 1.67 Hz . The wavelength is

λ=

v 2 m/s = = 1.2 m f 1.667 Hz

20.15.

Solve:

According to Equation 20.28, the phase difference between two points on a wave is

Δφ = φ2 − φ1 = 2π

Δr

λ

=

2π

λ

( r2 − r1 ) ⇒ ( 3π

rad − 0 rad ) =

2π

λ

( 80 cm − 20 cm ) ⇒ λ = 40 cm

20.16.

Solve:

According to Equation 20.28, the phase difference between two points on a wave is

Δφ = φ2 − φ1 =

2π

λ

( r2 − r1 )

If φ1 = π rad at r1 = 4.0 m, we can determine φ2 at any r value at the same instant using this equation. At r2 = 3.5 m,

φ2 = φ1 + At r2 = 4.5 m, φ = 32 π rad.

2π

λ

( r2 − r1 ) = π

rad +

2π π ( 3.5 m − 4.0 m ) = rad 2.0 m 2

20.17.

Solve:

Visualize:

For a sinusoidal wave, the phase difference between two points on the wave is given by Equation 20.28:

Δφ = φ2 − φ1 =

2π

λ

( r2 − r1 ) =

2π

λ

( 40 m − 30 m ) ⇒ λ =

2π (10 m ) Δφ

Δφ = 2π for two points on adjacent wavefronts and Δφ = 4π for two points separated by 2λ. Thus, λ = 10 m

when Δφ = 2π , and λ = 5 m when Δφ = 4π . The crests corresponding to these two wavelengths are shown in the figure. One can see that a crest of the wave passes the 40 m–listener and the 30 m–listener simultaneously. The lowest two possible frequencies will occur for the largest two possible wavelengths, which are 10 m and 5 m. Thus, the lowest frequency is f1 = The next highest frequency is f 2 = 68 Hz.

v

λ

=

340 m/s = 34 Hz 10 m

20.18. Visualize:

Solve:

(a) Because the same wavefront simultaneously reaches listeners at x = −7.0 m and x = + 3.0 m,

Δφ = 0 rad =

2π

λ

( r2 − r1 ) ⇒ r2 = r1

Thus, the source is at x = −2.0 m, so that it is equidistant from the two listeners. (b) The third person is also 5.0 m away from the source. Her y-coordinate is thus y = (5 m) 2 − (2 m) 2 = 4.6 m .

20.19. Solve: Two pulses of sound are detected because one pulse travels through the metal to the microphone while the other travels through the air to the microphone. The time interval for the sound pulse traveling through the air is Δtair =

Δx 4.0 m = = 0.01166 s = 11.66 ms vair 343 m/s

Sound travels faster through solids than gases, so the pulse traveling through the metal will reach the microphone before the pulse traveling through the air. Because the pulses are separated in time by 11.0 ms, the pulse traveling through the metal takes Δtmetal = 0.66 ms to travel the 4.0 m to the microphone. Thus, the speed of sound in the metal is vmetal =

Δx 4.0 m = = 6060 m/s ≈ 6100 m/s Δtmetal 0.00066 s

20.20.

Solve:

(a) In aluminum, the speed of sound is 6420 m/s. The wavelength is thus equal to

λ=

v 6420 m/s = = 3.21 × 10−3 m = 3.21 mm ≈ 3.2 mm f 2.0 × 106 Hz

(b) The speed of an electromagnetic wave is c. The frequency would be

f =

c

λ

=

3.0 × 108 m/s = 9.3 × 1010 Hz 3.21 × 10−3 m

20.21.

Solve:

(a) The frequency is

f =

vair

λ

=

343 m/s = 1715 Hz ≈ 1700 Hz 0.20 m

(b) The frequency is

f =

c

λ

=

3.0 × 108 m/s = 1.5 × 109 Hz = 1.5 GHz 0.20 m

(c) The speed of a sound wave in water is vwater = 1480 m/s. The wavelength of the sound wave would be v 1480 m/s λ = water = = 9.87 × 10−7 m ≈ 990 nm f 1.50 × 109 Hz

20.22. Solve:

Model: Light is an electromagnetic wave that travels with a speed of 3 × 108 m/s. (a) The frequency of the blue light is

f blue =

c

λ

=

3.0 × 108 m/s = 6.67 × 1014 Hz 450 × 10−9 m

(b) The frequency of the red light is

3.0 × 108 m/s = 4.62 × 1014 Hz 650 × 10−9 m (c) Using Equation 20.30 to calculate the index of refraction, f red =

λmaterial =

λvacuum n

⇒n=

λvacuum 650 nm = = 1.44 λmaterial 450 nm

20.23. Solve:

Model: Microwaves are electromagnetic waves that travel with a speed of 3 × 108 m/s. (a) The frequency of the microwave is

f microwaves =

c

λ

=

3.0 × 108 m/s = 1.0 × 1010 Hz = 10 GHz 3.0 × 10−2 m

(b) The refractive index of air is 1.0003, so the speed of microwaves in air is vair = c /1.00 ≈ c. The time for the microwave signal to travel is

t=

50 km 50 × 103 m = = 0.167 ms ≈ 0.17 ms vair ( 3.0 ×108 m 1.00 )

Assess: A small time of 0.17 ms for the microwaves to cover a distance of 50 km shows that the electromagnetic waves travel very fast.

20.24. Solve:

Model: Radio waves are electromagnetic waves that travel with speed c. (a) The wavelength is

λ=

c 3.0 × 108 m/s = = 2.96 m f 101.3 MHz

(b) The speed of sound in air at 20°C is 343 m/s. The frequency is

f =

vsound

λ

=

343 m/s = 116 Hz 2.96 m

20.25. Solve:

Model: Light is an electromagnetic wave. (a) The time light takes is

3.0 mm 3.0 × 10−3 m 3.0 × 10−3 m = = = 1.5 × 10−11 s 8 vglass cn 3.0 × 10 m/s 1.50 ( )

t= (b) The thickness of water is

d = vwatert =

c nwater

t=

3.0 × 108 m/s (1.5 ×10−11 s ) = 3.4 mm 1.33

20.26.

Solve:

(a) The speed of light in a material is given by Equation 20.29:

n=

c c ⇒ vmat = vmat n

The refractive index is

n=

λvac λ 420 nm ⇒ vsolid = c solid = ( 3.0 × 108 m/s ) = 1.88 × 108 m/s λmat λvac 670 nm

(b) The frequency is f =

vsolid

λsolid

=

1.88 × 108 m/s = 4.48 × 1014 Hz 420 nm

20.27. Model: Assume

that

the

glass

has

index

of

refraction

n = 1.5.

This

means

that

vglass = c/n = 2 × 108 m/s. Visualize: We apply v = λ f twice, once in air and then in the glass. The frequency will be the same in both cases. Solve: (a) In the air f air =

vair

λair

=

3.0 ×108 m/s = 8.57 × 108 Hz ≈ 8.6 × 108 Hz 0.35 m

The frequency is the same in both media, so f glass = 8.6 × 108 Hz. (b) Now that we know f glass and vglass , we can find λglass .

λglass = Assess:

vglass f glass

=

2.0 × 108 m/s = 23 cm 8.57 ×108 Hz

We get the same answer from λglass = λair /nglass = 35 cm/1.5 = 23 cm .

20.28.

Solve: The energy delivered to the eardrum in time t is E = Pt, where P is the power of the wave. The intensity of the wave is I = P / a where a is the area of the ear drum. Putting the above information together, we have

E = Pt = ( Ia ) t = Iπ r 2t = ( 2.0 × 10−3 W/m 2 )π ( 3.0 × 10−3 m ) ( 60 s ) = 3.4 × 10−6 J 2

20.29. Solve: The energy delivered to an area a in time t is E = Pt, where the power P is related to the intensity I as I = P / a. Thus, the energy received by your back is E = Pt = Iat = (0.80)(1400 W/m2)(0.30 × 0.50 m2)(3600 s) = 6.0 × 105 J

20.30. Solve: If a source of spherical waves radiates uniformly in all directions, the ratio of the intensities at distances r1 and r2 is 2

I1 r22 I ⎛ 2m ⎞ −3 = 2 ⇒ 50 m = ⎜ ⎟ = 1.6 × 10 I 2 m ⎝ 50 m ⎠ I 2 r1

⇒ I 50 m = I 2 m (1.6 × 10−3 ) = ( 2.0 W/m 2 )(1.6 × 10−3 ) = 3.2 × 10−3 W/m 2 Assess: The power generated by the sound source is P = I2m [4π(2 m)2] = (2.0 W/m2)(50.27) = 101 W. This is a significant amount of power.

20.31.

Solve: (a) The intensity of a uniform spherical source of power Psource a distance r away is I = Psource / 4π r 2 . Thus, the intensity at the position of the microphone is

I 50 m =

35 W 4π ( 50 m )

2

= 1.1 × 10−3 W/m 2

(b) The sound energy impinging on the microphone per second is

P = Ia = (1.1× 10−3 W/m 2 )(1.0 × 10−4 m 2 ) = 1.1× 10−7 W = 1.1× 10−7 J/s ⇒ Energy impinging on the microphone in 1 second = 1.1×10−7 J

20.32. Solve: Because the sun radiates waves uniformly in all directions, the intensity I of the sun’s rays when they impinge upon the earth is I=

Psun P 4 × 1026 W ⇒ I earth = sun2 = = 1400 W/m 2 2 4π r 4π rearth 4π (1.496 × 1011 m )2

With rsun-Venus = 1.082 × 1011 m and rsun-Mars = 2.279 × 1011 m, the intensities of electromagnetic waves at these planets are I venus = 2700 W/m 2 and I Mars = 610 W/m 2 .

20.33. Visualize: Equation 20.35 gives the sound intensity level as ⎛ I ⎞ ⎟ ⎝ I0 ⎠

β = (10 dB)log10 ⎜ where I 0 = 1.0 ×10−12 W/m 2 . Solve: (a)

⎛I ⎞ ⎛ 5.0 ×10−8 W/m 2 ⎞ = 47 dB ⎟ = (10 dB)log10 ⎜ −12 2 ⎟ ⎝ 1.0 ×10 W/m ⎠ ⎝ I0 ⎠

β = (10 dB)log10 ⎜ (b)

⎛I ⎞ ⎛ 5.0 ×10−2 W/m 2 ⎞ = 107 dB ⎟ = (10 dB)log10 ⎜ −12 2 ⎟ ⎝ 1.0 ×10 W/m ⎠ ⎝ I0 ⎠

β = (10 dB)log10 ⎜

Assess: As mentioned in the chapter, each factor of 10 in intensity changes the sound intensity level by 10 dB; between the first and second parts of this problem the intensity changed by a factor of 106 , so we expect the sound intensity level to change by 60 dB.

20.34. Visualize: We can solve Equation 20.35 for the sound intensity, finding I = I 0 ×10 β /10dB . Solve: (a)

I = I 0 ×10 β /10dB = (1.0 ×10−12 W/m 2 ) × 103.6 = 4.0 × 10−9 W/m 2 (b)

I = I 0 ×10 β /10dB = (1.0 ×10−12 W/m 2 ) × 109.6 = 4.0 × 10−3 W/m 2 Assess:

Since the sound intensity levels in the two parts of this problem differ by 60dB we expect the sound

intensities to differ by a factor of 106 .

20.35. Model: Assume the pole is tall enough that we don’t have to worry about the ground absorbing or reflecting sound. Visualize: The area of a sphere of radius R is A = 4π R 2 . Also recall that I = P/A ; we are given P = 5.0 W . We seek R for β = 90dB . Solve:

A=

P P 5.0 W = = = 5000 m 2 I I 0 ×10 β/10dB (1.0 ×10−12 W/m 2 ) ×1090dB/10dB R=

A 5000 m 2 = = 20 m 4π 4π

Assess: This is a reasonable distance from the loudspeaker for a moderately loud sound.

20.36. Model: The frequency of the opera singer’s note is altered by the Doppler effect. Solve:

(a) Using 90 km/h = 25 m/s, the frequency as her convertible approaches the stationary person is f+ =

f0 600 Hz = = 650 Hz 1 − vS v 1 − 25 m/s 343 m/s

(b) The frequency as her convertible recedes from the stationary person is f0 600 Hz f− = = = 560 Hz 1 + vS v 1 + 25 m/s 343 m/s

20.37. Model: Your friend’s frequency is altered by the Doppler effect. The frequency of your friend’s note increases as he races towards you (moving source and a stationary observer). The frequency of your note for your approaching friend is also higher (stationary source and a moving observer). Solve: (a) The frequency of your friend’s note as heard by you is f+ =

f0 400 Hz = = 432 Hz vS 25.0 m/s 1− 1− v 340 m/s

(b) The frequency heard by your friend of your note is

⎛ v ⎞ ⎛ 25.0 m/s ⎞ f + = f 0 ⎜1 + 0 ⎟ = ( 400 Hz ) ⎜1 + ⎟ = 429 Hz v⎠ 340 m/s ⎠ ⎝ ⎝

20.38.

Model: Sound frequency is altered by the Doppler effect. The frequency increases for an observer approaching the source and decreases for an observer receding from a source. Solve: You need to ride your bicycle away from your friend to lower the frequency of the whistle. The minimum speed you need to travel is calculated as follows:

v0 ⎞ ⎛ v ⎞ ⎛ f − = ⎜1 − 0 ⎟ f 0 ⇒ 20 kHz = ⎜ 1 − ⎟ ( 21 kHz ) ⇒ v0 = 16.3 m/s ≈ 16 m/s v⎠ ⎝ ⎝ 343 m/s ⎠ Assess: A speed of 16.3 m/s corresponds to approximately 35 mph. This is a possible but very fast speed on a bicycle.

20.39. Solve:

Model: The mother hawk’s frequency is altered by the Doppler effect. The frequency is f + as the hawk approaches you is

f+ =

Assess:

f0 800 Hz ⇒ 900 Hz = ⇒ vS = 38.1 m/s vS 1 − vS v 1− 343 m/s

The mother hawk’s speed of 38.1 m/s ≈ 80 mph is reasonable.

20.40. Visualize: changing shape. Solve:

The function D(x, t) represents a pulse that travels in the positive x-direction without

(a)

(b) The leading edge of the pulse moves forward 3 m each second. Thus, the wave speed is 3 m/s. (c) x − 3t is a function of the form D(x – vt), so the pulse moves to the right at v = 3 m/s.

20.41. Solve: (a) We see from the history graph that the period T = 0.20 s and the wave speed v = 4.0 m/s. Thus, the wavelength is λ=

v = vT = ( 4.0 m/s )( 0.20 s ) = 0.80 m f

(b) The phase constant φ0 is obtained as follows:

D ( 0 m, 0 s ) = A sin φ0 ⇒ −2 mm = ( 2 mm ) sin φ0 ⇒ sin φ0 = −1 ⇒ φ0 = − 12 π rad (c) The displacement equation for the wave is

2π t π⎞ ⎛ 2π x ⎞ ⎛ 2π x D ( x, t ) = A sin ⎜ − 2π ft + φ0 ⎟ = ( 2.0 mm ) sin ⎜ − − ⎟ = ( 2.0 mm ) sin ( 2.5π x − 10π t − 12 π ) 0.80 m 0.20 s 2⎠ λ ⎝ ⎠ ⎝ where x and t are in m and s, respectively.

20.42. Solve: (a) We can see from the graph that the wavelength is λ = 2.0 m. We are given that the wave’s frequency is f = 5.0 Hz. Thus, the wave speed is v = λf = 10 m/s. (b) The snapshot graph was made at t = 0 s. Reading the graph at x = 0 m, we see that the displacement is D ( x = 0 m, t = 0 s ) = D ( 0 m, 0 s ) = 0.5 mm = 12 A Thus D ( 0 m,0 s ) = 12 A = A sin φ0 ⇒ φ0 = sin −1 ( 12 ) =

π 6

rad or

5π rad 6

Note that the value of D(0 m, 0 s) alone gives two possible values of the phase constant. One of the values will cause the displacement to start at 0.5 mm and increase with distance—as the graph shows—while the other will cause the displacement to start at 0.5 mm but decrease with distance. Which is which? The wave equation for t = 0 s is

⎛ 2π x ⎞ D ( x, t = 0 ) = A sin ⎜ + φ0 ⎟ λ ⎝ ⎠ If x is a point just to the right of the origin and is very small, the angle (2π x / λ + φ0 ) is just slightly bigger than

the angle φ0 . Now sin 31° > sin 30°, but sin151° < sin150°, so the value φ0 = 16 π rad is the phase constant for which the displacement increases as x increases. (c) The equation for a sinusoidal traveling wave can be written as

⎡ ⎛x ⎤ ⎛ 2π x ⎞ ⎞ − 2π ft + φ0 ⎟ = A sin ⎢ 2π ⎜ − ft ⎟ + φ0 ⎥ D ( x, t ) = A sin ⎜ ⎝ λ ⎠ ⎠ ⎣ ⎝λ ⎦ Substituting in the values found above, ⎡ ⎛ x ⎞ π⎤ − ( 5.0 s −1 ) t ⎟ + ⎥ D ( x, t ) = (1.0 mm ) sin ⎢ 2π ⎜ 2.0 m ⎠ 6⎦ ⎣ ⎝

20.43.

Solve:

The time for the sound wave to travel down the tube and back is t = 440 μ s since 1 division

is equal to 100 μs. So, the speed of the sound wave in the liquid is

v=

2 × 25 cm = 1140 m/s ≈ 1100 m/s 440 μ s

20.44. Solve:

Model: The wave is a traveling wave on a stretched string. The wave speed on a string whose radius is R, length is L, and mass density is ρ is vstring = TS / μ with

μ=

m ρV ρπ R 2 L = = = ρπ R 2 L L L

If the string radius doubles, then ′ = vstring

TS

ρπ ( 2R )

2

=

vstring 2

=

280 m/s = 140 m/s 2

20.45. Solve:

Model: The wave pulse is a traveling wave on a stretched string. While the tension TS is the same in both the strings, the wave speeds in the two strings are not. We have

v1 =

TS

μ1

and

v2 =

TS

μ2

⇒ v12 μ1 = v22 μ 2 = TS

Because v1 = L1 / t1 and v2 = L2 / t2 , and because the pulses are to reach the ends of the string simultaneously, the above equation can be simplified to

L12 μ1 L22 μ 2 μ2 4.0 g/m L = 2 ⇒ 1= = = 2 ⇒ L1 = 2 L2 t2 t μ1 L2 2.0 g/m Since L1 + L2 = 4 m, 2 L2 + L2 = 4 m ⇒ L2 = 1.66 m ≈ 1.7 m

and

L1 = 2 (1.66 m ) = 2.34 m ≈ 2.3 m

20.46. Solve: Δt is the time the sound wave takes to travel down to the bottom of the ocean and then up to the ocean surface. The depth of the ocean is 2d = ( vsound in water ) Δt ⇒ d = ( 750 m/s ) Δt Using this relation and the data from Figure P20.46, we can generate the following table for the ocean depth (d ) at various positions (x) of the ship. x (km) d (km) Δt (s) 0 20 40 45 50 60

6 4 4 8 4 2

4.5 3.0 3.0 6.0 3.0 1.5

20.47. Visualize:

Solve: The explosive’s sound travels down the lake and into the granite, and then it is reflected by the oil surface. The echo time is thus equal to

techo = t water down + tgranite down + tgranite up + twater up 0.94 s =

d granite d granite 500 m 500 m + + + ⇒ d granite = 790 m 1480 m/s 6000 m/s 6000 m/s 1480 m/s

20.48. Model: Assume a room temperature of 20°C. Visualize:

Solve:

The distance between the source and the left ear (EL) is

d L = x 2 + ( y + 0.1 m ) = ⎡⎣( 5.0 m ) cos45° ⎤⎦ + ⎡⎣( 5.0 m ) sin45° + 0.1 m ⎤⎦ = 5.0712 m 2

2

2

Similarly d R = 4.9298 m. Thus, d L − d R = Δd = 0.1414 m

For the sound wave with a speed of 343 m/s, the difference in arrival times at your left and right ears is Δt =

Δd 0.1414 m = = 410 μ s 343 m/s 343 m/s

20.49. Solve:

Model: The laser beam is an electromagnetic wave that travels with the speed of light. The speed of light in the liquid is

vliquid =

30 × 10−2 m = 2.174 × 108 m/s 1.38 × 10−9 s

The liquid’s index of refraction is

n=

c vliquid

=

3.0 × 108 = 1.38 2.174 × 108

Thus the wavelength of the laser beam in the liquid is

λliquid =

λvac n

=

633 nm = 459 nm 1.38

20.50. Solve:

Model: The temperature is 20°C for both air and water. For the sound speed of vair = 343 m/s, the wavelength of the sound wave in air is

λair =

343 m/s = 1.340 m 256 Hz

On entering water the frequency does not change, so fwater = fair and fwater /fair = 1. The wave speed in water is vwater = 1480 m/s, so

vwater 1480 m/s = = 4.31 vair 343 m/s Finally, the wavelength in water is v 1480 m/s λ 5.781 m λwater = water = = 5.781 m ⇒ water = = 4.31 f water 256 Hz λair 1.340 m Assess: This last result is expected because v = f λ and the frequency remains unchanged as the wave enters from air into water.

20.51.

Solve:

The difference in the arrival times for the P and S waves is

Δt = tS − tP = Assess:

d d 1 1 ⎛ ⎞ 6 ⇒ 120 s = d ⎜ − − ⎟ ⇒ d = 1.23 × 10 m=1230 km vS vP ⎝ 4500 m/s 8000 m/s ⎠

d is approximately one-fifth of the radius of the earth and is reasonable.

20.52. Solve:

Model: This is a sinusoidal wave. (a) The equation is of the form D( y, t ) = A sin(ky + ω t + φ0 ), so the wave is traveling along the y-axis.

Because it is +ω t rather than −ω t the wave is traveling in the negative y-direction. (b) Sound is a longitudinal wave, meaning that the medium is displaced parallel to the direction of travel. So the air molecules are oscillating back and forth along the y-axis. (c) The wave number is k = 8.96 m -1 , so the wavelength is

λ=

2π 2π = = 0.701 m k 8.96 m −1

The angular frequency is ω = 3140 s −1 , so the wave’s frequency is f =

ω 3140 s −1 = = 500 Hz 2π 2π

Thus, the wave speed v = λ f = (0.70 m)(500 Hz) = 350 m/s. The period T = 1/f = 0.00200 s = 2.00 ms. (d) The interval t = 0 s to t = 4 ms is exactly 2 cycles of the wave. The initial value at y = 1 m is

D ( y = 1 m, t = 0 s ) = ( 0.02 mm ) sin ( 8.96 + 14 π ) = −0.0063 mm

Assess: The wave is a sound wave with speed v = 350 m/s. This is greater than the room-temperature speed of 343 m/s, so the air temperature must be greater than 20°.

20.53. Model: This is a sinusoidal wave. Solve: (a) The displacement of a wave traveling in the positive x-direction with wave speed v must be of the form D(x, t) = D(x− vt). Since the variables x and t in the given wave equation appear together as x + vt, the wave is traveling toward the left, that is, in the −x direction. (b) The speed of the wave is v=

ω k

=

2π 0.20 s = 12 m/s 2π rad 2.4 m

The frequency is f =

ω 2π rad 0.20 s = = 5.0 Hz 2π 2π

The wave number is k=

2π rad = 2.6 rad/m 2.4 m

(c) The displacement is

⎡ ⎛ 0.20 m 0.50 s ⎞ ⎤ D ( 0.20 m, 0.50 s ) = ( 3.0 cm ) sin ⎢ 2π ⎜ + + 1⎟ ⎥ = −1.5 cm ⎣ ⎝ 2.4 m 0.20 s ⎠ ⎦

20.54. Model: This is a sinusoidal wave traveling on a stretched string in the +x direction. Solve: (a) From the displacement equation of the wave, A = 2.0 cm, k = 12.57 rad/m, and ω = 638 rad/s. Using the equation for the wave speed in a stretched string, vstring =

2

2

⎛ω ⎞ ⎛ 638 rad/s ⎞ 2 ⇒ TS = μ vstring = μ ⎜ ⎟ = ( 5.00 × 10−3 kg/m3 ) ⎜ ⎟ = 12.6 N μ ⎝k⎠ ⎝ 12.57 rad/m ⎠

TS

(b) The maximum displacement is the amplitude Dmax ( x, t ) = 2.00 cm . (c) From Equation 20.17, v y max = ω A = ( 638 rad/s ) ( 2.0 × 10−2 m ) = 12.8 m/s

20.55.

Solve:

The wave number and frequency are calculated as follows:

2π rad = 4π rad/m ⇒ ω = vk = ( 4.0 m/s )( 4π rad/m ) = 16π rad/s 0.50 m Thus, the displacement equation for the wave is k=

2π

λ

=

D ( y, t ) = ( 5.0 cm ) sin ⎣⎡( 4π rad/m ) y + (16π rad/s ) t ⎦⎤ Assess:

The positive sign in the sine function’s argument indicates motion along the −y direction.

20.56.

Solve:

The angular frequency and wave number are calculated as follows:

ω = 2π f = 2π ( 200 Hz ) = 400π rad/s ⇒ k =

ω v

=

400π rad/s = π rad/m 400 m/s

The displacement equation for the wave is D ( x, t ) = ( 0.010 mm ) sin ⎣⎡(π rad/m ) x − ( 400π rad/s ) t + 12 π rad ⎦⎤ Assess: Note the negative sign with ω t in the sine function’s argument. This indicates motion along the +x direction.

20.57.

Solve:

A sinusoidal traveling wave is represented as D ( x, t ) = A sin ( kx − ω t ) . Replacing t with t + T

and using the relationship ω = 2π/T between the angular frequency and period, D ( x, t + T ) = A sin ( kx − ω ( t + T ) ) = A sin ( kx − ω t − ωT ) = A sin ( kx − ω t − 2π ) = A ⎡⎣sin ( kx − ω t ) cos ( 2π ) − cos ( kx − ω t ) sin ( 2π ) ⎤⎦ = A sin ( kx − ω t ) = D ( x, t )

20.58. Solve: According to Equation 20.28, the phase difference between two points on a wave is Δφ = 2π ( r2 − r1 ) λ . For the first point and second point, r1 =

(1.00 cm − 0 cm )

r2 =

( −1.00 cm − 0 cm ) + (1.50 cm − 0 cm )

2

+ ( 3.00 cm − 0 cm ) + ( 2.00 cm − 0 cm ) = 3.742 cm 2

2

2

2

+ ( 2.50 cm − 0 cm ) = 3.082 cm 2

The wavelength is

λ= ⇒ Δφ =

v 346 m/s = = 0.02641 m = 2.641 cm f 13,100 Hz

2π ( 3.742 cm − 3.082 cm ) π π 180° = rad = rad × = 90° 2.641 cm 2 2 π rad

20.59. Solve:

Model: We have a sinusoidal traveling wave on a stretched string. (a) The wave speed on a string and the wavelength are calculated as follows:

v=

TS

μ

=

20 N v 100.0 m/s = 100 m/s ⇒ λ = = = 1.0 m 0.002 kg/m f 100 Hz

(b) The amplitude is determined by the oscillator at the end of the string and is A = 1.0 mm. The phase constant can be obtained from Equation 20.15 as follows:

D ( 0 m, 0 s ) = A sin φ0 ⇒ −1.0 mm = (1.0 mm ) sin φ0 ⇒ φ0 = −

π 2

rad

(c) The wave (as distinct from the oscillator) is described by D ( x, t ) = A sin ( kx − ω t + φ0 ) . In this equation the

wave number and angular frequency are k=

2π

λ

=

2π = 2π rad/m 1.0 m

ω = vk = (100.0 m/s )( 2π rad/m ) = 200π rad/s

Thus, the wave’s displacement equation is D ( x, t ) = (1.0 mm ) sin ⎡⎣( 2π rad/m ) x − ( 200π rad/s ) t − 12 π rad ⎤⎦

(d) The displacement is

D ( 0.50 m, 0.015 s ) = (1.0 mm ) sin ⎡⎣( 2π rad/s )( 0.50 m ) − ( 200π rad/s )( 0.015 s ) − 12 π ⎤⎦ = −1.0 mm

20.60. Model: We have a wave traveling to the right on a string. Visualize:

Solve: The snapshot of the wave as it travels to the right for an infinitesimally small time Δt shows that the velocity at point 1 is downward, at point 3 is upward, and at point 2 is zero. Furthermore, the speed at points 1 and 3 is the maximum speed given by Equation 20.17: v1 = v3 = ω A . The frequency of the wave is

ω = 2π f = 2π

v

λ

=

2π ( 45 m/s ) = 300π rad/s ⇒ ω A = ( 300π rad/s ) ( 2.0 × 10−2 m ) = 19 m/s 0.30 m

Thus, v1 = −19 m/s, v2 = 0 m/s, and v3 = +19 m/s.

20.61. Model: This is a wave traveling to the left at a constant speed of 50 cm/s. Solve: The particles at positions between x = 2 cm and x = 7 cm have a speed of 10 cm/s, and the particles between x = 7 cm and x = 9 cm have a speed of −25 cm/s. That is, at the time the snapshot of the velocity is shown, the particles of the medium have upward motion for 2 cm ≤ x ≤ 7 cm, but downward motion for 7 cm ≤ x ≤ 9 cm. The width of the front section of the wave pulse is 7 cm – 2 cm = 5 cm and the width of the rear section is 9 cm – 7 cm = 2 cm. With a wave speed of 50 cm/s, the time taken by the front section to pass through a particular point is 5 cm 50 cm/s = 0.1 s and the time taken by the rear section of the wave to pass through a point is 2 cm 50 cm/s = 0.04 s . Thus the wave causes the upward moving particles to go through a displacement of

A = (10 cm/s )( 0.1 s ) = 1.0 cm . The downward moving particles have a maximum displacement of

( −25 cm/s )( 0.04 s ) = −1.0 cm .

20.62. Solve:

Model: The wave is traveling on a stretched string. The wave speed on the string is

v=

TS

μ

=

50 N = 100 m/s 0.005 kg/m

The speed of the particle on the string, however, is given by Equation 20.17. The maximum speed is calculated as follows:

v y = −ω A cos ( kx − ω t + φ0 ) ⇒ v y

max

= ω A = 2π fA = 2π

⎛ 100 m/s ⎞ A = 2π ⎜ ⎟ ( 0.030 m ) = 9.4 m/s λ ⎝ 2.0 m ⎠

v

20.63. Model: A sinusoidal wave is traveling along a stretched string. Solve: From Equation 20.17 and Equation 20.20, vy max = ωA and ay max = ω2A. These two equations can be combined to give ω=

a y max v y max

=

v 200 m/s 2 ω 2.0 m/s = 100 rad/s ⇒ f = = 15.9 Hz ≈ 16 Hz ⇒ A = y max = = 2.0 cm ω 100 rad/s 2.0 m/s 2π

20.64.

Solve: (a) At a distance r from the bulb, the 5 watts of visible light have spread out to cover the surface of a sphere of radius r. The surface area of a sphere is a = 4πr2. Thus, the intensity at a distance of 2 m is

I=

P P 5.0 W = = = 0.095 W/m 2 a 4π r 2 4π ( 2.0 m )2

Note that the presence of the wall has nothing to do with the intensity. The wall allows you to see the light, but the light wave has the same intensity at all points 2 m from the bulb whether it is striking a surface or moving through empty space. (b) Unlike the light from a light bulb, a laser beam does not spread out. We ignore the small diffraction spread of the beam. The laser beam creates a dot of light on the wall that is 2 mm in diameter. The full 5 watts of light is 2 concentrated in this dot of area a = π r 2 = π ( 0.001 m ) = 3.14 × 10−6 m 2 . The intensity is 5W P = = 1.6 MW/m 2 a 3.14 × 10−6 m 2 Although the power of the light source is the same in both cases, the laser produces light on the wall whose intensity is over 16 million times that of the light bulb. I=

20.65. Model: The radio wave is an electromagnetic wave. Solve: At a distance r, the 25 kW power station spreads out waves to cover the surface of a sphere of radius r. The surface area of a sphere is 4πr2. Thus, the intensity of the radio waves is I=

Psource 25 × 103 W = = 2.0 × 10−5 W/m 2 2 2 4π r 4π (10 × 103 m )

20.66.

Solve:

(a) The peak power of the light pulse is

Ppeak =

ΔE 500 mJ 0.50 J = = = 5.0 × 107 W Δt 10 ns 1.0 × 10−8 s

(b) The average power is Pavg =

Etotal 10 × 500 mJ 5.0 J = = = 5.0 W 1.0 s 1.0 s 1.0 s

The laser delivers pulses of very high power. But the laser spends most of its time “off,” so the average power is very much less than the peak power. (c) The intensity is

I laser =

5.0 × 107 W P 5.0 × 107 W = = = 6.4 × 1017 W/m 2 2 7.85 × 10−11 m 2 a π ( 5.0 μ m )

(d) The ratio is

I laser 6.4 × 1017 W/m 2 = = 5.8 × 1014 I sun 1.1 × 103 W/m 2

20.67. Solve:

Model: We have a traveling wave radiated by the tornado siren. (a) The power of the source is calculated as follows:

I 50 m = 0.10 W/m 2 =

Psource Psource 2 = ⇒ Psource = ( 0.10 W/m 2 ) 4π ( 50 m ) = (1000π ) W 2 2 4π r 4π ( 50 m )

The intensity at 1000 m is I1000 m =

Psource

4π (1000 m )

2

=

(1000π ) W = 250 μ W 2 4π (1000 m )

m

2

(b) The maximum distance is calculated as follows:

I=

Psource (1000π ) W ⇒ r = 16 km ⇒ 1.0 × 10−6 W/m 2 = 4π r 2 4π r 2

20.68. Model: Assume the saw is far enough off the ground that we don’t have to worry about reflected sound. Visualize:

First note that β1 − β 2 = 20dB ⇒ I1/I 2 = 10 ⋅ 10 = 100 (a change of 10 dB corresponds to a change in

intensity by a factor of 10). Then use I1 A1 = P and then P = I 2 A2 ⇒ A2 = P/I 2 , and finally solve for R2 = A2 / 4π . Solve:

Put all of the above together. P

R2 = Assess:

I1 A1

A2 I2 I2 = = = 4π 4π 4π

I1 I2

(4π R12 ) 4π

= R1

The scaling laws help and the answer is reasonable.

I1 = R1 100 = (5.0 m)(10) = 50 m I2

20.69.

Model: Assume the two loudspeakers broadcast the same power and that the platforms are high enough off the ground that we don’t have to worry about reflected sound. Visualize: Call the distance between the loudspeakers d . Call the intensity halfway between the speakers (at d / 2 ) I1 and the sound intensity level there β1 (= 75 dB ); call them I 2 and β 2 at 1/4 the distance from one pole

and 3/4 the distance from the other pole on the line between them. We seek β 2 . We first apply a general approach for different sound intensity levels: ⎡ ⎛I ⎞ ⎛ I ⎞⎤ ⎛ I /I ⎞ ⎛I ⎞ Δβ = β 2 − β1 = (10dB) ⎢ log10 ⎜ 2 ⎟ − log10 ⎜ 1 ⎟ ⎥ = (10 dB)log10 ⎜ 2 0 ⎟ = (10 dB)log10 ⎜ 2 ⎟ I I I I / ⎝ I1 ⎠ ⎝ 0⎠ ⎝ 0 ⎠ ⎦⎥ ⎝ 1 0⎠ ⎣⎢ Solve: Recall that for the general case of spherical symmetry I = P/A , where P is the power emitted by the source and A = 4π R 2 is the area of the sphere. Now we find the ratio of the intensities I 2 /I1 and then plug it in the formula above and add it to 75 dB.

I1 = I2 =

P P 2P + = 4π ( d / 2) 2 4π (d / 2) 2 π d 2

P P 4P 4P (36 + 4) P 40 P 20 + = + = = = I1 4π ( d / 4) 2 4π (3d / 4) 2 π d 2 9π d 2 9π d 2 9π d 2 9 ⎛I ⎞ ⎛ 20 ⎞ Δβ = (10 dB)log10 ⎜ 2 ⎟ = (10 dB)log10 ⎜ ⎟ = 3.48 dB I 9 ⎠ ⎝ ⎝ 1⎠

β 2 = β1 + Δβ = 75 dB + 3.48 dB = 78 dB Assess:

An increase of about 3dB corresponds to a doubling of the intensity. 20/9 is close to double.

20.70.

Model:

As suggested, model the bald head as a hemisphere with radius R = 0.080 m. This means the

surface area of the bald head (hemisphere) is A = 2π R 2 = 0.0402m 2 . Visualize: We are given β = 93 dB and ΔE = 0.10 J . We also know that I = I 0 × 10 β /10 dB and P = IA . Also recall P = ΔE Δt . Solve: Put all of the above together to find Δt .

Δt =

ΔE ΔE ΔE 0.10 J = = = = 1250 s ≈ 21 min P IA ( I 0 × 10 β /10 dB )(2π R 2 ) (10−12 W/m 2 × 109.3 )(0.0402 m 2 )

Assess: 21 min seems like quite a while to deliver 0.10 J of energy, but sounds waves don’t carry a lot of energy unless the intensity is high.

20.71.

Model: The bat’s chirping frequency is altered by the Doppler effect. The frequency is increased as the bat approaches and it decreases as the bat recedes away. Solve: The bat must fly away from you, so that the chirp frequency observed by you is less than 25 kHz. From Equation 20.38,

f− =

Assess:

f0 25,000 Hz ⇒ vS = 85.8 m/s ≈ 86 m/s ⇒ 20,000 Hz = 1 + vS / v ⎛ vS ⎞ 1+ ⎜ ⎟ ⎝ 343 m/s ⎠

This is a rather large speed: 85.8 m/s ≈ 180 mph. This is not possible for a bat.

20.72. Model: The sound generator’s frequency is altered by the Doppler effect. The frequency increases as the generator approaches the student, and it decreases as the generator recedes from the student. Solve: The generator’s speed is ⎛ 100 ⎞ vS = rω = r ( 2π f ) = (1.0 m ) 2π ⎜ rev/s ⎟ = 10.47 m/s ⎝ 60 ⎠ The frequency of the approaching generator is f0 600 Hz f+ = = = 619 Hz ≈ 620 Hz 1 − vS v 1 − 10.47 m/s 343 m/s Doppler effect for the receding generator, on the other hand, is

f0 600 Hz = = 582 Hz ≈ 580 Hz 1 + vS v 1 + 10.47 m/s 343 m/s Thus, the highest and the lowest frequencies heard by the student are 620 Hz and 580 Hz. f− =

20.73. Solve: We will closely follow the details of section 20.7 in the textbook. Figure 20.29 shows that the wave crests are stretched out behind the source. The wavelength detected by Pablo is λ− = 13 d , where d is the distance the wave has moved plus the distance the source has moved at time t = 3T. These distances are Δxwave = vt = 3vT and Δxsource = vSt = 3vST . The wavelength of the wave emitted by a receding source is thus

d Δxwave + Δxsource 3vT + 3vST = = = ( v + vS ) T 3 3 3 The frequency detected in Pablo’s direction is thus v v f0 = = f− = λ− ( v + vs ) T 1 + vS v

λ− =

20.74. Model: We are looking at the Doppler effect for the light of an approaching source. Solve:

(a) The time is

54 × 106 km = 180 s = 3.0 min 3 × 105 km/s (b) Using Equation 20.40, the observed wavelength is t=

λ=

Assess:

1 − vs c 1 − 0.1c c λ0 = ( 540 nm ) = (0.9045)(540 nm) = 488 nm ≈ 490 nm 1 + vs c 1 + 0.1c c

490 nm is slightly blue shifted from green.

20.75. Model: We are looking at the Doppler effect for the light of a receding source. Visualize:

Note that the daredevil’s tail lights are receding away from your rocket’s light detector with a relative speed of 0.2c. Solve: Using Equation 20.40, the observed wavelength is

λ=

1 + vS c 1 + 0.2c c λ0 = ( 650 nm ) = 796 nm ≈ 800 nm 1 − vS c 1 − 0.2c c

This wavelength is in the infrared region.

20.76. Model: The Doppler effect for light of a receding source yields an increased wavelength. Solve: Because the measured wavelengths are 5% longer, that is, λ = 1.05λ0, the distant galaxy is receding away from the earth. Using Equation 20.40,

λ = 1.05λ0 =

1 + vs c 1 + vs c 2 ⇒ vs = 0.049 c = 1.47 × 107 m/s λ0 ⇒ (1.05) = 1 − vs c 1 − vs c

20.77. Model: The Doppler effect for light of an approaching source leads to a decreased wavelength. Solve: The red wavelength (λ0 = 650 nm) is Doppler shifted to green (λ = 540 nm) due to the approaching light

source. In relativity theory, the distinction between the motion of the source and the motion of the observer disappears. What matters is the relative approaching or receding motion between the source and the observer. Thus, we can use Equation 20.40 as follows:

λ = λ0

1 − vs c 1 − vs c ⇒ 540 nm = ( 650 nm ) 1 + vs c 1 + vs c

⇒ vs = 5.5 × 104 km/s = 2.0 × 108 km/h The fine will be ⎛ 1$ ⎞ km/hr − 50 km/hr ) ⎜ ⎟ = $200 million ⎝ 1 km/hr ⎠ The police officer knew his physics.

( 2.0 ×10

8

Assess:

20.78. Model: The wave pulse is a traveling wave on a stretched string. The two masses hanging from the steel wire are in static equilibrium. Visualize:

Solve:

The wave speed along the wire is vwire =

4.0 m = 166.7 m/s 0.024 s

Using Equation 20.2, vwire = 166.7 m/s=

T1

=

T1

μ ( 0.060 kg 8.0 m ) G G Because point 1 is in static equilibrium, with Fnet = 0,

( Fnet ) x = T1 − T2 cos 40° ⇒ T2 =

⇒ T1 = 208.4 N

T1 = 2721 N cos 40°

( Fnet ) y = T2 sin 40° − w = 0 N ⇒ w = mg = T2 sin 40° ⇒ m =

( 272.1 N ) sin 40° = 17.8 kg 9.8 m/s 2

20.79. Solve: The time for the wave to travel from California to the South Pacific is t=

d 8.00 × 106 m = = 5405.4 s v 1480 m/s

A time decrease to 5404.4 s implies the speed has changed to v =

d = 1480.28 m/s. t

Since the 4.0 m/s increase in velocity is due to an increase of 1°C, an increase of 0.28 m/s occurs due to a temperature increase of ⎛ 1°C ⎞ ⎜ ⎟ ( 0.28 m/s ) = 0.07°C ⎝ 4.0 m/s ⎠

Thus, a temperature increase of approximately 0.07°C can be detected by the researchers.

20.80. Solve: The wave speeds along the two metal wires are v1 =

T

μ1

=

2250 N = 500 m/s 0.009 kg/m

T

v2 =

μ2

=

2250 N = 300 m/s 0.025 kg/m

The wavelengths along the two wires are

λ1 =

v1 500 m/s 1 = = m f 1500 Hz 3

λ2 =

v2 300 m/s 1 = = m f 1500 Hz 5

Thus, the number of wavelengths over two sections of the wire are 1.0 m

λ1

=

1.0 m

( 13 m )

=3

1.0 m

λ2

=

1.0 m

( 15 m )

The number of complete cycles of the wave in the 2.00-m-long wire is 8.

=5

20.81. Model: The wave pulse is a traveling wave on a stretched wire. Visualize:

Solve: (a) At a distance y above the lower end of the rope, the point P is in static equilibrium. The upward tension in the rope must balance the weight of the rope that hangs below this point. Thus, at this point

T = w = Mg = ( μ y ) g where μ = m/L is the linear density of the entire rope. Using Equation 20.2, we get

v=

T

μ

=

μ yg = gy μ

(b) The time to travel a distance dy at y, where the wave speed is v = gy , is

dt =

dy dy = v gy

Finding the time for a pulse to travel the length of the rope requires integrating from one end of the rope to the other: T L L dy 1 2 L Δt = ∫ dt = ∫ = L ⇒ Δt = 2 2 y = 0 g gy g g 0 0

(

)

20.82. Visualize:

Solve: as

(a) Using the graph, the refractive index n as a function of distance x can be mathematically expressed n = n1 +

n2 − n1 x L

At position x, the light speed is v = c / n. The time for the light to travel a distance dx at x is

dt =

dx n 1⎛ n −n ⎞ = dx = ⎜ n1 + 2 1 x ⎟ dx v c c⎝ L ⎠

To find the total time for the light to cover a thickness L of a glass we integrate as follows: T

T = ∫ dt = 0

2 n n 1 ⎛ n2 − n1 ⎞ (n − n ) ⎛n −n ⎞L ⎛n +n ⎞ x ⎟ dx = 1 ∫ dx + 2 1 ∫ x dx = 1 L + ⎜ 2 1 ⎟ = ⎜ 1 2 ⎟ L ⎜ n1 + ∫ c 0 cL 0 c c0⎝ L ⎝ cL ⎠ 2 ⎝ 2c ⎠ ⎠ L

L

(b) Substituting the given values into this equation, (1.50 + 1.60 ) × 0.010 m = 5.17 × 10−11 s T= 2 ( 3.0 × 108 m/s )

L

21.1. Model: The principle of superposition comes into play whenever the waves overlap. Visualize:

The graph at t = 1.0 s differs from the graph at t = 0.0 s in that the left wave has moved to the right by 1.0 m and the right wave has moved to the left by 1.0 m. This is because the distance covered by the wave pulse in 1.0 s is 1.0 m. The snapshot graphs at t = 2.0 s, 3.0 s, and 4.0 s are a superposition of the left and the right moving waves. The overlapping parts of the two waves are shown by the dotted lines.

21.2. Model: The principle of superposition comes into play whenever the waves overlap. Visualize:

The snapshot graph at t = 1.0 s differs from the graph t = 0.0 s in that the left wave has moved to the right by 1.0 m and the right wave has moved to the left by 1.0 m. This is because the distance covered by each wave in 1.0 s is 1.0 m. The snapshot graphs at t = 2.0 s, 3.0 s, and 4.0 s are a superposition of the left and the right moving waves. The overlapping parts of the two waves are shown by the dotted lines.

21.3. Model: The principle of superposition comes into play whenever the waves overlap. Visualize:

At t = 4.0 s the shorter pulses overlap and cancel. At t = 6.0 s the longer pulses overlap and cancel.

21.4. Model: The principle of superposition comes into play whenever the waves overlap. Solve: 4 s.

(b)

(a) As graphically illustrated in the figure below, the snapshot graph of Figure EX21.5b was taken at t =

21.5. Model: A wave pulse reflected from the string-wall boundary is inverted and its amplitude is unchanged. Visualize:

The graph at t = 2 s differs from the graph at t = 0 s in that both waves have moved to the right by 2 m. This is because the distance covered by the wave pulse in 2 s is 2 m. The shorter pulse wave encounters the boundary wall at 2.0 s and is inverted on reflection. This reflected pulse wave overlaps with the broader pulse wave, as shown in the snapshot graph at t = 4 s. At t = 6 s, only half of the broad pulse is reflected and hence inverted; the shorter pulse wave continues to move to the left with a speed of 1 m/s. Finally, at t = 8 s both the reflected pulse waves are inverted and they are both moving to the left.

21.6. Model: Reflections at both ends of the string cause the formation of a standing wave. Solve: Figure EX21.6 indicates 5/2 wavelengths on the 2.0-m-long string. Thus, the wavelength of the standing wave is λ = 52 ( 2.0 m ) = 0.80 m . The frequency of the standing wave is f =

v

λ

=

40 m/s = 50 Hz 0.80 m

21.7. Model: Reflections at the string boundaries cause a standing wave on the string. Solve: Figure EX21.7 indicates two full wavelengths on the string. Hence λ = 12 (60 cm) = 30 cm = 0.30 m. Thus

v = λ f = ( 0.30 m )(100 Hz ) = 30 m/s

21.8. Model: Reflections at the string boundaries cause a standing wave on the string. Solve: (a) When the frequency is doubled ( f ′ = 2 f 0 ) , the wavelength is halved ( λ ′ = 12 λ0 ) . This halving of the wavelength will increase the number of antinodes to six. (b) Increasing the tension by a factor of 4 means

v=

T

μ

⇒ v′ =

T′

μ

=

4T

μ

= 2v

For the string to continue to oscillate as a standing wave with three antinodes means λ ′ = λ0 . Hence, v′ = 2v ⇒ f ′λ ′ = 2 f 0λ0 ⇒ f ′λ0 = 2 f 0λ0 ⇒ f ′ = 2 f 0 That is, the new frequency is twice the original frequency.

21.9. Model: A string fixed at both ends supports standing waves.

Solve: (a) We have fa = 24 Hz = mf1 where f1 is the fundamental frequency that corresponds to m = 1. The next successive frequency is fb = 36 Hz = (m + 1) f1. Thus,

24 Hz f b ( m + 1) f1 m + 1 36 Hz = 12 Hz = = = = 1.5 ⇒ m + 1 = 1.5m ⇒ m = 2 ⇒ f1 = fa mf1 m 24 Hz 2 The wave speed is v = λ1 f1 =

2L f1 = ( 2.0 m )(12 Hz ) = 24 m/s 1

(b) The frequency of the third harmonic is 36 Hz. For m = 3, the wavelength is

λm =

2 L 2 (1 m ) 2 = = m 3 3 m

21.10. Model: A string fixed at both ends supports standing waves. Solve:

(a) A standing wave can exist on the string only if its wavelength is

λm =

2L m

m = 1, 2, 3, …

The three longest wavelengths for standing waves will therefore correspond to m = 1, 2, and 3. Thus,

λ1 =

2 ( 2.40 m ) = 4.80 m 1

λ2 =

2 ( 2.40 m ) = 2.40 m 2

λ3 =

2 ( 2.40 m ) = 1.60 m 3

(b) Because the wave speed on the string is unchanged from one m value to the other,

f 2λ2 = f3λ3 ⇒ f 3 =

f 2λ2

λ3

=

( 50 Hz )( 2.40 m ) = 75 Hz 1.60 m

21.11. Model: A string fixed at both ends forms standing waves. Solve:

(a) The wavelength of the third harmonic is calculated as follows:

λm =

2L 2 L 2.42 m ⇒ λ3 = = = 0.807 m ≈ 0.81 m m 3 3

(b) The speed of waves on the string is v = λ3 f3 = (0.807 m)(180 Hz) = 145.3 m/s. The speed is also given by

v = TS / μ , so the tension is TS = μ v 2 =

m 2 0.004 kg 2 v = (145.3 m/s ) = 69.7 N ≈ 70 m L 1.21 m

21.12. Model: For the stretched wire vibrating at its fundamental frequency, the wavelength of the standing wave is λ1 = 2 L. Visualize:

Solve:

The wave speed on the steel wire is vwire = f λ = f ( 2 L ) = ( 80 Hz )( 2 × 0.90 m ) = 144 m/s

and is also equal to

TS μ , where

μ=

mass 5.0 × 10−3 kg = = 5.555 × 10−3 kg/m length 0.90 m

The tension TS in the wire equals the weight of the sculpture or Mg. Thus, 2 2 5.555 × 10−3 kg/m ) (144 m/s ) ( Mg μ vwire vwire = ⇒M = = = 12 kg μ g 9.8 m/s 2

21.13. Model: The laser light forms a standing wave inside the cavity. Solve:

The wavelength of the laser beam is

λm =

2 ( 0.5300 m ) 2L ⇒ λ100,000 = = 10.60 μ m 100,000 m

The frequency is

f100,000 =

c

λ100,000

=

3.000 × 108 m/s = 2.830 × 1013 Hz 10.60 × 10−6 m

21.14. Solve: (a) For the open-open tube, the two open ends exhibit antinodes of a standing wave. The possible wavelengths for this case are 2L λm = m = 1, 2, 3, … m The three longest wavelengths are λ1 =

2 (1.21 m ) = 2.42 m 1

λ2 =

2 (1.21 m ) = 1.21 m 2

λ3 =

2 (1.21 m ) = 0.807 m 3

λ3 =

4 (1.21 m ) = 0.968 m 5

(b) In the case of an open-closed tube,

λm =

4L m

m = 1, 3, 5, …

The three longest wavelengths are

λ1 =

4 (1.21 m ) = 4.84 m 1

λ2 =

4 (1.21 m ) = 1.61 m 3

21.15. Model: We have an open-open tube that forms standing sound waves. Solve: The gas molecules at the ends of the tube exhibit maximum displacement, making antinodes at the ends. There is another antinode in the middle of the tube. Thus, this is the m = 2 mode and the wavelength of the standing wave is equal to the length of the tube, that is, λ = 0.80 m. Since the frequency f = 500 Hz, the speed of = sound in this case is v = fλ (500 Hz)(0.80 m) = 400 m/s. Assess: The experiment yields a reasonable value for the speed of sound.

21.16. Solve: For the open-open tube, the fundamental frequency of the standing wave is f1 = 1500 Hz when the tube is filled with helium gas at 0°C. Using λm = 2 L m , f1 helium =

vhelium

λ1

=

970 m/s 2L

Similarly, when the tube is filled with air, f1 air =

Assess:

vair

λ1

=

331 m/s f 331 m/s ⎛ 331 m/s ⎞ ⇒ 1 air = ⇒ f1 air = ⎜ ⎟ (1500 Hz ) = 512 Hz 2L f1 helium 970 m/s ⎝ 970 m/s ⎠

Note that the length of the tube is one-half the wavelength whether the tube is filled with helium or air.

21.17. Model: An organ pipe has a “sounding” hole where compressed air is blown across the edge of the pipe. This is one end of an open-open tube with the other end at the true “end” of the pipe. Solve: For an open-open tube, the fundamental frequency is f1 = 16.4 Hz. We have

λ1 =

⎞ 1 ⎛ 343 m/s ⎞ 2L λ 1⎛v ⇒ L = 1 = ⎜ sound ⎟ = ⎜ ⎟ = 10.5 m 1 2 2 ⎝ f1 ⎠ 2 ⎝ 16.4 Hz ⎠

Assess: The length of the organ pipe is ≈ 34.5 feet. That is actually somewhat of an overestimate since the antinodes of real tubes are slightly outside the tube. The actual length in a real organ is about 32 feet, and this is the tallest pipe in the so called “32 foot rank” of pipes.

21.18. Model: Reflections at the string boundaries cause a standing wave on a stretched string. Solve: Because the vibrating section of the string is 1.9 m long, the two ends of this vibrating wire are fixed, and the string is vibrating in the fundamental harmonic. The wavelength is

λm =

2L ⇒ λ1 = 2 L = 2 (1.90 m ) = 3.80 m m

The wave speed along the string is v = f1λ1 = (27.5 Hz)(3.80 m) = 104.5 m/s. The tension in the wire can be found as follows: ⎛ mass ⎞ 2 ⎛ 0.400 kg ⎞ T 2 v = S ⇒ TS = μ v 2 = ⎜ ⎟v = ⎜ ⎟ (104.5 m/s ) = 2180 N μ length 2.00 m ⎝ ⎠ ⎝ ⎠

21.19. Model: A string fixed at both ends forms standing waves.

Solve: A simple string sounds the fundamental frequency f1 = v/2L. Initially, when the string is of length LA = 30 cm, the note has the frequency f1A = v/2LA. For a different length, f1B = v/2LB. Taking the ratio of each side of these two equations gives

f1A v / 2 LA LB f = = ⇒ LB = 1A LA f1B v / 2 LB LA f1B We know that the second frequency is desired to be f1B = 523 Hz. The string length must be LB =

440 Hz ( 30 cm ) = 25.2 cm 523 Hz

The question is not how long the string must be, but where must the violinist place his finger. The full string is 30 cm long, so the violinist must place his finger 4.8 cm from the end. Assess: A fingering distance of 4.8 cm from the end is reasonable.

21.20. Model: Interference occurs according to the difference between the phases ( Δφ ) of the two waves. Solve: (a) A separation of 20 cm between the speakers leads to maximum intensity on the x-axis, but a separation of 60 cm leads to zero intensity. That is, the waves are in phase when (Δ x)1 = 20 cm but out of phase

when (Δ x) 2 = 60 cm. Thus,

( Δx)2 − ( Δ x)1 =

λ 2

⇒ λ = 2 ( 60 cm − 20 cm ) = 80 cm

(b) If the distance between the speakers continues to increase, the intensity will again be a maximum when the separation between the speakers that produced a maximum has increased by one wavelength. That is, when the separation between the speakers is 20 cm + 80 cm = 100 cm.

21.21. Model: The interference of two waves depends on the difference between the phases (Δφ ) of the two waves. Solve: (a) Because the speakers are in phase, Δφ0 = 0 rad. Let d represent the path-length difference. Using m = 0 for the smallest d and the condition for destructive interference, we get

Δφ = 2π ⇒ 2π

Δx

λ

Δx

λ

+ Δφ0 = 2 ( m + 12 )π rad

+ Δφ0 = π rad ⇒ 2π

d

λ

+ 0 rad = π rad ⇒ d =

m = 0, 1, 2, 3 …

λ 2

1 ⎛ v ⎞ 1 ⎛ 343 m/s ⎞ = ⎜ ⎟= ⎜ ⎟ = 0.25 m 2 ⎝ f ⎠ 2 ⎝ 686 Hz ⎠

(b) When the speakers are out of phase, Δφ0 = π . Using m = 1 for the smallest d and the condition for constructive interference, we get Δx m = 0, 1, 2, 3, … Δφ = 2π + Δφ0 = 2mπ

λ

λ 1 ⎛ v ⎞ 1 ⎛ 343 m/s ⎞ d ⇒ 2π + π = 2π ⇒ d = = ⎜ ⎟ = ⎜ ⎟ = 0.25 m λ 2 2 ⎝ f ⎠ 2 ⎝ 686 Hz ⎠

21.22. Model: We assume that the speakers are identical and that they are emitting in phase. Solve: Since you don’t hear anything, the separation between the two speakers corresponds to the condition of destructive interference. With Δφ0 = 0 rad, Equation 21.23 becomes

2π

d

λ

= 2 ( m + 12 )π rad ⇒ d = ( m + 12 ) λ ⇒ d =

Since the wavelength is

λ=

v 340 m/s = = 2.0 m f 170 Hz

three possible values for d are 1.0 m, 3.0 m, and 5.0 m.

λ 3λ 5λ 2

,

2

,

2

21.23. Model: Reflection is maximized if the two reflected waves interfere constructively. Solve: The film thickness that causes constructive interference at wavelength λ is given by Equation 21.32: λC =

−9 λ m ( 600 × 10 m ) (1) 2nd ⇒d = C = = 216 nm m 2n ( 2 )(1.39 )

where we have used m = 1 to calculate the thinnest film. Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicable because nair < nfilm < nglass.

21.24. Model: Reflection is maximized if the two reflected waves interfere constructively. Solve: The film thickness that causes constructive interference at wavelength λ is given by Equation 21.32: λC =

−9 λ m ( 500 × 10 m ) (1) 2nd ⇒d = C = = 200 nm m 2n ( 2 )(1.25)

where we have used m = 1 to calculate the thinnest film. Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicable because nair < noil < nwater.

21.25. Solve: (a) The circular wave fronts emitted by the two sources show that the two sources are in phase. This is because the wave fronts of each source have moved the same distance from their sources. (b) Let us label the top source as 1 and the bottom source as 2. Since the sources are in phase, Δφ0 = 0 rad. For

the point P, r1 = 3λ and r2 = 4λ. Thus, Δr = r2 − r1 = 4λ − 3λ = λ. The phase difference is Δφ =

2πΔr

λ

2π ( λ )

=

λ

= 2π

This corresponds to constructive interference. For the point Q, r1 = 72 λ and r2 = 2λ. The phase difference is

Δφ =

2πΔr

λ

=

2π ( 32 λ )

λ

= 3π

This corresponds to destructive interference. For the point R, r1 = 52 λ and r2 = 72 λ . The phase difference is Δφ =

2π ( λ )

λ

= 2π

This corresponds to constructive interference. P Q R

r1 3λ

λ 5 λ 2

7 2

r2 4λ 2λ 7 2

λ

Δr

λ λ

C/D C D

λ

C

3 2

21.26. Solve: (a) The circular wave fronts emitted by the two sources indicate the sources are out of phase. This is because the wave fronts of each source have not moved the same distance from their sources. (b) Let us label the top source as 1 and the bottom source as 2. The phase difference between the sources is Δφ0 = π . For the point P, r1 = 2λ and r2 = 3λ. The phase difference is Δφ =

2πΔr

λ

+ Δφ0 =

2π ( 3λ − 2λ )

λ

+ π = 3π

This corresponds to destructive interference. For the point Q, r1 = 3λ and r2 = 32 λ. The phase difference is

Δφ =

2π ( 32 λ )

λ

+ π = 4π

This corresponds to constructive interference. For the point R, r1 = 52 λ and r2 = 3λ. The phase difference is Δφ =

2π ( 12 λ )

λ

+ π = 2π

This corresponds to constructive interference.

Assess:

P Q

r1 2λ 3λ

r2 3λ

R

5 2

λ

3λ

3 2

λ

Δr

λ λ 1 λ 2 3 2

Note that it is not r1 or r2 that matter, but the difference Δr between them.

C/D D C C

21.27. Model: The two speakers are identical, and so they are emitting circular waves in phase. The overlap of these waves causes interference. Visualize:

Solve:

From the geometry of the figure,

r2 = r12 + ( 2.0 m ) = 2

( 4.0 m )

2

+ ( 2.0 m ) = 4.472 m 2

So, Δr = r2 − r1 = 4.472 m − 4.0 m = 0.472 m. The phase difference between the sources is Δφ0 = 0 rad and the wavelength of the sound waves is v 340 m/s λ= = = 0.1889 m f 1800 Hz

Thus, the phase difference of the waves at the point 4.0 m in front of one source is 2π ( 0.472 m ) + 0 rad = 5π rad = 2.5(2π rad) λ 0.1889 m This is a half-integer multiple of 2π rad, so the interference is perfect destructive. Δφ = 2π

Δr

+ Δφ0 =

21.28. Model: The two radio antennas are emitting out-of-phase, circular waves. The overlap of these waves causes interference. Visualize:

Solve:

From the geometry of the figure, r1 = 800 m and r2 =

( 800 m )

2

+ ( 600 m ) = 1000 m 2

So, Δr = r2 − r1 = 200 m and Δφ0 = π rad. The wavelength of the waves is

λ=

c 3.0 × 108 m/s = = 100 m f 3.0 × 106 Hz

Thus, the phase difference of the waves at the point (300 m, 800 m) is Δφ = 2π

Δr

λ

+ Δφ0 =

2π ( 200 m ) + π rad = 5π rad = 2.5(2π rad) 100 m

This is a half-integer multiple of 2π rad, so the interference is perfect destructive.

21.29. Solve: The beat frequency is f beat = f1 − f 2 ⇒ 3 Hz = f1 − 200 Hz ⇒ f1 = 203 Hz f1 is larger than f2 because the increased tension increases the wave speed and hence the frequency.

21.30. Solve: The flute player’s initial frequency is either 523 Hz + 4 Hz = 527 Hz or 523 Hz − 4 Hz = 519 Hz. Since she matches the tuning fork’s frequency by lengthening her flute, she is increasing the wavelength of the standing wave in the flute. A wavelength increase means a decrease of frequency because v = fλ. Thus, her initial frequency was 527 Hz.

21.31. Model: The superposition of two slightly different frequencies creates beats. Solve: Let λ1 = 780.54510 nm and λ2 > λ1 . This means f 2 < f1 and Δf = f1 − f 2 = ⇒ λ2 =

c

λ1

−

c

λ2

= 98.5 × 106 Hz

1 1 = (1/ λ1 ) − (Δf / c) 1/(780.54510 × 10−9 m) − (98.5 × 106 Hz) /(3.00 × 108 m/s)

= 780.54530 nm Assess:

A small difference in wavelengths, ( λ2 − λ1 ) = 0.00020 nm = 0.20 pm, can yield beats at a relatively

high frequency of 98.5 MHz.

21.32. Model: The principle of superposition applies to overlapping waves. Visualize:

Solve: Because the wave pulses travel along the string at a speed of 100 m/s, they move a distance of d = vt = (100 m/s)(0.05 s) = 5 m in 0.050 s. The front of the wave pulse moving left, which is located at x = 1 m at t = 0.050 s, was thus located at x = 6 m at t = 0 s. This helps us draw the snapshot of the wave pulse moving left at t = 0 s (shown as a dashed line). Subtracting this wave snapshot from the resultant at t = 0 s (shown as a solid line) yields the right-traveling wave’s snapshot at t = 0 s (shown as a dotted line). Finally, the snapshot graph of the wave pulse moving right at t = 0.050 s is the same as at t = 0 s (shown as a dotted line) except that it is shifted to the right by 5 m.

21.33. Model: The wavelength of the standing wave on a string vibrating at its second-harmonic frequency is equal to the string’s length. Visualize:

Solve:

The length of the string L = 2.0 m, so λ = L = 2.0 m. This means the wave number is k=

2π

λ

=

2π = π rad m 2.0 m

According to Equation 21.5, the displacement of a medium when two sinusoidal waves superpose to give a standing wave is D ( x, t ) = A ( x ) cos ω t , where A ( x ) = 2a sin kx = Amax sin kx. The amplitude function gives the

amplitude of oscillation from point to point in the medium. For x = 10 cm, A ( x = 10 cm ) = ( 2.0 cm ) sin ⎡⎣(π rad m )( 0.10 m ) ⎤⎦ = 0.62 cm Similarly,

A( x = 20 cm) = 1.18 cm,

A ( x = 30 cm ) = 1.62 cm,

A ( x = 40 cm ) = 1.90 cm,

A ( x = 50 cm ) = 2.00 cm. Assess:

Consistent with the above figure, the amplitude of oscillation is a maximum at x = 0.50 m.

and

21.34. Model: The wavelength of the standing wave on a string is λm = 2 L m , where m = 1, 2, 3, … We assume that 30 cm is the first place from the left end of the string where A = Amax 2. Visualize:

Solve:

The amplitude of oscillation on the string is A ( x ) = Amax sin kx . Since the string is vibrating in the third

harmonic, the wave number is k=

2π

λ

=

2π

( 2 L 3)

=3

π L

Substituting into the equation for the amplitude, 1 3π π ⎛ 3π ⎞ ⎛ 3π ⎞ 1 Amax = Amax sin ⎜ ( 0.30 m ) ⎟ ⇒ sin ⎜ ( 0.30 m ) ⎟ = ⇒ ( 0.30 m ) = rad ⇒ L = 5.4 m 2 L L 2 L 6 ⎝ ⎠ ⎝ ⎠

21.35. Model: The wavelength of the standing wave on a string vibrating at its fundamental frequency is equal to 2L. Solve: The amplitude of oscillation on the string is A ( x ) = 2a sin kx, where a is the amplitude of the traveling wave and the wave number is

k=

2π

λ

=

2π π = 2L L

Substituting into the above equation,

⎡⎛ π ⎞⎛ L ⎞ ⎤ ⎛ 1 ⎞ A ( x = 14 L ) = 2.0 cm = 2a sin ⎢⎜ ⎟⎜ ⎟ ⎥ ⇒ 1.0 cm = a ⎜ ⎟ ⇒ a = 2 cm = 1.4 cm ⎝ 2⎠ ⎣⎝ L ⎠⎝ 4 ⎠ ⎦

21.36. Solve: You can see in Figure 21.4 that the time between two successive instants when the antinodes are at maximum height is half the period, or 12 T . Thus T = 2(0.25 s) = 0.50 s, and so

f =

v 3.0 m/s 1 1 = = 2.0 Hz ⇒ λ = = = 1.5 m T 0.50 s f 2.0 Hz

21.37. Model: The wave on a stretched string with both ends fixed is a standing wave. For vibration at its fundamental frequency, λ = 2L.

Solve: The wavelength of the wave reaching your ear is 39.1 cm = 0.391 m. So the frequency of the sound wave is

f =

vair

λ

=

344 m/s = 879.8 Hz 0.391 m

This is also the frequency emitted by the wave on the string. Thus,

879.8 Hz =

vstring

λ

=

1 TS

λ

μ

=

1

150 N

λ 0.0006 kg/m

⇒ L = 12 λ = 0.284 m = 28.4 cm

⇒ λ = 0.568 m

21.38. Model: The wave on a stretched string with both ends fixed is a standing wave. Solve: We must distinguish between the sound wave in the air and the wave on the string. The listener hears a sound wave of wavelength λsound = 40 cm = 0.40 m. Thus, the frequency is

f =

vsound

λsound

=

343 m/s = 857.5 Hz 0.40 m

The violin string oscillates at the same frequency, because each oscillation of the string causes one oscillation of the air. But the wavelength of the standing wave on the string is very different because the wave speed on the string is not the same as the wave speed in air. Bowing a string produces sound at the string’s fundamental frequency, so the wavelength of the string is

λstring = λ1 = 2 L = 0.60 m ⇒ vstring = λstring f = ( 0.60 m )( 857.5 Hz ) = 514.5 m/s The tension is the string is found as follows: vstring =

TS

μ

⇒ TS = μ ( vstring ) = ( 0.001 kg/m )( 514.5 m/s ) = 260 N 2

2

21.39. Model: A string fixed at both ends forms standing waves.

Solve: (a) Three antinodes means the string is vibrating as the m = 3 standing wave. The frequency is f3 = 3f1, so the fundamental frequency is f1 = 13 (420 Hz) = 140 Hz. The fifth harmonic will have the frequency f5 = 5f1 = 700 Hz. (b) The wavelength of the fundamental mode is λ1 = 2L = 1.20 m. The wave speed on the string is v = λ1 f1 = (1.20 m) (140 Hz) = 168 m/s. Alternatively, the wavelength of the n = 3 mode is λ3 =

1 3

(2L) = 0.40 m, from

which v = λ3 f3 = (0.40 m)(420 Hz) = 168 m/s. The wave speed on the string is given by

v=

TS

μ

⇒ TS = μ v 2 = ( 0.0020 kg/m )(168 m/s ) = 56 N 2

Assess: You must remember to use the linear density in SI units of kg/m. Also, the speed is the same for all modes, but you must use a matching λ and f to calculate the speed.

21.40. Model: Assume that the extra kilogram doesn’t stretch the wire longer (so L stays the same) nor thinner (so μ stays the same). Also assume that because the wire is thin its own weight is negligible, so Ts is constant throughout the wire and is equal to Mg. Visualize: The wire is fixed at both ends so in the second harmonic L = λ . We are given f 2 = 200 Hz and

f 2′ = 245 Hz and M ′ = M + 1.0 kg. Apply v = λ f and v = Ts /μ . Solve:

Cancel off λ ,

μ , and

g in turn.

T ′/μ f 2′ v′/λ = = s = f 2 v/λ Ts /μ

M ′g Mg

=

( M + 1.0 kg) g Mg

=

M + 1.0 kg M

2

⎛ f 2′ ⎞ M + 1.0 kg ⎜ ⎟ = M ⎝ f2 ⎠

⎛ ⎛ f ′ ⎞2 ⎞ M ⎜ ⎜ 2 ⎟ − 1⎟ = 1.0 kg ⎜ ⎝ f2 ⎠ ⎟ ⎝ ⎠ M=

1.0 kg 2

⎛ f 2′ ⎞ ⎜ ⎟ −1 ⎝ f2 ⎠

=

1.0 kg 2

⎛ 245 Hz ⎞ ⎜ ⎟ −1 ⎝ 200 Hz ⎠

= 2.0 kg

Assess: We did not expect M to be really huge or a) it would have broken the wire, and b) adding one more kilogram wouldn’t have made as big a difference in f 2 as it did.

21.41. Model: The stretched string with both ends fixed forms standing waves. Visualize:

Solve: The astronauts have created a stretched string whose vibrating length is L = 2.0 m. The weight of the hanging mass creates a tension TS = Mg in the string, where M = 1.0 kg. As a consequence, the wave speed on the string is

v=

TS

μ

=

Mg

μ

where μ = (0.0050 kg)/(2.5 m) = 0.0020 kg/m is the linear density. The astronauts then observe standing waves at frequencies of 64 Hz and 80 Hz. The first is not the fundamental frequency of the string because 80 Hz ≠ 2 × 64 Hz. But we can easily show that both are multiples of 16 Hz: 64 Hz = 4 f1 and 80 Hz = 5 f1 . Both frequencies are also multiples of 8 Hz. But 8 Hz cannot be the fundamental frequency because, if it were, there would be a standing wave resonance at 9(8 Hz) = 72 Hz. So the fundamental frequency is f1 = 16 Hz. The fundamental wavelength is λ1 = 2L = 4.0 m. Thus, the wave speed on the string is v = λ1 f1 = 64.0 m/s . Now we can find g on Planet X:

v=

Mg

μ

⇒g=

μ M

v2 =

0.0020 kg/m 2 ( 64 m/s ) = 8.2 m/s 2 1.0 kg

21.42. Model: The stretched bungee cord that forms a standing wave with two antinodes is vibrating at the second harmonic frequency. Visualize:

Solve:

Because the vibrating cord has two antinodes, λ2 = L = 1.80 m. The wave speed on the cord is vcord = f λ = ( 20 Hz )(1.80 m ) = 36 m/s

The tension TS in the cord is equal to kΔL, where k is the bungee’s spring constant and ΔL is the 0.60 m the bungee has been stretched. Thus, T k ΔL vcord = S =

μ

μ

21.43. Solve: (a) Because the frequency of the standing wave on the copper wire is the same as the frequency on the aluminum wire, v v f Cu = f Al ⇒ Cu = Al λCu

λCu

Let nCu be the number of half-wavelength antinodes on the copper wire and nAl be the number of half-wavelength antinodes on the aluminum wire. Thus,

0.44 m ⎛λ ⎞ nCu ⎜ Cu ⎟ = 0.22 m ⇒ λCu = nCu ⎝ 2 ⎠

1.20 m ⎛λ ⎞ nAl ⎜ Al ⎟ = 0.60 m ⇒ λAl = nAl ⎝ 2 ⎠

nCu ⎛ vAl ⎞⎛ 0.44 m ⎞ = ⎜ ⎟ nAl ⎜⎝ vCu ⎟⎠ ⎝ 1.20 m ⎠ We can find vCu and vAl by using the following equations for a stretched wire: ⇒

vCu

=

vAl

( 0.44 m nCu ) (1.20 m vCu =

T

μ Cu

nAl )

⇒

vAl =

T

μ Al

The linear densities are calculated as follows:

μ Cu =

ρ V mCu ⎛ ρ Cu ⎞ 2 = Cu Cu = ⎜ ⎟π r ( 0.22 m ) 0.22 m 0.22 m ⎝ 0.22 m ⎠

= ( 8920 kg/m3 )π ( 5.0 × 10−4 m ) = 7.006 × 10−3 kg/m3 2

μ Al =

ρ V mAl ⎛ ρ Al ⎞ 2 = Al Al = ⎜ ⎟π r ( 0.60 m ) 0.60 m 0.60 m ⎝ 0.60 m ⎠

= ( 2700 kg/m3 )π ( 5.0 × 10−4 m ) = 2.121 × 10−3 kg/m3 2

⇒ vCu =

20 N 20 N = 53.43 m/s vAl = = 97.12 m/s 7.006 × 10−3 kg/m3 2.121 × 10−3 kg/m3

Going back to the nCu nAl equation, we have nCu ⎛ 97.12 m/s ⎞⎛ 0.44 m ⎞ 2 =⎜ ⎟⎜ ⎟ = 0.666 = ⇒ nCu = 2 and nAl = 3 nAl ⎝ 53.43 m/s ⎠⎝ 1.20 m ⎠ 3

Substituting into the expressions for wavelength and frequency,

λCu =

v 0.44 m 0.44 m 53.43 m/s = = 0.22 m ⇒ f Cu = Cu = = 243 Hz ≈ 240 Hz nCu 2 0.22 m λCu

λAl =

v 1.20 m 1.20 m 97.12 m/s = = 0.40 m ⇒ f Al = Al = = 243 Hz ≈ 240 Hz nAl 3 0.40 m λAl

(b) At this frequency of 240 Hz, there are 3 antinodes on the aluminum wire.

21.44. Visualize: Use primed quantities for when the sphere is submerged. We are given f5′ = f3 and M = 1.5 kg . We also know the density of water is ρ = 1000 kg/m3 . In the third mode before the sphere is

submerged L = 32 λ ⇒ λ = 32 L. Likewise, after the sphere is submerged L = 52 λ ′ ⇒ λ ′ = 52 L. The tension in the string before the sphere is submerged is Ts = Mg , but after the sphere is submerged, according to Archimedes’ principle, it is reduced by the weight of the water displaced by the sphere: Ts′ = Mg − ρVg , where V = 34 π R 3 . Solve:

We are looking for R so solve Ts′ = Mg − ρVg for ρVg and later we will isolate R from that.

ρVg = Mg − Ts′ Solve v′ = Ts /μ for Ts′. Also substitute for V.

⎛4⎞ ⎝ ⎠

ρ ⎜ ⎟ π R3 g = Mg − μ v′2 3 Now use v′ = λ ′ f ′. ⎛4⎞ ⎝ ⎠

ρ ⎜ ⎟ π R 3 g = Mg − μ (λ ′ f5′) 2 3 Recall that f 5′ = f3 and λ ′ = 52 L. ⎛ 4⎞ ⎝ ⎠

⎛

2

⎞

2

ρ ⎜ ⎟π R 3 g = Mg − μ ⎜⎜ Lf3 ⎟⎟ ⎜5 ⎟ 3 ⎝

⎠

⎛2 ⎝

v⎞

Substitute f 3 = v/λ . ⎛ 4⎞ ⎝ ⎠

2

ρ ⎜ ⎟π R 3 g = Mg − μ ⎜ L ⎟ 3 5 λ ⎠

Now use λ = 32 L and v = Ts /μ . ⎛2 T /μ ⎞ ⎛4⎞ ρ ⎜ ⎟ π R 3 g = Mg − μ ⎜ L 2 s ⎟ ⎜ ⎟ 3 5 ⎝ ⎠ 3L ⎠ ⎝

2

The 2’s, μ ’s, and L’s cancel. ⎛ 4⎞ ⎝ ⎠

⎛

⎛ 4⎞ ⎝ ⎠

⎛ 3⎞ ⎝ ⎠

⎞

3

2

ρ ⎜ ⎟π R 3 g = Mg − ⎜⎜ Ts ⎟⎟ ⎜5 ⎟ 3 ⎝

⎠

Ts = Mg . 2

ρ ⎜ ⎟π R 3 g = Mg − ⎜ ⎟ Mg 3 5 Cancel g and factor M out on the right side. ⎛

⎛ 4⎞ ⎝ ⎠

2 ⎛ 3⎞ ⎞ ⎝ ⎠ ⎠

⎛ ⎝

9 ⎞ ⎠

⎛ 16 ⎞ ⎝ ⎠

ρ ⎜ ⎟π R 3 = M ⎜1 − ⎜ ⎟ ⎟ = M ⎜ 1 − ⎟ = M ⎜ ⎟ ⎜ 3 5 ⎟ 25 25 ⎝

Now solve for R. ⎛ 3 ⎞⎛ 16 ⎞ M R 3 = ⎜ ⎟⎜ ⎟ ⎝ 4 ⎠⎝ 25 ⎠ πρ

R=

3

12 M 12 (1.5 kg) =3 = 6.1 cm 25 πρ 25 π (1000 kg/m3 )

Assess: The density of the sphere turns out to be about 1.5 × the density of water, which means it sinks and is in a reasonable range for densities.

21.45. Visualize: First compute the length L of the wire from the Pythagorean theorem. L = 2 2 m. Now μ = M/L = 0.075 kg/ 2 2 m = 0.02652 kg/m. Also, in the fundamental mode λ = 2 L; here λ = 4 2 m = 5.657 m. Solve: Apply the principles of statics to the point at the end of the bar. ΣFy = Ts sin 45° − (8 kg)( g ) = 0 N ⇒ Ts =

(8 kg)(9.8 m/s 2 ) = 110.87 N 2/ 2

Use these values for λ , μ , and Ts to find f. f = Assess:

v

λ

=

Ts /μ

λ

=

(110.87 N) / (0.02652 kg/m) = 11 Hz 5.657 m

This seems like a reasonable frequency for a mechanical system like this.

21.46. Model: Assume that while the spring provides the same tension to both strings it also acts as a fixed point for the end of each string so an integral number of half wavelengths fit in each string. Visualize: Use a subscript L for the left string and R for the right string. We are given μ L = 2.0 g/m. From the

assumption above we know (Ts ) L = (Ts ) R = Ts . We are also given f L = f R = f and LL = LR = L. Notice from the diagram that λL = L and λR = 23 L. Solve:

From v = λ f and v = Ts /μ eliminate v and solve for μ : μ = T / (λ f ) 2 .

1 9 μ R (Ts ) R / (λR f R ) 2 Ts / ( 23 Lf ) 2 = = = = μ L (Ts ) L / (λL f L ) 2 Ts / ( Lf ) 2 ( 32 ) 2 4 9 4

9 4

μ R = μ L = (2.0 g/m) = 4.5 g/m Assess:

We expect a slower wave speed in the right string to correspond to a larger mass density.

21.47. Model: The microwave forms a standing wave between the two reflectors. Solve: (a) There are reflectors at both ends, so the electromagnetic standing wave acts just like the standing wave on a string that is tied at both ends. The frequencies of the standing waves are

fm = m

vlight 2L

=m

3.0 × 108 m/s c =m = m (1.5 × 109 Hz ) = 1.5m GHz 2L 2 ( 0.10 m )

where we have noted that electromagnetic waves of all frequencies travel with the speed of light c. The generator can produce standing waves at any frequency between 10 GHz and 20 GHz. These are m 7 8 9 10 11 12 13

fm (GHz) 10.5 12.0 13.5 15.0 16.5 18.0 19.5

(b) There are 7 different standing wave frequencies. Even-numbered values of n create a node at the center, and odd-numbered values of n create an antinode at the center. So the frequencies where the midpoint is an antinode are 10.5, 13.5, 16.5, and 19.5 GHz.

21.48. Model: The fundamental wavelength of an open-open tube is 2L and that of an open-closed tube is 4L. Solve:

We are given that f1 open-closed = f3 open-open = 3f1 open-open

⇒

vair

λ1 open-closed

=3

⇒ Lopen-closed =

vair

λ1 open-open

2 Lopen-open 12

⇒ =

1 3 = 4 Lopen-closed 2 Lopen-open

2 ( 78.0 cm ) = 13.0 cm 12

21.49. Model: A tube forms standing waves. Solve: (a) The fundamental frequency cannot be 390 Hz because 520 Hz and 650 Hz are not integer multiples of it. But we note that the difference between 390 Hz and 520 Hz is 130 Hz as is the difference between 520 Hz and 650 Hz. We see that 390 Hz = 3 × 130 Hz = 3f1, 520 Hz = 4f1, and 650 Hz = 5f1. So we are seeing the third, fourth, and fifth harmonics of a tube whose fundamental frequency is 130 Hz. According to Equation 21.17, this is an open-open tube because fm = mf1 with m = 1, 2, 3, 4, … For an open-closed tube m has only odd values. (b) Knowing f1, we can now find the length of the tube:

L=

v 343 m/s = = 1.32 m 2 f1 2 (130 Hz )

(c) 520 Hz is the fourth harmonic. This is a sound wave, not a wave on a string, so the wave will have four nodes and will have antinodes at the ends, as shown.

(d) With carbon dioxide, the new fundamental frequency is

f1 =

v 280 m/s = = 106 Hz 2 L 2 (1.32 m )

Thus the frequencies of the n = 3, 4, and 5 modes are f3 = 3f1 = 318 Hz, f4 = 4f1 = 424 Hz, and f5 = 5f1 = 530 Hz.

21.50. Model: Particles of the medium at the nodes of a standing wave have zero displacement. Solve: The cork dust settles at the nodes of the sound wave where there is no motion of the air molecules. The separation between the centers of two adjacent piles is 12 λ . Thus, 123 cm λ = ⇒ λ = 82 cm 3 2

Because the piston is driven at a frequency of 400 Hz, the speed of the sound wave in oxygen is

v = f λ = ( 400 Hz )( 0.82 m ) = 328 m/s Assess:

A speed of 328 m/s in oxygen is close to the speed of sound in air, which is 343 m/s at 20°C.

21.51. Model: The nodes of a standing wave are spaced λ/2 apart. Visualize:

Solve: The wavelength of the mth mode of an open-open tube is λm = 2L/m. Or, equivalently, the length of the tube that generates the mth mode is L = m(λ/2). Here λ is the same for all modes because the frequency of the tuning fork is unchanged. Increasing the length of the tube to go from mode m to mode m + 1 requires a length change ∆L = (m + 1)(λ/2) – mλ/2 = λ/2 That is, lengthening the tube by λ/2 adds an additional antinode and creates the next standing wave. This is consistent with the idea that the nodes of a standing wave are spaced λ/2 apart. This tube is first increased ΔL = 56.7 cm − 42.5 cm = 14.2 cm, then by ∆L = 70.9 cm – 56.7 cm = 14.2 cm. Thus λ/2 = 14.2 cm and thus λ = 28.4 cm = 0.284 m. Therefore the frequency of the tuning fork is f =

v

λ

=

343 m/s = 1208 Hz ≈ 1210 Hz 0.284 m

21.52. Model: The open-closed tube forms standing waves. Visualize:

Solve: When the air column length L is the proper length for a 580 Hz standing wave, a standing wave resonance will be created and the sound will be loud. From Equation 21.18, the standing wave frequencies of an open-closed tube are fm = m(v/4L), where v is the speed of sound in air and m is an odd integer: m = 1, 3, 5, … The frequency is fixed at 580 Hz, but as the length L changes, 580 Hz standing waves will occur for different values of m. The length that causes the mth standing wave mode to be at 580 Hz is

L=

m ( 343 m/s ) ( 4 )( 580 Hz )

We can place the values of L, and corresponding values of h = 1 m − L, in a table: m 1 3 5 7

L 0.148 m 0.444 m 0.739 m 1.035 m

h=1m−L 0.852 m = 85.2 cm 0.556 m = 55.6 cm 0.261 m = 26.1 cm h can’t be negative

So water heights of 26 cm, 56 cm, and 85 cm will cause a standing wave resonance at 580 Hz. The figure shows the m = 3 standing wave at h = 56 cm.

21.53. Model: A stretched wire, which is fixed at both ends, forms a standing wave whose fundamental frequency f1 wire is the same as the fundamental frequency f1 open-closed of the open-closed tube. The two frequencies are the same because the oscillations in the wire drive oscillations of the air in the tube. Visualize:

Solve:

The fundamental frequency in the wire is

f1 wire =

vwire 1 TS 1 = = 2 Lwire 2 Lwire μ (1.0 m )

440 N = 469 Hz ( 0.0010 kg/0.050 m )

The fundamental frequency in the open-closed tube is

f1 open-closed = 469 Hz =

vair 340 m/s 340 m/s = ⇒ Ltube = = 0.181 m ≈ 18 cm 4 Ltube 4 Ltube 4 ( 469 Hz )

21.54. Model: A stretched wire, which is fixed at both ends, creates a standing wave whose fundamental frequency is f1 wire . The second vibrational mode of an open-closed tube is f 3 open-closed . These two frequencies are equal because the wire’s vibrations generate the sound wave in the open-closed tube. Visualize:

Solve:

The frequency in the tube is

f 3 open-closed =

3 ( 340 m/s ) 3vair = = 300 Hz 4 Ltube 4 ( 0.85 cm )

⇒ f1 wire = 300 Hz =

1 vwire TS = 2 Lwire 2 Lwire μ

⇒ TS = ( 300 Hz ) ( 2 Lwire ) μ = (300 Hz)2 (2 × 0.25 m)2(0.020 kg/m) = 450 N 2

2

21.55. Model: The standing waves in the tube will have a displacement antinode at the top where the gas molecules are free to move and a node at the water where they are not. Visualize: For the wire, we are given Ts = 400 N. In the fundamental mode with both ends of the wire fixed

Lwire = λwire / 2. Hence, λwire = 2Lwire = 2(50.0 cm) = 1.00 m. We also know μ wire = 0.00100 kg/0.500 m = 0.00200 kg/m. The given information is sufficient to compute the frequency. From v = λ f and v = Ts /μ eliminate v and solve for f .

f =

(Ts ) wire /μ wire

λwire

=

400 N/ (0.00200 kg/m) = 447.2 Hz 1.00 m

Solve: Now we turn our attention to the gas, realizing that the frequency of the wave in the wire will be the same as the frequency of the sound in the gas. There is initially a node of the standing sound wave at the water level in the tube. The water is then lowered until the next standing wave is achieved; this is the next time there is a node at the water level. The distance between adjacent nodes in a standing wave is 12 λ , so

Δh = 12 λgas = 30.5 cm ⇒ λgas = 61.0 cm. vgas = λgas f gas = (0.610 m)(447.2 Hz) = 273 m/s Assess: We were told the gas is more dense than air so it will stay in the tube; for more dense gases we expect a slower sound speed. Our answer bears this out, but is still in the range of the speed of sound for typical gases.

21.56. Model: In a rod in which a longitudinal standing wave can be created, the standing wave is equivalent to a sound standing wave in an open-open tube. Both ends of the rod are antinodes, and the rod is vibrating in the fundamental mode. Solve: Since the rod is in the fundamental mode, λ1 = 2L = 2(2.0 m) = 4.0 m. Using the speed of sound in aluminum, the frequency is v 6420 m/s = 1605 Hz ≈ 1600 Hz f1 = Al = 4.0 m λ1

21.57. Model: Model the tunnel as an open-closed tube. v (m = odd) to find L, but we need to 4L know m first. Since m takes on only odd values for the open-closed tube the next resonance after m is m + 2. We are given f m = 4.5 Hz and f m + 2 = 6.3 Hz. Solve: v f m + 2 (m + 2) 4 L m + 2 = = v fm m ( m) 4L Visualize:

We are given v = 335 m/s. We would like to use f m = m

⎛ f ⎞ m ⎜ m+2 ⎟ = m + 2 ⎝ fm ⎠ ⎛ f ⎞ m ⎜ m + 2 − 1⎟ = 2 ⎝ fm ⎠

m=

2 2 = =5 6.3 Hz fm+2 −1 −1 4.5 Hz fm

Now that we know m we can finish up.

fm = m Assess:

v v 335 m/s ⇒L=m = (5) = 93 m 4L 4 fm 4(4.5 Hz)

93 m seems like a reasonable length for a tunnel.

21.58. Model: A standing wave in an open-closed tube must have a node at the closed end of the tube and an antinode at the open end. Visualize:

Solve: We first draw a series of pictures showing all the possible standing waves. By examination, we see that the first standing wave mode is 14 of a wavelength, so the tube’s length is L = 14 λ . The next mode is 34 of a

wavelength. The tube’s length hasn’t changed, so in this mode L = 34 λ . The next mode is now slightly more than a wavelength. That is, L = 54 λ . The next mode is 74 of a wavelength, so L = 74 λ . We see that there is a pattern. The length of the tube and the possible standing wave wavelengths are related by mλ L= m = 1, 3, 5, 7, … = odd integers 4 Solving for λ, we find that the wavelengths and frequencies of standing waves in an open-closed tube are 4L ⎫ λm = ⎪⎪ m m = 1, 3, 5, 7, … = odd integers v v ⎬ =m ⎪ fm = λm 4 L ⎭⎪

21.59. Model: The amplitude is determined by the interference of the two waves.

Solve: For interference in one dimension, where the speakers are separated by a distance Δx, the amplitude of the net wave is A = 2a cos ( 12 Δφ ) , where a is the amplitude of each wave and Δφ = 2πΔx λ + Δφ0 is the phase

difference between the two waves. The speakers are emitting identical waves so they have identical phase constants and Δφ0 = 0 . Thus,

λ ⎛ πΔx ⎞ −1 ⎛ 1.5 ⎞ A = 1.5a = 2a cos ⎜ ⎟ ⇒ Δx = cos ⎜ ⎟ π ⎝ λ ⎠ ⎝ 2 ⎠ The wavelength of a 1000 Hz tone is λ = vsound f = 0.343 m. Thus the separation must be

Δx =

0.343 m

π

cos −1 ( 0.75 ) = 0.0789 m ≈ 7.9 cm

It is essential to note that the argument of the arccosine is in radians, not in degrees.

21.60. Model: Constructive or destructive interference occurs according to the phases of the two waves. Visualize:

Solve:

(a) To go from destructive to constructive interference requires moving the speaker Δx = 12 λ ,

equivalent to a phase change of π rad. Since Δx = 40 cm, we find λ = 80 cm. (b) Destructive interference at Δx = 10 cm requires 2π

Δx

λ

3π ⎛ 10 cm ⎞ rad + Δφ0 = 2π ⎜ ⎟ rad + Δφ0 = π rad ⇒ Δφ0 = 4 80 cm ⎝ ⎠

(c) When side by side, with Δx = 0, the phase difference is Δφ = Δφ0 = 3π/4 rad. The amplitude of the superposition of the two waves is 3π ⎛ Δφ ⎞ a = 2a cos ⎜ = 0.77a ⎟ = 2a cos 8 ⎝ 2 ⎠

21.61. Model: Interference occurs according to the difference between the phases of the two waves. Visualize:

Solve:

(a) The phase difference between the sound waves from the two speakers is Δφ = 2π

Δx

λ

+ Δφ0

We have a maximum intensity when Δx = 0.50 m and Δx = 0.90 m. This means

2π

( 0.50 m ) + Δφ λ

0

= 2mπ rad

⎛ 0.90 m ⎞ 2π ⎜ ⎟ + Δφ0 = 2 ( m + 1)π rad ⎝ λ ⎠

Taking the difference of the above two equations, vsound 340 m/s ⎛ 0.40 m ⎞ 2π ⎜ = = 850 Hz ⎟ = 2π ⇒ λ = 0.40 m ⇒ f = λ 0.40 m ⎝ λ ⎠

(b) Using again the equations that correspond to constructive interference,

π ⎛ 0.50 m ⎞ 2π ⎜ ⎟ + Δφ0 = 2mπ rad ⇒ Δφ0 = φ20 − φ10 = − rad 2 ⎝ 0.40 m ⎠ We have taken m = 1 in the last equation. This is because we always specify phase constants in the range –π rad to π rad (or 0 rad to 2π rad). m = 1 gives − 12 π rad (or equivalently, m = 2 will give 32 π rad).

21.62. Model: Constructive or destructive interference occurs according to the phases of the two waves. Solve:

The phase difference between the sound waves from the two speakers is

Δφ = 2π

Δx

λ

+ Δφ0

With no delay between the two signals, Δφ0 = 0 rad and

Δφ =

2π ( 2.0 m ) ⎛ 340 Hz ⎞ = 2π ( 2.0 m ) ⎜ ⎟ = 4π rad v f ⎝ 340 m/s ⎠

According to Equation 21.22, this corresponds to constructive interference. A delay of 1.47 ms corresponds to an inherent phase difference of ⎛ 2π ⎞ Δφ 0 = ⎜ ⎟ (1.47 ms ) rad = ⎝ T ⎠

( 2π f )(1.47 ms ) rad = 2π (1.47 ms )( 340 Hz ) rad = π

The phase difference Δφ between the signals is then ⎛ Δx ⎞ Δφ = 2π ⎜ ⎟ + Δφ0 = 4π rad + π rad = 5π rad ⎝ λ ⎠

Thus, the interference along the x-axis will be perfect destructive.

rad

21.63. Model: Reflection is maximized for constructive interference of the two reflected waves, but minimized for destructive interference. Solve: (a) Constructive interference of the reflected waves occurs for wavelengths given by Equation 21.32:

λm =

2nd 2 (1.42 )( 500 nm ) (1420 nm ) = = m m m

Thus, λ1 = 1420 nm, λ2 = 12 (1420 nm ) = 710 nm, λ3 = 473 nm, λ4 = 355 nm, … Only the wavelength of 473

nm is in the visible range. (b) For destructive interference of the reflected waves,

λ=

2 (1.42 )( 500 nm ) 1420 nm 2nd = = m − 12 m − 12 m − 12

Thus, λ1 = 2 × 1420 nm = 2840 nm, λ2 =

2 3

(1420 nm ) = 947 nm, λ3 = 568 nm, λ4 = 406 nm, …

The wavelengths

of 406 nm and 568 nm are in the visible range. (c) Beyond the limits 430 nm and 690 nm the eye’s sensitivity drops to about 1 percent of its maximum value. The reflected light is enhanced in blue (473 nm). The transmitted light at mostly 568 nm will be yellowish green.

21.64. Solve: (a) The intensity of reflected light from the uncoated glass is I 0 = ca 2 , where a is the amplitude of the reflected light. We will assume that the amplitude of the reflected light from both the bottom and the top of the coated film is a. The interference of the two reflected waves determines the amplitude of the resultant wave which is given by A = 2a cos ( Δφ 2 )

where

Δφ = 2π

Δx

λcoat

+ Δφ0

With Δφ0 = 0 rad, Δx = 2d , and λcoat = λair n , we have Δφ =

⇒ A = 2a cos

2π ( 2d ) 4π dn 4π ( 92 nm )(1.39 ) 1607 nm + 0 rad = = = λair n λ λ λ

803.5 nm

λ

Iλ ⎛ 803.5 nm ⎞ 2 ⎛ 803.5 nm ⎞ ⇒ I λ = cA2 = c 4a 2 cos 2 ⎜ ⎟ ⇒ = 4cos ⎜ ⎟ λ λ I0 ⎝ ⎠ ⎝ ⎠

(b) The values of ( I λ I 0 ) at λ = 400, 450, 500, 550, 600, 650, and 700 nm are 0.719, 0.182, 0.005, 0.048, 0.211,

0.431, and 0.674, respectively. (c)

21.65. Model: Reflection is minimized when the two reflected waves interfere destructively. Solve:

Equation 21.2 gives the condition for perfect destructive interference between the two waves:

Δφ = 2π

Δx

λ

+ Δφ0 = 2 ( m + 12 )π rad

The wavelength of the sound is

λ=

v 343 m/s = = 0.2858 m f 1200 Hz

Let d be the separation between the mesh and the wall. Substituting Δφ0 = 0 rad, Δx = 2d , m = 0, and the above value for the wavelength, 2π ( 2d ) 0.2858 m + 0 rad = π rad ⇒ d = = 0.0715 m = 7.15 cm 4 0.2858 m

21.66. Model: A light wave that reflects from a boundary at which the index of refraction increases has a phase shift of π rad. Solve: (a) Because nfilm > nair, the wave reflected from the outer surface of the film (called 1) is inverted due to the phase shift of π rad. The second reflected wave does not go through any phase shift of π rad because the index of refraction decreases at the boundary where this wave is reflected, which is on the inside of the soap film. We can write for the phases

φ1 = kx1 + φ10 + π rad

φ2 = kx2 + φ20 + 0 rad

⇒ Δφ = φ2 − φ1 = k ( x2 − x1 ) + (φ20 − φ10 ) − π rad = k Δx + Δφ0 − π rad = k Δx − π rad Δφ0 = 0 rad because the sources are identical. For constructive interference, ⎛ 2π ⎞ Δφ = 2mπ rad ⇒ k Δx − π rad = 2mπ rad ⇒ ⎜ ⎟ ( 2d ) = ( 2m + 1) π rad ⎝ λfilm ⎠ ⇒ λfilm =

λC n

=

2d 2nd 2.66d m = 0, 1, 2, 3, … ⇒ λC = = 1 1 m+ 2 m + 2 m + 12

(b) For m = 0 the wavelength for constructive interference is

( 2.66 )( 390 nm ) = 2075 nm ( 12 ) λC = 415 nm ( ~violet ) . Red and violet together give a purplish color.

λC = For m = 1 and 2, λC = 692 nm ( ~red ) and

21.67. Model: The two radio antennas are sources of in-phase, circular waves. The overlap of these waves causes interference. Visualize:

Solve: Maxima occur along lines such that the path difference to the two antennas is Δr = mλ . The 750 MHz = 7.50 × 108 Hz wave has a wavelength λ = c f = 0.40 m. Thus, the antenna spacing d = 2.0 m is exactly 5λ. The maximum possible intensity is on the line connecting the antennas, where Δr = d = 5λ . So this is a line of maximum intensity. Similarly, the line that bisects the two antennas is the Δr = 0 line of maximum intensity. In between, in each of the four quadrants, are four lines of maximum intensity with Δr = λ, 2λ, 3λ, and 4λ. Although we have drawn a fairly accurate picture, you do not need to know precisely where these lines are located to know that you have to cross them if you walk all the way around the antennas. Thus, you will cross 20 lines where Δr = mλ and will detect 20 maxima.

21.68. Model: The changing sound intensity is due to the interference of two overlapped sound waves. Solve: Minimum intensity implies destructive interference. Destructive interference occurs where the path length difference for the two waves is Δr = ( m + 12 ) λ . We have assumed Δφ0 = 0 rad for two speakers playing

“exactly the same” tone. The wavelength of the sound is λ = vsound f = ( 343 m/s ) 686 Hz = 0.500 m . Consider a

point that is a distance x in front of the top speaker. Let r1 be the distance from the top speaker to the point and r2 the distance from the bottom speaker to the point. We have r2 = x 2 + ( 3 m )

r1 = x

2

Destructive interference occurs at distances x such that

Δr = x 2 + 9 m 2 − x = ( m + 12 ) λ To solve for x, isolate the square root on one side of the equation and then square: x 2 + 9 m = ⎡⎣ x + ( m + 12 ) λ ⎤⎦ = x 2 + 2 ( m + 12 ) λ x + ( m + 12 ) λ 2 ⇒ x = 2

2

9 m − ( m + 12 ) λ 2 2

2 ( m + 12 ) λ

Evaluating x for different values of m: m 0 1 2 3

x (m) 17.88 5.62 2.98 1.79

Because you start at x = 2.5 m and walk away from the speakers, you will only hear minima for values x > 2.5 m. Thus, to correct significant figures, minima will occur at distances of 3.0 m, 5.6 m, and 18 m.

21.69. Model: The changing sound intensity is due to the interference of two overlapped sound waves. Visualize:

The listener moving relative to the speakers changes the phase difference between the waves.

Solve: (a) Initially when you are at P, equidistant from the speakers, you hear a sound of maximum intensity. This implies that the two speakers are in phase (Δφ0 = 0). However, on moving to Q you hear a minimum of sound intensity implying that the path length difference from the two speakers to Q is λ /2. Thus, 1 2

λ = Δr =

( r1 )

2

+ ( 5.0 m ) − r1 = 2

(12.0 m )

⇒ λ = 2.0 m ⇒ f =

v

λ

=

2

+ ( 5.0 m ) − 12.0 m = 1.0 m 2

340 m/s = 170 Hz 2.0 m

(b) At Q, the condition for perfect destructive interference is

Δφ =

2π ( Δr )

λ

+ 0 rad = 2 ( m − 12 )π rad ⇒

⇒ f = ( m − 12 )

2πΔr = 2 ( m − 12 )π rad v f

v ⎛ 340 m/s ⎞ = ( m − 12 ) ⎜ ⎟ Δr ⎝ 1.0 m ⎠

For m = 1, 2, and 3, f1 = 170 Hz, f 2 = 510 Hz, and f 3 = 850 Hz.

21.70. Model: The amplitude is determined by the interference of the two waves. Visualize:

Solve:

The amplitude of the sound wave is A = 2a cos ( 12 Δφ ) . With Δφ0 = 0 rad, the phase difference between

the waves is Δφ = φ2 − φ1 = 2π

Δr

λ

= 2π

Δr ⎛ πΔr ⎞ ⇒ A = 2a cos ⎜ ⎟ 2.0 m ⎝ 2.0 m ⎠

At the coordinates (0.0 m, 0.0 m), Δr = 0 m, so A = 2a. At the coordinates (0.0 m, 0.5 m),

Δr =

( 3.0 m )

2

+ ( 2.5 m ) − 2

( 3.0 m )

2

+ (1.5 m ) = 0.551 m ⇒ A = 2a cos 2

( 0.551 m )π 2.0 m

= 1.30a

At the coordinates (0.0 m, 1.0 m), Δr =

( 3.0 m )

2

+ ( 3.0 m ) − 2

( 3.0 m )

2

⎛ (1.08 m )π ⎞ 2 + (1.0 m ) = 1.08 m ⇒ A = 2a cos ⎜ ⎟ = 0.25a ⎝ 2.0 m ⎠

At the coordinates (0.0 m, 1.5 m), Δr = 1.568 m and A = 1.56a. At the coordinates (0.0 m, 2.0 m), Δr = 2.0 m and A = 2a.

21.71. Model: The two radio transmitters are sources of out-of-phase, circular waves. The overlap of these waves causes interference. Visualize:

Solve:

The phase difference of the waves at point P is given by

Δφ = 2π Δr =

( 3000 m )

2

+ ( 85 m ) − 2

Δr

λ

+ Δφ0

( 3000 m )

2

+ ( 35 m ) = 0.99976 m 2

The intensity at P is a maximum. Using m = 1 for the first maximum, and Δφ0 = π rad since the transmitters are out of phase, the condition for constructive interference is Δφ = 2mπ = 2π . Thus, 2π rad = 2π

Δr

λ

+ π rad ⇒ λ = 2Δr = 2 ( 0.99976 m ) ⇒ f =

c

λ

=

3 × 108 m/s = 150 MHz 2 ( 0.99976 m )

21.72. Model: The two radio antennas are sources of in-phase waves. The overlap of these waves causes interference. Visualize:

Solve:

(a) The phase difference of the two waves at point P is given by

Δφ = 2π

Δr

λ

+ Δφ0

Δr =

(800 m )

2

+ ( 650 m ) − 2

(800 m )

2

+ ( 550 m ) = 59.96 m 2

The wavelength of the radio wave is

λ=

c 3.0 × 108 m/s = = 100 m f 3.0 × 106 m

Since the sources are identical, Δφ0 = 0 rad. The phase difference at P due to the two waves is

⎛ 59.96 m ⎞ Δφ = 2π ⎜ ⎟ + 0 rad = 1.2π rad ⎝ 100 m ⎠ (b) Since Δφ = 1.20π = 0.6 ( 2π ) , which is neither m2π nor ( m + 12 ) 2π , the interference at P is somewhere in

between maximum constructive and perfect destructive. (c) At a point 10 m further north we have Δr =

( 800 m )

2

+ ( 660 m ) − 2

( 800 m )

2

+ ( 560 m ) = 60.58 m

⎛ 60.58 m ⎞ ⇒ Δφ = 2π ⎜ ⎟ + 0 rad = 1.21π rad = ⎝ 100 m ⎠

2

( 0.605) 2π

Because the phase difference is increasing as you move north, you are moving from a destructive interference condition Δφ = ( m + 12 ) 2π with m = 0 toward a constructive interference condition Δφ = m ( 2π ) with m = 1. The signal strength will therefore increase.

21.73. Model: The amplitude is determined by the interference of the two waves.

Solve: (a) We have three identical loudspeakers as sources. Δr between speakers 1 and 2 is 1.0 m and λ = 2.0 m. Thus Δr = 12 λ , which gives perfect destructive interference for in-phase sources. That is, the interference of the waves from loudspeakers 1 and 2 is perfect destructive, leaving only the contribution due to speaker 3. Thus the amplitude is a. (b) If loudspeaker 2 is moved away by one-half of a wavelength or 1.0 m, then all three waves will reach you in phase. The amplitude of the superposed waves will therefore be maximum and equal to A = 3a. (c) The maximum intensity is I max = CA2 = 9Ca 2 . The ratio of the intensity to the intensity of a single speaker is

I max I single speaker

=

9Ca 2 =9 Ca 2

21.74. Model: The superposition of two slightly different frequencies gives rise to beats. Solve:

The third harmonic of note A and the second harmonic of note E are

f3A = 3 f1A = 3( 440 Hz ) = 1320 Hz

f 2E = 2 f1E = 2 ( 659 Hz ) = 1318 Hz

⇒ f 3A − f 2E = 1320 Hz − 1318 Hz = 2 Hz (b) The beat frequency between the first harmonics is f1E − f1A = 659 Hz − 440 Hz = 219 Hz The beat frequency between the second harmonics is f2E − f2A = 1318 Hz − 880 Hz = 438 Hz

The beat frequency between f3A and f2E is 2 Hz. It therefore emerges that the tuner looks for a beat frequency of 2 Hz. (c) If the beat frequency is 4 Hz, then the second harmonic frequency of the E string is

f 2E = 1320 Hz − 4 Hz = 1316 Hz ⇒ f1E =

1 2

(1316 Hz )

= 658 Hz

Note that the second harmonic frequency of the E string could also be f 2E = 1320 Hz + 4 Hz = 1324 Hz ⇒ f1E = 662 Hz

This higher frequency can be ruled out because the tuner started with low tension in the E string and we know that T ⇒ f ∝ T vstring = λ f =

μ

21.75. Model: The superposition of two slightly different frequencies creates beats. Solve:

(a) The wavelength of the sound initially created by the flutist is

λ=

342 m/s = 0.77727 m 440 Hz

When the speed of sound inside her flute has increased due to the warming up of the air, the new frequency of the A note is f′=

346 m/s = 445 Hz 0.77727 m

Thus the flutist will hear beats at the following frequency: f ′ − f = 445 Hz − 440 Hz = 5 beats/s

Note that the wavelength of the A note is determined by the length of the flute rather than the temperature of air or the increased sound speed. (b) The initial length of the flute is L = 12 λ = 12 ( 0.77727 m ) = 0.3886 m. The new length to eliminate beats needs

to be L′ =

λ′ 2

1 ⎛ v′ ⎞ 1 ⎛ 346 m/s ⎞ = ⎜ ⎟= ⎜ ⎟ = 0.3932 m 2 ⎝ f ⎠ 2 ⎝ 440 Hz ⎠

Thus, she will have to extend the “tuning joint” of her flute by 0.3932 m − 0.3886 m = 0.0046 m = 4.6 mm

21.76. Solve: (a) Yvette’s speed is the width of the room divided by time. This means vY =

n ( 12 λ ) t

⇒

n 2vY = t λ

Note that λ is the distance between two consecutive antinodes, and n is the number of such half wavelengths that fill the entire width of the room. (b) Yvette observes a higher frequency f+ of the source she is moving toward and a lower frequency f− of the source she is receding from. If v is the speed of sound and f is the sound wave’s frequency, we have 1 2

⎛ v ⎞ f + = f ⎜1 + Y ⎟ v ⎠ ⎝

⎛ v ⎞ f − = f ⎜1 − Y ⎟ v ⎠ ⎝

The expression for the beat frequency is

⎛ v ⎞ f + − f − = f ⎜1 + Y ⎟ − v ⎠ ⎝

v vv 2v ⎛ v ⎞ f ⎜1 − Y ⎟ = 2 f Y = 2 Y = Y v v v λ λ ⎝ ⎠

(c) The answers to part (a) and (b) are the same. Even though you and Yvette have different perspectives, you should agree as to how many modulations per second she hears.

21.77. Model: The frequency of the loudspeaker’s sound in the back of the pick-up truck is Doppler shifted. As the truck moves away from you, its frequency is decreased. Solve: Because you hear 8 beats per second as the truck drives away from you, the frequency of the sound from the speaker in the pick-up truck is f − = 400 Hz − 8 Hz = 392 Hz. This frequency is

f− =

f0 vS 400 Hz ⇒1+ = = 1.020408 ⇒ vS = 7.0 m/s 343 m/s 392 Hz 1 + vS v

That is, the velocity of the source vS and hence the pick-up truck is 7.0 m/s.

21.78. Model: A stretched string under tension supports standing waves. Solve:

(a) The wave speed on a stretched string is

vstring =

T

μ

= fλ ⇒ f =

1 T

λ μ

The wavelength λ cannot change if the length of the string does not change. So, df 1 1 = dT λ μ

1 2

(T )

−1/ 2

=

1 2λ

1

μT

=

Δf ΔT 1 ⎛1 T ⎞ 1 = f ⇒ ⎜⎜ ⎟⎟ = f 2T 2T ⎝ λ μ ⎠ 2T

(b) Since there are 5 beats per second, Δf = 5 Hz =

f ΔT ΔT 10 Hz 10 Hz ⇒ = = = 0.020 = 2.0% 2 T T f 500 Hz

That is, an increase of 2.0% in the tension of one of the strings will cause 5 beats per second.

21.79. Model: The microphone will detect a loud sound only if there is a standing wave resonance in the tube. The sound frequency does not change, but changing the length of the tube can create a standing wave. Solve: The standing wave condition is f = 280 Hz = m

v 2L

m = 1, 2, 3,…

where L is the total length of the tube. When the slide is extended a distance s, the tube has two straight sides, each of length s + 80 cm, plus a semicircular bend of length 12 ( 2π r ) . The radius is r = 12 (10 cm ) = 5.0 cm. The

tube’s total length is

L = 2 ( s + 80 cm ) + 12 ( 2π × 5.0 cm ) = 175.7 cm + 2s = 1.757 m + 2s A standing wave resonance will be created if ⎡

⎤ v 343 m/s =m = 0.6125m ⎥ 2 ( 280 Hz ) ⎢⎣ 2 f ⎥⎦

[ L = 1.757 + 2s ] = ⎢ m

⇒ s = 0.3063m − 0.8785 meters We can tabulate the different extensions s that correspond to standing wave modes m = 1, m = 2, m = 3, and so on. m 1 2 3 4 5 6

s −0.572 m −0.266 m 0.040 m = 4.0 cm 0.347 m = 34.7 cm 0.653 m = 65.3 cm 1.959 m

Physically, the extension must be greater than 0 cm and less than 80 cm. Thus, the three slide extensions that create a standing wave resonance at 280 Hz are 4.0 cm, 35 cm, and 65 cm to two significant figures.

21.80. Model: The stretched wire is vibrating at its second harmonic frequency. Visualize: Let l be the full length of the wire, and L be the vibrating length of the wire. That is, L = ( 12 ) l .

Solve:

The wave speed on a stretched wire is

Ts

vwire =

μ

= fλ

The frequency f = 100 Hz and the wavelength λ = 12 l because it is a second harmonic wave. The tension Ts =

(1.25 kg)g because the hanging mass is in static equilibrium and μ = 1.00 × 10−3 kg/m. Substituting in these values, −3 2 kg/m ) l 2 ⎛ l ⎞ (1.00 × 10 g = 100 Hz ⇒ = 100 Hz = ( 2.00 m −1s −2 ) l 2 ( ) ( ) ⎜ ⎟ −3 2 (1.25 kg ) ⎝ 2⎠ (1.00 ×10 kg/m )

(1.25 kg ) g

To find l we can use the equation for the time period of a simple pendulum: l T2 ( 314 s/100 ) g = 0.250 s 2 g ⇒l = 2 g = ( ) 4π 4π 2 g 2

T = 2π

Substituting this expression for l into the equation for g, we get g = ( 2.000 m −1s −2 )( 0.025 s 2 ) g 2 ⇒ ( 0.125 m −1s 2 ) g 2 − g = 0 2

⇒ ⎡⎣( 0.125 m −1s 2 ) g − 1⎤⎦ g = 0 ⇒ g = 8.00 m/s 2 Assess:

A value of 8.0 m/s2 is reasonable for the information given in the problem.

21.81. Model: The steel wire is under tension and it vibrates with three antinodes. Solve: When the spring is stretched 8.0 cm, the standing wave on the wire has three antinodes. This means λ3 = 32 L and the tension TS in the wire is TS = k (0.080 m), where k is the spring constant. For this tension,

vwire =

TS

μ

⇒ f λ3 =

TS

μ

⇒ f =

3 k ( 0.08 m ) μ 2L

We will let the stretching of the spring be Δx when the standing wave on the wire displays two antinodes. This means λ2 = L and TS′ = kx. For the tension TS′, v′wire =

TS′

μ

⇒ f λ2 =

TS′

μ

⇒ f =

1 k Δx L μ

The frequency f is the same in the above two situations because the wire is driven by the same oscillating magnetic field. Now, equating the two frequency equations, 1 k Δx 3 k ( 0.080 m ) ⇒ Δx = 0.18 m = 18 cm = μ L μ 2L

21.82. Model: The frequency is Doppler shifted to higher values for a detector moving toward the source. The frequency is also shifted to higher values for a source moving toward the detector. Visualize:

Solve: (a) We will derive the formula in two steps. First, the object acts like a moving detector and “observes” a frequency that is given by f 0′ = f 0 (1 + v0 v ) . Second, as this moving object reflects (or acts as a “source” of −1 ultrasound waves), the frequency fecho as observed by the original source So is f echo = f 0′ (1 − v0 v ) . Combining

these two equations gives f echo =

f (1 + v0 v ) v + v0 f 0′ = 0 = f0 v − v0 1 − v0 v 1 − v0 v

(b) If v0 << v, then

⎛ v ⎞⎛ v ⎞ f echo = f 0 ⎜1 + 0 ⎟⎜1 − 0 ⎟ v ⎠⎝ v⎠ ⎝

−1

⎛ v ⎞⎛ v ⎞ ⎛ 2v ⎞ = f 0 ⎜ 1 + 0 ⎟⎜ 1 + 0 + … ⎟ = f 0 ⎜1 + 0 + … ⎟ v ⎠⎝ v v ⎝ ⎠ ⎝ ⎠

⇒ f beat = f echo − f 0 ≈

2v0 f0 v

(c) Using part (b) for the beat frequency, ⎛ 2v0 ⎞ 6 65 Hz = ⎜ ⎟ ( 2.40 × 10 Hz ) ⇒ v0 = 2.09 cm/s ⎝ 1540 m/s ⎠

(d) Assuming the heart rate is 90 beats per minute the angular frequency is

ω = 2π f = 2π (1.5 beats/s ) = 9.425 rad/s Using v0 = vmax = ω A , A=

T=

v0

ω

=

2.09 cm/s = 2.2 mm 9.425 rad/s

(1.50 + 1.60 )

2 ( 3.0 × 108 m/s )

× 0.010 m = 5.17 × 10−11 s

21.83. Solve: (a) The wavelengths of the standing wave modes are λm = ⇒ λ1 =

2 (10.0 m ) = 20.0 m 1

2L m = 1, 2, 3, …. m

λ2 =

2 (10.0 m ) = 10.0 m 2

λ3 =

2 (10.0 m ) = 6.67 m 3

The depth of the pool is 5.0 m. Clearly the standing waves with λ2 and λ3 are “deep water waves” because the 20 m depth is larger than one-quarter of the wavelength. The wave with λ1 barely qualifies to be a deep water standing wave.

(b) The wave speed for the first standing wave mode is

v1 =

g λ1 = 2π

( 9.8 m/s ) ( 20.0 m ) = 5.59 m/s 2

2π

Likewise, v2 = 3.95 m/s and v3 = 3.22 m/s . (c) We have v=

g λm g mg = f m λm ⇒ f m = = 2π 2πλm 4π L

Please note that m is the mode and not the mass. (d) The period of oscillation for the first standing wave mode is calculated as follows f1 = Likewise, T2 = 2.53 s and T3 = 2.07 s.

(1) ( 9.8 m/s 2 ) = 0.279 Hz 4π (10.0 m )

⇒ T1 = 3.58 s

21.84. Model: The overlap of the waves causes interference. Solve:

(a) The waves traveling to the left are

⎡ ⎛ x t ⎞⎤ D1 = a sin ⎢ 2π ⎜ − − ⎟ ⎥ ⎣ ⎝ λ T ⎠⎦

⎡ ⎛ ( x − L) t ⎞ ⎤ D2 = a sin ⎢ 2π ⎜ − − ⎟ + φ20 ⎥ λ T⎠ ⎣⎢ ⎝ ⎦⎥

The phase difference between the waves on the left side of the antenna is thus

⎛ ( x − L) t ⎞ L ⎛ x t⎞ ΔφL = φ2 − φ1 = 2π ⎜ − − ⎟ + φ20 − 2π ⎜ − − ⎟ = 2π + φ20 λ λ T⎠ ⎝ λ T⎠ ⎝ On the right side of the antennas, where x1 = x2 + L, the two waves are ⎡ ⎛ x t ⎞⎤ D1 = a sin ⎢ 2π ⎜ − ⎟ ⎥ ⎣ ⎝ λ T ⎠⎦

⎡ ⎛ ( x − L) t ⎞ ⎤ − ⎟ + φ20 ⎥ D2 = a sin ⎢ 2π ⎜ T⎠ ⎣⎢ ⎝ λ ⎦⎥

Thus, the phase difference between the waves on the right is

⎛ ( x − L) t ⎞ 2π L ⎛x t⎞ + φ20 ΔφR = φ2 − φ1 = 2π ⎜ − ⎟ + φ20 − 2π ⎜ − ⎟ = − λ T T λ λ ⎝ ⎠ ⎝ ⎠ We want to have destructive interference on the country side or on the left and constructive interference on the right. This requires ΔφL = φ20 +

2π L

λ

= 2π ( m + 12 ) ΔφR = φ20 −

2π L

λ

= 2π n

These are two simultaneous equations, and we can satisfy them both if L and φ20 are properly chosen. Subtracting the second equation from the first to eliminate φ20 , 4π L

λ

= 2π ( m + 12 − n ) ⇒ L = ( m + 12 − n )

λ 2

The smallest value of L that works is for n = m, in which case L = 14 λ . (b) From the ΔφR equation,

⎡ 2π ( 14 λ ) π 2π L π⎤ = φ20 − = φ20 − ⎥ = 2π n ⇒ φ20 = + 2π n rad ⎢φ20 − 2 λ λ 2 ⎣ ⎦ Adding integer multiples of 2π to the phase constant doesn’t really change the wave, so the physically significant phase constant is for n = 0, namely φ20 = 12 π rad.

(c) We have φ20 = 12 π rad = 14 ( 2π ) . If the wave from antenna 2 was delayed by one full period T, it would shift the wave by

one full cycle. We would describe this by a phase constant of 2π rad. So a phase constant of

1 4

( 2π )

rad can be achieved by

delaying the wave by Δt = T . 1 4

(d) A wave with frequency f = 1000 kHz = 1.00 × 106 Hz has a period T = 1.00 × 10−6 s = 1.00 μ s and wavelength

λ = c / f = 300 m. So this broadcast scheme will work if the antennas are spaced L = 75 m apart and if the broadcast from antenna 2 is delayed by Δt = 0.25 μ s = 250 ns.

21-1

22.1. Model: Two closely spaced slits produce a double-slit interference pattern.

Visualize: The interference pattern looks like the photograph of Figure 22.3(b). It is symmetrical with the m = 2 fringes on both sides of and equally distant from the central maximum. Solve: The bright fringes occur at angles θm such that

d sin θ m = mλ ⇒ sin θ 2 =

2 ( 500 ×10−9 m )

( 50 ×10

−6

m)

m = 0, 1, 2, 3, …

= 0.02 ⇒ θ 2 = 0.020 rad = 0.020 rad ×

180°

π rad

= 1.15°

22.2. Model: Two closely spaced slits produce a double-slit interference pattern.

Visualize: The interference pattern looks like the photograph of Figure 22.3(b). It is symmetrical, with the m = 2 fringes on both sides of and equally distant from the central maximum. Solve: The two paths from the two slits to the m = 2 bright fringe differ by Δr = r2 − r1 , where

Δr = mλ = 2λ = 2 ( 500 nm ) = 1000 nm Thus, the position of the m = 2 bright fringe is 1000 nm farther away from the more distant slit than from the nearer slit.

22.3. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The bright fringes are located at positions given by Equation 22.4, d sin θ m = mλ . For the m = 3 bright

orange fringe, the interference condition is d sinθ 3 = 3 ( 600 ×10−9 m ) . For the m = 4 bright fringe the condition

is d sinθ 4 = 4λ . Because the position of the fringes is the same, d sin θ 3 = d sin θ 4 = 4λ = 3 ( 600 × 10−9 m ) ⇒ λ =

3 4

( 600 × 10

−9

m ) = 450 nm

22.4. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The formula for fringe spacing is L λL L Δy = ⇒ 1.8 × 10−3 m = ( 600 × 10−9 m ) ⇒ = 3000 d d d The wavelength is now changed to 400 nm, and L d , being a part of the experimental setup, stays the same. Applying the above equation once again, λL Δy = = ( 400 ×10−9 m ) ( 3000 ) = 1.2 mm d

22.5. Visualize: The fringe spacing for a double slit pattern is Δy = λ = 600 nm. We also see from the figure that Δy = 13 cm. Solve:

d

Solve the equation for d. d=

Assess:

λL

λL Δy

=

0.36 mm is a typical slit spacing.

(600 × 10−9 m)(2.0 m) = 0.36 mm 1 × 10 −2 m 3

. We are given L = 2.0 m and

22.6. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The fringe spacing is λL λ L (589 × 10−9 m)(150 × 10−2 m) Δy = ⇒d = = = 0.22 mm d Δy 4.0 × 10−3 m

22.7. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The dark fringes are located at positions given by Equation 22.9: λL m = 0, 1, 2, 3, … ym′ = ( m + 12 ) d

⇒ y5′ − y1′ = ( 5 + 12 )

λL d

− (1 + 12 )

λL d

⇒ 6.0 × 10−3 m =

4λ (60 × 10−2 m) ⇒ λ = 500 nm 0.20 × 10−3 m

22.8. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: In a span of 12 fringes, there are 11 gaps between them. The formula for the fringe spacing is Δy =

Assess:

⎛ 52 × 10−3 m ⎞ (633 × 10 −9 m)(3.0 m) ⇒ d = 0.40 mm ⇒ ⎜ ⎟= d 11 d ⎝ ⎠

λL

This is a reasonable distance between the slits, ensuring d L = 1.34 ×10−4 << 1 .

22.9. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: The bright constructive-interference fringes are given by Equation 22.15:

d sin θ m = mλ ⇒ sin θ1 =

Likewise, sin θ 2 = 0.110 and θ 2 = 6.3°.

m = 0, 1, 2, … −9

(1)(550 × 10 m) = 0.055 ⇒ θ1 = 3.2° (1.0 × 10−2 m) /1000

22.10. Model: A diffraction grating produces a series of constructive-interference fringes at values of θ m determined by Equation 22.15. Solve: We have d sin θ m = mλ

m = 0, 1, 2, 3, … ⇒ d sin 20.0° = 1λ and d sin θ 2 = 2λ

Dividing these two equations,

sin θ 2 = 2sin 20.0° = 0.6840 ⇒ θ 2 = 43.2°

22.11. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: The bright constructive-interference fringes are given by Equation 22.15:

d sin θ m = mλ ⇒ d =

mλ (2)(600 × 10−9 m) = = 1.89 × 10−6 m sin θ m sin(39.5°)

The number of lines in per millimeter is (1 × 10−3 m) (1.89 × 10−6 m) = 530.

22.12. Model: A diffraction grating produces an interference pattern. Visualize:

Solve:

The interference pattern looks like the diagram in Figure 22.8.

The bright fringes are given by Equation 22.15:

d sin θ m = mλ

m = 0, 1, 2, 3, … ⇒ d sinθ1 = (1) λ ⇒ d = λ sin θ1

The angle θ1 can be obtained from geometry as follows: tanθ1 =

( 0.32 m ) 2.0 m

2

= 0.080 ⇒ θ1 = tan–1 (0.080) 4.57°

Using sin θ1 = sin 4.57° = 0.07968 , d=

633 ×10−9 m = 7.9 μ m 0.07968

22.13. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure 22.8. Solve: The bright interference fringes are given by

d sin θ m = mλ The slit spacing is d = 1 mm 500 = 2.00 ×10

−6

m and m = 1. For the red and blue light,

⎛ 656 ×10−9 m ⎞ ⎟ = 19.15° −6 ⎝ 2.00 ×10 m ⎠

θ1 red = sin −1 ⎜

m = 0, 1, 2, 3, …

⎛ 486 ×10−9 m ⎞ ⎟ = 14.06° −6 ⎝ 2.00 ×10 m ⎠

θ1 blue = sin −1 ⎜

The distance between the fringes, then, is Δy = y1 red − y1 blue where y1 red = (1.5 m ) tan19.15° = 0.521 m y1 blue = (1.5 m ) tan14.06° = 0.376 m

So, Δy = 0.145 m = 14.5 cm.

22.14. Model: Assume the screen is centered behind the slit. We actually want to solve for m, but given the other data, it is unlikely that we will get an integer from the equations for the edge of the screen, so we will have to truncate our answer to get the largest order fringe on the screen. Visualize: Refer to Figure 22.7. Use Equation 22.15: d sin θ m = mλ , and Equation 22.16: ym = L tanθ m . We 1 mm. As mentioned above, we are not guaranteed that a bright are given λ = 510 nm, L = 2.0 m, and d = 500

fringe will occur exactly at the edge of the screen, but we will kind of assume that one does and set ym = 1.0 m; if we do not get an integer for m then the fringe was not quite at the edge of the screen and we will truncate our answer to get an integer m. Solve: Solve Equation 22.16 for θ m and insert it in Equation 22.15.

θ m = tan −1

ym L

Solve Equation 22.15 for m. m=

d

λ

sin θ m =

1 mm 1. 0 m ⎞ y ⎞ ⎛ ⎛ sin ⎜ tan −1 m ⎟ = 500 sin ⎜ tan −1 ⎟ = 1.8 2.0 m ⎠ λ ⎝ L ⎠ 510 nm ⎝

d

Indeed, we did not get an integer, so truncate 1.8 to get m = 1. This means we will see three fringes, one for m = 0, and one on each side for m = ±1. Assess: Our answer fits with the statement in the text: “Practical gratings, with very small values for d , display only a few orders.”

22.15. Model: A narrow single slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The minima occur at positions yp = p So Δy = y2 − y1 =

Lλ a

−9 2λ L 1λ L λ L λ L ( 633 ×10 m ) (1.5 m ) − = ⇒ a= = = 2.0 ×10−4 m = 0.20 mm a a a Δy 0.00475 m

22.16. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The width of the central maximum for a slit of width a = 200λ is

w=

2λ L 2λ ( 2.0 m ) = = 20 mm 200λ a

22.17. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: Angle θ = 0.70° = 0.0122 rad is a small angle (<< 1 rad). Thus we use Equation 22.20 to find the wavelength of light. The angles of the minima of intensity are

θp = p ⇒λ =

aθ p p

=

λ

p = 1, 2, 3, κ a ( 0.10 ×10−3 m ) ( 0.0122 rad ) 2

= 610 nm

22.18. Visualize: Use Equation 22.22: w = the figure that w = 1.0 cm. Solve: Solve the equation for a. a= Assess:

2λ L . We are given λ = 600 nm and L = 2.0 m. We see from a

2λ L 2(600 × 10−9 m)(2.0 m) = = 0.24 mm w 0.010 m

0.24 mm is in the range of typical slit widths.

22.19. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The width of the central maximum for a slit of width a = 200λ is

w=

−9 2λ L 2 ( 500 ×10 m ) ( 2.0 m ) = = 0.0040 m = 4.0 mm 0.0005 m a

22.20. Model: The spacing between the two buildings is like a single slit and will cause the radio waves to be diffracted. Solve: Radio waves are electromagnetic waves that travel with the speed of light. The wavelength of the 800 MHz waves is

λ=

3 ×108 m/s = 0.375 m 800 ×106 Hz

To investigate the diffraction of these waves through the spacing between the two buildings, we can use the general condition for complete destructive interference: a sin θ p = pλ (p = 1, 2, 3, …) where a is the spacing

between the buildings. Because the width of the central maximum is defined as the distance between the two p = 1 minima on either side of the central maximum, we will use p = 1 and obtain the angular width Δθ = 2θ1 from

⎛λ⎞ ⎛ 0.375 m ⎞ a sin θ1 = λ ⇒ θ1 = sin −1 ⎜ ⎟ = sin −1 ⎜ ⎟ = 1.43° a ⎝ ⎠ ⎝ 15 m ⎠ Thus, the angular width of the wave after it emerges from between the buildings is Δθ = 2 (1.43° ) = 2.86° ≈ 2.9°.

22.21. Model: The crack in the cave is like a single slit that causes the ultrasonic sound beam to diffract. Visualize:

Solve:

The wavelength of the ultrasound wave is

λ=

340 m/s = 0.0113 m 30 kHz

Using the condition for complete destructive interference with p = 1,

⎛ 0.0113 m ⎞ a sin θ1 = λ ⇒ θ1 = sin −1 ⎜ ⎟ = 2.165° ⎝ 0.30 m ⎠ From the geometry of the diagram, the width of the sound beam is

w = 2 y1 = 2 (100 m × tanθ1 ) = 200 m × tan 2.165° = 7.6 m Assess: The small-angle approximation is almost always valid for the diffraction of light, but may not be valid for the diffraction of sound waves, which have a much larger wavelength.

22.22. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: The width of the central maximum for a circular aperture of diameter D is

w=

2.44λ L (2.44)(500 × 10−9 m)(2.0 m) = = 4.9 mm D 0.50 × 10−3 m

22.23. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: According to Equation 22.23, the angle that locates the first minimum in intensity is

θ1 =

1.22λ 1.22(2.5 × 10−6 m) = = 0.01525 rad = 0.874° D 0.20 × 10−3 m

These should be rounded to 0.015 rad = 0.87°.

22.24. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: From Equation 22.24, the diameter of the circular aperture is

D=

2.44λ L 2.44(633 × 10−9 m)(4.0 m) = = 0.25 mm w 2.5 × 10−2 m

22.25. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: From Equation 22.24, L=

Dw (0.12 × 10−3 m)(1.0 × 10−2 m) = = 78 cm 2.44λ 2.44(633 × 10−9 m)

22.26. Model: An interferometer produces a new maximum each time L2 increases by 12 λ causing the pathlength difference Δr to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the wavelength is 2ΔL2 2(100 × 10−6 m) λ= = = 4.0 × 10−7 m = 400 nm Δm 500

22.27. Model: An interferometer produces a new maximum each time L2 increases by 12 λ causing the pathlength difference Δr to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the number of fringe shifts is

Δm =

2ΔL2

λ

=

2(1.00 × 10−2 m) = 30,467 656.45 × 10−9 m

22.28. Model: An interferometer produces a new maximum each time L2 increases by 12 λ causing the pathlength difference Δr to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the distance the mirror moves is

ΔL2 =

Δmλ (33,198)(602.446 × 10−9 m) = = 0.0100000 m = 1.00000 cm 2 2

Assess: Because the wavelength is known to 6 significant figures and the fringes are counted exactly, we can determine ΔL to 6 significant figures.

22.29. Model: An interferometer produces a new maximum each time L2 increases by 12 λ causing the pathlength difference Δr to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: For sodium light of the longer wavelength (λ1) and of the shorter wavelength (λ2), ΔL = m

λ1 2

ΔL = ( m + 1)

λ2 2

We want the same path difference 2(L2 − L1) to correspond to one extra wavelength for the sodium light of shorter wavelength (λ2) . Thus, we combine the two equations to obtain: m

λ1 2

= ( m + 1)

λ2 2

⇒ m ( λ1 − λ2 ) = λ2 ⇒ m =

λ2

λ1 − λ2

=

589.0 nm = 981.67 ≅ 982 589.6 nm − 589.0 nm

Thus, the distance by which M2 is to be moved is

ΔL = m

λ1 2

⎛ 589.6 nm ⎞ = 982 ⎜ ⎟ = 0.2895 mm 2 ⎝ ⎠

22.30.

Model: Two closely spaced slits produce a double-slit interference pattern with the intensity graph looking like Figure 22.3(b). The intensity pattern due to a single slit diffraction looks like Figure 22.14. Both the spectra consist of a central maximum flanked by a series of secondary maxima and dark fringes. Solve: (a) The light intensity shown in Figure P22.30 corresponds to a double-slit aperture. This is because the fringes are equally spaced and the decrease in intensity with increasing fringe order occurs slowly. (b) From Figure P22.30, the fringe spacing is Δy = 1.0 cm = 1.0 × 10−2 m. Therefore,

Δy =

λL

d λ L (6.00 ×10−9 m)(2.5 m) ⇒d = = = 0.15 mm 0.010 m Δy

22.31. Model: Two closely spaced slits produce a double-slit interference pattern with the intensity graph looking like Figure 22.3(b). The intensity pattern due to a single slit diffraction looks like Figure 22.14. Both the spectra consist of a central maximum flanked by a series of secondary maxima and dark fringes. Solve: (a) The light intensity shown in Figure P22.31 corresponds to a single slit aperture. This is because the central maximum is twice the width and much brighter than the secondary maximum. (b) From Figure P22.31, the separation between the central maximum and the first minimum is y1 = 1.0 cm = 1.0 × 10−2 m. Therefore, using the small-angle approximation, Equation 22.21 gives the condition for the dark minimum: yp =

−9 pLλ Lλ ( 2.5 m ) ( 600 × 10 m ) ⇒a= = = 0.15 mm −2 1.0 ×10 m d y1

22.32. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference fringes are equally spaced on both sides of the central maximum. The interference pattern looks like Figure 22.3(b). Solve: In the small-angle approximation

Δθ = θ m +1 − θ m = (m + 1) Since d = 200λ , we have Δθ =

λ d

=

λ d

−m

λ d

1 rad = 0.286° 200

=

λ d

22.33. Model: Two closely spaced slits produce a double-slit interference pattern. Solve:

The light intensity of a double-slit interference pattern at a position y on the screen is

⎛πd I double = 4 I1 cos 2 ⎜ ⎝ λL

⎛ y ⎞ ⎞ y ⎟ = 4 I1 cos 2 ⎜ π ⎟ ⎠ ⎝ Δy ⎠

where Δy = λ L / d = 4.0 mm is the fringe spacing. Using this value for λ L d , we can find the position on the interference pattern where Idouble = I1 as follows:

π π π 4.0 ×10−3 m ⎛ ⎞ ⎛ ⎞ −1 ⎛ 1 ⎞ y cos y = = 1.3 mm 4 I1 cos 2 ⎜ y I ⇒ = = rad ⇒ = ⎜ ⎟ ⎜ ⎟ ⎟ 1 −3 −3 3 ⎝ 4.0 ×10 m ⎠ ⎝ 4.0 ×10 m ⎠ ⎝ 2⎠ 3

22.34. Solve: According to Equation 22.7, the fringe spacing between the m fringe and the m + 1 fringe is Δy = λ L d . Δy can be obtained from Figure P22.34. The separation between the m = 2 fringes is 2.0 cm implying that the separation between the two consecutive fringes is

Δy = 0.50 × 10−2 m = Assess:

λL d

⇒ L=

d Δy

λ

=

( 0.20 ×10

−3

1 4

(2.0 cm) = 0.50 cm. Thus,

m )( 0.50 ×10−2 m )

600 ×10−9 m

A distance of 167 cm from the slits to the screen is reasonable.

= 167 cm

22.35. Solve: According to Equation 22.7, the fringe spacing between the m fringe and the m + 1 fringe is Δy = λ L d . Δy can be obtained from Figure P22.34. Because the separation between the m = 2 fringes is 2.0 cm, two consecutive fringes are Δy = 14 ( 2.0 cm ) = 0.50 cm apart. Thus, Δy = 0.50 × 10−2 m =

λL d

⇒ λ=

−3 −2 d Δy ( 0.20 ×10 m )( 0.50 ×10 m ) = = 500 nm L 2.0 m

22.36. Solve: The intensity of light of a double-slit interference pattern at a position y on the screen is ⎛πd I double = 4 I1 cos 2 ⎜ ⎝ λL

⎞ y⎟ ⎠

where I1 is the intensity of the light from each slit alone. At the center of the screen, that is, at y = 0 m, I1 = 14 I double . From Figure P22.34, Idouble at the central maximum is 12 mW/m2. So, the intensity due to a single slit is I1 = 3 mW/m 2 .

22.37. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: 500 lines per mm on the diffraction grating gives a spacing between the two lines of d = 1 mm 500 =

(1×10−3 m) 500 = 2.0 ×10−6 m . The wavelength diffracted at angle θ m = 30° in order m is

λ=

−6 d sinθ m ( 2.0 ×10 m ) sin 30° 1000 nm = = m m m

We’re told it is visible light that is diffracted at 30°, and the wavelength range for visible light is 400–700 nm. Only m = 2 gives a visible light wavelength, so λ = 500 nm.

22.38. Model: Assume the screen is centered behind the slit. Visualize: Refer to Figure 22.7. Think carefully about the situation. The longest wavelength that will show three fringes on each side occurs when y3 = 0.50 m; but the shortest wavelength that will show three fringes on each side

occurs when we almost let the fourth fringe on the screen, i.e., when y4 = 0.50 m. Use Equation 22.15: d sin θ m = mλ , and Equation 22.16: ym = L tan θ m . We are given L = 1.0 m, and d = Solve:

1 200

mm.

Solve Equation 22.16 for θ m and insert it in Equation 22.15.

θ m = tan −1

ym L

Solve Equation 22.15 for λ.

λ=

d d y ⎞ ⎛ sin θ m = sin ⎜ tan −1 m ⎟ m m ⎝ L ⎠

λmax occurs when y3 = 0.50 m; and λmin occurs when y4 = 0.50 m.

Assess:

λmin =

d d y ⎞ ⎛ sin θ 4 = sin ⎜ tan −1 4 ⎟ = m m ⎝ L⎠

1 200

mm ⎛ −1 0.50 m ⎞ sin ⎜ tan ⎟ = 560 nm 4 1.0 m ⎠ ⎝

λmax =

d d y ⎞ ⎛ sin θ 3 = sin ⎜ tan −1 3 ⎟ = m m ⎝ L⎠

1 200

mm ⎛ −1 0.50 m ⎞ sin ⎜ tan ⎟ = 750 nm 3 1.0 m ⎠ ⎝

These are reasonable wavelengths, either in or close to the visible range.

22.39. Model: Each wavelength of light is diffracted at a different angle by a diffraction grating. Solve: Light with a wavelength of 501.5 nm creates a first-order fringe at y = 21.90 cm. This light is diffracted at angle ⎛ 21.90 cm ⎞ ⎟ = 23.65° ⎝ 50.00 cm ⎠

θ1 = tan −1 ⎜

We can then use the diffraction equation dsinθm = mλ, with m = 1, to find the slit spacing: d=

λ sin θ1

=

501.5 nm = 1250 nm sin(23.65°)

The unknown wavelength creates a first order fringe at y = 31.60 cm, or at angle

⎛ 31.60 cm ⎞ ⎟ = 32.29° ⎝ 50.00 cm ⎠

θ1 = tan −1 ⎜

With the split spacing now known, we find that the wavelength is

λ = d sinθ1 = (1250 nm)sin(32.29°) = 667.8 nm Assess: The distances to the fringes and the first wavelength were given to 4 significant figures. Consequently, we can determine the unknown wavelength to 4 significant figures.

22.40. Model: A diffraction grating produces an interference pattern like the diagram of Figure 22.8. Visualize:

Solve: (a) A key statement is that the lines are seen on the screen. This means that the light is visible light, in the range 400 nm–700 nm. We can determine where the entire visible spectrum falls on the screen for different values of m. We do this by finding the angles θm at which 400 nm light and 700 nm light are diffracted. We then use ym = L tan θ m to find their positions on the screen which is at a distance L = 75 cm. The slit spacing is

d = 1 mm 1200 = 8.333 × 10−7 m. For m = 1,

λ = 400 nm: θ1 = sin −1 ( 400 nm/d ) = 28.7° ⇒ y1 = 75 cm tan 28.7° = 41 cm λ = 700 nm: θ1 = sin −1 ( 700 nm/d ) = 57.1° ⇒ y1 = 75 cm tan 57.1° = 116 cm For m = 2,

λ = 400 nm: θ1 = sin −1 ( 2 ⋅ 400 nm/d ) = 73.8° ⇒ y2 = 75 cm tan 73.8° = 257 cm For the 700 nm wavelength at m = 2, θ 2 = sin −1 ( 2 ⋅ 700 nm/d ) = sin −1 (1.68 ) is not defined, so y2 → ∞ . We see that visible light diffracted at m = 1 will fall in the range 41 cm ≤ y ≤ 116 and that visible light diffracted at m = 2 will fall in the range y ≥ 257 cm. These ranges do not overlap, so we can conclude with certainty that the observed diffraction lines are all m = 1. (b) To determine the wavelengths, we first find the diffraction angle from the observed position by using

⎛ y⎞ ⎝ ⎠

⎛ ⎝

y

⎞ ⎠

θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ L 75 cm This angle is then used in the diffraction grating equation for the wavelength with m = 1, λ = d sin θ1 . Line

θ

λ

56.2 cm 65.9 cm 93.5 cm

36.85° 41.30° 51.27°

500 nm 550 nm 650 nm

22.41.

Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure 22.8. Solve: (a) A grating diffracts light at angles sin θ m = mλ d . The distance between adjacent slits is

d = 1 mm 600 = 1.667 ×10−6 m = 1667 nm. The angle of the m = 1 fringe is

⎛ 500 nm ⎞ ⎟ = 17.46° ⎝ 1667 nm ⎠

θ1 = sin −1 ⎜

The distance from the central maximum to the m = 1 bright fringe on a screen at distance L is Error! Objects cannot be created from editing field codes.

(Note that the small angle approximation is not valid for the maxima of diffraction gratings, which almost always have angles > 10°. ) There are two m = 1 bright fringes, one on either side of the central maximum. The distance between them is Δy = 2 y1 = 1.258 m ≈ 1.3 m. (b) The maximum number of fringes is determined by the maximum value of m for which sinθm does not exceed 1 because there are no physical angles for which sinθ > 1. In this case,

sinθ m =

mλ m ( 500 nm ) = d 1667 nm

We can see by inspection that m = 1, m = 2, and m = 3 are acceptable, but m = 4 would require a physically impossible sinθ 4 > 1. Thus, there are three bright fringes on either side of the central maximum plus the central maximum itself for a total of seven bright fringes.

22.42. Model: A diffraction grating produces a series of constructive-interference fringes at values of θ m that are determined by Equation 22.15. Solve: For the m = 3 maximum of the red light and the m = 5 maximum of the unknown wavelength, Equation 22.15 gives

d sinθ 3 = 3 ( 660 ×10−9 m ) d sin θ 5 = 5λunknown The m = 5 fringe and the m = 3 fringe have the same angular positions. This means θ5 = θ3. Dividing the two equations, 5λunknown = 3 ( 660 ×10−9 m ) ⇒ λunknown = 396 nm

22.43. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. The visible spectrum spans the wavelengths 400 nm to 700 nm. Solve: According to Equation 22.16, the distance ym from the center to the mth maximum is ym = L tan θ m . The

angle of diffraction is determined by the constructive-interference condition d sin θ m = mλ , where m = 0, 1, 2, 3, … The width of the rainbow for a given fringe order is thus w = yred − yviolet. The slit spacing is d=

1 mm 1.0 ×10−3 m = = 1.6667 ×10−6 m 600 600

For the red wavelength and for the m = 1 order, d sinθ1 = (1) λ ⇒ θ1 = sin −1

λ d

= sin −1

700 ×10−9 m = 24.83° 1.6667 ×10−6 m

From the equation for the distance of the fringe,

yred = L tanθ1 = ( 2.0 m ) tan ( 24.83° ) = 92.56 cm Likewise for the violet wavelength,

⎛ 400 ×10−9 m ⎞ ⎟ = 13.88° ⇒ yviolet = ( 2.0 m ) tan (13.88° ) = 49.42 cm −6 ⎝ 1.6667 ×10 m ⎠

θ1 = sin −1 ⎜

The width of the rainbow is thus 92.56 cm − 49.42 cm = 43.14 cm ≈ 43 cm .

22.44. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: (a) If blue light (the shortest wavelengths) is diffracted at angle θ, then red light (the longest wavelengths) is diffracted at angle θ + 30°. In the first order, the equations for the blue and red wavelengths are

sin θ =

λb d

d sin (θ + 30° ) = λr

Combining the two equations we get for the red wavelength, ⎛ λb2 ⎞ ⎛ λb ⎞ ⎟ 0.8660 + d ( 0.50 ) ⎜1 − 2 ⎟ ⎝d ⎠ ⎝ d ⎠

λr = d ( sin θ cos30° + cosθ sin 30° ) = d ( 0.8660sinθ + 0.50cosθ ) = d ⎜ ⇒ ( 0.50 ) d 1 −

λb2 d

2

= λr − 0.8660λb ⇒ ( 0.50 ) ( d 2 − λb2 ) = ( λr − 0.8660λb ) 2

2

2

⎛ λ − 0.8660λb ⎞ 2 ⇒ d= ⎜ r ⎟ + λb 0.50 ⎝ ⎠ Using λb = 400 × 10−9 m and λr = 700 ×10−9 m , we get d = 8.125 × 10−7 m. This value of d corresponds to 1 mm 1.0 ×10−3 m = = 1230 lines/mm d 8.125 ×10−7 m (b) Using the value of d from part (a) and λ = 589 ×10−9 m , we can calculate the angle of diffraction as follows: d sin θ1 = (1) λ ⇒ ( 8.125 ×10−7 m ) sin θ1 = 589 ×10−9 m ⇒ θ1 = 46.5°

22.45. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: An 800 line/mm diffraction grating has a slit spacing d = (1.0 × 10−3 m) 800 = 1.25 × 10−6 m. Referring to Figure P22.45, the angle of diffraction is given by

tan θ1 =

y1 0.436 m = = 0.436 ⇒ θ1 = 23.557° ⇒ sin θ1 = 0.400 L 1.0 m

Using the constructive-interference condition d sin θ m = mλ ,

λ=

d sin θ1 = (1.25 × 10−6 m ) ( 0.400 ) = 500 nm 1

We can obtain the same value of λ by using the second-order interference fringe. We first obtain θ2: tan θ 2 =

y2 0.436 m + 0.897 m = = 1.333 ⇒ θ 2 = 53.12° ⇒ sinθ 2 = 0.800 L 1.0 m

Using the constructive-interference condition,

λ=

−6 d sin θ 2 (1.25 ×10 m ) ( 0.800 ) = = 500 nm 2 2

Assess: Calculations with the first-order and second-order fringes of the interference pattern give the same value for the wavelength.

22.46. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: From Figure P22.45, tan θ1 =

0.436 m = 0.436 ⇒ θ1 = 23.557° ⇒ sin θ1 = 0.400 1.0 m

Using the constructive-interference condition d sin θ m = mλ ,

d sin 23.557° = (1) ( 600 ×10−9 m ) ⇒ d =

600 ×10−9 m = 1.50 × 10−6 m sin ( 23.557° )

Thus, the number of lines per millimeter is 1.0 ×10−3 m = 670 lines/mm 1.50 ×10−6 m Assess: The same answer is obtained if we perform calculations using information about the second-order bright constructive-interference fringe.

22.47. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: The dark fringes are located at

yp =

pλ L p = 0, 1, 2, 3, … a

The locations of the first and third dark fringes are y1 =

λL a

y3 =

3λ L a

Subtracting the two equations,

( y3 − y1 ) =

2 ( 589 ×10−9 m ) ( 0.75 m ) 2λ L 2λ L ⇒ a= = = 0.12 mm a y3 − y1 7.5 ×10−3 m

22.48. Visualize: The relationship between the diffraction grating spacing d, the angle at which a particular order of constructive interference occurs θ m , the wavelength of the light, and the order of the constructive interference m is d sin θ m = mλ . Also note N = 1/d . Solve: The first order diffraction angle for green light is

θ1 = sin −1 (λ/d ) = sin −1 (5.5 × 10−7 m/ 2.0 × 10−6 m) = sin −1 (0.275) = 0.278 rad = 16° Assess: This is a reasonable angle for a first order maximum.

22.49. Visualize: We are given L = 1.50 m and m = 2. We also know that for λ = 610 nm, y2 = 1.611 m, and we seek λ for y2 = 1.606 m. We will need to solve for d and use it to find the new λ. Use Equation 22.15: d sin θ m = mλ , and Equation 22.16: ym = L tanθ m . Solve:

d=

mλ = sin θ m

mλ (2)(610 nm) = = 1.667 mm ⎛ −1 ym ⎞ ⎛ −1 1.611 m ⎞ sin ⎜ tan sin tan ⎟ ⎜ ⎟ L ⎠ 1.50 m ⎠ ⎝ ⎝

Now that we know d we can use it to find the new wavelength.

λ= Assess:

d d y ⎞ 1.667 mm ⎛ −1 1.606 m ⎞ ⎛ sin θ m = sin ⎜ tan −1 m ⎟ = sin ⎜ tan ⎟ = 609 nm 2 1.50 m ⎠ m m ⎝ L ⎠ ⎝

The two bright peaks are caused by specific wavelengths of light that differ by only 1 nm.

22.50. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: These are not small angles, so we can’t use equations based on the small-angle approximation. As given by Equation 22.19, the dark fringes in the pattern are located at a sin θ p = pλ , where p = 1, 2, 3, … For the first minimum of the pattern, p = 1. Thus, a

λ

=

p 1 = sin θ p sin θ1

For the three given angles the slit width to wavelength ratios are

1 ⎛a⎞ = 2, (a): ⎜ ⎟ = ⎝ λ ⎠30° sin 30°

1 ⎛a⎞ (b): ⎜ ⎟ = = 1.15, ⎝ λ ⎠60° sin 60°

1 ⎛a⎞ =1 (c): ⎜ ⎟ = ⎝ λ ⎠90° sin 90°

Assess: It is clear that the smaller the a/λ ratio, the wider the diffraction pattern. This is a conclusion that is contrary to what one might expect.

22.51. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: 45 ° is not a small angle, so we can’t use equations based on the small-angle approximation. As given by Equation 22.19, the dark fringes in the pattern are located at a sin θ p = pλ , where p = 1, 2, 3, … For the first minimum to be at 45°, p = 1 and the width of the slit is

a=

pλ 633 ×10−9 m = = 895 nm sin θ p sin 45°

22.52. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: As given by Equation 22.19, the dark fringes in the pattern are located at a sinθ p = pλ , where p = 1, 2, 3, … For the diffraction pattern to have no minima, the first minimum must be located at least at θ1 = 90°. From the constructive-interference condition a sin θ p = pλ , we have

a=

pλ λ ⇒a= = λ = 633 nm sin θ p sin 90°

22.53. Model: A narrow slit produces a single-slit diffraction pattern. Visualize:

The dark fringes in this diffraction pattern are given by Equation 22.21:

yp =

pλ L a

p = 1, 2, 3, …

We note that the first minimum in the figure is 0.50 cm away from the central maximum. We are given a = 0.02 nm and L = 1.5 m. Solve: Solve the above equation for λ .

λ=

y pa pL

Assess: 670 nm is in the visible range.

=

(0.50 × 10−2 m)(0.20 × 10−3 m) = 670 nm (1)(1.5 m)

22.54. Model: A narrow slit produces a single-slit diffraction pattern. Solve:

The dark fringes in this diffraction pattern are given by Equation 22.21:

pλ L p = 1, 2, 3, … a We note from Figure P22.53 that the first minimum is 0.50 cm away from the central maximum. Thus, yp =

L= Assess:

ay p pλ

( 0.15 ×10 m )( 0.50 ×10 (1) ( 600 ×10 m ) −3

=

This is a typical slit to screen separation.

−9

−2

m)

= 1.3 m

22.55. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: Within the small angle approximation, which is almost always valid for the diffraction of light, the width of the central maximum is λL w = 2 y1 = 2.44 D

From Figure P22.53, w = 1.0 cm, so

D= Assess:

2.44λ L 2.44(500 × 10−9 m)(1.0 m) = = 0.12 mm w (1.0 × 10−2 m)

This is a typical size for an aperture to show diffraction.

22.56. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: Within the small-angle approximation, the width of the central maximum is w = 2.44

λL D

Because w = D, we have

D = 2.44

λL D

⇒ D = 2.44λ L ⇒ D = (2.44)(633 × 10−9 m)(0.50 m) = 0.88 mm

22.57. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: (a) Because the visible spectrum spans wavelengths from 400 nm to 700 nm, we take the average wavelength of sunlight to be 550 nm. (b) Within the small-angle approximation, the width of the central maximum is

w = 2.44

λL D

⇒ (1 × 10−2 m) = (2.44)

(550 × 10−9 m)(3 m) ⇒ D = 4.03 × 10−4 m = 0.40 mm D

22.58. Model: The antenna is a circular aperture through which the microwaves diffract. Solve: (a) Within the small-angle approximation, the width of the central maximum of the diffraction pattern is w = 2.44λL/D. The wavelength of the radiation is

λ=

c 3 × 108 m/s 2.44(0.025 m)(30 × 103 m) = = 0.025 m ⇒ w = = 920 m 9 f 12 × 10 Hz 2.0 m

That is, the diameter of the beam has increased from 2.0 m to 915 m, a factor of 458. (b) The average microwave intensity is

I=

100 ×103 W P = = 0.15 W/m 2 area π ⎡ 12 ( 915 m ) ⎤ 2 ⎣ ⎦

22.59. Model: The laser beam is diffracted through a circular aperture. Visualize:

Solve: (a) No. The laser light emerges through a circular aperture at the end of the laser. This aperture causes diffraction, hence the laser beam must gradually spread out. The diffraction angle is small enough that the laser beam appears to be parallel over short distances. But if you observe the laser beam at a large distance it is easy to see that the diameter of the beam is slowly increasing. (b) The position of the first minimum in the diffraction pattern is more or less the “edge” of the laser beam. For diffraction through a circular aperture, the first minimum is at an angle

θ1 =

−9 1.22λ 1.22 ( 633 × 10 m ) = = 5.15 × 10−4 rad = 0.029° D 0.0015 m

(c) The diameter of the laser beam is the width of the diffraction pattern:

w=

−9 2.44λ L 2.44 ( 633 × 10 m ) ( 3 m ) = = 0.00309 m = 0.31 cm D 0.0015 m

(d) At L = 1 km = 1000 m, the diameter is

w=

−9 2.44λ L 2.44 ( 633 × 10 m ) (1000 m ) = = 1.03 m ≈ 1.0 m D 0.0015 m

22.60. Model: The laser light is diffracted by the circular opening of the laser from which the beam emerges. Solve:

The diameter of the laser beam is the width of the central maximum. We have

w=

−9 8 2.44λ L 2.44 ( 532 × 10 m )( 3.84 × 10 m ) 2.44λ L = = 0.50 m ⇒ D= D w 1000 m

In other words, the laser beam must emerge from a laser of diameter 50 cm.

22.61. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize:

Solve:

(a) The m = 1 bright fringes are separated from the m = 0 central maximum by

Δy =

λL d

=

( 600 × 10

−9

m ) (1.0 m )

0.0002 m

= 0.0030 m = 3.0 mm

(b) The light’s frequency is f = c/λ = 5.00 × 1014 Hz. Thus, the period is T = 1/f = 2.00 × 10−15 s. A delay of 5.0 × 10−16 s = 0.50 × 10−15 s is 14 T . (c) The wave passing through the glass is delayed by

1 4

of a cycle. Consequently, the two waves are not in phase as

they emerge from the slits. The slits are the sources of the waves, so there is now a phase difference Δφ0 between the two sources. A delay of a full cycle (Δt = T) would have no effect at all on the interference because it corresponds to a phase difference Δφ0 = 2π rad. Thus a delay of 14 of a cycle introduces a phase difference Δφ0 = 1 4

( 2π ) = 12 π

rad.

(d) The text’s analysis of the double-slit interference experiment assumed that the waves emerging from the two slits were in phase, with Δφ0 = 0 rad. Thus, there is a central maximum at a point on the screen exactly halfway

between the two slits, where Δφ = 0 rad and Δr = 0 m. Now that there is a phase difference between the sources, the central maximum—still defined as the point of constructive interference where Δφ = 0 rad—will shift to one side. The wave leaving the slit with the glass was delayed by

1 4

of a period. If it travels a shorter distance to the

screen, taking of a period less than the wave coming from the other slit, it will make up for the previous delay and the two waves will arrive in phase for constructive interference. Thus, the central maximum will shift toward the slit with the glass. How far? A phase difference Δφ0 = 2π would shift the fringe pattern by Δy = 3.0 mm, 1 4

making the central maximum fall exactly where the m = 1 bright fringe had been previously. This is the point where Δr = (1)λ, exactly compensating for a phase shift of 2π at the slits. Thus, a phase shift of Δφ0 = 12 π = 14 ( 2π ) will shift the fringe pattern by 14 ( 3 mm ) = 0.75 mm . The net effect of placing the glass in the

slit is that the central maximum (and the entire fringe pattern) will shift 0.75 mm toward the slit with the glass.

22.62. Model: A diffraction grating produces an interference pattern, which looks like the diagram of Figure 22.8. Solve: (a) Nothing has changed while the aquarium is empty. The order of a bright (constructive interference) fringe is related to the diffraction angle θm by d sin θ m = mλ , where m = 0, 1, 2, 3, … The space between the slits is d=

1.0 mm = 1.6667 × 10−6 m 600

For m = 1, sin θ1 =

λ

⎛ 633 × 10−9 m ⎞ ⇒ θ1 = sin −1 ⎜ ⎟ = 22.3° −6 d ⎝ 1.6667 ×10 m ⎠

(b) The path-difference between the waves that leads to constructive interference is an integral multiple of the wavelength in the medium in which the waves are traveling, that is, water. Thus, 633 nm 633 nm λ 4.759 ×10−7 m λ= = = 4.759 ×10−7 m ⇒ sinθ1 = = = 0.2855 ⇒ θ1 = 16.6° nwater 1.33 d 1.6667 ×10−6 m

22.63. Model: An interferometer produces a new maximum each time L2 increases by 12 λ causing the pathlength difference Δr to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: The path-length difference between the two waves is Δr = 2L2 − 2L1. The condition for constructive interference is Δr = mλ, hence constructive interference occurs when

2 ( L2 − L1 ) = mλ ⇒ L2 − L1 = 12 mλ2 = 1200 ( 12 λ ) = 600λ where λ = 632.8 nm is the wavelength of the helium-neon laser. When the mirror M2 is moved back and a hydrogen discharge lamp is used, 1200 fringes shift again. Thus, L2′ − L1 = 1200 ( 12 λ ′ ) = 600λ ′ where λ ′ = 656.5 nm. Subtracting the two equations,

( L2 − L1 ) − ( L2′ − L1 ) = 600 ( λ − λ ′ ) = 600 ( 632.8 ×10−9 m − 656.5 ×10−9 m ) ⇒ L2′ = L2 + 14.2 ×10−6 m That is, M2 is now 14.2 μm closer to the beam splitter.

22.64.

Model: The gas increases the number of wavelengths in one arm of the interferometer. Each additional wavelength causes one bright-dark-bright fringe shift. Solve: From Equation 22.36, the number of fringe shifts is

Δm = m2 − m1 = ( n − 1)

2d

λvac

= (1.00028 − 1)

( 2 ) ( 2.00 ×10−2 m ) 600 ×10−9 m

= 19

22.65. Model: The water increases the number of wavelengths in one arm of the interferometer. Solve: (a) The incident light has λvac = 500 nm and f = c / λvac = 6.00 × 1014 Hz. Water doesn’t affect the frequency, which is still f w = f = 6.00 ×1014 Hz . The wavelength changes to λw = λvac / nw = 376 nm. (b) Light travels in a layer of thickness L twice—once to the right and then, after reflecting, once to the left—for a total distance 2L. The number of wavelengths in distance 2L is N = 2L/λ. If the 1 mm layer is a vacuum, the number of wavelengths is N vac =

2L

λvac

=

2 ( 0.001 m ) = 4000 500 ×10−9 m

If the vacuum is replaced by 1 mm of water, the number of wavelengths is Nw =

2L

λw

=

2 ( 0.001 m ) = 5319 376 × 10−9 m

So with the water added, the light travels ≈ 1320 extra wavelengths. (c) Each additional wavelength of travel shifts the pattern by 1 fringe. So the addition of the water shifts the interference pattern by ≈ 1320 fringes.

22.66. Model: The piece of glass increases the number of wavelengths in one arm of the interferometer. Each additional wavelength causes one bright-dark-bright fringe shift. Solve: We can rearrange Equation 22.36 to find that the index of refraction of glass is

n = 1+

λvac Δm 2d

( 500 ×10 nm ) ( 200 ) = 1.50 = 1+ 2 ( 0.10 ×10 m ) −9

−3

22.67. Model: The arms of the interferometer are of equal length, so without the crystal the output would be bright. Visualize: We need to consider how many more wavelengths fit in the electro-optic crystal than would have occupied that space (6.70 μ m ) without the crystal; if it is an integer then the interferometer will produce a bright output; if it is a half-integer then the interferometer will produce a dark output. But the wavelength we need to consider is the wavelength inside the crystal, not the wavelength in air.

λn =

λ n

We are told the initial n with no applied voltage is 1.522, and the wavelength in air is λ = 1.000 μ m. Solve: The number of wavelengths that would have been in that space without the crystal is

6.70 μ m = 6.70 1.000 μ m (a) With the crystal in place (and n = 1.522) the number of wavelengths in the crystal is

6.70 μ m = 10.20 1.000 μ m/1.522 10.20 – 6.70 = 3.50 which shows there are a half-integer number more wavelengths with the crystal in place than if it weren't there. Consequently the output is dark with the crystal in place but no applied voltage. (b) Since the output was dark in the previous part, we want it to be bright in the new case with the voltage on. That means we want to have just one half more extra wavelengths in the crystal (than if it weren't there) than we did in the previous part. That is, we want 4.00 extra wavelengths in the crystal instead of 3.5, so we want 6.70 + 4.00 = 10.70 wavelengths in the crystal.

6.70 μ m = 10.70 1.000 μ m/n

⇒

n=

10.70(1.000 μ m) = 1.597 6.70 μ m

Assess: It seems reasonable to be able to change the index of refraction of a crystal from 1.522 to 1.597 by applying a voltage.

22.68. Model: A diffraction grating produces an interference pattern like the one shown in Figure 22.8. We also assume that the small-angle approximation is valid for this grating. Solve: (a) The general condition for constructive-interference fringes is

d sin θ m = mλ

m = 0, 1, 2, 3, …

When this happens, we say that the light is diffracted at an angle θm. Since it is usually easier to measure distances rather than angles, we will consider the distance ym from the center to the mth maximum. This distance is ym = L tan θ m . In the small-angle approximation, sinθ m ≈ tanθ m , so we can write the condition for constructive interference as d

ym mλ L = mλ ⇒ ym = L d

The fringe separation is ym +1 − ym = Δy =

λL d

(b) Now the laser light falls on a film that has a series of “slits” (i.e., bright and dark stripes), with spacing

d′ =

λL d

Applying once again the condition for constructive interference: d ′ sin θ m = mλ ⇒ d ′

mλ L mλ L y′m = mλ ⇒ y′m = = = md d′ λL d L

The fringe separation is ym′ +1 − ym′ = Δy′ = d . That is, using the film as a diffraction grating produces a diffraction pattern whose fringe spacing is d, the spacing of the original slits.

22.69. Model: Two closely spaced slits produce a double-slit interference pattern. The interference pattern is symmetrical on both sides of the central maximum. Visualize:

In figure (a), the interference of laser light from the two slits forms a constructive-interference central (m = 0) fringe at P. The paths 1P and 2P are equal. When a glass piece of thickness t is inserted in the path 1P, the interference between the two waves yields the 10th dark fringe at P. Note that the glass piece is not present in figure (a). Solve: The number of wavelengths in the air-segment of thickness t is t m1 =

λ

The number of wavelengths in the glass piece of thickness t is t t nt m2 = = = λglass λ n λ The path length has thus increased by Δm wavelengths, where

Δm = m2 − m1 = ( n − 1)

t

λ

From the 10th dark fringe to the 1st dark fringe is 9 fringes and from the 1st dark fringe to the 0th bright fringe is one-half of a fringe. Hence,

1 19 19 t 19 λ 19 ( 633 ×10 m ) = ⇒ = ( n − 1) ⇒ t = = = 12.0 μm 2 2 2 2 ( n − 1) 2 1.5 − 1.0 λ −9

Δm = 9 +

22.70. Visualize: To I double

is I1 use Equation 22.14: 1 λL ⎛πd ⎞ to get the fraction desired. y ⎟ . Then divide by the distance to the first minimum y0 = = 4 I1 cos 2 ⎜ 2 d ⎝ λL ⎠

Solve:

find

the

y

location

where

the

intensity

First set I double = I1 .

⎛πd ⎞ I double = 4 I1 cos 2 ⎜ y⎟ ⎝ λL ⎠ ⎛πd I1 = 4 I1 cos 2 ⎜ ⎝ λL

⎞ y⎟ ⎠

1 ⎛πd = cos 2 ⎜ 4 ⎝ λL

⎞ y⎟ ⎠

1 ⎛πd = cos ⎜ 2 ⎝ λL

⎞ y⎟ ⎠

y=

λ L −1 ⎛ 1 ⎞ cos ⎜ ⎟ πd ⎝ 2⎠

Now set up the ratio that will give the desired fraction.

λ L −1 ⎛ 1 ⎞ cos ⎜ ⎟ y πd ⎝ 2 ⎠ = 2 cos −1 ⎛ 1 ⎞ = 2 π = 2 = ⎜ ⎟ 1 λL π y0′ ⎝ 2⎠ π 3 3 2 d Assess: The fraction must be less than 1, and 23 seems reasonable.

22.71. Model: A diffraction grating produces a series of constructive-interference fringes at values of θm that are determined by Equation 22.15. Solve: (a) The condition for bright fringes is d sin θ = mλ . If λ changes by a very small amount Δλ, such that θ changes by Δθ, then we can approximate Δλ/Δθ as the derivative dλ/dθ: 2

λ=

2

d d d Δλ d λ d ⎛ mλ ⎞ ⎛d⎞ 2 ≈ = cosθ = sin θ ⇒ 1 − sin 2 θ = 1− ⎜ ⎟ = ⎜ ⎟ −λ m m m Δθ dθ m ⎝ d ⎠ ⎝m⎠ Δλ ⇒ Δθ = 2 ⎛d⎞ 2 ⎜ ⎟ −λ ⎝m⎠

(b) We can now obtain the first-order and second-order angular separations for the wavelengths λ = 589.0 nm and λ + Δλ = 589.6 nm. The slit spacing is

d=

1.0 ×10−3 m = 1.6667 × 10−6 m 600

The first-order (m = 1) angular separation is Δθ =

0.6 ×10−9 m

2.7778 ×10−12 m 2 − 0.3476 ×10−12 m 2 = 3.85 ×10−4 rad = 0.022°

=

The second order (m = 2) angular separation is 0.6 × 10−9 m Δθ = = 1.02 × 10 −3 rad = 0.058° −12 2 −12 2 0.6945 × 10 m − 0.3476 × 10 m

0.6 ×10−9 m 1.5589 ×10−6 m

22.72. Model: The intensity in a double-slit interference pattern is determined by diffraction effects from the slits. Solve: (a) For the two wavelengths λ and λ + Δλ passing simultaneously through the grating, their first-order peaks are at y1 =

λL d

y1′ =

(λ + Δλ )L d

Subtracting the two equations gives an expression for the separation of the peaks: Δy = y1′ − y1 =

Δλ L d

(b) For a double-slit, the intensity pattern is

⎛πd ⎞ I double = 4 I1 cos 2 ⎜ y⎟ ⎝ λL ⎠ The intensity oscillates between zero and 4I1, so the maximum intensity is 4I1. The width is measured at the point where the intensity is half of its maximum value. For the intensity to be 12 I max = 2 I1 for the m = 1 peak:

π λL ⎛πd ⎞ ⎛πd ⎞ 1 πd 2 I1 = 4 I1 cos 2 ⎜ yhalf ⎟ ⇒ cos 2 ⎜ yhalf ⎟ = ⇒ yhalf = ⇒ yhalf = λ λ λ L L L 2 4 4d ⎝ ⎠ ⎝ ⎠ The width of the fringe is twice yhalf. This means w = 2 yhalf =

λL 2d

But the location of the m = 1 peak is y1 = λ L d , so we get w = 12 y1 . (c) We can extend the result obtained in (b) for two slits to w = y1 N . The condition for barely resolving two

diffraction fringes or peaks is w = Δymin. From part (a) we have an expression for the separation of the first-order peaks and from part (b) we have an expression for the width. Thus, combining these two pieces of information,

y1 Δλ L y d ⎛ λL ⎞ d λ = Δymin = min ⇒ Δλmin = 1 = ⎜ = ⎟ N d LN ⎝ d ⎠ NL N (d) Using the result of part (c), N=

λ 656.27 ×10−9 m = = 3646 lines Δλmin 0.18 ×10−9 m

22.73. Model: A diffraction grating produces an interference pattern like the one shown in Figure 22.8, when the incident light is normal to the grating. The equation for constructive interference will change when the light is incident at a nonzero angle.

Solve:

(a) The path difference between waves 1 and 2 on the incident side is Δr1 = d sin φ . The path

difference between the waves 1 and 2 on the diffracted side, however, is Δr2 = d sin θ . The total path difference between waves 1 and 2 is thus Δr1 + Δr2 = d ( sinθ + sin φ ) . Because the path difference for

constructive interference must be equal to mλ, d ( sin θ + sin φ ) = mλ

m = 0, ±1, ±2, …

(b) For φ = 30° the angles of diffraction are sin θ1 =

sin θ −1 =

(1) ( 500 ×10−9 m )

(1.0 ×10

−3

m ) 600

( −1) ( 500 ×10−9 m )

(1.0 ×10

−3

m ) 600

− sin 30° = −0.20 ⇒ θ1 = −11.5°

− sin 30° = −0.80 ⇒ θ −1 = −53.1°

22.74. Solve:

(a)

We have two incoming and two diffracted light rays at angles φ and θ and two wave fronts perpendicular to the rays. We can see from the figure that the wave 1 travels an extra distance Δr = d sin φ to reach the reflection spot. Wave 2 travels an extra distance Δr = d sin θ from the reflection spot to the outgoing wave front. The path difference between the two waves is

Δr = Δr1 − Δr2 = d ( sin θ − sin φ ) (b) The condition for diffraction, with all the waves in phase, is still Δr = mλ. Using the results from part (a), the diffraction condition is d sinθ m = mλ + d sin φ m = … −2, −1, 0, 1, 2, …

Negative values of m will give a different diffraction angle than the corresponding positive values. (c) The “zeroth order” diffraction from the reflection grating is m = 0. From the diffraction condition of part (b), this implies d sin θ 0 = d sin φ and hence θ 0 = φ . That is, the zeroth order diffraction obeys the law of reflection—the angle of reflection equals the angle of incidence. 1 (d) A 700 lines per millimeter grating has spacing d = 700 mm = 1.429 × 10−6 m = 1429 nm. The diffraction angles are given by

⎛ m ( 500 nm ) ⎞ ⎛ mλ ⎞ + sin φ ⎟ = sin −1 ⎜ + sin 40° ⎟ 1429 nm ⎝ d ⎠ ⎝ ⎠

θ m = sin −1 ⎜

(e)

M

θm

≤−5 −4 −3 −2 −1 0 1 ≥2

not defined −49.2° −24.0° −3.3° 17.0° 40.0° 83.1° not defined

22.75. Model: Diffraction patterns from two objects can just barely be resolved if the central maximum of one image falls on the first dark fringe of the other image. Solve: (a) Using Equation 22.24 with the width equal to the aperture diameter,

w=D=

2.44λ L ⇒ D = 2.44λ L = D

( 2.44 ) ( 550 ×10−7

m ) ( 0.20 m ) = 0.52 mm

(b) We can now use Equation 22.23 to find the angle between two distant sources that can be resolved. The angle is

α = 1.22

λ D

=

1.22 ( 550 ×10−9 m ) 0.52 ×10−3 m

= 1.29 ×10 −3 rad = 0.074°

(c) The distance that can be resolved is

(1000 m )α = (1000 m ) (1.29 × 10−3 rad ) = 1.3 m

22-1

23.1. Model: Light rays travel in straight lines. Solve:

(a) The time is t=

Δx 1.0 m = = 3.3 × 10−9 s = 3.3 ns c 3 ×108 m/s

(b) The refractive indices for water, glass, and cubic zirconia are 1.33, 1.50, and 1.96, respectively. In a time of 3.33 ns, light will travel the following distance in water:

⎛ c ⎞ ⎛ 3 × 108 m/s ⎞ −9 Δxwater = vwatert = ⎜ ⎟t = ⎜ ⎟ ( 3.33 × 10 s ) = 0.75 m 1.33 n ⎠ ⎝ water ⎠ ⎝ Likewise, the distances traveled in the glass and cubic zirconia are Δxglass = 0.667 m and Δxcubic zirconia = 0.458 m. Assess: The higher the refractive index of a medium, the slower the speed of light and hence smaller the distance it travels in that medium in a given time.

23.2. Model: Light rays travel in straight lines. Solve: Let tglass, toil, and tplastic be the times light takes to pass through the layers of glass, oil, and plastic. The time for glass is

tglass =

−2 Δx nglass (1.0 ×10 m ) (1.50 ) Δx Δx = = = = 0.050 ns vglass c nglass c 3.0 × 108 m/s

Likewise, toil = 0.243 ns and tplastic = 0.106 ns. Thus, ttotal = tglass + toil + tplastic = 0.050 ns + 0.243 ns + 0.106 ns = 0.399 ns = 0.40 ns. Assess: The small time is due to the high value for the speed of light.

23.3. Model: Light rays travel in straight lines. The light source is a point source. Visualize:

Solve:

Let w be the width of the aperture. Then from the geometry of the figure,

w 12.0 cm = ⇒ w = 8.0 cm 2.0 m 2.0 m + 1.0 m

23.4. Model: Light rays travel in straight lines. Also, the red and green light bulbs are point sources. Visualize:

Solve:

The width of the aperture is w = 1 m. From the geometry of the figure for red light,

w2 x′ = ⇒ x′ = 2w = 2 (1.0 m ) = 2.0 m 1m 3m + 1m

The red light illuminates the wall from x = 0.50 m to x = 4.50 m. For the green light, w4 x1 = ⇒ x1 = 1.0 m 1m 3m + 1m

3w 4 x2 = ⇒ x2 = 3.0 m 1m 3m + 1m

Because the back wall exists only for 2.75 m to the left of the green light source, the green light has a range from x = 0 m to x = 3.75 m.

23.5.

Visualize:

Note the similar triangles in this figure.

Solve:

15 cm d = 5 cm 180 cm d=

15 cm (180 cm) = 540 cm = 5.4 m 5 cm

Assess: This is a typical distance for photographs of people.

23.6. Model: Use the ray model of light. Visualize:

According to the law of reflection, θ r = θ i . Solve: From the geometry of the diagram,

θ i + φ = 90° Using the law of reflection, we get

θ r + ( 60° − φ ) = 90°

90° − φ = 90° − ( 60° − φ ) ⇒ φ = 30°

Assess: The above result leads to a general result for plane mirrors: If a plane mirror rotates by an angle φ relative to the horizontal, the reflected ray makes an angle of 2φ with the horizontal.

23.7. Model: Light rays travel in straight lines and follow the law of reflection. Visualize:

Solve:

We are asked to obtain the distance h = x1 + 5.0 cm. From the geometry of the diagram,

tan θ i =

x1 10 cm

tan θ r =

x2 15 cm

x1 + x2 = 10 cm

Because θ r = θ i , we have x1 x2 10 cm − x1 = = ⇒ (15 cm ) x1 = 100 cm 2 − 10 x1 ⇒ x1 = 4.0 cm 10 cm 15 cm 15 cm Thus, the ray strikes a distance 9.0 cm below the top edge of the mirror.

23.8. Model: Think of the view in the figure as a horizontal view of a vertical wall and the laser beam and hexagonal mirror in a vertical plane for ease of labeling. The laser beam will strike the highest spot on the wall when a new corner rotates into the laser beam and the angle the laser makes with the normal is greatest. We will compute how high on the wall this highest spot is from the center spot behind the laser; then we will multiply by two because symmetry says the reflected beam will hit the lowest spot just as the face rotates out of the laser beam (and the beam makes the largest angle with the normal in the downward direction), and then a new corner enters the beam with the reflection at the top again. Visualize: From the small right triangle inside the hexagon we deduce d = 0.20 m / sin 60°. Therefore, the distance from the wall to the corner of the hexagon just as it enters the laser beam is 2.0 m − d . This becomes the base of a large right triangle whose side on the wall is x and whose angle opposite x is 60°.

Solve:

Solve the large right triangle for x. tan 60° =

x 2.0 m − d

⎛ 0.20 m ⎞ x = (tan 60°)(2.0 m − d ) = (tan 60°) ⎜ 2.0 m − ⎟ = 3.06 m sin 60° ⎠ ⎝ Now, because of symmetry, double x to get the total length of the streak of laser light: 2 x = 6.1 m. Assess: The 50 cm distance from the laser to the center of the hexagon is irrelevant.

23.9. Model: Light rays travel in straight lines and follow the law of reflection. Visualize:

To determine the angle φ , we must know the point P on the mirror where the ray is incident. P is a distance x2 from the far wall and a horizontal distance x1 from the laser source. The ray from the source must strike P so that the angle of incidence θi is equal to the angle of reflection θr. Solve: From the geometry of the diagram, 1.5 m 3 m tan φ = = x1 + x2 = 5 m x2 x1 ⇒

⇒ tan φ =

1.5 m 3m 10 m = ⇒ (1.5 m ) x1 = 15 m 2 − ( 3 m ) x1 ⇒ x1 = 5 m − x1 x1 3

3m 9 = = 0.90 ⇒ φ = 42° x1 10

23.10. Model: Use the ray model of light. Visualize:

The arrow is at a distance s from the mirror, so its image is at a distance s′ behind the mirror. When you are at x = 0 m, a ray from the arrow’s head, after reflection from the mirror, is able to enter your eye. Similarly, a ray from the arrow’s tail, after normal incidence, is reflected into the eye. That is, the eye is able to see the arrow’s head and tail. While walking toward the right, a ray from the arrow’s head will reflect from the mirror’s right edge and enter your eye at P. A ray starting from the arrow’s tail will also enter your eye when you are at P. That is, while at P you will be able to see the entire image of the arrow. However, the light from the arrow’s head can never reach beyond point P. Solve: Point P is a distance x from the origin. From the geometry of the diagram, tan φ =

1 m 1 m x−2 m x−2 m 1 x−2m = = = ⇒ = ⇒ x = 3.5 m s 2 m s +1 m 3m 2 3m

Thus, the range of x over which you can see the entire arrow in the mirror is 0 m to 3.5 m.

23.11. Model: Use the ray model of light and the law of reflection. Visualize:

We only need one ray of light that leaves your toes and reflects in your eye. Solve: From the geometry of the diagram, the distance from your eye to the toes’ image is

2d = (400 cm) 2 + (165 cm) 2 = 433 cm Assess:

The light appears to come from your toes’ image.

23.12. Model: Use the ray model of light and Snell’s law. Visualize:

Solve:

According to Snell’s law for the air-water and water-glass boundaries,

nair sinθ air = nwater sin θ water

nwater sin θ water = nglass sin θ glass

From these two equations, we have nair sinθ air = nglass sin θ glass ⇒ sin θ glass =

nair ⎛ 1.0 ⎞ −1 ⎛ sin 60° ⎞ sin θ air = ⎜ ⎟ sin 60° ⇒ θ glass = sin ⎜ ⎟ = 35° nglass ⎝ 1.50 ⎠ ⎝ 1.5 ⎠

23.13. Visualize: Use Snell’s law n1 sinθ1 = n2 sinθ 2 . We are given θ 2 = θ CZ = 25°. We look up in Table 23.1 n1 = noil = 1.46 and n2 = nCZ = 1.96. Solve:

Solve the equation for θ1.

⎛ n2 ⎞ ⎛ 1.96 ⎞ sinθ 2 ⎟ = sin −1 ⎜ sin 25° ⎟ = 35° n 1.46 ⎝ ⎠ ⎝ 1 ⎠

θ1 = sin −1 ⎜

Assess: Since the ray goes into a material with higher index of refraction we know it bends toward the normal, so we expect θ1 > θ 2 ; this is the case.

23.14. Model: Use the ray model of light. The sun is a point source of light. Visualize:

A ray that arrives at the diver 50° above horizontal refracted into the water at θwater = 40°. Solve: Using Snell’s law at the water-air boundary

nair sinθ air = nwater sin θ water ⇒ sin θ air =

nwater ⎛ 1.33 ⎞ sin θ water = ⎜ ⎟ sin 40° nair ⎝ 1.0 ⎠

⇒ θair = 58.7° Thus the height above the horizon is θ = 90° − θair = 31.3° ≈ 31°. Because the sun is far away from the fisherman (and the diver), the fisherman will see the sun at the same angle of 31° above the horizon.

23.15. Model: Represent the laser beam with a single ray and use the ray model of light. Solve:

Using Snell’s law at the air-water boundary,

nair sinθ air = nliquid sinθ liquid ⇒ nliquid = nair Assess:

As expected, nliquid is larger than nair.

sin θ air ⎛ sin 37° ⎞ = 1.0 ⎜ ⎟ = 1.37 sin θ liquid ⎝ sin 26° ⎠

23.16. Model: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection. Visualize:

Solve:

The critical angle of incidence is given by Equation 23.9:

⎛ ncladding ⎞ −1 ⎛ 1.48 ⎞ ⎟ = sin ⎜ ⎟ = 67.7° ⎝ 1.60 ⎠ ⎝ ncore ⎠

θ c = sin −1 ⎜

Thus, the maximum angle a light ray can make with the wall of the core to remain inside the fiber is 90° − 67.7° = 23.3°. Assess: We can have total internal reflection because ncore > ncladding.

23.17. Model: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection. Visualize:

Solve:

The critical angle of incidence is given by Equation 23.9:

⎛ noil ⎞ ⎛ 1.46 ⎞ = sin −1 ⎜ ⎟ = 76.7° ⎜ n ⎟⎟ ⎝ 1.50 ⎠ ⎝ glass ⎠

θ c = sin −1 ⎜ Assess:

The critical angle exists because noil < nglass.

23.18. Model: Represent the can as a point source and use the ray model of light. Visualize:

Paraxial rays from the can refract into the water and enter into the fish’s eye. Solve: The object distance from the edge of the aquarium is s. From the water side, the can appears to be at an image distance s′ = 30 cm. Using Equation 23.13,

s′ =

30 cm n2 n ⎛ 1.33 ⎞ = 23 cm s = water s = ⎜ ⎟s ⇒ s = 1.33 n1 nair ⎝ 1.0 ⎠

23.19. Model: Represent the beetle as a point source and use the ray model of light. Visualize:

Paraxial rays from the beetle refract into the air and then enter into the observer’s eye. The rays in the air when extended into the plastic appear to be coming from the beetle at a shallower location, a distance s′ from the plastic-air boundary. Solve: The actual object distance is s and the image distance is s′ = 2.0 cm. Using Equation 23.13,

s′ = Assess:

n2 n 1.0 s = air s ⇒ 2.0 cm = s ⇒ s = 3.2 cm n1 nplastic 1.59

The beetle is much deeper in the plastic than it appears to be.

23.20. Model: Represent the diver’s head and toes as point sources. Use the ray model of light. Visualize:

Paraxial rays from the head and the toes of the diver refract into the air and then enter into your eyes. When these refracted rays are extended into the water, the head and the toes appear elevated toward you. Solve: Using Equation 23.13, n n n sT′ = 2 sT = air sT sH′ = air sH n1 nwater nwater Subtracting the two equations, her apparent height is

s′H − sT′ =

nair 1.0 ( sH − sT ) = (150 cm ) = 113 cm nwater 1.33

23.21. Model: Represent the aquarium’s wall as a point source, and use the ray model of light. Visualize:

Paraxial rays from the outer edge (O) are refracted into the water and then enter into the fish’s eye. When extended into the wall, these rays will appear to be coming from O′ rather from O. The point on the inside edge (I) of the wall will not change its apparent location. Solve: We are given that sO − sI = 4.00 mm and sO′ − sI′ = 3.50 mm. Using Equation 23.13,

sO′ = ⇒ sO′ − s′I =

nwater sO nwall

sI′ =

nwater sI nwall

nwater 1.33 4.00 mm ⎞ ( sO − sI ) ⇒ 3.50 mm = ( 4.00 mm ) ⇒ nwall = (1.33) ⎛⎜ ⎟ = 1.52 nwall nwall ⎝ 3.50 mm ⎠

23.22. Model: Use the ray model of light. Visualize:

Solve:

Using Snell’s law,

⎛ sin 30° ⎞ nair sin 30° = nred sin θ red ⇒ θ red = sin −1 ⎜ ⎟ = 19.2° ⎝ 1.52 ⎠ ⎛ sin 30° ⎞ nair sin 30° = nviolet sin θ violet ⇒ θ violet = sin −1 ⎜ ⎟ = 18.8° ⎝ 1.55 ⎠ Thus the angular spread is Δθ = θ red − θ violet = 19.2° − 18.8° = 0.4°

23.23. Model: Use the ray model of light and the phenomenon of dispersion. Visualize:

Solve: (a) From the graph in Figure 23.29, we estimate the index of refraction for the red light (656 nm) to be nred = 1.572 and for the blue light (456 nm) to be nblue = 1.587. (b) The angle of incidence onto the rear of the prism is 35°. Using these values for the refractive index and Snell’s law,

⎛ 1.572sin 35° ⎞ nred sin 35° = nair sin θ red ⇒ θ red = sin −1 ⎜ ⎟ = 64.4° 1.0 ⎝ ⎠

⎛ 1.587sin 35° ⎞ nblue sin 35° = nair sin θ blue ⇒ θ blue = sin −1 ⎜ ⎟ = 65.5° 1.0 ⎝ ⎠ ⇒ Δθ = θ blue − θ red = 1.1°

23.24. Model: Use the ray model of light and the phenomenon of dispersion. Visualize:

Solve:

Using Snell’s law for the red light,

nair sinθ air = nred sin θ red ⇒ 1.0sin θ air = 1.45sin 26.3° ⇒ θ air = sin −1 (1.45sin 26.3° ) = 40.0° Now using Snell’s law for the violet light, nair sin θ air = nviolet sinθ violet ⇒ 1.0sin 40.0° = nviolet sin 25.7° ⇒ nviolet = 1.48 Assess:

As expected, nviolet is slightly larger than nred.

23.25. Model: The intensity of scattered light is inversely proportional to the fourth power of the wavelength. −4 Solve: We want to find the wavelength of infrared light such that I IR = 0.01I 500 . Because I 500 ∝ ( 500 nm ) and I IR ∝ λ −4 , we have 4

I 500 ⎛ λ ⎞ =⎜ ⎟ = 100 ⇒ λ = 1580 nm I IR ⎝ 500 nm ⎠

23.26. Model: Use ray tracing to locate the image. Solve:

The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. You can see from the diagram that the image is in the plane where the three special rays converge. The image is inverted and is located at s′ = 20.0 cm to the right of the converging lens.

23.27. Model: Use ray tracing to locate the image. Solve:

The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. You can see from the diagram that the image is in the plane where the three special rays converge. The image is located at s′ = 15 cm to the right of the converging lens, and is inverted.

23.28. Model: Use ray tracing to locate the image. Solve:

The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. You can see that the rays after refraction do not converge at a point on the refraction side of the lens. On the other hand, the three special rays, when extrapolated backward toward the incidence side of the lens, meet at P′, which is 15 cm from the lens. That is, s′ = −15 cm. The image is upright.

23.29. Model: Use ray tracing to locate the image. Solve:

The figure shows the ray-tracing diagram using the steps of Tactics Box 23.3. The three rays after refraction do not converge at a point, but they appear to come from P′. P′ is 6 cm from the diverging lens, so s′ = −6 cm. The image is upright.

23.30. Model: Assume the biconvex lens is a thin lens.

Solve: If the object is on the left, then the first surface has R1 = +40 cm (convex toward the object) and the second surface has R2 = −40 cm (concave toward the object). The index of refraction of glass is n = 1.50, so the lensmaker’s equation is

⎛1 1 ⎞ 1 1 ⎞ ⎛ 1 = ( n − 1) ⎜ − ⎟ = (1.50 − 1) ⎜ − ⎟ ⇒ f = 40 cm f R R 40 cm 40 cm ⎠ − ⎝ 2 ⎠ ⎝ 1

23.31. Model: Assume the planoconvex lens is a thin lens.

Solve: If the object is on the left, then the first surface has R1 = ∞ and the second surface has R2 = −40 cm (concave toward the object). The index of refraction of polystyrene plastic is 1.59, so the lensmaker’s equation is

⎛1 1 ⎞ 1 1 ⎞ 1 0.59 ⎛1 ⇒ f = 68 cm = ( n − 1) ⎜ − ⎟ = (1.59 − 1) ⎜ − ⎟⇒ = f 40 cm f R R 40 cm ∞ − ⎝ ⎠ 2 ⎠ ⎝ 1

23.32. Model: Assume the biconcave lens is a thin lens.

Solve: If the object is on the left, then the first surface has R1 = −40 cm (concave toward the object) and the second surface has R2 = +40 cm (convex toward the object). The index of refraction of glass is 1.50, so the lensmaker’s equation is

⎛1 1 ⎞ 1 1 ⎞ 1 ⎞ ⎛ 1 ⎛ = ( n − 1) ⎜ − ⎟ = (1.50 − 1) ⎜ − ⎟ = ( 0.50 ) ⎜ − ⎟ ⇒ f = −40 cm f R R 40 cm 40 cm 20 cm ⎠ − + ⎝ ⎠ ⎝ 2 ⎠ ⎝ 1

23.33. Model: Assume the meniscus lens is a thin lens.

Solve: If the object is on the left, then the first surface has R1 = 30 cm (convex toward the object) and the second surface has R2 = 40 cm (convex toward the object). The index of refraction of polystyrene plastic is 1.59, so the lensmaker’s equation is

⎛1 1 ⎞ 1 1 ⎞ ⎛ 1 = ( n − 1) ⎜ − ⎟ = (1.59 − 1) ⎜ − ⎟ ⇒ f = 203 cm ≈ 200 cm f R R 30 cm 40 cm ⎠ ⎝ 2 ⎠ ⎝ 1

23.34. Model: The water is a spherical refracting surface. Consider the paraxial rays that refract from the air into the water. Solve: If the cat’s face is 20 cm from the edge of the bowl, then s = +20 cm. The spherical fish bowl surface has R = +25 cm, because it is the convex surface that is toward the object. Also n1 = 1 (air) and n2 = 1.33 (water). Using Equation 23.21,

n1 n2 n2 − n1 1 1.33 1.33 − 1 0.33 ⇒ + = = = 0.0132 cm −1 + = 20 cm s′ 25 cm 25 cm s s′ R

⇒

1.33 = ( 0.0132 − 0.050 ) cm −1 ⇒ s′ = −36 cm s′

This is a virtual image located 36 cm outside the fishbowl. The fish, inside the bowl, sees the virtual image. That is, the fish sees the cat’s face 36 cm from the bowl.

23.35. Model: Model the bubble as a point source and consider the paraxial rays that refract from the plastic into the air. The edge of the plastic is a spherical refracting surface. Visualize:

Solve: The bubble is at P, a distance of 2.0 cm from the surface. So, s = 2.0 cm. A ray from P after refracting from the plastic-air boundary bends away from the normal axis and enters the eye. This ray appears to come from P′, so the image of P is at P′ and it is a virtual image. Because P faces the concave side of the refracting surface, R = −4.0 cm. Furthermore, n1 = 1.59 and n2 = 1.0. Using Equation 23.21, n1 n2 n2 − n1 1.59 1.0 1.0 − 1.59 0.59 + = ⇒ + = =+ = 0.1475 cm −1 −4.0 cm s s′ R 2.0 cm s′ 4.0 cm

⇒

1 = 0.1475 cm −1 − 0.795 cm −1 ⇒ s′ = −1.54 cm s′

That is, the bubble appears 1.54 cm ≈ 1.5 cm beneath the surface.

23.36. Model: Assume the lens is thin. Visualize:

1 1 1 fs + = ⇒ s′ = s s′ f s− f Solve:

s′ =

fs (20 cm)(60 cm) = = 30 cm s− f 60 cm − 20 cm

The magnification is m = − s′ s = − 30 cm 60 cm = −1 2. This means the image is inverted and has a height of 0.50 cm. Assess: Ray tracing will confirm these results.

23.37. Solve: The image is at 40 cm as seen in the figure. It is inverted.

Assess:

When the object is outside the focal length we get an inverted image.

23.38. Solve: The image is at −30 cm as seen in the figure. It is upright.

Assess:

When the object is within the focal length we get a magnified upright image.

23.39. Solve: The image is at −12 cm as seen in the figure. It is upright.

Assess:

We expected an upright virtual image from the convex mirror.

23.40. Model: The speed of light in a material is determined by the refractive index as v = c n . Solve: To acquire data from memory, a total time of only 2.0 ns is allowed. This time includes 0.5 ns that the memory unit takes to process a request. Thus, the travel time for an infrared light pulse from the central processing unit to the memory unit and back is 1.5 ns. Let d be the distance between the central processing unit and the memory unit. The refractive index of silicon for infrared light is nSi = 3.5. Then,

1.5 ns =

(1.5 ns ) c = (1.5 ×10−9 s )( 3.0 ×108 m/s ) ⇒ d = 6.4 cm 2d 2d 2dnSi = = ⇒d = vSi c nSi c 2nSi 2 ( 3.5 )

23.41.

Model: Treat the red ball as a point source and use the ray model of light. Solve: (a) Using the law of reflection, we can obtain 3 images of the red ball. (b) The images of the ball are located at B, C, and D. Relative to the intersection point of the two mirrors, the coordinates of B, C, and D are: B(+1 m, −2 m), C(−1 m, +2 m), and D(+1 m, +2 m). (c)

23.42. Model: Treat the laser beam as a ray and use the ray model of light. Visualize:

Solve:

From the geometry of the mirrors and the rays, β = 50°, α =30°, and φ = 20°.

23.43. Model: For a mirror, the image distance behind the mirror equals the object’s distance in front of the mirror. Visualize:

Solve: Your face is 2.0 from the mirror into which you are looking. The image of your face (image 1) is 2.0 m behind the mirror, or 4.0 m away. Behind you, the image of the back of your head (image 2) is 3.0 m behind the mirror on the other wall. You can’t see this image because you’re looking to the right. However, the reflected rays that appear to come from image 2 (a virtual image) act just like the rays from an object—that is, just as the rays would if the back of your head were really at the position of image 2. These rays reflect from the mirror 2.0 m in front of you into which you’re staring and form an image (image 3) 8.0 m behind the mirror. This is the image of the back of your head that you see in the mirror in front of you. Since you’re 2.0 m from the mirror, the image of the back of your head is 10 m away.

23.44. Model: Treat the laser beam as a ray and use the ray model of light. Visualize:

As the cylinder rotates by an angle θ, the path of the reflected laser beam changes by an angle 2θ relative to the direction of incidence. Solve: Because the angle 2θ is very small,

tan 2θ ≅ 2θ =

1 180 2.0 ×10−3 m rad = degrees = 0.011° ⇒θ= 5.0 m 5000 π ( 5000 )

23.45. Model: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection. Visualize:

For angles θwater that are less than the critical angle, light will be refracted into the air. Solve: Snell’s law at the water-air boundary is nair sinθ air = nwater sinθ water . Because the maximum angle of θair is 90°, we have 1 ⎞ (1.0 ) sin 90° = 1.33sinθ water ⇒ θ water = sin −1 ⎛⎜ ⎟ = 48.75° 1.33 ⎝ ⎠

Applying Snell’s law again to the glass-water boundary, ⎛n ⎞ ⎛ 1.33 ( sin 48.75° ) ⎞ nglass sinθ glass = nwater sinθ water ⇒ θ glass = sin −1 ⎜ water sin θ water ⎟ = sin −1 ⎜ ⎟ = 42° ⎜ nglass ⎟ 1.50 ⎝ ⎠ ⎝ ⎠

Thus 42° is the maximum angle of incidence onto the glass for which the ray emerges into the air.

23.46. Model: Use the ray model of light. Visualize:

Solve: When the plastic is in place, the microscope focuses on the virtual image of the dot. From the figure, we note that s = 1.0 cm and s′ = 1.0 cm − 0.4 cm = 0.6 cm. The rays are paraxial, and the object and image distances are measured relative to the plastic-air boundary. Using Equation 23.13,

s′ =

nair 1.0 1.0 cm = 1.67 s ⇒ 0.6 cm = (1.0 cm ) ⇒ nplastic = nplastic nplastic 0.6 cm

23.47. Model: Use the ray model of light and the law of refraction. Solve: Snell’s law at the air-glass boundary is nair sinθ air = nglass sin θ glass . We require θ glass = 12 θ air , so nair sin ( 2θ glass ) = nglass sin θ glass ⇒ nair ( 2sin θ glass cosθ glass ) = nglass sin θ glass

⎛n ⎞ ⎛ 1.50 ⎞ ⇒ θ glass = cos −1 ⎜ glass ⎟ = cos −1 ⎜ ⎟ = 41.4° ⇒ θ air = 82.8° ⎝ 2 × 1.0 ⎠ ⎝ 2nair ⎠

23.48. Model: Use the ray model of light and the law of refraction. Visualize:

Solve: (a) The ray of light strikes the meter stick at Pempty , which is a distance L from the zero mark of the meter

stick. So, tan 60° =

L ⇒ L = ( 50 cm ) tan 60° = 86.6 cm 50 cm

(b) The ray of light refracts at Phalf and strikes the meter stick a distance x1 + x2 from the zero of the meter stick.

We can find x1 from the triangle PfullPhalf O′: tan 60° =

x1 ⇒ x1 = ( 25 cm ) tan 60° = 43.30 cm 25 cm

We also have x2 = ( 25 cm ) tan φhalf . Using Snell’s law,

⎛ sin 60° ⎞ nair sin 60° = nwater sin φhalf ⇒ φhalf = sin −1 ⎜ ⎟ = 40.63° ⎝ 1.33 ⎠

⇒ x2 = ( 25 cm ) tan 40.63° = 21.45 cm ⇒ x1 + x2 = 43.30 cm + 21.45 cm = 64.8 cm (c) The ray of light experiences refraction at Pfull and the angle of refraction is the same as in part (b). We get

tan φfull =

x3 ⇒ x3 = ( 50 cm ) tan 40.63° = 42.9 cm 50 cm

23.49.

Model: Use the ray model of light. Light undergoes total internal reflection if it is incident on a boundary at an angle greater than the critical angle. Visualize:

Solve: (a) To reach your eye, a light ray must refract through the top surface of the water and into the air. You can see in the figure that rays coming from the bottom of the tank are incident on the top surface at fairly small angles, but rays from the marks near the top of the tank are incident at very large angles—greater than the critical angle. These rays undergo total internal reflection in the water and do not exit into the air where they can be seen. Thus you can see the marks from the bottom of the tank upward. (b) The highest point you can see is the one from which the ray reaches the top surface at the critical angle θc. For a water-air boundary, the critical angle is θc = sin–1(1/1.33) = 48.75°. You can see from the figure that the depth of this point is such that L L 65.0 cm = tanθ c ⇒ d = = = 57.0 cm d tanθ c tan(48.75°)

Since the marks are every 10 cm, the high mark you can see is the one at 60 cm.

23.50. Model: Use the ray model of light and the law of refraction. Assume the sun is a point source of light. Visualize:

When the bottom of the pool becomes completely shaded, a ray of light that is incident at the top edge of the swimming pool does not reach the bottom of the pool after refraction. Solve: The depth of the swimming pool is d = 4.0 m tan θ water . We will find the angle by using Snell’s law. We have 4.0 m ⎛ sin 70° ⎞ nwater sin θ water = nair sin 70° ⇒ θ water = sin −1 ⎜ = 4.0 m ⎟ = 44.95° ⇒ d = tan 44.95° ⎝ 1.33 ⎠

23.51. Model: Use the ray model of light and the law of refraction. Assume that the laser beam is a ray of light. Visualize:

The laser beam enters the water 2.0 m from the edge, undergoes refraction, and illuminates the goggles. The ray of light from the goggles then retraces its path and enters your eyes. Solve: From the geometry of the diagram, tan φ =

1.0 m ⇒ φ = tan −1 ( 0.50 ) = 26.57° ⇒ θ air = 90° − 26.57° = 63.43° 2.0 m

Snell’s law at the air-water boundary is nair sinθ air = nwater sinθ water . Using the above result,

(1.0 ) sin 63.43° = 1.33sinθ water

⎛ sin 63.43° ⎞ ⇒ θ water = sin −1 ⎜ ⎟ = 42.26° ⎝ 1.33 ⎠

Taking advantage of the geometry in the diagram again, x = tan θ water ⇒ x = ( 3.0 m ) tan 42.26° = 2.73 m 3.0 m

The distance of the goggles from the edge of the pool is 2.73 m + 2.0 m = 4.73 m ≈ 4.7 m.

23.52. Model: Use the ray model of light and the law of refraction. Assume that the laser beam is a ray of light. Visualize:

Solve:

(a) From the geometry of the diagram at side A,

tan φ =

10 cm ⎛ 10 ⎞ ⇒ φ = tan −1 ⎜ ⎟ ⇒ φ = 33.69° 15 cm ⎝ 15 ⎠

This means the angle of incidence at side A is θair = 90° − 33.69° = 56.31°. Using Snell’s law at side A, ⎛ 1.0sin 56.31° ⎞ nair sin θ air = nwater sinθ water A ⇒ θ water A = sin −1 ⎜ ⎟ = 38.73° 1.330 ⎝ ⎠ This ray of light now strikes side B. The angle of incidence at this water-air boundary is θ water B = 90° − θ water A = 51.3°. The critical angle for the water-air boundary is

⎛ nair ⎞ −1 ⎛ 1.0 ⎞ ⎟ = sin ⎜ ⎟ = 48.8° ⎝ 1.33 ⎠ ⎝ nwater ⎠

θ c = sin −1 ⎜

Because the angle θwater B is larger than θc, the ray will experience total internal reflection. (b) We will now repeat the above calculation with x = 25 cm. From the geometry of the diagram at side A, φ = 21.80° and θ air = 68.20°. Using Snell’s law at the air-water boundary, θ water A = 44.28° and θ water B = 45.72°. Because θ water B < θ c , the ray will be refracted into the air. The angle of refraction is calculated as follows: ⎛ 1.33sin 45.72° ⎞ nair sin θ air B = nwater sinθ water B ⇒ θ air B = sin −1 ⎜ ⎟ = 72.2° 1 ⎝ ⎠ (c) Using the critical angle for the water-air boundary found in part (a), θwater A = 90° − 48.75° = 41.25°. According to Snell’s law, ⎛ 1.33sin 41.25° ⎞ nair sin θ air = nwater sinθ water A ⇒ θ air = sin −1 ⎜ ⎟ = 61.27° 1.0 ⎝ ⎠

⇒ φ = 90° − 61.27° = 28.73° The minimum value of x for which the laser beam passes through side B and emerges into the air is calculated as follows: tan φ =

10 cm 10 cm ⇒ x= = 18.2 cm ≈ 18 cm tan 28.73° x

23.53. Model: Use the ray model of light. Assume the bonfire is a point source right at the corner of the lake. Visualize:

Solve: (a) Light rays from the fire enter the lake right at the edge. Even though the rays in air are incident on the surface at a range of angles from ≈0° up to 90°, the larger index of refraction of water causes the rays to travel downward in the water with angles ≤θc, the critical angle. Some of these rays can reach a fish that is deep in the lake, but a shallow fish out from shore is in the “exclusion zone” that is not reached by any rays from the fire. Thus a fish needs to be deep to see the light from the fire. (b) The shallowest fish that can see the fire is one that receives light rays refracting into the water at the critical angle θc. These are rays that were incident on the water’s surface at ≈90°. The critical angle for a water-air boundary is

⎛ 1.00 ⎞ ⎟ = 48.75° ⎝ 1.33 ⎠

θ c = sin −1 ⎜ The fish is 20 m from shore, so its depth is d=

20 m = 17.5 m ≈ 18 m tan(48.75°)

That is, a fish 20 m from shore must be at least 18 m deep to see the fire.

23.54. Model: Use the ray model of light. Assume that the target is a point source of light. Visualize:

Solve:

From the geometry of the figure with θair = 60°,

tan θ air =

x1 ⇒ x1 = ( 2.0 m )( tan 60° ) = 3.464 m 2.0 m

Let us find the horizontal distance x2 by applying Snell’s law to the air-water boundary. We have

⎛ sin 60° ⎞ nwater sinθ water = nair sin θ air ⇒ θ water = sin −1 ⎜ ⎟ = 40.63° ⎝ 1.33 ⎠ Using the geometry of the diagram, x2 = tanθ water ⇒ x2 = (1.0 m ) tan 40.63° = 0.858 m 1.0 m

To determine θtarget, we note that tanθ target =

3.0 m 3.0 m = = 0.6941 ⇒ θtarget = 35° x1 + x2 3.464 m + 0.858 m

23.55. Model: Use the ray model of light and the phenomena of refraction and dispersion. Visualize:

The refractive index of violet light is greater than the refractive index of red light. The violet wavelength thus gets refracted more than the red wavelength. Solve: Using Snell’s law for the red light at the air-glass boundary,

⎛ n sinθ air ⎞ −1 ⎛ 1.0sin 30° ⎞ nair sinθ air = nred sin θ red ⇒ θ red = sin −1 ⎜ air ⎟ = sin ⎜ ⎟ = 19.30° n ⎝ 1.513 ⎠ red ⎝ ⎠ From the geometry of the diagram, d red = tan θ red 10.0 cm

d violet = tanθ violet 10.0 cm

⇒ d red = (10.0 cm ) tan (19.30° ) = 3.502 cm ⇒ d violet = 3.502 cm − 0.1 cm = 3.402 cm

⎛ d ⎞ ⎛ 3.402 cm ⎞ ⇒ θ violet = tan −1 ⎜ violet ⎟ = tan −1 ⎜ ⎟ = 18.79° ⎝ 10.0 cm ⎠ ⎝ 10.0 cm ⎠ That is, white light is incident on a piece of glass at 30°, and it gets dispersed. The violet light makes an angle of 18.79° with the vertical. Using Snell’s law, nviolet sinθ violet = nair sin θ air ⇒ nviolet =

(1.0 ) sin 30° = 1.552 sin18.79°

23.56. Model: Use the ray model of light and the phenomena of refraction and dispersion. Visualize:

Solve: Since violet light is perpendicular to the second surface, it must reflect at θviolet = 30° at the first surface. Using Snell’s law at the air-glass boundary where the ray is incident,

nair sin θ air = nviolet sinθ violet ⇒ nviolet =

nair sin θ air (1.0 ) sin 50° = = 1.5321 sin θ violet sin 30°

Since nviolet = 1.02 nred, nred = 1.5021. Using Snell’s law for the red light at the first surface

⎛ 1.0sin 50° ⎞ nred sin θ red = nair sin θ air ⇒ θ red = sin −1 ⎜ ⎟ = 30.664° ⎝ 1.5021 ⎠ The angle of incidence on the rear face of the prism is thus θr glass = 30.664° − 30° = 0.664°. Using Snell’s law once again for the rear face and for the red wavelength, ⎛ n sin θ r glass ⎞ −1 ⎛ 1.5021sin 0.664° ⎞ nred sin θ r glass = nair sinθ r air ⇒ θ r air = sin −1 ⎜ red ⎟ = sin ⎜ ⎟ = 0.997° n 1.0 ⎝ ⎠ air ⎝ ⎠

Because θv air = 0° and θr air = 0.997°, φ = θ r air − θ v air = 0.997° ≅ 1.00°.

23.57. Model: Use the ray model of light and the phenomenon of refraction. Visualize:

Solve:

(a) The critical angle θc for the glass-air boundary is

⎛ 1.0 ⎞ nglass sin θ c = nair sin 90° ⇒ θ c = sin −1 ⎜ ⎟ = 41.81° ⎝ 1.50 ⎠ For the triangle ABC,

θ glass 1 + 120° + ( 90° − θ c ) = 180° ⇒ θ glass 1 = 180° − 120° − ( 90° − 41.81° ) = 11.81° Having determined θglass 1, we can now find θair 1 by using Snell’s law: ⎛ 1.50 × sin11.81° ⎞ nair sin θ air 1 = nglass sinθ glass 1 ⇒ θ air 1 = sin −1 ⎜ ⎟ = 17.88° 1.0 ⎝ ⎠ Thus, the smallest angle θ1 for which a laser beam will undergo TIR on the hypotenuse of this glass prism is 17.9°. (b) After reflecting from the hypotenuse (face 3) the ray of light strikes the base (face 2) and refracts into the air. From the triangle BDE,

( 90° − θ

glass 2

) + 60° + ( 90° − θ ) = 180° ⇒ θ c

glass 2

= 90° + 60° + 90° − 41.81° − 180° = 18.19°

Snell’s law at the glass-air boundary of face 2 is

⎛ n sinθ glass 2 ⎞ −1 ⎛ 1.50sin18.19° ⎞ nglass sinθ glass 2 = nair sin θ air 2 ⇒ θ air 2 = sin −1 ⎜ glass ⎟ = sin ⎜ ⎟ = 27.9° 1.0 n ⎝ ⎠ air ⎝ ⎠ Thus the ray exits 27.9° left of the normal.

23.58. Model: Use the ray model of light. Solve:

(a) Using Snell’s law at the air-glass boundary, with φ being the angle of refraction inside the prism,

nair sin β = n sin φ ⇒ sin β = n sin φ Considering the triangle made by the apex angle and the refracted ray,

( 90° − φ ) + ( 90° − φ ) + α = 180° ⇒ φ = 12 α Thus sin β = n sin ( 12 α ) ⇒ β = sin −1 ( n sin( 12 )α )

(b) Using the above expression, we obtain

n=

sin β sin 52.2° = = 1.58 sin( 12 α ) sin 30°

23.59. Model: The bubble is a point source of light. The surface is a spherical refracting surface.

Solve: The bubble is in zircon, so n1 = 1.96 and n2 = 1.00. The surface is concave (object facing into a “cave”) as seen from the bubble (the object), so R = −3.0 cm. Equation 23.21 is n1 n2 n2 − n1 1.96 1.00 1.00 − 1.96 + = ⇒ + = ⇒ s′ = −3.0 cm s s′ R 3.0 cm s′ −3.0 cm

Thus, seen from outside, the bubble appears to be 3.0 cm beneath the surface. That is, a bubble at the center actually appears to be at the center.

23.60. Model: Use the ray model of light. The surface is a spherically refracting surface. Visualize:

Solve: Because the rays are parallel, s = ∞. The rays come to focus on the rear surface of the sphere, so s′ = 2R, where R is the radius of curvature of the sphere. Using Equation 23.21,

n1 n2 n2 − n1 1 n n −1 + = ⇒ + = ⇒ n = 2.00 s s′ R ∞ 2R R

23.61. Model: Assume that the converging lens is a thin lens. Use ray tracing to locate the image. Solve:

(a)

The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three rays after refraction converge to give an image at s′ = 40 cm. The height of the image is h′ = 2 cm. (b) Using the thin-lens formula,

1 1 1 1 1 1 1 1 + = ⇒ + = ⇒ = ⇒ s′ = 40 cm s s′ f 40 cm s′ 20 cm s′ 40 cm The image height is obtained from M =−

s′ 40 cm =− = −1 s 40 cm

The image is inverted and as tall as the object, that is, h′ = 2.0 cm. The values for h′ and s′ obtained in parts (a) and (b) agree.

23.62. Model: Use ray tracing to locate the image. Assume that the converging lens is a thin lens. Solve:

(a)

The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three special rays that experience refraction do not converge at a point. Instead they appear to come from a point that is 15 cm on the same side as the object itself. Thus s′ = −15 cm. The image is upright and has a height of h′ = 1.5 cm. (b) Using the thin-lens formula,

1 1 1 1 1 1 1 1 =− ⇒ s′ = −15 cm + = ⇒ + = ⇒ ′ ′ ′ s s f 10 cm s 30 cm s 15 m The image height is obtained from M =−

s′ −15 cm =− = +1.5 s 10 cm

The image is upright and 1.5 times the object, that is, 1.5 cm high. These values agree with those obtained in part (a).

23.63. Model: Use ray tracing to locate the image. Assume that the converging lens is a thin lens. Solve:

(a)

The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three special rays after refracting do not converge. Instead the rays appear to come from a point that is 60 cm on the same side of the lens as the object, so s′ = −60 cm. The image is upright and has a height of 8.0 cm. (b) Using the thin-lens formula,

1 1 1 1 1 1 1 1 + = ⇒ =− ⇒ s′ = −60 cm + = ⇒ s s′ f 15 cm s′ 20 cm s′ 60 cm The image height is obtained from M =−

s′ −60 cm =− = +4 s 15 cm

Thus, the image is 4 times larger than the object or h′ = Mh = 4h = 4 ( 2.0 cm ) = 8.0 cm. The image is upright. These values agree with those obtained in part (a).

23.64. Model: Use ray tracing to locate the image. Assume the converging lens is a thin lens. Solve:

(a)

The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. After refraction, the three special rays converge and give an image 50 cm away from the converging lens. Thus, s′ = +50 cm. The image is inverted and its height is 0.65 cm. (b) Using the thin-lens formula,

1 1 1 1 1 1 1 1 ⇒ = + = ⇒ + = ⇒ s′ = 50 cm ′ ′ ′ s s f 75 cm s 30 cm s 50 cm The image height is obtained from M =−

s′ 50 cm 2 =− =− 75 cm 3 s

The image height is h′ = Mh = ( −2 / 3)(1 cm ) = −0.67 cm. Because of the negative sign, the image is inverted.

These results agree with those obtained in part (a).

23.65. Model: Use ray tracing to locate the image. Assume the diverging lens is a thin lens. Solve:

(a)

The figure shows the ray-tracing diagram using the steps of Tactics Box 23.3. After refraction, the three special rays do not converge. The rays, on the other hand, appear to meet at a point that is 8.5 cm on the same side of the lens as the object. So s′ = −8.5 cm. The image is upright and has a height of 1.1 cm. (b) Using the thin-lens formula, 1 7 60 1 1 1 1 1 1 ⇒ =− ⇒ s′ = − cm = −8.6 cm + = ⇒ + = 15 cm s′ −20 cm s s′ f s′ 60 cm 7

The image height is obtained from

( −60/7 cm ) = + 4 = 0.57 s′ =− 15 cm 7 s Thus, the image is 0.57 times larger than the object, or h′ = Mh = (0.57)(2.0 cm) = 1.14 cm. The image is upright because M is positive. These values agree, within measurement accuracy, with those obtained in part (a). M =−

23.66. Model: Use ray tracing to locate the image. Assume the diverging lens is a thin lens. Solve:

(a)

The figure shows the ray-tracing diagram using the steps of Tactics Box 23.3. After refraction from the diverging lens, the three special rays do not converge. However, the rays appear to meet at a point that is 20 cm on the same side as the object. So s′ = −20 cm. The image is upright and has a height of 0.3 cm. (b) Using the thin-lens formula,

1 1 1 1 1 1 = − = − =− ⇒ s′ = −20 cm ′ s f s −30 cm 60 cm 20 cm The image height is obtained from M =−

s′ −20 cm 1 =− = = 0.33 s 60 cm 3

Thus, h′ = Mh = ( 0.33)(1.0 cm ) = 0.33 cm, and the image is upright because M is positive. These values for s′ and h′ agree with those obtained in part (a).

23.67. Visualize: Refer to Figure 23.57. 1 1 1 fs + = ⇒ s′ = s s′ f s− f We are given f = 60 cm, s = 20 cm, and h = 1.0 cm. Solve: fs (60 cm)(20 cm) s′ = = = −30 cm s− f 20 cm − 60 cm The negative sign means the image is behind the mirror; it is a virtual image. The magnification is m = − s′ s = 30 cm 20 cm = 1.5. This means the image is upright and has a height of h′ = mh = (1.5)(1.0 cm) = 1.5 cm. Assess: Ray tracing will confirm these results.

23.68. Visualize: Refer to Figure 23.55. 1 1 1 fs + = ⇒ s′ = s s′ f s− f We are given f = −60 cm, s = 20 cm, and h = 1.0 cm. Solve:

s′ =

fs (−60 cm)(20 cm) = = −15 cm s− f 20 cm + 60 cm

The negative sign means the image is behind the mirror; it is a virtual image. The magnification is m = − s′ s = 15 cm 20 cm = 0.75. This means the image is upright and has a height of h′ = mh = (0.75)(1.0 cm) = 0.75 cm. Assess: Ray tracing will confirm these results.

23.69. Model: Assume the lens is a thin lens and the thin-lens formula applies. Solve: Because we want to form an image of the spider on the wall, the image is real and we need a converging lens. That is, both s′ and s are positive. This also implies that the spider’s image is inverted, so M = − s′ s = − 12 . Using the thin-lens formula with s′ = 12 s,

1 1 1 1 1 1 3 1 s + = ⇒ +1 = ⇒ = ⇒ f = s f 3 s s′ f s 2s f We also know that the spider is 2.0 m from the wall, so s + s′ = 2.0 m = s +

1 2

s ⇒ s = 13 ( 4.0 m ) = 133.3 cm

Thus, f = 13 s = 44 cm and s′ = 2.0 m − 1.33 m = 0.67 m = 67 cm. We need a 44 cm focal length lens placed 67 cm from the wall.

23.70. Model: Assume the lens to be a thin lens. Solve: Because we want to form an image of the candle on the wall, we need a converging lens. We have s + s′ = 200 cm. Using the thin-lens formula,

1 1 1 1 1 1 + = ⇒ + = ⇒ s 2 − ( 200 cm ) s + 6400 cm 2 = 0 ′ s s f s 200 cm − s 32 cm The two solutions to this equation are s = 160 cm and 40 cm. When s = 160 cm, then s′ = 200 cm − 160 cm = 40 cm. The magnification is s′ 40 cm M =− =− = −0.25 s 160 cm

so the image is inverted and its height is (2.0 cm)(0.25) = 0.50 cm. When s = 40 cm, then s′ = 200 cm − 40 cm = 160 cm. The magnification is s′ 160 cm M =− =− = −4 s 40 cm so the image is again inverted and its height is (2.0 cm)(4) = 8.0 cm.

23.71. Model: The eye is a converging lens and assume it is a thin lens. Solve: (a) The diameter of an adult eyeball is typically 4.0 cm. (b) The near point distance is approximately 10 inches ≈ 25 cm. (c) Using the thin-lens formula,

1 1 1 1 1 1 1 29 + = ⇒ = ⇒ f ≈ 3.4 cm + = ⇒ 25 cm 4.0 cm f f 100 cm s s′ f

23.72. Model: Assume the projector lens is a thin lens. Solve:

(a) The absolute value of the magnification of the lens is

M =

h′ 98 cm = = 49 h 2 cm

Because the projector forms a real image of a real object, the image will be inverted. Thus, M = −49 = −

s′ ⇒ s′ = 49 s s

We also have s + s′ = 300 cm ⇒ s + 49s = 300 cm ⇒ s = 6.0 cm ⇒ s′ = 294 cm

Using these values of s and s′, we can find the focal length of the lens:

1 1 1 1 1 = + = + ⇒ f = 5.9 cm ′ f s s 6.0 cm 294 cm (b) From part (a) the lens should be 6.0 cm from the slide.

23.73. Visualize: The lens must be a converging lens for this scenario to happen, so we expect f to be positive. In the first case the upright image is virtual ( s′2 < 0) and the object must be closer to the lens than the focal point. The lens is then moved backward past the focal point and the image becomes real ( s′2 > 0).

1 1 1 ss′ + = ⇒ f = s s′ f s + s′ We are given s1 = 10cm and m1 = 2. Solve: Since the first image is virtual, s′ < 0. We are told ′ m1 = 2 = − s1 s1 ⇒ s1′ = −20 cm. We can now find the focal length of the lens. f =

the

first

magnification

is

s1s1′ (10 cm)(−20 cm) = = 20 cm s1 + s1′ 10 cm − 20 cm

After the lens is moved, m2 = −2 = − s2′ s2 . Start with the thin lens equation again.

1 1 1 + = s2 s′2 f Replace s′2 with − m2 s2 . 1 1 1 + = s2 − m2 s2 f Now solve for s2 . −m2 s2 + s2 1 = s2 (−m2 s2 ) f −m2 s2 + s2 1 = f − m2 s22 Cancel one s2 . m2 − 1 1 = m2 s2 f Multiply both sides by fs2 . ⎛ m −1⎞ ⎛ −2 − 1 ⎞ s2 = f ⎜ 2 ⎟ = (20 cm) ⎜ ⎟ = 30 cm ⎝ −2 ⎠ ⎝ m2 ⎠ The distance the lens moved is s2 − s1 = 30 cm − 10 cm = 20 cm. Assess: We knew s2 needed to be bigger than f; it is, and is in a reasonable range. The final answer for the distance the lens moved also seems reasonable.

23.74. Model: Assume the symmetric converging lens is a thin lens. Solve: Because the lens forms a real image on the screen of a real object, the image is inverted. Thus, M = −2 = − s′ s . Also, s + s′ = 60 cm ⇒ s +2s = 60 cm ⇒ s = 20 cm ⇒ s′ = 40 cm

We can use the thin-lens formula to determine the radius of curvature of the symmetric converging lens ( R1 = R2 ) as follows:

⎛1 1 ⎞ 1 1 1 + = = ( n − 1) ⎜ − ⎟ s s′ f ⎝ R1 R2 ⎠ Using R1 = + R (convex toward the object), R2 = − R (concave toward the object), and n = 1.59,

⎛ 1 1 1 1 + = (1.59 − 1) ⎜⎜ − R − R 20 cm 40 cm ⎝

⎞ 3 1.18 ⇒ R = 15.7 cm ≈ 16 cm = ⎟⎟ ⇒ 40 cm R ⎠

23.75. Visualize: We are given f = R 2 = 40 cm 2 = 20 cm. We are also given m = − s′ s = 3. Solve:

Solve the thin lens equation for s′.

1 1 1 + = s s′ f s′ =

fs s− f

Plug this into the magnification equation, s = − s′ m . s=−

1 fs m s− f

Cancel an s from the numerator of each side, and multiply both sides by s − f . s− f =− s= f −

f m

f 1⎞ ⎛ ⎛ 1⎞ = f ⎜1 − ⎟ = ( 20 cm ) ⎜ 1 − ⎟ = 13.3 cm m ⎝ m⎠ ⎝ 3⎠

Assess: This answer is within the focal length of the concave mirror as we expect for an upright, magnified, virtual image.

23.76. Visualize: We are given h = 2.0 cm and h′ = 1.0 cm so we know m = 0.5. Solve:

Solve m = − s′ s for s . s=

Plug this result into the thin lens equation.

− s′ m

1 1 1 ss′ + = ⇒ f = s s′ f s + s′

⎛ − s′ ⎞ ′ ⎛ − s′ ⎞ ′ ⎜ ⎟( s ) ⎜ ⎟(s ) ss′ m ⎠ m ⎠ f = =⎝ =⎝ − s′ 1⎞ ⎛ s + s′ + s′ s′ ⎜ 1 − ⎟ m ⎝ m⎠ Cancel one s′ and distribute the m in the denominator. − s′ − s′ 150 cm = = = −300 cm f = 1 ⎞ m − 1 0.5 − 1 ⎛ m ⎜1 − ⎟ ⎝ m⎠ Assess: The negative value for f tells us this is a convex mirror.

23.77. Visualize: We are given f = 25 cm. From the thin mirror equation we know that s′ =

fs . The five s− f

points under consideration are s = 50 cm, 75 cm, 100 cm, 125 cm, 150 cm. fs Solve: (a) Use s′ = for each of the five points. s− f s (cm) s′ (cm) h (cm) h′ (cm) 50 50 10 10 75 37.5 10 5.0 100 33.3 10 3.3 125 31.3 10 2.5 150 30 10 2.0 (b)

(c) The image is curved and not parallel to the axis. Assess: For further reading, see “Longitudinal Magnification Drawing Mistake” by Héctor Rabal, Nelly Cam, and Marcelo Trivi in The Physics Teacher, vol. 42, January 2004, pp. 31–33, but know that Equation 4 there is missing a couple of minus signs.

23.78. Visualize: First concentrate on the optic axis and the ray parallel to it. Geometry says if parallel lines are both cut by a diagonal (in this case the line through the center of curvature and normal to the mirror at the point of incidence) the interior angles are equal; so φ = θ i . The law of reflection says that θ i = θ r , so we conclude φ = θ r . Now concentrate on the triangle whose sides are R, a, and b. Because two of the angles are equal then it is isosceles; therefore b = a. Apply the law of cosines to this triangle. Solve:

b 2 = a 2 + R 2 − 2aR cos φ Because a = b, they drop out. R 2 = 2aR cos φ R = 2a cos φ

We want to know how big a is in terms of R, so solve for a. a=

R 2cos φ

If φ  1 then cos φ ≈ 1, so in the limit of small φ , a = R 2, and then since f = R − a it must also be that f = Assess:

R 2

Many textbooks forget to stress that f = R 2 only in the limit of small φ , i.e., for paraxial rays.

23.79. Model: Use the ray model of light and assume the lens is a thin lens. Visualize: Please refer to Figure 23.47. Solve: Let n1 be the refractive index of the fluid and n2 the refractive index of the lens. The lens consists of two spherical surfaces having radii of curvature R1 and R2 and the lens thickness t → 0. For the refraction from the surface with radius R1, we use Equation 23.21:

n1 n2 n2 − n1 + = s1 s1′ R1 For the refraction from surface with radius R2,

n2 n1 n1 − n2 + = R2 − s1′ s′2 A negative sign is used with s1′ because the image formed by the first surface of the lens is a virtual image. This virtual image is the object for the second surface. Adding the two equations,

⎛1 1 ⎞ n1 n1 1 1 1 (n − n ) ⎛ 1 1 ⎞ + = n2 − n1 ⎜ − ⎟ ⇒ + = = 2 1 ⎜ − ⎟ ′ s1 s2 s1 s′2 f n1 ⎝ R1 R2 ⎠ ⎝ R1 R2 ⎠ (b) In air, R1 = +40 cm (convex toward the object), R2 = −40 cm (concave toward the object), n1 = 1.0, and n2 = 1.50. So,

1 ⎛ 1.50 − 1.0 ⎞⎛ 1 1 ⎞ =⎜ − ⎟⎜ ⎟ ⇒ f = 40 cm f ⎝ 1.0 ⎠⎝ 40 cm −40 cm ⎠ In water, n1 = 1.33 and n2 = 1.50. So, 1 ⎛ 1.50 − 1.33 ⎞⎛ 1 1 ⎞ =⎜ − ⎟⎜ ⎟ ⇒ f = 156 cm f ⎝ 1.33 ⎠⎝ 40 cm −40 cm ⎠

23.80. Model: Use the ray model of light. Solve:

(a) The time (t) is the time to travel from A to the interface (t1) and from the interface to B (t2). That is,

t = t1 + t2 =

d1 d 2 d d nd n d n n + = 1 + 2 = 1 1 + 2 2 = 1 x2 + a2 + 2 v1 v2 c n1 c n2 c c c c

( w − x)

2

+ b2

(b) Because t depends on x and there is only one value of x for which the light travels from A to B in the least possible amount of time, we have

n2 ( w − x ) dt n1 x =0= − 2 2 2 dx c x +a c ( w − x ) + b2 The solution (hard to do!) would give xmin. (c) From the geometry of the figure, x x2 + a2

=

x = sin θ1 d1

w− x

( w − x)

2

+b

2

=

w− x = sin θ 2 d2

Thus, the condition of part (b) becomes n1 n sin θ1 − 2 sinθ 2 = 0 ⇒ n1 sin θ1 = n2 sinθ 2 c c

23.81. Model: Assume the ray model of light. The ball is not a thin lens. However, the image due to refraction from the first surface is the object for the second surface. Visualize:

Solve:

(a) For refraction from the first surface, R = +5 cm (convex toward the object). Thus,

n1 n2 n2 − n1 1.0 1.50 0.50 1.50 1 + = ⇒ + = ⇒ =− ⇒ s1′ = −22.5 cm s1 s1′ R 6 cm s1′ 5 cm s1′ 15 cm The image is virtual (to the left of the surface) and upright. For refraction from the second surface, s2 = 22.5 cm + 10 cm = 32.5 cm and R = −5.0 cm (concave toward the object). Thus,

1.50 1.0 1 − 1.50 1 1 1 1.50 + = = = − ⇒ s2′ = 18.6 cm ⇒ 32.5 cm s′2 −5.0 cm 10 cm s2′ 10 cm 32.5 cm The image is 18.6 cm from the right edge of the ball and thus 23.6 cm from the center. (b) The ray diagram showing the formation of the image is shown above. (c) Using the thin-lens equation, 1 1 1 1 1 1 + = ⇒ + = ⇒ f = 7.5 cm 6 cm + 5 cm 18.6 cm + 5 cm f s s′ f

23.82. Model: Use the ray model of light. Visualize:

The angle of refraction is θ 2 + δθ for those wavelengths that have a refractive index of n + δn. Solve: (a) Applying Snell’s law to the diagrams,

(1) sinθ1 = n sinθ 2

(1) sinθ1 = ( n + δ n ) sin (θ 2 + δθ )

Equating the right hand sides of the above two equations and using the formula for the sine of a sum, n sin θ 2 = ( n + δ n )( sin θ 2 cos δθ + cosθ 2 sin δθ ) = ( n + δ n )( sin θ 2 + cosθ 2δθ ) where we have assumed that δθ  θ . Multiplying the expressions, n sinθ 2 = n sinθ 2 + n cosθ 2δθ + δ n sinθ 2 + δ nδθ cosθ 2 We can ignore the last term on the right-hand side because it is the product of two small terms. The equation becomes

⎛δn⎞ n cosθ 2δθ = −δ n sin θ 2 ⇒ δθ = − tan θ 2 ⎜ ⎟ ⎝ n ⎠ Note that δθ has to be in radians. We can obtain the same result in the following way as well. From Snell’s law, sin θ 2 =

sin θ1 n

Differentiating relative to n

δθ n sin θ 2 − sinθ 2 ⎛ sinθ 2 ⎞ ⎛ 1⎞ = ( sinθ1 ) ⎜ − 2 ⎟ = − = ⎟ = cosθ 2 δ n δ n n n2 n ⎝ ⎠ ⎝ ⎠ ⎛δn⎞ ⇒ δθ = − tanθ 2 ⎜ ⎟ ⎝ n ⎠

δ⎜

(b) We have θ1 = 30° and nred = 1.552. Because the red wavelength is larger than the violet wavelength, nred < nviolet. Also, if the refraction angle for the red light is θ2, the refraction angle for the violet is less than θ2. Thus, δθ = −0.28°. From the formula obtained in part (a),

n ⎛δn⎞ δθ ⎟ ⇒ δn = − θ2 tan n ⎝ ⎠

δθ = − tanθ 2 ⎜ To determine tanθ2, we note that

⎛ sin 30° ⎞ nred sinθ 2 = nair sin 30° ⇒ θ 2 = sin −1 ⎜ ⎟ = 18.794° ⇒ tan θ 2 = 0.3403 ⎝ 1.552 ⎠ Thus, the expression for the change in the index of refraction is

δn = −

1.552 π rad ⎞ ( −0.28° ) ⎛⎜ ⎟ = 0.0223 ⇒ nviolet = nred + δn = 1.552 + 0.022 = 1.574 0.3403 ⎝ 180° ⎠

23.83. Visualize: The lateral magnification is m = − s′ s . Solve: We need to solve the thin-lens equation for s′.

1 1 1 + = s s′ f 1 1 1 = − s′ f s

s′ =

1 fs = 1 1 s− f − f s

Now insert this expression for s′ into the expression for m. fs s′ f s− f m=− =− =− s s s− f Now define the longitudinal magnification as the rate of change of s′ with respect to s. M=

ds′ d d ⎛ fs ⎞ = ( s′ ) = ⎜ ⎟ ds ds ds ⎝ s − f ⎠

Use the quotient rule of differentiation. M=

d ⎛ fs ⎞ ( s − f ) f − fs −f2 = ⎜ ⎟= 2 2 ds ⎝ s − f ⎠ (s − f ) (s − f )

This last result is equal to − m2 , so M = − m 2 . Assess: For further reading, see “Longitudinal Magnification Drawing Mistake” by Héctor Rabal, Nelly Cam, and Marcelo Trivi in The Physics Teacher, vol. 42, January 2004, pp. 31–33, but know that Equation 4 there is missing a couple of minus signs.

23.84. Model: Assume the lens is thin so we can solve the thin-lens equation for s′ = Visualize:

fs . s− f

We are given f = 0.150 m and s = 10 m. We are also given ds dt = 5.0 m s. Refer to Problem

23.83 to learn about longitudinal magnification: M = ds′ ds = −m 2 . Solve:

The speed of the image is ds′ dt . Use the chain rule. Then apply m = − s′ s . ⎛ fs ⎜ s− f ′ ds′ ds′ ds ds d s s d s ⎛ ⎞ = =M = −m2 = − ⎜ ⎟ = −⎜ dt ds dt dt dt ⎝ s ⎝ s ⎠ dt 2

2

⎛ f ⎞ ds ds ′ = −⎜ ⎟ dt s =10 m ⎝ s − f ⎠ dt

Assess:

2

⎞ 2 ⎟ ds ⎛ f ⎞ ds ⎟ = −⎜ ⎟ ⎠ dt ⎝ s − f ⎠ dt

2

s =10 m

0.15 m ⎛ ⎞ =⎜ ⎟ ( 5.0 m s ) = −1.16 mm s ⎝ 10 m − 0.15 m ⎠

The negative answer indicates the image is moving toward the lens.

24.1. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize:

The figure shows the two lenses and a ray-tracing diagram. The ray-tracing shows that the lens combination will produce a real, inverted image behind the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is ≈ 50 cm from the second lens and the height of the final image is 4.5 cm. (b) s1 = 15 cm is the object distance of the first lens. Its image, which is a virtual image, is found from the thinlens equation:

1 1 1 1 1 5 = − = − =− ⇒ s1′ = −24 cm s1′ f1 s1 40 cm 15 cm 120 cm The magnification of the first lens is m1 = −

s1′ ( −24 cm ) = 1.6 =− s1 15 cm

The image of the first lens is now the object for the second lens. The object distance is s2 = 24 cm + 10 cm = 34 cm. A second application of the thin-lens equation yields: 1 1 1 1 1 680 cm = − = − ⇒ s2′ = = 48.6 cm s2′ f 2 s2 20 cm 34 cm 14 The magnification of the second lens is m2 = −

s′2 48.6 cm =− = −1.429 s2 34 cm

The combined magnification is m = m1m2 = (1.6 )( −1.429 ) = −2.286 . The height of the final image is (2.286)(2.0 cm) =

4.57 cm. These calculated values are in agreement with those found in part (a).

24.2. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize:

The figure shows the two lenses and a ray-tracing diagram. The ray tracing shows that the lens combination will produce a virtual, inverted image in front of the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is 20 cm in front of the second lens and the height of the final image is 2.0 cm. (b) s1 = 60 cm is the object distance of the first lens. Its image, which is a real image, is found from the thin-lens equation:

1 1 1 1 1 1 = − = − = ⇒ s1′ = 120 cm s1′ f s s1 40 cm 60 cm 120 cm The magnification of the first lens is

m1 = −

s1′ 120 cm =− = −2 s1 60 cm

The image of the first lens is now the object for the second lens. The object distance is s2 = 160 cm −120 cm = 40 cm. A second application of the thin-lens equation yields: 1 1 1 −1 1 =− + = + ⇒ s′2 = −20 cm s2′ s2 f 2 +40 cm −40 cm The magnification of the second lens is m2 = −

s′2 −20 cm =− = +0.5 s2 40 cm

The overall magnification is m = m1m2 = ( −2 )( 0.5 ) = −1.0 . The height of the final image is (+1.0)(2.0 cm) = 2.0

cm. The image is inverted because m has a negative sign. These calculated values are in agreement with those found in part (a).

24.3. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize:

The figure shows the two lenses and a ray-tracing diagram. The ray tracing shows that the lens combination will produce a real, upright image behind the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is 10 cm behind the second lens and the height of the final image is 2 cm. (b) s1 = 20 cm is the object distance of the first lens. Its image, which is real and inverted, is found from the thin lens equation:

1 1 1 fs (10 cm)(20 cm) + = ⇒ s1′ = 1 1 = = 20 cm s1 s1′ f1 s1 − f1 20 cm − 10 cm The magnification of the first lens is m1 = −

s1′ 20 cm =− = −1 s1 20 cm

The image of the first lens is now the object for the second lens. The object distance is s2 = 30 cm − 20 cm = 10 cm. A second application of the thin-lens equation yields 1 1 1 fs (5 cm)(10 cm) + = ⇒ s2′ = 2 2 = = 10 cm s2 s2′ f 2 s2 − f 2 10 cm − 5 cm The magnification of the second lens is

m2 = −

s′2 10 cm =− = −1 s2 10 cm

The combined magnification is m = m1m2 = ( −1)(−1) = 1. The height of the final image is (1)(2.0 cm) = 2.0 cm. These calculated values are in agreement with those found in part (a). Assess: The thin-lens equation agrees with the ray tracing.

24.4. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize:

The figure shows the two lenses and a ray-tracing diagram. The ray tracing shows that the lens combination will produce a virtual, inverted image at the first lens. Solve: (a) From the ray-tracing diagram, we find that the image is 30 cm in front of the second lens and the height of the final image is 6 cm. (b) s1 = 20 cm is the object distance of the first lens. Its image, which is real and inverted, is found from the thin lens equation:

1 1 1 fs (10 cm)(20 cm) + = ⇒ s1′ = 1 1 = = 20 cm s1 s1′ f1 s1 − f1 20 cm − 10 cm The magnification of the first lens is m1 = −

s1′ 20 cm =− = −1 20 cm s1

The image of the first lens is now the object for the second lens. The object distance is s2 = 30 cm − 20 cm = 10 cm. A second application of the thin lens equation yields 1 1 1 fs (15 cm)(10 cm) + = ⇒ s′2 = 2 2 = = −30 cm s2 s2′ f 2 s2 − f 2 10 cm − 15 cm The magnification of the second lens is m2 = −

s′2 ( −30 cm) =− =3 10 cm s2

The combined magnification is m = m1m2 = ( −1)(3) = −3. The height of the final image is (3)(2.0 cm) = 6.0 cm. The image is inverted because m has a negative sign. These calculated values are in agreement with those found in part (a). Assess: The thin lens equation agrees with the ray tracing.

24.5. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize:

The figure shows the two lenses and a ray-tracing diagram. The ray tracing shows that the lens combination will produce a virtual, inverted image in front of the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is 3.3 cm behind the second lens and the height of the final image is ≈ 0.7cm. (b) s1 = 20 cm is the object distance of the first lens. Its image, which is real and inverted, is found from the thin-lens equation:

1 1 1 fs (10 cm)(20 cm) + = ⇒ s1′ = 1 1 = = 20 cm s1 s1′ f1 s1 − f1 20 cm − 10 cm The magnification of the first lens is m1 = −

s1′ 20 cm =− = −1 s1 20 cm

The image of the first lens is now the object for the second lens. The object distance is s2 = 30 cm − 20 cm = 10 cm. A second application of the thin-lens equation yields 1 1 1 fs ( −5 cm)(10 cm) + = ⇒ s2′ = 2 2 = = −3.33 cm s2 s2′ f 2 s2 − f 2 10 cm + 5 cm The magnification of the second lens is m2 = −

s′2 (−3.3 cm) =− = 0.33 s2 10 cm

The combined magnification is m = m1m2 = (−1)(0.33) = −0.33. The height of the final image is (0.33)(2.0 cm) = 0.66 cm. The image is inverted because m has a negative sign. These calculated values are in agreement with those found in part (a). Assess: The thin-lens equation agrees with the ray tracing.

24.6. Model:

s W f so we can use Equation 24.1: m = − f / s.

Solve:

h′ = mh = −

f 15 mm h=− (2.0 m) = −3.0 mm s 10 m

The height of the image on the detector is 3.0 mm. Assess: This seems reasonable given typical focal lengths and detector sizes.

24.7. Visualize: Equation 24.2 gives f-number = f / D. Solve:

f-number = Assess:

f 35 mm = = 5.0 D 7.0 mm

This is in the range of f-numbers for typical camera lenses.

24.8. Visualize: Solve Equation 24.2 for D. Solve:

D=

f 12 mm = = 3.0 mm f-number 4.0

Assess: This is in the same ballpark as the example after Equation 24.2.

24.9. Visualize: First we compute the f-number of the first lens and then the diameter of the second. Solve:

f-number =

f 12 mm = = 3.0 D 4.0 mm

Now for the new lens.

D= Assess:

f 18 mm = = 6.0 mm f-number 3.0

Given the same f-number, the longer focal length lens has a larger diameter.

24.10. Visualize: We want the same exposure in both cases. The exposure depends on I Δtshutter . We'll also use Equation 24.3. Solve:

exposure = I Δt ∝

1 Δt ( f -number) 2

1 1 Δt = Δt ′ 2 ( f -number) ( f -number)′2 Δt ′ =

( f-number)′2 (4.0) 2 ⎛ 1 ⎞ 1 1 Δt = s⎟ = s≈ s ⎜ 2 ( f-number) (5.6) 2 ⎝ 125 ⎠ 245 250

Assess: An alternate approach without a lot of calculation is that since we changed the lens (opened) by one f stop that doubles the intensity so we need half the time interval to achieve the same exposure.

24.11. Visualize: We want the same exposure in both cases. The exposure depends on I Δtshutter . We'll also use Equation 24.3. The lens is the same lens in both cases, so f = f ′. Solve: exposure = I Δt ∝

D2 Δt f2

D2 D′2 Δ = Δt ′ t f2 f ′2

Solve for D′; then simplify. 2

⎛ f ′ ⎞ ⎛ Δt ⎞ Δt 1 125s = (3.0 mm) = (3.0 mm) 4 = 6.0 mm D′ = D 2 ⎜ ⎟ ⎜ ⎟=D ′ ′ Δt 1 500s ⎝ f ⎠ ⎝ Δt ⎠ Assess: Since we decreased the shutter speed by a factor of 4 we need to increase the aperture area by a factor of 4, and this means increase the diameter by a factor of 2.

24.12. Model: Ignore the small space between the lens and the eye.

Visualize: Refer to Example 24.4, but we want to solve for s′, the near point. Solve: (a) The power of the lens is positive which means the focal length is positive, so Ramon wears converging lenses. This is the remedy for hyperopia. (b) We want to know where the image should be for an object s = 25 cm given 1 f = 2.0 m −1. f =

s′ =

1 = 0.50 m P

fs (0.50 m)(0.25 m) = = −0.50 m s− f 0.25 m + 0.50 m

So the near point is 50 cm. Assess: The negative sign on s′ is expected because we need the image to be virtual.

24.13. Model: Ignore the small space between the lens and the eye.

Visualize: Refer to Example 24.5, but we want to solve for s′, the far point. Solve: (a) The power of the lens is negative which means the focal length is negative, so Ellen wears diverging lenses. This is the remedy for myopia. (b) We want to know where the image should be for an object s = ∞ m given 1 f = −1.0 m −1. f =

1 = −1.0 m P

1 1 1 + = s s′ f When s = ∞ m, 1 1 1 + = ⇒ s′ = f = −1.0 m ∞ m s′ f So the far point is 100 cm. Assess: The negative sign on s′ is expected because we need the image to be virtual.

24.14. Model: With normal vision the farthest distance at which a relaxed eye can focus (the far point) is infinity. Rays coming from infinity are nearly parallel, so the focal length of the lens would be 24 mm, the same as the length of the eye. However, the far point is less than infinity for many people, and most sources quote the focal length of the eye as 17 mm–22 mm. However, for simplicity of calculation, assume the vision is normal and the focal length of the lens/cornea combination is 24 mm Visualize: Equation 24.2 gives f-number = f D . Solve: (a) For the fully dilated pupil (dark-adapted eye):

f-number =

f 24 mm = = 3.0 D 8.0 mm

(b) For the fully contracted pupil (eye in bright light):

f 24 mm = = 16 D 1.5 mm These answers correspond to the values given in the text. f-number =

Assess:

24.15. Model: The angle subtended by the image is 8 × the angle subtended by the object. Visualize: The angle subtended by the object is h s . Solve:

h s

θ = (8×) = (8×) Assess: The binoculars do indeed help.

14 cm = 0.0622 rad = 3.6° 1800 cm

24.16. Visualize: Equation 24.10 relates the variables in question: M =−

L 25 cm fobj feye

We are given M = 500×, L = 20 cm, and feye = 5.0 cm Solve:

Solve for f obj.

fobj = − Assess:

L 25 cm 20 cm 25 cm =− = 0.20 cm = 2.0 mm −500 5.0 cm M feye

This is in the same ball park as the case in the book.

24.17. Visualize: Equation 24.10 relates the variables in question: M =−

L 25 cm fobj feye

We are given M = 100 ×, L = 160 mm, and fobj = 8.0 cm. Solve:

Solve for f obj. feye = −

Assess:

L 25 cm 160 mm 25 cm =− = 5.0 cm f obj M 8.0 mm ( −100)

This is the same feye as in the previous exercise.

24.18. Model: Assume the thin-lens equation is valid. For part (b) refer to Equation 24.11 and the definition of α . Visualize:

We are given fobj = 9.0 mm and m=−

Solve:

s′ = −40 ⇒ s′ = 40 s s

(a) Use the thin-lens equation.

1 1 1 + = s s′ fobj

1 1 1 + = s 40s fobj 41 1 = 40s fobj s=

41 41 f obj = (9.0 mm) = 9.2 mm 40 40

(b) We are given n = 1.00. Refer to Figure 24.14a to determine α .

α = tan −1 (3.0 mm / 9.0 mm) = tan -1 (1/ 3) ⎛ ⎛ 1 ⎞⎞ NA = n sin α = (1.00) sin ⎜ tan −1 ⎜ ⎟ ⎟ = 0.32 ⎝ 3 ⎠⎠ ⎝ Assess: We used s = f in this calculation as suggested in the text. If we had used s = 9.2 mm (from part (a)) we would get NA = 0.31 , only a little different. These values of NA are typical for a simple microscope.

24.19. Visualize: We are given NA = 0.90, L = 160 mm, mobj = −20, and the book says n = 1.46. Solve:

Start with Equation 24.11. NA = n sin α ⎛ NA ⎞ ⎟ ⎝ n ⎠ ⎛ ⎛ NA ⎞ ⎞ tan α = tan ⎜ sin −1 ⎜ ⎟⎟ ⎝ n ⎠⎠ ⎝

α = sin −1 ⎜

⎛1⎞ But from Figure 24.14a we also have tan α = ⎜ ⎟ D / f obj , where D is the diameter of the lens. So combine those ⎝ 2⎠ two expressions for tan α and solve for D. ⎛ ⎛ NA ⎞ ⎞ D = 2 f obj tan ⎜ sin −1 ⎜ ⎟⎟ ⎝ n ⎠⎠ ⎝ We need the side calculation using Equation 24.9: mobj = −

L L ⇒ f obj = − f obj mobj

Insert this back in the equation for D. ⎛ L ⎞ ⎛ −1 ⎛ NA ⎞ ⎞ D = 2⎜ − tan sin ⎜ ⎟⎟ ⎜ mobj ⎟⎟ ⎜⎝ ⎝ n ⎠⎠ ⎝ ⎠ ⎛ 160 mm ⎞ ⎛ −1 ⎛ 0.90 ⎞ ⎞ D = 2⎜ − ⎟ tan ⎜ sin ⎜ ⎟ ⎟ = 13 mm −20 ⎠ ⎝ ⎝ ⎝ 1.46 ⎠ ⎠ Assess:

The answer seems to be in a reasonable range for objective lens diameter.

24.20. Visualize: Figure 24.15 shows from similar triangles that for the eyepiece lens to collect all the light Dobj f obj

=

Deye f eye

We also see from Equation 24.12 that M = − f obj / f eye . We are given M = −20 and Dobj = 12 cm. Solve:

Deye = Dobj

f eye f obj

=

Dobj −M

=

12cm = 0.60 cm = 6.0 mm 20

Assess: The answer is almost as wide as a dark-adapted eye.

24.21. Model: Assume the eyepiece is a simple magnifier with M eye = 25 cm / f eye . Visualize:

f eye = 25 cm /10 = 2.5 cm.

Solve: (a) The magnification of a telescope is

M =−

f obj f eye

=

100 cm = 40 2.5 cm

(b)

f-number =

f 1.00 m = = 5.0 D 0.20 m

Assess: These results are in reasonable ranges for magnification and f-number.

24.22. Model: Diffraction prevents focusing light to an arbitrarily small point. Model the lens of diameter D as an aperture in front of an ideal lens with an 8.0 cm focal length. Solve: Assuming that the incoming laser beam is parallel, the focal length of the lens should be 8.0 cm. From Equation 24.13, the minimum spot size in the focal plane of this lens is w=

2.44 ( 633 ×10−9 m )( 8.0 ×10−2 m ) 2.44λ f ⇒ 10 × 10−6 m = ⇒ D = 0.012 m = 1.2 cm D D

24.23. Model: Two objects are marginally resolvable if the angular separation between the objects, as seen from the lens, is α = 1.22λ / D. Solve:

Let Δy be the separation between the two light bulbs, and let L be their distance from a telescope. Thus,

α=

−2 Δy D (1.0 m ) ( 4.0 ×10 m ) Δy 1.22λ = = 55 km ⇒ L= = 1.22λ L D 1.22 ( 600 ×10−9 m )

24.24. Visualize: Equation 24.15 gives the smallest resolvable distance: d min = 0.61λ / NA. We are given λ = 500 nm and NA = 1.0. d min =

Assess:

0.61λ (0.61)(500 nm) = = 305 nm ≈ 310 nm NA 1.0

The smallest object one can see is on the order of the wavelength.

24.25. Visualize: Equation 24.15 gives the smallest resolvable distance: d min = 0.61λ / NA. We are given λ = 600 nm and d min = 0.75μ m = 750 nm. Solve:

NA = Assess:

0.61λ (0.61)(600 nm) = = 0.49 d min 750 nm

This is in the normal range for NA.

24.26. Visualize: We are given h1 = 1.0 cm, s1 = 4.0 cm, f1 = 5.0 cm, and f 2 = −8.0 cm. Solve:

First compute the image from the first lens. s1′ =

f1s1 ( 5.0 cm )( 4.0 cm ) = −20 cm = s1 − f1 4.0 cm − 5.0 cm

h1′ = −h1

s1′ −20 cm = −(1.0 cm) = 5.0 cm s1 4.0 cm

This is a virtual, upright image 20 cm to the left of the first lens. The second lens is 12 cm to the right of the first one, so s2 = 32 cm. s′2 =

( −8.0 cm )( 32 cm ) = −6.4 cm f 2 s2 = s2 − f 2 32 cm − ( −8.0 cm) h2 = h1′ = 5.0 cm

h2′ = − h2

s′2 −6.4 cm = − ( 5.0 cm ) = 1.0 cm s2 32 cm

This is a virtual, upright image 6.4 cm to the left of the second lens (5.6 cm to the right of the first lens). The image is 1.0 cm tall (the same size as the object). Assess: Ray tracing confirms these results.

24.27. Model: The parallel rays can be considered to come from an object infinitely far away: s1 = ∞ . The lens is a diverging lens. Visualize: If s1 = ∞ the thin lens equation tells us that s1′ = f1′; we are given that f1 = −10 cm. We are also given for the mirror f 2 = 10 cm. Solve: Since s1′ = −10 cm the image is virtual 10 cm to the left of the lens. The image from the lens becomes the

object for the mirror ⇒ s2 = 30 cm; this is three times the mirror's focal length, or s2 = 3 f 2 . s2′ =

f ( 3 f2 ) 3 f 2 s2 = 2 = f 2 = 15 cm s2 − f 2 3 f 2 − f 2 2

Therefore the initial parallel rays are brought to a focus 15 cm to the left of the mirror, or 5 cm to the right of the lens. Assess: The answer is reasonable and can be verified by ray tracing.

24.28. Visualize: The object is within the focal length of the converging lens, so we expect the image to be upright, virtual, and to the left of the lens. The image of the lens becomes the object for the mirror, and we expect the second image to be upright and virtual behind (to the right of) the mirror. Solve: s1′ =

f1s1 (10 cm )( 5 cm ) = −10 cm = s1 − f1 5 cm − 10 cm

The image of the lens becomes the object for the mirror ⇒ s2 = 15 cm, and we expect the second image to be upright and virtual behind (to the right of) the mirror.

s1′ =

h′ = h

( −30 cm )(15 cm ) = −10 cm f 2 s2 = s2 − f 2 15 cm − ( −30 cm )

s1′ s′2 ⎛ −10 cm ⎞⎛ −10 cm ⎞ = (1.0 cm ) ⎜ ⎟⎜ ⎟ = 1.3 cm s1 s2 ⎝ 5.0 cm ⎠⎝ 15 cm ⎠

The final image is 10 cm to the right of the mirror, or 15 cm to the right of the lens. It is upright with a height of 1.3 cm. Assess: Ray tracing will verify the answer.

24.29. Solve: (a) The image location from the first lens is s1′ =

f1s1 ( −2.5 cm)(2.5 cm) = = −1.25 cm s1 − f1 2.5 cm − ( −2.5 cm)

So the image from the first lens is 1.25 cm to the left of the first lens, upright and virtual. Now, s2 = d + 1.25 cm. We are told the final image is at infinity: s′2 = ∞ ⇒ s2 = f 2 ⇒ f 2 = d + 1.25 cm d = f 2 − 1.25cm = 3.75 cm (b)

(c)

h′ = −h

s1′ = 0.50 cm s1

The angular size is

θ = tan

h′ h′ 0.50 cm ≈ = = 0.10 rad f 2 f 2 5.0 cm

(d) If the object were held at the eye’s near point, it would subtend:

θ NP =

h 1.0 cm = = 0.040 rad 25 cm 25 cm

The angular magnification is M= Assess:

θ 0.10 rad = = 2.5 θ NP 0.040 rad

The numerical answers seem to agree with the drawing.

24.30. Visualize: See Figure 24.15. Parallel rays coming into the first lens will focus at the focal point of the first lens. If that position is also the focal point of the second lens then the rays will also leave the second lens parallel.

Solve: (a) This is similar to a telescope.

d = f1 + f 2 (b) Looking at the similar triangles in the diagram shows that

w1 w2 = f1 f2 w2 =

f2 w1 f1

Assess: Figure P24.30 says f 2 > f1 and our answer then shows that w2 > w1 which is the goal of a beam expander.

24.31. Visualize: Hard thought shows that if the left focal points for both lenses coincide then the parallel rays before and after the beam splitter are reproduced. The first lens diverges the rays as if they had come from the focal point of the converging lens.

Solve: (a)

d = f 2 − | f1 | But since we are given f1 < 0, this is equivalent to d = f 2 + f1 (b) Looking at the similar triangles in the diagram shows that

w1 w2 = | f1 | f 2 w2 =

f2 w1 | f1 |

Assess: Figure P24.31 says f 2 > | f1 | and our answer then shows that w2 > w1 which is the goal of a beam expander.

24.32. Visualize: We simply need to work backwards. We are given f1 = 7.0 cm and f 2 = 15 cm. We are also given s2′ = −10 cm. We use this to find s2 . Solve: (a)

s2 =

f 2 s′2 (15 cm)(−10 cm) = = 6.0 cm s2′ − f 2 −10 cm − 15 cm

So the final image is 6.0 cm to the left of the second lens, or 14 cm to the right of the first lens. That is, the object for the second lens is the image from the first lens, so s1′ = 20 cm = −6.0 cm = 14 cm. s1 =

f1s1′ (7.0 cm)(14 cm) = = 14 cm s1′ − f1 14 cm − 7.0 cm

Thus, L = 14 cm. (b) To find the height and orientation we need to look at the magnification. ⎛ s′ ⎞⎛ s′ ⎞ ⎛ 14 cm ⎞⎛ −10 cm ⎞ m = m1m2 = ⎜ − 1 ⎟⎜ − 2 ⎟ = ⎜ − ⎟⎜ − ⎟ = −1.7 ⎝ s1 ⎠⎝ s2 ⎠ ⎝ 14 cm ⎠⎝ 6.0 cm ⎠ h′ = hm = (1.0 cm)( −1.7) = −1.7 cm The negative sign indicates that the image is inverted. Assess: Ray tracing would verify the answers.

24.33. Model: The plane faces have an infinite radius of curvature. We are assuming the lens(es) are thin. Visualize: Use Equation 23.27, the lens makers’ equation: ⎛1 1 ⎞ 1 = (n − 1) ⎜ − ⎟ f ⎝ R1 R2 ⎠ Solve:

(a) For a symmetric convex lens call R = | R1 | = | R2 | where the sign convention says R1 > 0 and R2 < 0.

⎛1 1 ⎞ 1 ⎛1 1⎞ ⎛2⎞ = ( n − 1) ⎜ − ⎟ = (n − 1) ⎜ + ⎟ = (n − 1) ⎜ ⎟ f R R R R ⎝ ⎠ ⎝ R⎠ 2 ⎠ ⎝ 1 Invert both sides.

f =

1 R ( n − 1) 2

Now for the plano-convex halves: ⎛ 1 1⎞ 1 ⎛1 ⎞ ⎛1⎞ = ( n − 1) ⎜ − ⎟ = ( n − 1) ⎜ + 0 ⎟ = (n − 1) ⎜ ⎟ f1 R ∞ R ⎝ ⎠ ⎝R⎠ ⎝ 1 ⎠ Invert both sides.

f1 =

1 R ( n − 1)

⎛1 1 ⎞ 1 1⎞ ⎛ ⎛1⎞ = ( n − 1) ⎜ − ⎟ = (n − 1) ⎜ 0 + ⎟ = (n − 1) ⎜ ⎟ ∞ f2 R R ⎝ ⎠ ⎝ R⎠ 2 ⎠ ⎝ Invert both sides. f2 =

1 R (n − 1)

From the three results we see that f1 = f 2 = 2 f . (b) Use the thin-lens equation. As a single lens with s = 12 f : s′ =

fs f (1 f ) =1 2 =−f s− f 2 f − f

As a two-lens system with f1 = f 2 = 2 f and s1 = 12 f = 14 f1 : s1′ =

f1s1 (2 f )( 12 f ) f2 2 = 1 = 3 =− f s1 − f1 −2 f 3 2 f −2f

Since this is to the left of the lens s2 = 23 f s′2 =

(2 f )( 23 f ) 43 f 2 f 2 s2 = 2 = 4 =−f s2 − f 2 −3 f 3 f −2f

We get the same result treating it as one symmetric convex lens or as two plano-convex halves (with zero separation). Assess: We can think of each plano-convex half as providing half the refraction. The answers are consistent.

24.34. Model: Use the ray model of light. Assume both the lenses are thin lenses. Visualize:

Solve:

Begin by finding the image of the diverging lens,

1 1 1 1 1 1 + = + = ⇒ ⇒ s1′ = −10 cm 20 cm s1′ −20 cm s1 s1′ f1 This image is the object for the second lens. Its distance from the screen is s2 + s′2 = 110 cm − 10 cm = 100 cm. The overall magnification is M = m1m2 = −

h′ = −2 h

The magnification of the diverging lens is m1 = −

s1′ ( −10 cm ) = 1 =− s1 20 cm 2

Thus the magnification of the converging lens needs to be

m2 = −

s′2 = −4 ⇒ s′2 = 4s2 s2

Substituting this result into s2 + s2′ = 100 cm, we have s2 + 4s2 = 100 cm, which means s2 = 20 cm and s′2 = 80 cm. We can find the focal length by using the thin-lens equation for the converging lens:

1 1 1 1 1 1 + = ⇒ = + ⇒ f2 = 16 cm s2 s′2 f 2 f 2 20 cm 80 cm Hence, the second lens is a converging lens of focal length 16 cm. It must be placed 10 cm in front of the diverging lens, toward the screen, or 80 cm from the screen.

24.35. Model: Yang has myopia. Normal vision will allow Yang to focus on a very distant object. In measuring distances, we'll ignore the small space between the lens and her eye. Solve: Because Yang can see objects at 150 cm with a fully relaxed eye, we want a lens that creates a virtual image at s′ = −150 cm (negative because it's a virtual image) of an object at s = ∞ cm. From the thin-lens equation,

1 1 1 1 1 = + = + = −0.67 D f s s′ ∞ m −1.5 m So Yang gets a prescription for a −0.67 D lens which has f = −150 cm. Since Yang can accommodate to see things as close as 20 cm we need to create a virtual image at 20 cm of objects that are at s = new near point. That is, we want to solve the thin-lens equation for s when s′ = −20 cm and f = −150 cm. s= Assess:

fs′ ( −150 cm )( −20 cm ) = 23 cm = s′ − f −20 cm − ( −150 cm )

Diverging lenses are always used to correct myopia.

24.36. Visualize: Use Equation 23.21: n1 n2 n2 − n1 + = s s′ R where n1 = 1.00 for air and n2 = 1.34 for aqueous humor. If we think of incoming parallel rays coming to a focus in the humor then we have s = ∞ and s′ = f . Solve:

1.0 n2 n2 − n1 + = f R ∞ Solve for R. R= f

n2 − n1 1.34 − 1.00 = ( 3.0 cm ) = 0.76 cm n2 1.34

Assess: If you think about the dimensions of an eye, this answer seems physically possible.

24.37. Visualize: Use Equation 23.27, the lens makers' equation: ⎛1 1 ⎞ 1 = ( n − 1) ⎜ − ⎟ f ⎝ R1 R2 ⎠ For a symmetric lens R1 = R2 and f =

R and R = 2 ( n − 1) f 2 ( n − 1)

Also needed will be the magnification of a telescope: M = − f obj / f eye ⇒ f eye = − f obj / M (but we will drop the negative sign). We are given Robj = 100 cm and M = 20. Solve:

Reye = 2 ( n − 1) f eye Assess:

Robj 2 ( n − 1) Robj 100 cm = 2(n − 1) = 2 ( n − 1) = = = 5.0 cm 20 M M M f obj

We expect a short focal length and small radius of curvature for telescope eyepieces.

24.38. Model: Assume that each lens is a simple magnifier with M = 25 cm / f . Visualize:

M obj =

25 cm 25 cm ⇒ f obj = f obj M obj

M eye =

25 cm 25 cm ⇒ f eye = f eye M eye

Solve: (a) The magnification of a telescope is

M =−

f obj f eye

25 cm M obj M =− = − eye 25 cm M obj M eye

The way to maximize the magnitude of this is to have M eye > M obj .

M =−

5.0 = −2.5 2.0

The magnification is usually given without the negative sign, so it is 2.5 × . (b) To achieve this we used the 2.0× lens as the objective, which coincides with the text which says the objective should have a long focal length and the eyepiece a short focal length. (c)

L = f obj + f eye = Assess:

25 cm 25 cm 25 cm 25 cm + = + = 17.5 cm M obj M eye 2.0 5.0

This is not a very powerful telescope.

24.39. Model: To make a telescope you need an objective with a long focal length and an eyepiece with a short focal length. Visualize:

f =

1 P

Solve: (a) The lens with the smaller refractive power has the longer focal length and should be used as the object— that’s the lens with P = +3.0 D. The +4.5 D lens should be used as the eyepiece. (b)

M =−

f obj f eye

=−

Peye Pobj

=

4.5 D = −1.5 3.0 D

(c) In a telescope the lenses should be a distance apart equal to the sum of their focal lengths.

d = f obj + f eye =

1 1 1 1 + = + = 0.56m Pobj Peye 3.0 D 4.5 D

Assess: These numbers are reasonable, although a +4.5 D lens is fairly strong. You really could make yourself a telescope by holding the lenses a half-meter apart and get a little (1.5×) magnification. This cannot be done with the glasses of nearsighted people since they wear diverging lenses.

24.40. Model: Assume thin lenses and treat each as a simple magnifier with M = 25cm/f . Visualize:

Equation 24.10 gives the magnification of a microscope.

M = mobjM eye = −

L 25cm f obj f eye

Solve: (a) The more powerful lens (4×) with the shorter focal length should be used as the objective. (b) Solve the equation above for L (drop the negative sign). Mf f (12)( 254cm )( 252cm ) L = obj eye = = 37.5 cm 25 cm 25 cm Assess: This is a long microscope tube.

24.41. Visualize: We’ll use the thin-lens equation and also Equation 24.10: M =− Also recall that P = 1/f . Solve: The power of each lens is 1 1 P1 = = = 50 D f1 0.020 m

L 25 cm f obj f eye

P2 =

1 2 = = 100 D f1 0.010 m

Since we want to use the more powerful lens as the objective, the lens labeled P2 will be the objective. This means the focal lengths of the objective and eyepiece are f obj = 1.0 cm and f eye = 2.0 cm. (a) We want the eyepiece to L = 16 cm from the objective, so s′ = 16 cm − f eye = 16 cm − 2.0 cm = 14 cm. The

object distance for the objective is

s= (b) M = mobjM eye = Assess: answer:

fs′ (1.0 cm)(14 cm) = = 1.1 cm ′ s−f 14 cm − 1.0 cm

− s′ 25 cm (14 cm) 25 cm =− = −160 s f eye (1.08 cm) (2.0 cm)

As expected, s is just beyond the focal point. We can use approximation in Equation 24.9 to get a similar M=

L 25 cm (16 cm) 25 cm =− = −200 f obj f eye (1.0 cm) (2.0 cm)

but the approximation isn’t very good for this microscope.

24.42. Model: While s ≈ f obj we will not assume they are equal. Visualize:

Equation 24.9 says mobj ≈ − L / f obj. We are given L = 180 mm and mobj = −40, where the negative

sign means the image is inverted. Solve: Solve for f obj.

f obj = −

L 180 mm = = 4.5 mm mobj 40

From Equation 24.8, M eye = (25 cm) / f eye ⇒ f eye = 25 cm / 20 = 1.25 cm. For relaxed eye viewing the image of the objective must be 1.25 cm = 12.5 mm from the eyepiece, so s′ = 180 mm − 12.5 mm = 167.5 mm. Thus the sample distance is −1

1 ⎛ 1 ⎞ − s=⎜ ⎟ = 4.6 mm ⎝ 4.5 mm 167.5 mm ⎠ Assess:

You need a short focal length to achieve 800× magnification. We can also verify that s ≈ f obj .

24.43. Model: The width of the central maximum that accounts for a significant amount of diffracted light intensity is inversely proportional to the size of the aperture. The lens is an aperture that focuses light. Solve: To focus a laser beam, which consists of parallel rays from s = ∞, the focal length needs to match the distance to the target: f = L = 5.0 cm. The minimum spot size to which a lens can focus is

w=

2.44 (1.06 ×10−6 m )( 5.0 ×10−2 m ) 2.44λ f ⇒ 5.0 ×10−6 m = ⇒ D = 2.6 cm. D D

24.44. Model: Two objects are marginally resolved if the angular separation between the objects, as seen from your eye lens, is α = 1.22λ D , but the λ we want to use is the λ in the eye: λ = λair / n. Let Δx be the separation between the two headlights of the oncoming car and let L be the distance of these lights from your eyes. For small angles, Δx = α L. We are given D = 7 × 10−3 m, Δx = 1.2 m, and λ = λair / n = 600 nm /1.33 = 450 nm. Solve: Let Δy be the separation between the two headlights of the incoming car and let L be the distance of these lights from your eyes. Then,

α=

(1.20 m ) ( 7.0 × 10−3 m ) Δx 1.20 m 1.22λ 1.22 ( 450 nm ) ⇒ = = = L = = 15 km L L D (1.22 ) ( 450 × 10−9 m ) ( 7.0 ×10−3 m )

Assess: The two headlights are not resolvable if L > 15 km, marginally resolvable at 15 km, and resolvable at L < 15 km.

24.45. Visualize: The angle subtended at the eye due to a circle of diameter d at a near point of 25 cm is α = d / 25 cm. The angle to the first dark minimum in a circular diffraction pattern is θ1 = 1.22λ / D, where λ = λair / n is the wavelength of the light in the eye and D is the pupil diameter. To just barely see the circle as a circle the condition α = θ1 must be met. We are given D = 2.0 ×10−3 m and λ = λair / n = 600 nm /1.33 = 450 mm. Solve:

Equating α and θ1 we have d / 25 cm = 1.22λ / D, which may be solved for the diameter of the circle.

(25 cm)(1.22λ) (25 cm)(1.22)(450 ×10−9 m) = = 6.9 ×10−5 m = 0.069 mm D 2.0 ×10−3 m Assess: The above number is about the size of a needle point. d=

24.46. Visualize: For telescopes the angular resolution is 1.22λ θ= D

And for small angles, s = θ r. We want to know s. We are given λ = 650 nm, D = 2.4 m, and r = 33,000 ly = 2.8 × 1017 km. Solve: Combine the two equations. (a)

s =θr =

1.22λ (1.22)(650 × 10−9 m) r= 2.8 × 1017 km = 9.4 × 1010 km D 2.4 m

(b) The distance from the sun to Jupiter is 7.8 ×1011 m. So we divide the answer from part (a) by this number.

9.4 × 1010 km = 120 7.8 × 1011 m Assess: The HST is good but not good enough to resolve two objects as close together as the sun and Jupiter from a distance of 30,000 ly.

24.47. Model: α = 1.22λ / D.

Two objects are marginally resolved if the angular separation between the objects is

Visualize:

Solve:

(a) The angular separation between the sun and Jupiter is 780 × 109 m 780 × 109 m α= = = 1.92 × 10 −5 rad 4.3 light years 4.3 × ( 3.0 × 108 ) × ( 365 × 24 × 3600 ) m

−9 1.22λ 1.22 ( 600 ×10 m ) = ⇒ D = 0.038 m = 3.8 cm D D (b) The sun is vastly brighter than Jupiter, which is much smaller and seen only dimly by reflected light. In theory it may be possible to resolve Jupiter and the sun, but in practice the extremely bright light from the sun will overwhelm the very dim light from Jupiter.

α=

24.48. Visualize: We’ll start with Equation 24.15 and substitute in Equation 24.11 and an expression for α from Figure 24.14a. We are given f obj = 1.6 mm, d min = 400 nm, n = 1.0, and λ = 550 nm. Solve:

Solve for D.

d min =

0.61λ 0.61λ = = NA n sin α

0.61λ = ⎛ −1 D / 2 ⎞ n sin ⎜ tan ⎟ ⎜ f obj ⎟⎠ ⎝

⎛ D / 2 ⎞ 0.61λ sin ⎜ tan −1 ⎟= ⎜ f obj ⎟⎠ n d min ⎝

tan −1

D/2 0.61λ = sin −1 f obj n d min

⎛ D/2 0.61λ ⎞ = tan ⎜ sin −1 ⎟ f obj n d min ⎠ ⎝

⎛ ⎛ −1 0.61(550 nm) ⎞ 0.61λ ⎞ D = 2 f obj tan ⎜ sin −1 ⎟ = 2(1.6 mm) tan ⎜ sin ⎟ = 4.9 mm n d (1.0)(400 nm) ⎠ ⎝ ⎝ min ⎠ The diameter of the lens must exceed 4.9 mm. Assess: A half centimeter is in the ballpark for microscope lens diameters.

24.49. Model: For a diffraction-limited lens, the minimum focal length is the same size as its diameter. The smallest spot diameter over which you can focus light is wmin ≈ 2.5λ. Solve:

(a) The smallest spot size is wmin ≈ 2.5λ = 2.5 ( 800 × 10−9 m ) = 2 μ m.

(b) The total usable area of the optical disk is 2 2 π ⎡⎢( 5.5 × 10−2 m ) − ( 2 × 10−2 m ) ⎤⎥ = 0.00825 m 2 ⎣ ⎦ The area of each pit is the area of one bit of information and is ⎡⎣(1.25 )( 2 μ m ) ⎤⎦ = [ 2.5 μ m ] . The area of 1 byte is 8 times this quantity and the area of 1 megabyte (MB) of information is 106 times more. This means the number of megabytes (MB) of data that can be stored on the disk is 0.00825 m 2 = 165 MB 2 8 × ( 2.5 × 10−6 m ) × 106 MB−1 2

Assess:

A memory storage capacity of 165 MB is reasonable.

2

24.50. Visualize: Physically, the light rays can either go directly through the lens or they can reflect from the mirror and then go through the lens. We can consider the image from the lens alone and then consider the image from mirror becoming the object for the lens. Solve: (a) First case: the lens gets the subscript 1’s and the mirror the 2’s. The location of the image from the lens is fs (10 cm)(5 cm) s1′ = 1 1 = = −10 cm s1 − f1 5 cm − 10 cm The image is right at the mirror plane and a calculation for a mirror shows that when s2 = 0 then s2′ = 0, too. So the final image is at the mirror, 10 cm to the left of the lens. s′ −10 cm m=− 1 =− = 2.0 s1 5 cm

so h′ = hm = (1.0 cm)(2.0) = 2.0 cm. Second case: the mirror gets the subscript 1’s and the lens the 2’s. The location of the image from the mirror is fs (10 cm)(5 cm) s1′ = 1 1 = = −10 cm s1 − f1 5 cm − 10 cm or 10 cm behind (to the left of) the mirror. This image now becomes the object for the lens and s2 = 20 cm.

s′2 =

f 2 s2 (10 cm)(20 cm) = = 20 cm s2 − f 2 20 cm − 10 cm

So the image is 20 cm to the right of the lens.

⎛ s′ ⎞⎛ s′ ⎞ ⎛ −10 cm ⎞⎛ 20 cm ⎞ m = ⎜ − 1 ⎟⎜ − 2 ⎟ = ⎜ − ⎟⎜ − ⎟ = −2.0 ⎝ s1 ⎠⎝ s2 ⎠ ⎝ 5 cm ⎠⎝ 20 cm ⎠ so h′ = hm = (1.0 cm)(−2.0) = −2.0 cm, where the negative sign indicates the image is inverted. In summary, both images are 2.0 cm tall; one is upright 10 cm left of the lens, the other is inverted 20 cm to the right of the lens. (b)

Assess: The ray tracing verifies the calculations.

24.51. Model: In the small angle approximation the angle subtended by Mars without the telescope is θ obj = D / d where D is the diameter and d is the distance from the earth. Visualize:

We are given f eye = 2.5 cm.

Solve:

M=

θ eye 0.50° 0.50° ⎛ π rad ⎞ = = = 141 θ obj D / d 6800 ×103 m /1.1×1011 m ⎜⎝ 180° ⎟⎠

Ignoring the negative sign, M=

f obj f eye

f obj = Mf eye = 141(2.5 cm) = 353 cm The length of the telescope is L = f obj + feye = 352.5 cm + 2.5 cm = 355 cm = 3.55 m ≈ 3.5 m Assess:

This is longer than most amateur telescopes.

24.52. Visualize: The plane left face will not refract any of the rays (which are parallel to each other and perpendicular to the face), so nothing happens until the rays hit the first curved surface between lens 1 and lens 2. We’ll need to twice (once for each curved surface) apply Equation 23.21: n1 n2 n2 − n1 + = s s′ R

Solve: (a) For the first curved surface we say s = ∞ because the incoming rays are parallel.

n1 n2 n2 − n1 n2 + = ⇒ s′ = R ∞ s′ R n2 − n1 For the second curved surface (where the rays exit into the air) n2 is on the left and nair = 1.0 is on the right (so those will take the place of n1 and n2, respectively). Since the distance between the lenses is zero, s′ from the previous result will be plugged in for s for the second case. The final thing to note is that the magnitude of the new s′ = f because the doublet brings parallel rays to a focus at f, but s′ is negative due to the sign convention in Table 23.3.

n2 n n −n + air = air 2 n2 f R − R n2 − n1 Now solve for f.

n2 − n1 1 1 − n2 + = −f R R 1 1 − n2 − n2 + n1 = −f R

f =

R 2n2 − n1 − 1

(b)

f blue =

R 2 ( n2 )blue − ( n1 )blue − 1

f red =

R 2 ( n2 )red − ( n1 )red − 1

In the condition we desire f blue = f red , so the two denominators must be equal.

2 ( n2 )blue − ( n1 )blue − 1 = 2 ( n2 )red − ( n1 )red − 1 2 ⎡⎣( n2 )blue − ( n1 )red ⎤⎦ = ( n1 )blue − ( n1 )red 1 Δn2 = Δn1 2

(c) Simply find the Δn for each type of glass and hope one is twice the other.

Δncrown = 1.525 − 1.517 = 0.008 Δnflint = 1.632 − 1.616 = 0.016 1 Since Δncrown = Δnflint then crown glass must be the second material, or the converging lens, while flint glass 2 must be the first material, or the diverging lens. (d) Solve the original focal length expression for R. R = f ( 2n2 − n1 − 1)

Since f blue = f red , it doesn’t matter which color we choose for the n’s (as long as we are consistent). Say we pick blue, so n1 = 1.632 and n2 = 1.525. We are given f = 10.0 cm. R = (10.0 cm ) ⎡⎣ 2 (1.525) − 1.632 − 1⎤⎦ = 4.18 cm Assess: The answers to the various parts fit together and the final result is reasonable.

24.53. Visualize: The effective focal length is defined as the distance from the midpoint between

the two lenses to the point that initially parallel rays come to a focus. Because d < f the image from the first lens is to the right of the second lens, so we will use a negative object distance when we analyze the second lens. Solve: (a) If the original object distance is very large ( s1 ≈ ∞) then s1′ = f . This image is to the right of the right lens by

an amount f − d , but since we are treating this as a negative object distance we will put in s2 = −( f − d ) = d − f . The thin-lens equation for the second lens (the diverging one, with a negative focal length) becomes:

1 1 1 + = d − f s′2 − f Solve for s′2 .

s2′ =

− f (d − f ) f − fd = 2 (d − f ) − (− f ) d

This is the distance of the focus to the right of the second lens, however, we want the distance from the midpoint between the lenses, so we add 12 d to the answer. 1 f 2 − fd 1 f − fd 12 d 2 f 2 − fd + 12 d 2 f eff = s′2 + d = + d= 2 + = 2 d 2 d d d (b) We'll plug d = 12 f and d = 14 f in turn into the previous result. Then Equation 24.1 shows that if we take the ratios of

the resulting f eff ’s we'll have the zoom. zoom =

( f eff ) d =1 4 f ( f eff ) d =1 2 f

( = (

f 2 − f ( 14 f ) + 12 ( 14 f ) 2 1 4

f

f 2 − f ( 12 f ) + 12 ( 12 f ) 2 1 2

f

) = ⎛⎜ ) ⎝

1 2 1 4

⎞ f 2 − 14 f 2 + 321 f 2 ⎟ 2 1 2 1 2 ⎠ f −2 f +8 f

Cancel f 2 from each term. 25 5 5 ⎛ 4 ⎞ 1 − 14 + 321 = (2) 325 − (2) = = 2.5 ⎜ ⎟ 1 1 4 2 ⎝ 2 ⎠ 1− 2 + 8 8

So the lens is a 2.5× zoom lens. Assess: This is a reasonable amount of zoom. The magnification spans a factor of 2.5.

24-1

25.1. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve:

Substituting into the formula for the Balmer series,

λ=

91.18 nm 91.18 nm ⇒λ = = 410.3 nm 1 1 ⎛ 1 1⎞ − − ⎜ 2 2 ⎟ 22 62 ⎝2 n ⎠

where n = 3, 4, 5, 6, … and where we have used n = 6. Likewise for n = 8 and n = 10, λ = 389.0 nm and λ = 379.9 nm.

25.2. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve:

Balmer’s formula is

λ=

91.18 nm ⎛ 1 1⎞ ⎜ 2 − 2⎟ ⎝2 n ⎠

n = 3, 4, 5, 6, …

As n → ∞, 1 n 2 → 0. Thus, λn→∞ = 4 ( 91.18 nm ) = 364.7 nm.

25.3. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve:

Using Balmer’s formula,

λ = 389.0 nm =

91.18 nm 1 1 ⇒ − 2 = 0.2344 ⇒ n = 8 4 n ⎛ 1 1⎞ ⎜ 2− 2⎟ ⎝2 n ⎠

25.4. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: The Bragg condition is 2d cosθ m = mλ , where m = 1, 2, 3, … For first and second order diffraction,

2d cosθ1 = (1) λ

2d cosθ 2 = ( 2 ) λ

Dividing these two equations,

cosθ 2 = 2 ⇒ θ 2 = cos −1 ( 2cosθ1 ) = cos −1 ( 2cos68° ) = 41° cosθ1

25.5. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: The Bragg condition is 2d cosθ m = mλ. For m = 1 and for two different wavelengths,

2d cosθ1 = (1) λ1

2d cosθ1′ = (1) λ1′

Dividing these two equations,

cosθ1′ λ1′ cosθ1′ 0.15 nm = ⇒ = ⇒ θ1′ = cos −1 ( 0.4408 ) = 64° cosθ1 λ1 cos54° 0.20 nm

25.6. Model: The angles corresponding to the various orders of diffraction satisfy the Bragg condition. Solve:

The Bragg condition for m = 1 and m = 2 gives

2d cosθ1 = (1) λ

2d cosθ 2 = ( 2 ) λ

Dividing these two equations, cosθ1 =

cosθ 2 cos 45° ⎛ cos 45° ⎞ = ⇒ θ1 = cos −1 ⎜ ⎟ = 69.3° 2 2 ⎝ 2 ⎠

25.7. Model: The angles corresponding to the various diffraction orders satisfy the Bragg condition. Solve: The Bragg condition is 2d cosθ m = mλ , where m = 1, 2, 3, … The maximum possible value of m is the number of possible diffraction orders. The maximum value of cosθm is 1. Thus, 2d = mλ ⇒ m = We can observe up to the fourth diffraction order.

2d

λ

=

2 ( 0.180 nm ) = 4.2 ( 0.085 nm )

25.8. Model: Use the photon model of light. Solve:

The energy of the photon is

Ephoton = hf = h Assess:

⎛ 3.0 ×108 m/s ⎞ −19 = ( 6.63 × 10−34 Js ) ⎜ ⎟ = 3.98 × 10 J −9 λ ⎝ 500 ×10 m ⎠ c

The energy of a single photon in the visible light region is extremely small.

25.9. Model: Use the photon model of light. Solve:

The energy of the single photon is −34 8 ⎛ c ⎞ ( 6.63 ×10 Js )( 3.0 ×10 m/s ) = 1.99 ×10−19 J Ephoton = hf = h ⎜ ⎟ = 1.0 ×10−6 m ⎝λ⎠

⇒ Emol = N A Ephoton = ( 6.023 ×1023 )(1.99 ×10−19 J ) = 1.2 ×105 J Assess: energy.

Although the energy of a single photon is very small, a mole of photons has a significant amount of

25.10. Model: Use the photon model of light. Solve: The energy of a photon with wavelength λ1 is E1 = hf1 = hc λ1 . Similarly, E2 = hc λ2 . Since E2 is equal to 2E1,

hc

λ2

=2

hc

λ1

⇒ λ2 =

λ1 2

=

600 nm = 300 nm 2

Assess: A photon with λ = 300 nm has twice the energy of a photon with λ = 600 nm. This is an expected result, because energy is inversely proportional to the wavelength.

25.11. Model: Use the photon model of light. Solve:

The energy of the x-ray photon is

⎛ 3.0 × 108 m/s ⎞ ⎛c⎞ −16 E = hf = h ⎜ ⎟ = ( 6.63 × 10−34 Js ) ⎜ ⎟ = 2.0 × 10 J −9 ⎝λ⎠ ⎝ 1.0 ×10 m ⎠ Assess: This is a very small amount of energy, but it is larger than the energy of a photon in the visible wavelength region.

25.12. Solve: Your mass is, say, m ≈ 70 kg and your velocity is 1 m/s. Thus, your momentum is p = mv ≈ (70 kg)(1 m/s) = 70 kg m/s. Your de Broglie wavelength is

λ=

h 6.63 ×10−34 Js = ≈ 9 × 10−36 m p 70 kg m/s

25.13. Solve: (a) The baseball’s momentum is p = mv = (0.200 kg)(30 m/s) = 6.0 kg m/s. The baseball’s de Broglie wavelength is λ=

h 6.63 ×10−34 Js = = 1.1×10−34 m p 6.0 kg m/s

(b) Using λ = h p = h mv , we have v=

h 6.63 ×10−34 Js = = 1.7 × 10−23 m/s mλ ( 0.200 kg ) ( 0.20 ×10−9 m )

25.14. Visualize: We'll employ Equations 25.8 (λ = h / p) and 25.9 ( E = p 2 /2m) to express the wavelength in terms of kinetic energy. Solve: First solve Equation 25.9 for p : p = 2mE .

λ=

h h 6.63 × 10−34 J ⋅ s = = = 1.0 nm p 2mE 2 ( 9.11× 10−31 kg )( 2.4 × 10−19 J )

Assess: The energy given is about 1.5 eV, which is a reasonable amount of energy. The resulting wavelength is a few to a few dozen times the size of an atom.

25.15. Solve: (a) For an electron, the momentum p = mv = ( 9.11×10−31 kg ) v . The de Broglie wavelength is λ=

h = 0.20 ×10−9 m ⇒ 0.20 ×10−9 m = p

6.63 ×10−34 Js ⇒ v = 3.6 × 106 m/s ( 9.11×10−31 kg ) v

(b) For a proton, p = mv = (1.67 × 10−9 kg ) v. The de Broglie wavelength is

λ=

h 6.63 ×10−34 Js = 0.20 ×10−9 m ⇒ 0.20 × 10−9 m = ⇒ v = 2.0 × 103 m/s p (1.67 ×10−27 kg ) v

25.16. Model: The momentum of a wave-like particle has discrete values given by pn = n ( h 2 L ) where n = 1, 2, 3, …. Solve: Because we want the smallest box and the momentum of the electron can not exceed a given value, n must be minimum. Thus, p1 = mv =

6.63 ×10−34 Js h h ⇒L= = = 0.036 mm 2L 2mv 2 ( 9.11× 10−31 kg ) (10 m/s )

25.17. Model: A confined particle can have only discrete values of energy. Solve:

From Equation 24.14, the energy of a confined electron is

En =

h2 2 n 8mL2

n = 1, 2, 3, 4, …

The minimum energy is E1 =

h2 h 6.63 ×10−34 Js ⇒L= = = 2.0 × 10−10 m = 0.20 nm 2 −31 −18 8mL 8mE1 8 ( 9.11×10 kg )(1.5 ×10 J )

25.18. Model: Model the 5.0-fm-diameter nucleus as a box of length L = 5.0 fm. Solve:

The proton’s energy is restricted to the discrete values

( 6.63 ×10−34 Js ) n2 h2 2 = = (1.316 × 10−12 J ) n 2 n 2 −27 −15 8mL2 8 (1.67 ×10 kg )( 5.0 ×10 m ) 2

En =

where n = 1, 2, 3, … For n = 1, E1 = 1.3 ×10−12 J , for n = 2, E2 = (1.316 × 10−12 J ) 4 = 5.3 ×10−12 J , and for n = 3,

E3 = 9 E1 = 1.2 ×10−11 J .

25.19.

Model: The generalized formula of Balmer predicts a series of spectral lines in the hydrogen spectrum. Solve: (a) The generalized formula of Balmer

λ=

91.18 m 1⎞ ⎛ 1 ⎜ 2− 2⎟ ⎝m n ⎠

with m = 1 and n > 1 accounts for a series of spectral lines. This series is called the Lyman series and the first two members are

λ1 =

91.18 m = 121.6 nm 1⎞ ⎛ 1 − ⎜ 2 ⎟ ⎝ 2 ⎠

λ2 =

91.18 nm = 102.6 nm 1⎞ ⎛ 1 − ⎜ 2 ⎟ ⎝ 3 ⎠

For n = 4 and n = 5, λ3 = 97.3 nm and λ4 = 95.0 nm . (b) The Lyman series converges when n → ∞ . This means 1 n 2 → 0 and λ → 91.18 nm. (c) For a diffraction grating, the condition for bright (constructive interference) fringes is d sin θ p = pλ , where p =

1, 2, 3, … For first-order diffraction, this equation simplifies to d sin θ = λ . For the first and second members of the Lyman series, the above condition is d sin θ1 = λ1 = 121.6 nm and d sin θ 2 = λ2 = 102.6 nm. Dividing these two equations yields ⎛ 102.6 nm ⎞ sin θ 2 = ⎜ ⎟ sin θ1 = ( 0.84375 ) sinθ1 ⎝ 121.6 nm ⎠ The distance from the center to the first maximum is y = L tan θ . Thus, tan θ1 =

y1 0.376 m = ⇒ θ1 = 14.072° ⇒ sin θ 2 = ( 0.84375 ) sin (14.072° ) ⇒ θ 2 = 11.84° L 1.5 m

Applying the position formula once again, y2 = L tan θ 2 = (1.5 m ) tan (11.84° ) = 0.314 m = 31.4 cm

25.20.

Model: The generalized formula of Balmer predicts a series of spectral lines in the hydrogen spectrum. Solve: (a) The generalized formula of Balmer

λ=

91.18 m 1⎞ ⎛ 1 ⎜ 2− 2⎟ ⎝m n ⎠

with m = 3, and n > 3 accounts for a series of spectral lines. This series is called the Paschen series and the wavelengths are

λ=

91.18 nm 820.62 n 2 = n2 − 9 ⎛1 1⎞ − ⎜ 2 2 ⎟ ⎝3 n ⎠

The first four members are λ1 = 1876 nm, λ2 = 1282 nm, λ3 = 1094 nm, and λ4 = 1005 nm (b) The Paschen series converges when n → ∞. This means

1 91.18 nm →0⇒λ → = 820.6 nm 2 2 n ( 13 ) (c) For a diffraction grating, the condition for bright (constructive interference) fringes is d sin θ p = pλ , where p =

1, 2, 3, … For first-order diffraction, this equation simplifies to d sin θ = λ . For the first and second members of the Paschen series, the condition is d sin θ1 = λ1 and d sinθ 2 = λ2 . Dividing these two equations yields

⎛λ ⎞ ⎛ 1282 nm ⎞ sin θ 2 = sin θ1 ⎜ 2 ⎟ = sin θ1 ⎜ ⎟ = ( 0.6834 ) sinθ1 λ ⎝ 1876 nm ⎠ ⎝ 1⎠ The distance from the center to the first maximum is y = L tan θ . Thus, tan θ1 =

y1 0.607 m = = 0.4047 ⇒ θ1 = 22.03° ⇒ sin θ 2 = ( 0.6834 ) sin 22.03° ⇒ θ 2 = 14.85° L 1.5 m

Applying the position formula once again, y2 = L tan θ 2 = (1.5 m ) tan14.85° = 0.398 m = 39.8 cm

25.21. Model: Use the photon model of light. Solve:

(a) The wavelength is calculated as follows:

( 6.63 ×10−34 Js )( 3.0 ×108 m/s ) = 2.0 ×10−12 m ⎛c⎞ Egamma = hf = h ⎜ ⎟ ⇒ λ = 1.0 ×10−13 J ⎝λ⎠ (b) The energy of a visible-light photon of wavelength 500 nm is −34 8 ⎛ c ⎞ ( 6.63 ×10 Js )( 3.0 ×10 m/s ) Evisible = h ⎜ ⎟ = = 3.978 × 10−19 J 500 ×10−9 m ⎝λ⎠

The number of photons n such that Egamma = nEvisible is n=

Egamma Evisible

=

1.0 ×10−13 J = 2.5 ×105 3.978 ×10−19 J

25.22. Model: Use the photon model. Solve:

The energy of a 1000 kHz photon is

Ephoton = hf = ( 6.63 ×10−34 Js )(1000 ×103 Hz ) = 6.63 × 10−28 J The energy transmitted each second is 20 × 103 J. The number of photons transmitted each second is 20 × 103 J/6.63 × 10 −28 J = 3.0 × 1031.

25.23. Model: Use the photon model for the laser light. Solve:

(a) The energy is

⎛ 3 × 108 m/s ⎞ ⎛c⎞ −19 Ephoton = hf = h ⎜ ⎟ = ( 6.63 × 10−34 Js ) ⎜ ⎟ = 3.1× 10 J −9 ⎝λ⎠ ⎝ 633 ×10 m ⎠ (b) The energy emitted each second is 1.0 ×10 −3 J. The number of photons emitted each second is 1.0 ×10 −3 J/3.14 ×10 −19 J = 3.2 × 1015.

25.24. Model: Use the photon model for the incandescent light. Solve: The photons travel in all directions. At a distance of r from the light bulb, the photons spread over a sphere of surface area 4πr2. The number of photons per second per unit area at the location of your retina is

3 ×1018 s −1

4π (10 ×10 m ) 3

2

= 2.387 ×109 s −1 m −2

The number of photons that enter your pupil per second is 2.387 ×109 s −1m −2 × π ( 3.5 ×10−3 m ) = 9.2 ×104 s −1 2

25.25. Model: Use the photon model of light and the Bragg condition for diffraction. Solve: The Bragg condition for the reflection of x-rays from a crystal is 2d cosθ m = mλ. To determine the angles of incidence θm, we need to first calculate λ. The wavelength is related to the photon’s energy as E = hc λ . Thus, λ=

−34 8 hc ( 6.63 ×10 Js )( 3.0 ×10 m/s ) = = 1.326 × 10−10 m E 1.50 ×10−15 J

From the Bragg condition, ⎡ 1.326 ×10−10 m ) m ⎤ ⎛ mλ ⎞ −1 ( ⎢ ⎥ = cos −1 ( 0.3157m ) ⇒ θ1 = cos −1 ( 0.3157 ) = 71.6° = cos ⎟ −9 ⎝ 2d ⎠ ⎢⎣ 2 ( 0.21×10 m ) ⎥⎦

θ m = cos −1 ⎜

Likewise, θ 2 = cos −1 ( 0.3157 × 2 ) = 50.8° and θ 3 = 18.7°. Note that θ 4 = cos −1 ( 0.3157 × 4 ) is not allowed because the cosθ cannot be larger than 1. Thus, the x-rays will be diffracted at angles of incidence equal to 18.7°, 50.8°, and 71.6°.

25.26. Model: The angles for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: We cannot assume that these are the first and second order diffractions. The Bragg condition is 2d cosθ m = mλ. We have 2d cos 45.6° = mλ

2d cos 21.0° = ( m + 1) λ

Notice that θm decreases as m increases, so 21.6° corresponds to the larger value of m. Dividing these two equations, cos 45.6° m = = 0.7494 ⇒ m = 3 cos 21.0° m + 1 Thus these are the third and fourth order diffractions. Substituting into the Bragg condition, d=

3 × 0.0700 × 10−9 m = 1.50 × 10−10 m = 0.150 nm 2cos 45.6°

25.27. Model: The x-ray diffraction angles satisfy the Bragg condition. Solve: (a) The Bragg condition ( 2d cosθ m = mλ ) for normal incidence, θm = 0°, simplifies to 2d = mλ. For a thin film of a material on a substrate where nair < nmaterial < nsubstrate, constructive interference between the two reflected waves occurs when 2d = mλ, where λ is the wavelength inside the material. (b) From a thin film with a period of 1.2 nm, that is, with d = 1.2 nm, the two longest x-ray wavelengths that will reflect at normal incidence are

λ1 =

2d 1

This means that λ1 = 2 (1.2 nm ) = 2.4 nm and λ2 = 1.2 nm.

λ2 =

2d 2

25.28. Solve: A small fraction of the light wave of an appropriate wavelength is reflected from each little “bump” in the refractive index. These little bumps act like the atomic planes in a crystal. The light will be strongly reflected (and hence blocked in transmission) if it satisfies the Bragg condition at normal incidence (θ = 0). 2d = mλglass =

−6 2dnglass 2 ( 0.45 ×10 m ) (1.50 ) mλ ⇒λ = = = 1.35 μm nglass m 1

25.29. Model: Particles have a de Broglie wavelength given by λ = h p . The wave nature of the particles causes an interference pattern in a double-slit apparatus. Solve: (a) Since the speed of the neutron and electron are the same, the neutron’s momentum is

pn = mn vn =

mn m m mevn = n meve = n pe me me me

where mn and me are the neutron’s and electron’s masses. The de Broglie wavelength for the neutron is

λn =

h h pe m = = λe e pn pe pn mn

From Section 22.2 on double-slit interference, the fringe spacing is Δy = λ L / d . Thus, the fringe spacing for the electron and neutron are related by

Δy n =

⎛ 9.11× 10−31 kg ⎞ λn m −3 −7 Δye = e Δye = ⎜ ⎟ (1.5 × 10 m ) = 8.18 × 10 m = 0.818 μ m −27 λe mn ⎝ 1.67 ×10 kg ⎠

(b) If the fringe spacing has to be the same for the neutrons and the electrons,

Δye = Δyn ⇒ λe = λn ⇒

⎛ 9.11×10−31 kg ⎞ h h m 3 = ⇒ vn = ve e = ( 2.0 × 106 m/s ) ⎜ ⎟ = 1.1× 10 m/s −27 meve mn vn mn ⎝ 1.67 ×10 kg ⎠

25.30. Model: Electrons have a de Broglie wavelength given by λ = h p . The wave nature of the electrons causes a diffraction pattern. Solve: The width of the central maximum of a single-slit diffraction pattern is given by Equation 22.22: w=

2 (1.0 m ) ( 6.63 × 10−34 Js ) 2λ L 2 Lh 2 Lh = = = = 9.7 × 10−4 m = 0.97 mm −6 a ap amv (1.0 ×10 m )( 9.11×10−31 kg )(1.5 ×106 m/s )

25.31. Model: Neutrons have a de Broglie wavelength given by λ = h p . The wave nature of the neutrons causes a double-slit interference pattern. Solve: Measurements show that the spacing between the m = 1 and m = –1 peaks is 1.4 times as long as the length of the reference bar, which gives the real fringe separation Δy = 70 μm. Similarly, the spacing between the m = 2 and m = –2 is 2.8 times as long as the length of the reference bar and yields Δy = 70 μm. The fringe separation in a double-slit experiment is Δy = λ L d . Hence,

λ=

( 6.63 ×10−34 Js ) ( 3.0 m ) Δy d Δy d h h hL ⇒ = = ⇒v= = = 170 m/s −6 Δy md ( 70 ×10 m )(1.67 × 10−27 kg )( 0.10 ×10−3 m ) L p mv L

25.32. Model: Electrons have a de Broglie wavelength given by λ = h p . Visualize: Please refer to Figure 25.11. Notice that a scattering angle φ = 60° corresponds to an angle of incidence θ = 30°. Solve: Equation 25.6 describes the Davisson-Germer experiment: D sin ( 2θ m ) = mλ . Assuming m = 1, this equation simplifies to D sin 2θ = λ . Using λ = h mv , we have D=

h 6.63 ×10−34 Js = = 1.95 × 10−10 m = 0.195 nm −31 mv sin 2θ ( 9.11×10 kg )( 4.30 × 106 m/s ) sin ( 60° )

25.33. Model: A confined particle can have only discrete values of energy. Solve:

(a) Equation 25.14 simplifies to

( 6.63 ×10−34 Js ) h2 2 En = n = = (1.231× 10−19 J ) n 2 2 2 8mL 8 ( 9.11× 10−31 kg )( 0.70 × 10−9 m ) 2

Thus, E1 = (1.231×10−19 J )(12 ) = 1.2 × 10−19 J, E2 = (1.231×10−19 J )( 22 ) = 4.9 ×10−19 J, and E3 = 1.1×10−18 J. (b) The energy is E2 − E1 = 4.9 × 10−19 J − 1.2 ×10−19 J = 3.7 ×10−19 J. (c) Because energy is conserved, the photon will carry an energy of E2 − E1 = 3.69 ×10−19 J . That is, E2 − E1 = Ephoton = hf =

hc

λ

⇒λ =

( 6.63 ×10−34 Js )( 3.0 ×108 m/s ) = 540 nm hc = 3.69 ×10−19 J E2 − E1

25.34. Model: A particle confined in a one-dimensional box has discrete energy levels. Solve:

(a) Equation 24.14 for the n = 1 state is

( 6.63 ×10−34 Js ) = 5.5 ×10−64 J h2 En = = 8mL2 8 (10 ×10−3 kg ) ( 0.10 m )2 2

The minimum energy of the Ping-Pong ball is E1 = 5.5 × 10−64 J. (b) The speed is calculated as follows: E1 = 5.50 ×10−64 J =

1 2

mv 2 =

1 2

(10 ×10

−3

kg ) v 2 ⇒ v =

2 ( 5.50 ×10−64 J ) 10 ×10−3 kg

= 3.3 × 10−31 m/s

25.35. Model: A particle confined in a one-dimensional box has discrete energy levels. Solve:

Using Equation 24.14 for n = 1 and 2,

( 6.63 ×10−34 Js ) ( 3) = 1.809 ×10−37 Jm2 h2 22 − 12 ) ⇒ 1.0 ×10−19 J = E2 − E1 = 2 ( L2 8mL 8 ( 9.11×10−31 kg ) L2 2

⇒L=

1.809 × 10−37 Jm 2 = 1.3 ×10−9 m = 1.3 nm 1.0 ×10−19 J

25.36. Visualize: From the figure we see that the wavelength is 2.0 nm. We'll employ Equations 25.8 (λ = h / p ) and 25.9 ( E = p 2 /2m) to express the kinetic energy in terms of wavelength. Solve:

2 ( 6.63 ×10−34 J ⋅ s 2.0 nm ) = 6.0 ×10−20 J p2 ( h λ ) = = 2m 2m 2 ( 9.11×10−31 Kg ) 2

E= Assess:

This energy is a little less than one eV, which is reasonable.

25.37. Visualize: The strategy is to take ratios to find n and then plug it back in to find L. Solve:

h2 (n + 1) 2 En +1 8mL2 ( n + 1) 2 6.4 × 10−13 J = = = 2 h ( n) 2 3.6 × 10−13 J En 2 ( ) n 2 8mL Cancel ×10−13 J and take square roots. n +1 6.4 4 = = n 3.6 3

⇒

n=3

So E3 = 3.6 ×10−13 J. Now solve for L.

L= Assess:

( 6.63 ×10−34 J ⋅ s ) ( 3) h2n2 hn = = = 29 fm 8mEn 8mEn 8 (1.67 × 10−27 kg )( 3.6 × 10−13 J )

This is not an atomic-sized box, but a nuclear-sized box; that’s OK for neutrons.

25.38. Model: The allowed energies of a particle of mass m in a two-dimensional square box of side L are Enm = Solve:

h2 ( n2 + m2 ) 8mL2

(a) The minimum energy for a particle is for n = m = 1:

Emin = E11 = (b) The five lowest allowed energies are Emin ,

5 2

5Emin (for n = 1, m = 3 and n = 3, m = 1), and

13 2

h2 h2 12 + 12 ) = 2 ( 8mL 4mL2

Emin (for n = 1, m = 2 and n = 2, m = 1), 4Emin (for n = 2, m = 2), Emin (for n = 2, m = 3 and n = 3, m = 2).

25.39. Model: A particle confined in a one-dimensional box of length L has the discrete energy levels given by Equation 24.14. Solve: (a) Since the energy is entirely kinetic energy,

En =

h2 2 p 2 1 2 h n = = mvn ⇒ vn = n 8mL2 2m 2 2mL

n = 1, 2, 3, …

(b) The first allowed velocity is

v1 =

6.63 ×10−34 Js = 1.82 ×106 m/s 2 ( 9.11×10−31 kg )( 0.20 ×10−9 m )

For n = 2 and n = 3, v2 = 3.64 × 106 m/s and v3 = 5.46 × 106 m/s.

25.40. Model: Sets of parallel planes in a crystal diffract x-rays. Visualize: Please refer to Figure CP25.40. Solve: The Bragg diffraction condition is 2d cosθ m = mλ , where d is the interplanar separation. Because smaller m values correspond to higher angles of incidence, the diffraction angle of 71.3° in the x-ray intensity plot must correspond to m = 1. This means 2d cos71.3° = 1( 0.10 × 10−9 m ) ⇒ d =

0.10 ×10−9 m = 1.56 ×10−10 m 2 ( cos71.3° )

The cosines of the three angles in the x-ray intensity plot are cos71.3° = 0.321, cos50.1° = 0.642, and cos15.8° = 0.962. These are in the ratio 1 : 2 : 3, which tells us that these are the m = 1, 2, and 3 diffraction peaks from a single set of planes with d = 0.156 nm. We can see from the figure that the atomic spacing D of this crystal is related to the interplanar separation d by D=

d 0.156 nm = = 0.18 nm sin 60° sin60°

25.41.

Model: Sets of parallel planes in a crystal diffract x-rays. Visualize: Please refer to Figure 25.7. Solve: (a) The Bragg diffraction condition is 2d cosθ m = mλ. The plane spacing is dA = 0.20 nm and the x-ray wavelength is λ = 0.12 nm. Thus

cosθ m =

−9 mλ m ( 0.12 ×10 m ) = = ( 0.3) m ⇒ θ A1 = cos −1 ( 0.3) = 72.5° 2d A 2 ( 0.20 ×10−9 m )

Likewise for m = 2 and m = 3, θ A2 = cos −1 ( 0.6 ) = 53.1° and θ A3 = cos −1 ( 0.9 ) = 25.8°. These three angles for the x-ray diffraction peaks match the peaks shown in Figure 25.7c. (b) The new interplaner spacing is d B = d A 2 = 0.141 nm (see Figure 25.7b). The Bragg condition for the tilted atomic planes becomes

cosθ m =

mλ = 0.4243m 2d B

For m = 1, θ B1 = cos −1 ( 0.4243) = 64.9°. For m = 2, θ B2 = cos −1 ( 0.8486 ) = 31.9°. (c) The crystal is already tipped by 45° to get the tilted planes (see Figure 25.7b). So, for m = 1, θ1 = 64.9° − 45° = 19.9°. θ = 64.9° + 45° = 109.9° also, but we can’t see beyond 90°. For m = 2,

θ 2 = 31.9° + 45° = 76.9°. These two angles match the angles in the diffraction peaks of the tilted planes.

25.42. Model: This is an integrated problem that uses concepts from Chapter 22. There are two L’s in the problem: L in Chapter 22 refers to the screen distance from the slits, and the L we want here is the length of the box. The wavelength of the neutron determined by the two-slit pattern is the same as the wavelength in the confined box. Visualize: The figure shows Lbox = 2λ .

We also need Equation 22.6: ym =

mλ Lscreen . Also from the figure we see that y2 = 0.20 ×10−3 m. We are given d

Lscreen = 2.0 m and d = 15 ×10−6 m. Solve: Solve Equation 22.6 for λ.

λ= Lbox = 2λ = 2 Assess:

dym mLscreen

(15 × 10−6 m )( 0.20 ×10−3 m ) = 1.5 nm dym =2 mLscreen ( 2 )( 2.0 m )

The two pieces of this problem fit together and make sense together.

Model: Electrons have a de Broglie wavelength given by λ = h p . Trapped electrons in the confinement layer behave like a de Broglie wave in a closed-closed tube or like a string fixed at both ends. Solve: (a) The four longest standing-wave wavelengths in the layer are λ = 2L , L, 32 L , and 12 L . This follows

25.43.

from the general relation for closed-closed tubes: λ = 2 L / n. Thus, λ = 10.0 nm, 5.00 nm, 3.33 nm, and 2.50 nm. (b) We have p = mv =

h

λ

⇒v=

h 6.63 × 10−34 Js 0.7278 ×10−3 m 2 /s = = λ mλ ( 9.11×10−31 kg ) λ

Using the above four longest values of λ we get the four smallest values of v. Thus, v1 =

0.7278 ×10−3 m 2 /s = 7.28 × 104 m/s 10.0 ×10−9 m

v2 = 1.46 ×105 m/s , v3 = 2.18 ×105 m/s , and v4 = 2.91×105 m/s .

25.44. Model: As light is diffracted by matter, matter can also be diffracted by light. Solve:

The de Broglie wavelength of the sodium atoms is

λ=

h h 6.63 ×10−34 Js = = = 3.45 × 10−10 m p mv ( 3.84 × 10−26 kg ) ( 50 m/s )

The slit spacing of the “diffraction grating” is d = 12 λlaser = 12 600 nm = 300 nm. Using the diffraction grating equation with m = 1, we have

d sin θ = (1) λ ⇒ sinθ =

λ d

= 1.151×10−3 ≈ 1.2 ×10−3

In the small-angle approximation, sin θ ≅ tanθ = y L . We get

y = L sinθ = (1.0 m ) (1.151× 10−3 ) = 1.2 mm

25-1

26.1. Model: Use the charge model. Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred to the other because they are relatively free to move. Protons, on the other hand, are tightly bound in nuclei. So, electrons have been removed from the glass rod to make it positively charged. (b) Because each electron has a charge of 1.60 × 10−19 C, the number of electrons removed is

8.0 ×10−9 C = 5.0 ×1010 1.60 × 10−19 C

26.2. Model: Use the charge model. Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred into the other because they are relatively free to move. Protons, on the other hand, are tightly bound in the nuclei. So, electrons have been added to the plastic rod to make it negatively charged. (b) Because each electron has a charge of 1.60 × 10−19 C, the number of electrons added is

12 ×10−9 C = 7.5 ×1010 1.60 ×10−19 C

26.3. Model: Use the charge model and the model of a conductor as material through which electrons move.

Solve: (a) The charge of a plastic rod decreases from −15 nC to −10 nC. That is, −5 nC charge has been removed from the plastic. Because it is the negatively charged electrons that are transferred, −5 nC has been added to the metal sphere. (b) Because each electron has a charge of 1.60 ×10−19 C and a charge of 5 nC was transferred, the number of electrons transferred from the plastic rod to the metal sphere is

5 ×10−9 C = 3.1×1010 1.60 ×10−19 C

26.4. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve: (a) The charge of the glass rod decreases from +12 nC to +8.0 nC. Because it is the electrons that are transferred, −4.0 nC of electrons has been added to the glass rod. Thus, electrons are removed from the metal sphere and added to the glass rod. (b) Because each electron has a charge of 1.60 ×10−19 C and a charge of 4.0 nC was transferred, number of electrons transferred from the metal sphere to the glass rod is 4.0 ×10−9 C = 2.5 ×1010 1.60 × 10−19 C

26.5. Model: Use the charge model. Solve: Each oxygen molecule has 16 protons (8 per atom), and there are 6.02 ×1023 oxygen molecules in 1.0 mole of oxygen. Because one proton has a charge of +1.60 × 10−19 C, the amount of positive charge in 1.0 mole of oxygen is

6.022 ×1023 ×16 ×1.6 × 10−19 C = 1.54 ×106 C

26.6. Model: Use the charge model.

Since the density of water is 1 g/cm3, the mass of 1 L of water is 1000 mL ⎞ ⎛ 1 cm3 ⎞ ⎛ 1 g ⎞ (1.0 L ) ⎜⎛ ⎟ ⎜ 3 ⎟ = 1.0 kg. Each water molecule (H20) has 10 protons (8 in the oxygen atom and ⎟⎜ L ⎝ ⎠ ⎝ mL ⎠ ⎝ cm ⎠

Solve:

one per hydrogen atom), and thus 10 electrons. The number of moles is

1.0 ×103 g = 100 mole. There are 6.02 ×1023 10 g/mole

water molecules in 1.0 mole of water. Because one electron has a charge magnitude of 1.60 × 10−19 C, the amount of negative charge in 100 mole of water is 100 × 6.022 ×1023 ×10 × 1.6 × 10−19 C = 9.6 × 107 C

26.7. Model: move. Visualize:

Use the charge model and the model of a conductor as a material through which electrons

The charge carriers in a metal electroscope are the negative electrons. As the positive rod is brought near, electrons are attracted toward it and move to the top of the electroscope. The electroscope leaves now have a net positive charge, due to the missing electrons, and thus repel each other. At this point, the electroscope as a whole is still neutral (no net charge) but has been polarized. On contact, some of the electrons move to the positive rod to neutralize some (but not necessarily all) of the rod’s positive charge. After contact, the electroscope does have a net positive charge. When the rod is removed, the net positive charge on the electroscope quickly spreads to cover the entire electroscope. The net positive charge on the leaves causes them to continue to repel.

26.8. Model: Use the charge model. Solve: (a) No, we cannot conclude that the wall is charged. Attractive electric forces occur between (i) two opposite charges, or (ii) a charge and a neutral object that is polarized by the charge. Rubbing the balloon does charge the balloon. Since the balloon is rubber, its charge is negative. As the balloon is brought near the wall, the wall becomes polarized. The positive side of the wall is closer to the balloon than the negative side, so there is a net attractive electric force between the wall and the balloon. This causes the balloon to stick to the wall, with a normal force balancing the attractive electric force and an upward frictional force balancing the very small gravitational force on the balloon. (b)

26.9. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve:

The first step shows two neutral metal spheres touching each other. In the second step, the negative rod repels the negative charges which will retreat as far as possible from the top of the left sphere. Note that the two spheres are touching and the net charge on these two spheres is still zero. While the rod is there on top of the left sphere, the right sphere is moved away from the left sphere. Because the right sphere has an excess negative charge by the same amount as the left sphere has an excess positive charge, the separated right sphere is negatively charged as shown in the third step. As the two spheres are moved apart further and the negatively charged rod is moved away from the spheres, the charges on the two spheres redistribute uniformly over the entire sphere surface. Thus, we have oppositely charged the two spheres.

26.10. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve:

Charging two neutral spheres with opposite but equal charges can be done through the following four steps. (i) Touch the two neutral metal spheres together. (ii) Bring a charged rod (say, positive) close (but not touching) to one of the spheres (say, the left sphere). Note that the two spheres are still touching and the net charge on them is zero. The right sphere has an excess positive charge of exactly the same amount as the left sphere’s negative charge. (iii) Separate the spheres while the charged rod remains close to the left sphere. The charge separation remains on the spheres. (iv) Take the charged rod away from the two spheres. The separated charges redistribute uniformly over the metal sphere surfaces.

26.11. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve:

Charging two neutral spheres with like charges of exactly equal magnitude can be achieved through the following six steps. (i) Bring a charged rod (say, negative) near a neutral metal sphere. (ii) Touch the neutral sphere with the negatively charged rod, so that the rod-sphere system has a net negative charge. (iii) Move the rod away from the sphere. The sphere is now negatively charged. (iv) Bring this negatively charged sphere close to the second neutral sphere. (v) Touch these two spheres. The excess negative charge is distributed evenly over the two spheres. (vi) Separate the spheres. The excess charge will have the same sign as the charge on the charging rod and will be evenly distributed between the two spheres.

26.12. Model: Use the charge model. Solve:

(a)

(b) A charged object causes the charges inside a molecule to move apart while still being a part of the molecule. This is because electrons, that is, the negative charges in an insulator such as a piece of paper cannot move freely. Thus, the negative part of each molecule is slightly closer to a positively charged object and the positive part of each molecule is slightly farther from the positively charged object. This charge polarization leads to a slightly greater positive-to-negative attractive force than the positive-to-positive repulsive force. Because there are many molecules in a piece of paper, the total force of attraction is enough to cause a piece of paper to be attracted to the object. The test will also work for a negatively charged object because then the charges inside the molecules of paper will separate in the other direction.

26.13. Model: Model the charged masses as point charges. Visualize:

G G Solve: (a) The charge q1 exerts a force F1 on 2 on q2 to the right, and the charge q2 exerts a force F2 on 1 on q1 to the left. Using Coulomb’s law, F1 on 2 = F2 on 1 =

9 2 2 −6 −6 K q1 q2 ( 9.0 ×10 N m /C )(10 ×10 C )(10 ×10 C ) = = 0.90 N 2 r122 (1.0 m )

(b) Newton’s second law on either q1 or q2 is

F1 on 2 = m1a1 ⇒ a1 = Assess:

0.90 N = 0.90 m/s 2 1.0 kg

Even a micro-Couomb is a lot of charge. That is why F1 on 2 ( or F2 on 1 ) is a measurable force.

26.14. Model: Model the plastic spheres as point charges. Visualize:

G (a) The charge q1 = −50.0 nC exerts a force F1 on 2 on q2 = −50.0 nC to the right, and the charge q2 exerts G a force F2 on 1 on q1 to the left. Using Coulomb’s law, Solve:

F1 on 2 = F2 on 1 =

9 2 2 −9 −9 K q1 q2 ( 9.0 ×10 N m /C )( 50.0 × 10 C )( 50.0 ×10 C ) = = 0.056 N 2 r122 ( 2.0 ×10−2 m )

(b) The ratio is

0.056 N F1 on 2 = = 2.9 mg ( 2.0 ×10−3 kg )( 9.8 m/s2 )

26.15. Model: Model the glass bead and the ball bearing as point charges. Visualize:

G The ball bearing experiences a downward electric force F1 on 2 . By Newton’s third law, F2 on 1 = F1 on 2 . Solve: Using Coulomb’s law, F1 on 2 = K

q1 q2 ⇒ 0.018 N = r122

( 9.0 ×10

9

N m 2 /C 2 )( 20 ×10−9 C ) q2

(1.0 ×10−2 m )

2

⇒ q2 = 1.0 ×10−8 C

Because the force F1 on 2 is attractive and q1 is a positive charge, the charge q2 is a negative charge. Thus, q2 = −1.0 ×10−8 C = −10 nC.

26.16. Model: Charges A, B, and C are point charges.

G Please refer to Figure EX26.16. Charge A experiences an electric force FB on A due to charge B and G G G an electric force FC on A due to charge C. The force FB on A is directed to the right and the force FC on A is directed to the left. Solve: Coulomb’s law yields: Visualize:

FB on A =

FC on A = The net force on A is

( 9.0 ×10

9

N m 2 /C2 )(1.0 ×10−9 C )(1.0 × 10−9 C )

(1.0 ×10 m ) N m /C )(1.0 ×10 C )( 4.0 ×10 ( 2.0 ×10 m ) 2

−2

( 9.0 ×10

9

2

−9

2

−2

2

−9

C)

= 9.0 ×10−5 N

= 9.0 ×10−5 N

G G G Fon A = FB on A + FC on A = ( 9.0 × 10−5 N ) iˆ + ( 9.0 ×10−5 N ) −iˆ = 0 N

( )

26.17. Model: Charges A, B, and C are point charges. Visualize: Please refer to Figure EX26.17. Solve: The force on A from charge B is directed upward since the two negative charges repel.

FB on A =

1.0 ×10−9 C )( 2.0 ×10−9 C ) q A qB 9 2 2 ( = 9.0 × 10 Nm /C = 4.5 ×10−5 N ( ) 2 −2 4πε 0 r 2 ( 2.0 ×10 m ) 1

G So FB on A = ( 4.5 × 10−5 N, up ) . The force on A from charge C is directed downwards since the two opposite charges attract.

FC on A = ( 9.0 ×109 Nm 2 /C 2 )

(1.0 ×10 C )( 2.0 ×10 (3.0 ×10 m ) −9

−2

2

−9

C)

= 2.0 ×10−5 N

G So FC on A = (2.0 ×10−5 N, down). The net electric force on charge A is G G G FA = FB on A + FC on A = ( 4.5 ×10−5 N − 2.0 × 10−5 N, up )

= ( 2.5 ×10−5 N, up )

26.18. Model: Objects A and B are point charges. Visualize:

Because there are only two charges A and B, the force on charge A is due to charge B only, and the force on B is due to charge A only. Solve: Coulomb’s law gives the magnitude of the forces between the charge. Thus,

FA on B = FB on A =

( 9.0 ×10

9

N m 2 /C 2 )( 20.0 ×10−9 C )(10.0 ×10−9 C )

( 2.0 ×10−2 m )

2

= 4.5 ×10−3 N

Because the charge on object A is positive and on object B is negative, FB on A is upward and FA on B is downward. Thus, G G FB on A = +4.5 ×10−3 ˆj N FA on B = −4.5 × 10−3 ˆj N

26.19. Model: Assume the glass bead, the proton, and the electron are point charges. Visualize:

Solve:

Coulomb’s law gives

Fbead on electron = Fbead on proton =

( 9 ×10

9

N m 2 /C2 )( 20 × 10−9 C )(1.60 ×10−19 C )

(1.0 ×10

−2

m)

2

(a) Newton’s second law is F = ma, so

aproton = In vector form

Fbead on proton mproton

=

2.88 × 10−13 N = 1.72 ×1014 m/s 2 1.67 ×10−27 kg

G aproton = (1.72 × 1014 m/s 2, away from bead )

(b) Similarly, aelectron =

Fbead on electron 2.88 ×10−13 N = = 3.2 × 1017 m/s 2 melectron 9.11×10−31 kg

G Thus aelectron = ( 3.2 ×1017 m/s 2 , toward bead ) .

= 2.88 ×10−13 N

26.20. Model: Protons and electrons produce electric fields. Solve:

(a) The electric field of the proton is

⎡ ⎤ G +1.60 ×10−19 C ⎥ 1 q E= rˆ = ( 9.0 ×109 N m 2 /C 2 ) ⎢ rˆ = (1.44 × 10−3 N/C, away from proton ) 2 2 ⎢ (1.0 ×10−3 m ) ⎥ 4πε 0 r ⎣ ⎦ (b) The electric field of the electron is G E = (1.44 ×10−3 N/C ) ( − rˆ ) = (1.44 ×10−3 N/C, toward electron )

26.21. Model: Model the proton and theGelectron as point charges.G Solve:

G (a) The force that an electric field E exerts on a charge q is F = qE. A proton has q = e. Thus, G Fproton = e 200iˆ + 400ˆj N/C = 3.20iˆ + 6.40ˆj × 10−17 N

(

where we used e = 1.60 ×10

−19

)

(

C.

)

G G (b) The charge on an electron is q = −e. Thus, Felectron = − Fproton = −3.20iˆ − 6.40 ˆj × 10−17 N

(

)

(c) From Newton’s second law, aproton =

Fproton mproton

=

Fx2 + Fy2 mproton

=

7.15 × 10−17 N = 4.28 ×1010 m/s 2 1.67 ×10−27 kg

(d) The electron experiences a force of the same magnitude but it has a different mass. Thus,

aelectron =

7.15 ×10−17 N = 7.85 ×1013 m/s 2 9.11×10−31 kg

The forces may be the same, but the electron has a much larger acceleration due to its much smaller mass. Assess: The two forces in parts (a) and (b) are equal in magnitude but opposite in direction.

26.22. Model: The electric field is due to a charge and extends to all points in space. Solve:

The magnitude of the electric field at a distance r from a charge q is

E=

q q ⇒ 1.0 N/C = ( 9.0 ×109 N m 2 /C 2 ) ⇒ q = 1.11×10−10 C = 0.111 nC 2 2 4πε 0 r (1.0 m ) 1

26.23. Model: The electric field is that of a positive charge on the glass bead. The charge is assumed to be a point charge. Solve: The electric field is

G 8.0 × 10−9 C E = ( 9.0 ×109 N m 2 /C 2 ) rˆ = 1.80 ×105 rˆ N/C 2 −2 ( 2.0 ×10 m ) where rˆ is the unit vector from the charge to the point at which we calculated the field. That is, the direction of the electric field is away from the bead.

26.24. Model: The electric field is that of the charge on the object. Assume the charge on the object is a point charge. Solve: The electric field at a distance r from a point charge q is

G 1 q E= rˆ 4πε 0 r 2 G Because the electric field points toward the object, E = (180,000 N/C )( − rˆ ) . Thus, − (180,000 N/C ) = ( 9.0 ×109 N m 2 /C2 )

q

( 2.0 ×10−2 m )

2

⇒ q = −8.0 × 10−9 C = −8.0 nC

26.25. Model: A field is the agent that exerts an electric force on a charge. Visualize:

Solve:

G Newton’s second law on the plastic ball is Σ Fnet

( )

the electric force,

Fon q = FG ⇒ q E = mg ⇒ E =

y

G G = Fon q − FG . To balance the gravitational force with

−3 mg (1.0 × 10 kg ) ( 9.8 N/kg ) = = 3.3 × 106 N/C q 3.0 ×10−9 C

Because Fon q must be upward and the charge is negative, the electric field at the location of the plastic ball must G be pointing downward. Thus E = ( 3.3 ×106 N/C, downward ). G G G G G G Assess: F = qE means the sign of the charge q determines the direction of F or E. For positive q, E and F G G are pointing in the same direction. But E and F point in opposite directions when q is negative.

26.26. Model: A field is the agent that exerts an electric force on a charge. Visualize:

Solve:

G G (a) To balance the gravitational force on a proton Σ ( Fnet ) y = Fon p − FG

( )

Fon p = ( FG )p ⇒ q E = mg ⇒ E =

p

= 0 N. This means

−27 mg (1.67 ×10 kg ) ( 9.8 N/kg ) = = 1.02 × 10−7 N/C q 1.60 ×10−19 C

Because Fon p must be upward and the proton charge is positive, the electric field at the location of the proton G must also be pointing upward. Thus E = (1.02 × 10−7 N/C, downward ) . (b) In the case of the electron,

E=

−31 mg ( 9.11×10 kg ) ( 9.8 N/kg ) = = 5.58 ×10−11 N/C q 1.60 ×10−19 C

Because Fon e must be upward and the electron has a negative charge, the electric field at the location of the G electron must be pointing downward. Thus E = ( 5.58 × 10−11 N/C, downward ) .

26.27. Model: The electric field is that of a positive point charge located at the origin. Visualize:

The positions (5.0 cm, 0.0 cm), (−5.0 cm, 5.0 cm), and (−5.0 cm, −5.0 cm) are denoted by A, B, and C, respectively. Solve: (a) The electric field for a positive charge is G ⎛ 1 q ⎞ E =⎜ , away from q ⎟ 2 ⎝ 4πε 0 r ⎠

Using 1 4πε 0 = 9.0 ×109 N m 2 /C2 and q = 12 ×10−9 C, G ⎛ 108 N m 2 /C ⎞ E =⎜ , away from q ⎟ 2 r ⎝ ⎠ The electric fields at points A, B, and C are G 108 N m 2 /C ˆ EA = i = 4.3 ×104 iˆ N/C 2 −2 (5.0 ×10 m ) G EB =

EC =

( −5.0 ×10

( −5.0 ×10

108 N m 2 /C −2

m ) + ( 5.0 ×10 2

−2

108 N m 2 /C −2

m ) + ( −5.0 ×10 2

−2

⎡ 1 4 ˆ 4 ˆ ˆ ˆ ⎤ ⎢ 2 −i + j ⎥ = −1.53 ×10 i + 1.53 × 10 j N/C ⎦ m) ⎣ 2

(

) (

)

⎡ 1 4 ˆ 4 ˆ ˆ ˆ ⎤ ⎢ 2 −i − j ⎥ = −1.53 ×10 i − 1.53 ×10 j N/C ⎦ m) ⎣ 2

(

) (

(b) The three vectors are shown in the diagram. G G G Assess: The vectors EA , EB , and EC are pointing away from the positive charge.

)

26.28. Model: The electric field is that of a negative charge located at the origin. Visualize:

The positions (5.0 cm, 0.0 cm), (−5.0 cm, −5.0 cm), and (−5.0 cm, 5.0 cm) are denoted by A, B, and C, respectively. Solve: (a) The electric field for a positive charge is

G ⎛ 1 q ⎞ E =⎜ , away from q ⎟ 2 ⎝ 4πε 0 r ⎠

Using 1 4πε 0 = 9.0 ×109 N m 2 /C2 and q = −12 ×10−9 C, G ⎛ 108 N m 2 /C ⎞ E =⎜ , toward q ⎟ 2 r ⎝ ⎠ The electric fields at points A, B, and C are G 108 N m 2 /C EA = − ˆj = −4.3 ×104 ˆj N/C 2 ( 5.0 ×10−2 m )

( )

G EB =

( −5.0 ×10

G EC =

108 N m 2 /C −2

( −5.0 ×10

m ) + ( −5.0 ×10 2

−2

108 N m 2 /C −2

m ) + ( 5.0 ×10 2

−2

⎡ 1 ˆ ˆ ⎤ 4 ˆ 4 ˆ ⎢ 2 i + j ⎥ = 1.53 × 10 i + 1.53 × 10 j N/C ⎣ ⎦ m) 2

(

) (

)

⎡ 1 ˆ ˆ ⎤ i − j ⎥ = 1.53 × 104 iˆ − 1.53 × 104 ˆj N/C ⎢ ⎦ m) ⎣ 2 2

(

) (

(b) The three vectors are shown in the diagram. G G G Assess: Note that the vectors EA , EB , and EC are pointing toward the negative charge.

)

26.29. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve: Because the metal spheres are identical, the total charge is split equally between the two spheres. That is, qA = qB = 5.0 × 1011 electrons. Thus, the charge on metal spheres A and B is

( 5.0 ×10 )( −1.60 ×10 11

−19

C ) = −80 nC.

26.30. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve: Plastic is an insulator and does not transfer charge from one sphere to the other. The charge of metal sphere A is (1.0 ×1012 )( −1.60 ×10−19 C ) = −160 nC and the charge of metal sphere B is 0 C.

26.31. Model: Use the charge model. Solve:

The number of moles in the penny is M 3.1 g n= = = 0.04882 mol A 63.5 g/mol

The number of copper atoms in the penny is N = nN A = ( 0.04882 mol ) ( 6.02 ×1023 mol−1 ) = 2.939 ×1022 Since each copper atom has 29 electrons and 29 protons, the total positive charge in the copper penny is

( 29 × 2.939 ×10 )(1.60 ×10 22

−19

C ) = 1.36 ×105 C

Similarly, the total negative charge is −1.36 × 105 C. Assess: Total positive and negative charges are equal in magnitude.

26.32. Model: The beads are point charges. Visualize:

Solve: The beads are oppositely charged so are attracted to one another. The force on each is the same by Newton’s third law, and is

F=K

4.0 ×10−9 C )( 8.0 ×10−9 C ) q1 q2 −9 2 2 ( = 9.0 × 10 Nm /C = 7.2 ×10−4 N ( ) 2 −2 r2 2.0 × 10 m ( )

The beads accelerate at different rates because their masses are different. By Newton’s second law, the acceleration of the bead on the left is aleft =

−4 F ( 7.2 ×10 N ) = = 0.36 m/s 2 to the right. m ( 2.0 ×10−3 kg )

For the bead on the right, the acceleration is aright =

( 7.2 ×10 ( 4.0 ×10

−4

−3

N)

kg )

= 0.180 m/s 2 to the left.

26.33. Model: The protons are point charges. Solve:

(a) The electric force between the protons is

FE = K

9 2 2 −19 −19 q1 q2 ( 9.0 ×10 N m /C )(1.60 × 10 C )(1.60 ×10 C ) = = 58 N 2 r2 ( 2.0 ×10−15 m )

(b) The gravitational force between the protons is

FG =

G m1m2 ( 6.67 ×10 = r2

−11

N m 2 /kg 2 )(1.67 ×10−27 kg )(1.67 ×10−27 kg )

( 2.0 ×10

−15

m)

2

(c) The ratio of the electric force to the gravitational force is

FE 58 N = = 1.23 × 1036 FG 4.7 ×10−35 N

= 4.7 ×10−35 N

26.34. Model: The 125Xe nucleus and the proton are point charges. That is, all the charge on the Xe nucleus is assumed to be at its center. Visualize:

Solve:

(a) The magnitude of the force between the nucleus and the proton is given by Coulomb’s law:

Fnucleus on proton =

K qnucleus qproton r

2

=

( 9.0 ×10

9

N m 2 /C2 )( 54 ×1.60 ×10−19 C )(1.60 ×10−19 C )

( 5.0 ×10

−15

m)

2

(b) Applying Newton’s second law to the proton,

Fon proton = mproton aproton ⇒ aproton =

5.0 ×102 N = 3.0 × 1029 m/s 2 1.67 × 10−27 kg

= 5.0 ×102 N

26.35. Model: The two charged spheres are point charges. Solve: The electric force on one charged sphere due to the other charged sphere is equal to the sphere’s mass times its acceleration. Because the spheres are identical and equally charged, m1 = m2 = m and q1 = q2 = q. We have

F2 on 1 = F1 on 2 =

Kq1q2 Kq 2 = 2 = ma r2 r

−3 −2 2 mar 2 (1.0 × 10 kg )(150 m/s )( 2.0 × 10 m ) ⇒q = = = 6.7 × 10−15 C 2 K 9.0 ×109 N m 2 /C 2 ⇒ q = 8.2 × 10−8 C = 82 nC 2

2

26.36. Model: Objects A and B are point charges. Visualize:

Solve: (a) It is given that FA on B = 0.45 N. By Newton’s third law, FB on A = FA on B = 0.45 N. (b) Coulomb’s law is

FB on A = FA on B = 0.45 N = ⇒ qB =

( 0.45 N ) r 2 2K

=

KqA qB K ( 2qB )( qB ) = r2 r2

( 0.45 N ) (10 ×10−2 m )

2

2 ( 9.0 ×109 N m 2 /C 2 )

= 5.0 × 10−7 C ⇒ qA = 2qB = 1.0 × 10−6 C

(c) Newton’s second law is FB on A = mAaA. Hence,

aA =

FB on A FA on B 0.45 N = = = 4.5 m/s 2 mA mB 0.100 kg

26.37. Model: The charges are point charges. Visualize: Please refer to Figure P26.37. G G Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1 , where q1 is the 1.0 nC charge, q2 is the 2.0 nC charge, and q3 is the other 2.0 nC charge. We have

G ⎛ K q1 q2 ⎞ , away from q2 ⎟ F2 on 1 = ⎜ 2 r ⎝ ⎠ ⎛ ( 9.0 ×109 N m 2 /C 2 )(1.0 ×10−9 C )( 2.0 ×10−9 C ) ⎞ ⎟ , away from q =⎜ 2 2 −2 ⎜ ⎟ (1.0 ×10 m ) ⎝ ⎠ = (1.8 ×10−4 N, away from q2 ) = (1.8 ×10−4 N ) cos60°iˆ + sin 60° ˆj

(

)

G ⎛ K q1 q3 ⎞ F3 on 1 = ⎜ , away from q3 ⎟ = (1.8 × 10−4 N, away from q3 ) = (1.8 × 10−4 N ) − cos60°iˆ + sin 60 ˆj r2 ⎝ ⎠ G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 = 2 (1.8 ×10−4 N ) sin 60° ˆj = 3.1×10−4 ˆj N

(

So the force on the 1.0 nC charge is 3.1×10−4 N directed upward.

)

26.38. Model: The charges are point charges. Visualize: Please refer to Figure P26.38. G G Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1 , where q1 is the 1.0 nC charge, q2 is the 2.0 nC charge, and q3 is the −2.0 nC charges. We have

G ⎛ K q1 q2 ⎞ , away from q2 ⎟ F2 on 1 = ⎜ 2 r ⎝ ⎠ 9 2 2 ⎛ ( 9.0 ×10 N m /C )(1.0 × 10−19 C )( 2.0 × 10−19 C ) ⎞ =⎜ , away from q2 ⎟ 2 ⎜ ⎟ (1.0 ×10−2 m ) ⎝ ⎠

(

= (1.80 × 10−4 N, away from q2 ) = (1.80 × 10−4 N ) cos60°iˆ + sin 60° ˆj

)

G ⎛ K q1 q2 ⎞ F3 on 1 = ⎜ , toward q3 ⎟ = (1.80 ×10−4 N, toward q3 ) = (1.80 × 10−4 N ) cos60°iˆ − sin 60° ˆj . 2 ⎝ r ⎠ G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 = (1.80 ×10−4 N ) ( 2cos60° ) iˆ = 1.80 ×10−4 iˆ N

(

So, the force on the 1.0 nC charge is 1.80 × 10–4 N and it is directed to the right.

)

26.39. Model: The charges are point charges. Visualize:

G G The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have

Solve:

G ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ , away from q2 ⎟ 2 r ⎝ ⎠ ⎛ ( 9.0 ×109 N m 2 /C 2 )(10 × 10−9 C )( 5.0 ×10−9 C ) ⎞ ⎟ , away from =⎜ q 2 2 −2 ⎜ ⎟ 1.0 10 m × ( ) ⎝ ⎠ = ( 4.5 ×10−3 N, away from q2 ) = −4.5 × 10−3 ˆj N

G ⎛ K q1 q2 ⎞ , toward q3 ⎟ F3 on 1 = ⎜ 2 r ⎝ ⎠ 9 2 2 ⎛ ( 9.0 ×10 N m /C )(10 ×10−9 C )(15 ×10−9 C ) ⎞ , toward q3 ⎟ =⎜ 2 2 ⎜ ⎟ ( 3.0 ×10−2 m ) + (1.0 ×10−2 m ) ⎝ ⎠

(

= (1.35 × 10−3 N, toward q3 ) = (1.35 ×10−3 N ) − cosθ iˆ + sin θ ˆj

From the geometry of the figure,

⎛ 1.0 cm ⎞ ⎟ = 18.4° ⎝ 3.0 cm ⎠

θ = tan −1 ⎜

This means cos θ = 0.949 and sin θ = 0.316. Therefore, G F3 on 1 = −1.28 ×10−3 iˆ + 0.43 ×10−3 ˆj N

(

)

G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 = −1.28 × 10−3 iˆ − 4.07 × 10−3 ˆj N

(

The magnitude and direction of the resultant force vector are

( −1.28 ×10

Fon 1 =

N ) + ( −4.07 × 10−3 N ) = 4.3 × 10−3 N 2

2

4.07 ×10−3 N = 3.180 ⇒ φ = tan −1 ( 3.180 ) = 72.5° below the − x axis, 1.28 ×10−3 N points 252.5° counterclockwise from the +x-axis.

tanφ = or Fon 1

−3

)

)

26.40. Model: The charges are point charges. Visualize:

Solve:

G G The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have G ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ , toward q2 ⎟ 2 ⎝ r ⎠ 9 2 2 ⎛ ( 9.0 × 10 N m /C )(10 × 10−9 C )(10 × 10−9 C ) ⎞ , toward q2 ⎟ =⎜ 2 ⎜ ⎟ (1.0 ×10−2 m ) ⎝ ⎠ −3 −3 ˆ = ( 0.90 × 10 N, toward q2 ) = 0.90 × 10 j N G ⎛ K q1 q3 ⎞ , toward q3 ⎟ F3 on 1 = ⎜ 2 ⎝ r ⎠ ⎛ ( 9.0 × 109 N m 2 /C 2 )(10 × 10−9 C ) ( 8.0 × 10−9 C ) ⎞ , toward q3 ⎟ =⎜ 2 ⎜ ⎟ ( 3.0 ×10−2 m ) ⎝ ⎠ −4 −3 ˆ = ( 8.0 × 10 N, toward q3 ) = −0.80 × 10 i N

G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 = −0.80 × 10−3 iˆ + 0.90 × 10−3 ˆj N

(

)

The magnitude and direction of the resultant force vector are Fon 1 = tan φ =

( −0.80 ×10

−3

N ) + ( 0.90 × 10−3 N ) = 1.20 × 10−3 N 2

2

0.90 × 10−3 N ⇒ φ = tan −1 (1.13) = 48° above the − x -axis, 0.80 × 10−3 N

or Fon 1 points 132° counterclockwise from the +x-axis.

26.41. Model: The charges are point charges. Visualize:

Solve:

G G The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have G ⎛ K q1 q2 ⎞ , away from q2 ⎟ F2 on 1 = ⎜ 2 r ⎝ ⎠ ⎛ ( 9.0 ×109 N m 2 /C2 )( 5 ×10−9 C )(10 ×10−9 C ) ⎞ ⎟ q , away from =⎜ 2 2 2 −2 −2 ⎜ ⎟ 4.0 10 m 3.0 10 m × + × ( ) ( ) ⎝ ⎠ = (1.8 ×10−4 N, away from q2 ) = (1.8 × 10−4 N ) − cosθ iˆ − sin θ ˆj

(

)

From the geometry of the figure, tan θ =

G 4.0 cm ⇒ θ = 53.13° ⇒ F2 on 1 = (1.8 ×10−4 N ) −0.6iˆ − 0.8 ˆj 3.0 cm

(

)

⎛ ( 9.0 ×109 N m 2 /C 2 )( 5 ×10−9 C )( 5 × 10−9 C ) ⎞ G −4 ⎟ F3 on 1 = ⎜ , toward q N, toward q3 ) = 2.5 × 10−4 iˆ N 3 = ( 2.5 × 10 2 −2 ⎜ ⎟ ( 3.0 ×10 m ) ⎝ ⎠ G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 = 1.42 × 10−4 iˆ − 1.44 × 10−4 ˆj N

(

)

The magnitude and direction of the resultant force vector are Fon 1 =

(1.42 ×10

−4

N ) + ( −1.44 ×10−4 N ) = 2.0 × 10−4 N 2

2

⎛ 1.44 ×10−4 N ⎞ ⎟ = 45° clockwise from the +x-axis. −4 ⎝ 1.42 ×10 N ⎠

φ = tan −1 ⎜

26.42. Model: The charges are point charges. Visualize:

Solve:

G G The electric force on q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have Error! Objects cannot be created from editing field codes.

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From the geometry of the figure, 4.0 cm ⇒ cosθ = 0.6 and sinθ = 0.8 3.0 cm G G G G ⇒ F3 on 1 = −1.08 × 10−4 iˆ − 1.44 ×10−4 ˆj N ⇒ Fon 1 = F2 on 1 + F3 on 1 = −1.08 × 10−4 iˆ + 1.36 × 10−4 ˆj N tan θ =

(

)

(

The magnitude and direction of the resultant force vector are Fon 1 = tan φ =

(1.08 ×10

−4

N ) + ( −1.36 × 10−4 N ) = 1.74 × 10−4 N 2

2

1.36 ×10−4 N ⇒ φ = 52° clockwise from the −x-axis, or 128° CCW from +x-axis. 1.08 ×10−4 N

)

26.43. Model: The charges are point charges. Visualize: Please refer to Figure P26.43. Solve: Placing the 1.0 nC charge at the origin and calling it q1, the q2 charge is in the first quadrant, the q3 charge is in the fourth quadrant, the q4 charge is in the third quadrant, and the q5 charge is in the second quadrant. G G G G The electric force on q1 is the vector sum of the forces F2 on 1 , F3 on 1 , F4 on 1 , and F5 on 1. The magnitude of these four forces is the same because all four charges are equal and equidistant from q1. So,

( 9.0 ×10 N m /C )( 2.0 ×10 C )(1.0 ×10 ( 0.50 ×10 m ) + ( 0.50 ×10 m ) 9

F2 on 1 = F3 on 1 = F4 on 1 = F5 on 1 =

2

−9

2

−2

2

−2

2

−9

C)

= 3.6 ×10−4 N

G Thus, F on 1 = (3.6 × 10−4 N, toward q2) + (3.6 × 10−4 N, toward q3) + (3.6 × 10−4 N, toward q4) + (3.6 × 10−4 N, toward q5). In component form, G K Fon 1 = Fon 1 ⎡ cos 45°iˆ + sin 45° ˆj + cos 45°iˆ − sin 45° ˆj + − cos 45°iˆ − sin 45° ˆj + − cos 45°iˆ + sin 45° ˆj ⎤ = 0 N ⎣ ⎦

(

) (

) (

) (

)

26.44. Model: The charges are point charges. Visualize: Please refer to Figure P26.44. Solve: Placing the 1.0 nC charge at the origin and calling it q1, the q2 charge is in the first quadrant, the q3 charge is in the fourth quadrant, the q4 charge is in the third quadrant, and the q5 charge is in the second quadrant. The electric force on q1 is the vector sum of the electric forces from the other four charges q2, q3, q4, and q5. The magnitude of these four forces is the same because all four charges are equal in magnitude and are equidistant from q1. So,

( 9.0 ×10 N m /C )( 2.0 ×10 C )(1.0 ×10 ( 0.50 ×10 m ) + ( 0.50 ×10 m ) 9

F2 on 1 = F3 on 1 = F4 on 1 = F5 on 1 =

2

−9

2

−2

2

−2

−9

C)

2

= 3.6 ×10−4 N

G Thus, Fon 1 = (3.6 × 10−4 N, away from q2) + (3.6 × 10−4 N, away from q3) + (3.6 × 10−4 N, toward q4) + (3.6 ×

10−4 N, toward q5). In component form, K Fon 1 = Fon 1 ⎡ − cos 45°iˆ − sin 45° ˆj + − cos 45°iˆ + sin 45° ˆj + − cos 45°iˆ − sin 45° ˆj + − cos 45°iˆ + sin 45° ˆj ⎤ ⎣ ⎦ 3 −4 − = ( 3.6 ×10 N ) −4cos 45°iˆ = −1.02 × 10 iˆ N

(

(

) ( )

) (

) (

)

26.45. Model: The charges are point charges. Visualize: Please refer to Figure P26.45. Solve: Placing the 1.0 nC charge at the origin and calling it q1, the −6.0 nC is q3, the q2 charge is in the first quadrant, and the q4 charge is in the second quadrant. The net electric force on q1 is the vector sum of the electric forces from the other three charges q2, q3, and q4. We have G ⎛ K q1 q2 ⎞ , away from q2 ⎟ F2 on 1 = ⎜ 2 ⎝ r ⎠

⎛ ( 9.0 ×109 N m 2 /C 2 )(1.0 ×10−9 C )( 2.0 ×10−9 C ) ⎞ =⎜ , away from q2 ⎟ 2 ⎜ ⎟ ( 5.0 ×10−2 m ) ⎝ ⎠ −5 −5 ˆ = ( 0.72 ×10 N, away from q2 ) = ( 0.72 ×10 N ) − cos 45°i − sin 45° ˆj

(

)

G ⎛ K q1 q3 ⎞ F3 on 1 = ⎜ , toward q3 ⎟ 2 ⎝ r ⎠ ⎛ ( 9.0 ×109 N m 2 /C2 )(1.0 ×10−9 C )( 6.0 ×10−9 C ) ⎞ ⎟ =⎜ , toward q 3 2 −2 ⎜ ⎟ 5.0 10 m × ( ) ⎝ ⎠ −5 −5 ˆ = ( 2.16 ×10 N, away from q3 ) = 2.16 ×10 j N G ⎛ K q1 q4 ⎞ F4 on 1 = ⎜ , away from q4 ⎟ = ( 0.72 × 10−5 N ) cos 45°iˆ − sin 45° ˆj 2 r ⎝ ⎠ G G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 + F4 on 1 = ⎡⎣( 2.16 ×10 −5 N ) − ( 0.72 ×10 −5 N ) ( 2sin 45° ) ⎤⎦ ˆj = 1.14 × 10−5 ˆj N

(

)

26.46. Model: The charges are point charges. Visualize: Please refer to Figure P26.46. Solve: Placing the 1.0 nC charge at the origin and calling it q1, the −6.0 nC is q3, the q2 charge is in the first quadrant, and the q4 charge is in the second quadrant. The net electric force on q1 is the vector sum of the electric forces from the other three charges q2, q3, and q4. We have

G ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ , toward q2 ⎟ 2 ⎝ r ⎠ ⎛ ( 9.0 ×109 N m 2 /C 2 )(1.0 ×10−9 C )( 2.0 ×10−9 C ) ⎞ =⎜ , toward q2 ⎟ 2 ⎜ ⎟ ( 5.0 ×10−2 m ) ⎝ ⎠ −5 −5 ˆ = ( 0.72 × 10 N, toward q2 ) = ( 0.72 × 10 N ) cos 45°i + sin 45° ˆj

(

)

G ⎛ K q1 q3 ⎞ F3 on 1 = ⎜ , toward q3 ⎟ = ( 2.16 ×10−5 N, toward q3 ) = 2.16 × 10−5 ˆj N 2 r ⎝ ⎠ G ⎛ K q1 q4 ⎞ F4 on 1 = ⎜ , away from q4 ⎟ = ( 0.72 × 10−5 N ) cos 45°iˆ − sin 45° ˆj 2 ⎝ r ⎠ G G G G ⇒ Fon 1 = F2 on 1 + F3 on 1 + F4 on 1 = ( 0.72 ×10−5 N ) ( 2cos 45° ) iˆ + ( 2.16 ×10−5 N ) j

(

(

)

= 1.02 × 10−5 iˆ + 2.2 × 10−5 ˆj N

)

26.47. Model: The charged particles are point charges. Visualize:

Solve:

G G (a) The mathematical problem is to find the position for which the forces F1 on p and F2 on p are equal in

magnitude and opposite in direction. If the proton is at position x, it is a distance x from q1 and d − x from q2, where d = 1.0 cm. The magnitudes of the forces are

F1 on p =

K q1 qp r1p2

=

K q1 qp x2

F2 on p =

K q2 qp

(d − x)

2

Equating the two forces and using d = 1.0 cm, K q1 qp x2

=

K q2 qp

(d − x)

2

⇒

2.0 nC 4.0 nC ⇒ 1 + x2 − 2x = 2x2 ⇒ x2 + 2x − 1 = 0 = 2 x2 (d − x)

The solutions to the equation are x = +0.414 cm and −2.41 cm. Both are points where the magnitudes of the two forces are equal, but +0.414 cm is a point where the magnitudes are equal but the directions are the same. The solution we want is that the proton should be placed at x = 2.4 cm. (b) Yes, the net force on the electron located at x = −2.4 cm will be zero. This is because the solution in part (a) does not depend specifically on the type of the charge that experiences zero force from the other two charges. G Furthermore, if Fon p is zero for a proton, the electric field at that point must be zero. Thus, there will be no force on any charged particle at that point.

26.48. Model:

The charged particles are point charges.

Visualize:

Solve: The two 2.0 nC charges exert an upward force on the 1.0 nC charge. Since the net force on the 1 nC charge is zero, the unknown charge must exert a downward force of equal magnitude. This implies that q is a positive charge. The force of charges 2 on charge 1 is

G ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ , away from q2 ⎟ 2 ⎝ r12 ⎠ 9 2 2 (9.0 ×10 N m / C )(1.0 × 10−9 C)(2.0 ×10−9 C) = (cosθ iˆ + sinθ ˆj ) (0.020 m) 2 + (0.030 m) 2 From the figure, θ = tan–1(2/3) = 33.69°. Thus G F2 on 1 = (1.152 ×10−5 iˆ + 0.768 ×10−5 ˆj ) N G G G From symmetry, F3 on 1 is the same except the x-component is reversed. When we add F2 on 1 and F3 on 1 , the xcomponents cancel and the y-components add to give JG JG F 2 on 1 + F 3 on 1 = 1.536 × 10−5 ˆj N G Fq on 1 must have the same magnitude, pointing in the − ˆj direction, so G K q q1 (1.536 ×10−5 N)(0.020 m) 2 ⇒q= = 0.68 nC Fq on 1 = 1.536 ×10−5 N = 2 (9.0 ×109 N m 2 / C 2 )(1.0 × 10−9 C) r

A positive charge q = 0.68 nC will cause the net force on the 1.0 nC charge to be zero.

26.49. Model: The charged particles are point charges. Visualize: Please refer to Figure P26.49. Solve: The charge q2 is in static equilibrium, so the net electric field at the location of q2 is zero. We have

G G G 3.0 ×10−9 C ) q1 1 ˆ + 1 ( Enet = Eq1 + E−3 nC = i ± −iˆ = 0 N/C 4πε 0 ( 0.20 m )2 4πε 0 ( 0.10 m )2

( )

( )

We have used the ± sign to indicate that a positive charge on q1 leads to an electric field along +iˆ and a negative charge on q1 leads to an electric field along −iˆ. Because the above equation can only be satisfied if we use +iˆ, we infer that the charge q1 is a positive charge. Thus, 3.0 ×10−9 C ˆ q1 +iˆ − i = 0 N/C ⇒ q1 = 12.0 nC 2 2 ( 0.20 m ) ( 0.10 m )

( )

()

26.50. Model: The charged particles are point charges. Visualize:

Solve:

The force on q is the vector sum of the force from −Q and +Q. We have

G ⎛ −Q + q ⎞ KQq , toward − Q ⎟ = 2 F−Q on + q = ⎜ K 2 − cosθ iˆ − sinθ ˆj 2 a + y a + y2 ⎝ ⎠

(

)

G ⎛ K +Q + q ⎞ KQq , away from + Q ⎟ = 2 F+ Q on + q = ⎜ 2 − cosθ iˆ + sin θ ˆj 2 a + y a + y2 ⎝ ⎠

(

)

G KQq ⇒ Fnet = 2 ( −2cosθ ) iˆ + 0 ˆj N a + y2 From the figure we see that cosθ = a

a 2 + y 2 . Thus

( Fnet ) x = Assess:

−2 KQqa

(a

2

+ y2 )

3/ 2

Note that (Fnet)x = Fnet because the y-components of the two forces cancel each other out.

26.51. Model: The charged particles are point charges. Visualize:

Solve:

(a) The force on q is the vector sum of the force from −Q and +Q. We have

⎛ K +Q + q ⎞ G KQq F+ Q on + q = ⎜ , away from + Q ⎟ = −iˆ 2 ⎜ (a − x) ⎟ ( a − x )2 ⎝ ⎠

( )

⎛ K −Q + q ⎞ G KQq F− Q on + q = ⎜ , toward − Q ⎟ = −iˆ ⎜ ( a + x )2 ⎟ ( a + x )2 ⎝ ⎠

( )

⎡ 1 2 KQq ( a 2 + x 2 ) 1 ⎤ ⇒ ( Fnet ) x = − KQq ⎢ + = − ⎥ 2 2 2 ⎢⎣ ( a − x ) ( a + x ) ⎥⎦ ( a2 − x2 ) To arrive at the final expression we used ( a − x ) ( a + x ) = ⎡⎣( a − x )( a + x ) ⎤⎦ = ( a 2 − x 2 ) . (b) There are two cases when x > a. For x > a, 2

2

2

2

⎛ K +Q + q ⎞ G KQq +iˆ , away from + Q ⎟ = F+ Q on + q = ⎜ 2 ⎜ ( x − a) ⎟ ( x − a )2 ⎝ ⎠

( )

⎛ K −Q + q ⎞ G KQq F− Q on + q = ⎜ −iˆ , toward − Q ⎟ = ⎜ ( x + a )2 ⎟ ( x + a )2 ⎝ ⎠

( )

⎡ 1 1 ⎤ −4 KQqax ⇒ ( Fnet ) x = KQq ⎢ − = 2 2⎥ 2 ⎢⎣ ( x − a ) ( x + a ) ⎥⎦ ( x 2 − a 2 ) For x < − a (that is, for negative values of x), G KQq F+ Q on + q = −iˆ 2 x ( − a)

G KQq F− Q on + q = +iˆ 2 a ( + x)

( )

⇒ ( Fnet ) x = −

( )

4 KQqax

(x

2

− a2 )

2

That is, the net force is to the right when x > a and to the right when x < − a. We can combine these two cases into a single equation for x > a :

( Fnet ) x = Here, the force is always to the right when x > a.

4 KQqa x

(x

2

− a2 )

2

26.52. Model: The charges are point charges. Solve:

We will denote the charges −Q, 4Q and −Q by 1, 2, and 3, respectively.

G ⎛ K −Q q ⎞ KQq F1 on q = ⎜ , toward − Q ⎟ = 2 −iˆ 2 L L ⎝ ⎠

( )

⎛ ⎞ G K 4Q q 2 KQq ⎟ = 4 KQq +cos 45°iˆ + sin 45° ˆj ⇒ F = 2 F1 on q F2 on q = ⎜⎜ , away from 4 Q 2 on q = 2 2 ⎟ L2 2 L ⎜ 2L ⎟ ⎝ ⎠

(

(

)

)

G ⎛ K −Q q ⎞ KQq , toward − Q ⎟ = 2 − ˆj F3 on q = ⎜ 2 L L ⎝ ⎠

( )

The net electric force on the charge +q is the vector sum of the electric forces from the other three charges. The net force is G ˆj ⎞ KQq 2 KQq ⎛ iˆ KQq KQq KQq Fnet = 2 −iˆ + + ⎜+ ⎟ + 2 − ˆj = − 2 iˆ 1 − 2 − 2 ˆj 1 − 2 L L2 ⎝ L L L 2 2⎠

( )

(

)

2

( )

(

)

2

(

)

KQq ⎡ KQq ⎤ ⎡ KQq ⎤ ⇒ Fnet = ⎢ 2 1 − 2 ⎥ + ⎢ 2 1 − 2 ⎥ = 2 − 2 L2 ⎣ L ⎦ ⎣ L ⎦

(

)

(

)

26.53. Model: The charges are point charges. Visualize:

We must first identify the region of space where the third charge q3 is located. You can see from the figure that the forces can’t possibly add to zero if q3 is above or below the axis or outside the charges. However, at some point on the x-axis between the two charges the forces from the two charges will be oppositely directed. G G Solve: The mathematical problem is to find the position for which the forces F1 on 3 and F2 on 3 are equal in magnitude. If q3 is the distance x from q1, it is the distance L − x from q2. The magnitudes of the forces are F1 on 3 =

K q1 q3 Kq q3 = r132 x2

F2 on 3 =

K q2 q3 K ( 4q ) q3 = 2 r232 ( L − x)

Equating the two forces, Kq q3 K ( 4q ) q3 L 2 = ⇒ ( L − x ) = 4 x 2 ⇒ x = and − L 2 2 x 3 ( L − x) The solution x = −L is not allowed as you can see from the figure. To find the magnitude of the charge q3, we apply the equilibrium condition to charge q1: F2 on 1 = F3 on 1 ⇒

K q2 q1 K q3 q1 4 = ⇒ 4q = 9 q3 ⇒ q3 = q 2 1 L2 9 ( 3 L)

We are now able to check the static equilibrium condition for the charge 4q (or q2): F1 on 2 = F3 on 2 ⇒ K

4 q1 q2 K q3 q2 q q q = ⇒ 2= 9 2= 2 2 2 2 L L L L − x L ( ) (3 )

The sign of the third charge q3 must be negative. A positive sign on q3 will not have a net force of zero either on the charge q or the charge 4q. In summary, a charge of − 94 q placed x = 13 L from the charge q will cause the 3charge system to be in static equilibrium.

26.54. Model: Use the charge model and assume the copper spheres are point objects with point charges. Solve:

(a) The mass of the copper sphere is 3⎤ ⎛ 4π 3 ⎞ ⎡ 4π M = ρV = ⎜ r ⎟ = ( 8920 kg/m3 ) ⎢ (1.0 ×10−3 m ) ⎥ = 3.736 ×10−5 kg = 0.03736 g ⎝ 3 ⎠ ⎣ 3 ⎦

The number of moles in the sphere is n=

M 0.03742 g = = 5.884 ×10−4 mol A 63.5 g/mol

The number of copper atoms in the sphere is

N = nN A = ( 5.884 × 10−4 mol )( 6.02 ×1023 mol−1 ) = 5.542 ×1020 The number of electrons in the copper sphere is thus 29 × 3.542 × 1020 = 1.027 × 1022. The total positive or

negative charge in the sphere is (1.027 × 1022 )(1.60 × 10−19 C ) = 1643 C. Hence, the spheres will have a net charge of 10−9 × 1643 C = 1.643 × 10−6 C. The force between two such spheres is F=

Kq1q2

(1.0 ×10

−2

m)

2

(9.0 ×10 =

9

N m 2 /C2 )(1.643 × 10−6 C )

(1.0 ×10

−2

m)

2

2

= 2.4 × 102 N

(b) This is a force that is easily detectable. Since we don’t observe such forces, any difference between the proton charge and the electron charge must be smaller than 1 part in 109.

26.55. Model: The electron and the proton are point charges. Solve: The electric Coulomb force between the electron and the proton provides the centripetal acceleration for the electron’s circular motion. Thus, K ( e )( e ) mv 2 = = mrω 2 r2 r

⇒ f =

Ke 2 = mr 3

( 9.0 ×10 N m /C )(1.60 ×10 C ) ( 9.11×10 kg )( 5.3 ×10 ) 9

2

−31

2

−19

−11 3

2

= 4.12 ×1016 rad/s ×

1 rev = 6.6 × 1015 rev/s 2π rad

26.56. Model: Model the metal plate and yourself as point charges. Solve: At the beginning, both you and the metal plates are neutral. As the electrons are pumped from the metal plate into you, the plate becomes as much positively charged as you become negatively charged. When enough charge difference builds up between you and the metal plate, the gravitational force on you will be counterbalanced by the upward electrical force. You will slowly begin to hang suspended in the air when

Fplate on you = FG ⇒ mg ( 2.0 m ) = K 2

⇒q =

K q −q

( 2.0 m )

2

= mg

( 60 kg )( 9.8 N/kg )( 2.0 m ) 9.0 ×10 N m /C 9

2

2

2

Dividing this charge by the charge on an electron yields 3.2 ×1015 electrons.

= 5.11× 10−4 C

26.57. Model: The charged plastic beads are point charges and the spring is an ideal spring that obeys Hooke’s law. Solve: Let q be the charge on each plastic bead. The repulsive force between the beads pushes the beads apart. The spring is stretched until the restoring spring force on either bead is equal to the repulsive Coulomb force. That is,

Kq 2 k Δx r 2 = k Δx ⇒ q = 2 r K The spring constant k is obtained by noting that the weight of a 1.0 g mass stretches the spring 1.0 cm. Thus

mg = k (1.0 × 10−2 m) ⇒ k =

⇒q=

(1.0 ×10

−3

kg ) ( 9.8 N/kg )

1.0 ×10−2 m

= 0.98 N/m

( 0.98 N/m ) ( 4.5 ×10−2 m − 4.0 ×10−2 m )( 4.5 ×10−2 m ) 9.0 ×109 N m 2 /C2

2

= 33 nC

26.58. Solve: (a) Kinetic energy is K = 12 mv 2 , so the velocity squared is v2 = 2K/m. From kinematics, a particle moving through distance Δx with acceleration a, starting from rest, finishes with v2 = 2aΔx. To gain K = 2 × 10−18 J of kinetic energy in Δx = 2.0 μm requires an acceleration

a=

2K / m 2.0 × 10−18 J v2 K = = = = 1.10 × 1018 m/s 2 2Δx 2 Δx mΔx ( 9.11×10−31 kg )( 2.0 ×10−6 m )

(b) The force that produces this acceleration is

F = ma = ( 9.11×10−31 kg )(1.10 ×1018 m/s 2 ) = 1.00 × 10−12 N (c) The electric field is

E=

F 1.00 ×10−12 N = = 6.3 × 106 N/C e 1.6 ×10−19 C

(d) The force on an electron due to charge q is F = K q e r 2. To have a breakdown, the force on the electron

must be at least 1.00 × 10−12 N. The minimum charge that could cause a breakdown will be the charge that causes exactly a force of 1.00 × 10−12 N:

G K qe ( 0.010 m ) (1.00 ×10−12 N ) r 2F F = 2 = 1.00 ×10−12 N ⇒ q = = = 6.9 × 10−8 C = 68 nC r Ke ( 9.0 ×109 N m 2 / C 2 )(1.6 ×10−19 C ) 2

26.59. Model: The charged spheres are point charges. Visualize:

Each sphere is in static equilibrium and the string makes an angle θ with the vertical. The three forces acting on each sphere are the electric force, the gravitational force on the sphere, and the tension force. G G G G G Solve: In static equilibrium, Newton’s first law is Fnet = T + FG + Fe = 0. In component form,

( Fnet ) x = Tx + ( FG ) x + ( Fe ) x = 0 N ⇒ −T sin θ + 0 N +

⇒ T sin θ =

( Fnet ) = Ty + ( FG ) y + ( Fe ) y = 0 N

Kq 2 =0 N d2

Kq 2 Kq 2 = 2 d2 ( 2 L sinθ )

T cosθ − mg + 0 N = 0 N T cosθ = + mg

Dividing the two equations,

( 9.0 ×109 N m2/C2 )(100 ×10−9 C ) = 4.59 ×10−4 Kq 2 sin θ tan θ = 2 = 4 L mg 4 (1.0 m )2 ( 5.0 ×10−3 kg ) ( 9.8 N/kg ) 2

2

For small-angles, tan θ ≈ sin θ . With this approximation we obtain sinθ = 0.07714 rad and θ = 4.4°.

26.60. Model: The charged spheres are point charges. Visualize:

Each sphere is in static equilibrium when the string makes an angle of 20° with the vertical. The three forces acting on each sphere are the electric force, the gravitational force on the sphere, and the tension force. G G G G G Solve: In the static equilibrium, Newton’s first law is Fnet = T + FG + Fe = 0. In component form,

( )

( Fnet ) = Tx + ( FG ) x + ( Fe ) x = 0 N

⇒ −T sinθ + 0 N + ⇒ T sin θ =

( Fnet ) y = Ty + ( FG ) + ( Fe ) y = 0 N

Kq 2 =0N d2

T cosθ − mg + 0 N = 0 N

+ Kq 2 Kq 2 = 2 d2 ( 2 L sinθ )

T cosθ = + mg

Dividing the two equations and solving for q, 4 ( sin 2 20° tan 20° ) (1.0 m ) ( 3.0 ×10−3 kg ) ( 9.8 N/kg ) 4sin 2 θ tanθ L2 mg = = 0.75 μ C 9.0 ×109 N m 2 /C 2 K 2

q=

26.61. Model: The electric field is that of a positive point charge located at the origin. Visualize: Please refer to Figure P26.61. Place the 10 nC charge at the origin. Solve: The electric field is 9 2 2 −9 ⎞ G ⎛ 1 q ⎞ ⎛ ( 9.0 ×10 N m /C )(10 ×10 C ) ⎜ E =⎜ q = , away from , away from q ⎟ ⎟ 2 2 ⎜ ⎟ r ⎝ 4πε 0 r ⎠ ⎝ ⎠

⎛ 90.0 N m 2 /C ⎞ =⎜ , away from q ⎟ 2 r ⎝ ⎠ At each of the three points, ⎞ G ⎛ 90.0 N m 2 /C E1 = ⎜ , away from q ⎟ = 1.0 ×105 ˆj N/C 2 ⎜ ( 3.0 ×10−2 m ) ⎟ ⎝ ⎠ ⎞ G ⎛ 90.0 N m 2 /C , away from q ⎟ = ( 3.6 × 104 N/C ) cosθ iˆ + sinθ ˆj E2 = ⎜ 2 ⎜ ( 5.0 ×10−2 m ) ⎟ ⎝ ⎠ 4 3 ˆ 4 ˆ = ( 3.6 ×10 N/C ) 5 i + 5 j = 2.9 × 104iˆ + 2.2 × 104 ˆj N/C

(

(

) (

⎞ G ⎛ 90.0 N m 2 /C ⎟ = 5.6 × 104iˆ N/C , away from q E3 = ⎜ ⎜ ( 4.0 ×10−2 m )2 ⎟ ⎝ ⎠

)

)

26.62. Model: The electric field is that of a positive point charge. Visualize: Please refer to Figure P26.62. Place point 1 at the origin. Solve: The electric field is 9 2 2 −9 ⎞ G ⎛ 1 q ⎞ ⎛ ( 9.0 ×10 N m /C )( 2 × 10 C ) ⎜ E =⎜ q = , toward , toward q ⎟ ⎟ 2 2 ⎜ ⎟ r ⎝ 4πε 0 r ⎠ ⎝ ⎠

⎛ 18.0 N m 2 /C ⎞ =⎜ , toward q ⎟ 2 r ⎝ ⎠ The electric fields at the two points are ⎞ G ⎛ 18.0 N m 2 /C ⎟ E1 = ⎜ q , toward ⎜ (1.0 ×10−2 m )2 ⎟ ⎝ ⎠ = (1.8 × 105 N/C, 60° counter clockwise from the +x-axis or 60° north of east) ⎞ G ⎛ 18.0 N m 2 /C E2 = ⎜ , toward q ⎟ 2 ⎜ (1.0 ×10−2 m ) ⎟ ⎝ ⎠ = (1.8 × 105 N/C, 60° clockwise from the −x-axis or 60° north of west)

26.63. Model: The electric field is that of a positive point charge located at the origin. Visualize: Please refer to Figure P26.63. Place the 5.0 nC charge at the origin. Solve: The electric field is 9 2 2 −9 ⎞ G ⎛ 1 q ⎞ ⎛ ( 9 ×10 N m / C )( 5.0 × 10 C ) ⎜ E =⎜ q = , away from , away from q ⎟ ⎟ 2 2 ⎜ ⎟ r ⎝ 4πε 0 r ⎠ ⎝ ⎠

⎛ 45 N m 2 /C ⎞ =⎜ , away from q ⎟ 2 r ⎝ ⎠ At each of the three points, ⎞ G ⎛ 45 N m 2 /C ⎟ = ( 9.0 × 104 N/C ) cosθ iˆ + sin θ ˆj E1 = ⎜ q , away from ⎜ ( 2.0 ×10−2 m )2 + (1.0 × 10−2 m )2 ⎟ ⎝ ⎠ ⎛ 1 ˆ 2 ˆ⎞ 4 4ˆ = ( 9.0 ×10 N/C ) ⎜ i+ j ⎟ = 4.0 × 10 i + 8.0 × 104 ˆj N/C 5 ⎠ ⎝ 5

(

(

)

⎞ G ⎛ 45 N m 2 /C ⎟ = 4.5 ×105iˆ N/C E2 = ⎜ , away from q ⎜ (1.0 ×10−2 m )2 ⎟ ⎝ ⎠ ⎛ ⎞ 2 G 45 N m /C ⎟ = 4.0 × 104 iˆ − 8.0 × 104 ˆj N/C E3 = ⎜ q , away from ⎜ ( 2.0 × 10−2 m )2 + (1.0 × 10−2 m )2 ⎟ ⎝ ⎠

(

)

)

26.64. Model: The electric field is that of a negative charge at (x, y) = (2.0 cm, 1.0 cm). Visualize:

Solve: (a) The electric field of a negative charge points toward the charge, so we can roughly locate where the field has a particular value by inspecting the signs of Ex and Ey. At point a, the electric field has no y-component and the x-component points to the left, so its location must be to the right of the charge along a horizontal line. Using the equation for the field of a point charge, Ex = E =

Kq ⇒ ra = ra2

( 9.0 ×10

Kq = Ex

9

N m 2 /C 2 )(10.0 ×10−9 C ) 225,000 N/C

= 0.0200 m = 2.00 cm

Thus, point a is at the position (xa, ya) = (4 cm, 1 cm). (b) Point b is above and to the left of the charge. The magnitude of the field at this point is E = Ex2 + E y2 =

(161,000 N/C ) + ( 80,500 N/C ) 2

2

= 180,000 N/C

Using the equation for the field of a point charge, E=

Kq ⇒ rb = rb2

Kq = E

( 9.0 ×10

9

N m 2 /C2 )(10 ×10−9 C ) 180,000 N/C

= 2.236 cm

This gives the total distance but not the horizontal and vertical components. However, we can determine the G angle θ because Eb points straight toward the negative charge. Thus, ⎛ Ey ⎜ E ⎝ x

θ = tan −1 ⎜

⎞ ⎛ 80,500 ⎞ −1 ⎛ 1 ⎞ ⎟ = tan −1 ⎜ ⎟ = tan ⎜ ⎟ = 26.57° ⎟ ⎝ 2⎠ ⎝ 161,000 ⎠ ⎠

The horizontal and vertical distances are then d x = rb cosθ = 2.00 cm and d y = rb sin θ = 1.00 cm. Thus, point b is

at the position (xb, yb) = (0 cm, 2 cm). (c) Point c, which is below and to the left of the charge, is calculated by following a similar procedure. We first find that E = 36,000 N/C . From this we find that the total distance rc = 5.00 cm. The angle φ is ⎛ Ey ⎜ Ex ⎝

φ = tan −1 ⎜

⎞ ⎛ 21,600 ⎞ ⎟ = tan −1 ⎜ ⎟ = 36.87° ⎟ ⎝ 28,000 ⎠ ⎠

which gives the distances d x = rc cos φ = 4.00 cm and d y = rc sin φ = 3.00 cm. Thus point c is at position (xc, yc) = (−2 cm, − 2 cm).

26.65. Model: Visualize:

The electric field is that of a positive charge at (x, y) = (1.0 cm, 2.0 cm).

Solve: (a) The electric field of a positive charge points straight away from the charge, so we can roughly locate the points of interest based simply on whether the signs of Ex and Ey are positive or negative. For point a, the electric field has no y-component and the x-component points to the left, so point a must be to the left of the charge along a horizontal line. Using the field of a point charge,

Ex = E =

Kq ⇒ ra = ra2

Kq = Ex

( 9.0 ×10

9

N m 2 /C 2 )(10.0 ×10−9 C ) 225,000 N/C

= 0.0200 m = 2.00 cm

Thus, (xa, ya) = (−1 cm, 2 cm). (b) Point b is above and to the right of the charge. The magnitude of the field at this point is E = Ex2 + E y2 =

(161,000 N/C ) + (80,500 N/C ) 2

2

= 180,000 N/C

Using the field of a point charge, E=

Kq ⇒ rb = rb2

Kq G = E

( 9.0 ×10

9

N m 2 /C2 )(10.0 ×10−9 C ) 180,000 N/C

= 2.236 cm

This gives the total distance but not the horizontal and vertical components. However, we can determine the G angle θ because Eb points straight away from the positive charge. Thus, ⎛ Ey ⎞ ⎛ 80,500 N/C ⎞ −1 ⎛ 1 ⎞ ⎟ = tan −1 ⎜ ⎟ = tan ⎜ ⎟ = 26.57° ⎜ Ex ⎟ 161,000 N/C ⎠ ⎝ 2⎠ ⎝ ⎝ ⎠ The horizontal and vertical distances are then d x = rb cosθ = ( 2.236 cm ) cos 26.57° = 2.00 cm and d y = rb sin θ = 1.00 cm. Thus, point b is at position (xb, yb) = (3 cm, 3 cm). (c) To calculate point c, which is below and to the right of the charge, a similar procedure is followed. We first find E = 36,000 N/C from which we find the total distance rc = 5.00 cm. The angle φ is

θ = tan −1 ⎜

⎛ Ey ⎞ ⎛ 28,800 ⎞ ⎟ = tan −1 ⎜ ⎟ = 53.13° ⎜ Ex ⎟ ⎝ 21,600 ⎠ ⎝ ⎠ which gives distances d x = rc cos φ = 3.00 cm and d y = rc sin φ = 4.00 cm. Thus, point c is at position (xc, yc) = (4 cm, − 2 cm).

φ = tan −1 ⎜

26.66. Model: The electric field is that of three point charges. Visualize:

Solve:

(a) In the figure, the distances are r1 = r3 =

(1 cm ) + ( 3 cm ) 2

2

= 3.162 cm and the angle is

θ = tan −1 (1/ 3) = 18.43°. Using the equation for the field of a point charge, E1 = E3 =

9 2 2 −9 K q1 ( 9.0 ×10 N m /C )(1.0 ×10 C ) = = 9.0 kN/C 2 r12 ( 0.03162 m )

We now use the angle θ to find the components of the field vectors:

( ) N/C = (8.5iˆ − 2.8 ˆj ) kN/C = E cosθ iˆ + E sinθ ˆj = ( 8540iˆ + 2840 ˆj ) N/C = ( 8.5iˆ + 2.8 ˆj ) kN/C

E1 = E1 cosθ iˆ − E1 sin θ ˆj = 8540iˆ − 2840 ˆj E3

3

3

E2 is easier since it has only an x-component. Its magnitude is E2 =

9 2 2 −9 K q2 ( 9.0 ×10 N m /C )(1.0 ×10 C ) = = 10,000 N/C ⇒ E2 = E2 iˆ = 10.0 iˆ kN/C 2 r22 ( 0.0300 m )

(b) The electric field is defined in terms of an electric force acting on charge q: E = F q . Since forces obey a principle of superposition ( Fnet = F1 + F2 + … ) it follows that the electric field due to several charges also obeys a

principle of superposition. (c) The net electric field at a point 3 cm to the right of q2 is Enet = E1 + E2 + E3 = 27 iˆ kN/C. The y-components of

E1 and E2 cancel, giving a net field pointing along the x-axis.

26.67. Model: The charged ball attached to the string is a point charge. Visualize:

The ball is in static equilibrium in the external electric field when the string makes an angle θ = 20° with the vertical. The three forces acting on the charged ball are the electric force due to the field, the gravitational force on the ball, and the tension force. G G G G G Solve: In static equilibrium, Newton’s second law for the ball is Fnet = T + FG + Fe = 0. In component form,

( Fnet ) x = Tx + 0 N + qE = 0 N

( Fnet ) y = Ty − mg + 0 N = 0 N

The above two equations simplify to T sin θ = qE Dividing both equations, we get

tan θ =

T cosθ = mg

−3 qE mg tan θ ( 5.0 × 10 kg ) ( 9.8 N/kg ) tan 20° ⇒q= = = 1.78 × 10−7 C = 178 nC 100,000 N/C mg E

26.68. Model: The charged ball attached to the string is the point charge. Visualize:

The charged ball is in static equilibrium in the external electric field when the string makes an angle θ with the vertical. The three forces acting on the charge are the electric force due to the electric field, the gravitational force on the ball, and the tension force. G G G G G Solve: In static equilibrium, Newton’s second law for the charged ball is Fnet = T + FG + F3 = 0. In component form,

( Fnet ) x = Tx + 0 N + qE = 0 N

( Fnet ) y = Ty − mg + 0 N = 0 N

These two equations become T sin θ = qE and T cosθ = mg . Dividing the equations gives qE ( 25 ×10 C ) ( 200,000 N/C ) = = 0.255 ⇒ θ = 14.3° mg ( 2.0 ×10−3 kg ) ( 9.8 N/kg ) −9

tan θ =

26.69. Solve: (a) How many excess electrons on a dust particle produce an electric field of magnitude 1.0 ×

106 N/C a distance of 1.0 μm from the dust particle? (b) The number of electrons is

N=

(1.5 ×10

6

N/C )(1.0 × 10−6 m )

2

(9.0 ×109 N m 2/C2 )(1.60 ×10−19 C )

= 1.04 ×103

26.70. Solve: (a) Two equal charges separated by 1.50 cm exert repulsive forces of 0.020 N on each other. What is the magnitude of the charge? (b) The charge is q=

( 0.020 N )( 0.0150 m ) 9.0 ×109 N m 2 /C 2

2

= 22 nC

The problem does not give the direction of the force. So, the charges could be both positive or both negative.

26.71. Solve: (a) At what distance from a 15 nC charge is the electric field strength 54,000 N/C? (b) The distance is r=

( 9.0 ×10

9

N m 2 /C 2 )(15 ×10−9 C ) 54,000 N/C

= 0.050 m = 5.0 cm

26.72. Solve: (a) A 1.0 nC charge is placed at (0 cm, 2 cm) and another 1.0 nC charge is placed at (0 cm, –2 cm). A third charge +q is placed along a line halfway between q1 and q2 such that the angle between the forces on q due to each of the other two charges is 60°. The resultant force on q is 5.0 × 10−5 N, iˆ . What is the magnitude

(

)

of the charge q? (b)

We have ⎛ ( 9.0 ×109 N m 2 /C 2 )(1.0 ×10−9 C ) q ⎞ G , away from q3 ⎟ = ( 5625 N/C ) q cos30°iˆ − sin 30° ˆj F1 on 3 = ⎜ 2 ⎜ ⎟ ( 0.020 m sin 30°) ⎝ ⎠ G F2 on 3 = ( 5625 N/C ) q cos30°iˆ + sin 30 ˆj

(

(

)

G ⇒ Fon 3 = 2 × ( 5625 N/C ) q cos30°iˆ 2 ( 5625 N/C ) q cos30° = 5.0 ×10−5 N ⇒ q =

( 5.0 ×10

−5

N)

2 ( 5625 N/C ) cos30°

= 5.1 nC

)

26.73. Model: Use the charge model. Solve:

The mass of copper in a 2.0-mm-diameter copper ball is 3⎤ ⎛ 4π 3 ⎞ ⎡ 4π M = ρV = ρ ⎜ r ⎟ = ( 8920 kg/m3 ) ⎢ (1.0 ×10−3 m ) ⎥ = 3.736 × 10−5 kg = 0.03736 g 3 3 ⎝ ⎠ ⎣ ⎦

The number of moles in the ball is n=

M 0.03736 g = = 5.884 ×10−4 mol A 63.5 g/mol

The number of copper atoms in the ball is

N = nN A = ( 5.884 ×10−4 mol )( 6.02 ×1023 mol−1 ) = 3.542 ×1020 We note that the number of electrons per atom is the atomic number, and both the atomic number (29) and the average atomic mass (63.5 g) are taken from the periodic table in the textbook. The number of electrons in the copper ball is thus 29 × 3.542 × 1020 = 1.03 × 1022. The number of electrons removed from the copper ball is 50 ×10−9 C = 3.13 ×1011 1.60 ×10−19 C So, the fraction of electrons removed from the copper ball is 3.13 × 1011 = 3.0 ×10−11 1.03 × 1022 Assess:

This is indeed a very small fraction of the available number of electrons in the copper ball.

26.74. Model: The charged balls are point charges. Visualize:

Because of symmetry and the fact that the three balls have the same charge, the magnitude of the electric force on each ball is the same. The other forces acting on each ball are the gravitational force on the ball and the tension force. Solve: The force on ball 3 is the sum of the force from ball 1 and ball 2. We have

G ⎛ K q1 q3 ⎞ Kq 2 F1 on 3 = ⎜ , away from q ⎟ = 2 sin 30°iˆ + cos30° ˆj 1 2 r ⎝ r13 ⎠

(

G Kq 2 F2 on 3 = 2 − sin 30°iˆ + cos30 ˆj r

(

)

)

G 2 Kq 2 2 Kq 2 ⇒ Fon 3 = 2 cos30° ˆj ⇒ Fon 3 = 2 cos30° = Fon 2 = Fon 1 = Fe r r

⇒ Fe =

2 ( 9.0 ×109 N m 2 /C2 ) q 2 cos30°

( 0.20 m )

2

= ( 3.897 ×1011 q 2 ) N/C 2

The distance l, between one of the balls and the center of the equilateral triangle, is l cos30° =

r 0.10 m = 0.10 m ⇒ l = = 0.1155 m 2 cos30°

Thus, the angle made by the string with the plane containing the three balls is cosθ =

l 0.1155 m = ⇒ θ = 81.70° L 0.80 m

From the free-body diagram, we have T sin θ − mg = 0 N −T cosθ + Fe = 0 N

⇒ tanθ =

⇒ q=

mg = 11 3.897 10 N/C 2 ) tanθ × (

mg mg = Fe ( 3.897 ×1011 q 2 ) N/C2

( 3.0 ×10

−3

kg ) ( 9.8 N/kg )

( 3.897 ×10 N/C2 ) tan81.70° 11

= 1.05 ×10−7 C = 105 nC

26.75. Model: The charged spheres are point charges. Visualize:

The figure shows the free-body diagram of the forces on the sphere with the negative charge that is shown in Figure CP26.75. The force FE is due to the external electric field. The force Fe is the attractive force between the positive and the negative spheres. The tension in the string and the gravitational force are the remaining two forces on the spheres. Solve: The two electrical forces are calculated as follows:

FE = q E = (100 ×10−9 C )(105 N/C ) = 1.00 × 10−2 N −9 9 2 2 K q1 q2 ( 9.0 ×10 N m /C )(100 ×10 C ) 9.0 ×10−5 N m 2 /C Fe = = = 2 2 r r r2 2

From the geometry of Figure CP26.75,

r = 2 ( L sin10° ) = 2 ( 0.50 m ) sin10° = 0.174 m ⇒ Fe =

9.0 ×10−5 N m 2 /C

( 0.174 m )

2

= 3.0 ×10−3 N

From the free-body diagram, T cos10° = mg

T sin10° + Fe = FE

Rearranging and dividing the two equations, sin10° F − Fe = tan10° = E mg cos10° m=

FE − Fe 1.00 ×10−2 N − 0.30 ×10−2 N = = 0.41× 10−2 kg = 4.1 g g tan10° ( 9.8 N/kg ) tan10°

26.76.

Model: Visualize:

The charges are point charges.

Solve: The forces on the –1.0 nC charge lie along the line connecting the pairs of charges. Since the 10 nC G G charge is positive, F10 points as shown. The angle formed by the dashed lines where they meet is 90°, so Fq G G G must point towards q in order for F = F10 + Fq to hold. Therefore F=

F10 cos30°

The distance between the 10 nC and –1.0 nC charges is r = 5.0sin 30° cm = 2.5 cm. Thus

F10 = ( 9.0 ×109 N m 2 /C2 )

(10 ×10 C )(1.0 ×10 ( 2.5 ×10 m ) −9

−2

−9

C)

2

Hence F=

1.44 ×10−4 N = 1.66 ×10−4 N cos30°

= 1.44 ×10−4 N

26.77. Model: Charge Q and the dipole charges (q and −q) are point charges. Visualize: Please refer to Figure CP26.77. Solve: (a) The force on the dipole is the vector sum of the force on q and –q. We have ⎛ KQ q ⎞ KQq , away from Q ⎟ = −iˆ FQ on + q = ⎜ ⎜ ( r + s / 2 )2 ⎟ ( r + s / 2 )2 ⎝ ⎠

( )

FQ on −q =

KQq

( r − s / 2)

2

( +iˆ ) ⇒ F

net

⎛ ⎞ 1 1 = KQq ⎜ − ⎟ iˆ ⎜ ( r − s / 2 )2 ( r + s / 2 )2 ⎟ ⎝ ⎠

(b) The net force Fnet is toward the charge Q, because the attractive force due to Q on the negative charge of the dipole is more than the repulsive force of Q on the positive charge. (c) Using the binomial approximation, we get

( r ± s / 2)

−2

−2

s ⎞ ⎛ ⎛ 2s = r −2 ⎜1 ± ⎟ ≅ r −2 ⎜1 ∓ + ⎝ 2r ⎠ ⎝ 2r

KQq ⎛ ⎛ 2s ⎞ ⎛ 2s ⎞ ⎞ 2 KQqs ⎞ ⎟ ⇒ Fnet = 2 ⎜ ⎜1 + ⎟ − ⎜ 1 − ⎟ ⎟ = r ⎝ ⎝ 2r ⎠ ⎝ 2r ⎠ ⎠ r3 ⎠

(d) Coulomb’s law applies only to the force between two point charges. A dipole is not a point charge, so there’s no reason that the force between a dipole and a point charge should be an inverse-square force. Assess: Note that when s → 0 m, Fnet → 0 N. In this limit the dipole is a point with zero charge.

26-1

27.1.

Model: The electric field is that of the two charges placed on the y-axis. Visualize: Please refer to Figure EX27.1. We denote the upper charge by q1 and the lower charge by q2. Because both the charges are positive, their electric fields at P are directed away from the charges. Solve: The electric field from q1 is 9 2 2 −9 G ⎛ 1 q1 ⎞ ( 9.0 × 10 N m /C )( 3.0 × 10 C ) = cosθ iˆ − sin θ ˆj θ + E1 = ⎜ , below x -axis ⎟ 2 2 2 ( 0.050 m ) + ( 0.050 m ) ⎝ 4πε 0 r1 ⎠

(

)

Because tan θ = 5 cm 5 cm = 1, the angle θ = 45°. Hence, G ⎛ 1 ˆ 1 E1 = ( 5400 N/C ) ⎜ i− 2 ⎝ 2

ˆj ⎞ ⎟ ⎠

Similarly, the electric field from q2 is

G ⎛ 1 q2 ⎞ ⎛ 1 ˆ 1 ˆ⎞ E2 = ⎜ i+ j⎟ , θ above + x -axis ⎟ = ( 5400 N/C ) ⎜ 2 r πε 4 2 ⎠ ⎝ 2 0 2 ⎝ ⎠

G G G ⎛ 1 ⎞ˆ 3ˆ ⇒ Enet at P = E1 + E2 = 2 ( 5400 N/C ) ⎜ ⎟ i = 7.6 ×10 i N/C ⎝ 2⎠ Thus, the strength of the electric field is 7.6×103 N/C and its direction is horizontal. Assess: Because the charges are located symmetrically on either side of the x-axis and are of equal value, the ycomponents of their fields will cancel when added.

27.2. Model: The electric field at the point is found by superposition of the fields due to the two charges located on the y-axis. Visualize: Please refer to Figure EX27.2. The electric field due to the positive charge q1 at the point is away from q1. On the other hand, the electric field due to the negative charge q2 at the point is toward q2. These two electric fields are then added vertically to obtain the net electric field at the point. Solve: The electric field from q1 is 9 2 2 −9 G ⎛ 1 q1 ⎞ ⎛ ( 9.0 × 10 N m /C )( 3.0 × 10 C ) ⎞ ⎜ ⎟ cosθ iˆ − sin θ ˆj = E1 = ⎜ θ x , below + -axis ⎟ 2 2 2 ⎟ ( 0.050 m ) + ( 0.050 m ) ⎝ 4πε 0 r1 ⎠ ⎜⎝ ⎠

(

)

Because tan θ = 5 cm 5 cm, θ = 45°. So, G ⎛ 1 ˆ 1 ˆ⎞ E1 = ( 5400 N/C ) ⎜ + i− j⎟ 2 2 ⎠ ⎝ Similarly, the electric field from q2 is

G ⎛ 1 q2 ⎞ ⎛ 1 ˆ 1 i− E2 = ⎜ , θ below − x -axis ⎟ = ( 5400 N/C ) ⎜ − 2 2 2 ⎝ ⎝ 4πε 0 r2 ⎠

ˆj ⎞ ⎟ ⎠

G G G ⎛ 1 ⎞ˆ 3 ˆ ⇒ Enet = E1 + E2 = 2 ( 5400 N/C ) ⎜ − ⎟ j = −7637 j N/C = −7.6 × 10 N/C 2 ⎝ ⎠ Thus, the strength of the electric field is 7.6×103 N/C and its direction is vertically downward. Assess: A quick visualization of the components of the two electric fields shows that the horizontal components cancel.

27.3. Model: The electric field is that due to superposition of the fields of the two 3.0 nC charges located on the y-axis. Visualize: Please refer to Figure EX27.3. We denote the top 3.0 nC charge by q1 and the bottom 3.0 nC charge G G by q2. The electric fields ( E1 and E2 ) of both the positive charges are directed away from their respective G charges. With vector addition, they yield the net electric field Enet at the point P indicated by the dot. Solve: The electric fields from q1 and q2 are 9 2 2 −9 G ⎛ 1 q1 ⎞ ( 9.0 ×10 N m /C )( 3.0 ×10 C ) ˆ i = 10,800iˆ N/C , along + x -axis ⎟ = E1 = ⎜ 2 2 ( 0.05 m ) ⎝ 4πε 0 r1 ⎠

G ⎛ 1 q2 ⎞ E2 = ⎜ , θ above + x -axis ⎟ 2 ⎝ 4πε 0 r2 ⎠ Because tan θ = 10 cm 5 cm, θ = tan −1 ( 2 ) = 63.43°. So, G ( 9.0 × 109 N m 2 /C2 )( 3.0 × 10−9 C ) cos63.43° iˆ + sin 63.43° ˆj = 966iˆ + 1127 ˆj N/C E2 = 2 2 ( 0.10 m ) + ( 0.050 m )

(

) (

)

The net electric field is thus

G G G Enet at P = E1 + E2 = 11,766iˆ + 1127 ˆj N/C

(

)

To find the angle this net vector makes with the x-axis, we calculate

tan φ =

1127 N/C ⇒ φ = 5.5° 11,766 N/C

Thus, the strength of the electric field at P is Enet =

(11,766 N/C )

2

+ (1127 N/C ) = 11,820 N/C = 1.18×103 N/C 2

G and Enet makes an angle of 5.5° above the +x-axis. Assess: Because of the inverse square dependence on distance, E2 < E1. Additionally, because the point P has no special symmetry relative to the charges, we expected the net field to be at an angle relative to the x-axis.

27.4. Model: The electric field at the point is the superposition of the fields due to the two charges located on the y-axis. Visualize: Please refer to Figure EX27.4. We denote the positive charge by q1 and the negative charge by q2. G G The electric field E1 of the positive charge q1 is directed away from q1, but the field E2 is toward the negative G G charge q2. We will add E1 and E2 vectorially to find the strength and the direction of the net electric field vector. Solve: The electric fields from q1 and q2 are 9 2 2 −9 G ⎛ 1 q1 ⎞ ( 9.0 × 10 N m /C )( 3.0 × 10 C ) ˆ E1 = ⎜ , away from q along + x -axis = i = 10,800iˆ N/C ⎟ 1 2 2 ( 0.050 m ) ⎝ 4πε 0 r1 ⎠

G ⎛ 1 q2 ⎞ , θ below − x-axis ⎟ E2 = ⎜ 2 πε 4 r 0 2 ⎝ ⎠ From the geometry of the figure, tan θ =

10 cm ⇒ θ = 63.43° 5 cm

G ( 9.0 × 109 N m 2 /C 2 )( 3.0 × 10−9 C ) ⇒ E2 = − cos 63.43°iˆ − sin 63.43° ˆj = − 966iˆ + 1127 ˆj N/C 2 2 ( 0.10 m ) + ( 0.050 m )

(

) (

G G G ⇒ Enet = E1 + E2 = 9834iˆ − 1127 ˆj N/C ⇒ Enet =

(

)

( 9834 N/C ) + ( −1127 N/C ) 2

)

2

= 9.9×103 N/C

To find the angle this net vector makes with the horizontal, we calculate

tan φ =

( Enet ) y ( Enet ) x

=

1127 N/C ⇒ φ = 6.5° 9834 N/C

G Thus, the strength of the net electric field at the point is 9.9×103 N/C and Enet makes an angle of 6.5° below the +x-axis.

27.5. Model: The distances to the observation points are large compared to the size of the dipole, so model the field as that of a dipole moment. Visualize:

The dipole consists of charges ±q along the y-axis. The electric field in (a) points down. The field in (b) points up. Solve: (a) The dipole moment is G p = (qs, from – to +) = (1.0 × 10−9 C) ( 0.0020 m ) ˆj = 2.0 × 10−12 ˆj C m

The electric field at (10 cm, 0 cm), which is at distance r = 0.10 m in the plane perpendicular to the electric dipole, is G G 1 p 2.0 × 10−12 ˆj C m E=− = −(9.0 × 109 N m 2 /C 2 ) = −18.0 ˆj N/C 3 4π ε 0 r (0.10 m)3

The field strength, which is all we’re asked for, is 18.0 N/C. (b) The electric field at (0 cm, 10 cm), which is at r = 0.10 m along the axis of the dipole, is G −12 ˆ G 1 2p j C m) 9 2 2 2(2.0 × 10 = × = 36 ˆj N/C E= (9.0 10 N m /C ) 3 4π ε 0 r (0.10 m)3 The field strength at this point is 36 N/C.

27.6. Model: The distances to the observation points are large compared to the size of the dipole, so model the field as that of a dipole moment. Visualize:

The dipole consists of charges ±q along the y-axis. The electric field in (a) points up. The field in (b) points down. Solve: (a) The electric field at (0 cm, 10 cm), which is at r = 0.10 m along the axis of the dipole, is G G 1 2p E = 360 ˆj N/C = 4π ε 0 r 3 G r 3E (0.10 m)3 (360 ˆj N/C ) G ⇒ p= = = (2.0 × 10−11 ˆj C m) 2 (1/4π ε 0 ) 2(9.0 × 109 N m 2 /C 2 )

G By definition, the dipole moment is p = 2.0 × 10−11 ˆj C m = (qs, from – to +) = q (0.010 m) ˆj. Thus q=

2.0 × 10−11 C m = 2.0 ×10−9 C = 2.0 nC 0.010 m

G G (b) Point (10 cm, 0 cm) is in the plane perpendicular to the dipole. The electric field E = –x(1/4πε0) p /r3 is half the strength of the field at an equal distance r on the axis of the dipole. Hence the field strength at this point is 180 N/C.

27.7. Model: We will assume that the wire is thin and that the charge lies on the wire along a line. Solve: From Equation 27.15, the electric field strength of an infinitely long line of charge having linear charge density λ is

Eline = ⇒ Eline ( r = 5.0 cm ) =

2λ 4πε 0 5.0 ×10−2 m 1

1 2λ 4πε 0 r Eline ( r = 10.0 cm ) =

2λ 4πε 0 10.0 ×10−2 m 1

Dividing the above two equations gives

⎛ 5.0 ×10−2 m ⎞ 1 3 Eline ( r = 10.0 cm ) = ⎜ ⎟ Eline ( r = 5.0 cm ) = ( 2000 N/C ) = 1.00 × 10 N/C = 1000 N/C −2 10.0 10 m × 2 ⎝ ⎠

27.8. Model: The rods are thin. Assume that the charge lies along a line. Visualize:

The electric field of the positively charged glass rod points away from the glass rod, whereas the electric field of the negatively charged plastic rod points toward the plastic rod. The electric field strength is the magnitude of the electric field and is always positive. Solve: Example 27.3 shows that the electric field strength in the plane that bisects a charged rod is

Erod =

Q 1 4πε 0 r r 2 + ( L / 2 )2

The electric field from the glass rod at r = 1 cm from the glass rod is Eglass = ( 9.0 ×109 N m 2 /C2 )

10 ×10−9 C

( 0.01 m ) ( 0.01 m ) + ( 0.05 m ) 2

2

= 1.765 ×105 N/C

The electric fields from the glass rod at r = 2 cm and r = 3 cm are 0.835 × 105 N/C and 0.514 × 105 N/C. The electric field from the plastic rod at distances 1 cm, 2 cm, and 3 cm from the plastic rod are the same as for the glass rod. Point P1 is 1.0 cm from the glass rod and is 3.0 cm from the plastic rod, point P2 is 2 cm from both rods, and point P3 is 3 cm from the glass rod and 1 cm from the plastic rod. Because the direction of the electric fields at P1 is the same, the net electric field strength 1 cm from the glass rod is the sum of the fields from the glass rod at 1 cm and the plastic rod at 3 cm. Thus At 1.0 cm

E = 1.765 ×105 N/C + 0.514 ×105 N/C = 2.3 × 105 N/C

At 2.0 cm

E = 0.835 × 105 N/C + 0.835 ×105 N/C = 1.67 × 105 N/C

At 3.0 cm

E = 0.514 × 105 N/C + 1.765 × 105 N/C = 2.3 × 105 N/C

Assess: The electric field strength in the space between the two rods goes through a minimum. This point is exactly in the middle of the line connecting the two rods. Also, note that the arrows shown in the figure are not to scale.

27.9. Model: The rods are thin. Assume that the charge lies along a line. Visualize:

Because both the rods are positively charged, the electric field from each rod points away from the rod. Because the electric fields from the two rods are in opposite directions at P1, P2, and P3, the net field strength at each point is the difference of the field strengths from the two rods. Solve: Example 27.3 gives the electric field strength in the plane that bisects a charged rod:

Erod =

1

Q

4πε 0 r r 2 + ( L / 2 )2

The electric field from the rod on the right at a distance of 1 cm from the rod is

Eright = ( 9.0 ×109 N m 2 /C2 )

10 × 10−9 C

( 0.01 m ) ( 0.01 m ) + ( 0.05 m ) 2

2

= 1.765 ×105 N/C

The electric field from the rod on the right at distances 2 cm and 3 cm from the rod are 0.835 × 105 N/C and 0.514 × 105 N/C. The electric fields produced by the rod on the left at the same distances are the same. Point P1 is 1.0 cm from the rod on the left and is 3.0 cm from the rod on the right. Because the electric fields at P1 have opposite directions, the net electric field strengths are At 1.0 cm

E = 1.765 ×105 N/C − 0.514 ×105 N/C = 1.25 × 105 N/C

At 2.0 cm

E = 0.835 × 105 N/C − 0.835 × 105 N/C = 0 N/C

At 3.0 cm

E = 1.765 × 105 N/C − 0.514 × 105 N/C = 1.25 × 105 N/C

27.10. Model: The rod is thin, so assume the charge lies along a line. Visualize:

G G Solve: The force on charge q is F = qErod . From Example 27.3, the electric field a distance r from the center of a charged rod is G ( 9.0 ×109 N m 2/C2 )( 40 ×10−9 C ) iˆ = 1.406 ×105 iˆ N/C 1 Q Erod = iˆ = 2 5 4πε 0 r r 2 + ( L / 2 )2 ( 0.04 m ) ( 0.04 m ) + ( 0.05 m ) Thus, the force is

G F = ( 6.0 ×10−9 C )(1.406 × 105 N/C ) iˆ = 8.4 × 10−4 iˆ N G More generally, F = ( 8.4 ×10−4 N, away from the rod ) .

27.11. Model: Assume that the rings are thin and that the charge lies along circle of radius R. Visualize:

The rings are centered on the z-axis. Solve: (a) According to Example 27.5, the field of the left (negative) ring at z = 10 cm is

( E1 ) z =

zQ1

4πε 0 ( z + R 2

)

2 3/ 2

=

( 9.0 ×10

9

N m 2 /C 2 ) ( 0.10 m ) ( −20 × 10−9 C ) 2 3/ 2

⎡( 0.10 m ) + ( 0.050 m ) ⎤ ⎣ ⎦ 2

= −1.288 × 104 N/C

G That is, the field is E1 = (1.288 × 104 N/C, left ) . Ring 2 has the same quantity of charge and is at the same distance, G so it will produce a field of the same strength. Because Q2 is positive, E2 will also point to the left. The net field at the midpoint is G G G E = E1 + E2 = ( 2.6 × 104 N/C, left ) (b) The force is

G G F = qE = ( −1.0 × 10−9 C )( 2.6 × 104 N/C, left ) = ( 2.6 × 10−5 N, right )

27.12. Model: Assume that the rings are thin and that the charge lies along circle of radius R. Visualize:

Solve: cm is

(a) Let the rings be centered on the z-axis. According to Example 27.5, the field of the left ring at z = 10

( E1 ) z =

zQ1

4πε 0 ( z + R 2

)

2 3/ 2

=

( 9.0 ×10

9

N m 2 /C 2 ) ( 0.10 m ) ( 20 ×10−9 C ) 2 3/ 2

⎡( 0.10 m ) + ( 0.050 m ) ⎤ ⎣ ⎦ 2

= 1.29 ×104 N/C

G That is, E1 = (1.29 × 104 N/C, right ) . Ring 2 has the same quantity of charge and is at the same distance, so it will G produce a field of the same strength. Because Q2 is positive, E2 will point to the left. The net field at the G G G midpoint between the two rings is E = E1 + E2 = 0 N/C. (b) The field of the left ring at z = 0 cm is ( E1 ) z = 0 N/C. The field of the right ring at z = 20 cm to its left is

( E2 ) z =

( 9.0 ×10

9

N m 2 /C2 ) ( 0.20 m ) ( 20 ×10−9 C ) 2 32

= 4.1×103 N/C

⎡( 0.20 m ) + ( 0.050 ) ⎤ ⎣ ⎦ G G G ⇒ E = E1 + E2 = 0 N/C + (4.1×103 N/C, left)

So the electric field strength is 4.1×103 N/C.

2

27.13. Model: Each disk is a uniformly charged disk. When the disk is charged negatively, the on-axis electric field of the disk points toward the disk. The electric field points away from the disk for a positively charged disk. Visualize:

Solve:

(a) The surface charge density on the disk is Q Q 50 ×10−9 C η = = = = 6.366 × 10−6 C/m 2 2 2 A πR π ( 0.050 m )

From Equation 27.23, the electric field of the left disk at z = 0.10 m is

η ⎡ 1 −6.366 × 10−6 C/m 2 ⎡ 1 ⎤ ⎤ 1− = 1− = −38,000 N/C ⎢ ⎥ −12 2 2 ⎢ 2 2 2 ⎥ 2ε 0 ⎣ 2 8.85 × 10 C /N m ( ) ⎣⎢ 1 + ( 0.050 m /0.10 m ) ⎦⎥ 1 + R /z ⎦ G In other words, E1 = (38,000 N/C, left). Similarly, the electric field of the right disk at z = 0.10 m (to its left) is G G G G E2 = (38,000 N/C, left). The net field at the midpoint between the two rings is E = E1 + E2 = (7.6×104 N/C, left).

( E1 ) z =

(b) The force on the charge is G G F = qE = ( −1.0 ×10−9 C )( 7.6 ×104 N/C, left ) = ( 7.6 × 10−5 N, right ) Assess:

Note that the force on the negative charge is to the right because the electric field is to the left.

27.14. Model: Model each disk as a uniformly charged disk. When the disk is positively charged, the onaxis electric field of the disk points away from the disk. Visualize:

Solve:

(a) The surface charge density on the disk is

η=

50 ×10−9 C Q Q = = = 6.366 ×10−6 C/m 2 A π R 2 π ( 0.050 m )2

From Equation 27.23, the electric field of the left disk at z = 0.10 m is

( E1 ) z =

η ⎡ 1 6.366 × 10−6 C/m 2 1 ⎤ ⎡ ⎤ 1 − = 1− = 38,000 N/C −12 2 2 ⎢ 2 2 ⎥ 2 ⎥ 2ε 0 ⎢ 2 8.85 10 C /N m × ( ) R z 1 + 1 + ( 0.050 m /0.10 m ) ⎥⎦ ⎣ ⎦ ⎢⎣

G Hence, E1 = (38,000 N/C, right) . Similarly, the electric field of the right disk at z = 0.10 m (to its left) is G G G G E2 = (38,000 N/C, left) . The net field at the midpoint between the two disks is E = E1 + E2 = 0 N/C . (b) The electric field of the left disk at z = 0.050 m is G 6.366 × 10−6 C/m 2 1 ⎡ ⎤ 1− = 1.05 × 105 N/C ⇒ E1 = (1.05 × 105 N/C, right ) ⎥ 2 2 ⎢ −12 2 2 ( 8.85 × 10 C /N m ) ⎢ 1 + ( 0.050 m /0.10 m ) ⎥⎦ ⎣ G Similarly, the electric field of the right disk at z = 0.15 m (to its left) is E2 = (1.85 ×104 N/C, left ) . The net field

( E1 ) z =

is thus The field strength is 8.7 ×104 N/C.

G G G E = E1 + E2 = ( 8.7 × 104 N/C, right )

27.15. Model: The distance 2.0 mm is very small in comparison to the size of the electrode, so we can model the electrode as a plane of charge. Solve: From Equation 27.26, the electric field of a plane of charge is Eplane =

η 80 × 10−9 C Q = = = 1.13 × 105 N/C 2ε 0 2ε 0 A 2 ( 8.85 × 10−12 C 2 /N m 2 ) ( 0.20 × 0.20 m )

27.16. Model: A spherical shell of charge Q and radius R has an electric field outside the sphere that is exactly the same as that of a point charge Q located at the center of the sphere. Visualize: In the case of a metal ball, the charge resides on its surface. This can then be visualized as a charged spherical shell of radius R. Solve: Equation 27.28 gives the electric field of a charged spherical shell at a distance r > R: G Q Eball = rˆ 4πε 0 r 2 In the present case, Eball = 50,000 N/C at r = Q = 4πε 0 r 2 Eball =

1 2

(10 cm ) + 2.0 cm = 7.0 cm = 0.070 m.

( 0.070 m )

2

So,

50,000 N/C = 2.7 × 10−8 C = 27 nC 9.0 ×10 N m 2 /C 2 9

27.17. Model: The electric field is uniform in a region of space between closely spaced capacitor plates. Solve: The electric field inside a capacitor is E = Q ε 0 A . Thus, the charge needed to produce a field of strength E is Q = ε 0 AE = ( 8.85 ×10−12 C 2 /N m 2 ) ( 0.04 m × 0.04 m ) (1.0 ×106 N/C ) = 14.2 nC Thus, one plate has a charge of 14.2 nC and the other has a charge of −14.2 nC. Assess: Note that the capacitor as a whole has no net charge.

27.18. Model: The electric field in a region of space between two charged circular disks is uniform. Solve: The electric field strength inside the capacitor is E = Q ε 0 A . Thus, the area is A=

( 3.0 ×109 )(1.6 ×10−19 C ) Q D2 = = 2.71× 10−4 m 2 = π −12 2 2 5 ε 0 E ( 8.85 ×10 C /N m )( 2.0 ×10 N/C ) 4 ⇒D=

4A

π

= 1.86 cm

Assess: As long as the spacing is much less than the plate dimensions, the electric field is independent of the spacing and depends only on the diameter of the plates.

27.19. Model: The electric field in a region of space between the plates of a parallel-plate capacitor is uniform. Solve: The electric field inside a capacitor is E = Q ε 0 A. Thus, the charge needed to produce a field of strength E is 2 Q = ε 0 AE = ( 8.85 ×10−12 C 2 /N m 2 ) ⎡π ( 0.020 m ) ⎤ ( 3.0 ×106 N/C ) = 33.4 nC ⎣ ⎦

The number of electrons transferred from one plate to the other is 33.4 ×10−9 C = 2.1×1011 1.60 ×10−19 C

27.20. Model: The charged plastic bead is a point charge. Visualize:

The bead hangs suspended in the air when the net force acting on the bead is zero. The two forces that act on the bead are the electric force and the gravitational force. Because the bead is negatively charged, the electric field must be pointed downward to cause an upward force, which will balance the gravitational force. Solve: For the bead to be in static equilibrium,

( Fnet ) y = qE − mg = 0 N ⇒ E =

−3 mg ( 0.10 ×10 kg ) ( 9.8 N/kg ) = = 6.1× 105 N/C q (1.0 ×1010 )(1.60 ×10−19 C )

G Thus the required field is E = ( 6.1×105 N/C, down ).

27.21. Model: The disks form a parallel-plate capacitor. The electric field inside a parallel-plate capacitor is a uniform field, so the proton will have a constant acceleration. Visualize:

Solve:

(a) The two disks form a parallel-plate capacitor with surface charge density

η=

Q Q 10 × 10−9 C = = = 3.18 × 10−5 C/m 2 2 A πR π (0.010 m)2

From Equation 27.29, the field strength inside a capacitor is

E=

η 3.18 ×10−5 C/m 2 = = 3.6 ×106 N/C ε 0 8.85 ×10−12 C2 / N m 2

(b) G The electric field points toward the Gnegative Gplate,G so in the coordinate system of the figure E = −3.6 × 106 ˆj N/C. The field exerts forces F = qproton E = eE on the proton, causing an acceleration with a y-

component that is ay =

F eE y (1.6 × 10−19 C)( −3.6 × 106 N/C) = = = −3.45 × 1014 m/s 2 1.67 × 10−27 kg m m

After the proton is launched, this acceleration will cause it to lose speed. To just barely reach the positive plate, it should reach v1 = 0 m/s at y1 = 1 mm. The kinematic equation of motion is v12 = 0 m 2 / s 2 = v02 + 2a y Δy

⇒ v0 = −2a y Δy = −2(−3.45 × 1014 m/s 2 )(0.0010 m) = 8.3 × 105 m/s

Assess: The acceleration of the proton in the electric field is enormous in comparison to the gravitation acceleration g. That is why we did not explicitly consider g in our calculations.

27.22. Model: A uniform electric field causes a charge to undergo constant acceleration. Solve:

Kinematics yields the acceleration of the electron.

v12 = v0 2 + 2aΔx ⇒ a =

v12 − v0 2 (4.0 × 107 m/s) 2 − (2.0 × 107 m/s) 2 = = 5.0 × 1016 m/s 2 2Δx 2 ( 0.012 m )

The magnitude of the electric field required to obtain this acceleration is

E=

Fnet me a (9.11 × 10−31 kg)(5.0 × 1016 m/s 2 ) = = = 2.8 × 105 N/C. e e 1.6 × 10−19 C

27.23. Model: The infinite charged plane produces a uniform electric field. Solve: (a) The electric field of a plane of charge with surface charge density η is E=

η 2ε 0

⎛F⎞ ⎛ ma ⎞ ⇒ η = 2ε 0 E = 2ε 0 ⎜ ⎟ = 2ε 0 ⎜ ⎟ q ⎝ ⎠ ⎝ q ⎠

where m is the mass, q is the charge, and a is the acceleration of the electron. To obtain η we must first find a. From the kinematic of motion equation v12 = v02 + 2aΔx, a=

⇒η =

v12 − v02 (1.0 × 107 m/s) 2 − (0 m/s) 2 = = 2.5 × 1015 m/s 2 2Δx 2(2.0 × 10−2 m)

2(8.85 × 10−12 C 2 /N m 2 )(9.11 × 10−31 kg)(2.5 × 1015 m/s 2 ) = 2.5 × 10−7 C/m 2 1.60 × 10−19 C

(b) Using the kinematic equation of motion v1 = v0 + aΔt ,

Δt =

v1 − v0 1.0 ×107 m/s − 0 m/s = = 4.0 ×10−9 s a 2.5 ×1015 m/s 2

27.24. Model: The infinite negatively charged plane produces a uniform electric field that is directed toward the plane. Visualize:

Solve:

From the kinematic equation of motion v12 = 0 = v02 + 2aΔx and F = qE = ma, a=

qE −v02 − mv02 = ⇒ Δx = 2qE m 2Δx

Furthermore, the electric field of a plane of charge with surface charge density η is E = η 2ε 0 . Thus, −27 −12 6 2 2 −mv02ε 0 − (1.67 ×10 kg )( 2.0 ×10 m/s ) ( 8.85 ×10 C /N m ) Δx = = = 0.185 m qη (1.60 ×10−19 C )( −2.0 ×10−6 C/m2 ) 2

27.25. Model: The external electric field exerts a torque on the dipole moment of the water molecule. Solve:

From Equation 27.34, the torque exerted on a dipole moment by an electric field is τ = pE sin θ . The

maximum torque is exerted when sinθ = 1 or θ = 90°. Thus,

τ max = pE = ( 6.2 ×10−30 C m )( 5.0 ×108 N/C ) = 3.1×10−21 N m

27.26. Model: Because r >> s, we can use Equation 27.12 for the electric field in the plane perpendicular to the dipole. Visualize:

Solve: (a) From Newton’s third law, the force of Q on the dipole is equal and opposite to the force of the G G G dipole on Q. You can see from the diagram that Fdipole on Q is down and FQ on dipole is up, in the direction of p. The magnitude of the dipole field at the position of Q is

Edipole = G The magnitude of Fdipole on Q is

1 p 1 qs = 4πε 0 r 3 4πε 0 r 3

Fdipole on Q = QEdipole = G FQ on dipole has the same magnitude, so

1 qQs 4πε 0 r 3

G ⎛ 1 qQs →⎞ FQ on dipole = ⎜ , direction of p ⎟ 3 πε 4 r 0 ⎝ ⎠ G G (b) The electric field of charge Q exerts a torque on the dipole. The field E is perpendicular to the dipole p, so

θ = 90°. The torque is given by Equation 27.34: ⎛ 1 Q⎞ 1 qQs sin 90° = 2 ⎟ 4πε 0 r 2 ⎝ 4πε 0 r ⎠

τ = pE sin θ = (qs ) ⎜

27.27. Model: The size of a molecule is ≈ 0.1 nm. The proton is 2.0 nm away, so r >> s and we can use Equation 27.12 for the electric field in the plane that bisects the dipole. Visualize:

Solve:

G G You can see from the diagram that Fdipole on proton is opposite to the direction of p. The magnitude of the

dipole field at the position of the proton is Edipole =

p 5.0 ×10−30 C m = (9.0 × 109 N m 2 /C2 ) = 5.624 × 105 N/C 3 4πε 0 r (2.0 ×10−9 m)3 1

G The magnitude of Fdipole on proton is

Fdipole on proton = eEdipole = (1.60 × 10−19 C)(5.624 × 105 N/C) = 9.0 × 10−13 N G G –13 Including the direction, the force is Fdipole on proton = (9.0 × 10 N, direction opposite p ).

27.28. Model: The electric field is that of three point charges q1, q2, and q3. Visualize: Please refer to Figure P27.28. Assume the charges are in the x-y plane. The 5.0 nC charge is q1, the G G G G 10 nC charge is q3, and the −5.0 nC charge is q2. The net electric field at the dot is Enet = E1 + E2 + E3 . The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components. Solve: (a) The electric field produced by q1 is 9 2 2 −9 q1 ( 9.0 ×10 N m /C )( 5.0 ×10 C ) = = 112,500 N/C 2 4πε 0 r12 ( 0.020 m ) G G E1 points away from q1, so in component form E1 = 112,500iˆ N/C. The electric field produced by q2 is E2 = G G 28,120 N/C. E2 points toward q2, so E2 = 28,120ˆj N/C . Finally, the electric field produced by q3 is

E1 =

E3 =

1

9 2 2 −9 q3 ( 9.0 ×10 N m /C )(10 ×10 C ) = = 45,000 N/C 2 2 4πε 0 r32 ( 0.020 m ) + ( 0.040 m )

1

G E3 points away from q3 and makes an angle φ = tan −1 ( 4 / 2 ) = 63.43° with the x-axis. So, G E3 = E3 cos φ iˆ − E3 sin φ ˆj = 20,130iˆ − 40,250ˆj N/C

(

)

Adding these three vectors gives G G G G Enet = E1 + E2 + E3 = 132,600iˆ − 12,130ˆj N/C = 1.33 × 105iˆ − 1.21×104 ˆj N/C

(

)

(

)

This is in component form. (b) The magnitude of the field is

Enet = Ex2 + E y2 =

(132,600 N/C ) + ( −12,130 N/C ) 2

(

and its angle from the x-axis is θ = tan −1 Ex E y + x-axis).

)

2

= 133,200 N/C = 1.33 × 105 N/C G = 5.2°. We can also write Enet = (1.33 ×105 N/C, 5.2° CW from the

27.29. Model: The electric field is that of three point charges q1, q2, and q3. Visualize: Please refer to Figure P27.29. Assume the charges are in the x-y plane. The −5.0 nC charge is q1, the G G G G bottom 10 nC charge is q3, and the top 10 nC charge is q2. The net electric field at the dot is Enet = E1 + E2 + E3 . The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components. Solve: (a) The electric field produced by q1 is 9 2 2 −9 q1 ( 9.0 ×10 N m /C )( 5.0 ×10 C ) = = 112,500 N/C 2 4πε 0 r12 ( 0.020 m ) G G E1 points toward q1, so in component form E1 = −112,500 ˆj N/C. The electric field produced by q2 G G is E2 = 56,250 N/C. E2 points away from q2, so E2 = −56,250iˆ N/C. Finally, the electric field produced by q3 is

1

E1 =

E3 =

9 2 2 −9 q3 ( 9.0 ×10 N m /C )(10 ×10 C ) = = 45,000 N/C 2 2 4πε 0 r32 ( 0.020 m ) + ( 0.040 m )

1

G E3 points away from q3 and makes an angle φ = tan −1 ( 2 / 4 ) = 26.6° with the x-axis. So, G E3 = − E3 cos φ iˆ + E3 sin φ ˆj = −40,250iˆ + 20,130 ˆj N/C

(

)

Adding these three vectors gives G G G G Enet = E1 + E2 + E3 = −96,500iˆ − 92,400 ˆj N/C = −9.7 × 104 iˆ − 9.2 × 104 ˆj N/C

(

)

(

)

This is in component form. (b) The magnitude of the field is Enet = Ex2 + E y2 =

( 96,500 N/C ) + ( 92,400 N/C ) 2

(

and its angle below the –x-axis is θ = tan −1 Ex E y from the + x -axis ) .

)

2

= 133,600 N/C = 1.34 × 105 N/C G = 44°. We can also write Enet = (1.34 ×105 N/C, 136° CW

27.30. Model: The electric field is that of three point charges q1, q2, and q3. Visualize: Please refer to Figure P27.30. Assume the charges are in the x-y plane. The 10 nC charge is q1, the G G G G −10 nC charge is q3, and the –5.0 nC charge is q2. The net electric field at the dot is Enet = E1 + E2 + E3 . The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components. Solve: (a) The electric field produced by q1 is 9 2 2 −9 q1 ( 9.0 ×10 N m /C )(10 ×10 C ) = = 100,000 N/C 2 4πε 0 r12 ( 0.030 m ) G G E1 points away from q1, so in component form E1 = −100,000 ˆj N/C . The electric field produced by q2 is G G E2 = 18,000 N/C . E2 points towards q2, so E2 = 18,000iˆ N/C. Finally, the electric field produced by q3 is

E1 =

E3 =

1

9 2 2 −9 q3 ( 9.0 ×10 N m /C )(10 ×10 C ) = = 26,471 N/C 2 2 2 4πε 0 r3 ( 0.030 m ) + ( 0.050 m )

1

G E3 points toward q3 and makes an angle φ = tan −1 ( 3/ 5 ) = 31° with the x-axis. So, G E3 = E3 cos φ iˆ + E3 sin φ ˆj = 22,700iˆ + 13,634 ˆj N/C

(

)

Adding these three vectors gives G G G G Enet = E1 + E2 + E3 = 40,700iˆ − 86,366 ˆj N/C = 4.1× 104 iˆ − 8.6 × 104 ˆj N/C

(

)

(

)

This is in component form. (b) The magnitude of the field is

Enet = Ex2 + E y2 =

( 40,700 N/C ) + ( 86,366 N/C ) 2

(

and its angle from the x-axis is θ = tan −1 Ex E y the + x-axis).

)

2

= 95,476 N/C = 9.5 × 104 N/C G = 25° . We can also write Enet = (9.5 × 104 N/C, 25° CW from

27.31. Model: The electric field is that of three point charges q1 = −Q, q2 = −Q, and q3 G= +4Q.G G G Visualize: Assume the charges are in the x-y plane. The net electric field at point P is Enet = E1 + E2 + E3 . The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components.

Solve:

(a) The electric field produced by q1 points toward q1 and is given by

G ⎛ 1 Q ⎞ˆ E1 = − ⎜ i 2 ⎟ ⎝ 4πε 0 L ⎠ The electric field produced by q2 points toward q2 and is given by G ⎛ 1 Q⎞ˆ E2 = − ⎜ j. 2 ⎟ ⎝ 4πε 0 L ⎠ The electric field produced by q3 is

E3 =

1 ⎛ 4Q ⎞ 1 ⎛ 2Q ⎞ ⎜ 2 2⎟= ⎜ ⎟ 4πε 0 ⎝ L + L ⎠ 4πε 0 ⎝ L2 ⎠

G E3 points away from q3 and makes an angle φ = tan −1 ( L L ) = 45° with the x-axis. So G 1 ⎛ 2Q ⎞ 1 2Q ⎛ 1 ˆ E3 = i + ⎜ ⎟ cos φ iˆ + sin φ ˆj = 4πε 0 ⎝ L2 ⎠ 4πε 0 L2 ⎜⎝ 2

(

)

1 2

ˆj ⎞ ⎟ ⎠

Adding these three vectors gives

G G G G 1 Q ⎡⎛ 2 1 Q ˆ ˆ ⎞ ⎛ 2 ⎞ ⎤ − 1⎟ iˆ + ⎜ − 1⎟ ˆj ⎥ = Enet = E1 + E2 + E3 = i + j 2 ⎢⎜ 4πε 0 L ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦ 4πε 0 L2

(

(b) The force on the charge is F = ma = qEnet . Therefore,

a=

q 1 Q m 4πε 0

(

)

2 −1 2

L

2

=

1 ⎛ 0.586 qQ ⎞ ⎜ ⎟ 4πε 0 ⎝ mL2 ⎠

)(

)

2 −1

27.32. Model: Visualize:

Solve:

The electric field is that of two charges −q and +2q located at x = ±a.

(a) At point 1, the electric field from −q is

−q

1

E− q =

=

1

q 4πε 0 5a 2

4πε 0 ( a ) + ( 2a ) G E− q points toward −q and makes an angle φ1 = tan −1 ( 2a a ) = 63.43° below the −x-axis, hence 2

2

G 1 ⎛ q ⎞ 1 ⎛ q ⎞ ⎛ 1 ˆ 2 ˆ⎞ 1 ⎛ q ⎞ ˆ ˆ E− q = i− j⎟ = ⎜ ⎟ − cos φ1iˆ − sin φ1 ˆj = ⎜ ⎟ − ⎜ ⎟ − i −2j 4πε 0 ⎝ 5a 2 ⎠ 4πε 0 ⎝ 5a 2 ⎠ ⎜⎝ 5 5 ⎠ 4πε 0 ⎝ 5 5a 2 ⎠

(

)

(

The electric field from the +2q is

E+2 q =

1 2q 1 2q = 4πε 0 a 2 + ( 2a )2 4πε 0 5a 2

G E+2q points away from +2q and makes an angle φ1 = tan −1 ( 2a a ) = 63.43° above the −x-axis. So, G 1 ⎛ 2q ⎞ 1 ⎛ q ⎞ E+2 q = −2iˆ + 4ˆj ⎜ ⎟ − cos φ2iˆ + sin φ2 ˆj = 4πε 0 ⎝ 5a 2 ⎠ 4πε 0 ⎜⎝ 5 5a 2 ⎟⎠

(

)

(

Adding these two vectors, G G G 1 ⎛ q ⎞ E1 net = E− q + E+2 q = −3iˆ + 2ˆj 4πε 0 ⎜⎝ 5 5a 2 ⎟⎠

(

At point 2, the electric field from −q points toward −q, so

G 1 ⎛ q ⎞ˆ E− q = ⎜ ⎟i 4πε 0 ⎝ a 2 ⎠ The electric field from +2q points away from +2q, so G 1 ⎛ 2q ⎞ ˆ E+2 q = − ⎜ ⎟i 4πε 0 ⎝ 9a 2 ⎠ Adding these two vectors,

)

)

)

G G G 1 ⎛ 7q ⎞ ˆ E2 net = E− q + E+2 q = ⎜ ⎟i 4πε 0 ⎝ 9a 2 ⎠ At point 3, the electric field from −q points toward −q, so G 1 ⎛ q ⎞ˆ E− q = − ⎜ ⎟i 4πε 0 ⎝ 9a 2 ⎠ The electric field from +2q points away from +2q, so G 1 ⎛ 2q ⎞ ˆ E+2 q = ⎜ ⎟i 4πε 0 ⎝ a 2 ⎠ Adding these vectors, G G G 1 ⎛ 17q ⎞ ˆ E3 net = E− q + E+2 q = ⎜ ⎟i 4πε 0 ⎝ 9a 2 ⎠ G G Point 4 is a mirror image of point 1. Since E1 net points to the left and up, E4 net has a reversed y-component and points to the left and down. Thus, G 1 ⎛ q ⎞ E4 net = −3iˆ − 2ˆj 4πε 0 ⎜⎝ 5 5a 2 ⎟⎠

(

(b) The four electric field vectors at points 1-4 are shown in the figure.

)

27.33. Model: The electric field is that of two positive charges. Visualize:

G G G G G The figure shows E1 and E2 due to the individual charges. The total field is E = E1 + E2 . Solve: (a) From symmetry, the y-components of the two electric fields cancel out. The x-components are equal and add. Thus, G E = 2 ( E1 ) x iˆ = 2E1 cosθ iˆ The field strength and angle due to q1 are

E1 =

q q = 4πε 0 r 2 4πε 0 ( x 2 + s 2 /4 )

G Thus the magnitude of the net field is E =

cosθ =

2qx

4πε 0 ( x 2 + s 2 /4 )

3/ 2

x = r

x x + s 2 /4 2

.

(b) The electric field strength at position x is E=

( 9.0 ×10

9

N m 2 /C 2 ) ( 2 ) (1.0 × 10−9 C ) x

⎡ x 2 + ( 0.003 m )2 ⎤ ⎣ ⎦

3/ 2

=

18 x ⎡ x 2 + ( 0.003 m )2 ⎤ ⎣ ⎦

3/ 2

N/C

where x has to be in meters. We can now evaluate E for different values of x: x (mm) 0 2 4 6 10

x (m) 0.000 0.002 0.004 0.006 0.010

E (N/C) 0 768,000 576,000 358,000 158,000

(c) We can use the values of part (b) as a start in drawing the graph. Also note that E → 0 N/C as x → 0 m and as x → ∞. The graph is shown in the figure above.

27.34. Model: The electric field of a dipole is that of two opposite charges ±q that make the dipole. Visualize: Please refer to Figure 27.7. The figure shows a dipole aligned on the y-axis, so the x-axis is the G G G bisecting axis. The field at a point on the x-axis is Edipole = E+ + E− . Solve: From the symmetry of the situation we can see that the x-components of the two contributions to the G electric field will cancel, that they have equal y-components, and that Edipole points in the −y-direction. Thus,

(E

)

dipole x

(E

= 0 N/C

)

dipole y

= −2 E+ sin θ

G where θ is the angle E+ makes with the x-axis. From the geometry of the figure, sin θ =

s/2 1/ 2

⎡ x + ( s / 2 )2 ⎤ ⎣ ⎦ 2

For a point charge +q, the field is

E+ =

1 q 4πε 0 x 2 + ( s / 2 )2

Combining these pieces gives the dipole field at distance x along the bisecting axis: G 1 q s/2 qs ˆj = − ˆj Edipole = − ( 2 ) 3/ 2 2 4πε 0 x 2 + ( s / 2 )2 ⎡ x 2 + ( s / 2 )2 ⎤1/ 2 4πε 0 ( x + s 2 / 4 ) ⎣ ⎦

If x >> s, then ( x 2 + s 2 / 4 )

3/ 2

≈ x3 . Thus G 1 qsjˆ Edipole ≈ − 4πε 0 x 3

G If we note that p = qsjˆ and if we replace x with a more general variable r to denote the distance from the dipole, then G G 1 p Edipole ≈ − 4πε 0 r 3 This is Equation 27.12.

27.35. Model: The electric field is that of three point charges. Visualize:

G G G G The field at points on the x-axis is Enet = E1 + E2 + E3 .

G G G Solve: (a) We note from the symmetry that the y-component of E1 and E3 cancel. Since E2 has no yG G component, the net field will have only an x-component. The x-components of E1 and E3 are equal, so

( Enet ) x = E2 − 2E1 cosθ

( Enet ) y = ( Enet ) z = 0 N/C

Note that the signs of q1 and q3 were used in writing this equation. The electric field strength due to q2 is E2 =

1 q2 1 2q = 4πε 0 r22 4πε 0 x 2

The electric field strength due to q1 is E1 =

1 q1 1 q = 4πε 0 r12 4πε 0 x 2 + d 2

From geometry, cosθ =

x = r1

x x2 + d 2

Assembling these pieces, the net field is

( Enet ) x =

1 2q 1 2q − 4πε 0 x 2 4πε 0 x 2 + d 2

x x2 + d 2

=

2q ⎡ 1 x ⎤ − 4πε 0 ⎢ x 2 ( x 2 + d 2 )3 / 2 ⎥ ⎢⎣ ⎥⎦

G 2q ⎡ 1 x ⎤ˆ ⇒ Enet = − i 4πε 0 ⎢ x 2 ( x 2 + d 2 )3 / 2 ⎥ ⎢⎣ ⎥⎦

G G (b) For x << d, the observation point is very close to q2 = +2q. Furthermore, at x ≈ 0 m the fields E1 and E3 are nearly opposite to each other and will nearly cancel. So for x << d we expect the field to be that of a point charge +2q at the origin. To test this prediction, we note that for x << d

(x

x 2

+d

)

2 3/ 2

≈

G x 1 2q ˆ ≈ 0 ⇒ Enet = i 3 d 4πε 0 x 2

This is, indeed, the field on the x-axis of point charge 2q at the origin. For x >> d, the three charges appear as a G single charge of value qnet = q1 + q2 + q3 = 0. So we expect E ≈ 0 when x >> d. In this limit,

1 x 1 x 1 1 − ≈ − = − =0 x 2 ( x 2 + d 2 )3/ 2 x 2 x 3 x 2 x 2

so the field does rapidly become zero, as expected. (c) A graph of Ex is shown in the figure above.

27.36. Model: The electric field is that of two infinite lines of charge extending out of the page. Visualize: Please refer to Figure P27.36. The line charges lie on the x-axis. Solve: (a) From Equation 27.15, the electric field strength due to an infinite line of charge at a distance r from the line charge is E=

1 2λ 4πε 0 r

For the left and right line charges, the electric fields are Eleft = Eright =

1

2λ

4πε 0

y + ( d / 2) 2

2

=

1

4λ

4πε 0

4 y2 + d 2

G G Eleft makes an angle φ above the +x-axis and Eright makes the same angle φ above the −x-axis. From the geometry of the figure,

cos φ = G ⇒ Eleft =

d /2

=

y +d /4 2

2

1 4πε 0

d 4y + d 2

2

4λ

sin φ =

d ⎛ iˆ + ⎜ 2 4y + d ⎝ 4y + d2 2

2

2y 4 y2 + d 2

ˆj ⎞ ⎟ 4y + d ⎠ 2y 2

2

G 1 4λ d 2y ⎛ ˆj ⎞ − Eright = iˆ + ⎟ 2 2 2 2 4πε 0 4 y 2 + d 2 ⎜⎝ 4y + d 4y + d ⎠ G G G We now add these two vectors to find Enet = Eleft + Eright . The x-components cancel to give

G 1 ⎛ 4λ 2y ⎞ ⎛ Enet = ( 2) ⎜ ⎜ ⎟ 2 2 4πε 0 ⎝ 4 y + d ⎠ ⎝ 4 y 2 + d 2

1 16λ y ˆ ⎞ˆ j ⎟ j = 4πε 2 2 0 (4y + d ) ⎠

Thus the field strength is

E=

1 16λ y 4πε 0 4 y 2 + d 2

(b) To draw a graph of Enet versus y, we calculated Enet at a few selected values of y.

y (in units of d)

0 ± 0.5 ±1 ±2 ±3 ±4

⎛ 1 λ⎞ Enet ⎜ in units of ⎟ 4πε 0 d ⎠ ⎝ 0 ±4 ± 3.2 ± 1.88 ± 1.30 ± 0.98

27.37. Model: The electric field is that of two infinite lines of charge extending out of the page. Visualize: Please refer to Figure P27.37. The line charges lie on the x-axis. Solve: (a) From Equation 27.15, the electric field strength due to an infinite line of charge at a distance r from the line charge is 1 2λ 4πε 0 r

E=

For the left and right line charges, the electric fields are

Eleft = Eright =

cos φ =

1

2λ

4πε 0

y + ( d / 2)

d /2 y + d /4 2

2

G 1 ⇒ Eleft = 4πε 0 G 1 Eright = 4πε 0

=

2

2

=

d 4y + d 2

2

1

4λ

4πε 0

4 y2 + d 2

4λ d ⎛ iˆ + ⎜ 4 y2 + d 2 ⎝ 4 y2 + d 2 4λ

d ⎛ iˆ − ⎜ 2 4y + d ⎝ 4y + d2 2

2

2y

sin φ =

4 y2 + d 2

2y ˆj ⎞ ⎟ 4 y2 + d 2 ⎠ ˆj ⎞ ⎟ 4y + d ⎠

G G G 1 ⎛ 4λ d ⎞ ⎛ ⇒ Enet = Eleft + Eright = ( 2) 4πε 0 ⎜⎝ 4 y 2 + d 2 ⎟⎠ ⎜⎝ 4 y 2 + d 2

2y 2

2

1 8λ d ⎞ˆ iˆ ⎟ i = 4πε 2 + d2) y 4 ( 0 ⎠

Thus the field strength is

E=

1 8λ d 4πε 0 4 y 2 + d 2

(b) To draw a graph of Enet versus y, we calculated Enet at a few selected values of y.

y (in units of d) 0 0.5 1.0 2.0 3.0 4.0

⎛ 1 λ⎞ Enet ⎜ in units of ⎟ 4πε 0 d ⎠ ⎝ 8 4 1.6 0.47 0.22 0.12

27.38. Model: The fields are those of two infinite lines of charge with linear charge density λ. Visualize:

Solve:

The field at distance r from an infinite line of charge is

G 1 2λ E= rˆ 4πε 0 r It points straight away from the line. With two perpendicular lines, the field due to the line along the x-axis points in the y-direction and depends inversely on distance y. Similarly, the field due to the line along the y-axis points in the x-direction and depends inversely on distance x. That is Ex =

1 2λ 4πε 0 x

Ey =

1 2λ 4πε 0 y

G 2λ ⎛ 1 ˆ 1 ˆ ⎞ ⇒ E = Exiˆ + E y ˆj = ⎜ i + j⎟ y ⎠ 4πε 0 ⎝ x The field strength at this point in space is E = Ex2 + E y2 =

2λ 1/ x 2 + 1/ y 2 4πε 0

27.39. Model: Assume that the wire is thin and that the charge lies on the wire along a line. Solve:

From Equation 27.15, the electric field for an infinite uniformly charged line is Eline =

1 2λ 4πε 0 r

where r is the distance from the line in the plane that bisects the line. Solving for the linear charge density,

λ =

rEline ( 0.050 m )( 2000 N/C ) = 5.56 ×10−9 C/m = 2 (1 4πε 0 ) 2 ( 9.0 ×109 Nm 2 / C2 )

The charge in 1.0 cm is

Q = L λ = (1.0 × 10−2 m )( 5.56 ×10−9 C/m ) = 5.56 × 10−11 C = 0.056 nC

Because the electric field is directed toward the line, Q is negative. Thus

Q = −0.056 nC.

27.40. Model: The rods are lines of charge with a uniform linear charge density. Visualize:

Solve:

From Example 27.3, the electric field strength E at distance d from the center of a charged rod is

E=

Q

1

4πε 0 d d 2 + ( L / 2 )2

Because point P is equidistant from the center for each of the three rods, the electric field strength is the same for each rod. We have 10 ×10−9 C E = E1 = E2 = E3 = ( 9.0 ×109 N m 2 /C2 ) 2 d d 2 + ( 0.050 m ) From the geometry in the figure,

d ⎛ L⎞ = tan 30° ⇒ d = ⎜ ⎟ tan 30° = ( 0.05 m ) tan 30° = 0.02887 m L ⎝2⎠

1 2

⇒E=

90 N m 2 /C

( 0.02887 m ) ( 0.02887 m ) + ( 0.050 m ) 2

2

= 54,000 N/C

G G G E1 is along the −y-direction, E2 makes an angle of 30° below the −x-axis, and E3 makes an angle of 30° below the +x-axis. In component form, G E1 = − ( 54,000 N/C ) ˆj

G E2 = ( 54,000 N/C ) − cos30°iˆ − sin 30° ˆj

(

G E3 = ( 54,000 N/C ) cos30°iˆ − sin 30° ˆj

(

)

)

Adding these three vectors, G Enet = −2 ( 54,000 N/C ) sin 30° ˆj − ( 54,000 N/C ) ˆj = −1.08 × 105 ˆj N/C Thus, the electric field strength at the center of the triangle is 1.08 × 105 N/C.

27.41. Model: The electric field is that of an infinite line charge. Visualize:

Solve: The wire must be negative to attract the proton. From Equation 27.15, the electric field strength due to an infinite line of charge at a distance r from the line charge is

E=

1 2λ 4πε 0 r

The force on the proton due to this electric field causes the centripetal acceleration: F = qE =

1 2q λ mv 2 mv 2 = ⇒ λ = ( 4πε 0 ) 4πε 0 r r 2q

Using v = 2π r T = 2π rf ,

λ = ( 4πε 0 )

m ( 4π 2 r 2 f 2 ) 2q

(1.67 ×10 kg ) 4π (1.0 ×10 m ) (1.0 ×10 = ( 9.0 ×10 N m /C ) 2 (1.60 ×10 C ) −27

−2

2

9

Because the wire has to be negative, λ = −2.3 nC/m .

2

2

2

−19

6

s −1 )

2

= 2.3 × 10−9 C/m

27.42. Model: Visualize:

The electric field is that of a line of charge of length L.

The origin of the coordinate system is at the center of the rod. Divide the rod into many small segments of charge Δ q and length Δ x′. G G Solve: (a) Segment i creates a small electric field Ei at point P that points to the right. The net field E will point to the right and have Ey = Ez = 0 N/C. The distance to segment i is x′, so

Ei = ( Ei ) x =

Δq 4πε 0 ( x − xi′ )

2

⇒ Ex = ∑ ( Ei ) x =

1 Δq ∑ 4πε 0 ( x − xi′ )2

Δ q is not a coordinate, so before converting the sum to an integral we must relate the charge Δ q to length Δ x′. This is done through the linear charge density λ = Q/L, from which Δq = λΔx′ =

Q Δx′ L

With this charge, the sum becomes

Ex =

(Q / L) 4πε 0

Δx′

∑ ( x − x′ ) i

2

i

Now we let Δ x′ → dx′ and replace the sum by an integral from x′ = − 12 L to x′ = + 12 L. Thus, Ex =

(Q / L ) L / 2 4πε 0

dx′

∫ ( x − x′)

−L / 2

2

=

(Q / L ) ⎡ 4πε 0

L/2 G 1 ⎤ 1 Q (Q / L ) L = ⇒E= iˆ 2 2 2 ⎢⎣ x − x′ ⎦⎥ − 4 x L 4 4 πε πε x − L2 4 −L / 2 0 0

The electric field strength at x is

E=

1

Q 4πε 0 x − L2 4 2

(b) For x >> L ,

E=

1 Q 4πε 0 x 2

That is, the line charge behaves like a point charge. (c) Substituting into the above formula 3.0 ×10−9 C E = ( 9.0 × 109 N m 2 /C2 ) = 9.8 × 104 N/C 2 2 ( 3.0 ×10−2 m ) − ( 12 × 5.0 ×10−2 m )

27.43. Model: The electric field is that of a line charge of length L. Visualize: Please refer to Figure P27.43. Let the bottom end of the rod be the origin of the coordinate system. Divide the rod into many small segments of charge Δ q and length Δ y′. Segment i creates a small electric field at the point P that makes an angle θ with the horizontal. The field has both x and y components, but Ez = 0 N/C. The distance to segment i from point P is ( x 2 + y′2 ) . 12

Solve:

The electric field created by segment i at point P is

G Ei =

x y′ Δq Δq ⎛ ˆj ⎞ cosθ iˆ − sin θ ˆj = iˆ − ⎜ ⎟ 4πε 0 ( x 2 + y′2 ) 4πε 0 ( x 2 + y′2 ) ⎝ x 2 + y′2 x 2 + y′2 ⎠ G G G The net field is the sum of all the Ei , which gives E = ∑ Ei . Δ q is not a coordinate, so before converting the

(

)

i

sum to an integral we must relate charge Δ q to length Δ y′. This is done through the linear charge density λ = Q/L, from which we have the relationship Δq = λΔy′ =

Q Δy′ L

With this charge, the sum becomes G Q/L ⎡ xΔy′ y′Δy′ E= iˆ − ∑ 3/ 2 2 4πε 0 i ⎢ ( x 2 + y′2 )3 / 2 ( x + y′2 ) ⎢⎣

ˆj ⎤ ⎥ ⎥⎦

Now we let Δ y′ → dy′ and replace the sum by an integral from y′ = 0 m to y′ = L . Thus, L G (Q / L) ⎛ L xdy′ y′dy′ E= iˆ − ∫ ⎜∫ 2 3 / 2 2 2 3/ 2 4πε 0 ⎜ 0 ( x + y′2 ) 0 ( x + y′ ) ⎝

=

L L ⎛ ⎞ y′ −1 ⎤ ⎤ ˆ ⎡ ˆj ⎟ = ( Q / L ) ⎜ x ⎡ − i ⎢ 2 2 ⎥ ⎟ 4πε 0 ⎜ ⎢⎣ x 2 x 2 + y′2 ⎥⎦ ⎣ x + y′ ⎦ 0 0 ⎝ ⎠

Q 1 1 ⎛ Q ⎞⎛ x iˆ − ⎜ ⎟ 1− 2 4πε 0 x x 2 + L2 4πε 0 ⎝ Lx ⎠ ⎜⎝ x + L2

⎞ˆ ⎟j ⎠

⎞ ˆj ⎟ ⎟ ⎠

27.44. Model: Assume that the ring of charge is thin and that the charge lies along circle of radius R. Visualize: Please refer to Figure 27.15. Solve: From Example 27.5, the on-axis field of a ring of charge Q and radius R is Ez =

zQ

4πε 0 ( z 2 + R 2 )

3/ 2

When z << R, this means we are near the center of the ring. At that point, segments of charge i and j that are G G 180° apart create fields Ei and E j that cancel each other. When the fields of all segments of charge around the ring are added, the net result is zero. This is indicated by the above expression because when z = 0 m, the electric field is zero. When z >> R, then

(z

2

+ R2 )

3/ 2

= ( z2 )

3/ 2

= z3

If this is used in the expression for Ez, we get

Ez ≈

zQ 1 Q = 4πε 0 z 3 4πε 0 z 2

This is the field of a point charge Q as seen along the z-axis.

27.45. Model: Assume that the ring of charge is thin and that the charge lies along circle of radius R. Solve:

(a) From Example 27.5, the on-axis field of a ring of charge Q and radius R is

Ez =

1 zQ 4πε 0 ( z 2 + R 2 )3/ 2

For the field to be maximum at a particular value of z, dE dz = 0. Taking the derivative, z ( 3/ 2 )( 2 z ) ⎤ dE Q ⎡ 1 1 3z 2 = − =0⇒ = ⎢ 2 ⎥ 3 / 2 5 / 2 3 2 5/ 2 dz 4πε 0 ⎢ ( z + R 2 ) ( z 2 + R 2 ) ⎥⎦ ( z 2 + R2 ) ( z 2 + R2 ) ⎣

⇒1= (b) The field strength at the point z = R

( Ez )max =

Q 4πε 0

(R 2) ⎡ R 2 +R ⎤ ) ⎥⎦ ⎢⎣( 2

2

3/ 2

=

2

2 is Q

2 3 3 4πε 0 R

3z 2 R ⇒z= 2 2 z +R 2

27.46. Model: Assume that the semicircular rod is thin and that the charge lies along the semicircle of radius R. Visualize:

The origin of the coordinate system is at the center of the circle. Divide the rod into many small segments of G charge Δ q and arc length Δ s. Segment i creates a small electric field Ei at the origin. The line from the origin to segment i makes an angle θ with the x-axis. Solve: Because every segment i at an angle θ above the axis is matched by segment j at angle θ below the axis, the y-components of the electric fields will cancel when the field is summed over all segments. This leads to a net field pointing to the right with

Ex = ∑ ( Ei ) x = ∑ Ei cosθ i i

E y = 0 N/C

i

Note that angle θ i depends on the location of segment i. Now all segments are at the same distance ri = R from the origin, so

Ei =

Δq Δq = 4πε 0 ri 2 4πε 0 R 2

The linear charge density on the rod is λ = Q/L, where L is the rod’s length. This allows us to relate charge Δ q to the arc length Δ s through Δ q = λ Δ s = (Q/ L)Δ s Thus, the net field at the origin is Ex = ∑ i

( Q / L ) Δ s cosθ 4πε 0 R

2

i

=

Q ∑ cosθi Δ s 4πε 0 LR 2 i

The sum is over all the segments on the rim of a semicircle, so it will be easier to use polar coordinates and integrate over θ rather than do a two-dimensional integral in x and y. We note that the arc length Δ s is related to the small angle Δθ by Δ s = RΔθ , so Ex =

Q ∑ cosθ i Δθ 4πε 0 LR i

With Δ θ → dθ , the sum becomes an integral over all angles forming the rod. θ varies from Δ θ = −π/2 to θ = +π/2. So we finally arrive at

Ex =

π /2 Q Q cosθ dθ = sin θ ∫ − π / 2 4πε 0 LR 4πε 0 LR

π /2 −π / 2

=

2Q 4πε 0 LR

Since we’re given the rod’s length L and not its radius R, it will be convenient to let R = L/π. So our final G expression for E , now including the vector information, is G 1 ⎛ 2π Q ⎞ ˆ E= ⎜ ⎟i 4πε 0 ⎝ L2 ⎠

(b) Substituting into the above expression,

E=

( 9.0 ×10

9

N m 2 /C2 ) 2π ( 30 ×10−9 C )

( 0.10 m )

2

= 1.70 ×105 N/C

27.47. Model: Assume that the quarter-circle plastic rod is thin and that the charge lies along the quartercircle of radius R. Visualize:

The origin of the coordinate system is at the center of the circle. Divide the rod into many small segments of charge Δ q and arc length Δ s. G Solve: (a) Segment i creates a small electric field Ei at the origin with two components:

( Ei ) x = Ei cosθ i

( Ei ) y = Ei sinθ i

Note that the angle θ i depends on the location of the segment i. Now all segments are at distance ri = R from the origin, so

Ei =

1 Δq 1 Δq = 2 4πε 0 ri 4πε 0 R 2

The linear charge density of the rod is λ = Q/L, where L is the rod’s length (L = quarter-circumference = π R/2). This allows us to relate charge Δ q to the arc length Δ s through ⎛Q⎞ ⎛ 2Q ⎞ Δq = λΔs = ⎜ ⎟ Δs = ⎜ ⎟ Δs ⎝L⎠ ⎝πR ⎠ Using Δ s = RΔ θ , the components of the electric field at the origin are

( Ei ) x =

1 Δq 1 1 ⎛ 2Q ⎞ 1 ⎛ 2Q ⎞ cosθi = ⎜ ⎟ RΔθ cosθ i = ⎜ ⎟ Δθ cosθ i 4πε 0 R 2 4π 0 R 2 ⎝ π R ⎠ 4πε 0 ⎝ π R 2 ⎠

( Ei ) y =

1 Δq 1 1 ⎛ 2Q ⎞ 1 ⎛ 2Q ⎞ sin θi = ⎟ RΔθ sin θi = ⎜ ⎟ Δθ sin θ i 2 2 ⎜ 4πε 0 R 4πε 0 R ⎝ π R ⎠ 4πε 0 ⎝ π R 2 ⎠

(b) The x- and y-components of the electric field for the entire rod are the integrals of the expressions in part (a) from θ = 0 rad to θ = π/2. We have

Ex =

1 ⎛ 2Q ⎞ π / 2 cosθ dθ ⎜ ⎟ 4πε 0 ⎝ π R 2 ⎠ ∫ 0

1 ⎛ 2Q ⎞ π / 2 sin θ dθ ⎜ ⎟ 4πε 0 ⎝ π R 2 ⎠ ∫ 0

Ey =

(c) The integrals are

∫

π /2 0

sin θ dθ = [ − cosθ ]0

π /2

π ⎛ ⎞ = − ⎜ cos − cos0 ⎟ = +1 2 ⎝ ⎠

∫

π /2 0

The electric field is

G 1 2Q ˆ ˆ E= i+j 4πε 0 π R 2

(

)

cosθ dθ = [sin θ ]0

π /2

= sin

π 2

− sin 0 = +1

27.48. Model: Assume that the plastic sheets are planes of charge. Visualize: Please refer to Figure P27.48. Solve: At point 1 the electric field due to the left sheet and the right sheet are

G ⎛η ⎞ η Eleft = ⎜ 0 , toward right ⎟ = 0 iˆ ε 2 ⎝ 0 ⎠ 2ε 0

G ⎛ 3η ⎞ 3η Eright = ⎜ 0 , toward left ⎟ = − 0 iˆ ε 2 2ε 0 ⎝ 0 ⎠

G G G η ⇒ Enet = Eleft + Eright = − 0 iˆ

ε0

G G G G At point 2, Eleft = − (η0 2ε 0 ) iˆ, Eright = − ( 3η0 2ε 0 ) iˆ, and Enet = − ( 2η0 ε 0 ) iˆ. At point 3, Enet = + (η0 ε 0 ) iˆ.

27.49. Motion: The very wide charged electrode is a plane of charge. Solve: For a plane of charge with surface charge density η, the electric field is

(E )

plane z

=

η 2ε 0

⇒ η = 2ε 0 ( Eplane ) = z

Q Q = A π R2

⇒ Q = 2π R 2ε 0 ( Eplane ) = 2π ( 0.0050 m ) (8.85 × 10−12 C 2 /N m 2 ) (1000 N/C ) = 1.39 × 10 −12 C = 1.39 × 10−3 nC 2

z

Assess: Note that the field strength of a plane of charge does not depend on z and is a constant. We did not have to use the distance z = 5.0 cm given in the problem.

27.50. Model: An insulating sphere of charge Q and radius R has an electric field outside the sphere that is exactly the same as that of a point charge Q located at the center of the sphere. Visualize:

Solve:

The electric field of a charged sphere at a distance r > R is given by Equation 27.28:

G Esphere =

Q 4πε 0 r 2

rˆ

G G G G G In the present case, E = E1 + E2 , where E1 and E2 are the fields of the individual spheres. The distance from the

center of each sphere to the midpoint between is r1 = r2 = 4 cm. Thus, E1 =

(9.0 ×10

9

( 0.040 m )

( 9.0 × 10 =

9

E2

N m 2 / C2 )(10 × 10−9 C ) 2

N m 2 / C2 )(15 × 10−9 C )

( 0.040 m )

2

= 5.625 × 104 N/C

= 8.438 × 104 N/C

The fields point in the same direction, so G E = ( 5.625 × 104 N/C + 8.438 × 104 N/C, right ) = (1.41×105 N/C, right ) .

The electric field will point left when Q1 and Q2 are interchanged. The electric field strength in both cases is 1.41 × 105 N/C.

27.51. Model: The parallel plates form a parallel-plate capacitor. The electric field inside a parallel-plate capacitor is a uniform field, so the electron and proton will have constant acceleration. Visualize:

The negative plate is at x = 0 m and the positive plate is at x = d = 1 cm. Solve: Both particles accelerate from rest (v0 = 0 m/s), so at time t their positions are

1 1 xe = xe0 + aet 2 = aet 2 2 2

1 1 xp = xp0 + apt 2 = d + apt 2 2 2

At some instant of time t1 the electron and proton have the same position: xe = xp = x1. This is the point where they pass each other. At this instant, 1 x1 = aet12 2

1 x1 = d + apt12 2

These are two equations in the two unknowns x1 and t1. From the first equation, 12 t12 = x1 ae . Using this in the second equation gives

x1 = d +

ap ae

x1 ⇒ x1 =

d 1 + ap /ae

To finish, we need to find the accelerations of the electron and proton. Both particles are in a parallel-plate capacitor with Ecap = Q/ε0A. The field points to the left, so Ex = −Q/ε0A. The proton’s acceleration is

ap =

Fp mp

=

qp E x mp

=

e ( −Q/ε 0 A ) eQ/ε 0 A =− mp mp

The proton’s acceleration is negative, as expected. For the electron, ae =

Fe qe Ex −e ( −Q/ε 0 A ) eQ/ε 0 A = = = me me me me

Consequently, the acceleration ratio is ap ae = me mp . Using this, the point where the two charges pass is x1 =

1 cm d = = 0.9995 cm 1 + me /mp 1 + 9.11 × 10−31 kg/1.67 × 10−27 kg

Assess: This is very close to where the proton starts. Since there is a factor of ~1800 difference in the masses of the proton and electron, the electron accelerates much faster than the proton.

27.52. Model: The electric field is uniform inside the capacitor, so constant-acceleration kinematic equations apply to the motion of the proton. Visualize:

G Solve: From Equation 27.29, the electric field between the parallel plates E = (η ε 0 ) ˆj. The force on the proton is G G G G G qE qη ˆ qη = F = ma = qE ⇒ a = j ⇒ ay = m mε 0 mε 0

Using the kinematic equation y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) , 2

1 ⎛ qη ⎞ 1 2 Δy = y1 − y0 = ( 0 m/s )( t1 − t0 ) + a y ( t1 − t0 ) = ⎜ ⎟ ( t1 − t0 ) 2 ⎝ mε 0 ⎠ 2 To determine t1 − t0, we consider the horizontal motion of the proton. The proton travels a distance of 2.0 cm at a constant speed of 1.0 × 106 m/s. The velocity is constant because the only force acting on the proton is due to the field between the plate along the y-direction. Using the same kinematic equation, Δx = 2.0 × 10−2 m = v0 x ( t1 − t0 ) + 0 m ⇒ ( t1 − t0 ) =

2.0 ×10−2 m = 2.0 ×10−8 s 1.0 ×106 m/s

−19 −6 −8 2 1 (1.60 × 10 C )(1.0 × 10 C/m )( 2.0 × 10 s ) = 2.2 mm 2 (1.67 ×10−27 kg )(8.85 ×10−12 C2/Nm2 ) 2

⇒ Δy =

27.53. Model: Assume that the electric field inside the capacitor is constant, so constant-acceleration kinematic equations apply. Visualize: Please refer to Figure P27.53. Solve: (a) The force on the electron inside the capacitor is G G G G G qE F = ma = qE ⇒ a = m G Because E is directed upward (from the positive plate to the negative plate) and q = −1.60 ×10−19 C , the

acceleration of the electron is downward. We can therefore write the above equation as simply ay = qE/m. To determine E, we must first find ay. From kinematics, 1 2 x1 = x0 + v0 x ( t1 − t0 ) + ax ( t1 − t0 ) ⇒ 0.040 m = 0 m + v0 cos 45° ( t1 − t0 ) + 0 m 2

⇒ ( t1 − t0 ) =

( 0.040 m )

( 5.0 ×10

6

m/s ) cos 45°

= 1.1314 × 10−8 s

Using the kinematic equations for the motion in the y-direction,

⎛ t −t ⎞ ⎛ qE ⎞⎛ t1 − t0 ⎞ v1 y = v0 y + a y ⎜ 1 0 ⎟ ⇒ 0 m/s = v0 sin 45° + ⎜ ⎟⎜ ⎟ 2 ⎝ ⎠ ⎝ m ⎠⎝ 2 ⎠ ⇒E=−

2 ( 9.1× 10−31 kg )( 5.0 ×106 m/s ) sin 45° 2 m v0 sin 45° =− = 3550 N/C = 3.6 × 103 N/C −19 −8 q ( t1 − t0 ) − × × 1.60 10 C 1.1314 10 s ( )( )

(b) To determine the separation between the two plates, we note that y0 = 0 m and v0 y = (5.0 ×106 m/s)sin 45°, but at

y = y1, the electron’s highest point, v1y = 0 m/s. From kinematics, v12y = v02y + 2a y ( y1 − y0 ) ⇒ 0 m 2 / s 2 = v02 sin 2 45° + 2a y ( y1 − y0 )

⇒ ( y1 − y0 ) = −

v02 sin 2 45° v2 =− 0 2a y 4a y

From part (a), ay =

−19 qE ( −1.60 ×10 C ) ( 3550 N/C ) = = −6.242 ×1014 m/s 2 9.1×10−31 kg m

⇒ y1 − y0 = −

( 5.0 × 10

6

m/s )

2

4 ( −6.242 × 1014 m/s 2 )

= 0.010 m = 1.0 cm

This is the height of the electron’s trajectory, so the minimum spacing is 1.0 cm.

27.54. Model: The parallel plates form a parallel-plate capacitor. The electric field inside a parallel-plate capacitor is a uniform field, so the electrons will have a constant acceleration. Visualize:

Solve: (a) The bottom plate should be positive. The electron needs to be repelled by the top plate, so the top plate must be negative and the bottom plate positive. In other words, the electric field needs to point away from G the bottom plate so the electron’s acceleration a is toward the bottom plate. (b) Choose an xy-coordinate system with the x-axis parallel to the bottom plate and the origin at the point of G entry. Then the electron’s acceleration, which is parallel to the electric field, is a = ajˆ . Consequently, the problem looks just like a projectile problem. The kinetic energy K = 12 mv02 = 3.0 ×10−17 J gives an initial speed

v0 = ( 2 K m )

1/ 2

= 8.115 ×106 m/s . Thus the initial components of the velocity are vx 0 = v0 cos 45° = 5.74 × 106 m/s

v y 0 = v0 sin 45° = 5.74 × 106 m/s

G What acceleration a will cause the electron to pass through the point (x1, y1) = (1.0 cm, 0 cm)? The kinematic equations of motion are 1 x1 = x0 + vx 0t1 + axt12 = vx 0t1 = 0.010 m 2

1 1 y1 = y0 + v y 0t1 + a yt1 = v y 0t1 + at12 = 0 m 2 2

From the x-equation, t1 = x1 vx 0 = 1.742 × 10−9 s. Using this in the y-equation gives

a=−

2v y 0t1 t12

= −6.59 × 1015 m/s 2

But the acceleration of an electron in an electric field is a=

( 9.11×10−31 kg )( −6.59 ×1015 m/s2 ) = 37,500 N/C = 3.8 ×104 N/C Felec qelec E −eE ma = = ⇒E=− =− m m m e 1.60 ×10−19 C

(c) The minimum separation dmin must equal the “height” ymax of the electron’s trajectory above the bottom plate. (If d were less than ymax, the electron would collide with the upper plate.) Maximum height occurs at t = 12 t1 = 8.71 × 10−10 s. At this instant, 1 ymax = v y 0t + at 2 = 0.0025 m = 2.5 mm 2 Thus, dmin = 2.5 mm.

27.55. Model: Assume that the plates form a parallel-plate capacitor. The electric field inside the capacitor is uniform, so that the protons will have a constant acceleration. Visualize:

Solve: (a) The acceleration needed to slow the protons to 2.0 × 105 m/s can be obtained using the kinematic equation of motion v12 = v02 + 2a ( x1 − x0 ) . We have

( 2.0 ×105 m/s ) − ( 2.0 ×106 m/s ) = −9.9 ×1013 m/s2 v2 − v2 a= 1 0 = 2 ( x1 − x0 ) 2 ( 0.020 m ) 2

2

The force on the proton due to the electric field between the plates is F = ma = qEx . The electric field is Ex =

−27 13 2 ma (1.67 ×10 kg )( −9.9 ×10 m/s ) = = −1.033 ×106 N/C +1.6 ×10−19 C q

The minus sign indicates that the field points to the left. Because E = η ε 0 for a parallel-plate capacitor, where E is the field strength, η = ε 0 E = (8.85 ×10−12 C2 /Nm 2 )(1.033 ×106 N/C ) = 9.1× 10−6 C/m 2 It is clear that the electric field must oppose the motion of the protons. This requires that the first plate be negatively charged and the second plate be positively charged. (b) Let us calculate E that makes protons come to rest completely. Combining the equations for E and a in part (a),

E=

m ( v12 − v02 )

2q ( x1 − x0 )

=

(1.67 ×10

2 2 kg ) ⎡( 0 m/s ) − ( 2 ×106 m/s ) ⎤ ⎥⎦ ⎣⎢ = 1.04 ×106 N/C 2 (1.6 ×10−19 C ) ( 0.020 m ) −27

⇒ η = ε 0 E = ( 8.85 ×10−12 C2 /Nm 2 )(1.04 × 106 N/C ) = 9.24 × 10−6 C/m 2 Because the surface charge density of ±1.0 × 10−5 C/m 2 in your device is larger than 9.24 × 10−6 C/m 2 , the protons will not reach the positive plate. The beam will in fact reverse its direction. The protons will not be stopped, so no, the device does not work.

27.56. Model: An orbiting electron experiences a force that causes centripetal acceleration. Visualize:

The electron orbits at a radius r = R + h = 12 ( 2.0 mm ) + 1.0 mm = 2.0 mm. The force that causes the G circular motion is simply the electric force F given by Coulomb’s law:

Solve:

F=

K Qsphere qelec r

2

=

KeQsphere r

2

=

( 9 ×10

9

N m 2 / C2 )(1.6 × 10−19 C )(1.0 × 10−9 C )

( 2.0 ×10

−3

m)

2

For circular motion, F = macentripetal =

mv 2 rF ⇒v= = r m

( 2.0 ×10

−3

m )( 3.6 ×10−13 N )

9.11×10−31 kg

= 2.8 × 107 m/s

= 3.60 × 10−13 N

27.57. Model: An orbiting proton experiences a force that causes centripetal acceleration. Visualize:

Solve: The proton orbits at a radius r = 5.0 mm + 1.0 mm = 6.0 mm = 6.0 × 10−3 m. The force that causes the G circular motion is simply the electric force F given by Coulomb’s law:

F=

1

Qball qproton

4πε 0

r2

= macentripetal =

mv 2 ( 4πε 0 ) mv 2r ⇒ Qball = r qproton

The speed of the proton is v=

−3 2π r 2π ( 6.0 ×10 m ) = = 3.77 ×104 m/s 1.0 ×10−6 s T

(1.67 ×10 kg )( 3.77 ×10 m/s ) ( 6.0 ×10 = ( 9.0 ×10 N m /C )(1.60 ×10 C ) −27

⇒ Qball

4

9

2

2

2

−19

−3

m)

= 9.9 ×10−12 C

The charge on the ball must be negative to hold the proton in its orbit, so Q = −9.9 ×10−12 C.

27.58. Model: Visualize:

Solve:

The electron orbiting the proton experiences a force given by Coulomb’s law.

The force that causes the circular motion is

F=

2 2 2 1 qp qe mev 2 me ( 4π r f ) = = 2 4πε 0 r r r

where we used v = 2π r T = 2π rf . The frequency is

⎛ 1 ⎞ qp qe f = ⎜ ⎟ 2 3 = ⎝ 4πε 0 ⎠ 4π me r

( 9.0 ×10 N m /C )(1.60 ×10 4π ( 9.11×10 kg )( 5.3 × 10 9

2

2

−31

2

−19

−11

C)

m)

2

3

= 6.56 ×1015 Hz

27.59. Model: Visualize:

Solve:

The electron orbiting the proton experiences a force given by Coulomb’s law.

The force that causes the circular motion is F=

2 2 2 1 qp qe mev 2 me ( 4π r f ) = = 2 4πε 0 r r r

where we used v = 2π r T = 2π rf . The radius is 1/ 3

qp qe ⎡ 1 ⎤ r=⎢ 2 2 ⎥ ⎢⎣ 4πε 0 me ( 4π f ) ⎥⎦

⎡ (1.60 ×10−19 C )(1.60 ×10−19 C ) ⎤⎥ = ⎢( 9.0 × 109 N m 2 /C 2 ) 2 ⎢ ( 9.1× 10−31 kg ) 4π 2 (1.0 × 1012 s−1 ) ⎥⎦ ⎣

1/ 3

= 1.86 × 10−8 m = 18.6 nm

27.60. Model: The electric field at the dipole’s location is that of the ion with charge q. Visualize:

Solve:

(a) We have p = α E. The units of α are the units of p/E and are

C m C 2 m C 2s 2 = = N/C N kg (b) The electric field due to the ion at the location of the dipole is

⎛ 1 q ⎞ 1 q ˆ Eat dipole = ⎜ i , away from q ⎟ = 2 2 ⎝ 4πε 0 r ⎠ 4πε 0 r

G G Because p = α E , the induced dipole moment is

⎛ 1 q ⎞ˆ G p =α ⎜ i 2 ⎟ ⎝ 4πε 0 r ⎠ From Equation 27.11, the electric field produced by the dipole at the location of the ion is 2 G G 1 2p 1 ⎛ 2 ⎞ ⎛ 1 q ⎞ ˆ ⎛ 1 ⎞ ⎛ 2qα ⎞ ˆ Edipole = α i = = ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎟i 4πε 0 r 3 4πε 0 ⎝ r 3 ⎠ ⎝ 4πε 0 r 2 ⎠ ⎝ 4πε 0 ⎠ ⎝ r 5 ⎠ The force the dipole exerts on the ion is 2

G G ⎛ 1 ⎞ ⎛ 2q 2α ⎞ ˆ Fdipole on ion = qEdipole = ⎜ ⎟ ⎜ 5 ⎟i ⎝ 4πε 0 ⎠ ⎝ r ⎠ G G According to Newton’s third law, Fdipole on ion = − Fion on dipole . Therefore,

⎛ ⎛ 1 ⎞ 2 2q 2α ⎞ G Fion on dipole = ⎜ ⎜ , toward ion ⎟ ⎟ 5 ⎜ ⎝ 4πε 0 ⎠ r ⎟ ⎝ ⎠

27.61. Model: density λ. Visualize:

Solve:

The electric field at the dipole’s location is that of the infinite line of charge of linear charge

The field at distance r from an infinite line of charge is G 1 2λ E= rˆ 4πε 0 r

It points straight away from the line. The dipole consists of charge +q at distance r + s/2 and charge –q at distance r – s/2. The net force on the dipole due to the field of the line is

G G G 2qλ ⎛ 1 1 ⎞ 2qλ ⎛ ( r − s/2 ) − (r + s/2 ) ⎞ F = qE+ − qE− = − ⎜ ⎟ rˆ ⎜ ⎟ rˆ = 4πε 0 ⎝ r + s/2 r − s/2 ⎠ 4πε 0 ⎝ (r + s/2 )(r − s/2 ) ⎠ =−

2qsλ 2 pλ rˆ = − rˆ 4πε 0 ( r 2 − ( s/2) 2 ) 4πε 0 ( r 2 − ( s/2) 2 )

where we used p = qs for the dipole moment. The minus sign shows that this is an attractive force, toward the line of charge. If r >> s, the (s/2)2 term in the denominator is negligible and we find that the line of charge exerts an attractive force of magnitude 2 pλ F= 4πε 0 r 2

27.62. Solve: (a) Charges ±2.0 nC form an electric dipole. The electric field strength 2.5 cm from the dipole in the plane perpendicular to the dipole is 1150 N/C. What is the charge separation? (b) Solving for s, 3 (1150 N/C )( 0.025 m ) = 9.98 × 10−4 m = 1.0 × 10−3 m = 1.0 mm s= ( 9 ×109 N m 2 /C2 )( 2.0 ×10−9 C )

27.63. Solve: (a) A very long charged wire has linear charge density 2.0×10–7 C/m. At what distance from the wire is the field strength 25,000 N/C? (b) Solving for r, ( 9 × 109 N m2 /C2 ) 2 ( 2.0 × 10−7 C/m ) = 0.144 m = 14.4 cm r= 25,000 N/C

27.64. Solve: (a) At what distance along the axis of a charged disk is the electric field strength half its value at the center of the disk? Give your answer in terms of the disk’s radius R. (b) Dividing both sides of the equation by η 2ε 0 , 1−

z z 2 + R2

=

1 R ⇒ 2z = z 2 + R 2 ⇒ 4z 2 = z 2 + R 2 ⇒ z = 2 3

27.65. Solve: (a) A proton is released from the positive plate of a parallel-plate capacitor and accelerates toward the negative plate at 2.0 × 1012 m/s2. If the capacitor plates are 2.0 cm × 2.0 cm squares, what is the magnitude of the charge on each? (b) Solve the first equation for E and substitute into the second equation. The charge is

Q=

(1.67 ×10

−27

kg )( 2.0 ×1012 m/s 2 )( 8.85 ×10−12 C2 /N m 2 ) ( 0.020 m ) 1.60 ×10−19 C

2

= 7.4 ×10−11 C

27.66. Model: The long charged wire is an infinite line of charge. The charges on the wire and the plastic rod are uniform. Visualize: Please refer to Figure P27.66. The plastic stirrer is located on the x-axis. Solve: The electric field of the infinite line of charge at a distance x from its axis is

Ex =

1 2λ 4πε 0 x

Because the electric field is a function of x, the plastic stirrer experiences an electric field that varies along the length of the stirrer. We handled such problems earlier. Take a small segment of the charge on the stirrer and calculate the electric force due to the line charge on this charge segment Δ q. A summation of all such forces on the charge segments Δ q will yield the net force on the stirrer. This procedure is equivalent to integrating the electric force on a small segment Δ q over the length of the stirrer. The force on charge Δ q of length Δ x at position x due to the infinite line of charge is ΔF = E Δq = Eλ ′Δx =

1 ⎛ 2λλ ′Δx ⎞ ⎜ ⎟ 4πε 0 ⎝ x ⎠

In the above expression, λ′ = Q/L is the linear charge density of the stirrer and λ is the linear charge density of the plastic rod. We have also used the relation Δq Δx = λ ′. Changing Δ x → dx and integrating x from x = 0.020 m to x = 0.080 m, the total force on the stirrer is ⎛ 1 ⎞ ⎛ 1 ⎞ 1 2λλ ′ 0.080 m ⎛ 0.080 m ⎞ dx = ⎜ ⎟ ( 2λλ ′ ) ln x 0.020 m = ⎜ ⎟ ( 2λλ ′ ) ln ⎜ ⎟ 4 πε x 4 πε 4 πε ⎝ 0.020 m ⎠ 0 0 ⎠ 0 ⎠ ⎝ ⎝ 0.020 m 0.080 m

F=

∫

⎛ 10 × 10−9 C ⎞ −4 = ( 9.0 × 109 N m 2 /C 2 ) 2 (1.0 × 10−7 C/m ) ⎜ ⎟ ln ( 4 ) = 4.2 × 10 N 0.060 m ⎝ ⎠

27.67. Model: The electric field is that of the ball. This electric field polarizes charge in the aluminum foil. Visualize:

Please refer to Figure CP27.67. Solve: (a) The electric field at the location of the piece of foil is

G ⎛ Q ⎞ Q ˆ Eball = ⎜ j , straight down ⎟ = − 2 4πε 0 h 2 ⎝ 4πε 0 h ⎠ (b) The electric field inside the foil is not G just the field due to the +q Gand −q charges on the surfaces of the foil. It is the superposition of the ball’s field Eball and the polarization field Epolar that arises when the foil is polarized by G G G G G the charged ball. That is, Ein = Eball + Epolar . In electrostatic equilibrium, the internal field must be Ein = 0 N/C.

(An internal field that wasn’t zero would cause the motion of electrons and thus prevent the foil from being in electrostatic the charged ball is brought near, the foil is polarized just to the point where G G When G G equilibrium.) Epolar = − Eball , causing Ein = 0 N/C. G G G G (c) In part (b) we noted that Ein = Eball + Epolar = 0 N/C. We know the ball’s field, from part (a). The polarization field is that of a parallel-plate capacitorGsince the polarized foil, with +q and −q surfaces, is just like a capacitor. The polarization field points upward, Epolar = (η ε 0 ) ˆj , where η is the surface charge density. Adding the two fields and setting the sum to zero gives G G Eball + Epolar = −

G ˆj + η ˆj = 0 N/C ⇒ η = Q ε0 4πε 0 h 4π h 2 Q

2

Once we know η, the actual charge on the foil surface of area A = πR2 is q = η A = ηπ R 2 =

R2 Q 4h 2

(d) The attractive and repulsive forces can each be separately calculated and combined, but it is easier to consider the polarized foil to be a dipole since it has two opposite charges ± q separated by a distance t. Because t << h, we can write the on-axis electric field of the foil as the on-axis field of a dipole. Using Equation 27.11, the electric field produced by the dipole is 2p 2qt = Edipole = 3 4πε 0 r 4πε 0 h3 G G The dipole exerts a force on the charged ball Ffoil on ball = QEdipole . From Newton’s third law,

Fball on foil = Ffoil on ball =

2Qqt 4πε 0 h3

The foil is lifted when Fball on foil ≥ mg = ρVg , where ρ is the mass density of the foil and V = πR2t is the foil’s volume. That is, when 2Qqt 2Qq ≥ ρπ R 2tg ⇒ ≥ ρπ R 2 g 4πε 0 h3 4πε 0 h3 since the foil thickness t cancels. Using the expression for q we found in part (c),

⎡ ⎤ R 2Q 2 Q2 ≥ ρπ R 2 g ⇒ h ≤ ⎢ ⎥ 5 2 ( 4πε 0 ) h 2 4 g π πε ρ 0) ⎣⎢ ( ⎦⎥

1/ 5

The minimum height for which the ball lifts the foil is

⎡ ( 9.0 × 109 N m 2 /C2 )( 50 × 10−9 C )2 ⎤ ⎥ h=⎢ ⎢ 2π ( 2700 kg/m3 )( 9.8 m/s 2 ) ⎥ ⎣ ⎦

1/ 5

= 0.0106 m = 1.06 cm

27.68. Model: The rod is thin and is assumed to be a line of charge of length L. Visualize:

Solve: (a) The λ-versus-y graph over the length of the rod is shown in the figure. (b) Consider a segment of charge Δ q of length Δ y at a distance y from the center of the rod. The amount of charge in this segment is Δq = λΔy = a y Δy Converting Δ q to dq, Δ y to dy, and integrating from y = −L/2 to y = +L/2, the total charge is

Q = ∫ dq =

+L/2

∫

−L / 2

L/2

a y dy = 2 ∫ ay dy = 2a 0

y2 2

L/2

= 0

aL2 4

Thus the constant a is a=

4Q L2

(c) With the origin of the coordinate system at the center of the rod, consider two symmetrically located charge segments Δ q with length Δ y. The electric field at P from the top charge segment makes an angle θ below the xaxis and the electric field of the bottom charge segment makes an angle θ above the x-axis. Because the charge density is symmetric about the origin (i.e., λ (at –y) = λ (at y)), the y-components of the two contributions cancel out. Thus, we have to calculate only the x-component of the electric field at P. Because the electric field strength of the lower half of the rod is the same as that of the upper half, we only need to obtain the electric field strength of half the rod, then multiply by two. The electric field along the x-direction due to a charge segment Δ q is xa y Δy 1 Δq 1 λΔy 1 x ΔE x = cosθ = = 4πε 0 ( x 2 + y 2 ) 4πε 0 ( x 2 + y 2 ) x 2 + y 2 4πε 0 ( x 2 + y 2 )3 / 2

Changing Δ E to dE, Δ y = dy, integrating y from y = 0 m to y = L/2, and multiplying by 2 to take into account the entire rod, the electric field is L/2

Ex = 2

∫ 0

=

ax y dy 2ax = 3/ 2 2 2 4πε 0 ( x + y ) 4πε 0

L/2

∫ 0

(x

L/2

y dy

2

+ y2 )

3/ 2

=

2ax ⎡ −1 ⎤ 4πε 0 ⎢⎣ x 2 + y 2 ⎥⎦ 0

2ax ⎡ 1 1 8Q ⎡ x ⎤ ⎤ − ⎥ = 4πε L2 ⎢1 − 2 ⎥ 2 2 2 4πε 0 ⎢⎣ x x +L 4⎦ x +L 4⎦ 0 ⎣

The last step used the expression for a from part (b).

27.69. Model: Model the infinitely long sheet of charge with width L as a uniformly charged sheet.

Solve: (a) Divide the sheet into many long strips parallel to the y-axis and of width Δ x. Each strip has a linear charge density λ = ηΔ x and acts like a long, charged wire. At the point of interest, strip i contributes a small field G Ei of strength

Ei =

2λ 2ηΔx = 4πε 0 ri 4πε 0 ri

By symmetry, the x-components of all the strips will add to zero and the net field will point straight away from the sheet along the z-axis. The net field’s z-component will be Ez = ∑ ( Ei ) z = ∑ Ei cosθ i i

The distance ri = ( xi2 + z

)

2 1/ 2

i

and cosθi = z ri = z ( xi2 + z

Ez = ∑ i

2ηΔx

4πε 0 ( x + z 2 i

) (x

2 1/ 2

)

2 1/ 2

2 i

. Thus,

z +z

)

2 1/ 2

=

2η z Δx ∑ 4πε 0 i xi2 + z 2

If we now let Δ x → dx, the sum becomes an integral ranging from x = Δ L/2 to x = L/2. This gives L/2

2η z dx η z ⎡ 1 −1 ⎛ x ⎞ ⎤ η ⎡ −1 ⎛ L ⎞ ⎛ −L ⎞⎤ = tan ⎜ ⎟ ⎥ = tan ⎜ ⎟ − tan −1 ⎜ ⎟⎥ ⎢ ⎢ 2 2 ∫ 4πε 0 − L / 2 x + z 2πε 0 ⎣ z ⎝ z ⎠ ⎦ − L / 2 2πε 0 ⎣ ⎝ 2z ⎠ ⎝ 2z ⎠⎦ L/2

Ez =

From trigonometry, tan −1 ( −φ ) = − tan −1 (φ ) . So finally,

Ez =

G η η ⎛ L⎞ ⎛ L⎞ tan −1 ⎜ ⎟ ⇒ E = tan −1 ⎜ ⎟ kˆ 2 z πε 0 πε ⎝ ⎠ ⎝ 2z ⎠ 0

(b) As z → 0 m,

G η π ˆ η ˆ ⎛ L⎞ π tan −1 ⎜ ⎟ → ⇒E→ k= k 2ε 0 2 πε 0 2 ⎝ 2z ⎠

This is the electric field due to a plane as you can see from Equation 27.26. We obtain this result because in the limit as z → 0 m, the dimension L becomes extremely large. As z → ∞,

G L η L ˆ 1 2λ ˆ ⎛ L⎞ tan −1 ⎜ ⎟ → ⇒E→ k= k 2 2 z z z πε 2 4 πε ⎝ ⎠ 0 0 z

where we have used ηL = λ as the charge per unit length of the sheet. This is the electric field due to a long, charged wire. We obtain this result because for z >> L, the infinitely long sheet “looks” like an infinite line charge. (c) The following table shows the field strength Ez in units of η/ε0 for selected values of z in units of L. A graph of Ez is shown in the figure above. z L 0 0.25 0.50 1.0 2.0 3.0 4.0

Ez

η ε0

0.5 0.35 0.25 0.15 0.08 0.05 0.04

27.70. Model: Model the infinitely long sheet of charge with width L as a uniformly charged sheet. Visualize:

Solve: (a) Consider a point on the x-axis at a distance d from the center of the sheet of charge. (We’ll call this distance d to begin with, rather than x, to avoid confusion with x as the integration variable.) Once again, let the sheet of charge be divided into small strips of width Δ x. Each strip has a linear charge density λ = ηΔ x and acts like a long, charged wire. Strip i is at distance ri = d − xi from the point of interest, so it contributes the small field

G 2λ ˆ 2ηΔx Ei = i = iˆ 4πε 0 ri 4πε 0 ( d − xi )

G In this situation all fields Ei point in the same direction. Their x-components all add to give a net field in the +x-direction:

Ex = ∑ ( Ei ) x = i

2η Δx ∑ 4πε 0 i ( d − xi )

We’ll replace the sum with an integral from x = −L/2 to x = +L/2, giving L/2 2η dx 2η 2η 2η ⎛ 2d + L ⎞ = ⎡ − ln ( d − x ) ⎦⎤ − L / 2 = ⎡ − ln ( d − L 2 ) + ln ( d + L 2 ) ⎦⎤ = ln ⎜ ⎟ 4πε 0 − L∫/ 2 ( d − x ) 4πε 0 ⎣ 4πε 0 ⎣ 4πε 0 ⎝ 2d − L ⎠ L/2

Ex =

Now that the integration is complete, we can note that d really is the x-coordinate of the point of interest. Substituting x for d and changing to vector form, we end up with G 2η ⎛ 2 x + L ⎞ ˆ ln ⎜ Enet = ⎟i 4πε 0 ⎝ 2 x − L ⎠ (b) If the point is very distant compared to width of the sheet of charge (x >> L), then the sheet of charge looks like a line of charge with linear density (charge per unit length) λ = ηL. Write

Ex =

2η ⎛ 1 + L/2 x ⎞ 2η ln ⎜ [ln(1 + L/2 x ) − ln(1 − L/2 x )] ⎟= 4πε 0 ⎝ 1 − L/2 x ⎠ 4πε 0

If x >> L, then L/2x << 1 and we can use the approximation ln(1 + u) ≈ u if u << 1. Thus Ex ≈

2η ⎡ L ⎛ L ⎞ ⎤ 2η L 2λ − ⎜ − ⎟⎥ = = ⎢ 4πε 0 ⎣ 2 x ⎝ 2 x ⎠ ⎦ 4πε 0 x 4πε 0 x

This is the field of a line of charge with λ = ηL, as we expected.

(c) The following table shows the field strength Ex in units of 2η/4πε0 for selected values of z in units of L. A graph of Ex is shown in the figure above.

z L

0.75 1.0 1.5 2.0 3.0 4.0

Ex

2η 4πε 0

1.61 1.10 0.69 0.51 0.34 0.25

27.71. Model: The electric field is uniform inside the capacitor, so constant-acceleration kinematic equations apply to the motion of the electron. Visualize: Please refer to Figure CP27.71. The condition for the electron to not hit the negative plate is that its vertical velocity should just become zero as the electron reaches the plate. Solve: The force on the electron inside the capacitor is G G G G G qE F = ma = qE ⇒ a = m

⇒ ay =

− (1.60 ×10−19 C )(1.0 ×104 N/C ) 9.11×10−31 kg

= −1.756 × 1015 m/s 2

The initial velocity v0 has two components: v0 x = v0 cos 45° and v0 y = v0 sin 45°. Because the electric field inside

the capacitor is along the +y-direction, the electron has a negative acceleration that reduces the vertical velocity. We require v1y = 0 m/s if it is not to hit the plate. Using kinematics,

v12y = v02y + 2a y ( y1 − y0 ) ⇒ ( 0 m/s ) = v02y + 2a y Δy 2

⇒ v0 y = −2a y Δy = −2 ( −1.756 × 1015 m/s 2 ) ( 0.02 m ) = 8.381 × 106 m/s

⇒ v0 =

8.381× 106 m/s = 1.19 ×107 m/s sin 45°

27.72. Model: The two electrodes form a parallel-plate capacitor. The electric field inside the electrodes is uniform, so the electrons have constant acceleration. Visualize:

Solve: (a) The ink drops are deflected up or down as they experience an electric field between the two parallel electrodes. The electric field exerts a force on the ink drops which is F = qE = may ⇒ E = may /q We therefore need to determine the mass m, the charge q, and the acceleration ay of the ink drops. We have 3⎤ ⎛ 4π 3 ⎞ ⎡ 4π m = ρV = ρ ⎜ r ⎟ = ( 800 kg/m3 ) ⎢ (15 ×10−6 m ) ⎥ = 1.131×10−11 kg ⎝ 3 ⎠ ⎣ 3 ⎦

q = ( 8 × 105 )(1.60 × 10−19 C ) = 1.28 × 10−13 C

At maximum deflection, the drop’s angle upon exiting the plates must be tan θ =

vy vx

⎛ 3 mm ⎞ =⎜ ⎟ = ±0.15 ⇒ v y = ( ±0.15) vx = ( ±0.15)( 20 m/s ) = ±3.0 m/s ⎝ 2.0 cm ⎠

From the kinematic equation v1 y = v0 y + a y ( t1 − t0 ) , ay =

v1 y − v0 y t1 − t0

⇒ ay =

±3.0 m/s − 0 m/s t1 − t0

We can obtain t1 − t0 from the x-motion between the plates as follows:

1 2 x1 = x0 + v0 x ( t1 − t0 ) + ax ( t1 − t0 ) ⇒ x1 − x0 = 6.0 mm = v0 x ( t1 − t0 ) + 0 m 2 ⇒ t1 − t0 =

±3.0 m/s 6.0 × 10 −3 m 6.0 × 10−3 m = ±1.0 ×104 m/s 2 = = 3.0 × 10−4 s ⇒ a y = 3.0 ×10−4 s 20 m/s v0 x

We are now in a position to obtain the field strength E from the equation

E=

ma y q

=

(1.131×10

−11

kg )(1.0 ×104 m/s 2 )

1.28 ×10−13 C

= 8.84 × 105 N/C

The electric field strength is 8.8×105 N/C. (b) The surface charge density is

η = ε 0 E = (8.85 ×10−12 C 2 /Nm 2 )( 8.84 ×105 N/C ) = 7.82 × 10−6 C/m 2 The area of each plate is

Aplate = ( 6.0 ×10−3 m )( 4.0 ×10−3 m ) = 24.0 × 10−6 m 2

Thus the charge on the plates is Q = ±η Aplate = ±1.88 × 10−10 C = ±0.188 nC. Assess: Because of the large acceleration due to the electric field, the acceleration due to gravity can be ignored in the calculations.

27.73. Model: The orbital motion of the positron-electron system about their center of mass is due to the electric force between the positron and the electron. Solve: The distance between the charges is twice the radius of their orbits about the center of mass. The force that causes the circular motion is

F=

2 1 qpositron qelectron melectron v 2 mpositron v m 4π 2 f 2 r 2 = = = electron 2 4πε 0 r r r ( 2r )

where we used v = 2πrf. The frequency is

⎛ 1 ⎞ qpositron qelectron f = ⎜ = ⎟ 2 3 ⎝ 4πε 0 ⎠ melectron 16π r

( 9.0 ×10 ( 9.1×10

9

N m 2 /C 2 )(1.60 ×10−19 C )

−31

2

kg )16π 2 ( 0.50 ×10−9 m )

3

= 1.13 × 1014 Hz

27.74. Model: radius R. Visualize:

Solve:

The electric field is that of a positively charged ring. The ring is thin and the charge lies along circle of

(a) From Example 27.5, the electric field on the axis of a ring of radius R is

Ez =

1

zQ

4πε 0 ( z + R 2 )3/ 2 2

The force on negative charge −q is Fz = −qEz = −

1 qQz 4π 0 ( z 2 + R 2 )3/ 2

The meaning of the negative sign is that the force on the electron points left when the displacement is to the right and to the right when the displacement is to the left. That is, the electric force tries to keep the charge at z = 0 m. (b) For small amplitude oscillations, that is, when z << R, the force is Fz = −

1 qQ ⎛ z2 ⎞ + z 1 ⎜ ⎟ 4πε 0 R 3 ⎝ R 2 ⎠

−3/ 2

=−

1 qQ ⎛ 3 z 2 …⎞ 1 qQ z ⎜1 − 2 + ⎟ ≈ − z 3 4πε 0 R ⎝ 2 R 4πε 0 R 3 ⎠

This is simply Hooke’s law Fz = − kz where the “spring constant” is

k=

qQ 4πε 0 R3

A particle that obeys Hooke’s law undergoes simple harmonic motion with a frequency given by f =

1 2π

k 1 = m 2π

qQ 4πε 0 mR 3

(c) Substituting into the above expression,

f =

1 2π

(1.0 ×10

−13

C )(1.6 × 10−19 C )( 9.0 × 109 N/m 2 C2 )

( 9.11×10

−31

kg )(1.0 × 10−6 m )

3

= 2.0 × 1012 Hz

27-1

28.1. Visualize:

As discussed in Section 28.1, the symmetry of the electric field must match the symmetry of the charge distribution. In particular, the electric field of a cylindrically symmetric charge distribution cannot have a component parallel to the cylinder axis. Also, the electric field of a cylindrically symmetric charge distribution cannot have a component tangent to the circular cross section. The only shape for the electric field that matches the symmetry of the charge distribution with respect to (i) translation parallel to the cylinder axis, (ii) rotation by an angle about the cylinder axis, and (iii) reflections in any plane containing or perpendicular to the cylinder axis is the one shown in the figure.

28.2. Visualize:

The object has spherical symmetry, so the electric field is radial.

28.3. Visualize:

Figure 28.6 shows the electric field for an infinite plane of charge. For two parallel planes, this is the only shape of the electric field vectors that matches the symmetry of the charge distribution.

28.4. Model: The electric flux “flows” out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure EX28.4. Let A be the area in m2 of each of the six faces of the cube. G G Solve: The electric flux is defined as Φ e = E ⋅ A = EA cosθ , where θ is the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is

Φ out = ( 20 N/C + 20 N/C + 10 N/C ) A cos0° = ( 50 A ) N m 2 /C Similarly, the electric flux into the closed cube surface is Φ in = (15 N/C + 15 N/C + 15 N/C ) A cos180° = − ( 45 A ) N m 2 /C The net electric flux is ( 50 A ) N m 2 /C − ( 45 A ) N m 2 /C = ( 5 A ) N m 2 /C. Since the net electric flux is positive (i.e., outward), the closed box contains a positive charge.

28.5. Model: The electric flux “flows” out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure EX28.5. Let A be the area of each of the six faces of the cube. G G Solve: The electric flux is defined as Φ e = E ⋅ A = EA cosθ , where θ is the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is

Φ out = (10 N/C + 10 N/C + 20 N/C + 5 N/C ) A cos0° = ( 45 A ) N m 2 /C Similarly, the electric flux into the closed cube surface is Φ in = (15 N/C + 20 N/C ) A cos (180° ) = − ( 35 A ) N m 2 /C Hence, Φout + Φin = 10 N m2/C. Since the net electric flux is positive (i.e., outward), the closed box contains a positive charge.

28.6. Model: The electric flux “flows” out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure EX28.6. Let A be the area of each of the six faces of the cube. G G Solve: The electric flux is defined as Φ e = E ⋅ A = EA cosθ , where θ is the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is

Φ out = ( 20 N/C + 10 N/C + 10 N/C ) A cos0° = ( 40 A ) N m 2 /C Similarly, the electric flux into the closed cube surface is Φ in = ( 20 N/C + 15 N/C ) A cos180° = − ( 35 A ) N m 2 /C

Because the cube contains negative charge, Φout + Φin must be negative. This means Φout + Φin + Φunknown < 0 N m2/C. Therefore,

( 40 A) N m 2 /C + ( −35 A) N m 2 /C + Φ unknown < 0 N m 2 / C ⇒ Φ unknown < ( −5 A ) N m 2 /C That is, the unknown vector points into the front face of the cube and its field strength is greater than 5 N/C.

28.7. Model: The electric flux “flows” out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure EX28.7. Let A be the area of each of the six faces of the cube. G G Solve: The electric flux is defined as Φ e = E ⋅ A = EA cosθ , where θ is the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is

Φ out = ( 20 N/C + 10 N/C ) A cos0° = ( 30 A ) N m 2 /C Similarly, the electric flux into the closed cube surface is Φ in = ( 20 N/C + 15 N/C + 10 N/C ) A cos180° = − ( 45 A ) N m 2 /C

Because the cube contains positive charge, Φout + Φin must be positive. This means Φout + Φin + Φunknown > 0 N m2/C. Therefore,

( 30 A) N m 2 /C + ( −45 A) N m 2 /C + Φ unknown > 0 N m 2 / C ⇒ Φ unknown > (15 A ) N m 2 /C That is, the unknown vector points out of the front face of the cube and its field strength is greater than 15 N/C.

28.8. Model: The electric flux “flows” out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure EX28.8. Let A be the area of each of the six faces of the cube. G G Solve: The electric flux is defined as Φ e = E ⋅ A = EA cosθ , where θ is the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is

Φ out = (15 N/C + 15 N/C + 10 N/C ) A cos0° = ( 40 A ) N m 2 /C Similarly, the electric flux into the closed cube surface is Φ in = ( 20 N/C + 15 N/C ) A cos180° = − ( 35 A ) N m 2 /C Because the cube contains no charge, the total flux is zero. This means Φin + Φout + Φunknown = 0 N m2/C ⇒ Φ unknown = − ( 5 A ) N m 2 /C

The unknown electric field vector on the front face points in, and the electric field strength is 5 N/C.

28.9. Model: The electric field is uniform over the entire surface. Visualize: Please refer to Figure EX28.9. The electric field vectors make an angle of 30° with the planar surface. Because the normal nˆ to the planar surface is at an angle of 90° with the surface, the angle between nˆ G and E is θ = 60°. Solve: The electric flux is G G Φ e = E ⋅ A = EA cosθ = ( 200 N/C ) (1.0 ×10−2 m 2 ) cos 60° = 1.0 N m 2 /C

28.10. Model: The electric field is uniform over the entire surface. Visualize: Please refer to Figure EX28.10. The electric field vectors make an angle of 30° below the surface. G Because the normal nˆ to the planar surface is at an angle of 90° relative to the surface, the angle between nˆ and E is θ = 120°. Solve: The electric flux is G G 2 Φ e = E ⋅ A = EA cosθ = ( 200 N/C ) (15 ×10−2 m ) cos120° = −2.3 N m 2 /C

28.11. Model: The electric field is uniform over the entire surface. Visualize: Please refer to Figure EX28.11. The electric field vectors make an angle of 60° above the surface. G Because the normal nˆ to the planar surface is at an angle of 90° relative to the surface, the angle between nˆ and E is θ = 30°. Solve: The electric flux is G G Φ e = E ⋅ A = EA cosθ ⇒ 25 N m 2 /C = E (1.0 ×10−2 m 2 ) cos30° ⇒ E = 2.9×103 N/C

28.12. Model: The electric field is uniform over the rectangle in the xy plane. Solve:

(a) The area vector is perpendicular to the xy plane. Thus G A = ( 2.0 cm × 3.0 cm ) kˆ = ( 6.0 × 10−4 m 2 ) kˆ

The electric flux through the rectangle is G G Φ e = E ⋅ A = 50iˆ + 100kˆ ⋅ 6.0 × 10−4 kˆ N m 2 /C = 6.0 × 10−2 N m 2 /C

(

(b) The electric flux is

)(

)

G G Φ e = E ⋅ A = 50iˆ + 100ˆj ⋅ 6.0 × 10−4 kˆ N m 2 /C = 0 N m 2 / C

(

)(

)

G Assess: In (b), E is in the plane of the rectangle. That is why the flux is zero.

28.13. Model: The electric field over the rectangle in the xz plane is uniform. Solve:

(a) The area vector is perpendicular to the xz plane. Thus G A = ( 2.0 cm × 3.0 cm ) ˆj = ( 6.0 × 10−4 m 2 ) ˆj

The electric flux through the rectangle is G G Φ e = E ⋅ A = 50iˆ + 100kˆ N/C ⋅ ( 6.0 × 10−4 m 2 ) ˆj

(

= ( 300 × 10 (b) The flux is

)

−4

( )

( )

N m /C ) iˆ ⋅ ˆj + ( 600 × 10−4 N m 2 /C ) kˆ ⋅ ˆj = 0 N m 2 /C 2

G G Φ e = E ⋅ A = 50iˆ + 100ˆj N/C ⋅ ( 6.0 × 10−4 m 2 ) ˆj

(

)

( )

= ( 300 ×10−4 N m 2 /C ) iˆ ⋅ ˆj + ( 600 × 10−4 N m 2 /C ) ˆj ⋅ ˆj = 0 N m /C+ ( 6.0 ×10 2

−2

N m 2 /C ) = 6.0 ×10−2 N m 2 /C

28.14. Model: The electric field over the circle in the xy plane is uniform. Solve:

The area vector of the circle is G 2 A = π r 2 kˆ = π ( 0.015 m ) kˆ =

( 7.07 ×10

−4

m 2 ) kˆ

Thus, the flux through the area of the circle is G G Φ e = E ⋅ A = 1500iˆ + 1500 ˆj + 1500 kˆ N/C ⋅ ( 7.07 ×10−4 m 2 ) kˆ

(

)

Using iˆ ⋅ kˆ = ˆj ⋅ kˆ = 0 and kˆ ⋅ kˆ = 1,

Φ e = (1500 N/C ) ( 7.07 ×10−4 m 2 ) = 1.06 N m 2 /C

G

28.15. Model: The electric field over the two faces of the box perpendicular to E is uniform. Visualize:

Solve: The electric field is parallel to four sides of the box, so the electric fluxes through these four surfaces are zero. The field is out of the top surface, so Φtop = EA. The field is into the bottom surface, so Φbottom = –EA. Thus the net flux is Φe = 0 N m2/C2.

28.16. Model: The electric field through the two cylinders is uniform. Visualize: Please refer to Figure EX28.16. Let A = π R2 be the area of the end of the cylinder and let E be the electric field strength. Solve:

→

(a) There’s no flux through the side walls of the cylinder because E is parallel to the wall. On the right →

end, where E points outward, Φright = EA = π R2E. The field points inward on the left, so Φleft = –EA = –π R2E. Altogether, the net flux is Φe = 0 N m2/C2. →

(b) The only difference from part (a) is that E points outward on the left end, making Φleft = EA = π R2E. Thus the net flux through the cylinder is Φe = 2π R2E.

28.17. Visualize:

For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e = Qin ε 0 .

28.18. Visualize:

For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e =

Qin ε 0 .

28.19. Visualize: Please refer to Figure EX28.19. Solve: For any closed surface that encloses a total charge Qin, the net electric flux through the surface is Φ e = Qin ε 0 . We can write three equations from the three closed surfaces in the figure:

ΦA = −

q

ε0

=

q1 + q3

ε0

⇒ q1 + q3 = −q ΦC =

−2q

ε0

=

q2 + q3

ε0

ΦB =

3q

ε0

q1 + q2

ε0

⇒ q2 + q3 = −2q

Subtracting third equation from the first, q1 – q2 = +q Adding second equation to this equation, 2q1 = +4q ⇒ q1 = 2q That is, q1 = +2q, q2 = +q, and q3 = −3q.

=

⇒ q1 + q2 = 3q

28.20. Visualize: Please refer to Figure EX28.20. For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e = Qin ε 0 . For the closed surface of the torus, Qin includes only the −1 nC charge. So, the net flux through the torus is due to this charge: Φe =

This is inward flux.

−1.0 ×10−9 C = −113 N m 2 /C 8.85 ×10−12 C2 /Nm 2

28.21. Visualize: Please refer to Figure EX28.21. For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e = Qin ε 0 . The cylinder encloses the +1 nC charge only as both the +100 nC and the −100 nC charges are outside the cylinder. Thus, Φe =

This is outward flux.

1.0 ×10−9 C = 113 N m 2 /C. 8.85 ×10−12 C2 /Nm 2

28.22. Solve: For any closed surface enclosing a total charge Qin, the net electric flux through the surface is Φe =

Qin

ε0

⇒ Qin = ε 0Φ e = ( 8.85 ×10−12 C2 /Nm 2 )( −1000 N m 2 /C ) = −8.85 nC

28.23. Solve: For any closed surface that encloses a total charge Qin, the net electric flux through the surface is

Φe =

Qin

ε0

( 55.3 ×10 )( −1.60 ×10 6

=

8.85 × 10

−12

2

−19

C /N m

2

C)

= −1.00 N m 2 /C

28.24. Model: A copper penny is a conductor. The excess charge on a conductor resides on the surface. Solve:

The electric field at the surface of a charged conductor is

G ⎛η ⎞ Esurface = ⎜ , perpendicular to surface ⎟ ⎝ ε0 ⎠

⇒ η = ε 0 Esurface = ( 8.85 ×10−12 C 2 /Nm 2 ) ( 2000 N/C ) = 17.7 ×10−9 C/m 2

28.25. Model: The excess charge on a conductor resides on the outer surface. Solve:

The electric field at the surface of a charged conductor is

G ⎛η ⎞ Esurface = ⎜ , perpendicular to surface ⎟ ⎝ ε0 ⎠

⇒ η = ε 0 Esurface = ( 8.85 ×10−12 C 2 /Nm 2 )( 3.0 ×106 N/C ) = 2.7 ×10−5 C/m 2 Assess: It is the air molecules just above the surface that “break down” when the E-field becomes strong enough to accelerate stray charges to approximately 15 eV between collisions, thus causing collisional ionization. It does not make any difference whether E points toward or away from the surface.

28.26. Model: The excess charge on a conductor resides on the outer surface. Visualize: Please refer to Figure EX28.26. Solve: Point 1 is at the surface of a charged conductor, hence

G ⎛η ⎞ ( 5.0 ×1010 )(1.60 ×10−19 C m2 ) = 904 N/C Esurface = ⎜ , perpendicular to surface ⎟ ⇒ Esurface = 8.85 ×10−12 C2 /N m 2 ⎝ ε0 ⎠ At point 2 the electric field strength is zero because this point lies inside the conductor. The electric field strength at point 3 is zero because there is no excess charge on the interior surface of the box. This can be quickly seen by considering a Gaussian surface just inside the interior surface of the box as shown in Figure 28.31.

28.27. Model: The copper plate is a conductor. The excess charge resides on the surface of the plate. Solve: (a) One-half of the electrons are located on the top surface and one-half on the bottom surface of the copper plate, so the surface charge density is

( 0.5 ×10 )(1.60 ×10 10

η=

−19

0.1 m × 0.1 m

C)

= 80 ×10−9 C/m 2

Thus, the electric field at the surface of the plate is E=

η 80 ×10−9 C/m 2 = = 9.0 × 103 N/C ε 0 8.85 ×10−12 C2 /N m 2

Because the charge on the plate is negative, the direction of the electric field is toward the plate. (b) E = 0 N/C because the electric field within a conductor is zero. (c) The electric field E = 9.0×103 N/C, toward the plate.

28.28. Visualize: Please refer to Figure EX28.28. Solve: For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e = Qin ε 0 . In the present case, the conductor is neutral and there is a point charge Q inside the cavity. Thus Qin = Q and the flux is

Φe =

Q

ε0

28.29. Model: The electric field over the four faces of the cube is uniform. Visualize: Please refer to Figure P28.29. G G G Solve: (a) The electric flux through a surface area A is Φ e = E ⋅ A = EA cosθ , where θ is the angle between G the electric field and the vector A that points outward from the surface. Thus

Φ1 = EA cosθ1 = ( 500 N/C ) ( 9.0 ×10−4 m 2 ) cos150° = −0.39 N m 2 /C Φ 2 = EA cosθ 2 = ( 500 N/C ) ( 9.0 ×10−4 m 2 ) cos60° = 0.23 N m 2 /C

Φ 3 = EA cosθ 3 = ( 500 N/C ) ( 9.0 ×10−4 m 2 ) cos30° = 0.39 N m 2 /C Φ 4 = EA cosθ 4 = ( 500 N/C ) ( 9.0 × 10−4 m 2 ) cos120° = −0.23 N m 2 /C (b) The net flux through these four sides is Φ e = Φ1 + Φ 2 + Φ 3 + Φ 4 = 0 N m 2 /C. Assess: The net flux through the four faces of the cube is zero because there is no enclosed charge.

28.30. Model: The electric field over the five surfaces is uniform. Visualize: Please refer to Figure P28.30. G G G Solve: The electric flux through a surface area A is Φ e = E ⋅ A = EA cosθ where θ is the angle between the electric field and a line perpendicular to the plane of the surface. The electric field is perpendicular to side 1 and is G G parallel to sides 2, 3, and 5. Also the angle between E and A4 is 60°. The electric fluxes through these five surfaces are

Φ1 = E1 A1 cosθ1 = ( 400 N/C )( 2 m × 4 m ) cos (180° ) = −3200 N m 2 /C Φ 2 = E2 A2 cos90° = Φ 3 = Φ 5 = 0 N m 2 /C

Φ 4 = E4 A4 cosθ 4 = ( 400 N/C ) ( ( 2 m sin 30° ) × 4 m ) cos60° = +6400 N m 2 /C Assess: Because the flux into these five faces is equal to the flux out of the five faces, the net flux is zero, as we found.

28.31. Model: Because the tetrahedron contains no charge, the net flux through the tetrahedron is zero. Visualize:

Solve: (a) The area of the base of the tetrahedron is 43 a 2 , where a = 20 cm is the length of one of the sides. Because the base of the tetrahedron is parallel to the ground and the vertical uniform electric field passes upward G G through the base, the angle between E and A is 180°. Thus,

G G 3 2 3 2 Φ base = E ⋅ Abase = EAbase cos180° = − E a = − ( 200 N/C ) ( 0.20 m ) = −3.46 N m 2 /C 4 4 The electric flux through the base is −3.5 N m2/C. G (b) Since E is perpendicular to the base, the other three sides of the tetrahedron share the flux equally. Because the tetrahedron contains no charge,

Φ net = Φ base + 3Φ side = 0 N m 2 / C ⇒ −3.46 N m 2 /C + 3Φ side = 0 N m 2 / C

⇒ Φ side = 1.15 N m 2 /C

28.32. Solve: (a) When centered at the origin the sphere encloses both q1 and q2. For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φe =

Qin

ε0

=

( −4Q + 2Q ) = −2Q ε0

ε0

(b) When centered at x = 2a, the sphere encloses only q2 located at x = +a. The net electric flux through the closed surface is thus

Φe =

Qin

ε0

=

2Q

ε0

28.33. Solve: For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e = Qin ε 0 . The flux through the top surface of the cube is one-sixth of the total: Φ e surface =

Qin 10 ×10−9 C = = 188 N m 2 /C 6ε 0 6 ( 8.85 ×10−12 C2 /N m 2 )

28.34. Solve: For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is Φ e = Qin ε 0 . The total flux through the cube is Φ e = 6 (100 N m 2 /C ) =

Qin

ε0

⇒ Qin = ( 600 N m 2 /C )( 8.85 × 10−12 C 2 /N m 2 ) = 5.31×10−9 C = 5.31 nC

28.35. Solve: (a) The electric field is G 2 E = ( 5000 r 2 ) rˆ N/C = 5000 ( 0.20 ) rˆ N/C = 200rˆ N/C So the electric field strength is 200 N/C. 2 (b) The area of the surface is Asphere = 4π ( 0.20 m ) = 0.5027 m 2 . Thus, the electric flux is

G G Φ e = AE ⋅ dA = EAsphere = ( 200 N/C ) ( 0.5027 m 2 ) = 100.5 N m 2 /C ≈ 101 N m 2 /C (c) Because Φ e = Qin ε 0 , Qin = ε0Φe = (8.85 × 10−12 C2/N m2)(100.5 N m2/C) = 8.9 × 10−10 C

28.36. Solve: (a) The electric field is G ⎛ 200 ⎞ E =⎜ ⎟ rˆ N/C = 2000rˆ N/C ⎝ 0.10 ⎠ So the electric field strength is 2000 N/C. 2 (b) The area of the spherical surface is Asphere = 4π ( 0.10 m ) = 0.1257 m 2 . Hence, the flux is G G Φ e = AE ⋅ dA = EAsphere = ( 2000 N/C ) ( 0.1257 m 2 ) = 250 N m 2 /C (c) Because Φ e = Qin ε 0 , Qin = ε0Φe = (8.85 × 10−12 C2/N m2)(250 N m2/C) = 2.2 × 10−9 C = 2.2 nC

28.37. Model: The excess charge on a conductor resides on the outer surface. Visualize:

Solve:

(a) Consider a Gaussian surface surrounding the cavity just inside the conductor. The electric field →

inside a conductor in electrostatic equilibrium is zero, so E is zero at all points on the Gaussian surface. Thus Φe = 0. Gauss’s law tells us that Φe = Qin/ε0, so the net charge enclosed by this Gaussian surface is Qin = Qpoint + Qwall = 0.

We know that Qpoint = +100 nC, so Qwall = –100 nC. The positive charge in the cavity attracts an equal negative charge to the inside surface. (b) The conductor started out neutral. If there is –100 nC on the wall of the cavity, then the exterior surface of the conductor was initially +100 nC. Transferring –50 nC to the conductor reduces the exterior surface charge by 50 nC, leaving it at +50 nC. Assess: The electric field inside the conductor stays zero.

28.38. Model: The ball is uniformly charged. The charge distribution is spherically symmetric. Visualize:

The figure shows spherical Gaussian surfaces with radii r = 5 cm, 10 cm, and 20 cm. These surfaces match the G symmetry of the charge distribution. So, E is perpendicular to the Gaussian surface and the value of the field strength is the same at all points on the surface. Solve: (a) Because the ball is uniformly charged, its charge density is

ρ=

3 ( 80 ×10−9 C ) Q Q =4 3= = 2.387 ×10−6 C/m3 ≈ 2.4 ×10−6 C/m3 3 V 3πr 4π ( 0.20 m )

(b) Once again, because the ball is uniformly charged,

⎛ 4π 3 ⎞ ⎛ 4π 3 ⎞ Q = ρV = ρ ⎜ r ⎟ = ( 2.387 ×10−6 C/m3 ) ⎜ r ⎟ = (1.00 ×10−5 C/m3 ) r 3 3 3 ⎝ ⎠ ⎝ ⎠ When r = 5 cm, Q5 = (1.00 ×10−5 C/m3 ) ( 0.050 m ) = 1.25 nC. When r = 10 cm and 20 cm, Q10 = 10.0 nC and 3

Q20 = 80 nC.

G G (c) Gauss’s law is Φ e = AE ⋅ dA = EA = Qin ε 0 . For the 5.0-cm-radius Gaussian spherical surface,

Q5

ε0

= E5 A5 ⇒ E5 =

Q5 1.25 × 10−9 C = = 4.5 × 103 N/C ε 0 A5 (8.85 ×10−12 C 2 /Nm 2 ) ⎡ 4π ( 0.050 m )2 ⎤ ⎣ ⎦

Similarly, we can apply Gauss’s law to the 10 cm-radius and 20 cm-radius spherical surfaces and obtain E10 = 9.0 ×103 N/C and E20 = 1.80 × 103 N/C. Assess: We note that E20 = 2E10 = 4E5. This result is consistent with the result of Example 28.4 according to which

Einside =

1

Q r 4πε 0 R3

28.39. Model: The excess charge on a conductor resides on the outer surface. The field inside, outside, and within the hollow metal sphere has spherical symmetry. Visualize:

The figure shows spherical Gaussian surfaces with radii r ≤ a, a < r < b, and r ≥ b. These surfaces match the symmetry of the charge distribution. Solve: (a) For r ≤ a, Gauss’s law is G G Q +Q Φ e = AE ⋅ dA = in =

ε0

ε0

Notice that the electric field is everywhere perpendicular to the spherical surface. Because of the spherical symmetry of the charge, the electric field magnitude E is the same at all points on the Gaussian surface. Thus,

Φ e = EAsphere = E ( 4π r 2 ) =

Q

ε0

⇒ E=

G 1 Q ⇒E= rˆ 4πε 0 r 4πε 0 r 2 Q

2

where we made use of the fact that E is directed radially outward. The field depends only on the enclosed charge, not on the charge on the outer sphere. For a < r < b, Gauss’s law is G G Q Φ e = AE ⋅ dA = EAsphere = E ( 4π r 2 ) = in

ε0

Here Qin = 0 C. It is not +Q, because the charge in the cavity polarizes the metal sphere in such a way that E = 0 in the metal. Thus a charge −Q moves to the inner surface. Because the hollow sphere has a net charge of +2Q, the exterior surface now has a charge of +3Q. Thus, the electric field E = 0 N/C. For r ≥ b, Qin = Qexterior + Qinterior + Qcavity = +3Q + (−Q) + (+Q) = +3Q Gauss’s law applied to the Gaussian surface at r ≥ b yields: G G G Q +3Q 1 ⎛ 3Q ⎞ 1 ⎛ 3Q ⎞ Φ e = AE ⋅ dA = EAsphere = E ( 4π r 2 ) = in = ⇒ E= ⎜ 2 ⎟ ⇒E= ⎜ ⎟ rˆ 4πε 0 ⎝ r ⎠ 4πε 0 ⎝ r 2 ⎠ ε0 ε0 (b) As determined in part (a), the inside surface of the hollow sphere has a charge of −Q, and the exterior surface of the hollow sphere has a charge of +3Q. Assess: The hollow sphere still has the same charge +2Q as given in the problem, although the sphere is polarized.

28.40. Model: The excess charge on a conductor resides on the outer surface. The charge distribution on the two spheres is assumed to have spherical symmetry. Visualize: Please refer to Figure P28.40. The Gaussian surfaces with radii r = 8 cm, 10 cm, and 17 cm match G the symmetry of the charge distribution. So, E is perpendicular to these Gaussian surfaces and the field strength has the same value at all points on the Gaussian surface. G G Solve: (a) Gauss’s law is Φ e = AE ⋅ dA = Qin ε 0 . Applying it to a Gaussian surface of radius 8 cm,

Qin = −ε 0 EAsphere = − ( 8.85 ×10−12 C2 /N m 2 ) (15,000 N/C ) ⎡ 4π ( 0.08 m ) ⎤ = −1.07 × 10−8 C ⎣ ⎦ Because the excess charge on a conductor resides on its outer surface and because we have a solid metal sphere inside our Gaussian surface, Qin is the charge that is located on the exterior surface of the inner sphere. (b) In electrostatics, the electric field within a conductor is zero. Applying Gauss’s law to a Gaussian surface just inside the inside surface of the hollow sphere at r = 10 cm, G G Q Φ e = AE ⋅ dA = in ⇒ Qin = 0 C 2

ε0

That is, there is no net charge. Because the inner sphere has a charge of −1.07 × 10−8 C, the inside surface of the hollow sphere must have a charge of +1.07 × 10−8 C. (c) Applying Gauss’s law to a Gaussian surface at r = 17 cm, G G 2 Qin = ε 0 AE ⋅ dA = ε 0 EAsphere = ( 8.85 ×10−12 C 2 /Nm 2 ) (15,000 N/C ) 4π ( 0.17 m ) = 4.82 ×10−8 C This value includes the charge on the inner sphere, the charge on the inside surface of the hollow sphere, and the charge on the exterior surface of the hollow sphere due to polarization. Thus, Qexterior hollow + (1.07 × 10−8 C ) + ( −1.07 ×10−8 C ) = 4.82 × 10−8 C ⇒ Qexterior hollow = 4.82 ×10−8 C

28.41. Model: The charge distribution at the surface of the earth is assumed to be uniform and to have spherical symmetry. Visualize:

G Due to the symmetry of the charge distribution, E is perpendicular to the Gaussian surface and the field strength has the same value at all points on the surface. G G Solve: Gauss’s law is Φ e = AE ⋅ dA = Qin ε 0 . The electric field points inward (negative flux), hence Qin = −ε 0 EAsphere = − ( 8.85 × 10−12 C 2 /N m 2 ) (100 N/C ) 4π ( 6.37 ×106 m ) = −4.51× 105 C 2

G

28.42. Model: The electric field inside the box is 0. Visualize: Please refer to Figure 28.32. Solve: The charge on the capacitor plates produces a uniform electric field pointing to the right. When the metal G G box is added, the field inside the box is Enet = 0 because there is no charge enclosed within the box. The principle G G G G of superposition tells us that Enet = Ecapacitor + Ebox , where Ebox is the electric field due to the charge on the

exterior surface of the box. The net surface charge is zero, because the box as a whole is neutral, but the box is polarized by the capacitor’s electric field to where the left end of the box is negative and the right end is positive. G G G Since Enet = 0, we’re led to conclude that the field due to the charge on the surface of the box is Ebox = – G Ecapacitor = a uniform field pointing to the left. Note that this is the electric field inside the box due to the surface

charge on the box. The field outside the box is much more complex. If we freeze the surface charges and remove the G G box from the capacitor, we’re left with Enet = Ebox = a uniform field pointing to the left. This is shown in the picture below.

28.43. Model: The hollow metal sphere is charged such that the charge distribution is spherically symmetric. Visualize:

The figure shows spherical Gaussian surfaces at r = 4 cm, at r = 8 cm, and at r = 12 cm. These surfaces match the G symmetry of the spherical charge distribution. So E is perpendicular to the Gaussian surface and the field strength has the same value at all points on the Gaussian surface. Solve: The charge on the inside surface is

Qinside = (−100 nC/m2) 4π (0.06 m)2 = −4.524 nC This charge is caused by polarization. That is, the inside surface can be charged only if there is a charge of +4.524 nC at the center which polarizes the metal sphere. Applying Gauss’s law to the 4.0-cm-radius Gaussian surface, which encloses the +4.524 nC charge,

G G Q Φ e = AE ⋅ dA = EAsphere = in

ε0

⇒E=

−9 9 2 2 Qin 4.524 ×10−9 C 4.524 ×10 C ( 9.0 × 10 N m /C ) = = = 2.54 × 104 N/C 2 Aε 0 4πε 0 ( 0.04 m )2 ( 0.04 m )

G At r = 4 cm, E = ( 2.54 ×104 N/C, outward ).

G G There is no electric field inside a conductor in electrostatic equilibrium. So at r = 8 cm, E = 0. Applying Gauss’s law to a 12.0-cm-radius sphere,

G G Q Φ e = AE ⋅ dA = EAsphere = in

ε0

The charge on the outside surface is

Qoutside = (+100 nC/m2)4π (0.10 m)2 = 1.257 × 10−8 C ⇒ Qin = Qoutside + Qinside + Qcenter = 1.257 × 10−8 C − 0.452 ×10−8 C + 0.452 × 10−8 C = 1.257 × 10−8 C ⇒E=

At r = 12 cm,

1.257 × 10−8 C Qin = = 7.86 × 103 N/C ε 0 A (8.85 ×10−12 C 2 /Nm 2 ) 4π ( 0.12 m )2

G E = ( 7.86 ×103 N/C, outward ) .

28.44. Model: The charged hollow spherical shell is assumed to have a spherically symmetric charge distribution. Visualize:

The figure shows two spherical Gaussian surfaces at r < R and r > R. These surfaces match the symmetry of the G spherical charge distribution. So, E is perpendicular to the Gaussian surface and the field strength has the same value at all points on the surface. Solve: (a) Gauss’s law applied to the Gaussian surface inside the sphere (r < R) is

G G Q +q 1 +q Φ e = AE ⋅ dA = EAsphere = in ⇒ E ( 4π r 2 ) = ⇒ E= 4πε 0 r 2 ε0 ε0

G ⎛ 1 +q ⎞ 1 q rˆ , outward ⎟ = ⇒E =⎜ 2 2 πε πε r 4 4 0 0 r ⎝ ⎠ (b) Gauss’s law applied to the Gaussian surface outside the sphere (r > R) is

G G Q −2q + q − q 1 +q ⇒ E= Φ e = AE ⋅ dA = − EAsphere = in = = 4πε 0 r 2 ε0 ε0 ε0 G ⎛ 1 q ⎞ 1 q ⇒E =⎜ rˆ , inward ⎟ = − 2 2 πε πε r 4 4 0 0 r ⎝ ⎠ Assess: A uniform spherical shell of charge has the same electric field at r > R as a point charge placed at the center of the shell. Additionally, the electric field for r < R is zero, meaning that the shell of charge exerts no electric force on a charged particle inside the shell.

28.45. Model: The hollow plastic ball has a charge uniformly distributed on its outer surface. This distribution leads to a spherically symmetric electric field. Visualize:

The figure shows Gaussian surfaces at r < R and r > R. Solve: (a) Gauss’s law for the Gaussian surface for r < R where Qin = 0 is G G Q Φ e = AE ⋅ dA = in = 0 N m2/C ⇒ E = 0 N/C

ε0

(b) Gauss’s law for the Gaussian surface for r > R is G G Q Q Q Φ e = AE ⋅ dA = EAsphere = in = ⇒ EAsphere = ⇒E=

ε0

ε0

ε0

Q

ε 0 Asphere

=

1 Q 4πε 0 r 2

Assess: A uniform spherical shell of charge has the same electric field at r > R as a point charge placed at the center of the sphere. Additionally, the shell of charge exerts no electric force on a charged particle inside the shell.

28.46. Model: The charge distributions of the ball and the metal shell are assumed to have spherical symmetry. Visualize:

The spherical symmetry of the charge distribution tells us that the electric field points radially inward or outward. We will therefore choose Gaussian surfaces to match the spherical symmetry of the charge distribution and the field. The figure shows four Gaussian surfaces in the four regions: r ≤ a, a < r < b, b ≤ r ≤ c and r > c. G G Solve: (a) Gauss’s law is Φ e = AE ⋅ dA = Qin ε 0 . Applying it to the region r ≤ a, where the charge is negative G so E points inward, we get

− EAsphere = − E ( 4π r 2 ) = Here ρ = −Q

4π 3

+ ρ ( 43π r 3 )

ε0

⇒E=

−ρr 3ε 0

a 3 is the charge density (C/m3). Thus

G ⎛ 1 Qr ⎞ 1 Qr E =⎜ rˆ , inward ⎟ = − 3 4πε 0 a 3 ⎝ 4πε 0 a ⎠ Applying Gauss’s law to the region a < r < b, G −Q 1 Q 1 Q − EAsphere = ⇒E= ⇒E=− rˆ 2 ε0 4πε 0 r 4πε 0 r 2 G G E = 0 in the region b ≤ r ≤ c because this is a conductor in electrostatic equilibrium. To apply Gauss’s law to the region r > c, we use Qin = −Q + 2Q = +Q. Thus,

EAsphere = (b)

Q

ε0

⇒E=

G 1 Q 1 Q ⇒E= rˆ 2 4πε 0 r 4πε 0 r 2

28.47. Model: The three planes of charge are infinite planes. Visualize:

From planar symmetry the electric field can point straight toward or away from the plane. The three planes are labeled as P (top), P′, and P″(bottom). Solve: From Example 28.6, the electric field of an infinite charged plane of charge density η is

Eplane =

η 2ε 0

⇒ EP = EP′′ =

η 4ε 0

=

EP′ 2

In region 1 the three electric fields are G G η ˆ η ˆ EP = − j EP ′ = j 4ε 0 2ε 0 G G Adding the three contributions, we get Enet = 0 N/C. In region 2 the three electric fields are

G Thus, Enet = (η 2ε 0 ) ˆj.

G η ˆ EP = j 4ε 0

G η ˆ EP ′ = j 2ε 0

G η ˆ EP′′ = − j 4ε 0

G −η ˆ EP′′ = j 4ε 0

In region 3,

G Thus, Enet = − (η 2ε 0 ) ˆj.

G η ˆ EP = j 4ε 0

G η ˆ EP ′ = − j 2ε 0

G η ˆ EP′′ = − j 4ε 0

In region 4,

G η ˆ EP = j 4ε 0 G G Thus Enet = 0 N/C.

G η ˆ EP ′ = − j 2ε 0

G η ˆ EP′′ = j 4ε 0

28.48. Model: The charge has planar symmetry, so the electric field must point toward or away from the slab. Furthermore, the field strength must be the same at equal distances on either side of the center of the slab. Visualize:

Choose Gaussian surfaces to be cylinders of length 2z centered on the z = 0 plane. The ends of the cylinders have area A. Solve: (a) For the Gaussian cylinder inside the slab, with z < z0, Gauss’s law is

G G G G G G G G Qin AE ⋅ dA = ∫ top E ⋅ dA + ∫ bottom E ⋅ dA + ∫ sides E ⋅ dA =

ε0

The field is parallel to the sides, so the third integral is zero. The field emerges from both ends, so the first two integrals are the same. The charge enclosed is the volume of the cylinder multiplied by the charge density, or Qin = ρV = ρ (2zA). Thus

Qin

ε0

=

G G G G ρ (2 zA) ρz E ⋅ dA + 0 = 2 EA ⇒ E = = ∫ E ⋅ dA + ∫ top bottom ε0 ε0

The field increases linearly with distance from the center. (b) The analysis is the same for the cylinder that extends outside the slab, with z > z0, except that the enclosed charge Q = ρ (2z0A) is that within a cylinder of length 2z0 rather than 2z. Thus Qin

ε0

=

G G G G ρ (2 z0 A) ρz E ⋅ dA + 0 = 2 EA ⇒ E = 0 = ∫ E ⋅ dA + ∫ top bottom ε0 ε0

The field strength outside the slab is constant, and it matches the result of part (a) at the boundary. (c)

28.49. Model: The infinitely wide plane of charge with surface charge density η polarizes the infinitely wide conductor. Visualize:

G G Because E = 0 in the metal there will be an induced charge polarization. The face of the conductor adjacent to the plane of charge is negatively charged. This makes the other face of the conductor positively charged. We thus have three infinite planes of charge. These are P (top conducting face), P′ (bottom conducting face), and P″(plane of charge). Solve: Let η1, η2, and η3 be the surface charge densities of the three surfaces with η2 a negative number. The electric field due to a plane of charge with surface charge density η is E = η 2ε 0 . Because the electric field inside a conductor is zero (region 2), G G G G G η η η EP + EP′ + EP′′ = 0 N/C ⇒ − 1 ˆj + 2 ˆj + 3 ˆj = 0 N/C ⇒ −η1 + η2 + η = 0 C/m 2 2ε 0 2ε 0 2ε 0

We have made the substitution η3 = η. Also note that the field inside the conductor is downward from planes P and P′ and upward from P″. Because η1 + η2 = 0 C/m2, because the conductor is neutral, η2 = −η1. The above equation becomes

−η1 − η1 + η = 0 C/m 2 ⇒ η1 = 12 η ⇒ η2 = − 12 η We are now in a position to find electric field in regions 1–4. For region 1, G G η ˆ η ˆ EP = j EP′ = − j 4ε 0 4ε 0 G G G G The electric field is Enet = EP + EP′ + EP′′ = (η 2ε 0 ) ˆj. G G In region 2, Enet = 0 N/C. In region 3, G G η ˆ η ˆ EP = − j EP ′ = j 4ε 0 4ε 0 G The electric field is Enet = (η 2ε 0 ) ˆj. In region 4, G The electric field is Enet

G η ˆ EP = − j 4ε 0 = − (η 2ε 0 ) ˆj.

G η ˆ EP ′ = j 4ε 0

G η ˆ EP′′ = j 2ε 0

G η ˆ EP′′ = j 2ε 0

G η ˆ EP′′ = − j 2ε 0

28.50. Model: Assume the metal slabs are large enough to model as infinite planes. Then the electric field has planar symmetry, pointing either toward or away from the planes. Visualize:

One Gaussian surface is a cylinder with end-areas a extending past the two metal slabs. A second Gaussian surface ends in region 3, the space between the slabs. G G G Solve: (a) Because these are metals, we immediately know that E2 = E4 = 0. To find the fields outside the slabs, consider the Gaussian surface on the left. The electric field everywhere points up or down, so there is no flux through the sides of this cylinder. Because both slabs are positive, the electric fields in regions 1 and 5 point outward. Gauss’s law is G G G G G G G G Qin AE ⋅ dA = ∫ top E ⋅ dA + ∫ bottom E ⋅ dA + ∫ sides E ⋅ dA = E1a + E5a + 0 =

ε0

The Gaussian surface encloses both the upper and lower surfaces of the top slab. The total charge per unit area on both surfaces of this slab is Q1/A = Q/A, so the charge enclosed within the cylinder is Qa/A. (For this calculation we don’t need to know how the charge is distributed between the two surfaces.) Similarly, the enclosed charge on the lower slab is Q2a/A = 2Qa/A. Thus E1a + E5 a =

Qa 2Qa 3Q + ⇒ E1 + E5 = ε0 A ε0 A ε0 A

G G Fields E1 and E5 are both a superposition of the fields of four sheets of surface charge. Because the field of a plane of charge is independent of distance from the plane, the superposition at points above the top plane must be the same magnitude, but opposite direction, as the superposition at points below the bottom plane. Consequently, E1 = E5 (same field strengths). This is a rather subtle point in the reasoning and one worth thinking about. Thus E1 = E5 = 3Q/2ε0A. Now consider the Gaussian surface on the right. The lower slab is more positive than the upper slab, so the electric field in region 3 must point upward, into the lower face of this cylinder. Thus G G G G G G G G Qin AE ⋅ dA = ∫ top E ⋅ dA + ∫ bottom E ⋅ dA + ∫ sides E ⋅ dA = E1a − E3a + 0 =

ε0

where the minus sign with E3 is because of the direction. We know E1, and we’ve already determined that Qin = Qa/A. Thus E3 = E1 −

Q 3Q Q Q = − = ε 0 A 2ε 0 A ε 0 A 2ε 0 A

Summarizing, E1 = 3Q/2ε0A, E2 = 0, E3 = Q/2ε0A, E4 = 0, and E5 = 3Q/2ε0A. (b) The electric field at the surface of a conductor is E = η/ε0. We can use the known fields and η = ε0E to find G the four surface charge densities. At surface a, E1 points away from the surface. Thus

ηa = +ε 0 E1 = +ε 0 G At surface b, E3 points toward the surface. Thus

3Q 3Q =+ 2ε 0 A 2A

ηb = −ε 0 E3 = −ε 0

Q 1Q =− 2ε 0 A 2A

Surface c is opposite to surface b, because the field points away from the surface, so ηc = +Q/2A. Finally, at G surface d, the field points away from the surface and has the same strength as E1 , hence ηd = +3Q/2A. Assess: Notice that ηa + ηb = Q/A and ηc + ηd = 2Q/A. This is the expected “net” surface charge density for slabs with total charge Q and 2Q.

28.51. Model: A long, charged wire can be modeled as an infinitely long line of charge. Visualize:

The figure shows an infinitely long line of charge that is surrounded by a hollow metal cylinder of radius R. The symmetry of the situation indicates that the only possible shape of the electric field is to point straight in or out from the wire. The shape of the field suggests that we choose our Gaussian surface to be a cylinder of radius r and length L, centered on the wire. Solve: (a) For the region r < R, Gauss’s law is

G G Q Φ e = AE ⋅ dA = in ⇒

ε0

∫

top

G G E ⋅ dA + ∫

bottom

G G E ⋅ dA + ∫

side

G G λL E ⋅ dA =

ε0

G G λL λL ⇒ 0 N m 2 / C + 0 N m 2 / C + E ⋅ Aside = ⇒ E ( 2π r ) L =

ε0

⇒E=

ε0

G ⎛ λ 1 ⎞ λ 1 λ rˆ ⇒E =⎜ , outward ⎟ = 2πε 0 r ⎝ 2πε 0 r ⎠ 2πε 0 r

(b) Applying Gauss’s law to the Gaussian surface at r > R,

G G G G G G G G Qin AE ⋅ dA = ∫ top E ⋅ dA + ∫ bottom E ⋅ dA + ∫ side E ⋅ dA =

ε0

G G Q ⇒ 0 N m 2 /C + 0 N m 2 /C + E ⋅ Awall = in

ε0

⇒ E ( 2π rL ) =

Qline + Qcylinder

ε0

=

G λ L + 2λ L 3λ 1 3λ rˆ ⇒E= ⇒E= ε0 2πε 0 r 2πε 0 r

28.52. Model: A long, charged cylinder is assumed to be infinite and to have linear symmetry. Visualize:

The cylindrical symmetry of the situation indicates that the only possible shape of the electric field is to point straight in or out from the cylinder. The shape of the field suggests that we choose our Gaussian surface to be a cylinder of radius r and length L, which is concentric with the charged cylinder. Solve: (a) For r ≥ R, Gauss’s law is G G G G G G G G Qin AE ⋅ dA = ∫ left E ⋅ dA + ∫ right E ⋅ dA + ∫ wall E ⋅ dA =

ε0

⇒ 0 N m 2 / C + 0 N m 2 / C + EA =

Qin

ε0

⇒E=

(b) For r ≤ R, Gauss’s law is

EA =

Qin

ε0

⇒ E=

G ⎛ λ 1⎞ λL λ = ⇒E =⎜ ⎟ rˆ ( 2π rL ) ε 0 2πε 0r ⎝ 2πε 0 r ⎠

ρ (π r 2 L ) Qin = Aε 0 ( 2π rL ) ε 0

An expression for the volume charge density ρ in terms of the linear charge density can be calculated by considering the charge on a cylinder of length d and radius R:

λ d = ρ (π R 2 d ) ⇒ ρ =

λ π R2

G 1 λr 1 λr ⎛ λ ⎞ πr L rˆ ⇒E =⎜ = ⇒E= 2 ⎟ 2 2πε 0 R 2 ⎝ π R ⎠ ( 2π rL ) ε 0 2πε 0 R 2

(c) At r = R, 1 λR 1 λ Er = R = rˆ = rˆ 2πε 0 R 2 2πε 0 R

28.53. Model: The charge distribution in the shell has spherical symmetry. Visualize:

The spherical surfaces of radii r ≥ Rout, r ≤ Rin, and Rin ≤ r ≤ Rout, concentric with the spherical shell, are Gaussian surfaces. Solve: (a) Gauss’s law for the Gaussian surface r ≥ Rout is

G G Qin G 1 Q 1 Q Q ⇒ E ( 4π r 2 ) = ⇒ E = ⇒E= rˆ AE ⋅ dA = 2 ε0 ε0 4πε 0 r 4πε 0 r 2 The vector form comes from the fact that the field is directed radially outward. (b) For r ≤ Rin, Gauss’s law is

G G Qin 0 C G G = ⇒ E =0 AE ⋅ dA =

ε0

ε0

(c) For Rin ≤ r ≤ Rout, Gauss’s law is

G G Qin Q ⇒ E ( 4π r 2 ) = in AE ⋅ dA =

ε0

Q in =

4π 3 ( r − Rin3 ) ρ = 43π 3

ε0

( r − R )Q = Q ⎛ r ( R − R ) ⎜⎝ R 3

4π 3

3 in

3 out

3 in

3

− Rin3 ⎞ ⎟ − Rin3 ⎠

3 out

G 1 Q r 3 − Rin3 1 Q ⎛ r 3 − Rin3 ⎞ E ⇒ = ⎜ 3 ⎟ rˆ 3 − Rin3 − Rin3 ⎠ 4π r 2 ε 0 Rout 4πε 0 r 2 ⎝ Rout G G (d) The result obtained in part (c) for the electric field simplifies to E = 0, when r = Rin which is the result obtained in part (b). Furthermore, at r = Rout, the electric field obtained in part (c) becomes ⇒E=

G 1 Q E= rˆ 2 4πε 0 Rout

which is the same as the electric field obtained in part (a). (e)

28.54. Model: Assume that the negative charge uniformly distributed in the atom has spherical symmetry. Visualize:

The nucleus is a positive point charge +Ze at the center of a sphere of radius R. The spherical symmetry of the charge distribution tells us that the electric field must be radial. We choose a spherical Gaussian surface to match the spherical symmetry of the charge distribution and the field. The Gaussian surface is at r < R, which means that we will calculate the amount of charge contained in this surface. G G Solve: (a) Gauss’s law is AE ⋅ dA = Qin ε 0 . The amount of charge inside is

⎡ r3 ⎤ ( − Ze ) r3 ⎛ 4π 3 ⎞ Qin = ρ ⎜ r ⎟ + Ze = 4π 3 ( 43π r 3 ) + Ze = − ( Ze ) 3 + Ze = Ze ⎢1 − 3 ⎥ R ⎝ 3 ⎠ (3R) ⎣ R ⎦

⇒ Ein ( 4π r 2 ) =

Ze ⎡ 1 r ⎤ Ze ⎡ r 3 ⎤ ⎢1 − 3 ⎥ ⇒ Ein = 4πε ⎢ r 2 − R 3 ⎥ ε0 ⎣ R ⎦ ⎦ 0 ⎣

(b) At the surface of the atom, r = R. Thus,

Ein =

Ze ⎡ 1 R⎤ − = 0 N/C 4πε 0 ⎢⎣ R 2 R 3 ⎥⎦

This is an expected result, which can be quickly obtained from Gauss’s law. Applying Gauss’s law to a Gaussian surface just outside r = R. Because the atom is electrically neutral, Qin = 0. Thus G G Qin = 0 ⇒ E = 0 N/C AE ⋅ dA =

ε0

(c) The electric field strength at r = 12 R = 0.050 nm is

⎡ 1 ( 0.050 nm ) ⎤ = 4.6 ×1013 N/C − Ein = 92 (1.60 ×10−19 C )( 9.0 ×109 C 2 /Nm 2 ) ⎢ 2 3 ⎥ ⎢⎣ ( 0.050 nm ) ( 0.10 nm ) ⎥⎦

28.55. Model: The long thin wire is assumed to be an infinite line of charge. Visualize: Please refer to Figure CP28.55. The cube of edge length L is centered on the line charge with a linear charge density λ. Although the line charge has cylindrical symmetry, we will take the cube as our Gaussian surface. G G G G Solve: (a) The electric flux through an area dA in the yz plane is d Φ = E ⋅ dA. The electric field E due to an infinite line of charge at a distance s from the line charge is

G λ 1 λ E= rˆ = 2πε 0 r 2π 0

1

y 2 + ( L 2)

2

rˆ

G Also dA = L dyiˆ. Thus,

dΦ = =

λ 2πε 0

1 y + ( L 2) 2

2

λ L dy L dy rˆ ⋅ iˆ = 2πε 0

( )

λ L dy

L2

2πε 0 y 2 + ( L 2 )

2

y2 + ( L 2)

2

=

1 y + ( L 2) 2

2

cosθ

λ L2 dy 2 4πε 0 ⎡ y 2 + ( L 2 ) ⎤ ⎣ ⎦

(b) The expression for the flux dΦ can now be integrated to obtain the total flux through this face as follows: Φ = ∫ dΦ = =

L2

∫

−L 2

λ L2 dy 2 4πε 0 ⎡ y 2 + ( L 2 ) ⎤ ⎣ ⎦

L2

=

λ L2 ⎡ 2 −1 ⎛ 2 y ⎞ ⎤ tan ⎜ ⎟ ⎥ 4πε 0 ⎢⎣ L ⎝ L ⎠⎦ −L 2

λL λ L ⎡π ⎛ π ⎞⎤ λ L ⎡⎣ tan −1 (1) − tan −1 ( −1) ⎤⎦ = −⎜− ⎟ = 2πε 0 2πε 0 ⎢⎣ 4 ⎝ 4 ⎠ ⎥⎦ 4ε 0

(c) Because there are four faces through which the flux flows, the net flux through the cube is Φ e = 4Φ =

λ L ( Qin L ) L Qin = = ε0 ε0 ε0

28.56.

Model: Visualize:

Solve:

The field has cylindrical symmetry.

(a) The volume charge density ρ ( r ) = r ρ0 R is linearly proportional to r. The graph is shown above.

(b) Consider the cylindrical shell of length L, radius r, and thickness dr shown in the diagram. The charge within a small volume dV is

dq = ρ dV =

ρ0r R

( 2π r ) drL =

2πρ 0 L 2 r dr R

Integrating this expression to obtain the total charge in the cylinder: R

R 2πρ 0 L 2 2πρ 0 L ⎡ r 3 ⎤ 2πρ 0 LR 3 3λ = λ L ⇒ ρ0 = Q = ∫ dq = r dr = ⎢ ⎥ = ∫ 3R 2π R 2 R 0 R ⎣ 3 ⎦0

(c) Consider the cylindrical Gaussian surface of length l at r < R shown in the figure. Gauss’s law is G G Φ e = AE ⋅ dA = Qin / ε 0 . The charge on the inside of the Gaussian surface is

⎛ r3 ⎞ 2πρ 0l 2 2πρ0lr 3 2π ⎛ 3λ ⎞ 3 = r dr = lr = λl ⎜ 3 ⎟ ⎜ 2 ⎟ 3R 3R ⎝ 2π R ⎠ R ⎝R ⎠ 0 r

Qin = ∫ dq = ∫ ρ dV = ∫

G G λl ⎛ r 3 ⎞ λ ⎛ r2 ⎞ Q ⇒ AE ⋅ dA = E ( 2π rl ) = in = ⎜ 3 ⎟ ⇒ E = ⎜ ⎟ 2πε 0 ⎝ R 3 ⎠ ε0 ε0 ⎝ R ⎠ The last step is justified because the electric field has radial dependence. (d) The expression for the electric field strength simplifies at r = R to E=

λ R2 λ = 2πε 0 R 3 2πε 0 R

This is the same result as obtained in example 28.5 for a long, charged wire.

28.57.

Visualize:

Solve: (a) Consider the spherical shell of radius r and thickness dr shown in the figure. The charge dq within a small volume dV is C dq = ρ dV = 2 ( 4π r 2 ) dr = 4π Cdr r Integrating this expression to obtain the total charge in the sphere: R

Q = ∫ dq = ∫ 4π Cdr = 4π CR ⇒ C = 0

Q 4π R

(b) Consider the spherical Gaussian surface at r < R shown in the figure. Gauss’s law applied to this surface is G G Q Qin Q Φ e = AE ⋅ dA = in ⇒ E ( 4π r 2 ) = in ⇒ E = ε0 ε0 4πε 0 r 2

Using the results from part (a), r

r

C Q ⎛ Q ⎞ 4π r 2 dr = ∫ ⎜ ⎟ 4π dr = r 2 r R R 4 π ⎝ ⎠ 0 0

Qin = ∫ dq = ∫ ρ dV = ∫

G 1 Q ⎛Q ⎞ 1 rˆ ⇒ E = ⎜ r⎟ ⇒E= 2 4πε 0 Rr ⎝ R ⎠ 4πε 0 r (c) At r = R, the equation for the electric field obtained in part (b) simplifies to G 1 Q E= rˆ 4πε 0 R 2

This is the same result as obtained in Example 28.3. The result was expected because a spherical charge behaves, for r ≥ R, as if the entire charge were at the center.

28.58. Model: The charge density within the sphere is nonuniform. We will assume that the distribution has spherical symmetry. Visualize:

Solve: (a) Consider the thin shell of width dr at a distance r from the center shown in the figure. The volume of this thin shell is dV = ( 4π r 2 ) dr. The charge contained in this volume is R R R⎛ r⎞ r3 ⎞ ⎛ Q = ∫ dq = ∫ 4π r 2 ρ dr = ∫ 4π r 2 ρ0 ⎜1 − ⎟ dr = 4πρ 0 ∫ ⎜ r 2 − ⎟ dr 0 R⎠ ⎝ R⎠ ⎝ 0 0 R

⎡ r3 r 4 ⎤ 3Q R3 = 4πρ 0 ⎢ − ⇒ ρ0 = ⎥ = 4πρ0 π R3 12 ⎣ 3 4R ⎦0 (b) Gauss’s law for the Gaussian surface shown in the figure at r < R is G G Qin Q ⇒ E ( 4π r 2 ) = in AE ⋅ dA =

ε0

ε0

The charge inside the Gaussian surface is r r r ⎛ ⎡ r3 r 4 ⎤ r⎞ r3 ⎞ ⎛ Qin = ∫ dq = ∫ 4π r 2 ρ dr = ∫ 4π r 2 ρ0 ⎜1 − ⎟ dr = 4πρ 0 ∫ ⎜ r 2 − ⎟ dr = 4πρ 0 ⎢ − ⎥ R⎠ ⎝ R⎠ ⎣ 3 4R ⎦ 0 0 0⎝

Gauss’s law becomes E ( 4π r 2 ) =

4πρ0 ⎡ r 3 r 4 ⎤ ρ0 ⎡ r r 2 ⎤ ⎛ 3Q ⎞ 1 = − = − E ⇒ ε 0 ⎢⎣ 3 4 R ⎥⎦ ⎜⎝ π R3 ⎟⎠ ε 0 ε 0 ⎢⎣ 3 4 R ⎥⎦

r ⎡ 3r ⎤ Qr ⎛ 3r ⎞ = 1− 4− ⎟ 3⎜ ⎢ ⎥ 3 ⎣ 4 R ⎦ 4πε 0 R ⎝ R⎠

The direction of the field is radially outward. (c) At r = R, the above expression for the electric field reduces to E=

QR ⎛ 3R ⎞ 1 Q ⎜4− ⎟= 4πε 0 R 3 ⎝ R ⎠ 4πε 0 R 2

This is an expected result, because all charge Q is inside R and the field looks like that of a point charge.

28.59.

Visualize:

Solve:

(a) It is clear that E ( r ) ∝ r 4 up to r = R. That is, the maximum value of E(r) occurs at r = R. At this point,

E=

1

Q 4πε 0 R 2

because all the charge is inside R and the charge is spherically distributed about a point at the center. Thus, E=

⎛ R4 ⎞ Q Q = Emax ⎜ 4 ⎟ ⇒ Emax = 2 2 4 πε 4πε 0 R R ⎝ ⎠ 0R 1

(b) Applying Gauss’s law to the surface at r < R shown in the figure,

G G Qin ⎛ Q ⎞ r4 ⎛ r6 ⎞ ⎛ r4 ⎞ Q ⇒ ⎜ Emax 4 ⎟ 4π r 2 = in ⇒ Qin = ⎜ ε 4 4π r 2 = Q ⎜ 6 ⎟ AE ⋅ dA = 2 ⎟ 0 ε0 R ⎠ ε0 ⎝R ⎠ ⎝ ⎝ 4πε 0 R ⎠ R To find Qin, we consider the thin spherical shell of width dr and charge dq at a distance r from the center shown in the figure. Thus,

dq = ρ dV = ρ ( 4π r 2 ) dr ⇒ ∫ dq = Qin = ∫ ρ ( r ) ( 4π r 2 ) dr Using this form of Qin in the equation obtained from Gauss’s law, Qin = Q

r6 = ρ ( r )4π r 2 dr ⇒ R6 ∫

Q r6

∫ ρ ( r ) r dr = 4π R 2

6

Taking the position derivative on both sides 3Q r 3 d Q d 6 Q 2 2 5 r r r dr r ρ = ⇒ ρ r r = 6 r ⇒ ρ = ( ) ( ) ( ) ( ) ( ) 2π R 6 dr ∫ 4π R 6 dr 4π R 6 (c) Consider a spherical shell of width dr and charge dq at a distance r from the center. Then

dq = ρ ( r ) dV =

R

3Qr 3 6Q 5 6Q R 6 r5 2 4 r dr π ⇒ dq = 6 Q dr ⇒ = = = ⋅ =Q Q dq r dr ∫ 2π R 6 R6 R 6 ∫0 R6 6

28-1

29.1. Model: The mechanical energy of the proton is conserved. A parallel-plate capacitor has a uniform electric field. Visualize:

The figure shows the before-and-after pictorial representation. The proton has an initial speed vi = 0 m/s and a final speed vf after traveling a distance d = 2.0 mm. Solve: The proton loses potential energy and gains kinetic energy as it moves toward the negative plate. The potential energy U is defined as U = U0 + qEx, where x is the distance from the negative plate and U0 is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton is ΔUp = Uf – Ui = (U0 + 0 J) – (U0 + qEd) = −qEd The change in the kinetic energy of the proton is

ΔK = Kf − Ki = 12 mvf2 − 12 mvi2 = 12 mvf2 The law of conservation of energy is ΔK + ΔUp = 0 J. This means 1 2

⇒ vf =

mvf2 + ( −qEd ) = 0 J

2 ( +1.60 ×10−19 C ) ( 50,000 N/C ) ( 2.0 ×10−3 m ) 2qEd = = 1.38 ×105 m/s m (1.67 ×10−27 kg )

Assess: As described in Section 29.1, the potential energy for a charge q in an electric field E is U = U0 + qEx, where x is the distance measured from the negative plate. Having U = U0 at the negative plate (with x = 0 m) is completely arbitrary. We could have taken it to be zero. Note that only ΔU, and not U, has physical consequences.

29.2. Model: The mechanical energy of the electron is conserved. A parallel-plate capacitor has a uniform electric field. Visualize:

The figure shows the before-and-after pictorial representation. The electron has an initial speed vi = 0 m/s and a final speed vf after traveling a distance d = 1.0 mm. Solve: The electron loses potential energy and gains kinetic energy as it moves toward the positive plate. The potential energy U is defined as U = U0 + qEx, where x is the distance from the negative plate and U0 is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the electron is ΔUe = Uf – Ui = (U0 + qEd) – (U0 + 0 J) = qEd The change in the kinetic energy of the electron is

ΔK = Kf − Ki = 12 mvf2 − 12 mvi2 = 12 mvf2 Now, the law of conservation of mechanical energy gives ΔK + ΔU = 0 J. This means 1 2

⇒ vf =

−2qEd = m

mvf2 + qEd = 0 J

( −2 ) ( −1.60 ×10−19 C ) ( 20,000 N/C ) (1.0 ×10−3 m ) 9.11×10−31 kg

= 2.7 × 106 m/s

Assess: Note that ΔUe = qEd is the change in the potential energy of the electron. It is negative because q = − e for the electron. Thus, the potential energy becomes more negative as d increases, that is, the potential energy of the electron decreases with an increase in d (or x).

29.3. Model: The mechanical energy of the proton is conserved. A parallel-plate capacitor has a uniform electric field. Visualize:

The figure shows the before-and-after pictorial representation. Solve: The proton loses potential energy and gains kinetic energy as it moves toward the negative plate. The potential energy is defined as U = U0 + qEx, where x is the distance from the negative plate and U0 is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton is

ΔU p = U f − U i = (U 0 + 0 J ) − (U 0 + qEd ) = −qEd The change in the kinetic energy of the proton is

ΔK = Kf − Ki = 12 mvf2 − 12 mvi2 = 12 mvf2 Applying the law of conservation of energy ΔK + ΔUp = 0 J, we have 1 2

mvf2 + ( − qEd ) = 0 J ⇒ vf2 =

2qEd m

When the amount of charge on each plate is doubled, then the final velocity of the proton is

vf′2 =

2qE′d m

Dividing these equations, vf′2 E ′ = ⇒ vf′ = vf2 E

E′ vf E

For a parallel-plate capacitor E = η ε 0 = Q Aε 0 . Therefore,

vf′ =

Q′ vf = 2 ( 50,000 m/s ) = 70,711 m/s Q

Assess: The proton’s velocity is expected to increase because an increased charge on the capacitor plates leads to a higher electric field between the plates and hence to an increased force on the proton.

29.4. Model: The mechanical energy of the charged particles is conserved. A parallel-plate capacitor has a uniform electric field. Visualize:

The figure shows the before-and-after pictorial representation. Solve: The potential energy is defined as U = U0 + qEx, where x is the distance from the negative plate and U0 is the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton as it moves from the positive plate to the negative plate is ΔU p = U f − U i = (U 0 + 0 J ) − (U 0 + eEd ) = −eEd

This decrease in potential energy appears as an increase in the proton’s kinetic energy: ΔK = K f − K i = 12 mpvf2 p − 12 mpvi2p = 12 mpvf2 p

Applying the law of conservation of mechanical energy ΔK + ΔUp = 0 J, we have 1 2

mvf2p + ( − eEd ) = 0 J ⇒ vf2 p =

2eEd mp

When the proton is replaced with a helium ion and the same experiment is repeated,

vf2ion =

2eEd mion

Dividing the two equations,

vf ion =

mp mion

vf p =

1 4

( 50,000 m/s ) = 25,000 m/s

Assess: Being a heavier particle, the helium ion’s velocity is expected to be smaller compared to the proton’s velocity.

29.5. Model: The charges are point charges. Visualize: Please refer to Figure EX29.5. Solve: The electric potential energy of the electron is

U electron = U13 + U 23 ⎡ −19 −19 ⎢ (1.60 ×10 C )( −1.60 ×10 C ) = ( 9.0 ×109 N m 2 /C2 ) ⎢ + 2 2 −9 −9 ⎢⎣ ( 2.0 ×10 m ) + ( 0.50 ×10 m ) = −1.12 ×10 −19 J − 1.12 ×10−19 J = −2.24 ×10−19 J

(1.60 ×10 ( 2.0 ×10

⎤ C )( −1.60 × 10−19 C ) ⎥ 2 2 ⎥ m ) + ( 0.50 × 10−9 m ) ⎥ ⎦

−19

−9

29.6. Model: The charges are point charges. Visualize: Please refer to Figure EX29.6. Solve: For a system of point charges, the potential energy is the sum of the potential energies due to all pairs of charges:

U elec = ∑ i,j

Kqi q j rij

= U12 + U13 + U 23 = K

q1q2 qq qq +K 1 3 +K 2 3 r12 r13 r23

⎡ ( 2.0 nC )( −1.0 nC ) ( 2.0 nC )( −1.0 nC ) ( −1.0 nC )( −1.0 nC ) ⎤ = ( 9.0 × 109 N m 2 /C2 ) ⎢ + + ⎥ 0.030 m 0.030 m 0.030 m ⎣ ⎦ −6 −6 −6 −6 = −0.60 × 10 J − 0.60 × 10 J + 0.30 × 10 J = −0.90 × 10 J Assess:

Note that U12 = U21, U13 = U31, and U23 = U32.

29.7. Model: The charges are point charges. Visualize: Please refer to Figure EX29.7. Solve: For a system of point charges, the potential energy is the sum of the potential energies due to all distinct pairs of charges:

U elec = ∑ i,j

Kqi q j rij

= U12 + U13 + U 23

⎡ ( 3.0 ×10−9 C )( 2.0 × 10−9 C ) ( 3.0 × 10−9 C )( 2.0 ×10−9 C ) ( 2.0 ×10−9 C )( 2.0 × 10−9 C ) ⎤ ⎥ = ( 9.0 ×109 N m 2 /C2 ) ⎢ + + 2 2 ⎥ ⎢ 0.030 m 0.040 m 0.030 m 0.040 m + ( ) ( ) ⎦ ⎣ −6 −6 −6 −6 = 1.80 ×10 J + 1.35 × 10 J + 0.72 × 10 J = 3.87 ×10 J Assess:

Note that U12 = U21, U13 = U31, and U23 = U32.

29.8. Model: A water molecule is at the point of minimum potential energy when it is aligned with an electric field. However, an external force can rotate the water molecule causing its dipole moment to make an angle with the field. G G Solve: The potential energy of an electric dipole moment in a uniform electric field is U dipole = − p ⋅ E = − pE cosθ . This means

U dipole parallel = − pE cos0° = − pE

U dipole perpendicular = − pE cos90° = 0 J

⇒ Udipole perpendicular – Udipole parallel = 1.0 × 10−21 J = 0 J – (–pE) ⇒E=

Assess:

1.0 ×10−21 J 1.0 × 10−21 J = = 1.61× 108 N/C p 6.2 ×10−30 C m

Note that the units with E are J/C m. Because 1 J/C m = 1 N m/C m = 1 N/C, the units of E are N/C.

29.9. Model: An external electric field supplies energy to a dipole. Visualize: On an energy diagram, the oscillation occurs between the points where the potential-energy curve crosses the total energy line.

Solve: (a) The potential energy of an electric dipole moment in a uniform electric field is G G U = − p ⋅ E = − pE cos φ . This means

Uφ = 0° = − pE = −2 μ J

U φ = 60° = − pE cos60° = − 12 pE = − 12 ( 2 μ J ) = −1 μ J

The mechanical energy Emech = U + K. We know that at φ = 60°, K φ = 60° = 0 J. So, Emech = U φ = 60° + Kφ = 60° = −1 μ J + 0 J = − 1 μ J

(b) Conservation of mechanical energy gives

Uφ = 60° + Kφ = 60° = U φ = 0° + Kι = 0° ⇒ −1 μ J + 0 J = −2 μ J + Kφ = 0° ⇒ Kφ = 0° = 1 μ J

29.10. Model: Mechanical energy is conserved. The potential energy is determined by the electric potential. Visualize:

The figure shows a before-and-after pictorial representation of an electron moving through a potential difference of 1000 V. A negative charge speeds up as it moves into a region of higher potential (U → K). Solve: The potential energy of charge q is U = qV. Conservation of energy, expressed in terms of the electric potential V, is Kf + qVf = Ki + qVi ⇒ Kf = Ki + q(Vi – Vf) = Ki – q(Vf – Vi)

⇒ 12 mvf2 = 0 J − ( −e )( ΔV ) ⇒ vf =

2 (1.60 ×10−19 C ) (1000 V ) 2eΔV = = 1.874 × 107 m/s m 9.11×10−31 kg

Assess: Note that the electric potential difference of 1000 V already existed in space due to other charges or sources. The electron of our problem has nothing to do with creating the potential.

29.11. Model: Energy is conserved. The potential energy is determined by the electric potential. Visualize:

The figure shows a before-and-after pictorial representation of a proton moving through a potential difference of −1000 V. A positive charge speeds up as it moves into a region of lower potential (U → K). Solve: The potential energy of charge q is U = qV. Conservation of energy, expressed in terms of the electric potential V, is Kf + qVf = Ki + qVi, ⇒ Kf = Ki + q(Vi − Vf) = 0 J − qΔV

⇒ 12 mvf2 = − q ( −1000 V ) ⇒ vf = Assess: V.

2 (1.60 × 10−19 C ) (1000 V ) 1.67 ×10−27 kg

= 4.38 × 105 m/s

Note that the proton of our problem has nothing to do with creating the potential difference of −1000

29.12. Model: Energy is conserved. The potential energy is determined by the electric potential. Visualize:

The figure shows a before-and-after pictorial representation of a He+ ion moving through a potential difference. The ion’s initial speed is zero and its final speed is 2.0 × 106 m/s. A positive charge speeds up as it moves into a region of lower potential (U → K). Solve: The potential energy of charge q is U = qV. Conservation of energy, expressed in terms of the electric potential V, is Kf + qVf = Ki + qVi ⇒ q(Vf – Vi) = Ki − Kf

4 (1.67 ×10−27 kg )( 2.0 ×106 m/s ) K − K f 0 J − 12 mvf2 ⇒ ΔV = i = =− = −8.4 ×104 V q q 2 (1.60 ×10−19 C ) 2

Assess:

This result implies that the helium ion moves from a higher potential toward a lower potential.

29.13. Model: Energy is conserved. The potential energy is determined by the electric potential. Visualize:

The figure shows a before-and-after pictorial representation of an electron moving through a potential difference. Because the negative electron gains speed as it travels, it moves into a region of higher potential (U → K). Solve: The potential energy of charge q is U = qV. Using the conservation of energy equation, Kf + qVf = Ki + qVi ⇒ Vf − Vi = ΔV = 1 ( K i − K f ) = 1 ( 0 J − 12 mvf2 ) q ( −e ) −31 6 mv 2 ( 9.11×10 kg )( 2.0 ×10 m/s ) ⇒ ΔV = f = = 11.4 V 2e 2 (1.60 ×10−19 C ) 2

Assess: A positive value of ΔV shows that the electron moved from a region of lower potential to a region of higher potential.

29.14. Model: Energy is conserved. The potential energy is determined by the electric potential. Visualize:

The figure shows a before-and-after pictorial representation of an electron moving through a potential difference. Solve: (a) Because the electron is a negative charge and it slows down as it travels, it must be moving from a region of higher potential to a region of lower potential. (b) Using the conservation of energy equation, Kf + Uf = Ki + Ui ⇒ Kf + qVf = Ki + qVi 1 1 1 2 ⇒ Vf − Vi = ( Ki − K f ) = ( mv − 0 J ) q ( −e ) 2 i

(9.11×10−31 kg ) ( 500,000 m/s ) = −0.712 V mvi2 =− 2e 2 (1.60 ×10−19 C ) 2

⇒ ΔV = −

Assess: The negative sign with ΔV verifies that the electron moves from a higher potential region to a lower potential region.

29.15. Model: Energy is conserved. The potential energy is determined by the electric potential. Visualize:

The figure shows a before-and-after pictorial representation of a proton moving through a potential difference. Solve: (a) Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential. (b) Using the conservation of energy equation, Kf + Uf = Ki + Ui ⇒ Kf + qVf = Ki + qVi

⇒ Vf − Vi =

1 1 ( Ki − K f ) = ( 12 mvi2 − 0 J ) q (e)

−27 mvi2 (1.67 ×10 kg ) ( 800,000 m/s ) = = 3340 V 2e 2 (1.60 ×10−19 C ) 2

⇒ ΔV = Assess:

A positive ΔV confirms that the proton moves into a higher potential region.

29.16. Model: Energy is conserved. Solve:

(a) The conservation of energy equation Kf + Uf = Ki + Ui is Kf + qVf = 0 J + qVi ⇒ Kf = q(Vi – Vf)

The above equation can be rewritten separately for the proton and the electron as K fp = 12 mpvfp2 = ( +e ) ( ΔVp )

K fe = 12 mevfe2 = ( −e )( −ΔVe )

Note that we have written Vi – Vf as ΔVp for the proton, but Vi – Vf = –ΔVe for the electron. This is because Vi > Vf to speed up a proton, and Vf > Vi to speed up an electron. With vfp = vfe, the above two equations give ΔVp ΔVe

=

mp me

=

1.67 ×10−27 kg = 1833 9.11× 10−31 kg

(b) The first two equations of part (a) can be divided to give

K fp K fe

=

ΔVp ΔVe

For the same final kinetic energy of the proton and the electron ΔVp ΔVe = 1.

29.17. Solve: By definition 1 V = 1 J/C and 1 J = 1 N m. Consequently, 1

V J Nm N =1 =1 =1 m Cm Cm C

29.18. Model: The electric potential difference between the plates is determined by the uniform electric field in the parallel-plate capacitor. Solve: (a) The potential of an ordinary AA or AAA battery is 1.5 V. Actually, this is the potential difference between the two terminals of the battery. If the electric potential of the negative terminal is taken to be zero, then the positive terminal is at a potential of 1.5 V. (b) If a battery with a potential difference of 1.5 V is connected to a parallel-plate capacitor, the potential difference between the two capacitor plates is also 1.5 V. Thus, ΔVC = 1.5 V = V+ − V− = Ed where d is the separation between the two plates. The electric field inside a parallel-plate capacitor is

E= ⇒Q=

⎛ Q ⎞ η Q = ⇒ 1.5 V = ⎜ ⎟d ε 0 Aε 0 ⎝ Aε 0 ⎠

(1.5 V )( Aε 0 ) = (1.5 V )π ( 2.0 ×10−2 m ) d

2

(8.85 ×10

2.0 ×10−3 m

−12

C2 /Nm 2 )

= 8.3 ×10−12 C

Thus, the battery moves 8.3 × 10−12 C of electron charge from the positive to the negative plate of the capacitor.

29.19. Model: The electric potential difference between the plates is determined by the uniform electric field in the parallel-plate capacitor. Solve: (a) The potential difference ΔVC across a capacitor of spacing d is related to the electric field inside by

E=

ΔVC ⇒ ΔVC = Ed = (1.0 ×105 V/m ) ( 0.0020 m ) = 200 V d

(b) The electric field of a capacitor is related to the surface charge density by

E=

η Q A = ε0 ε0

⇒ Q = ε 0 AE = ( 8.85 × 10−12 C 2 /N m 2 )( 4.0 × 10−4 m 2 )(1.0 × 105 V/m ) = 3.5 × 10−10 C

29.20. Model: The electric potential between the plates of a parallel-plate capacitor is determined by the uniform electric field between the plates. Solve: (a) The potential difference across the plates of a capacitor is

⎛η ⎞ ( Q A) d = Qd = ( 0.708 ×10−9 C )(1.00 ×10−3 m ) = 200 V ΔVC = Ed = ⎜ ⎟ d = ε0 Aε 0 ( 4.00 ×10−4 m 2 )(8.85 ×10−12 C 2 /N m 2 ) ⎝ ε0 ⎠ (b) For d = 2.00 mm, ΔVC = 400 V. Assess: Note that the units in part (a) are N m/C. But Exercise 29.17 showed that 1 N/C = 1 V/m, so 1 N m/C = 1 V. We also see that the potential difference across a parallel-plate capacitor is directly proportional to the plate separation.

29.21. Model: Energy is conserved. The electron’s potential energy inside the capacitor can be found from the capacitor’s electric potential. Solve: (a) The voltage across the capacitor is ΔVC = Ed = ( 5.0 ×105 V/m )( 2.0 × 10−3 m ) = 1000 V (b) Because E = η ε 0 for a parallel-plate capacitor, with η = Q A , the charge on each plate is

Q = π R 2 Eε 0 = π (1.0 ×10−2 m ) ( 5.0 ×105 V/m )( 8.85 ×10−12 C2 /N m 2 ) = 1.39 ×10−9 C 2

(c) The electron has charge q = −e, and its potential energy at a point where the capacitor’s potential is V is U = −eV. Since the electron is launched from the negative (lower potential) plate toward the positive (higher potential) plate, its potential energy becomes more negative (because of the negative sign of the electron charge). That is, the potential energy decreases, which must lead to an increase in the kinetic energy. Conversely, the electron’s speed as it is launched is smaller than 2.0 × 107 m/s. The conservation of energy equation is

Kf + qVf = Ki + qVi ⇒

⇒ vi2 = vf2 + ⇒ vi =

( 2.0 ×10

7

m/s ) − 2

1 2

mvi2 = 12 mvf2 + q (Vf − Vi )

2 ( −e )(1000 V ) m

2 (1.60 ×10−19 C ) (1000 V ) 9.11×10−31 kg

= 7.0 ×106 m/s

29.22. Model: The charge is a point charge. Solve:

(a) The electric potential of a charge q is V=

1 q 1 q 25 × 10−9 C 225 N m 2 /C ⇒r= = ( 9.0 ×109 N m 2 /C 2 ) = 4πε 0 r 4πε 0 V V V

For V = 1000 V,

r1000 =

225 N m 2 /C = 0.23 m 1000 V

For V = 2000 V, r2000 = 0.113 m ; for V = 3000 V, r3000 = 0.075 m ; for V = 4000 V, r4000 = 0.056 m . (b)

29.23. Model: The charge is a point charge. Visualize: Please refer to Figure EX29.23. Solve: (a) The electric potential of the point charge q is

V=

⎛ 2.0 ×10−9 C ⎞ 18.0 N m 2 /C 1 q = ( 9.0 ×109 N m 2 /C2 ) ⎜ ⎟= 4πε 0 r r r ⎝ ⎠

For points A and B, r = 0.010 m. Thus,

VA = VB =

18.0 N m 2 /C Nm ⎛V⎞ = 1800 = 1800 ⎜ ⎟ m = 1.80 kV 0.010 m C ⎝m⎠

For point C, r = 0.020 m and VC = 900 V. (b) The potential differences are

ΔVAB = VB − VA = 1.80 kV −1.80 kV = 0 V

ΔVBC = VC − VB = 0.90 kV − 1.80 kV = −0.90 kV

29.24. Solve: Outside a charged sphere of radius R, the electric potential is identical to that of a point charge Q at the center. That is, V=

1 Q 4πε 0 r

r≥R

The potential of the ball bearing is the potential right on the surface of the ball bearing. Thus, ( 9.0 ×109 N m2/C2 )( 2.0 ×109 )( −1.60 ×10−19 C ) = −5.8 kV V= 0.50 ×10−3 m

29.25. Model: Outside a charged sphere the electric potential is identical to that of a point charge at the center. Solve: (a) For a proton, assumed to be a point charge, the electric potential is V=

1 4πε 0

( +e ) = r

( 9.0 ×10

9

N m 2 /C 2 )

1.60 × 10−19 C = 27 V 0.053 ×10−9 m

(b) The potential energy of a charge q at a point where the potential is V is U = qV. The potential energy of the electron in the proton’s potential is U = (−1.60 × 10−19 C) (27 V) = −4.3 × 10−18 J

29.26. Model: The net potential is the sum of the potentials due to each charge. Visualize: Please refer to Figure EX29.26. Solve: The potential at the dot is V=

1 q1 1 q2 1 q3 + + 4πε 0 r1 4πε 0 r2 4πε 0 r3

⎡ 2.0 × 10−9 C 2.0 × 10−9 C 2.0 × 10−9 C ⎤ 3 = ( 9.0 ×109 N m 2 /C 2 ) ⎢ + + ⎥ = + 1.41× 10 V 0.040 m 0.050 m 0.030 m ⎣ ⎦

Assess:

Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.

29.27. Model: The net potential is the sum of the potentials due to each charge. Visualize: Please refer to Figure EX29.27. Solve: From the geometry in the figure,

1.5 cm 1.5 cm 1.5 cm 1.5 cm = = = cos30° ⇒ r1 = r2 = r3 = = 1.732 cm cos30° r1 r2 r3 The potential at the dot is

V=

1 q1 1 q2 1 q3 + + 4πε 0 r1 4πε 0 r2 4πε 0 r3

⎡ 2.0 × 10−9 C 1.0 ×10−9 C 1.0 × 10−9 C ⎤ = ( 9.0 × 109 N m 2 /C 2 ) ⎢ − − ⎥= 0V ⎣ 0.01732 m 0.01732 m 0.01732 m ⎦ Assess:

Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.

29.28. Model: The electric potential at the dot is the sum of the potentials due to each charge. Visualize: Please refer to Figure EX29.28. Solve: The electric potential at the dot is

V=

1 q1 1 q2 1 q3 + + 4πε 0 r1 4πε 0 r2 4πε 0 r3

⎡ −5.0 ×10−9 C 5.0 ×10−9 C ⎤ q = ( 9.0 ×109 Nm 2 /C2 ) ⎢ + + ⎥ = 3140 V 2 2 0.040 m 0.020 m ⎦⎥ ⎣⎢ ( 0.020 m ) + ( 0.040 m ) Solving yields q = 1.00×10–8 C = 10.0 nC. Assess: Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.

29.29. Model: The net potential is the sum of the scalar potentials due to each charge. Visualize:

Solve: Let the point on the x-axis where the electric potential is zero be at a distance x from the origin. At this point, V1 + V2 = 0 V. This means

1 ⎧⎪ 3.0 ×10−9 C −1.0 ×10−9 C ⎫⎪ + ⎨ ⎬ = 0 V ⇒ − x + 3 x − 4.0 cm = 0 cm 4πε 0 ⎪⎩ x x − 4.0 cm ⎪⎭ Either − x + 3( x − 4.0 cm ) = 0 cm, or − x + 3( 4.0 cm − x ) = 0 cm. In the first case, x = 6.0 cm and, in the second case, x 3 cm. That is, we have two points on the x-axis where the potential is zero.

=

29.30. Model: The net potential is the sum of the scalar potentials due to each charge. Visualize:

Solve: Let the point on the y-axis where the electric potential is zero be at a distance y from the origin. At this point, V1 + V2 = 0 V. This means 1 ⎧ q1 q2 ⎫ ⎨ + ⎬=0 V⇒ 4πε 0 ⎩ r2 r2 ⎭

⇒ 3 (16 cm ) + y 2 = 4 2

−3.0 × 10−9 C

( −9.0 cm )

( −9 cm )

2

2

+ y2

+

4.0 × 10−9 C

(16.0 cm )

2

+ y2

=0

+ y 2 ⇒ 9 ( 256 cm2 + y 2 ) = 16 ( 81 cm2 + y 2 )

⇒ 7y2 = 1008 cm2 ⇒ y = ±12 cm.

29.31. Model: While the potential is the sum of the scalar potentials due to each charge, the electric field is the vector sum of the electric fields due to each charge. Visualize: Please refer to Figure EX29.31. Solve: (a) Ex points left for x > b because Ex is negative. So, the charge at x = b is negative. For x < a, Ex points left, so the charge at x = a is positive. For a < x < b, Ex is positive which is consistent with the charge choices. (b) By the symmetry of the drawing about the middle, it appears that the magnitudes of the charges are the same. q = 1. Thus a qb (c)

The graph of the electric potential shown in the figure is consistent with the electric field as well as the charges.

29.32. Model: While the potential is the sum of the scalar potentials due to each charge, the electric field is the vector sum of the electric fields due to each charge. Visualize: Please refer to Figure EX29.32. Solve: (a) As V is always positive, both charges must be positive. The positive signs for the two equal charges at x = a and at x = b are also consistent with the behavior of the potential in the range a < x < b. (b) By the symmetry of the drawing about the middle, we infer that the magnitudes of the charges are the same. Thus qa = 1. qb (c)

The graph of Ex, the x-component of the electric field, as a function of x is shown in the figure.

29.33. Model: Consider the rod to be a line of charge. Visualize:

Divide the rod into small segments, each with charge ±Δq.

Solve: Consider two segments, one positive and one negative, equally distant from the center of the rod. These segments are the same distance r from the dot. Thus the contribution of this pair of segments to the potential at the dot is 1 +Δq 1 −Δq V+ + V– = + =0 V 4πε 0 r 4πε 0 r

Since we can divide the entire rod into pairs of symmetrically placed segments, the net result of adding the potentials due to each pair is V = 0 V. Assess: This conclusion depends on the dot being directly outward from the midpoint of the rod. The potential is not zero at other points.

29.34. Solve: Let the unknown charges be Q1 and Q2. Then Q1 + Q2 = 30 × 10−9 C and U=

1 Q1Q2 1 Q1Q2 = = −180 × 10−6 J 4πε 0 r12 4πε 0 2.0 × 10−2 m

− (180 ×10−6 J )( 2.0 ×10−2 m )

= −4.0 × 10−16 C 2 9.0 ×109 N m 2 /C2 Solving the first equation for Q2 and substituting into the second equation, ⇒ Q1Q2 =

Q1 ( 30 ×10−9 C − Q1 ) = −4.0 ×10−16 C 2 ⇒ Q12 − ( 30 ×10−9 C ) Q1 − ( 4.0 ×10−16 C2 ) = 0

( 30 ×10 ⇒Q = 1

−9

C) ±

( −30 ×10

−9

2

C ) + 4 ( 4.0 ×10−16 C 2 ) 2

⇒ Q1 = +40 nC and −10 nC

That is, the two charges are −10 nC and 40 nC. Assess: As they must, the two charges when added yield a total charge of 30 nC, and when substituted into the potential energy equation yield U = −180 × 10−6 J .

29.35. Solve: Let the two unknown, positive charges be Q1 and Q2. They are separated by a distance r12 = 5.0 × 10−2 m. Their electric potential energy is U=

( 72 ×10−6 J )( 5.0 ×10−2 m ) = 4.0 ×10−16 C2 1 Q1Q2 = 72 ×10−6 J ⇒ Q1Q2 = 4πε 0 r12 9.0 ×109 N m 2 /C2

The electric force between these two charges is F= Assess:

1 Q1Q2 4.0 ×10−16 C 2 9 2 2 = 9.0 × 10 N m /C = 1.44 × 10−3 N ( ) 2 4πε 0 r122 ( 5.0 ×10−2 m )

As far as the magnitudes are concerned, F=

U 72 ×10−6 J = = 1.44 ×10−3 N r 5.0 ×10−2 m

29.36. Model: While the net potential is the sum of the scalar potentials due to each charge, the net electric field is the vector sum of the electric fields. Visualize:

Solve: (a) Let the two charges be labeled as Q1 = −2.0 nC and Q2 = +2.0 nC. As pictorially shown in the figure, the net electric field will not become zero anywhere except at infinity. We will now show this result mathematically. Let the point of zero electric field be at a distance x from the origin. Then, G G G Enet = E1 − E2 = 0 N/C ⇒ E1 = E2 ⇒

1

2.0 ×10−9 C

4πε 0 ( x + 1.0 cm )

2

=

1

2.0 × 10−9 C

4πε 0 ( x − 1.0 cm )2

⇒ (x + 1.0 cm)2 = (x – 1.0 cm)2 Obviously, this equation can only be satisfied if x = ∞. That is, the electric field is zero at +∞ and at −∞. (b) Because potential is a scalar quantity, the potential of charge Q1 is always negative and the potential of charge Q2 is always positive. These two scalar numbers will add to give zero potential at x = 0 m and at ±∞. Mathematically this can be shown as follows. When −1.0 cm ≤ x ≤ 1.0 cm, let the point of zero potential be at x. Then, the condition V1 + V2 = 0 V/m is

1

( −2 ×10

−9

C)

4πε 0 1.0 cm + x

+

1

( 2 ×10

−9

C)

4πε 0 (1.0 cm − x )

= 0 ⇒ −(1.0 cm + x) + (1.0 cm – x) = 0 ⇒ x = 0 cm.

When x > 1.0 cm, 1

( −2 ×10

−9

C)

4πε 0 ( x + 1.0 cm )

+

1

( 2 ×10

−9

C)

4πε 0 ( x − 1.0 cm )

This equation can only be satisfied at x = ±∞. (c) The graphs are shown in the figure above.

= 0 ⇒ (x + 1.0 cm) = (x – 1.0 cm)

29.37. Model: While the net potential is the sum of the potentials due to each charge, the net electric field is the vector sum of the electric fields. Visualize:

The charge Q1 = 20.0 nC is at the origin. The charge Q2 = −10.0 nC is 15.0 cm to the right of the charge Q1 on the x-axis. Solve: (a) As the pictorial representation shows, the point P on the x-axis where the electric field is zero can only be on the right side of the charge Q2, that is, at x ≥ 15.0 cm. At this point E1 = E2 , so we have

1 20.0 ×10−9 C 1 10.0 ×10−9 C = ⇒ x2 = 2(x – 15.0 cm)2 4πε 0 x2 4πε 0 ( x − 15.0 cm )2 ⇒ x2 − (60.0 cm)x + 450 cm2 = 0 ⇒ x =

( 60.0 cm ) ±

3600 cm 2 − 1800 cm 2 2

⇒ x = 51.2 cm and 8.8 cm. The root x = 8.8 cm is not possible physically. So, the electric fields cancel out at x = 51.2 cm. The electric potential at this point is

V=

⎡ 20.0 × 10−9 C 1 Q1 1 Q2 −10.0 ×10−9 C ⎤ + = ( 9.0 ×109 N m 2 /C 2 ) ⎢ + ⎥ = +103 V 4πε 0 r1 4πε 0 r2 ( 0.512 m − 0.150 m ) ⎥⎦ ⎢⎣ 0.512 m

(b) The point on the x-axis where the potential is zero can be obtained from the condition V1 + V2 = 0 V, which is

1 Q1 1 Q2 20.0 × 10−9 C −10.0 ×10−9 C + =0 V⇒ + =0 4πε 0 r1 4πε 0 r2 x (15.0 cm − x )

⇒ 2(15.0 cm – x) – x = 0 ⇒ x = 10.0 cm The electric field 10.0 cm away from charge Q1 is −9 G G G 1 20.0 ×10−9 C ˆ 1 (10.0 ×10 C ) ˆ Enet = E1 + E2 = i + i = 5.40 ×104iˆ N/C 2 4πε 0 ( 0.100 m ) 4πε 0 ( 0.050 m )2

29.38. Model: Mechanical energy is conserved. Visualize:

Solve:

Before the charges are released, the total mechanical energy is the potential energy of the three charges. 1 ⎡ q1q2 q1q3 q2 q3 ⎤ Ei = U12 + U13 + U 23 = + + ⎢ ⎥ r13 r23 ⎦ 4πε 0 ⎣ r12

⎡ ( 5.0 × 10−9 C )( −1.0 × 10−9 C ) ( 5.0 × 10−9 C )( 5.0 × 10−9 C ) ( −1.0 × 10−9 C )( 5.0 × 10−9 C ) ⎤ ⎥ = ( 9.0 × 109 Nm 2 /C2 ) ⎢ + + 0.010 m 0.020 m 0.010 m ⎢⎣ ⎥⎦ −6 = 2.25 × 10 J

After the charges are released and move very far (infinitely) apart, the total potential energy is zero. By the symmetry of the original configuration, the middle charge does not move when released, and the two 5.0 nC charges are moving at the same speed. The final energy is thus 1 1 1 ⎛1 ⎞ Ef = m1v12 + m2v2 2 + m3v32 = 2 ⎜ m1v 2 ⎟ = Ei = 2.25 × 10−6 J 2 2 2 ⎝2 ⎠ ⇒v=

2.25 ×10−6 J = 4.7 × 10−2 m/s = 4.7 cm/s 1.0 ×10−3 kg

29.39. Model: The charged beads are point charges. Visualize:

Solve: To find the maximum speeds vfA and vfB it is easier to use the conservation of momentum and conservation of energy equations than kinematics, as suggested in the hint. The beads achieve their maximum speeds when they are infinitely separated. The momentum conservation equation along x-direction pafter = pbefore = 0 kg m/s means −mAvfA + mBvfB = 0 kg m/s ⇒ −(0.015 kg)vfA + (0.025 kg)vfB = 0 kg m/s = 0 kg m/s ⇒ vfB = 53 vfA

The conservation of energy conservation is Uf + Kf = Ui + Ki. Noting that Uf → 0 J as the masses are infinitely separated and Ui =

9 2 2 −9 −9 1 QAQB ( 9.0 ×10 N m /C )( 5.0 ×10 C )(10.0 ×10 C ) = = 3.75 ×10−6 J 4πε 0 rAB 0.12 m

⇒0J +

⇒

1 2

(

1 2

2 2 mA vfA + 12 mBvfB ) = 3.75 × 10−6 J + 0 J

( 0.015 kg ) vfA2 + 12 ( 0.025 kg ) ( 53 vfA )

2

2 = 3.75 ×10−6 J ⇒ 0.012vfA = 3.75 ×10−6 J

Solving these equations yields vfA = 1.77 cm/s and vfB = 53 (1.77 cm/s ) = 1.06 cm/s. These are the maximum speeds of the two beads and occur when they are infinitely separated from each other.

29.40. Model: Mechanical energy is conserved. Metal spheres are point particles and they have point charges. Visualize:

Solve: (a) The system could have both kinetic and potential energy, although here K = 0 J. The energy of the system is

E0 = K 0 + U 0 = 0 J +

( 9.0 ×109 N m2 /C2 )( 2.0 ×10−6 C )( 2.0 ×10−6 C ) = 0.72 J qA qB = 4πε 0 r0 0.050 m

(b) In static equilibrium, the net force on sphere A is zero. Thus T = FB on A =

9 2 2 −6 −6 qA qB ( 9.0 ×10 N m /C )( 2.0 × 10 C )( 2.0 × 10 C ) = = 14.4 N 2 2 4πε 0 r0 ( 0.050 m )

(c) The spheres move apart due to the repulsive electric force between them. The surface is frictionless, so they continue to slide without stopping. When they are very far apart (r1 → ∞), their potential energy U1 → 0 J. Energy is conserved, so we have 2 2 E1 = K1 + U1 = 12 mAvA1 + 12 mBvB1 + 0 J = E0

Momentum is also conserved: Pafter = mAvA1 + mBvB1 = Pbefore = 0 kg m/s. Note that these are velocities and that vA1 is a negative number. From the momentum equation, mv vA1 = − B B1 mA Substituting this into the energy equation, 2 ⎛ mv ⎞ ⎛m ⎞ 2 2 E0 = 12 mA ⎜ − B B1 ⎟ + 12 mBvB1 = 12 mB ⎜ B + 1⎟ vB1 ⎝ mA ⎠ ⎝ mA ⎠

⇒ vB1 =

2 E0 = mB ( mB mA + 1)

2 ( 0.720 J ) = 10.95 m/s ( 0.004 kg )( 4 g 2 g + 1)

Using this result, we can then find vA1 = − mBvB1 mA = −21.9 m/s. These are the velocities, so the final speeds are 21.9 m/s for the 2 g sphere and 11.0 m/s for the 4 g sphere.

29.41. Model: Energy is conserved. The charged spheres are point charges. Visualize: Please refer to Figure P29.41. Label the spheres 1, 2, 3, and 4 in a clockwise manner, with the sphere in the upper left-hand corner being sphere 1. Solve: The total potential energy of the four spheres before they are allowed to move is Ui = (U12 + U23 + U34 + U41) + (U13 + U24) All the charges are identical, r12 = r23 = r34 = r41 = 1.0 cm, and r13 = r24 = 2.0 cm. The initial potential energy is

⎛ 1 10 × 10−9 C)(10 × 10−9 C) ⎞ ⎛ 1 10 × 10−9 C)(10 × 10−9 C) ⎞ −5 Ui = 4⎜ ⎟ + 2⎜ ⎟ = 48.73 × 10 J 0.010 m 0.01414 m ⎝ 4π e0 ⎠ ⎝ 4π e0 ⎠ Since all charges are at rest, Ki = 0 J. As the spheres are allowed to move away from one another and they are far apart, Uf = 0 J. The final kinetic energy is K f = 4( 12 mvf2 ) = 2(1.0 × 10−3 kg)vf2

From the energy conservation equation Uf + Kf = Ui + Ki,

2 (1.0 × 10−3 kg ) vf2 = 48.73 × 10−5 J ⇒ vf = 0.49 m/s

29.42. Model: Energy is conserved. The potential energy is determined by the electric potential. Visualize: Please refer to Figure P29.42. Solve: The proton at point A is at a potential of 30 V and its speed is 50,000 m/s. At point B, the proton is at a potential of −10 V and we are asked to find its speed. Clearly, the proton moves into a lower potential region, so its speed will increase. The conservation of energy equation Kf + Uf = Ki + Ui is 1 2

⇒ vf =

Assess:

mvf2 + ( +e )( −10 V ) = 12 m ( 50,000 m/s ) + ( + e )( 30 V ) 2

( 50,000 m/s )

2

+

2 (1.60 × 10−19 C ) ( 40 V ) 1.67 ×10−27 kg

The speed of the proton is higher, as expected.

= 1.01× 105 m/s

29.43. Model: Energy is conserved. Solve:

(a)

(b) The potential energy of the positive charge as a function of x is U(x) = 5000 qx2. This is analogous to the potential energy of a mass on a spring. Thus the motion is simple harmonic motion. (c) Turning points at ±8.0 cm mean that xmax = ±8.0 cm. At the turning points, the energy is all potential energy and the kinetic energy is zero. The mechanical energy of the charged particle is

E = U + K = U max + 0 J = 5000 q ( xmax ) = 5000 (10.0 ×10 −9 C ) ( 0.080 m ) = 3.2 ×10−7 J 2

2

(d) The conservation of energy equation K1 + U1 = K2 + U2 is 2 = 3.20 ×10−7 J ⇒ vmax = Kmax + 0 J = 0 J + Umax ⇒ 12 mvmax

2 ( 3.20 ×10−7 J )

(1.0 ×10

−3

kg )

= 2.5 cm/s

29.44. Model: Energy is conserved. Solve:

(a)

(b) The potential energy of the positively charged proton as a function of x is

U ( x ) = eV ( x ) = 6000ex 2 = 6000 (1.60 ×10−19 C ) x 2 = 9.6 × 10−16 x 2 J. This is analogous to the potential energy of a mass on a spring. Thus the motion is simple harmonic motion. 1 (c) The form for the potential energy of a mass on a spring is U ( x ) = kx 2 . Comparing that to the expression 2 above yields 1 keff = 9.6 ×10−16 J/m 2 = 9.6 × 10−16 N/m ⇒ keff = 1.92 × 10−15 N/m 2 Assess: The constant in the given expression for V carry units of V/m2. This must be carefully accounted for in later expressions.

29.45. Model: Energy is conserved. The proton’s potential energy inside the capacitor can be found from the capacitor’s potential difference. Visualize: Please refer to Figure P29.45. Solve: (a) The electric potential at the midpoint of the capacitor is 250 V. This is because the potential inside a parallel-plate capacitor is V = Es where s is the distance from the negative electron. The proton has charge q = e and its potential energy at a point where the capacitor’s potential is V is U = eV. The proton will gain potential energy ΔU = eΔV = e(250 V) = 1.60 × 10−19 C (250 V) = 4.00 × 10−17 J if it moves all the way to the positive plate. This increase in potential energy comes at the expense of kinetic energy which is

K = 12 mv 2 =

1 2

(1.67 ×10

−27

kg ) ( 200,000 m/s ) = 3.34 × 10−17 J 2

This available kinetic energy is not enough to provide for the increase in potential energy if the proton is to reach the positive plate. Thus the proton does not reach the plate because K < ΔU. (b) The energy-conservation equation Kf + Uf = Ki + Ui is 1 2

⇒ vf = vi2 +

mvf2 + qVf = 12 mvi2 + qVi ⇒ 12 mvf2 = 12 mvi2 + q (Vi − Vf )

2q (Vi − Vf ) = m

( 2.0 × 10

5

m/s ) + 2

2 (1.60 × 10−19 C ) ( 250 V − 0 V ) 1.67 × 10−27 kg

= 2.96 × 105 m/s

29.46.

Model:

Energy is conserved.

Visualize:

Solve:

(a) The electric field inside a parallel-plate capacitor is constant with strength

E=

3 ΔV ( 25 ×10 V ) = = 2.1×106 V/m. 0.012 m d

(b) Assuming the initial velocity is zero, energy conservation yields

Ui = Kf + U f 1 0 = mevf 2 + ( −e )( Ed ) 2 ⇒ vf = Assess:

2 (1.60 ×10−19 C )( 2.1× 106 V/m ) ( 0.012 m ) 9.11×10−31 kg

= 9.4 ×107 m/s

This speed is about 31% the speed of light. At that speed, relativity must be taken into account.

29.47. Model: Mechanical energy is conserved. Visualize:

Solve: (a) Both gravitational and electric potential energy are involved. The electric potential between the plates is V = Es, where s is measured from the more negative plate. The electric field between the plates is

E=

ΔV 3.0 ×106 V = = 1.00 ×106 V/m. 3.0 m d

Conservation of energy is used to find the height. K i + U gi + U ei = K f + U gf + U ef 1 mvi 2 + 0 J + q ( 0 V ) = 0 J + mgh + q ( Eh ) 2

1 1 2 mvi 2 1.0 ×10−3 kg ) ( 5.0 m/s ) ( 2 ⇒h= 2 = = 0.85 m mg + qE (1.0 ×10−3 kg )( 9.8 m/s 2 ) + ( 4.0 × 10−9 C )(1.00 × 106 V/m ) (b) The top plate is at a negative potential, so the charge is attracted to it. The electric potential is now zero at the top plate and positive at the bottom plate. Does the charge make it all the way to the top plate? If we assume it does not and obtain a height more than 3.0 m then we will know that it does. Conservation of energy for the charge requires 1 1 mvi 2 + 0 J + qE ( 3.0 m ) = mgh + qE ( 3.0 m − h ) ⇒ mvi 2 = ( mg − qE ) 2 2

1 mvi 2 ⇒h= 2 = 2.6 m mg − qE Assess: Since h < 3.0 m the charge did not hit the top plate.

29.48.

Model: Visualize:

Solve:

Mechanical energy is conserved.

The initial energy of the electron is the same as the energy at the turning point, when its speed is zero.

1 Ei = Ef ⇒ mevi 2 + U1i + U 2i = U1f + U 2f 2 ⎛ 1 1 ( e )( −e ) ⎞ = 2 ⎛ 1 ( e )( −e ) ⎞ ⇒ mevi 2 + 2 ⎜ ⎟ ⎜ ⎟ −11 r 2 ⎝ 4πε 0 5.5 ×10 m ⎠ ⎝ 4πε 0 ⎠ 2 (1.6 ×10 C ) 1 ⇒ ( 9.11×10−31 kg )(1.5 × 106 m/s ) − 2 ( 9.0 ×109 Nm 2 /C2 ) 2 5.5 ×10−11 m −19

= −2 ( 9.0 ×109 Nm 2 /C2 )

(1.6 ×10

−19

C)

2

2

r

⇒ 1.025 ×10−18 J − 8.38 ×10−18 J = −

4.61×10−28 Nm 2 r

⇒ r = 6.27 ×10−11 m = 0.0627 nm Since r 2 = ( 0.055 nm ) + y 2 , y = 2

( 0.0627 nm ) − ( 0.055 nm ) 2

2

= 0.030 nm.

Assess: The electron moves a distance outward that is less than the distance to each of the protons.

29.49. Model: Energy is conserved. The electron ends up so far away from the glass sphere that we can consider its potential energy to be zero. Visualize:

The minimum speed to escape is the speed that allows the electron to reach rf = ∞ when vf = 0 m/s. Solve: The conservation of energy equation Kf + Uf = Ki + Ui is

⎛ 1 q⎞ 0 J + 0 J = 12 mvi2 + qVi ⇒ 0 J = 12 mvi2 + ( −e ) ⎜ ⎟ ⎝ 4πε 0 R ⎠ ⇒ vi =

2 (1.60 ×10−19 C ) ⎛ 10 ×10−9 C ⎞ 2e 1 q 7 = 9.0 × 109 N m 2 /C2 ) ⎜ ( ⎟ = 8.0 × 10 m/s −31 −2 m 4πε 0 R 9.11× 10 kg 0.50 × 10 m ⎝ ⎠

29.50. Model: Energy is conserved. The potential energy is determined by the electric potential. Solve:

The conservation of energy Kf + Uf = Ki + Ui is

Kf + qVf = Ki + qVi ⇒ Kf = Ki + q(Vi – Vf) In this equation, K i = mv = 0 J , the final proton potential is zero, and the initial proton potential is V. We have 1 2

2 i

Kf = eV. That is, the maximum kinetic energy or maximum speed is directly proportional to the potential of the surface the proton is launched from. From Equation 29.32, Example 29.11, and Example 29.12, when z → 0 for the ring and disk:

Vsphere =

1 Q 4πε 0 R

Vring =

1 Q 4πε 0 R

Vdisk =

1 2Q 4πε 0 R

Because the potential on the disk is a factor of 2 larger than the potential on the sphere or the ring, the proton’s kinetic energy or the speed will be maximum in the case of the disk.

29.51. Model: Energy is conserved. Visualize:

The dipole initially is at the higher potential energy. As it moves to its minimum energy state, the decrease in the potential energy appears as rotational kinetic energy. G G Solve: The potential energy of a dipole is U dipole = − p ⋅ E = − pE cosθ , where θ is the angle between the dipole

and the electric field. The energy conservation equation Kf + Uf = Ki + Ui is 1 2

I ω 2 + U f = 0 J + U i ⇒ 12 I ω 2 = U i − U f = − pE cos90° − ( − pE cos 0° ) ⇒ 12 I ω 2 = pE ⇒ ω =

2 pE I

The moment of inertia of the dipole is I dipole = 2m ( 12 s ) . Substituting into the above equation, 2

ω=

2 ( qs ) E

2m ( 12 s )

2

=

4 ( 2.0 × 10−9 C ) ( 0.10 m )(1000 V/m )

(1.0 × 10

−3

kg ) ( 0.10 m )

2

= 0.28 rad/s

29.52. Model: The electrons and the proton are point charges. Visualize:

Solve:

We are given that r12 = r23 = r13 = 1.0 ×10−9 m . From the geometry of the figure,

r r23 = cos30° ⇒ r24 = = 0.5774 × 10−9 m = r14 = r34 r24 2cos30°

1 2 23

The contributions to the total potential energy are

U12 = U13 = U 23 = U14 = U 24 = U 34 =

( 9.0 ×10

9

N m 2 /C2 )( −1.60 ×10−19 C )( −1.60 ×10−19 C ) 1.0 ×10−9 m

( 9.0 ×10

9

N m 2 /C2 )( −1.60 ×10−19 C )(1.60 ×10−19 C ) 0.5774 ×10−9 m

= 2.304 ×10−19 J

= −3.990 ×10−19 J

Summing all of the contributions,

U elec = U12 + U13 + U 23 + U14 + U 24 + U 34

= 3( 2.304 ×10−19 J ) + 3( −3.990 ×10−19 J ) = −5.1×10−19 J

Assess:

Note that U12 = U21, U13 = U31, and U23 = U32, U14 = U41, U24 = U42, and U34 = U43.

29.53. Model: Outside a charged sphere the electric potential is identical to that of a point charge at the center. Visualize:

Solve:

For r ≥ R, V = Q 4πε 0 r . The potentials at the two points are

V2 mm =

( 9.0 ×10

9

N m 2 /C2 ) Q

(1 mm + 2 mm )

( 9.0 ×10

9

=

N m 2 /C2 ) Q −3

3.0 ×10 m

( 9.0 ×10

9

V4 mm =

N m 2 /C2 ) Q

5.0 ×10−3 m

1 1 ⎛ ⎞ ⇒ V2 mm − V4 mm = +500 V = ( 9.0 ×109 N m 2 /C2 ) Q ⎜ − ⎟ −3 −3 × × 3.0 10 m 5.0 10 m ⎝ ⎠ ⇒Q= Assess:

⎛ 15.0 ×10−6 m 2 ⎞ 500 V = 4.2 ×10−10 C ( 9.0 ×109 N m2/C2 ) ⎜⎝ 2.0 ×10−3 m ⎟⎠

Do not forget to include the radius of the glass bead in r.

29.54. Model: Energy is conserved. Because the iron nucleus is very large compared to the proton, we will assume that the nucleus does not move (no recoil) and that the proton is essentially a point particle with no diameter. Visualize:

The proton is fired from a distance much greater than the nuclear diameter, so ri ≈ ∞ and Ui ≈ 0 J. Because the nucleus is so small, a proton that is even a few atoms away is, for all practical purposes, at infinity. As the proton approaches the nucleus, it is slowed by the repulsive electric force. At the end point, the proton has just reached the surface of the nucleus (rf = nuclear diameter) with vf = 0 m/s. (The proton won’t remain at this point but will be pushed back out again, but the subsequent motion is not part of this problem.) Solve: Initially, the proton has kinetic energy but no potential energy. At the point of closest approach, where vf = 0 m/s, the proton has potential energy but no kinetic energy. Energy is conserved, so Kf + Uf = Ki + Ui. This equation is

0 J+

( e )( 26e ) = 1 m 4πε 0 rf

2

v +0 J

2 proton i

where rf is half the nuclear diameter. The initial speed of the proton is

vi =

2 (1.6 ×10−19 C )( 26 ×1.6 ×10−19 C )( 9.0 ×109 N m 2 / C 2 ) 2 ( e )( 26e ) = = 4.0 ×107 m/s 4πε 0 rf mproton (1.67 ×10−27 kg )( 4.5 ×10−15 m )

29.55. Model: Energy is conserved. Because the mercury nucleus is very large compared to the proton, we will assume that the nucleus does not move (no recoil) and that the proton is essentially a point particle with no diameter. Visualize:

The proton is fired from a distance much greater than the nuclear diameter, so ri = ∞ and Ui = 0 J. Because the nucleus is so small, a proton that is even a few atoms away is, for all practical purposes, at infinity. As the proton approaches the nucleus, it is slowed by the repulsive electric force. At the end point of the problem, the proton reaches the distance of closest approach (d) with vf = 0 m/s. (The proton won’t remain at this point but will be pushed back out again, but the subsequent motion is not part of the problem.) Solve: Initially, the proton has kinetic energy but no potential energy. At the point of closest approach, where vf = 0 m/s, the proton has potential energy but no kinetic energy. Energy is conserved, so Kf + Uf = Ki + Ui . This equation is

0 J+

e ( 80e ) 1 = mproton vi2 + 0 J 4πε 0 d 2

( 9.0 ×109 N m2 / C2 ) (160 ) (1.6 × 10−19 C ) = 1.38 × 10−14 m = 13.8 fm 160e2 ⇒d = = 2 2 4πε 0 mproton vi (1.67 × 10−27 kg )( 4.0 ×107 m/s ) 2

The radius of the nucleus is 7.0 fm, so the proton’s closest approach to the surface is 6.8 fm.

29.56. Model: Energy is conserved. Visualize:

The alpha particle is initially at rest (vi alpha = 0 m/s) at the surface of the thorium nucleus. The potential energy of the alpha particle is U i alpha . After the decay, the alpha particle is far away from the thorium nucleus, Uf alpha = 0 J,

and moving with speed vf alpha . Solve: Initially, the alpha particle has potential energy and no kinetic energy. As the alpha particle is detected in the laboratory, the alpha particle has kinetic energy but no potential energy. Energy is conserved, so Kf alpha + Uf = alpha Ki alpha + Ui alpha. This equation is 1 2

⇒ vf alpha =

1 4πε 0

mvf2alpha + 0 J = 0 J +

1

( 2e )( 90e )

4πε 0

ri

( 360e ) = ( 9.0 ×10 N m /C ) 360 (1.60 ×10 C ) mr 4 (1.67 ×10 kg )( 7.5 ×10 m ) 2

9

2

−27

i

−19

2

−15

2

= 4.1× 107 m/s

29.57. Model: Energy is conserved. Visualize:

Solve: (a) The neutron has zero charge. In the decay n → p+ + e– + v, where the neutrino v is neutral, the net charge of the final state is e + (−e) + 0 = 0. So charge is conserved. (b) As shown in the figure, the tritium nucleus has one proton and two neutrons. It is the single proton that makes this an isotope of hydrogen, the Z = 1 element in the periodic table. One of the neutrons undergoes the decay n → p + e + v. The electron and neutrino are ejected from the nucleus as beta radiation, but the new proton remains in the nucleus. The nucleus still has three nucleons, but now two are protons and only one is a neutron. An atom whose nucleus contains two protons is the Z = 2 element of the periodic table, namely helium. (c) As shown in the figure, the electron is “launched” from the surface of the 3He nucleus with speed vi. The initial energy is

Ei = K i + U i = 12 mevi2 +

qelec qnuc 1 ( −e )( 2e ) = 2 mevi2 + 4πε 0 ri 4πε 0 ri

If the electron just barely escapes, it slows to vf → 0 m/s as rf → ∞. Its final energy is Ef = Kf + Uf = 0 J + 0 J = 0 J. Energy is conserved, Ef = Ei, so 1 2

⇒ vi = −

2 ( −e )( 2e ) = me 4πε 0 ri

mevi2 +

( −e )( 2e ) = 0 J 4πε 0 ri

( 2 ) ( 9.0 ×109 N m 2 /C2 ) ( 2 ) (1.60 ×10−19 C )

( 9.11×10

−31

kg )(1.5 ×10−15 m )

2

= 8.21× 108 m/s

Assess: Our “classical” analysis gives an answer that exceeds the speed of light (3 × 108 m/s), which is forbidden by Einstein’s theory of relativity. We would need to perform a relativistic analysis, which is beyond the scope of this chapter, to get the correct minimum escape speed.

29.58.

Model: Energy is conserved in the collision. Solve: (a) The centers of the protons are 2.4 fm apart when they touch and fusion occurs. Assume they start very far apart so that the initial electric potential energy between them may be ignored. The minimum temperature required is found if it is assumed they have zero velocity just as they touch. Energy conservation requires

K1i + K 2i + U12i = K1f + K 2f + U12f 1 1 1 e2 mp v1i 2 + mp v2i 2 + 0 J = 0 J + 0 J + 2 2 4πε 0 2.4 × 10−15 m Since the protons both have exactly the same average kinetic energy of 3/2 kBT,

1 e2 ⎛3 ⎞ 2 ⎜ k BT ⎟ = −15 ⎝2 ⎠ 4πε 0 2.4 ×10 m

( 9.0 ×10 Nm /C ) (1.6 ×10 3(1.38 ×10 J/K ) 2.4 ×10 9

⇒T =

2

2

−23

−19 −15

C) m

2

= 2.3 ×109 K

(b) The ratio of the kinetic energy of the protons in part (a) to the average kinetic energy of protons with a temperature of 15 million K is 3

2

3

k B ( 2.3 ×109 k )

2

k B (15 ×106 k )

= 153.

The energy of the proton undergoing fusion exceeds the average proton energy by a factor of 153. Assess: Then how can fusion take place in the sun? The answer lies in the fact that there is a distribution of proton velocities. Only protons with velocities in the long high-velocity tail of the Maxwell-Boltzman velocity distribution participate in fusion in the sun’s core.

29.59. Model: The electric field inside a capacitor is uniform. Solve: (a) Because the parallel-plate capacitor was connected to the terminals of a 15 V battery for a long time, the potential difference across the capacitor right after the battery is disconnected is ΔV= 15 V. The electric field strength inside the capacitor is

E=

ΔVC 15 V = = 3000 V/m = 3.0 kV/m d 0.50 ×10−2 m

Because E = η ε 0 for a parallel-plate capacitor and η = Q A , the total charge on each plate is Q = EAε 0 = ( 3000 V/m ) π ( 0.050 m ) (8.85 ×10 −12 C 2 /Nm 2 ) = 2.1× 10−10 C 2

(b) After the electrodes are pulled away to a separation of d ′ = 1.0 cm, the charges on the plates are unchanged. That is, Q′= Q. Because A′ = A, the electric field inside the capacitor is also unchanged. So, E′ = E. The potential difference across the capacitor is ΔVC′ = E ′d ′. Because d increases from 0.50 cm to 1.0 cm (d′ = 1.0

cm), the potential difference ΔVC′ increases from 15 V to 30 V. (c) When the electrodes are expanded, the new area is A′ = π(r′)2 = π(2r)2 = 4A. The charge Q′ on the capacitor plates, however, stays the same as before (Q′ = Q). The electric field is

E′ =

η ′ Q′ Q E 3000 V/m = = = = = 750 V/m = 0.75 kV/m ε 0 A′ε 0 4 Aε 0 4 4

The potential difference across the capacitor plates ΔVC′ = E′d ′ = ( 750 V )( 0.050 m ) = 3.8 V.

29.60. Model: The electric field inside a capacitor is uniform. Solve: (a) While the capacitor is attached to the battery, the plates are at the same potentials as the terminals of the battery. Thus, the potential difference across the capacitor is ΔVC = 15 V. The electric field strength inside the capacitor is

E=

ΔVC 15 V = = 3000 V/m = 3.0 kV/m d 0.50 ×10−2 m

Because E = η ε 0 = Q Aε 0 , the charge on each plate is

Q = EAε 0 = (3000 V/m) π (0.050 m)2 (8.85 × 10−12 C2/N m2) = 2.1 × 10−10 C (b) After the electrodes are pulled away to a new separation of d′ = 1.0 cm, the potential difference across the capacitor stays the same as before. That is, ΔVC′ = ΔVC = 15 V. The electric field strength inside the capacitor is

E′ =

ΔV ′ 15 V = = 1500 V/m = 1.50 kV/m ′ d 0.010 m

The charge on each plate is

Q′ = E ′Aε 0 = (1500 V)π (0.05 m) 2 (8.85 × 10−12 C2 / N m 2 ) = 1.04 × 10−10 C (c) The potential difference ΔVC′ = ΔVC = 15 V is unchanged when the electrodes are expanded to area A′ = 4A . The electric field between the plates is

E′ = E =

ΔVC′ 15 V = = 3000 V/m = 3.0 kV/m d′ 0.50 ×10−2 m

The new charge is Q′ = E ′A′ε 0 = 4 EAε 0 = 4Q = 8.3 × 10−12 C

29.61. Solve: (a) Outside a uniformly charged sphere, the electric field is identical to that of a point charge Q at the center. For r ≥ R, E=

1 Q 4πε 0 r 2

Evaluating this at r = R and using Equation 29.32, E0 =

1 Q V0 = 4πε 0 R 2 R

(b) The field strength is E0 =

500 V = 100,000 V/m = 100 kV/m 0.50 ×10−2 m

29.62. Visualize:

Solve: A sphere of radius R and charge Q has a surface potential V0 = Q 4πε 0 R . When the two drops A and B merge to form drop C, two quantities are conserved: the total charge and the total volume of mercury. The conservation of charge yields

QC = Qtotal = QA + QB = 0.10 nC + 0.10 nC = 0.20 nC Since the quantity of mercury does not change, the volumes of drops A, B, and C are related by ⎡⎣ VolC = 34 π RC3 ⎤⎦ = ⎡⎣ VolA + Vol B = 34 π RA3 + 43 π RB2 ⎤⎦ ⇒ RC = ( RA3 + RB3 )

1/ 3

We’re not given the radii of drops A and B, but we know their surface potentials (V0)A and (V0)B. Thus,

(V0 )A =

( 0.10 ×10 QA QA = ⇒ RA = 4πε 0 (V0 )A 4πε 0 RA

−9

C )( 9.0 ×109 N m 2 /C 2 ) 300 V

= 3.0 × 10−3 m = RB

Combining these radii gives RC = 3.8 × 10−3 m, and

(V0 )C =

( 0.20 ×10−9 C )( 9.0 ×109 N m 2/C2 ) = 0.47 kV QC = 4πε 0 RC 3.8 × 10−3 m

29.63. Solve: (a) Because the excess charge resides on the outer surface of a conductor, the charge placed on the inside surface of a hollow metal sphere will move rapidly to the outside surface. (b) Because a spherical shell of charge has the same electric potential as a point charge Q at the center, the potential on the surface is V=

1 Q ⎛ Q ⎞ ⇒ 500,000 V = ( 9.0 × 109 N m 2 /C 2 ) ⎜ ⎟ ⇒ Q = 8.3 μ C 4πε 0 R ⎝ 0.15 m ⎠

(c) The electric field inside a conductor is zero, so the electric field just inside the sphere is zero. From Gauss’s Law, the electric field outside a charged conductor is

E=

η Q 8.33 ×10−6 C = = = 3.3 ×106 V/m ε 0 Aε 0 4π ( 0.15 m )2 ( 8.85 ×10−12 C 2 /Nm 2 )

29.64. Solve: A charged particle placed inside a uniformly charged spherical shell experiences no electric force. That is, E = 0 V/m inside the shell. We know from Section 29.5 that a difference in potential between two points or two plates is the source of an electric field. Since the potential on the surface of the shell is V = Q 4πε 0 R , the potential inside must be the same. This ensures that the potential difference is zero and hence the electric field is zero inside the shell. The potential at the center of the spherical shell is thus the same as at the surface. That is, Vcenter = Vsurface = Q 4πε 0 R .

29.65. Model: The potential at any point is the superposition of the potentials due to all charges. Outside a uniformly charged sphere, the electric potential is identical to that of a point charge Q at the center. Visualize: Please refer to Figure P29.65. Sphere A is the sphere on the left and sphere B is the one on the right. Solve: The potential at point a is the sum of the potentials due to the spheres A and B:

Va = VA at a + VB at a =

1 QA 1 QB + 4πε 0 RA 4πε 0 0.70 m

100 ×10−9 C 25 ×10−9 C + ( 9.0 × 109 N m 2 /C2 ) 0.30 m 0.70 m = 3000 V + 321 V = 3321 V = ( 9.0 ×109 N m 2 /C 2 )

Similarly, the potential at point b is the sum of the potentials due to the spheres A and B: Vb = VB at b + VA at b =

1 QB 1 QA + 4πε 0 RB 4πε 0 0.95 m

⎛ 25 ×10−9 C 100 ×10−9 C ⎞ = ( 9.0 ×109 N m 2 /C2 ) ⎜ + ⎟ 0.95 m ⎠ ⎝ 0.05 m = 4500 V + 947 V = 5447 V Thus, the potential at point b is higher than the potential at a. The difference in potential is Vb – Va = 5447 V – 3321 V = 2126 V = 2.1 kV. Assess: VA at a = 3000 V and the sphere B has a potential of 225 V at point a. The spherical symmetry dictates that the potential on a sphere’s surface be the same everywhere. So, in calculating the potential at point a due to the sphere B we used the center-to-center separation of 1.0 m rather than a separation of 100 cm – 30 cm = 70 cm from the center of sphere B to the point a. The former choice leads to the same potential everywhere on the surface whereas the latter choice will lead to a distribution of potentials depending upon the location of the point a. Similar reasoning also applies to the potential at point b.

29.66. Model: The charges making the dipole are point charges. The potential is the sum of potentials from each charge. Visualize:

Point P is far away compared to the separation s of the dipole charges, that is, y >> s. Solve: (a) The potentials at P due to the positive and negative charges of the dipole are

V+ =

⇒ Vdipole = V+ + V− =

1 q 4πε 0 y − s 2

V− =

1 ( −q ) 4πε 0 y + s 2

⎞ q ⎛ 1 1 ⎞ q ⎛ s 1 p 1 p − ≅ ⎜ ⎟= ⎜ 2 2 ⎟= 2 2 4πε 0 ⎝ y − s 2 y + s 2 ⎠ 4πε 0 ⎝ y − s 4 ⎠ 4πε 0 y − s 4 4πε 0 y 2

(b) The potential due to the dipole moment of water is

Vdipole = ( 9.0 ×109 N m 2 /C 2 )

6.2 ×10−30 C m

(1.0 ×10

−9

m)

2

= 0.056 V

29.67. Model: The potential is the sum of potentials from each charge. Visualize:

Point P is located on the x-axis and the two positive charges are located on the y-axis. The separation between the two charges is s. Solve: (a) The potentials at P due to the two positive charges are

Vtop =

1 4πε 0

( +q ) x +s 4 2

⇒ Vq + q = Vtop + Vbottom =

2

Vbottom =

1 4πε 0

2q x +s 4 2

2

1 4πε 0 =

2q

( +q ) x2 + s2 4 1

4πε 0 x 1 + s 2 4 x 2

(b) The V-versus-x graph is shown in the figure. The potential due to a charge 2q located at the origin is V2 q = 2q 4πε 0 x , and is also plotted. This expression differs from the expression obtained in part (a) by the factor

(1 + s

2

4 x 2 ) 2 . For x >> s , this factor becomes 1 and the two potentials become the same. −1

29.68. Model: The net potential is the sum of potentials from all the charges. Visualize: Please refer to Figure P29.68. Point P at which we want the net potential due to the linear electric quadrupole is far away compared to the separation s, that is, y >> s. Solve: The net potential at P is

Vnet = V1 + V2 + V3 = =

1

q1 1 q2 1 q3 + + 4πε 0 y − s 4πε 0 y 4πε 0 y + s

1 ⎡ ( + q ) ( −2q ) ( + q ) ⎤ q ⎡ y 2 + ys − 2 y 2 + 2 s 2 + y 2 − ys ⎤ q 2s 2 ⎢ ⎥ + + = = ⎢ ⎥ 2 2 4πε 0 ⎣ y − s y y + s ⎦ 4πε 0 ⎢ y ( y2 − s2 ) ⎥⎦ 4πε 0 y ( y − s ) ⎣

At distances y >> s , Vnet =

1 4πε 0

( 2qs ) = 2

y3

1 Q 4πε 0 y 3

where Q = 2qs2 is called the electric quadrupole moment of the charge distribution. Assess: This charge distribution is in fact a combination of two dipoles. As seen above their effects do not completely cancel.

29.69. Model: Assume the thin rod is a line of charge with uniform linear charge density. Visualize: Please refer to Figure P29.69. The point P is a distance d from the origin. Divide the charged rod into N small segments, each of length Δx and with charge Δq. Segment i, located at position xi, contributes a small amount of potential Vi at point P. Solve: The contribution of the ith segment is

Vi =

Δq Δq Q Δx L = = 4πε 0 ri 4πε 0 ( d − xi ) 4πε 0 ( d − xi )

where Δq = λΔx and the linear charge density is λ = Q L . We are placing the point P at a distance d rather than x from the origin to avoid confusion with xi. The Vi are now summed and the sum is converted to an integral giving V=

L/2 ⎛d +L 2⎞ dx Q Q = ⎡⎣ − ln ( d − x ) ⎤⎦ − L / 2 = ln ⎜ ⎟ 4πε 0 L − L/ 2 d − x 4πε 0 L 4πε 0 L ⎝ d − L 2 ⎠

Q

L/ 2

∫

Replacing d with x, the potential due to a line charge of length L at a distance x along the axis is V=

⎛ x+L 2⎞ Q ln ⎜ ⎟ 4πε 0 L ⎝ x − L 2 ⎠

29.70. Model: Assume the thin rod is a line of charge with uniform linear charge density. Visualize: Please refer to Figure P29.69. The point P is located a distance z from the origin. Divide the charged rod into N small segments, each of length Δx and with charge ΔQ. Segment i, located at position xi, contributes a small amount of potential Vi at point P. Solve: The contribution of the ith segment is

Vi =

ΔQ ΔQ = 4πε 0ri 4πε 0 xi2 + z 2

Charge ΔQ is related to length Δx through the linear charge density: ΔQ = λΔx = ( Q L ) Δx . Thus

Vi =

QΔx 4πε 0 L xi2 + z 2

Potential is a scalar, so we can find the net potential at P by summing all the Vi: V = ∑Vi = i

Q Δx ∑ 4πε 0 L i xi2 + z 2

If we now let Δx → dx, the sum becomes an integral over the length of the rod, from xmin = −L/2 to xmax = +L/2:

V=

Q

4πε L ∫ 0

L/2

−L / 2

dx x2 + z 2

=

Q

⎡ln 4πε 0 L ⎣⎢

(

⎡ z 2 + ( L 2 )2 + L L/2 Q x2 + z 2 + x ⎤ ln ⎢ = ⎦⎥ − L / 2 4πε 0 L ⎢ z 2 + ( L 2 )2 − L ⎣

)

2⎤ ⎥ 2 ⎥⎦

We can also get a simpler expression by using symmetry to integrate from 0 to L/2 and multiplying by 2. This gives

V=

2 ⎡⎛ L ⎞ ⎛ L⎞ ⎤ ln ⎢⎜ ⎟ + 1 + ⎜ ⎟ ⎥ 2πε 0 L ⎢⎝ 2z ⎠ ⎝ 2 z ⎠ ⎥⎦ ⎣

Q

This expression can be shown to be equivalent to the one obtained earlier.

29.71. Model: Because the rod is thin, assume the charge lies along the semicircle of radius R. Visualize: Please refer to Figure P29.71. The bent rod lies in the xy-plane with point P as the center of the semicircle. Divide the semicircle into N small segments of length Δs and of charge ΔQ = ( Q π R ) Δs, each of which can be modeled as a point charge. The potential V at P is the sum of the potentials due to each segment of charge. Solve: The total potential is

V = ∑Vi = ∑ i

i

1 ΔQ 1 1 Q 1 Q ⎛ Q ⎞ = R Δθ = ∑ ∑ Δθ . ⎜ ⎟ Δs = 2 ∑ 4πε 0 R 4πε 0 R i ⎝ π R ⎠ 4πε 0 π R i 4πε 0 π R i

All of the terms come to the front of the summation because these quantities did not change as far as the summation is concerned. The summation does not have to convert to an integral because the sum of all the Δθ around the semicircle is π. Hence, the potential at the center of a charged semicircle is

Vcenter =

1 Qπ 1 Q = 4πε 0 π R 4πε 0 R

2 29.72. Model: The disk has a uniform surface charge density η = Q A = Q π ( Rout − Rin2 ) .

Visualize: Please refer to Figure 29.31. Orient the disk in the xy-plane, with point P at distance z. Divide the disk into rings of equal width Δr. Ring i has radius ri and charge ΔQi. Solve: Using the result of Example 29.11, we write the potential at distance z of ring i as

Vi =

1 4πε 0

ΔQi ri + z 2

2

⇒ V = ∑Vi = i

1 ΔQi ∑ 4πε 0 i ri 2 + z 2

Noting that ΔQi = ηΔAi = η2πriΔr, V= =

1 4πε 0

η 2π ∑ i

ri Δr ri + z 2

2

=

η 2ε 0

Rout

∫

Rin

r dr r +z 2

2

=

η ⎡ 2 2 ⎤R η ⎡ 2 r +z Rout + z 2 − Rin2 + z 2 ⎤ = ⎣ ⎦ ⎦ R 2ε 0 2ε 0 ⎣

Q ⎡ R2 + z 2 − R2 + z 2 ⎤ out in 2 2 ⎣ ⎦ 2πε 0 ( Rout − Rin )

In the limit Rin → 0 m,

V=

Q ⎡ 2 Rout + z 2 − z ⎤ 2 ⎣ ⎦ 2πε 0 Rout

This is the same result obtained for a disk of charge in Example 29.12.

out

in

29.73. Solve: (a) A charge of 40 nC is separated into two charges which are placed 3.0 cm apart. The potential energy of the charge arrangement is 90 μJ. What are the two charges?

(b) We have q1q2 = 3.0 × 10−16 C2 and q1 + q2 = 40 × 10−9 C. Substituting into the first equation an expression for q1 from the second equation,

( 40 ×10

−9

C − q2 ) q2 = 3.0 × 10−16 C 2 ⇒ q22 − ( 40 × 10−9 C ) q2 + 3.0 × 10−16 C 2 = 0

The solutions to this equation are q2 = 30 nC and 10 nC. This means the two charges are 10 nC and 30 nC.

29.74. Solve: (a) A proton is shot toward a 2.0 nC point charge with a speed of 2.5 × 106 m/s. What is the proton’s speed when it is 1.0 mm from the charge? (b) Solving the equation yields vi = 1.67 × 106 m/s.

29.75. Solve: (a) Two 3.0 nC charges are distance d apart. At a point 3.0 cm from one charge, on the side opposite the other charge, the potential is 1200 V. Find the separation d between the charges. (b) The given equation simplifies to 27 N m 2 /C 27 N m 2 /C 900 N m/C + = 1200 V ⇒ = 300 V ⇒ d = 0.060 m = 6.0 cm 0.030 m + d 0.030 m + d

29.76. Model: The charges are point charges. The potential at a point is the sum of the potentials from all the charges. Visualize:

The two charges are in the xy-plane. Solve: Let the potential be zero at the point P(x, y). Then, VP =

1 4πε 0

Q1 x2 + y2

+

1 4πε 0

⇒ 3 x2 + y 2 =

Q2

( 4.0 cm − x )

(4 − x)

2

2

+ y2

=0 V=

1.0 ×10−9 C x2 + y2

+

−3.0 × 10 −9 C

(4 − x)

2

+ y2

+ y 2 ⇒ 9x2 + 9y2 = 16 + x2 + y2 – 8x

⇒ 8x2 + 8x – 16 + 8y2 = 0 ⇒ 8 ( x + 12 ) − 2 − 16 + 8 y 2 = 0 ⇒ ( x + 12 ) + y 2 = 188 = 2

2

9 4

This is the equation of a circle, x′2 + y′2 = R2, whose center is at ( x = − 12 cm, y = 0 cm ) and whose radius is

R=

9 4

= 32 = 1.5 cm. All values of (x, y) that satisfy this equation are the points with zero potential. This zero

potential contour map is shown as a dotted line in the figure. Assess: It is clear from the contour map that two zero-potential points lying on the x-axis are at x = 1.0 cm and x = −2.0 cm. Each position is nearer to the smaller charge and away from the bigger charge, as it must be.

29.77. Model: Apply the principles of conservation of energy and conservation of momentum. Visualize:

The figure shows a before-and-after pictorial representation of the charged particles moving toward each other. The proton’s physical quantities are denoted by the subscript p and that of the alpha particle by the subscript a. Solve: The conservation of energy equation is Kf + Uf = Ki + Ui. Initially when the charges are far away, Ui = 0 J and K i = 12 mp vip2 + 12 ma via2 . At the distance of closest approach, Uf =

1 ( e )( 2e ) 4πε 0 d

K f = 12 ma vfa2 + 12 mpvfp2

The particles are not at rest at closest approach. That would violate conservation of momentum. Instead, the particles are instantaneously moving with the same velocity vfa = vfp = vf so that the distance between them is (instantaneously) not changing. The conservation of energy equation simplifies to 1 2

(m

a

+ mp ) vf2 +

1 2e 2 = 4πε 0 d

1 2

(m

p

+ ma ) ( 0.010c )

2

The conservation of momentum equation Pafter = Pbefore is mavf + mpvf = ma(0.010c) – mp(0.01c)

⇒ vf =

( 0.010c ) ( ma − mp ) ( 0.010 ) ( 3 × 108 m/s ) ( 4 u − 1 u ) ma + mp

=

4 u +1 u

= 1.80 × 106 m/s

Substituting into the energy-conservation equation, 1 2

( 4 u + 1 u ) (1.80 ×106 m/s )

2

+ ( 9.0 × 109 N m 2 /C2 )

⇒

⇒d =

2 (1.60 × 10−19 C ) d

2

= 12 ( 4 u + 1 u ) ( 3.0 × 106 m/s )

4.608 × 10−28 N m 2 = 1.44 × 1013 u m 2 /s 2 d

4.608 ×10−28 N m 2 = 1.93 ×10−14 m (1.44 ) (1.661×10−27 kg ) ×1013 ( m2 /s2 )

2

29.78. Model: Energy and momentum are conserved. Visualize: Please refer to Figure CP29.78. Let the two spheres have masses mA and mB, and speeds vA and vB when they are very far apart. Solve: The energy-conservation equation Kf + Uf = Ki + Ui is

( ⇒

1 2

1 2

mA vA2 + 12 mBvB2 ) + 0 J = 0 J +

1 qA qB 4πε 0 rAB

( 0.002 kg ) vA2 + 12 ( 0.001 kg ) vB2 = ( 9.0 ×109 N m 2 /C2 )

( 2.0 ×10

−9

C )(1.0 × 10−9 C )

2.0 ×10−3 m

⇒ vA2 + 0.5 vB2 = 0.0090 m 2 /s 2 To solve for vA and vB, we need another equation relating vA and vB. From the momentum conservation equation Pafter = Pbefore we get mAvA + mBvB = 0 kg m/s ⇒ (2.0 g)vA + (1.0 g)vB = 0 kg m/s ⇒ vB = −2vA Substituting this expression into the energy-conservation equation,

vA2 + 0.5 ( 2vA ) = 0.009 m 2 /s 2 ⇒ 3vA2 = 0.009 m2 /s 2 2

Solving these equations, we get vA = –0.0548 m/s and vB = +0.110 m/s. Thus the speeds are 0.055 m/s for A and 0.110 m/s for B.

29.79. Model: Energy and momentum are conserved. Visualize: Please refer to Figure CP29.79. Let the two spheres have masses mC and mD, and speeds vC and vD when the two spheres collide. Solve: The energy-conservation equation Kf + Uf = Ki + Ui is

(

1 2

⎛ 1 qC qD ⎞ qC qD 1 mCvC2 + 12 mDvD2 ) + ⎜ ⎟=0 J+ 4πε 0 0.010 m ⎝ 4πε 0 0.002 m ⎠

The momentum conservation equation Pafter = Pbefore is

⎛ 2.0 g ⎞ mCvC + mDvD = 0 kg m/s ⇒ vC = − ⎜ ⎟ vD = −2vD ⎝ 1.0 g ⎠ Substituting this expression in the energy equation, 1 2

( 0.001 kg ) 4vD2 + 12 ( 0.002 kg ) vD2 = ( 9.0 ×109 N m2 /C2 )( 2.0 ×10−9 C )( −1.0 ×10−9 C ) ⎡⎢

1 1 ⎤ − ⎥⎦ 0.010 m 0.002 m ⎣

⇒ ( 0.0030 kg ) vD2 = 7.2 ×10−6 N m Solving these equations gives vD = 0.049 m/s = 4.9 cm/s and hence vC = −9.8 cm/s. The speed of C is +9.8 cm/s.

29.80. Model: The charges making the dipole are point charges. The potential is the sum of potentials from each charge. Visualize:

Point P is located at a distance r from the center of the dipole and it makes an angle θ with the dipole axis. Solve: The potential at P due to the dipole is

⎡ ⎤ 1 Q 1 ( −Q ) Q ⎢ 1 1 ⎥ + = − 2 ⎥ 2 4πε 0 r+ 4πε 0 r− 4πε 0 ⎢ x 2 + ( y − s 2 )2 + + x y s 2 ( ) ⎣ ⎦ ⎡ ⎤ Q ⎢ 1 1 ⎥. = − 2 ⎥ 2 2 4πε 0 ⎢ x 2 + y 2 − ys + ( s 2 )2 x y ys s 2 + + + ( ) ⎣ ⎦

V=

Because x, y >> s , (s/2)2 can be ignored compared with x2, y2, and ys. The potential is

V=

−1/ 2 −1/ 2 ⎤ 1 1 1 ⎛ ys ⎞ ⎤ Q ⎡ Q ⎡ 1 ⎛ ys ⎞ − = − − + 1 1 ⎢ 2 ⎥ ⎢ ⎜ ⎜ ⎟ ⎥. 2 ⎟ 4πε 0 ⎣⎢ r − ys r ⎝ r 2 ⎠ ⎦⎥ r 2 + ys ⎦⎥ 4πε 0 ⎣⎢ r ⎝ r ⎠

Using the binomial expansion, V= =

Q 4πε 0 1 4πε 0

1 ⎡⎛ ys … ⎞ ⎛ ys Q 1 2 ys ⎞⎤ + ⎟ − ⎜1 − 2 + … ⎟ ⎥ = ⎜1 + 2 πε r ⎢⎣⎝ 2r 2 2 r 4 ⎠ ⎝ ⎠⎦ 0 r 2r

( Qs ) y = r

3

1 4πε 0

where p = Qs is the dipole moment of the dipole.

( Qs ) r cosθ r

3

=

p cosθ 4πε 0 r 2 1

29.81. Model: The potential energy is determined by the electric potential. Solve:

(a) The electric field in the parallel-plate capacitor when the plates have charges ±q is η q qd E= = ⇒ ΔVC = Ed = ε 0 Aε 0 Aε 0

The increase in the potential energy dU due to moving charge dq through potential difference ΔVC is

dU = dq ΔVC =

qd dq Aε 0

(b) Integrating the expression for dU from q = 0 (uncharged) to q = Q (fully charged), Q

U = ∫ dU = ∫ 0

Q

Q qd d d ⎡ q2 ⎤ 1 ⎛ Qd ⎞ 1 dq = q dq = ⎟ Q = QΔVC ⎢ ⎥ = ⎜ ∫ Aε 0 Aε 0 0 Aε 0 ⎣ 2 ⎦ 0 2 ⎝ Aε 0 ⎠ 2

where we have used the expression for ΔVC from part (a).

29.82. Model: The potential energy is determined by the electric potential. Solve: (a) A sphere of radius R that is uniformly charged with a charge q is at a potential V0 (see Equation 29.32): 1 q V0 = 4πε 0 R The increase in the potential energy dU when an infinitesimal amount of charge dq is moved from infinity (where the potential is zero) to the surface of the sphere (where the potential is V0) is

dU = dqV0 =

1 q dq 4πε 0 R

(b) Integrating the expression for U from q = 0 (uncharged) to q = Q (fully charged), Q

Q

q 1 1 Q2 dq = qdq = ∫ 4πε 0 R 4πε 0 R 0 4πε 0 2 R 0

U = ∫ dU = ∫

1

(c) The self-energy of a proton is

(1.60 ×10−19 C) = 2.3 ×10−13 J 1 Q2 U= = ( 9.0 ×109 N m 2 /C2 ) 4πε 0 2 R ( 2 ) ( 0.50 ×10−15 m ) 2

29.83. Model: Assume the wire is a line of charge with uniform linear charge density. The electric potential at the point is the sum of contributions from all charges present. Visualize: Please refer to Figure CP29.83. Let the origin be the center of our coordinate system. Divide the charged wire into two straight sections and the central semicircle. Divide each section into N small segments, each of length Δx and with charge Δq. Segment i, located at position xi, contributes a small amount of potential Vi at the semicircle center. Solve: For the right hand straight section, the contribution of the ith segment is

Vi =

Δq 4πε 0 ri

=

Δq 4πε 0 xi

=

λΔx 4πε 0 xi

where Δq = λΔx. The Vi are now summed and the sum is converted to an integral giving

Vstraight section =

3R λ 3 R dx λ λ = ln ( 3) ⎡⎣ln ( x ) ⎤⎦ R = ∫ 4πε 0 R x 4πε 0 R 4πε 0 R

The integration limits are set by the physical location of the straight section. By symmetry, the left hand straight section adds the same amount to the total potential at the semicircle center. The potential due to the ith segment of the semicircular section is Vi =

λ ( R Δθ ) Δq Δq λ = = = Δθ 4πε 0 ri 4πε 0 R 4πε 0 R 4πε 0 R

where we have used the arc length Δx = R Δθ . The Vi are now summed and the sum converted to an integral giving

Vsemicircle =

λ

π

4πε 0 R ∫ 0

dθ =

λπ 4πε 0 R

=

λ 4ε 0 R

The total potential is the sum of the potentials due to the three sections: ⎛ λ ⎞ λ λ ⎛ 2ln(3) ⎞ V = Vsemicircle + 2Vstraight section = ln ( 3) ⎟ = + 2⎜ ⎜1 + 4ε 0 R 4 R 4 πε ε π ⎟⎠ 0 0R ⎝ ⎝ ⎠

29.84.

Model: The electric potential at a point is the sum of contributions from all charges present. Visualize: Please refer to Figure 29.32 in Example 29.12. Divide the disk into rings of equal width Δr. Ring i has radius ri and charge Qi. We can use the result of Example 29.11 to write the potential at distance z of ring i as

Vi =

ΔQi

1 4πε 0

ri 2 + z 2

The potential at the point is the sum of all rings. V = ΣVi =

1

N

4πε 0

∑ i =1

ΔQi ri 2 + z 2

The charge on each ring is related to the area of each ring by

ΔQi = ηΔAi = cri 2π rΔr = 2π cri 2Δr With this substitution and converting the sum to an integral, the potential is

2π c r 2 dr ∫ 4πε 0 0 r 2 + z 2 R

V= Using an integral table, V=

)

(

R

⎛ z 1 c ⎛1 c ⎛ 2 2 2 2 ⎞ 2 2 ⎜ R R + z + ln ⎜ ⎜ r r + z − ln r + r + z ⎟ = 2 2 2ε 0 ⎝ 2 2 ⎠ 0 4ε 0 ⎜⎝ ⎝R+ R +z

The total charge on the disk is R

Q = ∫η dA = ∫ ( cr )2π rdr = 0

3Q Thus c = . With this substitution, 2π R3 z ⎛ 3Q ⎛ V= R R 2 + z 2 + ln ⎜ 3⎜ ⎜ 2 2 8πε 0 R ⎝ ⎝R+ R +z

⎞⎞ ⎟ ⎟⎟ ⎠⎠

2π c 3 R 3

⎞⎞ ⎟ ⎟⎟ ⎠⎠

29.85. Model: Assume the cylinder is uniformly charged. Visualize:

Divide the cylinder into narrow rings of width dz, radius R, and charge dq.

Solve: The potential at the center is the sum of the potential of all the rings. The potential at distance z on the axis of a ring of charge was found in Example 29.11 to be 1 dq Vring = 4πε 0 R 2 + z 2

The surface area of the ring is dA = (2πR)dz, so the amount of charge is dq = η dA = (2πRη)dz. The surface area of the cylinder is A = (2πR)L, so the surface charge density is η = Q/2πRL. Thus dq = (Q/L)dz and the potential of the ring is Vring =

Q/L 4πε 0

dz R2 + z 2

The potential at the center of the cylinder is found by summing (i.e., integrating) the potential due to all rings. That is,

Vring = =

Q/L L / 2 dz 2 Q/L = 4πε 0 ∫− L / 2 R 2 + z 2 4πε 0

∫

2 2 Q/L ⎛⎜ L ⎛ L ⎞ ⎞⎟ + 1+ ⎜ ln ⎟ 4πε 0 ⎜ 2 R ⎝ 2 R ⎠ ⎟⎠ ⎝

L/2 0

dz R2 + z 2

=

(

)

L/2 2 Q/L ⎡ ln z + R 2 + z 2 ⎤ ⎦⎥ 0 4πε 0 ⎣⎢

30.1. Solve: The potential difference ΔV between two points in space is xf

ΔV = V ( xf ) − V ( xi ) = − ∫ Ex dx xi

where x is the position along a line from point i to point f. When the electric field is uniform, xf

ΔV = − Ex ∫ dx = − Ex Δx = − (1000 V/m )( 0.30 m − 0.10 m ) = −200 V xi

30.2. Visualize:

Solve:

The potential difference ΔV between two points on the y-axis is yf

ΔV = − ∫ E y dy yi

When the electric field is uniform, the above result simplifies to ΔV = − E y Δy. In the present problem,

Δy = yf – yi = 0.05 m – (−0.05 m) = 0.10 m The y-component of the electric field is

E y = −50,000 V/m ⇒ ΔV = – (–50,000 V/m)(0.10 m) = +5.0 kV Assess: Vf – Vi = 5.0 kV shows that the potential at point f is higher than at point i. This is because the electric field is directed from the point at the higher potential to the point at the lower potential.

30.3. Model: The potential difference is the negative of the area under the Ex vs x curve. Visualize: Please refer to Figure EX30.3. Solve: The potential difference between x = 1.0 m and x = 3.0 m is 1 ⎛ ⎞ ΔV = − ⎜ ( 200 V/m )( 2.0 m − 1.0 m ) + ( 200 V )( 3.0 m − 2.0 m ) ⎟ = −300 V. 2 ⎝ ⎠ Assess: The potential difference is negative since the electric field points in the direction of decreasing potential.

30.4. Model: The potential difference is the negative of the area under the Ex vs x curve. Visualize: Please refer to Figure EX30.4. Solve: The potential difference between the origin and x = 3.0 m is 1 ⎛1 ⎞ ΔV = V ( x = 3.0 m ) − V ( x = 0 m ) = − ⎜ ( −100 V )(1.0 m − 0 m ) + ( 200 V )( 3.0 m − 1.0 m ) ⎟ 2 2 ⎝ ⎠ = −150 V Thus V ( 3.0 m ) = V ( 0 m ) − 150 V = −50 V − 150 V = −200 V Assess:

The potential decreases from the origin to x = 3.0 m.

30.5. Solve: The work done is exactly equal to the increase in the potential energy of the charge. That is, W = ΔU = qΔV = q (Vf − Vi ) = (1.0 ×10−6 C ) (1.5 V ) = 1.5 × 10−6 J Assess:

The work done by the escalator on the charge is stored as electric potential energy of the charge.

30.6. Solve: The Van de Graaff generator or the motor that runs the belt does work in lifting a positive ion (q = e) against the downward force on the positive charge that is moving up the belt. The work done is

W = ΔU = qΔV = q (1.0 × 106 V − 0 V ) = (1.60 × 10−19 C )(1.0 × 106 V ) = 1.6 × 10−13 J Assess:

The work done by the generator in lifting the charge is stored as electric potential energy of the charge.

30.7. Solve: The emf is defined as the work done per unit charge by the charge escalator or the battery. That is, ε=

Wchem 0.60 J = = 12 V q 0.050 C

30.8.

Solve: The work done is the potential energy gained by the electron.

W = U = qΔV = −eΔV

( 2.4 ×10 − (1.6 ×10

−19

⇒ ΔV =

−19

J)

C)

= −1.5 V

The potential difference from the positive to the negative terminal is –1.5 V, so the emf of the solar cell is 1.5 V.

30.9. Model: The electric field points in the direction of decreasing potential and is perpendicular to the equipotential lines. Visualize: Please refer to Figure EX30.9. The three equipotential surfaces correspond to potentials of 0 V, 100 V, and 200 V. Solve: The electric field component along a direction of constant potential is Es = − dV ds = 0 V/m. But, the electric field component perpendicular to the equipotential surface is G ΔV 100 V E = = = 10 kV/m Δ s 0.01 m The direction of the electric field vector is “downhill,” perpendicular to the equipotential surfaces, and directed to the left. That is, the electric field is 10 kV/m to the left.

30.10. Model: The electric field is perpendicular to the equipotential lines and points “downhill.”

Visualize: Please refer to Figure EX30.10. Three equipotential surfaces at potentials of −200 V, 0 V, and 200 V are shown. Solve: The electric field component perpendicular to the equipotential surface is

E=−

ΔV 200 V =− = −20 kV/m Δs 0.01 m

The electric field vector is in the third quadrant, 45° below the negative x-axis. That is, G E = ( 20 kV/m, 45° below − x -axis )

30.11. Model: The electric field is the negative of the slope of the graph of the potential function. Visualize: Please refer to Figure EX30.11. Solve: There are three regions of different slope. For 0 cm < x < 10 cm and 20 cm < x < 30 cm,

ΔV = 0 V/m ⇒ Ex = 0 V/m Δx For 10 cm < x < 20 cm, ΔV −100 V − (100 V ) = = −2000 V/m ⇒ Ex = +2000 V/m Δx 0.20 m − 0.10 m

Assess:

Because Es = − dV ds , the electric field is zero where the potential is not changing.

30.12. Model: The electric field is the negative of the slope of the graph of the potential function. Visualize: Please refer to Figure EX30.12. Solve: There are three regions of different slope. For 0 cm < x < 1 cm,

ΔV 50 V − 0 V = = 5000 V/m ⇒ Ex = −5000 V/m Δ x 0.01 m − 0 m For 1 cm < x < 2 cm, ΔV −50 V − ( 50 V ) = = −10,000 V/m ⇒ Ex = 10,000 V/m dx 0.02 m − 0.01 m For 2 cm < x < 3 cm, 0 V − ( −50 V ) ΔV = = 5000 V/m ⇒ Ex = −5000 V/m Δ x 0.03 m − 0.02 m

Assess:

ΔV Δ x and Ex are the negative of each other.

30.13. Visualize:

Solve:

The electric potential difference ΔV between two points in a uniform electric field is

V ( x f ) − V ( xi ) = − ∫ Ex dx = − Ex ( x f − xi ) Choosing xi = −1.0 m and xf = +1.0 m, +1000 V − (−1000 V) = −Ex[1.0 m − (−1.0 m)] ⇒ Ex = −1.0 kV/m Alternatively, xi = 1.0 m and xf = −1.0 m. For this choice,

−1000 V − (+1000 V) = −Ex [−1.0 m − (1.0 m)] ⇒ Ex = −1.0 kV/m Assess: The choice of initial and final positions does not change the physical nature of the electric field or the potential difference.

30.14. Model: The electric field is the negative of the derivative of the potential function. Solve: (a) From Equation 30.11, the component of the electric field in the s-direction is Es = − dV ds . For the given potential,

dV d V = (100 x 2 V ) = 200 x ⇒ Ex = −200 x V m dx dx m At x = 0 m, Ex = 0 V/m. (b) At x = 1 m, Ex = −200 (1) V/m = −200 V/m. Assess: The potential increases with x, so the electric field must point in the −x-direction.

30.15.

Solve:

(a) Since Ex = −

dV , we have dx

Ex = −

d ( 50 x − 100 / x ) V/m = − ( 50 + 100 / x 2 ) V/m dx

At x = 1.0 m,

(

Ex = − 50 + 100 / (1.0 )

2

) V/m = −150 V/m

(b) At x = 2.0 m, ⎛ 100 ⎞ Ex = −⎜ 50 + ⎟ V/m = −75 V 2 ⎜ ( 2.0 ) ⎟⎠ ⎝

30.16. Model: Use Kirchhoff ’s loop law. Visualize: Please refer to Figure EX30.16. Solve: For any path that starts and ends at the same point, ΔVloop = ∑ ( ΔV )i = 0 V. Therefore, i

ΔV12 + ΔV23 + ΔV34 + ΔV41 = 30 V + 50 V + ΔV34 – 60 V = 0 V ⇒ ΔV34 = −20 V

That is, the potential at point 4 is less than the potential at point 3.

30.17. Model: Assume that the battery is ideal and that the capacitor is a parallel-plate capacitor. Solve:

(a) From Equation 30.17, the capacitance is

C=

ε0 A d

=

(8.85 × 10

−12

C 2 / N m 2 ) ( 0.020 m × 0.020 m ) 0.00050 m

= 7.1 × 10−12 F = 7.1 pF

(b) The charges on the electrodes are

Q = ±C ΔVC = ± ( 7.1×10−12 F ) (100 V ) = ±7.1× 10−10 C = ±0.71 nC

30.18. Solve: From Equation 30.17, the capacitance is C= Assess:

ε0 A d

=

ε 0 L2 d

⇒L=

Cd

ε0

=

(100 × 10

−12

F )( 0.20 × 10−3 m )

8.85 × 10−12 C 2 / N m 2

The capacitance depends only on the geometry of the capacitor.

= 0.048 m = 4.8 cm

30.19. Model: Assume that the battery is ideal and the connecting wires have zero resistance. Solve:

The capacitance of a capacitor is the ratio of the charge and the potential difference ΔVC:

C=

Q Q 30 ×10−6 C ⇒ ΔVC = = = 3.0 V ΔVC C 10 ×10−6 F

For an ideal battery, E = ΔVbat = ΔVC . So, E = 3.0 V.

30.20. Model: Assume that the battery is ideal and the wires connecting the battery with the capacitor have zero resistance. Solve: Using Equation 30.19,

E = ΔVbat = ΔVC =

48 ×10−6 C Q = = 24 V 2.0 ×10−6 F C

30.21. Model: Assume that the battery is ideal. Solve:

Equation 30.19 is ΔVC =

Q Q 45 ×10−9 C =E ⇒ C = = = 5.0 ×10−9 F C ΔVC 9.0 V

30.22. Solve: According to Equation 30.21, Ceq = C1 + C2 + C3 = 6 μF + 10 μF + 16 μF = 32 μF

30.23. Solve: According to Equation 30.23, −1

−1

⎛ 1 ⎛ 1 1 1 ⎞ 1 1 ⎞ −6 Ceq = ⎜ + + ⎟ =⎜ + + ⎟ = 3.0 × 10 F = 3.0 μ F μ μ μ C C C 6 F 10 F 16 F ⎝ ⎠ 2 3 ⎠ ⎝ 1

30.24. Model: Two capacitors in parallel combine to give greater capacitance. Solve: Since we want a capacitance of 50 μF and we have a 30 μF capacitor, we must connect the second capacitor in parallel with the 30 μF capacitor. That is, C + 30 μF = 50 μF ⇒ C = 50 μF − 30 μF = 20 μF

30.25. Model: Two capacitors in series combine to give less capacitance. Solve: Since we have a 75 μF capacitor and we want a 50 μF capacitance, we must connect the second capacitor in series with the 75 μF capacitor. The capacitance of the second capacitor is calculated as follows: 1 1 1 + = ⇒ C = 150 μF 75 μ F C 50 μ F

30.26. Visualize: Please refer to Figure EX30.26. Solve: Any two electrodes, regardless of their shape, form a capacitor whose capacitance is defined as C = Q ΔVC . The capacitance is C=

20 ×10−9 C = 2.0 ×10−10 F = 200 pF 100 V

30.27. Model: Assume the battery is ideal. Visualize: Please refer to Figure EX30.27. Solve: For an ideal battery, the potential difference across the capacitor is the same as the emf of the battery. Thus,

E = ΔVC =

Q ⇒ Q = CE = (10 × 10−6 F)(1.5 V) = 15.0 × 10−6 C C

30.28. Solve: From Equation 30.26, the energy stored in a capacitor is U C = 12 C ( ΔVC ) ⇒ ΔVC = 2

2 (1.0 J ) 2U C = = 1.41 kV C 1.0 ×10−6 F

30.29. Solve: The graph of Figure EX30.29 shows that the capacitor is fully charged at 3.0 s and that its charge is 200 μC. From t = 0 s to t = 3.0 s, the charge on the capacitor follows the equation Q = ( 200 μ C 3.0 s ) t. Using Equation 30.26, the energy stored in the capacitor as a function of time is 2

UC =

1 Q 2 1 ⎛ 200 μ C ⎞ t2 = ⎜ = (1.111×10−3 C2 /s F ) t 2 ⎟ 2 C 2 ⎝ 3.0 s ⎠ 2.0 μ F

30.30. Solve: Using Equation 30.26, U C1 = 12 C1 ( ΔVC1 )

2

U C2 = 12 C2 ( ΔVC2 )

2

Because C2 = 12 C1 and ΔVC2 = 2ΔVC1, we have C1 ( ΔVC1 ) U C1 12 C1 ( ΔVC1 ) 1 = = = U C2 12 C2 ( ΔVC2 )2 ( 12 C1 ) 4 ( ΔVC1 )2 2 2

2

30.31. Solve: The energy density is uE = ( 5.0 ×10−12 J ) ( 0.020 m × 0.020 m × 0.020 m ) = 6.25 ×10−7 J m3 . uE = 12 ε 0 E 2 ⇒ E =

2uE

ε0

=

2 ( 6.25 ×10−7 J/m3 ) 8.85 ×10−12 C2 /N m 2

= 1.19 kV/m

30.32. Solve: (a) From Equation 30.26, the energy stored in the charged capacitor is 1 1 ⎛ Aε ⎞ 2 2 U C = C ( ΔVC ) = ⎜ 0 ⎟ ( ΔVC ) 2 2⎝ d ⎠

2 2 −12 1 π ( 0.010 m ) (8.85 ×10 C /N m ) 2 ( 200 V ) = 1.11×10−7 J 2 0.50 ×10−3 m 2

=

(b) From Equation 30.28, the energy density in the electric field is 2

2

1 1 ⎛ ΔV ⎞ 1 200 V ⎛ ⎞ 3 uE = ε 0 E 2 = ε 0 ⎜ C ⎟ = ( 8.85 ×10−12 C2 /N m 2 ) ⎜ ⎟ = 0.71 J/m −3 2 2 ⎝ d ⎠ 2 0.50 × 10 m ⎝ ⎠

30.33.

Model: Assume the capacitor is a parallel-plate capacitor. Solve: (a) The capacitance is increased by a factor of the dielectric constant κ over the equivalent vacuumfilled capacitor.

C = κ C0 = κ

ε0 A d

(8.85 ×10 = ( 3.7 )

−12

C2 / N m 2 ) π ( 0.050 m )

0.20 ×10−3 m

2

= 1.29 × 10−9 F = 1.29 nF

We have used κ = 3.7 for paper from Table 30.1. (b) The dielectric strength is the maximum possible electric field in the capacitor before breakdown. In this capacitor,

E=

ΔV ΔV ⇒ 16 ×106 V/m = ⇒ ΔV = 3.2 kV d ( 0.20 ×10−3 m )

30.34. Solve:

Model: The electrodes form a parallel-plate capacitor. (a) The capacitance of the equivalent vacuum-filled capacitor is

C0 =

ε0 A

=

d

(8.85 ×10

−12

C 2 /Nm 2 )( 5.0 ×10−3 m )

( 0.10 ×10−3 m )

2

= 2.21×10−12 F

From Table 30.1, the dielectric constant of mylar is κ = 3.1. With the battery attached, the potential difference across the plates is ΔVC = ΔVbatt = 9.0 V . The charge on the plates is

Q = C ΔV = κ C0 ΔV = ( 3.1) ( 2.21 × 10−12 F ) ( 9.0 V ) = 6.17 × 10−11 C = 62 pC The electric field inside the capacitor is

E=

E0

κ

=

1 ΔVC

κ d

9.0 V ⎛ 1 ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟ = 20 kV/m −3 3.1 0.10 × 10 m ⎝ ⎠⎝ ⎠

(b) With the battery connected, ΔVC = 9.0 V. The capacitor is now vacuum-insulated. E = E0 =

ΔVC 9.0 V = = 90 kV/m d 0.10 ×10−3 m

Q = C0 ΔVC = ( 2.21×10−12 F ) ( 9.0 V ) = 20 pC Assess: Since the battery remains connected as the mylar is withdrawn, the potential difference across the plates does not change.

30.35. Solve:

Model: The disks form a parallel-plate capacitor. (a) The charge on each of the metal disks is

Q = C ΔVC = κ Pyrex C0 ΔVC = κ Pyrex

ε0 A d

ΔVC

The charge density on a disk is

η=

(8.85 × 10−12 C2 /N m2 ) (1000 V ) Q ε ΔV = κ Pyrex 0 C = ( 4.7 ) A d ( 0.50 × 10−3 m )

= 8.3 × 10−5 C/m 2 = 83 μ C/m 2 (b) In the absence of the dielectric, the surface charge density on the disks is

η0 =

η κ Pyrex

= 1.77 × 10−5 C/m 2 .

The surface charge density on the Pyrex is ⎛ 1 ⎞ 1 ⎞ ⎛ = 1.77 ×10−5 C/m 2 ) ⎜1 − ⎟ ⎜ κ Pyrex ⎟⎟ ( 4.7 ⎝ ⎠ ⎝ ⎠ −5 2 2 = 1.39 × 10 C/m = 13.9 μ C/m

ηinduced = η0 ⎜1 −

Assess:

The induced charge density on the dielectric is less than the charge density on the plates.

30.36. Visualize: Please refer to Figure P30.36. Solve: (a) The electric field points “downhill.” So, point A is at a higher potential than point B. (b) In a region that has a uniform electric field, Equation 30.3 gives the potential difference between two points: ΔV = − ∫ Ex dx = − Ex ( xf − xi ) ⇒ VB – VA = −(1000 V/m)(0.07 m) = −70 V That is, the potential at point A is 70 V higher than the potential at point B.

30.37. Solve: (a)

(b) Equation 30.3 gives the potential difference between two points in space: xf

30 cm

xi

−20 cm

ΔV = − ∫ Ex dx = −

∫ ( −1000 x V/m ) dx

0.30 m

⎡ x2 ⎤ 1000 ⎡ 2 2 = + ⎢1000 ⎥ V=+ ( 0.30 m ) − ( −0.20 m ) ⎦⎤ V = +65 V ⎣ 2 2 ⎣ ⎦ −0.20 m Assess: E is positive for negative x but negative for positive x, so the potential difference depends on the square of the positions.

30.38. Solve: (a)

(b) Equation 30.3 gives the potential difference between two points in space: x

x

xf

f f ⎡ x2 ⎤ ΔV = V ( xf ) − V ( xi ) = − ∫ Ex dx = − ∫ ( 5000 x V/m ) dx = −5000 ⎢ ⎥ V = −2500 ( xf2 − xi2 ) V ⎣ 2 ⎦ xi xi xi

Taking V(xi) = 0 V at xi = 0 m, and replacing xf with simply x, V(x) = −(2500 x2) V. (c) A graph of V versus x over the region −1 m ≤ x ≤ 1 m is shown in part (a). Assess: As it must be, we have

−

dV d = − ( −2500 x 2 ) V = 5000x = Ex dx dx

30.39. Visualize:

Solve:

Equation 30.3 gives the potential difference between two points in space rf

ΔV = V ( rf ) − V ( ri ) = − ∫ Er dr ri

r

⇒ V ( r ) − V ( R ) = −∫

λ

2πε 0 r R

dr = −

λ 2πε 0

[ln r ]R = −

⇒ V ( r ) = V0 − Assess:

r

λ 2πε 0

[ln r − ln R ] = −

λ 2πε

ln

r R

r λ ln 2πε 0 R

At r = R, ln ( R R ) = 0 and V(R) = V0, as it must be due to the assumed constant of integration.

30.40. Model: The electric field is the negative of the slope of the graph of the potential function. Visualize: Please refer to Figure P30.40. Solve: We have

Ex = −

dV ⇒ dV = −Exdx dx

For x ≥ 1 cm, Ex = 1000 V/m. Integrating the above expression, V = − (1000 V/m)x + C Because V = 0 V at x = 0.03 m, C = (1000 V/m)(0.03 m) = 30 V. Thus, V = − (1000 V/m)x + 30 V (x ≥ 1 cm) This means V at x = 0.02 m is 10 V and V at x = 0.01 m is 20 V. For x < 1 cm, Ex = (105 V/m)x. Therefore, dV = − (105 V/m) x dx. Integrating again, 5 5 ∫ dV = − ∫ (10 V/m ) x dx + C ⇒ V = − (10 V/m )

Because V = 20 V at x = 0.01 m, C = 25 V. At x = 0 m, V = 25 V.

x2 + C (x < 1 cm) 2

30.41. Model: Assume the electrodes form parallel-plate capacitors with a uniform electric field between the plates. Visualize:

Please refer to Figure P30.41. The three metal electrodes serve as plates for two capacitors. On the middle electrode, half the charge is located on the left face and half on the right face, thus forming two capacitors. Each plate of the two capacitors carries a charge of ±50 nC. Solve: (a) In the space 0 cm < x < 1 cm, the electric field points to the left and its magnitude is

E=

η Q 50 × 10−9 C = = = 1.41 × 107 V/m ε 0 Aε 0 ( 0.02 m )2 ( 8.85 × 10−12 C2 / N m 2 )

G In the region 1 cm ≤ x ≤ 2 cm, E = 0 because in electrostatics the inside of a conductor has no free charge. The electric field in the region 2 cm < x < 3 cm points to the right and has the same magnitude as the electric field in the region 0 cm < x < 1 cm. (b) The potential difference between two points in space with a uniform electric field is

ΔV = Vf − Vi = E ( xf − xi ) Assuming that the negative plate at x = 0 m is at zero potential (Vi = 0 V at xi = 0 cm), Vf = xf E, or simply V = xE. Thus, the potential increases linearly with distance x from the negative plate in the region 0 ≤ x ≤ 1. At x = 1 cm, the potential is

V = xE = (1.0 × 10−2 m)(1.41 × 107 V/m) = 1.41 × 10−5 V The potential must be the same throughout the region 1 cm ≤ x ≤ 2 cm. If this were not the case, we would not have an electrostatic situation with the electric field E = 0 V/m. Using the previous reasoning, the potential decreases linearly in the region 2 cm < x < 3 cm.

30.42. Model: The electric field is the negative of the slope of the graph of the potential function. Visualize:

G Solve: (a) The x-component of E is Ex = −dV/dx. From Figure P30.42, the slope of the graph can be measured in five different regions. For x < −3 m and x > 3 m, the slope is 0 V/m. For −3 m < x < −1 m and 1 m < x < 3 m the slope is −5 V/m. For −1 m < x < 1 m the slope is +10 V/m. (b) The potential is independent of y, so the equipotential lines are parallel to the y-axis. The x-values of various equipotentials can be read from Figure P30.42. These give the potential map shown above. (c) The field vectors point in the direction of decreasing potential. Furthermore, the results of part (a) show that the field vectors in the inner region (−1 m < x < 1 m) are twice as long as the field vectors in the outer region. As you can see by comparing the vectors to the graph of the electric field, this information was used to add the field vectors to the potential map.

30.43. Model: The electric field is the negative of the slope of the potential function. Solve:

The on-axis potential of a charged disk was obtained in Chapter 29 to be

Vdisk =

Q ⎡ 2 2 R + z − z⎤ ⎦ 2πε 0 R 2 ⎣

where the charge on the disk of radius R is Q and the point is a distance z away from the center of the disk. Because E = − dV dz , we have Edisk = − Assess:

⎤ ⎤ dVdisk Q ⎡ 12 ( 2 z ) Q ⎡ z =− − 1⎥ = ⎢ ⎢1 − 2 2 ⎥ dz 2πε 0 R 2 ⎣ R 2 + z 2 ⎦ 2πε 0 R 2 ⎣ R +z ⎦

Using binomial expansion with z >> R, we have

Edisk =

⎡ ⎤ −1/ 2 z ⎢1 − ⎥ = Q ⎡1 − (1 + R 2 z 2 ) ⎤ = Q ⎡1 − 1 + 12 R 2 z 2 + "⎤ = Q 1/ 2 2 2 ⎢ ⎦ 4πε z 2 ⎥⎦ 2πε R 2 ⎣ 2 2 2πε 0 R ⎢ z (1 + R z ) ⎥ 2πε 0 R ⎣ 0 0 ⎣ ⎦ Q

That is, the disk behaves like a point charge. This is an expected result.

30.44. Model: Assume the charged rod is a line of charge of length L. Visualize: Please refer to Figure P30.44. Solve: (a) Divide the charged rod into N small segments, each of length Δx and with charge Δq. The segment i located at position xi contributes a small amount of potential Vi at point P:

Vi =

Δq 4πε 0 ri

=

Δq ( QΔ x L ) = 4πε 0 ( x0 − xi ) 4πε 0 ( x0 − xi )

Point P is at a distance x0 from the origin. This is done to avoid confusion with xi. The Vi are now summed and the sum is converted to an integral giving

V=

⎛ x +L 2⎞ L/2 dx Q Q ln ⎜ 0 = ⎡ − ln ( x0 − x ) ⎤⎦ − L / 2 = ⎟ 4πε 0 L − L∫/ 2 x0 − x 4πε 0 L ⎣ 4πε 0 L ⎝ x0 − L 2 ⎠ Q

L/2

Replacing x0 with x, the potential due to a line charge of length L at a distance x from the center is V=

Q ⎛ x + L/2 ⎞ ln ⎜ ⎟ 4πε 0 L ⎝ x − L/2 ⎠

(b) Because Ex = − dV dx ,

Ex = −

⎤ Q d Q ⎡ 1 1 Q 1 ⎡ln ( x + L/2 ) − ln ( x − L/2 ) ⎦⎤ = − − ⎢ ⎥= 4πε 0 L dx ⎣ 4πε 0 L ⎣⎢ ( x + L/2 ) ( x − L/2 ) ⎦⎥ 4πε 0 x 2 − L2 /4

Assess: When L = 0 m, Ex = Q 4πε 0 x 2. This is the electric field of a point charge Q a distance x away from a point charge, as expected.

30.45. Model: The electric field is the negative of the slope of the potential graph. Visualize: Please refer to Figure P30.45. Solve: Since the contours are uniformly spaced along the y-axis above and below the origin, the slope method is the easiest to apply. Point 1 is in the center of a 75 V change (25 V to 100 V) over a distance of 2 cm, so the slope ΔV/Δ s is 37.5 V/cm or 3750 V/m. Point 2 has the same potential difference in half the distance. Thus the slope at point 2 is 7500 V/m. The magnitudes of the electric fields at points 1 and 2 are 3750 V/m and 7500 V/m. The directions of the electric fields are downward at point 1 and upward at point 2, that is, from the higher potential to the lower potential. That is, G G E1 = ( 3750 V m, down ) E2 = ( 7000 V m, up )

30.46. Model: The electric field is the negative of the slope of the graph of the potential function. Visualize:

→

Solve: (a) At each point, E is perpendicular to the equipotential lines, points “downhill” toward lower potential, and has strength E = dV/ds ≈ ΔV/Δ s. We can read ΔV from the contour lines and use the 8-cm-scale at the bottom of the figure to estimate Δ s. Point 1, on the 100 V contour, is between the 200 V and the 0 V contours. The actual distance on the figure between the 200 V and the 0 V contours can be measured to be ≈6 mm. The length of the 8-cm-scale is ≈12 mm. Thus

200 V ⎛ 6 mm ⎞ Δ s1 ≈ ⎜ = 5000 V/m ⎟ (8 cm) ≈ 4 cm ⇒ E1 ≈ 12 mm 0.040 m ⎝ ⎠ From the symmetry of the figure, the field strengths are the same at points 2, 4, and 5. That is, E1 = E2 = E4 = E5 ≈ 5000 V/m. The contour lines are closer at point 3, so the field strength E3 is larger. The measured spacing between the 100 V and –100 V lines, which bracket the point, is ≈1.5 mm. Thus 200 V ⎛ 1.5 mm ⎞ Δ s3 ≈ ⎜ = 20,000 V/m ⎟ (8 cm) ≈ 1 cm ⇒ E3 ≈ 0.010 m ⎝ 12 mm ⎠

The field strength at 3 is four times that at the other points. (b) The field vectors are shown on the figure. Each is drawn perpendicular to the equipotential at that point.

30.47. Model: The electric field is the negative of the slope of the graph of the potential function.

Solve: The electric potential in a region of space is V = (150 x2 – 200 y2) V where x and y are in meters. The xand y-components are dV dV Ey = − = + ( 400 y ) V/m Ex = − = − ( 300 x ) V/m dy dx At (x,y) = (2.0 m, 2.0 m), Ex = −600 V/m and Ey = 800 V/m. The magnitude and direction of the electric field are E = E x2 + E y2 =

tan θ =

Ey Ex

=

( −600 V/m ) + ( 800 V/m ) 2

2

= 1000 V/m

800 V/m 4 = ⇒ θ = 53.1° above the −x-axis 600 V/m 3

The electric field points 180° − 53° = 127° counterclockwise (ccw) from the +x-axis.

30.48.

Solve:

The electric field is related to the electric potential by G G ⎛ ∂V ˆ ∂V ˆ ⎞ E ( x, y ) = −∇V ( x, y ) = − ⎜ i+ j⎟ dy ⎠ ⎝ dx Compute the partial derivatives. −3 200 x ∂V 200 y ∂V ∂ ⎛ 200 ⎞ ⎛ 1⎞ ⎟ = 200 ⎜ − ⎟ ( 2 x ) ( x 2 + y 2 ) 2 = − =− = ⎜ 3 3 2 2 2 2 2 2 ∂ ∂x ∂x ⎝⎜ x + y ⎟⎠ y ⎝ 2⎠ (x + y ) ( x + y2 ) 2 G G 200 Thus E = xiˆ + yjˆ . At (x,y) = (2.0 m, 1.0 m), E = 17.9 2.0iˆ + 1.0 ˆj V/m. The magnitude 3 ( x2 + y 2 ) 2

(

(

)

E = 17.9

( 2.0

2

+ 1.02 ) V/m = 40 V/m

⎛E ⎞ ⎛1⎞ The direction is θ = tan −1 ⎜ y ⎟ = tan −1 ⎜ ⎟ = 27D ccw from the +x-axis. E ⎝2⎠ ⎝ x⎠

)

30.49. Model: The field strength is the negative of the slope of the graph of the potential function. Solve:

(a)

EquipotentialsG are drawn by connecting points of equal potential. (b) To find E , first determine the direction perpendicular to the equipotential and pointing toward a lower potential. Call this direction the s-axis. The field strength is E = dV ds ≈ ΔV Δ s , where Δ s is the distance between equipotential lines that have potential difference ΔV. The electric field at point A points straight to the left, toward the 50 V equipotential. The distance between the 50 V line and the 75 V line is one grid spacing, or Δ s = 5 cm. Thus, 25 V EA ≈ = 500 V/m 0.05 m G As a vector, the field is EA = ( 500 V/m, left ) . Using the 50 V and 100 V equipotentials to calculate the field at point G B, EB ≈ ( 50 V ) ( 0.05 m ) = 1000 V/m. Thus EB = (1000 V/m, right ) . The field at point C points at a 45° angle toward the 50 V equipotential. The distance Δs between the 50 V and 100 V equipotentials is 5 cm × 2 = 7.07 cm, so G EC ≈ ( 50 V ) ( 0.0707 m ) = 707 V. As a vector, EC = ( 707 V/m, 45° above straight left ) . The field at point D is G symmetrical with point C, so ED = ( 707 V/m, 45° below straight left ) .

(c) The electric field vectors are shown in the figure.

30.50. Model: Conductors connected by a conducting wire are at the same potential. Visualize:

Charge flows from sphere 1 to sphere 2 when the wire is connected. Once connected, the entire system is one large conductor in electrostatic equilibrium. Consequently, the entire system must be an equipotential. That is, the two spheres must have equal potentials V1 = V2 after being connected by the wire. Solve: The potential of a sphere is V = Q 4πε 0 R , so the final charges Q1f and Q2f on the two spheres must be such that Q1f Q2f Q2f ⇒ Q1f = 12 Q2f = = 4πε 0 R1 4πε 0 R2 4πε 0 ( 2 R1 )

where we used the fact that R2 = 2R1. It is also true that charge must be conserved, so Q1f + Q2f = Q1i + Q2i = 6 nC Using Q2f = 2Q1f, we find Q1f + 2Q1f = 3Q1f = 6 nC from which we finally get Q1f = 2 nC and Q2f = 4 nC.

30.51.

Solve:

Because the spheres have equal and opposite charges and they have the same diameter, their potential is zero on a line that bisects the two spheres. On the left and right of this 0 V equipotential line, the equipotential lines show the curvature of the sphere the line is nearer to. This is because the potential nearer to one sphere is dominated by the potential of this sphere. The equipotential lines at 0 V, ±100 V and ±200 V are shown in the figure.

30.52. Model: The potential inside a conductor is the same as the potential on the surface. Solve:

The potential on the surface of the copper sphere must be 500 V. That is,

500 V =

1 Q ( 0.020 m )( 500 V ) = 1.11 nC ⇒ Q = ( 4πε 0 ) R ( 500 V ) = 4πε 0 R 9.0 ×109 N m 2 /C

30.53. Solve:

Model: The charged metal spheres are isolated and far from each other and anything else. (a) The charge on a sphere of radius r, charged to a potential V0 is given by equation 29.33:

Q1 = 4πε 0 rV 1 0

Q2 = 4πε 0 r2V0

The ratio of charge densities is ⎛ Q1 ⎞ ⎜ ⎟ η1 ⎝ A1 ⎠ Q1 A2 ⎛ r1 ⎞⎛ 4π r2 2 ⎞ r2 2 R = = = ⎜ ⎟⎜ =2 ⎟= = η 2 ⎛ Q2 ⎞ Q2 A1 ⎝ r2 ⎠⎝ 4π r12 ⎠ r1 R ⎜ ⎟ ⎝ A2 ⎠

(b) The electric field strengths at the surface of the conductors is E1 =

E η1 η and E2 = 2 , thus 1 = 2. E2 ε0 ε0

Assess: Halving the radius doubles the electric field strength as well as the charge density. Sharp corners (i.e., parts of conductors with a small radius of curvature, like a small sphere) have strong electric fields. This characteristic is why lightning rods work.

30.54. Model: The battery is assumed to be ideal. Visualize:

The pictorial representation shows the capacitor plates connected to a battery, the battery removed from the plates, and the plates moved apart with insulating handles. Solve: (a) The battery charges the plates through the wires. Once the wires are disconnected, the charge is trapped on the plates and will not change as their spacing is increased. The initial capacitance of the plates is Ci =

ε0 A di

=

(8.85 ×10

−12

C 2 /N m 2 ) ( 0.020 m × 0.020 m ) 0.0010 m

= 3.54 × 10−12 F

Consequently, the initial voltage ΔVi = 9 V charges the plates to

Q = ±C ΔV = ± ( 3.54 × 10−12 F ) ( 9.0 V ) = ±3.2 × 10−11 C

(b) The charge is still Q = ±3.2 × 10−11 C after the spacing is increased, but the final capacitance is

Cf =

ε0 A df

=

ε0 A 2d i

=

Ci 2

As a result, the final potential difference is

ΔVf =

Q Q Q = = 2 = 2ΔVi = 2 × 9 V = 18 V Cf Ci /2 Ci

30.55. Model: Assume the battery is ideal. Visualize:

The pictorial representation shows the capacitor plates connected to a battery, and the capacitor plates moved apart with insulating handles while they are connected to the battery. Solve: (a) The initial capacitance of the plates is Ci =

ε0 A di

=

(8.85 ×10

−12

C 2 /N m 2 ) ( 0.020 m )

1.0 × 10−3 m

2

= 3.54 × 10−12 F

Consequently, an initial voltage ΔVi = 9.0 V charges the plates to Q = ±Ci ΔVi = ± ( 3.54 × 10−12 F ) ( 9.0 V ) = ±31.9 × 10−12 C = ±32 pC (b) The new capacitance is Cf = ε 0 A 2di = 12 Ci . The potential difference across the plates is determined by the

battery and is unchanged: ΔVf = ΔVi = 9.0. Thus, the new charge on the plates is

Q = ±Ci ΔVf = ± 12 ( 3.54 ×10−12 F ) ( 9.0 V ) = 16.0 × 10−12 C = 16.0 pC

30.56. Model: Capacitance is a geometric property of two electrodes. Visualize:

Solve: The ratio of the charge to the potential difference is called the capacitance: C = Q ΔVC . The potential difference across the capacitor is

ΔVC =

1 Q 1 Q Q ⎡1 1⎤ − = ⎢ − ⎥ 4πε 0 R1 4πε 0 R2 4πε 0 ⎣ R1 R2 ⎦ −1

⎡1 1⎤ RR R1R2 ⇒ C = 4πε 0 ⎢ − ⎥ = 4πε 0 1 2 = 4πε 0 = 100 × 10−12 F − × R R R R 1.0 10−3 m ⎣ 1 2⎦ 2 1

⇒ R1R2 = (100 ×10−12 F )(1.0 ×10−3 m )( 9.0 ×109 N m 2 /C2 ) = 900 × 10−6 m 2 Using R2 = R1 + 1.0 mm, R1 ( R1 + 1.0 ×10−3 m ) = 900 ×10−6 m ⇒ R12 + (1.0 × 10−3 m ) R1 − 900 × 10−6 m = 0

⇒ R1 =

−1.0 ×10−3 m ±

(1.0 ×10 2

−3

m ) + 3600 × 10−6 m 2 2

= 0.0295 m = 2.95 cm

The outer radius is R2 = R1 + 0.001 m = 0.0305 m = 3.05 cm. So, the diameters are 5.9 cm and 6.1 cm.

30.57. Solve: (a) The equivalent capacitance of N capacitors that are in parallel and have capacitances C1, C2, C3, . . . CN is Ceq = C1 + C2 + C3 + " CN If the capacitors are identical with each having a capacitance C, then Ceq = NC. (b) The equivalent capacitance of N capacitors that are in series and have capacitances C1, C2, C3, … CN is

1 1 1 1 1 = + + +" Ceq C1 C2 C3 CN If the capacitors are identical with each capacitor having a capacitance C, then 1 N C = ⇒ Ceq = Ceq C N

30.58.

Visualize:

The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure. Solve: Because C1 and C2 are in series, their equivalent capacitance Ceq 12 is

1 1 1 1 1 1 = + = + = ⇒ Ceq 12 = 12 μF Ceq 12 C1 C2 20 μ F 30 μ F 12 μ F Then, Ceq 12 and C3 are in parallel. So, Ceq = Ceq 12 + C3 = 12 μF + 25 μF = 37 μF

30.59.

Visualize:

The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure. Solve: Because C1 and C2 are in parallel, their equivalent capacitance Ceq 12 is Ceq 12 = C1 + C2 = 20 μF + 60 μF = 80 μF Then, Ceq 12 and C3 are in series. So,

1 1 1 1 1 9 80 −1 = + = + = ( μ F ) ⇒ Ceq = μ F = 8.9 μ F C eq Ceq 12 C3 80 μ F 10 μ F 80 9

30.60. Model: Assume that the battery is ideal. Visualize:

The pictorial representation shows the equivalent capacitance of the three capacitors. Solve: Because C1, C2, and C3 are in series,

1 1 1 1 1 1 1 1 −1 = + + = + + = ( μ F ) ⇒ Ceq = 2 μF Ceq C1 C2 C3 12 μ F 4 μ F 6 μ F 2 A potential difference of ΔVC = ε = 30 V across a capacitor of equivalent capacitance 2 μF produces a charge Q = = CeqΔVC (2 μF)(30 V) = 60 μC. Because Ceq is a combination of three series capacitors, Q1 = Q2 = Q3 = 60 μC. We are now able to find the potential difference across each capacitor:

ΔV1 = Assess:

Q1 60 μ C = = 5.0 V C1 12 μ F

ΔV2 =

Q2 60 μ C = = 15.0 V 4 μF C2

ΔV1 + ΔV2 + ΔV3 = 30 V = ΔVbat , as it should.

ΔV3 =

Q3 60 μ C = = 10.0 V. 6 μF C3

30.61. Model: Assume that the battery is ideal. Visualize:

The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure. Solve: Because C1 and C2 are in parallel, Ceq 12 = C1 + C2 = 4 μF + 12 μF = 16 μF Ceq 12 and C3 are in series, so

1 1 1 1 1 18 −1 32 = + = + = ( μ F ) ⇒ Ceq = 18 μF Ceq Ceq 12 C3 16 μ F 2 μ F 32 A potential difference of ΔVC = 9 V across a capacitor of equivalent capacitance

32 18

μ F produces a charge

32 Q = Ceq ΔVC = ( 18 μ F ) 9 V = 16 μ C

Because Ceq is a series combination of two capacitors Ceq 12 and C3, Q3 = Qeq 12 = 16 μC. The potential difference across C3 is

ΔV3 =

Q3 16 μ C = = 8.0 V C3 2 μ F

Now, Qeq 12 = 16 μC is the charge on the equivalent capacitor with Ceq 12 = 16 μF. So, the potential difference across the equivalent capacitor Ceq 12 is ΔVeq 12 =

Qeq 12 Ceq 12

=

16 μ C = 1.0 V 16 μ F

Parallel capacitors C1 and C2 have the same potential difference as the equivalent capacitor Ceq 12, so ΔV1 = ΔV2 = 1.0 V. The charge on each is given by Q = CΔV, so Q1 = (4 µF)(1.0 V) = 4.0 µC and Q2 = (12 µF)(1.0 V) = 12.0 µC. In summary, Q1 = 4.0 µC, ΔV1 = 1.0 V; Q2 = 12.0 µC, ΔV2 = 1.0 V; and Q3 = 16.0 µC, ΔV3 = 8.0 V. Assess: Note that ΔV3 + ΔVeq 12 = 9.0 V = ΔVbat, as it should. Also that Q1 + Q2 = 16.0 µC = Qeq 12, as it should.

30.62. Model: Assume the battery is an ideal battery. Visualize:

The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure. Solve: Because C2 and C3 are in series,

1 1 1 1 1 10 −1 24 = + = + = ( μ F ) ⇒ Ceq 23 = 10 μ F = 2.4 μ F Ceq 23 C2 C3 4 μ F 6 μ F 24 Ceq 23 and C1 are in parallel, so Ceq = Ceq 23 + C1 = 2.4 μF + 5 μF = 7.4 μF A potential difference of ΔVC = 9 V across a capacitor of equivalent capacitance 7.4 μF produces a charge Q = CeqΔVC = (7.4 μF)(9 V) = 66.6 μC Because Ceq is a parallel combination of C1 and Ceq 23, these capacitors have ΔV1 = ΔVeq 23 = ΔVC = 9 V. Thus the charges on these two capacitors are Q1 = (5 μF)(9 V) = 45 μC

Qeq 23 = (2.4 μF)(9 V) = 21.6 μC

Because Qeq 23 is due to a series combination of C2 and C3, Q2 = Q3 = 21.6 μC. This means

ΔV2 =

Q2 21.6 μ C = = 5.4 V C2 4 μF

ΔV3 =

Q3 21.6 μ C = = 3.6 V C3 6 μF

In summary, Q1 = 45 μC, V1 = 9 V; Q2 = 21.6 μC, V2 = 5.4 V; and Q3 = 21.6 μC, V3 = 3.6 V.

30.63. Model: Capacitors in parallel add to a greater capacitance compared to individual capacitances. On the other hand, capacitors in series add to a smaller capacitance compared to individual capacitances. Visualize:

Solve:

(a) Three capacitors in series:

1 1 1 1 3 −1 = + + = ( μ F ) ⇒ Ceq = 4 μF Ceq 12 μ F 12 μ F 12 μ F 12 (b) Two capacitors in parallel and the third in series with this parallel combination: Ceq 12 = 12 μF + 12 μF = 24 μF ⇒

1 1 1 1 1 1 = + = + = ⇒ Ceq = 8 μF Ceq Ceq 12 12 μ F 24 μ F 12 μ F 8 μ F

(c) Two capacitors in series and the third in parallel with this series combination:

1 1 1 1 −1 = + = ( μ F ) ⇒ Ceq 12 = 6 μF Ceq 12 12 μ F 12 μ F 6

⇒ Ceq = Ceq 12 + 12 μF = 6 μF + 12 μF = 18 μF (d) Three capacitors in parallel: Ceq = 12 μF + 12 μF +12 μF = 36 μF

30.64. Model: Capacitance is a geometric property. Visualize: Please refer to Figure P30.64. Shells R1 and R2 are a spherical capacitor C. Shells R2 and R3 are a spherical capacitor C′. These two capacitors are in series. Solve: The ratio of the charge to the potential difference is called the capacitance: C = Q ΔVC . The potential

differences across the capacitors C and C′ are ΔVC =

1 Q Q Q ⎡1 1⎤ − = ⎢ − ⎥ 4πε 0 R1 4πε 0 R2 4πε 0 ⎣ R1 R2 ⎦

ΔVC′ =

Q Q ⎡1 1 Q 1⎤ − = ⎢ − ⎥ 4πε 0 R2 4πε 0 R3 4πε 0 ⎣ R2 R3 ⎦

1

1

⎛1 1 ⎞ ⇒ C = ( 4πε 0 ) ⎜ − ⎟ ⎝ R1 R2 ⎠

−1

⎛ 1 1 ⎞ C ′ = ( 4πε 0 ) ⎜ − ⎟ R R 3 ⎠ ⎝ 2

−1

Because these two capacitors are in series,

1 1 1 1 ⎛1 1 1 1 ⎞ 1 ⎛ R3 − R1 ⎞ = + = ⎜ − + − ⎟= ⎜ ⎟ Cnet C C ′ 4πε 0 ⎝ R1 R2 R2 R3 ⎠ 4πε 0 ⎝ R1R3 ⎠

⎛ RR ⎞ ⎡ ( 0.010 m )( 0.030 m ) ⎤ 1 −12 Cnet = 4πε 0 ⎜ 1 3 ⎟ = ⎥ = 1.67 ×10 F = 1.67 pF 9 2 2 ⎢ ⎝ R3 − R1 ⎠ 9.0 ×10 N m /C ⎣ 0.030 m − 0.010 m ⎦ Assess:

Cnet depends on only the inner and outer shells, not on R2.

30.65. Model: Assume the battery is ideal. Visualize:

The circuit in Figure P30.65 has been redrawn to show that the six capacitors are arranged in three parallel combinations, each combination being a series combination of two capacitors. Solve: (a) The equivalent capacitance of the two capacitors in series is 12 C. The equivalent capacitance of the

six capacitors is 32 C. (b) As points a and b are midpoints of identical capacitors, Va = Vb = 6.0 V. Therefore, the potential difference between points a and b is zero.

30.66. Model: Capacitance is a geometric property and does not depend on the voltage, charge, or any other external factors. We assume that the capacitors are parallel-plate capacitors. Visualize: Please refer to Figure P30.66. The two electrodes are equivalent to two capacitors in parallel. Solve: The total capacitance of the two capacitors is ⎛1 1 ⎞ ε A ε A Cnet = C1 + C2 = 0 + 0 = ε 0 A ⎜ + ⎟ d1 d2 ⎝ d1 d 2 ⎠

1 1 ⎛ ⎞ = ( 8.85 ×10−12 C2 /N m 2 ) ( 0.010 m × 0.010 m ) ⎜ + ⎟ −3 −3 ⎝ 1.0 ×10 m 2.0 ×10 m ⎠ = 1.33 ×10−12 F = 1.33 pF

30.67. Model: Assume the battery is ideal. Visualize: Please refer to Figure P30.67. While the switch is in position A, the capacitors C2 and C3 are uncharged. When the switch is placed in position B, the charged capacitor C1 is connected to C2 and C3. C2 and C3 are connected in series to form an equivalent capacitor Ceq 23. Solve: While the switch is in position A, a potential difference of V1 = 100 V across C1 charges it to

Q1 = C1V1 = (15 ×10−6 F ) (100 V ) = 1500 μ C When the switch is moved to position B, this initial charge Q1 is redistributed. The charge Q1′ goes on C1 and the charge Qeq 23 goes on Ceq

23.

The voltage across C1 and Ceq

23

is the same and Q1′ + Qeq 23 = Q1 = 1500 μ C.

Combining these two conditions, we get 1500 μ C − Qeq 23 Qeq 23 Q1′ Qeq 23 = ⇒ = C1 Ceq 23 C1 Ceq 23 Since Ceq 23 = ( 201μ F + 301μ F ) = 12 μ F, we can rewrite this equation as −1

1500 μ C − Qeq 23 15 μ F

=

Qeq 23 12 μ F

⇒ Qeq 23 = 0.67 mC ⇒ Q1′ = Q1 − Qeq 23 = 1.500 mC − 0.67 mC = 0.83 mC

Having found the charge Qeq 23, it is easy to see that Q2 = Q3 = 0.67 mC because Ceq 23 is a series combination of C2 and C3. Thus, Q 670 μ C Q 670 μ C Q′ 830 μ C ΔV2 = 2 = = 34 V ΔV3 = 3 = = 22 V ΔV1 = 1 = = 55 V 20 μ F 30 μ F C2 C3 C1 15 μ F

30.68. Model: Assume the battery is an ideal battery. Visualize: Please refer to Figure P30.68. The battery is connected to two series capacitors C1 and C2. Solve: The equivalent capacitance is

1 1 1 CC ⇒ Ceq = 1 2 = + Ceq C1 C2 C1 + C2 Because the charge on capacitor C2 is 450 μC, the charge on Ceq and C1 is also 450 μC. We have

ΔVeq = Assess:

Qeq Ceq

⇒ 60 V =

450 μ C ( 450 μ C )(12 μ F + C2 ) ⇒ C = ( 450 μ C )(12μ F ) = 20 μ F = 2 C1C2 ( C1 + C2 ) 720 μ C − 450 μ C (12 μ F) C2

Note that capacitors connected in series have the same charge.

30.69. Model: Assume the battery is ideal. Visualize:

Solve:

When the capacitors are individually charged, their charges are

Q1 = C1V = (10 μ F )(10 V ) = 100 μ C

Q2 = C2V = ( 20 μ F )(10 V ) = 200 μ C

These two capacitors are then connected with the positive plate of C1 connected with the negative plate of C2 as shown in the figure. Let the new charges on C1 and C2 be Q1′ and Q2′ . Then,

Q1′ + Q2′ = 200 μ C − 100 μ C = 100 μ C The voltages across parallel capacitors are the same, so

ΔV1′ = ΔV2′ =

Q1′ Q2′ C = ⇒ Q2′ = 2 Q1′ C1 C2 C1

Substituting this expression for Q2′ into the previous equation,

Q1′ +

⎡ 20 μ F ⎤ C2 Q1′ = 100 μ C ⇒ Q1′ ⎢1 + ⎥ = 100 μC C1 ⎣ 10 μ F ⎦

Solving these equations, we get Q1′ = 33 μ C and Q2′ = 67 μ C. Finally,

ΔV1′ =

Q1′ 33 μ C = = 3.3 V = ΔV2′ C1 10 μ F

30.70. Solve: The magnitude of the work done by the external force is equal to the change in the electric potential energy of the capacitor. The work done is Wforce = ΔU = U 2 − U1 =

1 Q2 1 Q2 − 2 C2 2 C1

Note that the charge on the plates is not changed as the distance between the electrodes is changed. Thus, 2⎡ ⎡1 1⎤ 1 1 ⎤ − = 2.4 J Wforce = 12 Q 2 ⎢ − ⎥ = 12 ( 4.0 × 10−3 C ) ⎢ μ μ F ⎥⎦ C C 2.0 F 5.0 ⎣ 1⎦ ⎣ 2

Assess:

The work done on the capacitor is stored as electric potential energy in the capacitor.

30.71. Solve: (a) The capacitance of the parallel-plate capacitor is C1 =

−12 2 2 Aε 0 ( 0.10 m × 0.10 m ) (8.85 ×10 C /N m ) = = 8.85 ×10−11 F d1 1.0 ×10−3 m

The electric potential energy stored in the capacitor is −9 1 Q 2 1 (10 ×10 C ) = = 5.7 ×10−7 J U1 = 2 C1 2 8.85 ×10−11 F 2

(b) There is no change in the charge. The energy change is due to the change in the capacitance. The new capacitance is

C2 = −7

ε0 A d2

=

ε0 A 2d1

=

C1 2

The amount of energy stored is U 2 = 11.4 ×10 J. (c) Work was done on the capacitor by the agent pulling the plates apart, thereby adding energy into the system.

30.72. Solve: The energy density in the electric field at the surface of the sphere is u = 12 ε 0 E 2, where E is the electric field strength at the surface. The electric field strength and the potential on the surface of the sphere are related as follows: E=

1000 V Q ⎛ 1 Q⎞1 V =⎜ = 2.0 × 105 N/m ⎟ = = 2 4πε 0 R ⎝ 4πε 0 R ⎠ R R 5.0 ×10-3 m 1

⇒ u = 12 ( 8.85 × 10−12 C2 /Nm 2 )( 2.0 × 105 V/m ) = 0.177 J/m3 2

30.73. Model: Conservation of energy. Solve: The energy stored in the capacitor is dissipated through the flash lamp. Power is the rate at which energy is dissipated/absorbed, so the energy dissipated by the flash lamp is PΔt = (10 W)(10 μs) = 1.0 × 10−4 J This is the electric potential energy in the capacitor. Using U C = 12 C ΔVC2 ,

1.0 ×10−4 J =

1 2

C ( 3.0 V ) ⇒ C = 22 μF 2

30.74. Model: The capacitor is a parallel-plate capacitor and energy is conserved. Solve:

The energy needed to melt a 0.50 kg block of ice to water at 0°C is

Q = McΔT + MLf = ( 0.50 kg )( 2090 J/kg K )(10 K ) + ( 0.50 kg ) ( 3.33 ×105 J/kg ) The value of Lf is taken from Table 17.3 and the value of c for ice is taken from Table 17.2. Thus, Q = 176,950 J. The capacitor must store electric potential of this amount. Hence, U C = 12 C ( ΔVC ) ⇒ C = 2

2U C

( ΔVC )

2

=

2 (176,950 J )

( 50 V )

2

= 141.56 F

Because C = Aε 0 d , the size of the capacitor is A=

Cd

ε0

=

(141.56 F ) ( 2.0 ×10−3 m ) 8.85 ×10−12 C 2 /N m 2

The sheet’s edge is ≈179 km which is not reasonable.

= 3.2 ×1010 m 2

30.75.

Model: The induced charge on the polarized dielectric material forms a capacitor. Visualize: Please refer to Figure 30.32. Solve: From equation 30.30,

Einduced = E0 − E. The direction of the fields is taken care of by the signs in the equation and are as shown in Figure 30.32. E Multiplying through by ε 0 and using κ = 0 , E

⎛

1⎞

⎛

1⎞

ε 0 Einduced = ε 0 E0 − ε 0 E = ε 0 E0 ⎜1 − ⎟ ⇒ ηinduced = η0 ⎜ 1 − ⎟ ⎝ κ⎠ ⎝ κ⎠

30.76. Solve:

Model: The Geiger counter is modeled as two very long concentric conducting cylinders. From Example 28.5, the electric field due to a long charged wire carrying charge density λ is

E=

λ 2πε 0 r

.

The potential difference between the wire and tube is 1.0 mm λ λ λ dr = − ln ( r ) 25 mm = ln ( 25 ) 2πε 0 r 2πε 0 2πε 0 25 mm

1.0 mm

ΔV = −

∫

Note that λ is taken to be positive and the integration is from the lower potential tube to the higher potential wire in order to get ΔV > 0. The dielectric strength defines the maximum allowed electric field. The maximum field occurs at the surface of the wire, at minimum r. Thus Emax = 1.0 ×106 V/m =

λmax

2πε 0 (1.0 mm )

⇒ λmax = 2πε 0 (1.0 ×10 m )(1.0 × 106 V/m ) = 2000πε 0 −3

Using this in the expression for the potential difference, ΔV = Assess:

2000πε 0 ln ( 25 ) = 3.2 kV 2πε 0

A voltage of 3.2 kV is about right for a Geiger counter.

30.77.

Model: Visualize:

Solve:

The dielectric is slipped in parallel to the parallel-plate capacitor.

The vacuum-insulated parallel-plate capacitor has a potential difference ΔVC = E0 d and a capacitance

Q Q = . When the dielectric is placed between the plates, the electric field strength in the dielectric is ΔVC E0 d reduced by a factor of κ . Thus C0 =

ΔVC′ = −

+ plate

G d E d d d E d Ed⎛ 1⎞ G E0 ⋅ d s = E0 + 0 + E0 = E0 + 0 = 0 ⎜1 + ⎟ . κ 4 2 4 2 κ 2 2 ⎝ κ⎠ − plate

∫

The capacitance is now C=

Q ΔVC′

=

2κ Q ⎛⎜ 2 ⎞⎟ = C0 E0 d ⎜ 1 + 1 ⎟ 1+ κ ⎝ κ⎠

Assess: For materials other than vacuum, κ > 1, and C > C0 as expected. In the case of vacuum, κ = 1 and we get C = C0, as expected. Note that the integral to find the potential difference is taken from the negative plate to the positive plate. While the dielectric placement was treated as though it were dead center between the plates, this is not necessary, since half the horizontal distance between the plates is taken up by dielectric and the other half by vacuum, so the integral has the same result.

30.78. Solve: (a) Find an expression for the electric potential in a region where Ez = 2z V/m. The potential at z = 0 is 10 V. (b) Integrating both sides relative to z, V = − ∫ 2 z dz + C = − z 2 + C At z = 0 m, Vz = 10 V. This means C = 10 V. Thus V = 10 V – z2.

30.79. Solve: (a) A capacitor is constructed with two 10 cm × 10 cm plates. When the capacitor is connected to a 100 V source, ±400 nC is observed on the plates. What is the separation of the plates? (b) From the first equation, C=

400 ×10−9 C = 4.0 ×10−9 F 100 V

Thus, the second equation becomes 4.0 ×10−9 F =

(8.85 ×10

−12

C2 /N m 2 ) ( 0.10 m × 0.10 m ) d

⇒ d = 2.2 × 10−5 m = 0.022 mm

30.80. Solve: (a) What capacitor should be placed parallel to 3 μF and 6 μF capacitors that are in series so that the equivalent capacitance is 4 μF? (b) We have 18 μF + C = 4 μF ⇒ C = 2 μF 9

30.81.

Solve:

(a)

The equipotentials are lines of constant potential. The 0 V equipotential requires that V = 100( x 2 − y 2 ) = 0 V ⇒ y 2 = x 2 ⇒ y = ± x These are straight lines at ±45° in the xy-plane. The 100 V equipotential is a bit trickier. There, V = 100(x2 – y2) = 100 V ⇒ y2 = x2 – 1 ⇒ y = ± x 2 − 1

We need to recognize this as the equation for two hyperbolas, one passing through (x, y) = (1, 0) and the other through (x, y) = (−1, 0). The other equipotentials are also hyperbolas, leading to the potential map shown in the figure. (b) The electric field is

Ex = −

dV = −200 x V/m dx

G Written as a vector, this is E = −200 xiˆ − yjˆ V/m.

(

)

(c) The electric field lines are shown on the potential map.

Ey = −

dV = +200 y V/m dy

30.82. Visualize: The distances from the point to the two charges are r+ and r–. These can be expressed in terms of the coordinates x and y by using the Pythagorean theorem.

Solve:

(a) Potential is a scalar, so the potential at (x, y) is the sum of the potentials due to each charge. This is

V = V+ + V− =

q 4πε 0 r+

+

−q 4πε 0 r−

=

⎡ ⎤ q ⎢ 1 1 ⎥ − 2 ⎥ 2 4πε 0 ⎢ x 2 + ( y − s 2 )2 + + 2 x y s ( ) ⎦ ⎣

(b) The binomial approximation is (1 + u)n ≈ 1 + nu if u << 1. To use this, we’ll need to rearrange the square roots of V to be in this form. The first term is

1 x + ( y − s/2 ) 2

2

=

1 x + y − sy + s /4 2

2

2

=

1

1

x +y 2

−1/ 2

2

1+

=

⎡ − sy + s 2 /4 ⎤ 1+ 2 ⎥ 2 2 ⎢ x + y2 ⎦ x +y ⎣

≈

⎡ 1 sy 1 s2 ⎤ − ⎢1 + 2 2 2 2⎥ x +y ⎣ 2x +y 8x +y ⎦

1

≈

− sy + s 2 /4 x2 + y2

⎡ 1 − sy + s 2 /4 ⎤ ⎢1 − 2 2 ⎥ x +y ⎣ 2 x +y ⎦ 1

2

2

1

2

2

The approximation for the second term yields the same expression except the term that is linear in y is negative. Substituting into the expression for the potential in part (a),

V=

qsy

4πε 0 ( x 2 + y 2 )

3/ 2

This was a lot of algebra, but the resulting expression for V is much simpler than the exact expression of part (a). (c) The electric field components are ⎡ ⎤ qs ( 3 xy ) dV qsy ⎢ 3 2x ⎥= =− − 5 / 2 2 2 2 2 5/ 2 ⎢ ⎥ dx 4πε 0 2 ( x + y ) ⎣ ⎦ 4πε 0 ( x + y ) ⎡ ⎤ qs ( 2 y 2 − x 2 ) dV qs qsy ⎢ 3 2y ⎥ =− − − = Ey = − 3/ 2 4πε 0 ⎢ 2 ( x 2 + y 2 )5 / 2 ⎥ 4πε 0 ( x 2 + y 2 )5 / 2 dy 4πε 0 ( x 2 + y 2 ) ⎣ ⎦ Ex = −

(d) On-axis means on the axis of the dipole, or at x = 0 m. In this case, Ex = 0 V/m and the electric field is in exact agreement with Equation 27.11: G 2qs ˆ Eon-axis = j 4πε 0 y 3 (e) The bisecting axis is the x-axis, so y = 0 m. In this case, Ex = 0 V/m and the electric field is in exact agreement with Equation 27.12:

G −qs ˆ Ebisecting axis = j 4πε 0 x3

30.83.

Visualize:

Solve: The charge density inside the cylinder is ρ. Consider a Gaussian surface with a radius r inside the cylinder. The charge enclosed by this closed surface is Qenclosed = (πr2L)ρ. Using Gauss’s law, 2 G ⎛ ρ ⎞ G G Qenclosed (π r L ) ρ π r 2 Lρ ⎛ ρ ⎞ E ⋅ dA = E 2 π rL = = ⇒ E = =⎜ ( ) ⎟r ⇒ E = ⎜ ⎟ rrˆ. ú ε0 ε0 ε 0 2π rL ⎝ 2ε 0 ⎠ ⎝ 2ε 0 ⎠

In the last step, we have taken into account the radial direction of the electric field pointing outward. The potential difference between the surface and the axis is R R G G ρ ρ −ρ r 2 G ΔV = − ∫ E ⋅ dr = − ∫ r rˆ ⋅ dr = − ∫ r dr = 2ε 0 2ε 0 2ε 0 2 0 0

R

=− 0

ρ R2 4ε 0

30.84.

Visualize:

Solve: (a) The E-versus-r graph is shown above. (b) From Equation 30.3, the potential difference between the center (V0) and a point at radius r is r

1 Qr Q r2 dr = − 3 4πε 0 R 4πε 0 R 3 2 0

ΔV = Vr − V0 = − ∫ Er dr = − ∫

We can find V0 from the reference point for the potential. At r = R the potential matches the known outside potential, so VR =

1 Q 4πε 0 R

Using the expression in the ΔV equation, VR − V0 = −

1 Q 1 ⎛ Q ⎞ 1 ⎛ 3Q ⎞ R2 ⇒ V0 = + ⎜ ⎟= ⎜ ⎟ 3 4πε 0 R 4πε 0 ⎝ 2 R ⎠ 4πε 0 ⎝ 2 R ⎠ 4πε 0 R 2 Q

Using this result for V0, the potential at radius r inside the cylinder is Vr =

1 ⎛ 3Q ⎞ Q ⎛ r2 ⎞ 1 Q ⎡ 3 r2 ⎤ ⎜ 3⎟= ⎜ ⎟− ⎢ − ⎥ 4πε 0 ⎝ 2 R ⎠ 4πε 0 ⎝ 2 R ⎠ 4πε 0 R ⎣ 2 2 R 2 ⎦

(c) The ratio is

Vcenter V0 ⎡ 1 ⎛ 3Q ⎞ ⎤ = =⎢ ⎜ ⎟⎥ Vsurface VR ⎣ 4πε 0 ⎝ 2 R ⎠ ⎦ (d) A V-versus-r graph is shown in the figure.

⎡ 1 ⎛ Q ⎞⎤ 3 ⎢ ⎜ ⎟⎥ = ⎣ 4πε 0 ⎝ R ⎠ ⎦ 2

30.85.

Visualize:

For infinitely long cylinders, the electric field is radial from the inner cylinder to the outer

cylinder.

Solve: (a) The electric field in the space between the cylinders is the electric field of the charged inner cylinder. (Gauss’s law tells us that the outer cylinder does not contribute to the electric field between the cylinders.) From Chapter 27, we know that the field of a cylinder with linear charge density λ is

G E=

λ 2πε 0 r

rˆ

The potential difference between the two cylinders is ΔV = −

∫

R2 R1

Er dr = −

λ 2πε 0

∫

R2 R1

dr λ λ R R =− ln 2 [ln(r )]R12 = − 2πε 0 2πε 0 R1 r

The minus sign tells us that the potential decreases upon moving outward, but we only need the absolute value of ΔV. Capacitance is defined as C = Q/ΔV. Consider a segment of the cylinders of length L. The charge on the inner cylinder is Q = λL. Consequently, the capacitance of this segment is

C=

Q λL 2πε 0 L = = ΔV (λ /2πε 0 ) ln( R2 / R1 ) ln( R2 / R1 )

The capacitance per meter is found by dividing out the L: capacitance per meter =

C 2πε 0 = L ln( R2 / R1 )

(b) Using the above expression, the capacitance per meter of cable is 4πε 0 1 = = 31 pF/m ( 2 ) ln ( R2 R1 ) 2 ( 9.0 ×109 N m 2 /C2 ) ln ⎛ 3.0 mm ⎞ ⎜ ⎟ ⎝ 0.50 mm ⎠

30.86.

Visualize:

Solve:

The circuit is redrawn to better visualize the series and parallel connections. The numbers help follow the 13 C is shown. rearrangement. The reduction to the equivalent capacitance of 7

30-1

31.1. Solve: The wire’s cross-sectional area is A = π r 2 = π (1.0 × 10−3 m ) = 3.1415 × 10−6 m 2 , and the electron 2

current through this wire is i =

Ne = 2.0 × 1019 s −1 . Using Table 31.1 for the electron density of iron and Equation Δt

31.3, the drift velocity is vd =

Assess:

i 2.0 × 1019 s −1 = = 7.5 × 10−5 m/s = 75 μm/s nA ( 8.5 × 1028 m −3 )( 3.1415 × 10−6 m 2 )

The drift speed of electrons in metals is small.

31.2. Solve: Using Equation 31.3 and Table 31.1, the electron current is i = nAvd = ( 5.9 × 1028 m −3 ) π ( 0.5 × 10−3 m ) ( 5.0 × 10−5 m/s ) = 2.3 × 1018 s −1 2

The time for 1 mole of electrons to pass through a cross section of the wire is

t= Assess:

N A × 1 mole 6.02 × 1023 = = 2.62 × 105 s ≈ 3.0 days i 2.3 × 1018 s −1

The drift speed is small, and Avogadro’s number is large. A time of the order of 3 days is reasonable.

31.3. Solve: Equation 31.2 is N e = nAvd Δt . Using Table 31.1 for the electron density, we get A=

⇒D=

π D2 4

=

Ne nvd Δt

4 (1.0 × 1016 ) 4 Ne = = 9.3 × 10−4 m = 0.93 mm π nvd Δt π ( 5.8 × 1028 m −3 )(8.0 × 10−4 m/s )( 320 × 10−6 s )

31.4. Solve: The number of electrons crossing a cross sectional area of a wire is the electron current i. 2

⎛ 1.6 × 10−3 m ⎞ −4 19 −1 i = nAvd = ( 6.0 × 1028 m −3 ) π ⎜ ⎟ ( 2.0 × 10 m/s ) = 2.4 × 10 s 2 ⎝ ⎠ The electron density n for aluminum is taken from Table 31.1. The number of electrons passing through the cross section in one day is Ne = iΔt = (2.4×1019 s–1)(365 days)(24 hr/day)(3600 s/hr) = 7.6×1026 electrons Assess: The large electron density compensates for the small drift velocity to deliver a huge number of electrons in current.

31.5. Solve: Using Equation 31.2, n=

Ne 1.44 × 1014 = = 6.0 × 1028 m −3 Avd Δt ( 4.00 × 10−6 m 2 )( 2.00 × 10−4 m/s )( 3.0 × 10−6 s )

From Table 31.1, the metal is aluminum.

31.6. Model: Use the conduction model to relate the drift speed to the electric field strength. Solve:

From Equation 31.7, the electric field is −31 −4 mv ( 9.11 × 10 kg )( 2.0 × 10 m/s ) E= d = = 0.023 N/C eτ (1.60 ×10−19 C)( 5.0 ×10−14 s )

31.7. Solve: For L = 1.0 cm = 1.0 × 10−2 m , the surface area of the wire is A = (2π r)L = πDL = π (1.0×10−3 m)(1.0×10−2 m) = (1.0×10−5 m2)π

The surface charge density of the wire is

η=

−1 −19 Q (1000 cm × 1 cm )1.60 × 10 C = = 5.1× 10−12 C/m 2 A (1.0 ×10−5 m2 )π

31.8. Solve: (a) Each gold atom has one conduction electron. Using Avogadro’s number and n as the number of moles, the number of atoms is N = nN A =

ρ (π r 2 L ) m ρV NA = NA = NA MA MA MA

The density of gold is ρ = 19,300 kg/m3, the atomic mass is MA = 197 g mol−1, r = 0.50 × 10−3 m, L = 0.10 m, and NA = 6.02 × 1023 mol−1. Substituting these values, we get N = 4.6 × 1021 electrons. (b) If all the electrons are transferred a charge of (4.6 × 1021)(−1.60 × 10−19 C) = −740 C will be delivered. To deliver a charge of −32 nC, however, the electrons within a length l have to be delivered. Thus, −32 × 10−9 C l= (10 cm ) = 4.3 × 10−10 cm = 4.3 × 10−12 m −740 C

31.9. Model: We will use the model of conduction to relate the electric field strength to the mean free time between collisions. Solve: From Equation 31.8, the electric field is ( 9.11× 10−31 kg )( 5.0 ×1019 s −1 ) mi E= = = 0.31 N/C neτ A ( 8.5 × 1028 m −3 )(1.6 × 10−19 C )( 4.2 × 10−15 s ) π ( 0.9 × 10−3 m )2

31.10. Solve: (a) The electron current is i = nAvd ⇒ vd =

( 3.5 × 1017 s−1 ) i = nA ( 6.0 × 1028 m −3 ) π ( 5.0 × 10−4 m )2

= 7.43 × 10−6 m/s The electron drift speed is 7.4×10–6 m/s. The electron density for aluminum is taken from Table 31.1. (b) The electron drift velocity is related to the electric field in the wire by ( 9.11×10−31 kg )( 7.43 ×10−6 m/s ) = 2.1×10−14 s eτ mv vd = E ⇒ τ = d = m eE (1.6 ×10−19 C)( 2.0 ×10−3 V/m ) Assess:

This is about the right order of magnitude based on the examples in the text.

31.11. Visualize:

The current density J in a wire, as given by Equation 31.13, does not depend on the thickness of the wire. Solve: (a) The current in the wire is 2

⎡1 ⎤ I wire = J wire Awire = ( 4.5 × 105 A/m 2 )π ⎢ (1.5 × 10−3 m ) ⎥ = 0.795 A 2 ⎣ ⎦

Because current is continuous, Iwire = Ifilament. Thus, Ifilament = 0.795 Α ≈ 0.80 A. (b) The current density in the filament is I 0.795 A = 7.0 × 107 A/m 2 J filament = filament = 2 Afilament ⎡1 ⎤ π ⎢ ( 0.12 × 10−3 m )⎥ ⎣2 ⎦

31.12. Solve: (a) The current density is J=

I I = = A π R2

0.85 A ⎡1 ⎤ π ⎢ ( 0.00025 m ) ⎥ 2 ⎣ ⎦

2

= 1.73 × 107 A/m 2

(b) The electron current, or number of electrons per second, is Ne I 0.85 A 0.85 C/s = = = = 5.3 × 1018 s −1 Δt e 1.60 × 10−19 C 1.60 × 10−19 C

31.13. Solve: (a) From Equation 31.13 and Table 31.1, the current density in the gold wire is J = nevd = ( 5.9 × 1028 m −3 )(1.60 × 10−19 C )( 3.0 × 10−4 m/s ) = 2.83 × 106 A/m 2 The current density is 2.8 × 106 A/m 2 . (b) The current is

I = JA = ( 2.83 × 106 A/m 2 ) π ( 0.25 × 10−3 m ) = 0.56 A 2

31.14. Solve: From Equation 31.13, the current in the wire is I = JA = ( 7.50 × 105 A/m 2 )( 2.5 × 10−6 m × 75 × 10−6 m ) = 0.141 mA

31.15. Solve: Equation 31.10 is Q = IΔt. The amount of charge delivered is 60 s ⎞ ⎛ 3 Q = (10.0 A ) ⎜ 5.0 min × ⎟ = 3.0 × 10 C 1 min ⎠ ⎝ The number of electrons that flow through the hair dryer is

N=

Q 3.0 × 103 C = = 1.88 × 1022 e 1.60 × 10−19 C

31.16. Solve: From Equation 31.10, I=

13 −19 Q Ne ( 2.0 × 10 )(1.60 × 10 C ) = = = 0.0032 C/s = 0.0032 A = 3.2 mA Δt Δt 1.0 × 10−3 s

31.17. Visualize:

G The direction of the current I in a material is opposite to the direction of motion of the negative charges and is the same as the direction of motion of positive charges. Solve: The charge due to positive ions moving to the right per second is q+ = N + ( 2e ) = ( 5.0 × 1015 )( 2 × 1.60 × 10−19 C ) = 1.60 × 10−3 C The charge due to negative ions moving to the left per second is

q− = N ( −e ) = ( 6.0 × 1015 s )( −1.60 × 10−19 C ) = −0.96 × 10−3 C Thus, the current in the solution is i=

−3 −3 q+ − q− 1.60 × 10 C − ( −0.96 × 10 C ) = = 2.56 × 10−3 A = 2.6 mA t 1s

31.18. Solve: (a) The current density is J=

2.5 A I = = 6.25 × 105 A/m 2 = 6.3 × 105 A/m 2 A 4.0 × 10−6 m 2

(b) Using Equation 31.13 and Table 31.1, the drift speed is vd =

J 6.25 × 105 A/m 2 = = 6.5 × 10−5 m/s ne ( 6.0 × 1028 m −3 )(1.6 × 10−19 C )

31.19. Visualize:

Solve:

The current-carrying cross section of the wire is 2 2 A = π r12 − π r22 = π ⎡( 0.0010 m ) − ( 0.00050 m ) ⎤ = 2.356 × 10−6 m 2 ⎣ ⎦

The current density is J=

10 A = 4.2 × 106 A/m 2 2.356 × 10−6 m 2

31.20. Model: Use the model of conduction to relate the mean time between collisions to conductivity. Solve:

From Equation 31.17, Table 31.1, and Table 31.2, the mean time between collisions for aluminum is

τ Al =

−31 7 −1 − 1 mσ Al ( 9.11 × 10 kg )( 3.5 × 10 Ω m ) = = 2.1 × 10−14 s 2 nAle 2 ( 6.0 × 1028 m −3 )(1.60 × 10−19 C )

Similarly, the mean time between collisions for iron is τ iron = 4.2 × 10−15 s. Assess:

The mean time between collisions in metals are of the order of 10−14 s.

31.21. Model: We will use the model of conduction to relate the mean time between collisions to conductivity. Solve: From Equation 31.17, Table 31.1, and Table 31.2, the mean time between collisions for silver is

τ silver =

−31 7 −1 − 1 mσ silver ( 9.11 × 10 kg )( 6.2 × 10 Ω m ) = = 3.8 × 10−14 s 2 28 −3 −19 nsilver e 2 ( 5.8 × 10 m )(1.60 × 10 C )

Similarly, the mean time between collisions for gold is τ gold = 2.5 × 10−14 s . Assess:

Mean free times in metals are of the order of ≈10−14 s.

31.22. Solve: The current density is J = σ E. Using Equation 31.18 and Table 31.2, the current in the wire is I = σ EA = ( 3.5 × 107 Ω −1m −1 ) ( 0.012 V/m ) ( 4 × 10−6 m 2 ) = 1.68 A

31.23. Solve: From Equation 31.18 and Table 31.2, the electric field is E=

J

σ

=

I

σA

=

(1.0 ×10

5.0 A

7

Ω m −1 ) π (1.0 × 10−3 m ) −1

2

= 0.159 N/C

31.24. Solve: From Equations 31.18 and 31.19, the resistivity is −3 E E EA Eπ r 2 ( 0.085 V/m ) π (1.5 × 10 m ) ρ= = = = = = 5.0 × 10−8 Ω m J I A I I 12 A 2

31.25. Solve: From Equations 31.18 and 31.19, the resistivity is −3 E E EA Eπ r 2 ( 0.0075 V/m ) π ( 0.50 × 10 m ) ρ= = = = = = 1.51 × 10−6 Ω m J I A I I 3.9 × 10−3 A 2

From Table 31.2, we see that the wire is made of nichrome.

31.26. Solve: (a) Since J = σ E and J = I A , the electric field is E=

I

=

I

σ A π r 2σ

=

0.020 A

π ( 0.25 × 10 m ) ( 6.2 × 10 Ω m −3

2

7

−1

−1

)

= 1.64 × 10−3 N/C

(b) Since the current density is related to vd by J = I A = nevd , the drift speed is

vd =

I

π r 2 ne

=

0.020 A

π ( 0.25 × 10 m ) ( 5.8 × 10 m −3

2

28

−3

)(1.60 × 10

−19

C)

= 1.10 × 10−5 m/s

Assess: The values of n and σ for silver have been taken from Table 31.1 and Table 31.2. The drift velocity is typical of metals.

31.27. Model: Because current is conserved, the currents in the two segments of the wire are the same. Visualize: Please refer to Figure P31.27. Solve: The currents in the two segments of the wire are related by I1 = I2 = I. But I = AJ, so we have A1J1 = A2J2. The wire’s diameter is constant, so J1 = J2 and σ1E1 = σ2E2. The ratio of the electric fields is

E2 σ 1 1 1 = = = E1 σ 2 σ 2 σ 1 2

31.28. Visualize:

Solve: The current density through the cube is J = σE, and the actual current is I = AJ. Combining these equations, the conductivity is I 9.0 A σ= = = 1.80 × 107 Ω −1m −1 AE (10−4 m 2 )( 5.0 × 10−3 V/m )

If all quantities entering the calculation are in SI units then the result for σ has to be in SI units. From the value for σ, we can identify the metal as being tungsten.

31.29. Model: Assume the battery is ideal. Solve: (a) Because the battery provides a current of 0.50 A and the current is defined as the amount of charge that passes through a cross section per second, the charge lifted by the escalator is 0.50 C/s. (b) The work done by the escalator in lifting charge Q is

W = U = QΔV = (1.0 C )(1.5 V ) = 1.5 J (c) The power output of the charge escalator is the work done by the escalator per unit time:

P=

W ΔU QΔV = = = I ΔV = ( 0.50 A )(1.5 V ) = 0.75 W Δt Δt Δt

31.30. Solve: (a) Resistivity depends only on the type of material, not the geometry of the wire. Wires 1 and 2 are made of the same material, so ρ2 = ρ1 and thus ρ2/ρ1 = 1.00. (b) The resistance of a wire of length L and radius r is given by Equation 31.23: R=

ρL A

=

ρL π r2

Because the two wires have the same resistivity, 2

2

2 R2 ρ L2 π r 2 ⎛ r1 ⎞ L2 ⎛ 1 ⎞ 2 1 = =⎜ ⎟ = ⎜ ⎟ = = 0.50 2 R1 ρ L1 π r1 ⎝ r2 ⎠ L1 ⎝ 2 ⎠ 1 2

31.31. Solve: (a) Using Table 31.2 and Equation 31.23, the resistance is R=

ρL A

ρ L (1.7 × 10 Ωm ) (1.0 m ) = = 0.087 Ω 2 π r2 π ( 2.5 × 10−4 m ) −8

=

(b) The resistance is

R=

ρL A

=

ρL d

2

=

( 3.5 × 10

−5

Ωm ) ( 0.10 m )

( 0.0010 m )

2

= 3.5 Ω

31.32. Solve: We can identify the material by its resistivity. Starting with the wire’s resistance, R=

ρL

π r 2 R π ( 0.0004 m ) (1.1 Ω ) ρL ⇒ ρ= = = 5.5 × 10−8 Ωm 2 L 10 m πr 2

=

A From Table 31.2, we can identify the wire as being made of tungsten.

31.33. Model: Assume the battery is an ideal battery. Solve: 31.20:

Connecting the wire to the battery leads to an electric field inside the wire that is given by Equation

Ewire =

ΔVwire ⇒ ΔVwire = LEwire = (0.30 m)(5.0×10−3 V/m) = 1.50×10−3 V = 1.50 mV L

31.34. Model: Assume the battery is an ideal battery.

Solve: (a) The current in a wire is related to its resistance R and to the potential difference ΔVwire between the ends of the wire as I=

ΔVwire ΔV 1.5 V ⇒ R = wire = = 3.0 Ω R 0.50 A I

From Equation 31.23,

R=

ρL A

⇒L=

RA

ρ

=

( 3.0 Ω )π ( 0.30 × 10−3 m ) 2.8 × 10−8 Ωm

2

= 30 m

(b) For a uniform wire of the same material, R ∝ L . If the wire in this problem is cut in half, its resistance will decrease to 12 ( 3 Ω ) = 1.5 Ω . Thus, using I = ΔVwire R , we have I = 1.0 A .

31.35. Model: Assume the battery is an ideal battery. Solve: We can find the current I from Equation 31.24 provided we know the resistance R of the gold wire. From Equation 31.23 and Table 31.2, the resistance of the wire is

R=

ρL A

ρ L ( 2.4 × 10 Ωm ) (100 m ) ΔVwire 0.70 V = = 305.6 Ω ⇒ I = = = 2.3 mA 2 −3 π r2 305.6 Ω R π ( 0.050 × 10 m ) −8

=

31.36. Solve: From Table 31.2, the resistivity of carbon is ρ = 3.5 × 10−5 Ω m. From Equation 31.23, the resistance of lead from a mechanical pencil is

R=

ρL A

ρ L ( 3.5 × 10 Ω m ) ( 0.060 m ) = = 5.5 Ω 2 π r2 π ( 0.35 × 10−3 m ) −5

=

31.37. Solve: (a) From Table 31.2, the resistivity of aluminum is ρ = 2.8 × 10−8 Ω m. From Equation 31.23, the length L of a wire with a cross-sectional area A and having a resistance R is L=

AR

ρ

(10 ×10 =

−6

m ) (1000 Ω ) 2

2.8 × 10−8 Ω m

= 3.57 m ≈ 3.6 m

(b) The number of turns is the length of the wire divided by the circumference of one turn. Thus,

3.57 m = 380 2π (1.5 × 10−3 m )

31.38. Solve: The slope of the I versus ΔV graph, according to Ohm’s law, is the resistance R. From the graph in Figure EX31.38, the slope of the material is ΔV 100 V = = 50 Ω I 2A

31.39. Solve: We need an aluminum wire whose resistance and length are the same as that of a 0.50-mmdiameter copper wire. That is, RCu =

ρCu L ACu

⇒ rAl =

= RAl =

ρ Al L AAl

⎛ρ ⎞ ⎛ρ ⎞ ⇒ AAl = ⎜ Al ⎟ ACu ⇒ π rAl2 = ⎜ Al ⎟ π rCu2 ρ ⎝ Cu ⎠ ⎝ ρ Cu ⎠

ρ Al 2.8 × 10−8 Ω m rCu = ( 0.25 mm ) = 0.32 mm ρ Cu 1.7 × 10−8 Ω m

We need a 0.64-mm-diameter aluminum wire.

31.40. Solve: Equation 31.18 will be used to relate electric field strength with the diameter. We have I I 4I σπ ED 2 J= =1 = =σE ⇒ I = 2 2 A 4πD πD 4 Because the current is the same in the two wires, 2

⎛σ ⎞ ⎛D ⎞ Enichrome = ⎜ aluminum ⎟ ⎜ aluminum ⎟ Ealuminum σ D ⎝ nichrome ⎠ ⎝ nichrome ⎠ Using the values of σ from Table 31.2, 2

⎛ 3.5 × 107 Ω −1m −1 ⎞ ⎛ 1.0 mm ⎞ Enichrome = ⎜ ⎟ ( 0.0080 V/m ) = 0.104 V/m −1 −1 ⎟ ⎜ 5 ⎝ 6.7 × 10 Ω m ⎠ ⎝ 2.0 mm ⎠

31.41. Solve: The density of aluminum is 2700 kg/m3, so 1.0 m3 of aluminum has a mass of 2700 kg. The

conduction-electron density is the number of electrons in 1.0 m3. Because the atomic mass of aluminum is 27 u, 27 g of aluminum contains NA = 6.02 × 1023 atoms. Thus, the number of aluminum atoms in 2700 kg of aluminum is 6.02 × 1023 ( 2700 kg ) = 6.0 × 1028 ( 27 g ) The number density of aluminum atoms is 6.0 × 1028 m −3 . Since each aluminum atom contributes one conduction electron to the metal, the number density of conduction electrons in aluminum is n = 6.0 × 1028 m −3 . Assess: The number density n obtained above agrees with that given in Table 31.1.

31.42. Solve: (a) The moving electrons are a current, even though they’re not confined inside a wire. The electron current is 50 × 10−6 A Ne I = = = 3.1 × 1014 s −1 Δt e 1.60 × 10−19 C This means during the time interval Δt = 1 s, 3.1 × 1014 electrons strike the screen. (b) The current density is I I 50 × 10−6 A = 4.0 × 102 A/m 2 J= = 2= 2 A πr π ( 0.00020 m ) (c) The acceleration can be found from kinematics:

v12 = ( 4.0 × 107 m/s ) = v02 + 2aΔx = 2aΔx ⇒ a = 2

( 4.0 × 10 m/s ) = 1.60 × 10 2 ( 5.0 × 10 m ) 2

7

17

−3

m/s 2

But the acceleration is a = F m = eE m . Consequently, the electric field must be E=

−31 17 2 ma ( 9.11 × 10 kg )(1.60 × 10 m/s ) = = 9.1 × 105 V/m e 1.6 × 10−19 C

(d) When they strike the screen, each electron has a kinetic energy

K = 12 mv12 =

1 2

( 9.11×10

−31

kg )( 4.0 × 107 m/s ) = 7.288 × 10−16 J

Power is the rate at which the screen absorbs this energy. The power of the beam is P=

N ΔE = K e = ( 7.288 × 10−16 J )( 3.1 × 1014 s −1 ) = 0.23 J/s = 0.23 W Δt Δt

Assess: Power delivered to the screen by the electron beam is reasonable because the screen over time becomes a little warm.

31.43. Visualize: Please refer to Figure P31.43. Solve: (a) The current associated with the moving film is the rate at which the charge on the film moves past a certain point. The tangential speed of the film is rev 1 min 2π rad × × × 1.0 cm = 9.425 cm/s min 60 s 1 rev In 1.0 s the film moves a distance of 9.425 cm. This means the area of the film that moves to the right in 1.0 s is (9.425 cm)(4.0 cm) = 37.7 cm2. The amount of charge that passes to the right in 1.0 s is v = ω r = ( 90 rpm )( 4.0 cm ) = 90

Q = (37.7 cm2)(−2.0 × 10−9 C/cm2) = −75.4 × 10−9 C

Since I = Q Δt , we have

I=

− ( 75.4 × 10−9 C ) 1s

= 75.4 nA

The current is 75 nA. (b) Having found the current in part (a), we can once again use I = Q Δt to obtain Δt:

Δt =

−6 Q −10 × 10 C = = 133 s I 75.4 × 10−9 A

31.44. Model: Current is the rate at which the charge moves across a certain cross section. Solve: (a) For the circular motion of the electron around the proton, Coulomb’s force between the electron and the proton causes the centripetal acceleration. Thus, 2 −e e mv 2 m ⎛ 2π r ⎞ ( 9.0 × 109 N m 2/C2 )(1.60 × 10−19 C ) = 6.56 × 1015 Hz 1 = = ⇒ = f = ⎜ ⎟ 3 4πε 0 r 2 r r⎝ T ⎠ T 4π 2 ( 0.053 × 10−9 m ) ( 9.11 × 10−31 kg ) 2

1

The frequency is 6.6 × 1015 Hz. (b) Charge Q = e passes any point on the orbit once every period. Thus the effective current is

I=

Q e = = ef = (1.6 × 10−19 C )( 6.56 × 1015 Hz ) = 1.05 × 10−3 A Δt T

31.45. Model: The current is the rate at which the charge of the ions moves through the ionic solution. Solve:

Because the atomic mass of gold is 197 g, the number of gold atoms is

N=

⎛ 0.50 g ⎞ M 6.02 × 1023 mol−1 = 1.53 × 1021 atoms NA = ⎜ −1 ⎟ MA ⎝ 197 g mol ⎠

We need to deposit N = 1.53 × 1021 gold ions, each with a charge of −1.60 × 10−19 C, in 3.0 hours on the statue. The current is I=

21 −19 Q (1.53 × 10 )(1.60 × 10 C ) = = 23 mA Δt 3.0 × 3600 s

31.46. Solve: (a) A current of 1.8 pA for the potassium ions means that a charge of 1.8 pC flows through the potassium ion channel per second. The number of potassium ions that pass through the ion channel per second is 1.8 × 10−12 C/s = 1.125 × 107 s −1 1.6 × 10−19 C Since the channel opens only for 1.0 ms, the total number of potassium ions that pass through the channel is (1.125 × 107 s−1 )(1.0 ×10−3 s ) = 1.13 × 104 atoms. (b) The current density in the ion channel is

J=

I 1.8 pA 1.8 × 10−12 A = = = 2.5 × 107 A/m 2 2 A π ( 0.30 nm 2 ) π ( 0.15 × 10−9 m )2

31.47. Solve: (a) Current is defined as I = Q/Δt, so the charge delivered in time Δt is Q = IΔt = (150 A)(0.80 s) = 120 C (b) The drift speed is

vd =

J I A I 150 A = = = = 5.617 × 10−4 m/s ne ne π r 2 ne π ( 0.0025 m )2 ( 8.5 × 1028 m −3 )(1.60 × 10−19 C )

At this speed the electrons drift a distance d = (5.617 × 10−4 m/s)(0.80 s) = 4.49 × 10−4 m = 0.45 mm

31.48. Solve: The total charge in the battery is Q = I Δt = (90 A)(3600 s) = 3.2 × 105 C

31.49. Model: We assume that the charge carriers are uniformly distributed throughout the wire. Solve:

Using Equation 31.16, we can write the current density as

J=

⎛ ne 2τ ⎞ I ne 2τ = E ⇒ I =⎜ E⎟A A m ⎝ m ⎠

For a given wire, the current is thus proportional to the area through which the current flows: I total ∝ R 2 and

I center ∝ ( 12 R ) . Therefore, 2

I center 1 = = 25% I total 4 That is, 25% of the total current flows in the part of the wire with radius r ≤ R/2.

31.50. Solve: Equation 31.13 defines the current density as J = I A . This means A=

Assess:

π D2 4

=

4 (1.0 A ) I 4I ⇒D= = = 0.050 cm = 0.50 mm J πJ π ( 500 A/cm 2 )

Fuse wires are usually thin.

31.51. Visualize:

Solve: (a) Consider a cylindrical surface inside the metal at a radial distance r from the center. The current is flowing through the walls of this cylinder, which have surface area A = ( 2π r ) L. Thus

I = JA = σ E ( 2π r ) L Thus the electric field strength at radius r is E=

(b) For iron, with σ = 1.0 × 107 Ω−1m−1,

I 2πσ Lr

⎛ ⎞⎛ 25 A 1 ⎞ ⎟ = 4.0 × 10−4 V/m Einner = ⎜ 7 −1 −1 ⎜ 2π ( 0.10 m ) (1.0 × 10 Ω m ) ⎟ ⎝⎜ 0.010 m ⎠⎟ ⎝ ⎠ ⎛ ⎞⎛ 25 A 1 ⎞ ⎟ = 1.59 × 10−4 V/m Eouter = ⎜ 7 −1 −1 ⎜ 2π ( 0.10 m ) (1.0 × 10 Ω m ) ⎟ ⎝⎜ 0.025 m ⎠⎟ ⎝ ⎠

31.52. Visualize:

Solve: (a) Consider a spherical surface inside the hollow sphere at a radial distance r from the center. The current is flowing outward through this surface, which has surface area A = 4π r 2 . Thus

I = JA = σ E ( 4π r 2 ) Thus the electric field strength at radius r is E=

(b) For copper, with σ = 6.0 × 107 Ω−1m−1,

I 4πσ r 2

Einner =

25 A 1 = 3.3 × 10−4 V/m 2 −1 −1 7 4π ( 6.0 × 10 Ω m ) ( 0.010 m )

Eouter =

25 A 1 = 5.3 × 10−5 N/C 2 −1 −1 7 4π ( 6.0 × 10 Ω m ) ( 0.025 m )

31.53. Solve: (a)

(b) Since I = ΔQ Δt , for infinitesimal changes

I=

dQ d = ( 4t − t 2 ) = 4 − 2t dt dt

(c)

(d) The value of I at t = 2.0 s is zero. This is because the charge, having reached its maximum value, has stopped entering the wire. The negative values of I mean that charge is flowing out of the wire.

31.54. Solve: (a) The value of Q (C) for selected values of t are 0 0

t (s) Q (C)

1 7.87

2 12.6

4 17.3

6 19.0

8 19.6

10 19.9

(b) Since I = ΔQ Δt , for infinitesimal changes I=

dQ d = ( 20 C ) (1 − e − t 2.0 s ) = ( 20 C ) ( −e − t 2.0 s ) ( −1 2.0 s ) = (10 A ) e − t 2.0 s dt dt

(c) The maximum value of the current occurs at t = 0 s and is 10.0 A. (d) The values of I (A) for selected values of t are t (s) I (A)

0 10

1 6.07

2 3.68

4 1.35

6 0.50

8 0.18

10 0.07

31.55. Solve: (a) The values of I (A) for selected values of t are t (μs) I (A)

0 2.0

1 1.21

2 0.74

4 0.27

6 0.10

8 0.036

10 0.014

(b) Because I = dQ dt , t

t

0

0

Q = ∫ I dt = ∫ ( 2.0 A ) e − t ( 2.0 μ s ) dt = ⎡⎣( −4.0 A μs ) e− t 2.0 μ s ⎤⎦ = ( 4.0 μ C ) ⎡⎣1 − e − t 2.0 μ s ⎤⎦ where we have used the condition Q = 0 C at t = 0 μs. (c) The values of Q (μC) for selected values of t are t (μs) Q (μC)

0 0

1 1.57

2 2.52

4 3.46

6 3.80

8 3.92

10 3.98

31.56. Model: Use the calculation of the electric field of a ring of charge from Chapter 27. Visualize:

Both the rings contribute equally to the field strength. The radius of each ring is R = 1.5 mm. The left ring is negatively charged and the right ring is positively charged because 20 electrons have been transferred from the right ring to the left ring. Solve: From Chapter 27, the electric field of a ring of charge +Q at z = −1.0 mm on the axis where the axis of each ring is kˆ is G ( −1.0 × 10−3 m )( 20 × 1.6 × 10−19 C ) kˆ 1 zQ E+ = kˆ = ( 9 × 109 N m 2 /C 2 ) 3 / 2 3/ 2 4πε 0 ( z 2 + R 2 ) ⎡( −1.0 × 10 −3 m )2 + (1.5 × 10−3 m )2 ⎤ ⎣⎢ ⎦⎥ −3 ˆ = −4.92 × 10 k V/m

The left ring with charge −Q makes an equal contribution G E− = −4.92 × 10−3 kˆ V/m G G G ⇒ Enet = E+ + E− = −9.84 × 10−3 kˆ V/m The negative sign with E+, E− , and Enet means these electric fields are in the −z direction. Using J = I A = σ E , the current is I = π (1.5 × 10 −3 m)2(3.5 × 107 Ω−1m−1)(9.84 × 10−3 V/m) = 2.4 A Assess: This result is consistent with the value given in Table 27.1 for the electric field strength in a currentcarrying wire.

31.57. Model: Because current is conserved, the current flowing in the 2.0-mm-diameter segment of the wire is the same as in the 1.0-mm-diameter segment. Visualize: Please refer to Figure P31.57. We will denote all quantities for the 1.0-mm-diameter wire with the subscript 1, and all quantities for the 2.0-mm-diameter wire with the subscript 2. Solve: Equation 31.13 is J = nevd. This means the current densities in the two segments are J2 = nevd2 J1 = nevd1 Dividing these equations, we get vd2 = ( J 2 J1 ) vd1. Because current is conserved, I1 = I2 = 2.0 A. So, 2

2

⎛D ⎞ J 2 I 2 A2 A1 A ⎛ 1.0 mm ⎞ −4 −5 = = ⇒ vd2 = 1 vd1 = ⎜ 1 ⎟ vd1 = ⎜ ⎟ ( 2.0 × 10 m/s ) = 5.0 × 10 m/s 2.0 mm J1 I1 A1 A2 A2 D ⎝ ⎠ ⎝ 2⎠ Assess:

A drift velocity which is small and only

reasonable.

( 14 )

of the drift velocity in the 1.0-mm-diameter wire is

31.58. Model: Because current is conserved, the current in the 3.0-mm-diameter end of the wire is the same as in the current in the 1.0-mm-diameter end of the wire. Visualize: Please refer to Figure P31.58. We will denote all quantities for the 1.0-mm-diameter end of the wire with the subscript 1, and all quantities for the 3.0-mm-diameter end of the wire with the subscript 3. Solve: Equation 31.13 is J = nevd. This means the current densities at the two ends are J3 = nevd3 J1 = nevd1 Dividing these equations, we obtain vd3 = ( J 3 J1 ) vd1 . Because current is conserved, I1 = I3. So, 2

vd3 = Assess:

2

⎛D ⎞ I 3 A3 A ⎛ 1.0 mm ⎞ −4 −6 vd1 = 1 vd1 = ⎜ 1 ⎟ vd1 = ⎜ ⎟ ( 0.50 × 10 m/s ) = 5.6 × 10 m/s 3.0 mm I1 A1 A3 D ⎝ ⎠ ⎝ 3⎠

The smallness of the drift velocity is physically reasonable.

31.59. Model: Because current is conserved, the currents in the aluminum and the nichrome segments of the wire are the same. Visualize: Please refer to Figure P31.59. Solve: From Equations 31.13 and 31.18 we have J = I A = σ E . This means

E=

⎛I ⎞⎛ A ⎞ ⎛σ ⎞ ⎛ D2 ⎞⎛σ ⎞ I E ⇒ nichrome = ⎜ nichrome ⎟ ⎜ aluminum ⎟ ⎜ aluminum ⎟ = ⎜ 2aluminum ⎟ ⎜ aluminum ⎟ Aσ Ealuminum ⎝ I aluminum ⎠ ⎝ Anichrome ⎠ ⎝ σ nichrome ⎠ ⎝ D nichrome ⎠ ⎝ σ nichrome ⎠

where we used the conservation of current and A = π / 4 D 2 . For Enichrome = Ealuminum, the above equation simplifies to Dnichrome =

σ aluminum 3.5 × 107 Ω −1m −1 Daluminum = (1.0 mm ) = 7.2 mm σ nichrome 6.7 × 105 Ω −1m −1

31.60. Model: Electric current is conserved. Visualize: Please refer to Figure P31.60. For the top, middle, and bottom segments the subscripts “top,” “mid,” and “bot” are used. Solve: (a) Since current is conserved, Itop = Imid = Ibot = 10 A. (b) The current density is J = I/A = I/πR2. Thus,

J top = J bot =

10 A

π ( 0.001 m )

2

= 3.18 × 106 A/m 2

J mid =

10 A

π ( 0.0005 m )

2

= 1.27 × 107 A/m 2

(c) The electric field is E = J/σ. Thus,

Etop = Ebot =

J top

σ

=

3.18 × 106 A/m 2 = 0.0909 V/m 3.5 × 107 Ω −1m −1

Emid =

J mid

σ

=

1.27 × 107 A/m 2 = 0.364 V/m 3.5 × 107 Ω −1m −1

(d) The drift speed is vd = J/ne. Thus

( vd )top = ( vd )bot = ( vd )mid =

J top

=

ne

3.18 × 106 A/m 2 = 3.31 × 10−4 m/s ( 6.0 × 1028 m−3 )(1.60 ×10−19 C )

J mid 1.27 × 107 A/m 2 = = 1.33 × 10−3 m/s ne ( 6.0 × 1028 m −3 )(1.60 × 10−19 C )

(e) The mean time between collisions is τ = mvd /eE. Thus,

(9.11× 10 kg )( 3.31× 10 m/s ) = 2.07 × 10 eE (1.60 ×10 C ) ( 0.0909 V/m ) m(v ) ( 9.11× 10 kg )(1.33 ×10 m/s ) = 2.08 × 10 s = = eE (1.60 × 10 C ) ( 0.364 V/m )

τ top = τ bot =

m ( vd ) top

−31

=

−4

−19

−14

s

top

−31

τ mid

−3

−14

d mid

−19

mid

The collision times are the same in all three segments because collisions are part of the microphysics of electrons inside the metal, independent of the macrophysics of fields and currents. (f ) The electron current is i = Ne /Δt = I/e. Because I1 = I2 = I3 = 10 A, 10 A ⎛ Ne ⎞ ⎛ Ne ⎞ ⎛ Ne ⎞ = 6.25 × 1019 s −1 ⎜ ⎟ =⎜ ⎟ =⎜ ⎟ = −19 ⎝ Δt ⎠ top ⎝ Δt ⎠ mid ⎝ Δt ⎠ bot 1.60 × 10 C Quantity

Top

Middle

Bottom

I J E vd

10 A 3.2 × 106 A/m 0.091 V/m 3.3 × 10−4 m/s 2.1 × 10−14 s 6.3 × 1019 s−1

10 A 1.27 × 107 A/m2 0.36 V/m 1.33 × 10−3 m/s 2.1 × 10−14 s 6.3 × 1019 s−1

10 A 3.2 × 106 A/m2 0.091 V/m 3.3 × 10−4 m/s 2.1 × 10−14 s 6.3 × 1019 s−1

τ i

31.61. Model: Assume the battery is ideal. Solve:

(a) The electric field inside the wire is E = ΔVwire L . Attaching the wire to the battery makes ΔVwire =

ΔVbat = 1.5 V . Thus, E=

1.5 V = 10 V/m 0.15 m

(b) Using Table 31.2, the current density is

J =σE =

E

ρ

=

10 V/m = 6.7 × 106 A/m 2 1.5 × 10−6 Ω m

(c) The current in a wire is related to the potential difference by I = ΔVwire R . Thus,

ΔVwire 1.5 V = = 0.75 Ω I 2A

R=

The resistance is related to the wire’s geometry by R=

ρL A

=

ρL ρL = ⇒ r= π r2 πR

(1.5 ×10

Thus, the wire’s diameter is d = 2r = 0.62 mm.

−6

Ω m ) ( 0.15 m )

π ( 0.75 Ω )

= 3.1 × 10−4 m = 0.31 mm

31.62. Model: Assume the battery is ideal. Visualize:

Solve:

The tube’s resistance is

R=

ρL A

=

ρL

π (r − r 2 2

2 1

)

=

(1.5 ×10

−6

Ωm ) ( 0.20 m )

2 2 π ⎡( 0.0015 m ) − ( 0.0014 m ) ⎤

⎣

When connected to the battery, ΔVtube = ΔVbat = 3.0 V. Thus, the current is I=

ΔVtube 3.0 V = = 9.1 A 0.329 Ω R

⎦

= 0.329 Ω

31.63. Model: Assume the battery is ideal.

Visualize: The current supplied by the battery and passing through the wire is I = ΔVbat/R. A graph of current versus time has exactly the same shape as the graph of ΔVbat with an initial value of I0 = (ΔVbat)0/R = (1.5 V)/(3.0 Ω) = 0.50 A. The horizontal axis has been changed to seconds.

Solve:

Current is I = dQ/dt. Thus the total charge supplied by the battery is

Q=

∫

∞

I dt = area under the current-versus-time graph

0

= 12 (7200 s)(0.50 A) = 1.80 × 103 C

31.64. Model: The charged metal plates form a parallel-plate capacitor. Solve: (a) When the charged plates are connected, the maximum current in the wire can be obtained from I max = ΔVmax R where ΔVmax is the potential difference across the capacitor just before the wire is connected. ΔVmax for the capacitor is ΔVmax = Q C , where C is the capacitance for a parallel-plate capacitor. Noting that C = ε 0 A d and R = ρL/πr2, the maximum current is

I max = =

ΔVmax Q Q Q π r2 = = = 2 R CR ( ε 0 A d ) ( ρ L π r ) ε 0 ρ A

(8.85 ×10

(12.5 ×10 −12

−9

C ) π ( 0.112 × 10−3 m )

2

C2 /N m 2 )(1.7 × 10−8 Ω m ) π ( 5.0 × 10−2 m )

2

= 4.2 × 105 A

In the above calculations, note that the length of the wire L is the same as the separation of the two plates d. Also, the resisitivity of copper is taken from Table 31.2. (b) The largest electric field in the wire is Emax =

ΔVmax L

Calculating ΔVmax separately, ΔVmax =

(12.5 × 10−9 C )(1.0 × 10−2 m ) = 1800 V Q Q Qd = = = C ε 0 A d ε 0 A ( 8.85 × 10−12 C 2 /N m 2 ) ⎡π ( 0.050 m )2 ⎤ ⎣ ⎦ 1800 V 5 ⇒ Emax = = 1.80 × 10 V/m 0.010 m

(c) The current I will decrease in time. As charge leaves one plate and moves to the other plate, the voltage across the capacitor and hence the current goes down. (d) The energy dissipated in the wire is the energy present in the capacitor before it was connected with the wire. This energy is

U dissipated =

1 Q2 1 Q 1 1 = Q = QΔVmax = (12.5 × 10−9 C ) (1800 V ) = 1.12 × 10−5 J 2 C 2 C 2 2

31.65. Model: The volume of the wire remains constant as it is stretched. The cross-sectional area of the wire changes uniformly as it stretches. Solve: The resistance of the wire before it is stretched is ρ L ρ L2 ρ L2 = = . R= A AL V The volume V remains constant as the wire is stretched. After stretching, the resistance is ρ L′ 2 R′ = . V Taking the ratio of these two equations and using the fact that ρ is a property of the material and therefore does not change, R L2 L2 1 = 2= = ⇒ R′ = 4 R 2 R′ L′ ( 2L ) 4 The wire’s resistance is 4R. Assess: Stretching a wire increases the one dimension of length but decreases the two dimensions of crosssectional area, so the resistance increases.

31.66. Model: The wire is uniform. The electric field in the wire is the negative of the slope of the V vs s curve. Solve: The electric field in the wire is

E=−

ΔV ( 3.0 V − 0.0 V ) = −10 V/m =− Δs ( 0.30 m − 0.0 m )

The negative sign indicates the electric field direction is along the negative s direction, and is not needed to find current density. From Table 31.2, the conductivity of tungsten is σ = 1.8 × 107 Ω −1m −1 . Thus the current density in the wire is J = σ E = (1.8 × 107 Ω −1m −1 ) (10 V/m ) = 1.80 × 108 A/m 2

31.67. Model: The copper wire is uniform. Solve:

Equation 31.22 relates the current in a wire to the potential difference across it: −8 ρ LI (1.7 × 10 Ωm ) ( 20 m )( 8.0 A ) A I= ΔV ⇒ ΔV = = = 0.87 V 2 ρL A π (1.0 × 10−3 m )

The resistivity ρ of copper is taken from Table 31.2. Assess: While voltage drops in household wires are small compared to the applied voltage, voltage drops in transmission wires between homes and power plants could be quite large. The power company transports energy in a way that minimizes the voltage drop, as we will learn in Chapter 36.

31.68. Solve: (a) The charge delivered is

( 50 × 10 A )( 50 × 10 s ) = 2.5 C . −6

3

(b) The current in the lightning rod and the potential drop across it are related by Equation 31.22. Using ρ for iron from Table 31.2,

−8 3 ρ LI ( 9.7 × 10 Ωm ) ( 5.0 m ) ( 50 × 10 A ) A ΔV ⇒ A = = = 2.43 × 10−4 m 2 ρL ΔV 100 V This is the area required for a maximum voltage drop of 100 V. The corresponding diameter of the lightning rod is

I=

r=

A

π

=

2.43 × 10−4 m 2

π

= 8.8 × 10−3 m = 8.8 mm

31.69. Solve: The density of copper ρ = 8940 kg/m3 = 8.94 g/cm3. The volume of the hollow copper cylinder is

V=

M

ρ

=

62.0 g = 6.935 cm3 8.94 g/cm3

Because cylinder’s length is 10 cm, the area of cross section of the hollow cylinder is A=

V 6.935 cm3 = = 0.6935 cm 2 L 10 cm

Thus, the current I = JA = (150,000 A/m2)(0.6935 × 10−4 m2) = 10.4 A.

31.70. Solve: The total charge delivered by the battery is Q = ∫ dQ =

∞

∞

0s

0s

∫ I dt = ∫ ( 0.75 A ) e

− t ( 6 hours )

= ( 0.75 A )( −6 hours ) ⎡⎣e − t ( 6 hours ) ⎤⎦

= ( 0.75 A )( 6 × 3600 s ) = 1.62 × 10 C 4

The number of electrons transported is

1.62 × 104 C = 1.01 × 1023 1.6 × 10−19 C

∞ 0

31.71. Model: Assume that the conduction electrons are point particles. Solve: (a) The rms velocity of the conduction electrons at room temperature is obtained from the relationship 2 1 mvrms = 32 kBT . We have 2 vrms =

3 (1.38 × 10−23 J/K ) ( 293 K ) 3kBT = = 1.15 × 105 m/s m 9.11 × 10−31 kg

(b) The mean free path of an electron depends on the number of atoms of copper per volume. The electron will collide with an atom if it comes within a distance r from the atom, where r is the radius of a copper atom. From Chapter 18, the mean free path is L L 1 λ= = = N collision ( N V ) π r 2 L ( N V ) π r 2 For atoms, r ≈ 0.5 × 10−10 m. Using N V = 8.5 × 1028 m −3 from Table 31.1,

λ=

(8.5 ×10

1

28

)π ( 0.5 ×10−10 )

2

= 1.5 × 10−9 m = 1.5 nm

31.72. Visualize:

The current density in the beam increases with distance from the center. We consider a thin circular shell of width dr at a distance r from the center to calculate the current density at the edge. Solve: (a) The beam current is 1.5 mA. This means the beam transports a charge of 1.5 × 10−3 C in 1 s. The number of protons delivered in one second is

1.5 × 10−3 C = 9.375 × 1015 = 9.4 × 1015 1.6 × 10−19 C (b) From J = I A , the current in the ring of width dr at a distance r from the center is

2π r 2 dr ⎛r⎞ dI = JdA = J edge ⎜ ⎟ ( 2π rdr ) = J edge R ⎝R⎠ The total current I = 1.5 mA is found by integrating this expression:

I total = ∫ dI = 1.5 × 10−3 A = ⇒ J edge = (1.5 × 10−3 A )

J edge R

R

∫ 2π r dr = 2

0

3

2π ( 2.5 × 10−3 m )

2

2π J edge R 2 3

= 115 A/m 2

31.73. Solve: (a) The charge Q experiences a change of the potential ΔVbat as it passes through the wire. Thus ΔU = QΔVbat. (b) The energy lost by the charge is transformed into internal energy of the wire—the wire is heated. (c) The power supplied by the battery is ΔU Q P= = ΔVbat Δt Δt (d) The power supplied is P = (1.2 A)(1.5 V) = 1.80 W.

31.74. Model: The currentsGin the two segments of the wire are the same. G

Visualize: The electric fields E1 and E2 point in the direction of the current. Establish a cylindrical Gaussian surface with end area a that extends into both segments of the wire.

Solve: (a) Because current is conserved, I1 = I2 = I. The cross-section areas of the two wires are the same, so the current densities are the same: J1 = J2 = I/A. Thus the electric fields in the two segments have strengths

E1 =

J1

σ1

=

I Aσ 1

E2 =

J2

σ2

=

I Aσ 2

The electric field enters the Gaussian surface on the left (negative flux) and exits on the right. No flux passes through the wall of cylinder, so the net flux is Φe = E2a – E1a. The Gaussian cylinder encloses charge Qin = ηa on the boundary between the segments. Gauss’s law is

Φe =

Qin

ε0

⇒ E2 a − E1a =

Ia ⎛ 1 1 ⎞ ηa ⎜ − ⎟= A ⎝ σ 2 σ1 ⎠ ε 0

Thus the surface charge density on the boundary is

η=

ε0I ⎛ 1 1 ⎞ ⎜ − ⎟ A ⎝ σ 2 σ1 ⎠

(b) From the expression obtained in part (a)

η=

Q Iε 0 ⎛ 1 1 ⎞ = − ⎜ ⎟ 2 2 ⎜ π R (π R ) ⎝ σ iron σ copper ⎟⎠

1 1 ⎛ ⎞ −18 ⇒ Q = ( 5 A ) ( 8.85 × 10−12 C 2 /Nm 2 ) ⎜ − ⎟ = 3.7 × 10 C 7 −1 −1 6.0 × 107 Ω −1m −1 ⎠ ⎝ 1.0 × 10 Ω m Assess:

This charge corresponds to a deficit of a mere 23 electrons on the boundary between the metals.

32.1. Solve:

From the circuit in Figure EX32.1, we see that 50 Ω and 100 Ω resistors are connected in series across the battery. Another resistor of 75 Ω is also connected across the battery.

32.2. Solve: In Figure EX32.2, the positive terminal of the battery is connected to a resistor. The other end of that resistor is connected to resistor and a capacitor in parallel.

32.3. Model: Assume that the connecting wires are ideal. Visualize: Please refer to Figure EX32.3. Solve: The current in the 2 Ω resistor is I1 = 6 V 2 Ω = 3 A to the left. The current in the 5 Ω resistor is

I 2 = 10 V 5 Ω = 2 A downward. Using Kirchhoff’s junction law, we see that I = I1 + I2 = 3 A + 2 A = 5 A

This current flows toward the junction, that is, downward.

32.4. Model: The batteries and the connecting wires are ideal. Visualize: Please refer to Figure EX32.4. Solve: (a) Choose the current I to be in the clockwise direction. If I ends up being a positive number, then the current really does flow in this direction. If I is negative, the current really flows counterclockwise. There are no junctions, so I is the same for all elements in the circuit. With the 9 V battery being labeled 1 and the 6 V battery being labeled 2. Kirchhoff’s loop law is

∑ ΔV = ΔV i

bat 1

⇒ I=

+ ΔVR + ΔVbat 2 = +E1 − IR − E2 = 0

E1 − E2 9 V − 6 V = = 0.10 A R 30 Ω

Note the signs: Potential is gained in battery 1, but potential is lost both in the resistor and in battery 2. Because I is positive, we can say that I = 0.100 A flows from left to right through the resistor. (b) The graph shows 9 V gained in battery 1, ΔVR = −IR = 3 V lost in the resistor, and another 6 V lost in battery 2. The final potential is the same as the initial potential, as required.

32.5. Model: Assume ideal connecting wires and an ideal battery for which ΔVbat = E .

Visualize: Please refer to Figure EX32.5. We will choose a clockwise direction for I. Note that the choice of the current’s direction is arbitrary because, with two batteries, we may not be sure of the actual current direction. The 3 V battery will be labeled 1 and the 6 V battery will be labeled 2. Solve: (a) Kirchhoff’s loop law, going clockwise from the negative terminal of the 3-V battery is ΔVclosed loop = ∑ ( ΔV )i = ΔVbat 1 + ΔVR + ΔVbat 2 = 0 i

⇒ +3 V – (18 Ω) I + 6 V = 0 ⇒ I =

9V = 0.5 A 18 Ω

Thus, the current through the 18 Ω resistor is 0.5 A. Because I is positive, the current is left to right (i.e., clockwise). (b)

Assess: The graph shows a 3 V gain in battery 1, a −9 V loss in the resistor, and a gain of 6 V in battery 2. The final potential is the same as the initial potential, as required.

32.6. Visualize: Please refer to Figure EX32.6. Define the current I as a clockwise flow. Solve: (a) There are no junctions, so conservation of current tells us that the same current flows through each circuit element. From Kirchhoff’s loop law, ΣΔVi = ΔVbat + ΔV10 + ΔV20 = 0 As we go around the circuit in the direction of the current, potential is gained in the battery (ΔVbat = Ebat = +15 V) and potential is lost in the resistors (ΔVresistor = −IR). The loop law is 0 = Ebat – IR1 – IR2 = Ebat – I(R1 + R2) ⇒ I =

Ebat 15 V = = 0.50 A R1 + R2 30 Ω

Now that we know the current, we can find the potential difference across each resistor: ΔV10 = IR1 = (0.50 A)(10 Ω) = 5.0 V (b)

ΔV20 = IR2 = (0.50 A)(20 Ω) = 10.0 V

32.7. Model: The 1500 W rating is for operating at 120 V. Solve:

The hair dryer dissipates 1500 W at ΔVR = 120 V. Thus, the hair dryer’s resistance is

( ΔVR ) R=

2

(120 V ) =

2

= 9.60 Ω PR 1500 W The current in the hair dryer when it is used is given by Ohm’s law: I=

ΔVR 120 V = = 12.5 A 9.60 Ω R

32.8. Model: Assume ideal connecting wires and an ideal battery. Visualize: Please refer to Figure EX32.8. Solve: The power dissipated by each resistor can be calculated from Equation 32.14, PR = I2R, provided we can find the current through the resistors. Let us choose a clockwise direction for the current and solve for the value of I by using Kirchhoff’s loop law. Going clockwise from the negative terminal of the battery,

∑ ( ΔV ) i

i

= ΔVbat + ΔVR1 + ΔVR2 = 0 ⇒ +12 V – IR1 – IR2 = 0

⇒ I=

12 V 12 V 2 = = A = 0.40 A R1 + R2 12 Ω + 18 Ω 5

The power dissipated by resistors R1 and R2 is: PR1 = I 2 R1 = ( 0.40 A ) (12 Ω ) = 1.92 W 2

PR2 = I 2 R2 = ( 0.40 A ) (18 Ω ) = 2.9 W 2

32.9. Model: The 100 W rating is for operating at 120 V.

Solve: A standard bulb uses ΔV = 120 V. We can use the power dissipation to find the resistance of the filament:

ΔV 2 ΔV 2 (120 V ) ⇒ R= = = 144 Ω P 100 W R 2

P=

But the resistance is related to the filament’s geometry: R=

ρL A

=

ρL ρL = ⇒ r= 2 πr πR

The filament’s diameter is d = 2r = 23.6 μm.

( 9.0 × 10

−7

Ω m ) ( 0.070 m )

π (144 Ω )

= 1.18 × 10−5 m = 11.8 μ m

32.10. Solve: Making use of the conversions 1 J 1 s = 1 W and 1 kW = 1000 J/s, we can write 1 kWh = (1000 J/s)(3600 s) = 3.6 × 106 J

32.11. Solve: (a) The average power consumed by a typical American family is Pavg = 1000

kWh kWh 1000 = 1000 = kW = 1.389 kW month 30 × 24 h 720

Because P = (ΔV )I with ΔV as the voltage of the power line to the house,

I avg =

Pavg ΔV

=

1.389 kW 1389 W = = 11.6 A 120 V 120 V

(b) Because P = ( ΔV ) /R, 2

Ravg =

( ΔV ) Pavg

2

=

(120 V ) 1389 W

2

= 10.4 Ω

32.12. Solve: The cost of running the waterbed 35% of the time for a year is ⎛ 24 hr ⎞ ⎛ kW ⎞⎛ $0.11 ⎞ (0.35)(450 W)(365 days) ⎜ ⎟⎜ ⎟⎜ ⎟ = $152 ⎝ day ⎠ ⎝ 1000 W ⎠⎝ kW hr ⎠

32.13. Visualize: Please refer to Figure EX32.13. Solve:

The three resistors are in series. The total resistance of this combination is Req = R + 50 Ω + R = 2R + 50 Ω

Thus, Req > 50 Ω as long as R > 0 Ω.

32.14. Model: Assume ideal connecting wires and an ideal battery. Solve: As shown in Figure EX32.14, a potential difference of 5.0 V causes a current of 100 mA through the three series resistors. The situation is the same if we replace the three resistors with an equivalent resistor Req. That is, a potential difference of 5.0 V across Req causes a current of 100 mA through it. From Ohm’s law,

Req =

ΔVR 5.0 V ⇒ R + 15 Ω + 10 Ω = ⇒ R + 25 Ω = 50 Ω ⇒ R = 25 Ω 100 mA I

32.15. Model: Assume ideal connecting wires and an ideal power supply. Visualize:

The two light bulbs are basically two resistors in series. Solve: A 75 W (120 V) light bulb has a resistance of ΔV 2 (120 V ) = = 192 Ω P 75 W 2

R=

The combined resistance of the two bulbs is Req = R1 + R2 = 192 Ω + 192 Ω = 384 Ω The current I flowing through Req is

I=

ΔV 120 V = = 0.3125 A Req 384 Ω

Because Req is a series combination of R1 and R2, the current 0.3125 A flows through R1 and R2. Thus, PR1 = I2R1 = (0.3125 A)2(192 Ω) = 18.8 W = PR2

32.16. Model: Assume ideal connecting wires and an ideal power supply. Visualize:

Solve: We have two resistors in series such that Req = Rbulb + Rcontacts. Rbulb can be found from the fact that we have a 100 W (120 V) bulb: ΔV 2 (120 V ) = = 144 Ω P 100 W 2

Rbulb =

We have a total resistance of Req = 144 Ω + 5.0 Ω = 149 Ω. The current flowing through Req is

I=

ΔV 120 V = = 0.8054 A Req 149 Ω

Because Req is a series combination of Rbulb and Rcontacts, this current flows through both bulbs. Thus, Pbulb = I2Rbulb = (0.8054 A)2(144 Ω) = 93.4 W Assess: The corroded leads change the circuit’s total resistance and reduce the current below that at which the bulb was rated. So, it makes sense for it to operate at less than full power.

32.17. Model: Assume ideal connecting wires and an ideal ammeter but not an ideal battery. Visualize: Please refer to Figure EX32.17. Solve: An ideal ammeter has zero resistance, so the battery is being short circuited. If I is the current flowing through the circuit, then

I=

E E 1.5 V ⇒ r= = = 0.65 Ω r I 2.3 A

The power dissipated by the internal resistance is P = I2r = (2.3 A)2(0.6522 Ω) = 3.5 W

32.18. Model: Assume ideal connecting wires but not an ideal battery.

Visualize: The circuit for an ideal battery is the same as the circuit in Figure EX32.18, except that the 1 Ω resistor is not present. Solve: In the case of an ideal battery, we have a battery with E = 15 V connected to two series resistors of 10 Ω and 20 Ω resistance. Because the equivalent resistance is Req = 10 Ω + 20 Ω = 30 Ω and the potential difference across Req is 15 V, the current in the circuit is

I=

ΔV E 15 V = = = 0.50 A Req Req 30 Ω

The potential difference across the 20 Ω resistor is ΔV20 = IR = (0.50 A)(20 Ω) = 10.0 V In the case of a real battery, we have a battery with E = 15 V connected to three series resistors: 10 Ω, 20 Ω, and an internal resistance of 1.0 Ω. Now the equivalent resistance is

′ = 10 Ω + 20 Ω + 1.0 Ω = 31 Ω Req The potential difference across Req is the same as before ( E = 15 V). Thus, I′ =

ΔV ′ E 15 V = = = 0.4839 A Req′ Req′ 31 Ω

Therefore, the potential difference across the 20 Ω resistor is ΔV20′ = I ′R = ( 0.4839 A )( 20 Ω ) = 9.68 V That is, the potential difference across the 20 Ω resistor is reduced from 10.0 V to 9.68 V due to the internal resistance of 1 Ω of the battery. The percentage change in the potential difference is ⎛ 10.0 V − 9.68 V ⎞ ⎜ ⎟ × 100 = 3.2% 10.0 V ⎝ ⎠

32.19. Model: Assume ideal connecting wires but not an ideal battery. Visualize: Please refer to Figure EX32.20. Solve: From Equation 32.21, the potential difference across the battery is

ΔVbat =

⎛ E ⎞ ER ⎛ 9.0 V ⎞ ⇒ r = R⎜ − 1⎟ = ( 20 Ω ) ⎜ − 1⎟ = 1.18 Ω R+r Δ V ⎝ 8.5 V ⎠ ⎝ bat ⎠

Assess: 1 Ω is a typical internal resistance for a battery. This causes the battery’s terminal voltage in the circuit to be 0.5 V less than its emf.

32.20.

Visualize:

The figure shows a metal wire of resistance R that is cut into two pieces of equal length. This produces two wires each of resistance R/2. Solve: Since these two wires are connected in parallel,

1 1 1 2 2 4 R = + = + = ⇒ Req = 4 Req R 2 R 2 R R R

32.21. Visualize: The three resistors in Figure EX32.21 are equivalent to a resistor of resistance Req = 75 Ω. Solve:

Because the three resistors are in parallel, 1 1 1 1 2 1 400 Ω + R ( 200 Ω ) R = ⇒ Req = 75 Ω = = + + = + = Req R 200 Ω R R 200 Ω ( 200 Ω ) R ( 400 Ω + R )

⇒R=

400 Ω = 240 Ω 200 Ω −1 75 Ω

200 Ω ⎛ 400 Ω ⎞ 1+ ⎜ ⎟ ⎝ R ⎠

32.22. Model: Assume ideal connecting wires. Visualize: Please refer to Figure EX32.22. Solve: The resistance R is given by Ohm’s law, R = ΔVR I R . To determine IR we use Kirchhoff’s junction law. The input current I splits into the three currents I10, I15, and IR. That is, 2.0 A = I10 + I15 + IR = 8 V + 8 V + I R ⇒ IR = 2.0 A – 0.80 A – 0.533 A = 0.667 A 10 Ω 15 Ω

Using this value of IR in Ohm’s law, R=

8V = 12 Ω 0.667 A

32.23. Model: The connecting wires are ideal with zero resistance. Solve:

For the first step, the 10 Ω and 30 Ω resistors are in series and the equivalent resistance is 40 Ω. For the second step, the 60 Ω and 40 Ω resistors are in parallel and the equivalent resistance is −1

1 ⎤ ⎡ 1 ⎢ 40 Ω + 60 Ω ⎥ = 24 Ω ⎣ ⎦ For the third step, the 24 Ω and 10 Ω resistors are in series and the equivalent resistance is 34 Ω.

32.24. Model: The connecting wires are ideal with zero resistance. Solve:

For the first step, the resistors 30 Ω and 45 Ω are in parallel. Their equivalent resistance is

1 1 1 = + ⇒ Req 1 = 18 Ω Req 1 30 Ω 45 Ω For the second step, resistors 42 Ω and Req 1 = 18 Ω are in series. Therefore, Req 2 = Req 1 + 42 Ω = 18 Ω + 42 Ω = 60 Ω

For the third step, the resistors 40 Ω and Req 2 = 60 Ω are in parallel. So, 1 1 1 = + ⇒ Req 3 = 24 Ω Req 3 60 Ω 40 Ω The equivalent resistance of the circuit is 24 Ω.

32.25. Model: The connecting wires are ideal with zero resistance. Solve:

In the first step, the resistors 100 Ω, 100 Ω, and 100 Ω in the top branch are in series. Their combined resistance is 300 Ω. In the middle branch, the two resistors, each 100 Ω, are in series. So, their equivalent resistance is 200 Ω. In the second step, the three resistors are in parallel. Their equivalent resistance is

1 1 1 1 ⇒ Req = 54.5 Ω = + + Req 300 Ω 200 Ω 100 Ω The equivalent resistance of the circuit is 54.5 Ω.

32.26. Model: The connecting wires are ideal with zero resistance. Solve:

For the first step, the two resistors in the middle of the circuit are in parallel, so their equivalent resistance is

1 1 1 = + ⇒ Req 1 = 50 Ω Req 1 100 Ω 100 Ω The three 100 Ω resistors at the end are in parallel. Their equivalent resistance is 1 1 1 1 100 Ω = 33.3 Ω = + + ⇒ Req 2 = Req 2 100 Ω 100 Ω 100 Ω 3 For the second step, the three resistors are in series, so their equivalent resistance is 100 Ω + 50 Ω + 33.3 Ω = 183 Ω The equivalent resistance of the circuit is 183 Ω.

32.27. Model: Grounding does not affect a circuit’s behavior. Visualize: Please refer to Figure EX32.27. Solve: Because the earth has Vearth = 0 V, point d has a potential of zero. In going from point d to point a, the potential increases by 9 V. Thus, point a is at a potential of 9 V. Let us calculate the current I in the circuit before calculating the potentials at points b and c. Applying Kirchhoff’s loop rule, starting clockwise from point d,

∑ ( ΔV ) i

i

= ΔV9 V bat + ΔV2 Ω + ΔV6 V bat + ΔV1 Ω = 0

⇒ +9 V – I(2 Ω) – 6 V – I(1 Ω) = 0 ⇒ I =

3V =1 A 3Ω

There is a drop in potential from point a to point b by an amount IR = (1 A)(2 Ω) = 2 V. Thus, the potential at point b is 9 V − 2 V = 7 V. The potential decreases from 7 V at point b to 7 V − 6 V = 1 V at point c. There is a further decrease in potential across the 1 Ω resistor of IR = (1 A)(1 Ω) = 1 V. That is, the potential of 1 V at c becomes 0 V at point d, as it must. In summary, the potentials at a, b, c, and d are 9 V, 7 V, 1 V, and 0 V.

32.28. Model: Grounding does not affect a circuit’s behavior. Visualize: Please refer to Figure EX32.28. Solve: Let us first obtain the value of the current I in the circuit. Applying Kirchhoff’s loop rule, starting clockwise from point c,

∑ ( ΔV ) i

i

= ΔV1 Ω + ΔV15 V bat + ΔV4 Ω + ΔV5 V bat = 0

⇒ −I(1 Ω) + 15 V – I(4 Ω) – 5 V = 0 ⇒ I =

10 V =2A 5Ω

Because the earth has Vearth = 0 V, point c is at zero potential. There is a potential drop of IR = (2 A)(1 Ω) = 2 V across the 1 Ω resistor, so the potential at point d is −2 V. From point d to point a, there is an increase in potential of 15 V, thus the potential at point a is 15 V – 2 V = 13 V. The potential decreases from point a to point b by IR = (2 A)(4 Ω) = 8 V, so the potential at point b is 13 V – 8 V = 5 V. The potential at point c is 5 V lower than the potential at b, so it is 0 V, as it must be. In summary, the potentials at a, b, c, and d are 13 V, 5 V, 0 V, and −2 V.

32.29. Solve: Noting that the unit of resistance is the ohm (V/A) and the unit of capacitance is the farad (C/V), the unit of RC is RC =

V C C C × = = =s A V A Cs

32.30. Model: Assume ideal wires as the capacitors discharge through the two 1 kΩ resistors. Visualize: The circuit in Figure EX32.30 has an equivalent circuit with resistance Req and capacitance Ceq. Solve: The equivalent capacitance is 1 1 1 = + ⇒ Ceq = 1 μF Ceq 2 μ F 2 μ F and the equivalent resistance is Req = 1 kΩ + 1 kΩ = 2 kΩ. Thus, the time constant for the discharge of the capacitors is

τ = ReqCeq = (2 kΩ)(1 μF) = 2 × 10−3 s = 2 ms Assess:

The capacitors will be almost entirely discharged 5τ = 5 × 2 ms = 10 ms after the switch is closed.

32.31. Model: Assume ideal wires as the capacitors discharge through the two 1 kΩ resistors. Visualize: The circuit in Figure EX32.31 has an equivalent circuit with resistance Req and capacitance Ceq. Solve: The equivalent capacitance is Ceq = 2 μF + 2 μF = 4 μF, and the equivalent resistance is 1 1 1 = + ⇒ Req = 0.5 kΩ Req 1 kΩ 1 kΩ Thus, the time constant for the discharge of the capacitors is

τ = ReqCeq = (0.5 kΩ)(4 μF) = 2 × 10−3 s = 2 ms

32.32. Model: The capacitor discharges through a resistor. Assume that the wires are ideal. Solve:

The decay of the capacitor charge is given by the Equation 32.34: Q = Q0 e−t/τ. The time constant is

τ = RC = (1.0 × 103 Ω)(10 × 10−6 F) = 0.010 s The initial charge on the capacitor is Q0 = 20 μC and it decays to 10 μC in time t. That is, ⎛ 10 μ C ⎞ t t ⇒− ⇒ t = (0.010 s)ln 2 = 6.9 ms 10μ C = (20μ C)e − t / 0.010 s ⇒ ln ⎜ ⎟=− 0.010 s 0.010 s ⎝ 20 μ C ⎠

32.33. Model: The capacitor discharges through a resistor. Assume ideal wires. Visualize: The switch in the circuit in Figure EX32.33 is in position a. When the switch is in position b the circuit consists of a capacitor and a resistor. Solve: (a) The switch has been in position a for a long time. That means the capacitor is fully charged to a charge Q0 = CΔV = C E = (4 μF)(9 V) = 36 μC Immediately after the switch is moved to the b position, the charge on the capacitor is Q0 = 36 μC. The current through the resistor is I0 =

ΔVR 9V = = 0.36 A 25 Ω R

Note that as soon as the switch is closed, the potential difference across the capacitor ΔVC appears across the 25 Ω resistor. (b) The charge Q0 decays as Q = Q0 e−t/τ, where

τ = RC = (25 Ω)(4 μF) = 100 μs Thus, the charge is

Q = ( 36 μ C ) e − 50 μs 100 μs = ( 36 μ C ) e−0.5 = 22 μ C The resistor current is I = I 0e − t τ = ( 0.36 A ) e − 50 μ s 100 μ s = 0.22 A (c) Likewise, at t = 200 μs, the charge is Q = 4.9 μC and the current is I = 49 mA.

32.34. Model: A capacitor discharges through a resistor. Assume ideal wires.

Solve: A capacitor initially charged to Q0 decays as Q = Q0 e−t/RC. We wish to find R so that a 1.0 μF capacitor will discharge to 10% of its initial value in 2.0 ms. That is,

( 0.10 ) Q0 = Q0e−2.0 ms R (1.0 μ F) ⇒ ln ( 0.10 ) = − Assess:

2.0 × 10−3 s −2.0 × 10−3 s ⇒ = = 0.87 kΩ R R (1.0 × 10−6 F ) (1.0 ×10−6 F) ln(0.10)

A time constant of τ = RC = (870 Ω)(1.0 × 10−6 F) = 0.87 ms is reasonable.

32.35. Model: A capacitor discharges through a resistor. Assume ideal wires. Solve: The discharge current or the resistor current follows Equation 32.35: I = I 0e− t RC . We wish to find the capacitance C so that the resistor current will decrease to 25% of its initial value in 2.5 ms. That is,

0.25 I 0 = I 0e − 2.5 ms (100 Ω )C ⇒ ln ( 0.25 ) = −

2.5 × 10−3 s ⇒ C = 18.0 μF (100 Ω ) C

32.36. Visualize: Please refer to Figure P32.36. Solve: Bulbs D and E are in series, so the same current will go through both and make them equally bright (D = E). Bulbs B and C are in parallel, so they have the same potential difference across them. Because they are identical bulbs with equal resistances, they will have equal currents and be equally bright (B = C). Now the equivalent resistance of B + C in parallel is less than the resistance of E, so the total resistance along the path through A is less than the total resistance along path through D. The two paths have the same total potential difference—the emf of the battery—so more current will flow through the A path than through the D path. Consequently, A will have more current than D and E and will be brighter than D and E (A > D = E). Bulbs B and C each have half the current of A, because the current splits at the junction, so A is also brighter than B and C (A > B = C). The remaining issue is how B and C compare to D and E. Suppose B and C were replaced by wires with zero resistance, leaving just bulb A in the middle path. Then the resistance of the path through A would be half of the resistance of the path through D. This would mean that the current through A would be twice the current through D, so IA = 2ID. When B and C are present, their resistance adds to the resistance of A to lower the current through the middle path. So in reality, IA < 2ID. We already know that I B = I C = 12 I A , so we can conclude that IB = IC < ID. Since the current through B and C is less than the current through D and E, D and E are brighter than B and C. The final result of our analysis is A > D = E > B = C.

32.37. Visualize: Please refer to Figure P32.77. Solve: Bulb A is in parallel with the battery and experiences the full potential difference E whereas the other 5 bulbs divide up the potential into smaller pieces. So A will be brightest. Stated another way, the resistance of the right path, with bulb D in series with several other bulbs, is greater than the resistance of the middle path, with only bulb A. Both paths experience the full potential difference of the battery, so the current starting down the middle path is larger than the current starting down the right path, causing A to be brighter than D (A > D). All of the current in the right path passes through D, then it divides up. So D is brighter than B, C, E, or F (D > B, C, E, F). C and E are in parallel and have the same potential difference. Because they are identical bulbs with equal resistances, they have the same current and are equally bright (C = E). The current through F is the sum of the currents through C and E, so it is brighter than they are (F > C = E). The three-resistor combination C + E + F is in parallel with B. The combination C + E + F has more resistance than B, so more current will flow through B than through C + E + F. Consequently, B is brighter than F (B > F). Putting all these pieces together, the final result is A > D > B > F > C = E.

32.38. Solve: The resistivity of aluminum is 2.8 × 10−8 Ωm and we want the wire to dissipate 7.5 W when connected to a 1.5 V battery. The resistance of the wire must be V2 V 2 (1.5 V ) ⇒R= = = 0.30 Ω R P 7.5 W 2

P=

Using the formula for the resistance of a wire,

R=ρ

L ⇒ 0.30 Ω = A

( 2.8 × 10

−8

Ωm )

L ⇒ L = ( 3.366 × 107 m −1 ) r 2 π r2

We need another relation connecting L and r. Making use of the mass density of aluminum, 1.0 × 10−3 kg = 2700 kg/m 3 ⇒ r 2 L = 1.179 × 10−7 m3 π r 2L

Using the value of L obtained above, r2(3.366 × 107 m−1)r2 = 1.179 × 10−7 m3⇒ r4 = 3.50 × 10−15 m4 ⇒ r = 2.43 × 10−4 m = 0.243 mm Thus, the diameter of the wire is 0.48 mm and the length is L = (3.366 × 107 m−1)(2.43 × 10−4 m)2 = 1.99 m Assess: It is reasonable to make a 1.99 m long wire with a diameter of 0.48 mm from an aluminum block of 1.0 g.

32.39. Solve: The copper wire and the iron wire are connected in series. The composite resistance is simply the equivalent resistance of RCu and RFe: R = RCu + RFe =

ρCu LCu ACu

+

ρ Fe LFe AFe

Using the resistivity data from Table 31.2,

(1.7 ×10 Ω m ) ( 0.20 m ) + ( 9.7 ×10 Ω m ) ( 0.60 m ) = 4.3 ×10 π ( 0.50 × 10 m ) π ( 0.50 × 10 m ) −8

R=

−8

−3

2

−3

2

−3

Ω + 74 × 10−3 Ω = 78 mΩ

32.40. Model: The wires and battery are ideal. Visualize:

Solve:

We can find the equivalent resistance necessary for the battery to deliver 9 W of power: P=

( ΔV )

2

⇒R=

( ΔV )

2

=

( 6.0 V )

2

= 4.0 Ω R P 9.0 W The combination of the 2.0 Ω, 3.0 Ω, and 6.0 Ω resistors that make 4.0 Ω is shown in the figure. The 3.0 Ω and 6.0 Ω parallel combination has an equivalent resistance of 2.0 Ω, which when added to the 2.0 Ω resistor in series totals 4.0 Ω equivalent resistance.

32.41.

Visualize:

Solve:

(a) The three resistors in parallel have an equivalent resistance of

1 1 1 1 = + + ⇒ Req = 4.0 Ω Req 12 Ω 12 Ω 12 Ω (b) One resistor in parallel with two series resistors has an equivalent resistance of

1 1 1 1 1 1 ⇒ Req = 8.0 Ω = + = + = Req 12 Ω + 12 Ω 12 Ω 24 Ω 12 Ω 8 Ω (c) One resistor in series with two parallel resistors has an equivalent resistance of −1

1 1 ⎞ ⎛ 1 = 12 Ω + ⎜ + ⎟ = 12 Ω + 6 Ω = 18.0 Ω Req ⎝ 12 Ω 12 Ω ⎠ (d) The three resistors in series have an equivalent resistance of 12 Ω + 12 Ω + 12 Ω = 36 Ω

32.42. Model: Use the laws of series and parallel resistances. Visualize:

Solve: Despite the diagonal orientation of the 12 Ω resistor, the 6 Ω, 12 Ω, and 4 Ω resistors are in parallel because they have a common connection at both the top end and at the bottom end. Their equivalent resistance is −1

1 1 ⎞ ⎛ 1 + + Req = ⎜ ⎟ = 2Ω ⎝ 6 Ω 12 Ω 4 Ω ⎠ The trickiest issue is the 10 Ω resistor. It is in parallel with a wire, which is the same thing as a resistor with R = 0 Ω . The equivalent resistance of 10 Ω in parallel with 0 Ω is −1

1 ⎞ 1 −1 ⎛ 1 + Req = ⎜ ⎟ = (∞) = = 0 Ω ∞ ⎝ 10 Ω 0 Ω ⎠ In other words, the wire is a short circuit around the 10 Ω, so all the current goes through the wire rather than the resistor. The 10 Ω resistor contributes nothing to the circuit. So the total circuit is equivalent to a 2 Ω resistor in series with the 2 Ω equivalent resistance in series with the final 3 Ω resistor. The equivalent resistance of these three series resistors is Rab = 2 Ω + 2 Ω + 3 Ω = 7 Ω

32.43. Model: Assume the batteries and the connecting wires are ideal. Visualize: Please refer to Figure P32.43. Solve: (a) The two batteries in this circuit are oriented to “oppose” each other. The curent will flow in the direction of the battery with the greater voltage. The direction of the current is counterclockwise because the 12 V battery is greater. (b) There are no junctions, so the same current I flows through all circuit elements. Applying Kirchhoff’s loop law in the counterclockwise direction and starting at the lower right corner, ∑ΔVi = 12 V − I(12 Ω) − I(6 Ω) − 6 V – IR = 0 Note that the IR terms are all negative because we’re applying the loop law in the direction of current flow, and the potential decreases as current flows through a resistor. We can easily solve to find the unknown resistance R: 6 V − I(18 Ω) – IR = 0 ⇒ R =

6 V − (18 Ω ) I 6 V − (18 Ω )( 0.25 A ) = =6Ω 0.25 A I

(c) The power is P = I2R = (0.25 A)2(6 Ω) = 0.38 W. (d)

The potential difference across a resistor is ΔV = IR, giving ΔV6 = 1.5 V, and ΔV12 = 3 V. Starting from the lower left corner, the graph goes around the circuit clockwise, opposite from the direction in which we applied the loop law. In this direction, we speak of potential as lost in the batteries and gained in the resistors.

32.44. Model: Assume that the connecting wires are ideal but the battery is not ideal. Visualize:

Solve: The figure shows a variable resistor R connected across the terminals of a battery that has an emf E and an internal resistance r. Using Kirchhoff’s loop law and starting from the lower left corner, + E − Ir − IR = 0 ⇒ E = I(r + R) From the point in Figure P32.44 that corresponds to R = 0 Ω, E = (6 A)(r + 0 Ω) = (6 A)r

From the point that corresponds to R = 10 Ω,

E = (3 A)(r + 10 Ω) Combining the two equations, (6 A)r = (3 A)(r + 10 Ω) ⇒ 2r = r + 10 Ω ⇒ r = 10 Ω Also, E = (3 A)(10 Ω + 10 Ω) = 60 V. Assess: With E = 60 V and r = 10 Ω, the equation E = I(r + R) is satisfied by all values of R and I on the graph in Figure P32.44.

32.45. Model: The connecting wires are ideal, but the battery is not.

Visualize: Please refer to Fig. P32.45. We will designate the current in the 5 Ω resistor I5 and the voltage drop ΔV5. Similar designations will be used for the other resistors. Solve: Since the 10 Ω resistor is dissipating 40 W,

P10 = I102 R10 = 40 W ⇒ I10 =

P10 40 W = = 2.0 A ⇒ ΔV10 = I10R10 = (2.0 A)(10 Ω) = 20 V R10 10 Ω

The 20 Ω resistor is in parallel with the 10 Ω resistor, so they have the same potential difference: ΔV20 = ΔV10 = 20 V. From Ohm’s law,

I 20 =

ΔV20 20 V = = 1.0 A R20 20 Ω

The combined current through the 10 Ω and 20 Ω resistors first passes through the 5 Ω resistor. Applying Kirchhoff’s junction law at the junction between the three resistors, I5 = I10 + I20 = 1.0 A + 2.0 A = 3.0 A ⇒ ΔV5 = I5R5 = (3.0 A)(5 Ω) = 15 V Knowing the currents and potential differences, we can now find the power dissipated: P5 = I5ΔV5 = (3.0 A)(15 V) = 45 W

P20 = I20ΔV20 = (1.0 A)(20 V) = 20 W

32.46. Model: Assume that the connecting wires are ideal, but the battery is not. The battery has internal resistance. Also assume that the ammeter does not have any resistance. Visualize: Please refer to Figure P32.46. Solve: When the switch is open,

E – Ir – I(5.0 Ω) = 0 V ⇒ E = (1.636 A)(r + 5.0 Ω) where we applied Kirchhoff’s loop law, starting from the lower left corner. When the switch is closed, the current I comes out of the battery and splits at the junction. The current I′ = 1.565 A flows through the 5.0 Ω resistor and the rest (I − I′) flows through the 10.0 Ω resistor. Because the potential differences across the two resistors are equal, I′(5.0 Ω) = (I − I′) (10.0 Ω ) ⇒(1.565 A)(5.0 Ω) = (I − 1.565 A)(10.0 Ω) ⇒ I = 2.348 A Applying Kirchhoff’s loop law to the left loop of the closed circuit,

E − Ir − I′(5.0 Ω) = 0 V ⇒ E = (2.348 A)r + (1.565 A)(5.0 Ω)= (2.348 A)r + 7.825 V Combining this equation for E with the equation obtained from the circuit when the switch was open, (2.348 A)r + 7.825 V = (1.636 A)r + 8.18 V ⇒ (0.712 A)r = 0.355 V ⇒ r = 0.50 Ω We also have E = (1.636 A)(0.50 Ω + 5.0 Ω) = 9.0 V.

32.47. Model: Assume that the connecting wire and the battery are ideal. Visualize: Please refer to Figure P32.47. Solve: The middle and right branches are in parallel, so the potential difference across these two branches must be the same. The currents are known, so these potential differences are ΔVmiddle = (3.0 A)R = ΔVright = (2.0 A)(R + 10 Ω) This is easily solved to give R = 20 Ω. The middle resistor R is connected directly across the battery, thus (for an ideal battery, with no internal resistance) the potential difference ΔVmiddle equals the emf of the battery. That is

E = ΔVmiddle = (3.0 A)(20 Ω) = 60 V

32.48. Model: The connecting wires are ideal. Visualize:

Solve: Let the current in the circuit be I. The terminal voltage of the 2.5 V battery is Va − Vb. This is also the terminal voltage of the 1.5 V battery. Va − Vb can be obtained by noting that Vb + 2.5 V – I(1 Ω) = Va ⇒ Va – Vb = 2.5 V – I(1 Ω) To determine I, we apply Kirchhoff’s loop law starting from the lower left corner:

+2.5 V – I(1 Ω) – I(1 Ω) – 1.5 V = 0 V ⇒ I = 0.5 A Thus, Va – Vb = 2.5 V – (0.5 A)(1 Ω) = 2.0 V. That is, the terminal voltage of the 1.5 V and the 2.5 V batteries is 2.0 V.

32.49. Model: The connecting wires are ideal. Visualize: Please refer to Figure 32.20. Solve: (a) The internal resistance r and the load resistance R are in series, so the total resistance is R + r and the current flowing in the circuit, due to the emf E , is I = E/(R + r). The power dissipated by the load resistance is

P = I 2R =

E 2R

(R + r)

2

This is not the power EI generated by the battery, but simply the power dissipated by the load R. The power is a function of R, so we can find the maximum power by setting dP/dR = 0: E E2 dP 2E 2 R = − = 2 3 dR ( R + r ) ( R + r )

2

( R + r ) − 2E 2 R = E 2 ( r − R ) = 0 3 3 (R + r) (R + r)

That is, the power dissipated by R is a maximum when R = r. (b) The load’s maximum power dissipation will occur when R = r = 1 Ω, in which case

P=

(R + r)

( 9 V ) (1 Ω ) = 20 W 2 ( 2 Ω) 2

E 2R 2

=

(c) When R is very small (R → 0 Ω), the current is a maximum (I → E/r) but the potential difference across the load is very small (ΔV = IR → 0 V). So the power dissipation of the load is also very small (P = IΔV → 0 W). When R is very large (R → ∞), the potential difference across the load is a maximum (ΔV → E) but the current is very small (I → 0 A). Once again, P is very small. If P is zero both for R → 0 and for R → ∞, there must be some intermediate value of R where P is a maximum.

32.50. Model: The batteries are ideal, the connecting wires are ideal, and the ammeter has a negligibly small resistance. Visualize: Please refer to Figure P32.50. Solve: Kirchhoff’s junction law tells us that the current flowing through the 2.0 Ω resistance in the middle branch is I1 + I2 = 3.0 A. We can therefore determine I1 by applying Kirchhoff’s loop law to the left loop. Starting clockwise from the lower left corner, +9.0 V – I1(3.0 Ω) – (3.0 A)(2.0 Ω) = 0 V ⇒ I1 = 1.0 A ⇒ I2 = (3.0 A – I1) = (3.0 A – 1.0 A) = 2.0 A Finally, to determine the emf E, we apply Kirchhoff’s loop law to the right loop and start counterclockwise from the lower right corner of the loop:

+E − I2(4.5 Ω) − (3.0 A)(2.0 Ω) = 0 V ⇒ E – (2.0 A)(4.5 Ω) – 6.0 V = 0 V ⇒ E = 15 V

32.51. Model: Assume an ideal battery and ideal connecting wires. Visualize:

Solve: (a) Grounding one point doesn’t affect the basic analysis of the circuit. In Figure P32.51, there is a single loop with a single current I flowing in the clockwise direction. Applying Kirchhoff’s loop law clockwise from the lower right corner gives

12 V = 0.50 A 24 Ω Knowing the current, we can use ΔV = IR to find the potential difference across each resistor: ∑ΔVi = −8I + 12 V – 4I – 12I = 0 V ⇒ I =

ΔV8 = 4 V

ΔV4 = 2 V

ΔV12 = 6 V

The purpose of grounding one point in the circuit is to establish that point as the specific potential V = 0 V. Grounding point d makes that potential there Vd = 0 V. Then we can use the known potential differences to find the potential at other points in the circuit. Point a is 4 V less than point d (because potential decreases in the direction of = Vd – 4 V = current flow), so Va –4 V. Point b is 12 V more than point a because of the battery. So Vb = Va + 12 V = 8 V. Point c is 2 V less than point b, so Vc = Vb – 2 V = 6 V. Point d is 6 V less than point c, so Vd = Vc – 6 V = 0 V. This is a consistency check—making one complete loop brings us back to the potential at which we started, namely 0 V. (b) The information about the potentials is shown in the graph above. (c) Moving the ground to point a doesn’t change the basic analysis of part (a) or the potential differences found there. All that changes is that now Va = 0 V. Point b is 12 V more than point a because of the battery. So, Vb = Va + 12 V = 12 V. Point c is 2 V less than point b, so Vc = Vb – 2 V = 10 V. Point d is 6 V less than point c, so Vd = Vc – 6 V = 4 V. Point a is 4 V less than point d, so Va = Vd – 4 V = 0 V. This brings us back to where we started. The information about the potentials is shown in the graph above.

32.52. Visualize: Please refer to Figure P32.52. Solve: Because the 4 Ω resistor is grounded at both ends, the potential difference across this resistor is zero. That is, no current flows through the 4 Ω resistor, and the negative terminals of both batteries are at zero potential. To determine the current in the 2 Ω resistor, we apply Kirchhoff’s loop law. We assume that current I flows clockwise through the 2 Ω resistor. Starting from the lower left corner, the sum of the potential differences across various elements in the circuit is +9 V – I(2 Ω) – 3 V = 0 V ⇒ I = 3 A

32.53. Solve: The cost of energy of a 60 W incandescent bulb over its lifetime is 1 kW $0.10 × 1000 h × = $6.00 1000 W 1 kWh This means the life-cycle cost of the incandescent bulb is $6.50. The cost of energy of a 15 W compact fluorescent bulb over its lifetime is 1 kW $0.10 15 W × × 10,000 h × = $15.00 1000 W 1 kWh With the bulb’s cost of $15, the life-cycle cost is $30.00. To make a comparison of the cost effectiveness of the two bulbs, we note that you need ten incandescent bulbs to last as long as one fluorescent bulb. Thus, it will cost a consumer $65.00 to use incandescent bulbs for the same time span as a single fluorescent bulb that will cost only $30.00. The fluorescent bulb is cheaper. 60 W ×

32.54. Solve: (a) The cost per month of the 1000 W refrigerator is 1000 W ×

1 kW $0.10 × 30 × 24 h × 0.20 × = $14.40 1000 W 1 kWh

(b) The cost per month of a refrigerator with a 800 W compressor is $11.52. The difference in the running cost of the two refrigerators is $2.88 per month. So, the number of months before you recover the additional cost of $100 (of the energy efficient refrigerator) is $100/$2.88 = 34.7 months.

32.55. Visualize: Please refer to Figure P32.55. Solve: (a) Only bulb A is in the circuit when the switch is open. The bulb’s resistance R is in series with the internal resistance r, giving a total resistance Req = R + r. The current is E 1.50 V I bat = = = 0.231 A R + r 6.50 Ω This is the current leaving the battery. But all of this current flows through bulb A, so IA = Ibat = 0.231 A. (b) With the switch closed, bulbs A and B are in parallel with an equivalent resistance Req = 12 R = 3.00 Ω. Their equivalent resistance is in series with the battery’s internal resistance, so the current flowing from the battery is E 1.50 V I bat = = = 0.428 A Req + r 3.50 Ω

But only half this current goes through bulb A, with the other half through bulb B, so I A = 12 I bat = 0.214 A . (c) The change in IA when the switch is closed is 0.017 A. This is a decrease of 7.4%. (d) If r = 0 Ω, the current when the switch is open would be IA = Ibat = 0.250 A. With the switch closed, the current would be Ibat = 0.500 A and the current through bulb A would be I A = 12 I bat = 0.250 A . The current through A would not change when the switch is closed.

32.56. Model: The battery and the connecting wires are ideal. Visualize:

The figure shows the two circuits formed from the circuit in Figure P32.56 when the switch is open and when the switch is closed. Solve: (a) Using the rules of series and parallel resistors, we have simplified the circuit in two steps as shown in figure (a). A battery with emf E = 24 V is connected to an equivalent resistor of 3 Ω. The current in this circuit is 24 V 3 Ω = 8 A . Thus, the current that flows through the battery is Ibat = 8 A. To determine the potential difference ΔVab, we will find the potentials at point a and point b and then take their difference. To do this, we need the currents Ia and Ib. We note that the potential difference across the 3 Ω-3 Ω branch is the same as the potential difference across the 5 Ω-1 Ω branch. So, E = 24 V = Ia(3 Ω + 3 Ω) ⇒ Ia = 4 A = Ib

Now, Vc – Ia(3 Ω) = Va, and Vc – Ib(5 Ω) = Vb. Subtracting these two equations give us ΔVab: Va – Vb = Ib(5 Ω) – Ia(3 Ω) = (4 A)(5 Ω) – (4 A)(3 Ω) = +8 V (b) Using the rules of the series and the parallel resistors, we have simplified the circuit as shown in figure (b). A battery with emf E = 24 V is connected to an equivalent resistor of 218 Ω . The current in this circuit is

24 V

21 8

Ω = 9.143 A. Thus, the current that flows through the battery is Ibat = 9 A. When the switch is closed,

points a and b are connected by an ideal wire and must therefore be at the same potential. Thus Vab = 0 V.

32.57. Model: The voltage source/battery and the connecting wires are ideal. Visualize: Please refer to Figure P32.57. Solve: Let us first apply Kirchhoff’s loop law starting clockwise from the lower left corner: +Vin – IR – I (100 Ω) = 0 V ⇒ I =

Vin R + 100 Ω

The output voltage is

Vin Vout 100 Ω ⎛ ⎞ Vout = (100 Ω ) I = (100 Ω ) ⎜ = ⎟⇒ R 100 + Ω V R + 100 Ω ⎝ ⎠ in

For Vout = Vin 10 , the above equation can be simplified to obtain R: Vin 10 100 Ω = ⇒ R + 100 Ω = 1000 Ω ⇒ R = 900 Ω Vin R + 100 Ω

32.58. Model: Assume ideal connecting wires. Visualize: Please refer to Figure P32.58. Because the ammeter we have shows a full-scale deflection with a current of 500 μA = 0.500 mA, we must not allow a current more than 0.500 mA to pass through the ammeter. Since we wish to measure a maximum current of 50 mA, we must split the current in such a way that 0.500 mA flows through the ammeter and 49.500 mA flows through the resistor R. Solve: (a) The potential difference across the ammeter and the resistor is the same. Thus, VR = Vammeter ⇒ (49.500 × 10−3 A)R = (0.500 × 10−3 A)(50.0 Ω) ⇒ R = 0.505 Ω (b) Effective resistance is

1 1 1 = + ⇒ Req = 0.500 Ω Req 0.505 Ω 50.0 Ω

32.59. Model: Assume ideal connecting wires. Visualize: Please refer to Figure P32.59. Because the ammeter we have shows a full-scale deflection with a current of 500 μA, we must not pass a current greater than this through the ammeter. Solve: The maximum potential difference is 5 V and the maximum current is 500 μA. Using Ohm’s law, ΔV = IA(R + Rammeter) ⇒ 5.0 V = (500 × 10−6 A)(R + 50.0 Ω) ⇒ R = 9.95 kΩ

32.60. Model: The battery and the connecting wires are ideal. Visualize:

The figure shows how to simplify the circuit in Figure P32.60 using the laws of series and parallel resistances. We have labeled the resistors as R1 = 6 Ω, R2 = 15 Ω, R3 = 6 Ω, and R4 = 4 Ω. Having reduced the circuit to a single equivalent resistance Req, we will reverse the procedure and “build up” the circuit using the loop law and the junction law to find the current and potential difference of each resistor. Solve: R3 and R4 are combined to get R34 = 10 Ω, and then R34 and R2 are combined to obtain R234:

1 1 1 1 1 ⇒ R234 = 6 Ω = + = + R234 R2 R34 15 Ω 10 Ω Next, R234 and R1 are combined to obtain Req = R234 + R1 = 6 Ω + 6 Ω = 12 Ω From the final circuit, E 24 V I= = = 2A Req 12 Ω Thus, the current through the battery and R1 is IR1 = 2 A and the potential difference across R1 is I(R1) = (2 A)(6 Ω) = 12 V As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference ΔV.

In Step 1 of the above figure, Req = 12 Ω is returned to R1 = 6 Ω and R234 = 6 Ω in series. Both resistors must have the same 2.0 A current as Req. We then use Ohm’s law to find ΔVR1 = (2 A)(6 Ω) = 12 V

ΔVR234 =(2 A)(6 Ω) = 12 V

As a check, 12 V + 12 V = 24 V, which was ΔV of the Req resistor. In Step 2, the resistance R234 is returned to R2 and R34 in parallel. Both resistors must have the same ΔV = 12 V as the resistor R234. Then from Ohm’s law,

I R2 =

12 V = 0.80 A 15 Ω

I R34 =

12 V = 1.2 A 10 Ω

As a check, IR2 + IR34 = 2.0 A, which was the current I of the R234 resistor. In Step 3, R34 is returned to R3 and R4 in series. Both resistors must have the same 1.2 A as the R34 resistor. We then use Ohm’s law to find (ΔV )R3 = (1.2 A)(6 Ω) = 7.2 V

(ΔV )R4 = (1.2 A) (4 Ω) = 4.8 V

As a check, 7.2 V + 4.8 V = 12 V, which was ΔV of the resistor R34. Resistor Potential difference (V) 6 Ω left 12 12 15 Ω 7.2 6 Ω right 4.8 4Ω

Current (A) 2 0.80 1.2 1.2

32.61. Model: The battery and the connecting wires are ideal. Visualize:

The figure shows how to simplify the circuit in Figure P32.61 using the laws of series and parallel resistances. Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and “build up” the circuit using the loop law and the junction law to find the current and potential difference of each resistor. Solve: From the last circuit in the diagram,

12 V E = =2A 6Ω 6Ω Thus, the current through the battery is 2 A. As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference ΔV. I=

In Step 1, the 6 Ω resistor is returned to a 3 Ω and 3 Ω resistor in series. Both resistors must have the same 2 A current as the 6 Ω resistance. We then use Ohm’s law to find ΔV3 = (2 A)(3 Ω) = 6 V

As a check, 6 V + 6 V = 12 V, which was ΔV of the 6 Ω resistor. In Step 2, one of the two 3 Ω resistances is returned to the 4 Ω, 48 Ω, and 16 Ω resistors in parallel. The three resistors must have the same ΔV = 6 V. From Ohm’s law, I4 =

6V = 1.5 A 4Ω

Resistor 3Ω 4Ω 48 Ω 16 Ω

I 48 =

6V 1 = A 48 Ω 8

Potential difference (V) 6 6 6 6

I16 =

6V 3 = A 16 Ω 8

Current (A) 2 1.5 1.2 3.8

32.62. Model: The battery and the connecting wires are ideal. Visualize:

The figure shows how to simplify the circuit in Figure P32.62 using the laws of series and parallel resistances. We will reverse the procedure and “build up” the circuit using the loop law and junction law to find the current and potential difference of each resistor. Solve: Having found Req = 12 Ω, the current from the battery is I = (24 V)/(12 Ω) = 2.0 A. As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference ΔV.

In Step 1 of the above figure, the 12 Ω resistor is returned to 4 Ω and 8 Ω resistors in series. Both resistors must have the same 2.0 A as the 12 Ω resistor. We use Ohm’s law to find ΔV4 = 8 V and ΔV8 = 16 V. As a check, 8 V + 16 V = 24 V, which was ΔV of the 12 Ω resistor. In Step 2, the 8 Ω resistor is returned to the 12 Ω and 24 Ω resistors in parallel. Both resistors must have the same ΔV = 16 V as the 8 Ω resistor. From Ohm’s law, I12 = (16 V ) (12 Ω ) = 43 A and I 24 = 23 A . As a check, I12 + I24 = 2.0 A, which was the current I of the 8 Ω

resistor. In Step 3, the 12 Ω resistor is returned to the two 6 Ω resistors in series. Both resistors must have the same 34 A as the 12 Ω resistor. We use Ohm’s law to find ΔV6 = 8 V and ΔV6 = 8 V. As a check, 8 V + 8 V = 16 V, which was ΔV of the 12 Ω resistor. Finally, in Step 4, the 6 Ω resistor is returned to the 8 Ω and 24 Ω resistors in parallel. Both resistors must have the same ΔV = 8 V as the 6 Ω resistor. From Ohm’s law, I8 = (8 V)/(8 Ω) = 1 A and I 24 = 13 A . As a check, I 8 + I 24 = 34 A , which was the current I of the 6 Ω resistor. Resistor 4Ω 6Ω 8Ω Bottom 24 Ω Right 24 Ω

Potential difference (V) 8 8 8 8 16

Current (A) 2 1.3 1 0.33 0.66

32.63. Model: The batteries and the connecting wires are ideal. Visualize:

The figure shows how to simplify the circuit in Figure P32.63 using the laws of series and parallel resistances. Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and “build up” the circuit using the loop law and the junction law to find the current and potential difference of each resistor. Solve: From the last circuit in the figure and from Kirchhoff’s loop law,

12 V − 3 V = 1.0 A 9Ω Thus, the current through the batteries is 1.0 A. As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference. I=

In Step 1 of the above figure, the 9 Ω resistor is returned to the 6 Ω and 3 Ω resistors in series. Both resistors must have the same 1.0 A current as the 6 Ω resistor. We use Ohm’s law to find ΔV3 = (1.0 A)(3 Ω) = 3.0 V

ΔV6 = 6.0 V

As a check, 3.0 V + 6.0 V = 9 V, which was ΔV = (12 V – 3 V) = 9 V of the 9 Ω resistor. In Step 2, the 6 Ω resistor is returned to the 24 Ω and 8 Ω resistors in parallel. The two resistors must have the same potential difference ΔV = 6.0 V. From Ohm’s law, I7 =

6.0 V 3 = A 8Ω 4

I 24 =

6.0 V 1 = A 24 Ω 4

As a check, 0.75 A + 0.25 A = 1.0 A which was the current I of the 6 Ω resistor. In Step 3, the 8 Ω resistor is returned to the 3 Ω and 5 Ω (right) resistors in series, so the two resistors must have the same current of 0.828 A. We use Ohm’s law to find ΔV3 = (3/4 A)(3 Ω) = 9/4 V

ΔV4 = (3/4 A)(5 Ω) = 15/4 V

As a check, 9/4 V + 15/4 V = 24/4 V = 6.0 V, which was ΔV of the 8 Ω resistor. In Step 4, the 3 Ω resistor is returned to 4 Ω (left) and 12 Ω resistors in parallel, so the two must have the same potential difference ΔV = 9/4 V. From Ohm’s law, I4 =

9/4 V = 9 /16 A = 0.56 A 4Ω

I12 =

9/4 V 9 = A = 0.19 A 12 Ω 48

As a check, 0.56 A + 0.19 A = 0.75 A, which was the same as the current through the 3 Ω resistor. Resistor Potential difference (V) Current (A) 6.0 0.25 24 Ω 3.0 1.0 3Ω 3.75 0.75 5Ω 2.25 0.56 4Ω 2.25 0.19 12 Ω

32.64. Model: The battery and the connecting wires are ideal. Visualize:

The figure shows how to simplify the circuit in Figure P32.64 using the laws of series and parallel resistances. Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and “build up” the circuit to find the current and potential difference of each resistor. Solve: (a) From the last circuit in the figure and from Kirchhoff’s law, I = 100 V 10 Ω = 10 A. Thus, the current through the battery is 10 A. Now as we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference.

In Step 1 of the above diagram, we return the 10 Ω resistor to the 4 Ω, 4 Ω, and 2 Ω resistors in series. These resistors must have the same 10 A current as the 10 Ω resistance. That is, the current through the 2 Ω and the 4 Ω resistors is 10 A. The potential differences are ΔV2 = (10 A)(2 Ω) = 20 V

ΔV4 (left) = (10 A)(4 Ω) = 40 V

ΔV4 (left) = (10 A)(4 Ω) = 40 V

In Step 2, we return the left 4 Ω resistor to the 20 Ω and 5 Ω resistors in parallel. The two resistors must have the same potential difference ΔV = 40 V. From Ohm’s law, I5 =

40 V = 8A 5Ω

I 20 =

40 V = 2.0 A 20 Ω

The currents through the various resistors are I2 = I4 = 10 A, I5 = 8 A, and I20 = 2.0 A. 2 (b) The power dissipated by the 20 Ω resistor is I 202 ( 20 Ω ) = ( 2.0 A ) ( 20 Ω ) = 80 W . (c) Starting with zero potential at the grounded point, we travel along the outside path to point a and add/subtract the potential differences on the way:

0 V + (20 Ω)I20 + (2 Ω)I2 = (20 Ω)(2.0 A) + (2 Ω)(10 A) = 60 V = Va

32.65. Model: The wires and batteries are ideal. Visualize:

Solve: Assign currents I1, I2, and I3 as shown in the figure. If I3 turns out to be negative, we’ll know it really flows right to left. Apply Kirchhoff’s loop rule counterclockwise to the top loop from the top right corner: – I1 (5 Ω) + 12 V – I3 (10 Ω) = 0. Apply the loop rule counterclockwise to the bottom loop starting at the lower left corner: – I2 (5 Ω) + 9 V + I3 (10 Ω) = 0. Note that since we went against the current direction through the (10 Ω) resistor the potential increased. Apply the junction rule to the right middle: I1 = I2 + I3. These three equations can be solved for the current I3 by subtracting the second equation from the first then 3 making the substitution I2 – I1 = –I3 which was derived from the third equation. The result is I3 = A = 0.12 25 A, left to right.

32.66. Model: The wires and batteries are ideal. Visualize:

Solve: The circuit has been redrawn for clarity. Assign the currents I1, I2, and I3 as shown in the figure. To find the power dissipated by the 2 Ω resistor, we must find the current through it. If I3 turns out to be negative, we’ll know it really flows bottom to top. Apply Kirchhoff’s loop rule clockwise to the left loop from the bottom left corner: + 12 V – I1 (4 Ω) – I3 (2 Ω) = 0. Apply the loop rule clockwise to the right loop starting at the top right corner: + 15 V – I2 (4 Ω) + I3 (2 Ω) = 0 Note that since we went against the current direction through the 2 Ω resistor the potential increased. Apply the junction rule to the right middle: I1 = I2 + I3 These three equations can be solved for the current I3 by subtracting the second equation from the first then 3 making the substitution I2 – I1 = –I3 which was derived from the third equation. The result is I3 = − A. So the 8 3 current in the 2 Ω resistor is A , bottom to top. The power dissipated in the resistor is 8 2

9 ⎛3 ⎞ P2 Ω = ⎜ A ⎟ ( 2 Ω ) = W = 0.28 W 8 32 ⎝ ⎠

32.67. Model: The wires and batteries are ideal. Visualize:

Solve: If no power is dissipated in the 200 Ω resistor, the current through it must be zero. To see if this is possible, set up Kirchhoff’s rules for the circuit, then assume the current through the 200 Ω resistor is zero and see if there is a solution. Assume the unknown battery is oriented with its positive terminal at the top and currents I1, I2, I3 defined as shown in the figure above. Apply Kirchhoff’s loop rule clockwise to the left loop: 50 V – I1 (100 Ω) – I3 (200 Ω) = 0 Again, counterclockwise to the right hand loop: E – I2 (300 Ω) – I3 (200 Ω) = 0 The junction rule yields I1 + I2 = I3. Now assume I3 = 0 and solve for E. In that case, the first equation gives 50 V 1 I1 = = A. 100 Ω 2 From the third equation, I2 = –I1, so the second equation gives us ⎛ 1 ⎞ E = I 2 ( 300 Ω ) = ⎜ − A ⎟ ( 300 Ω ) = −150 V ⎝ 2 ⎠ Thus E = 150 V and it is oriented with negative terminal on top, opposite to our guess.

32.68. Model: The wires are ideal, but the batteries are not. Visualize:

Solve:

(a) The good battery alone can drive a current through the starter motor 12 V I= = 200 A ( 0.01 Ω + 0.05 Ω )

(b) Alone, the dead battery drives a current

I=

8.0 V

( 0.50 Ω + 0.05 Ω )

= 14.5 A

(c) Let I1, I2, I3 be defined as shown in the figure above. Kirchhoff’s laws applied to the good battery and dead battery loop, good battery and starter motor loop, and the top middle junction yield three equations in the three unknown currents: 12 V − I1 ( 0.01 Ω ) − I 3 ( 0.05 Ω ) = 0

12 V − I1 ( 0.01 Ω ) − I 2 ( 0.50 Ω ) − 8.0 V = 0

I1 = I 2 + I 3 Substituting for I1 from the third equation into the first and second equations gives 12 V − I 2 ( 0.01 Ω ) − I 3 ( 0.06 Ω ) = 0 4 V − I 2 ( 0.51 Ω ) − I 3 ( 0.01 Ω ) = 0 Solving for I2 from the first equation,

I2 =

12 V − I 3 ( 0.06 Ω ) ( 0.01 Ω )

Substituting into the second equation and solving for I3 yields the current through the starter motor is 199 A. (d) Substituting the value for I3 into the expression for I2 yields the current through the dead battery is 3.9 A. Assess: The good battery is charging the dead battery as well as running the started motor. A total of 203 A flows through the good battery.

32.69. Model: The wires and batteries are ideal. Visualize:

Solve: The circuit is redrawn above for clarity, and currents added. We must find I5. Repeatedly apply Kirchhoff’s rules. The loop rule applied clockwise about the three triangles yields Left: 9 V – I1 (6 Ω) – I3 (12 Ω) = 0 ⇒ I1 = 1.5 A – 2I3

Center: –I4 (24 Ω) + I3 (12 Ω) = 0⇒I4 = ½ I3 Right: 15 V – I2 (10 Ω) – I4 (24 Ω) = 0 ⇒I2 = 1.5 A – 2.4 I4 The junction rule applied at the bottom corners gives equations into which the results above may be substituted: I1 = I3 + I5 ⇒1.5 A – 2I3 = I3 +I5 ⇒ I5 = 1.5 A – 3I3 I4 = I2 + I5 ⇒ I4 = 1.5 A – 2.4I4 + I5 ⇒ I5 = 3.4I4 – 1.5 A Using I4 = ½ I3 and solving for I3, 30 1.5 A – 3I3 = 3.4(½ I3) – 1.5 A ⇒ I3 = A 47 30 201 ⇒I5 = 1.5 A – 3( A) = − A = –2.1 A 47 94 Since the current is negative, 2.1 A flows from left to right through the bottom wire.

32.70. Model: Assume ideal wires. The capacitor discharges through the resistor. Solve:

(a) The capacitor discharges through the resistor R as Q = Q0e−t/τ. For Q= Q0/2, Q0 t ⎛1⎞ ⇒ t = −(0.010 s)ln(0.5) = 6.9 ms = Q0e − t 10 ms ⇒ ln ⎜ ⎟ = − 2 2 0.010 s ⎝ ⎠

(b) If the initial capacitor energy is U0, we want the time when the capacitor’s energy will be U = U 0 2 . Noting

that U 0 = Q02 2C , this means Q = Q0

2 . Applying the equation for the discharging capacitor,

Q0 t ⎛ 1 ⎞ ⎛ 1 ⎞ = Q0e − t 10 ms ⇒ ln ⎜ ⎟ = − 0.010 s ⇒ t = − ( 0.010 s ) ln ⎜ ⎟ = 3.5 ms 2 ⎝ 2⎠ ⎝ 2⎠

32.71. Model: The capacitor discharges through the resistor, and the wires are ideal.

Solve: In an RC circuit, the charge at a given time is related to the original charge as Q = Q0e−t/τ. For a capacitor Q = CΔV, so ΔV = ΔV0e−t/τ. From the Figure P32.71, we note that ΔV0 = 30 V and ΔV = 10 V at t = 4 ms. So, 4 × 10−3 s 4 × 10−3 s − 4 ms R ( 50×10−6 F ) ⎛ 10 V ⎞ R 10 V = ( 30 V ) e ⇒ ln ⎜ = − ⇒ = − = 73 Ω ⎟ R ( 50 × 10−6 F ) ⎝ 30 V ⎠ ( 50 ×10−6 F) ln ( 13 )

32.72. Model: The capacitor discharges through the resistors. The wires are ideal. Solve:

The charge Q on the capacitor when it is connected across a 50 V source is Q = CΔV = (0.25 μF)(50 V) = 12.5 μC

As this fully charged capacitor is connected in series with a 25 Ω resistor and a 100 Ω resistor, it will dissipate all its stored energy

Q 2 1 (12.5 μ C ) = = 312.5 × 10−6 J C 2 0.25 μ F 2

U C = 12

through the two resistors. Energy dissipated by the resistor is I2R, which means the 100 Ω resistor will dissipate four times more energy than the 25 Ω resistor at any given time. Thus, the energy dissipated by the 25 Ω resistor is 25 Ω ⎛ ⎞ −6 312.5 × 10−6 ⎜ ⎟ = 62.5 × 10 J = 63 μJ ⎝ 25 Ω + 100 Ω ⎠

32.73. Model: The battery and the connecting wires are ideal. Visualize: Please refer to Figure P32.73. Solve: (a) A very long time after the switch has closed, the potential difference ΔVC across the capacitor is E . This is because the capacitor charges until ΔVC = E , while the charging current approaches zero. (b) The full charge of the capacitor is Qmax = C(ΔVC)max = C E . (c) In this circuit, I = + dQ dt because the capacitor is charging, that is, because the charge on the capacitor is increasing. (d) From Equation 32.36, capacitor charge at time t is Q = Qmax(1 − e−t/τ). So,

I=

dQ d ⎛1⎞ ⎛ 1 ⎞ −t τ E −t τ = CE (1 − e− t τ ) = CE ⎜ ⎟ e− t τ = CE ⎜ ⎟e = e dt dt R τ ⎝ ⎠ ⎝ RC ⎠

A graph of I as a function of t is shown below.

32.74. Model: Assume the battery and the connecting wires are ideal. Visualize: Please refer to Figure P32.74. Solve: (a) If the switch has been closed for a long time, the capacitor is fully charged and there is no current flowing through the right branch that contains the capacitor. So, a voltage of 60 V appears across the 60 Ω resistor and a voltage of 40 V appears across the 40 Ω resistor. That is, maximum voltage across the capacitor is 40 V. Thus, the charge on the capacitor is Q0 = EC = (40 V)(2.0 × 10−6 F) = 80 μC (b) Once the switch is opened, the battery is disconnected from the capacitor. The capacitor C has two resistances (10 Ω and 40 Ω) in series and discharges according to Q = Q0e−t/RC. For Q = 0.10 Q0,

0.10Q0 = Q0e− t /(50 Ω )(2.0 μ F) ⇒ ln ( 0.10 ) = −

t ( 50 Ω )( 2.0 μ F )

⇒ t = − ( 50 Ω )( 2.0 μ F ) ln ( 0.10 ) = 0.23 ms

32.75. Model: Assume the battery and the connecting wires are ideal. Visualize: Please refer to Figure CP32.75. Solve: Because the switch has been in position a for a very long time, the capacitor is fully charged to Q0 = EC = (50 V)(20 × 10−6 F) = 1.00 × 10−3 C When the switch is placed in position b for 1.25 ms, its charge will decay through the resistor to

Q = Q0e − t RC = (1.00 × 10−3 C ) e

(

− 1.25×10−3 s

)

( 50 Ω ) ( 20×10−6

F

)

= 0.2865 × 10−6 C

The energy in the capacitor after 1.25 ms is −6 1 Q 2 1 ( 0.2865 × 10 C ) U= = = 2.05 mJ 2 C 2 20 × 10−6 F 2

The energy in the capacitor at t = 0 s (when the switch was just flipped to position b) is 1 Q02 1 (1.0 × 10 C ) U0 = = = 25 mJ 2 C 2 20 × 10−6 F −3

2

So, the energy dissipated by the 50 Ω resistor is 25 mJ – 2.05 mJ = 23 mJ.

32.76. Model: The connecting wires are ideal. The capacitors discharge through the resistors. Visualize:

The figure shows how to simplify the circuit in Figure P32.76 using the laws of series and parallel resistors and the laws of series and parallel capacitors. Solve: The 30 Ω and 20 Ω resistors are in parallel and are equivalent to a 12 Ω resistor. This 12 Ω resistor is in series with the 8 Ω resistor so the equivalent resistance of the circuit Req = 20 Ω. The two 60 μF capacitors are in series producing an equivalent capacitance of 30 μF. This 30 μF capacitor is in parallel with the 20 μF capacitor so the equivalent capacitance Ceq of the circuit is 50 μF. The time constant of this circuit is

τ = ReqCeq = (20 Ω)(50 μF) = 1.0 ms The current due to the three capacitors through the 20 Ω equivalent resistor is the same as through the 8 Ω resistor. So, the voltage across the 8 Ω resistor follows the decay equation V = V0e−t/τ. For V = V0/2, we get t ⎛1⎞ V0/2 = V0e−t/1.0 ms ⇒ ln ⎜⎝ 2 ⎟⎠ = − 1.0 ms ⇒ t = 0.69 ms

32.77. Model: The battery and the connecting wires are ideal. Visualize: Please refer to Figure 32.38a. Solve: After the switch closes at t = 0 s, the capacitor begins to charge. At time t, let the current and the charge in the circuit be i and q, respectively. Also, assume clockwise direction for the current i. Using Kirchhoff’s loop law and starting clockwise from the lower left corner of the loop,

+E − iR −

q dq q dq dt = =0⇒E = R + ⇒ RC dq = ( EC – q)dt ⇒ EC − q RC C dt C

Integrating both sides, Q

t

Q dq dt t t ∫0 EC − q = ∫0 RC ⇒ − ⎡⎣ln ( EC − q )⎤⎦ 0 = RC ⇒ − ln ( EC − Q ) + ln ( EC ) = RC

t EC − Q ⎛ EC − Q ⎞ ⇒ ln ⎜ ⇒ = e − t RC ⇒ Q = EC (1 − e − t RC ) ⎟=− E C RC E C ⎝ ⎠

Letting Qmax = EC and τ = RC, we get Q = Qmax (1 − e−t/τ).

32.78. Model: The battery and the connecting wires are ideal. Visualize: Please refer to Figure 32.38a. Solve: (a) According to Equation 32.36, during charging the charge on the capacitor increases according to Q = Qmax (1 − e − t τ ) . Therefore, the current in the circuit behaves as I=

dQ EC − t RC E − t RC ⎛ 1⎞ e = Qmax ⎜ − ⎟ ( −e − t τ ) = = e dt RC R ⎝ τ⎠

Using Equation 32.8, the power supplied by the battery as the capacitor is being charged is E2 ⎛E ⎞ Pbat = I E = ⎜ e − t RC ⎟ E = e − t RC R ⎝R ⎠ Because Pbat = dU dt , we have ∞

∞

∞ E 2 −t RC E2 e dt = ⎡⎣− RC e−t RC ⎤⎦ = E 2C 0 R 0 R

dU = Pbat dt ⇒ ∫ dU = ∫ Pbat dt = ∫ 0

That is, the total energy which has been supplied by the battery when the capacitor is fully charged is E2C. (b) The power dissipated by the resistor as the capacitor is being charged is E2 Presistor = I 2 R = 2 R e − 2t RC R Because Presistor = dU dt , we have ∞

∞

E 2 − 2t RC E 2 ⎡ RC − 2t RC ⎤ E 2C e dt ⇒ U resistor = e − = ⎢ ⎥ 2 R⎣ 2 ⎦0 0 R

dU = Presistordt ⇒ ∫ dU = ∫

(c) The energy stored in the capacitor when it is fully charged is 2 1 Qmax 1 C 2E 2 1 2 UC = = = CE 2 C 2 C 2 (d) For energy conservation, the energy delivered by battery is equal to the energy dissipated by the resistor R plus the energy stored in the capacitor C. This is indeed the case because Ubat = E2C, U resistor = 12 E 2C , and

U C = 12 E 2C .

32.79. Model: The battery and the connecting wires are ideal. Visualize: Please refer to Figure CP32.79. Solve: (a) During charging, when the neon gas behaves like an insulator, the charge on the capacitor increases according to Equation 32.36. That is, Q = Qmax (1 − e−t/τ). Because Q = CΔVC, ΔVC = ΔVC0(1 − e−t/τ) = E(1 − e−t/τ) Let us say that the period of oscillation begins when ΔV = Voff and it ends when ΔV = Von. Then,

Voff = E (1 − e− toff

τ

) ⇒ E −EV

off

= e −toff τ ⇒

toff

τ

⎛ E ⎞ = ln ⎜ ⎟ ⎝ E − Voff ⎠

Because the period T = ton – toff, we have

⎛ E ⎞ ⎛ E ⎞ ⎛ E − Voff ⎞ ⎛ E − Voff ⎞ T = τ ln ⎜ ⎟ − τ ln ⎜ ⎟ = τ ln ⎜ ⎟ = RC ln ⎜ ⎟ ⎝ E − Von ⎠ ⎝ E − Voff ⎠ ⎝ E − Von ⎠ ⎝ E − Von ⎠ (b) Substituting the given values into the above expression and noting that and T = 1 f = 0.10 s ,

0.10 s ⎛ 90 V − 20 V ⎞ = 5.1 kΩ 0.10 s = R (10 × 10−6 F ) ln ⎜ ⎟⇒ R= ⎝ 90 V − 80 V ⎠ (10 × 10−6 F) ln 7

32.80. Model: The wire can be thought of as a large number of small wire segments in series. Solve:

(a) The resistance ΔR of a segment of the wire of length Δx is ρΔx ρΔx ρΔ x 2 x l ΔR = = = e 2 x − A π r0 2 π re l

(

0

)

The resistance of the complete length of wire is L

R = ΣΔR → ∫ 0

ρ π r0

2

e

2x

l

ρ l 2x dx = 2 e l π r0 2

L

= 0

(

)

2L ρl e l −1 2 2π r0

(b) The exponential decrease in radius would seem flat if l W L . Or, put another way,

L V1 . l

(c) The Taylor series expansion of the exponential function is 2 2L 2L 1 ⎛ 2L ⎞ + ⎜ e l =1+ ⎟ +" l 2⎝ l ⎠ 2L L 2L Keeping the expansion to first order in , e l ≈ 1 + . Thus l l ρl ⎛ 2L ⎞ ρ L ρ L R≈ − 1⎟ = 2 = ⎜1 + 2π r02 ⎝ l A ⎠ π r0 Assess: This is the expected resistance for a wire of constant radius r0.

32-1

33.1. Model: A magnetic field is caused by an electric current. Visualize: Please refer to Figure EX33.1. Solve: The magnitude of the magnetic field at point 1 is 2.0 mT and its direction can be determined by using the right-hand rule. Grab the current carrying wire so that your thumb points in the direction of the current. G Because your fingers at point 1 point into the page, B1 = (2.0 mT, into the page). At point 2, the magnetic field due to the bottom wire is into the page. The right-hand rule tells us that the magnetic field from the top wire is G also into the page. At point 2, B2 = (4.0 mT, into the page).

33.2. Model: A magnetic field is caused by an electric current. Visualize: Please refer to Figure EX33.2. Solve: The current in the wire is directed to the right. B2 = 20 mT + 20 mT = 40 mT because the two overlapping wires are carrying current in the same direction and each wire produces a magnetic field having the same direction at point 2. B3 =20 mT – 20 mT = 0 mT, because the two overlapping wires carry currents in opposite directions and each wire produces a field having opposite directions at point 3. The currents at 4 are also in opposite directions, but the point is to the right of one wire and to the left of the other. From the right-hand rule, the field of both currents is out of the page. Thus B4 = 20 mT + 20 mT =40 mT.

33.3. Model: The magnetic field is that of a moving charged particle. Visualize:

The first point is on the x-axis, with θa = 90°. The second point is on the y-axis, with θb = 180°, and the third point is on the −y-axis with θc = 0°. Solve: (a) Using Equation 33.1, the Biot-Savart law, the magnetic field strength is

Ba =

−7 −19 +7 μ0 qv sin θ (10 T m/A )(1.60 × 10 C )(1.0 × 10 m/s ) sin 90° = = 1.60 × 10−15 T 2 −2 4π r 2 1.0 × 10 m ( )

G G To use the right-hand rule for finding the direction of B, point your thumb in the direction of v . The magnetic G G G field vector B is perpendicular to the plane of r and v and points in the same direction that your fingers point. G In the present case, the fingers point along the kˆ direction. Thus, Ba = 1.60 × 10−15 kˆ T. (b) Bb = 0 T because sin θb = sin 180° = 0. (c) Bc = 0 T because sin θc = sin 0° = 0.

33.4. Model: The magnetic field is that of a moving charged particle. Visualize:

The first point is on the x-axis, with θa = 90°. The second point is on the z-axis, with θb = 0°, and the third point is in the yz plane with θc = 45°. Solve: (a) Using Equation 33.1, the Biot-Savart law, the magnetic field strength is

Ba =

−7 −19 +7 μ0 qv sin θ (10 T m/A )(1.60 × 10 C )( 2.0 × 10 m/s ) sin 90° = = 3.2 × 10−15 T 2 −2 4π r 2 (1.0 ×10 m )

G G To use the right-hand rule for finding the direction of Ba , point your thumb in the direction of v . Your fingers G G point along the –y-axis, but since the charge is negative, Ba points along the +y-axis. Thus, Ba = 3.2 × 10−15 ˆj T. (b) Bb = 0 T because sin θ b = sin 0° = 0. (c) For the third point,

Bc =

(10

−7

T m/A )(1.60 × 10−19 C )( 2.0 × 107 m/s ) sin 45°

= 1.13 × 10−15 T

(1.0 ×10 m ) + (1.0 ×10 m ) G G G The direction of Bc is perpendicular to the plane formed by r and v . The right-hand rule gives the direction of G G G v × r along the –x-axis, but because the charge is negative, Bc = 1.13 × 10−15 iˆ T. −2

2

−2

2

33.5. Model: The magnetic field is that of a moving charged particle. Visualize: Please refer to Figure EX33.5. Solve: Using the Biot-Savart law,

B=

−7 −19 7 μ0 qv sin θ (10 T m/A )(1.60 × 10 C )( 2.0 × 10 m/s ) sin 45° = = 1.13 × 10−15 T 2 2 −2 −2 4π r 2 × + × 1.0 10 m 1.0 10 m ( ) ( )

G G The right-hand rule applied to the proton points B into the page. Thus, B = −1.13 × 10−15 kˆ T.

33.6. Model: The magnetic field is that of a moving charged particle. Visualize: Please refer to Figure EX33.6. Solve: The Biot-Savart law is

B=

−7 −19 7 μ0 qv sin θ (10 T m/A )(1.60 × 10 C )( 2.0 × 10 m/s ) sin135° = = 2.83 × 10−16 T 2 2 −2 −2 4π r 2 × + × 2.0 10 m 2.0 10 m ( ) ( )

G The right-hand rule for the positive charge indicates the field points out of the page. Thus, B = 2.83 × 10−16 kˆ T.

33.7. Model: The magnetic field is that of a moving proton. Visualize:

The magnetic field lies in the xy-plane.

G Solve: Using the right-hand rule, the charge is moving along the +z-direction. That is, v = vkˆ. Using the BiotSavart law,

B=

(10−7 T m/A )(1.60 × 10−19 C ) v sin 90° ⇒ v = 6.3 ×106 m/s in the +z-direction μ0 qv sin θ −13 ⇒ 1.0 × 10 T = 2 4π r 2 (1.0 × 10−3 m )

33.8. Model: The magnetic field is that of an electric current in a long straight wire. Solve: From Example 33.3, the magnetic field strength of a long, straight wire carrying current I at a distance d from the wire is μ I Bwire = 0 2π d The current needed to produce the earth’s magnetic field is calculated as follows:

Bearth surface = 5 × 10−5 T =

( 2 ×10

−7

T m/A ) I

0.010 m

⇒ I = 2.5 A

Likewise, the currents needed for a refrigerator magnet, a laboratory magnet, and a superconducting magnet are 250 A, 5000–50,000 A, and 500,000 A.

33.9. Model: The magnetic field is that of an electric current in a long straight wire. Solve: From Example 33.3, the magnetic field strength of a long, straight wire carrying current I at a distance d from the wire is B=

μ0 I 2π d

The distance d at which the magnetic field is equivalent to Earth’s magnetic field is calculated as follows: Bearth surface = 5 × 10−5 T =

( 2 × 10

−7

T m/A )

10 A ⇒ d = 4.0 cm d

Likewise, the corresponding distances for a refrigerator magnet, a laboratory magnet, and a superconducting magnet are 0.40 mm, 20 μm to 2.0 μm, and 0.20 μm.

33.10.

Solve:

From Example 33.5, the on-axis magnetic field of a current loop is

Bloop ( z ) =

μ0

IR 2

2 ( z2 + R2 )2

We want to find the value of z such that B ( z ) = 2 B ( 0 ) .

μ0

IR 2

2 ( z2 + R2 )

⇒ ( z2 + R2 )2 = 3

3 2

=2

μ0 IR 2

2 ( R2 ) 2 3

(

)

1 2 R3 R2 −2 ⇒ z 2 + R 2 = 2 ⇒ z = R 2 3 − 1 = 0.77 R z 23

3

.

33.11. Model: The magnetic field is that of a current loop. Solve:

(a) From Equation 33.7, the magnetic field strength at the center of a loop is Bloop center =

μ0 I 2R

⇒ I=

2RBloop center

μ0

=

2 ( 0.50 × 10−2 m )( 2.5 × 10−3 T ) 4π (10−7 T m/A )

= 20 A

(b) For a long, straight wire that carries a current I, the magnetic field strength is

Bwire =

4π (10−7 T m/A ) ( 20 A ) μ0 I ⇒ 2.5 × 10−3 T = ⇒ d = 1.60 × 10−3 m 2π d 2π d

33.12. Model: The magnetic field is the superposition of the magnetic fields of three wire segments. Visualize: Please refer to Figure EX33.12. Solve: The magnetic field of the horizontal wire, with current I, encircles the wire. Because the dot is on the axis of the wire, the input current creates no magnetic field at this point. The current divides at the junction, with I/2 traveling upward and I/2 traveling downward. The right-hand rule tells us that the upward current creates a field at the dot that is into the page; the downward current creates a field that is out of the page. Although we could calculate the strength of each field, the symmetry of the situation (the dot is the same distance and direction from the base of each wire) tells us that the fields of the upward and downward current must have the G G same strength. Since they are in opposite directions, their sum is 0 . Altogether, then, the field at the dot is B = G 0T.

33.13. Model: Assume the wires are infinitely long. Visualize:

The field vectors are tangent to circles around the currents. The net magnetic field is the vectorial sum of the G G fields Btop and Bbottom . Points a and c are at a distance d = 2 cm from both wires and point b is at a distance d =

1 cm. Solve:

The magnetic field at points a, b, and c are G G G μI μI Ba = Btop + Bbottom = 0 cos 45°iˆ − sin 45° ˆj + 0 cos 45°iˆ + sin 45° ˆj 2π d 2π d −7 × 2 10 T m/A 10 A ( ) ( ) μI ⎛ 1 ⎞ˆ −4 ˆ = 0 2cos 45°iˆ = 2⎜ ⎟ i = 2.0 × 10 i T 2π d 2 × 10−2 m ⎝ 2⎠

(

)

(

)

G ( 2 × 10−7 T m/A ) (10 A ) iˆ = 4.0 × 10−4iˆ T μI μI Bb = 0 iˆ + 0 iˆ = 2 2π d 2π d 1 × 10−2 m G μI μI Bc = 0 cos 45°iˆ + sin 45° ˆj + 0 cos 45°iˆ − sin 45° ˆj = 2.0 × 10−4 iˆ T 2π d 2π d

(

)

(

)

33.14. Model: Assume the wires are infinitely long. Visualize: Please refer to Figure EX33.14. Solve: The magnetic field strength at point a is

G G G ⎛ μI ⎞ ⎛μI ⎞ Bat a = Btop + Bbottom = ⎜ 0 , out of page ⎟ + ⎜ 0 , into page ⎟ ⎝ 2π d ⎠ top ⎝ 2π d ⎠ bottom ⇒ Bat a =

⎞ 1 1 1 μ0 I ⎛ 1 ⎛ ⎞ −7 − − ⎜⎜ ⎟ = ( 2 × 10 T m/A ) (10 A ) ⎜ ⎟ −2 −2 2π ⎝ 2.0 cm ( 4.0 + 2.0 ) cm ⎟⎠ ⎝ 2.0 × 10 m 6.0 × 10 m ⎠

G ⇒ Bat a = ( 6.7 × 10−5 T, out of page ) At points b and c,

G ⎛ μI ⎞ ⎛ μI ⎞ Bat 2 = ⎜ 0 , into page ⎟ + ⎜ 0 , into page ⎟ = ( 2.0 × 10−4 T, into page ) ⎝ 2π d ⎠ ⎝ 2π d ⎠ G ⎛μI ⎞ ⎛μI ⎞ Bat 3 = ⎜ 0 , into page ⎟ + ⎜ 0 , out of page ⎟ = ( 6.7 × 10−5 T, out of page ) ⎝ 2π d ⎠ ⎝ 2π d ⎠

33.15. Model: Assume that the 10 cm distance is much larger than the size of the small bar magnet. Solve:

(a) From Equation 33.9, the on-axis field of a magnetic dipole is

μ0 2μ 4π Bz 3 ( 5.0 × 10 T ) ( 0.10 m ) ⇒μ= = = 0.025 A m 2 3 μ0 2 4π z 2 (10−7 T m/A ) 3

−6

B=

(b) The on-axis field strength 15 cm from the magnet is B=

2 ( 0.025 A m μ0 2μ = (10−7 T m/A ) 3 3 4π z ( 0.15 m )

2

) = 1.48 × 10

−6

T = 1.48 μ T

33.16. Solve: (a) The magnetic dipole moment of the superconducting ring is μ = (π R 2 ) I = π (1.0 × 10−3 m ) (100 A ) = 3.1× 10−4 A m 2 2

(b) From Example 33.5, the on-axis magnetic field of the superconducting ring is

Bring =

μ0

IR 2

2 ( z 2 + R 2 )3 / 2

=

2π (10−7 T m/A ) (100 A ) (1.0 × 10−3 m ) ⎡ 0.050 m )2 + ( 0.0010 m )2 ⎤ ⎣( ⎦

32

2

= 5.0 × 10−7 T

33.17. Model: The size of the loop is much smaller than 50 cm, so that the magnetic dipole moment of the loop is μ = AI . Solve:

The magnitude of the on-axis magnetic field of the loop is 3 ( 0.50 m ) ( 7.5 × 10−9 T ) μ 2 ( AI ) −9 = 7.5 × 10 T ⇒ A = = 1.88 × 10−4 m 2 B= 0 4π z 3 2 ( 25 A ) (10−7 T m/A )

Let L be the edge-length of the equilateral triangle. Thus 1 1.88 × 10−4 m 2 = L × L sin 60° ⇒ L = 0.021 m = 2.1 cm. 2 Assess: The 2.1 cm edge-length is much smaller than 50 cm, as we assumed.

33.18. Model: The radius of the earth is much larger than the size of the current loop. Solve:

(a) From Equation 33.9, the magnetic field strength at the surface of the earth at the earth’s north pole is

B=

−7 22 2 μ0 μ ( 2 × 10 T m/A )( 8.0 × 10 A m ) = = 6.2 × 10−5 T 3 6 2π z 3 6.38 × 10 m ( )

This value is close to the value of 5 × 10−5 T given in Table 33.1. (b) The current required to produce a dipole moment like that on the earth is 2 μ = AI = (π Rearth ) I ⇒8.0 × 1022 A m2 = π (6.38 × 106 m)2I ⇒ I = 6.3 × 108 A

Assess:

This is an extremely large current to run through a wire around the equator.

33.19. Visualize:G Please refer to Figure EX33.19. G G G Solve: Because B is in the same direction as the integration path s from i to f, the dot product of B and d s is simply Bds. Hence the line integral G ∫ B ⋅ d s = ∫ Bds = B ∫ ds = B ( ( 0.50 m ) + ( 0.50 m ) f

i

G

f

f

i

i

2

2

) = (0.10 T)

2 ( 0.50 m ) = 0.071 T m

33.20. Visualize: Please refer to Figure EX33.20. G

Solve:

The line integral of B between points i and f is f

G

G

∫B⋅d s i

G Because B is perpendicular to the integration path from i to f, the dot product is zero at all points and the line integral is zero.

33.21. Model: The magnetic field is that of the three currents enclosed by the loop. Visualize: Please refer to Figure EX33.21. Solve: Ampere’s law gives the line integral of the magnetic field around the closed path: G G −7 −6 úB ⋅ d s = μ0 I through = 3.77 × 10 T m = μ0 ( I1 − I 2 + I3 ) = ( 4π × 10 T m/A ) ( 6.0 A − 4.0 A + I3 )

3.77 × 10−6 T m ⇒ I3 = 1.0 A 4π × 10−7 T m/A Assess: The right-hand rule was used above to assign positive signs to I1 and I3 and a negative sign to I2.

⇒ ( I 3 + 2.0 A ) =

33.22. Model: Only the two currents enclosed by the closed path contribute to the line integral. Visualize: Please refer to Figure EX33.22. Solve: Ampere’s law gives the line integral of the magnetic field around the closed path: G G úB ⋅ d s = μ0 I through = 1.38 ×10−5 T m = μ0 ( I 2 + I3 ) = ( 4π × 10−7 T m/A ) (8.0 A + I3 )

1.38 × 10−5 T m ⇒ I3 = 3.0 A, out of the page. 4π × 10−7 T m/A Assess: The right-hand rule was used above to assign a positive sign to I2. Since I3 is also positive, it is in the same direction as I2. ⇒ ( I 3 + 8.0 A ) =

33.23. Model: The magnetic field is that of a current flowing into the plane of the paper. The currentcarrying wire is very long. Visualize: Please refer to Figure EX33.23. Solve: Divide the line integral into three parts: f G G G G B⋅ ds = B⋅ ds +

∫

i

∫

left line

∫

semicircle

G G B⋅ ds +

∫

G G B⋅ ds

right line

G The magnetic field of the current-carrying wire is tangent to clockwise circles around the wire. B is everywhere perpendicular to the left line and to the right line, thus the first and third parts of the line integral are zero. Along G the semicircle, B is tangent to the path and has the same magnitude B = μ0I/2π d at every point. Thus f G μI μ I (4π × 10−7 T m/A)(2.0 A) G B ⋅ ds = 0 + BL + 0 = 0 (π d ) = 0 = = 1.26 × 10−6 T m i 2π d 2 2

∫

where L = π d is the length of the semicircle, which is half the circumference of a circle of radius d.

33.24. Model: Assume that the solenoid is an ideal solenoid. Solve:

We can use Equation 33.16 to find the current that will generate a 3.0 mT field inside the solenoid:

μ0 NI

Bsolenoid =

l

⇒ I=

Bsolenoidl μ0 N

Using l = 0.15 m and N = 0.15 m 0.0010 m = 150,

I= Assess:

( 3.0 × 10 ( 4π ) (10

−3

−7

T ) ( 0.15 m )

T m/A ) (150 )

= 2.4 A

This is a reasonable current to pass through a good conducting wire of diameter 1 mm.

33.25. Model: Assume that the solenoid is an ideal solenoid. Solve: We can use Equation 33.16 to find the current that will generate a magnetic field of 1.5 T inside the solenoid:

Bsolenoid =

μ0 NI l

⇒ I=

Bsolenoidl μ0 N

Using l = 1.8 m and N = 1.8 m 0.0020 = 900,

I= Assess:

(1.5 T )(1.8 m )

4π (10−7 T m/A ) 900

= 2.4 kA

Large currents can be passed through conducting wires. The current density through the

superconducting wire is superconducting wire.

2 ( 2400 A ) π ( 0.001 m ) = 7.6 ×108 A/m2

. This value is reasonable for a

33.26. Model: A magnetic field exerts a magnetic force on a moving charge. Visualize: Please refer to Figure EX33.26. Solve: (a) The force on the charge is G G G Fon q = qv × B = (1.60 × 10−19 C )(1.0 × 107 m/s ) cos 45°iˆ + sin 45°kˆ × 0.50iˆ T

(

= (1.60 × 10−19 C ) ( 0.50 T )

(1.0 ×10

7

2

m/s )

) (

( iˆ × iˆ + kˆ × iˆ ) = +5.7 ×10

G G (b) Because the cross product iˆ × iˆ in the equation for the force is zero, Fon q = 0 N.

−13

ˆj N

)

33.27. Model: A magnetic field exerts a magnetic force on a moving charge. Visualize: Please refer to Figure EX33.27. Solve: (a) The force is G G G Fon q = qv × B = ( −1.60 × 10−19 C ) −1.0 × 107 ˆj m/s × 0.50iˆ T = −8.0 × 10−13 kˆ N

(

) (

)

(b) The force is G Fon q = ( −1.60 × 10−19 C )(1.0 × 107 m/s ) − cos 45° ˆj + sin 45°kˆ × 0.50iˆ T = 5.7 × 10−13 − ˆj − kˆ N

(

) (

)

(

)

33.28. Solve: The cyclotron frequency of a charged particle in a magnetic field is fcyc = qB/2π m. An ion has charge q = −e. Its mass is the sum of the masses of the two atoms minus the mass of the missing electron. For example, the mass of N +2 is m = mN + mN – melec = 2(14.0031 u)(1.661 × 10−27 kg/u) – 9.11 × 10−31 kg = 4.65174 × 10−26 kg

Note that we’re given the atomic masses very accurately, and we need to retain that accuracy to tell the difference between N +2 and CO+. Calculating each mass and frequency: Mass (kg)

N +2

4.65174 × 10−26

1.6423

+ 2

−26

1.4378 1.6429

O

+

CO Assess:

f (MHz)

Ion

5.31341 × 10 4.64986 × 10−26

The difference between N +2 and CO+ is not large but is easily detectable.

33.29. Model: A charged particle moving perpendicular to a uniform magnetic field moves in a circle. Solve:

The frequency of revolution of a charge moving at right angles to the magnetic field is

f cyc =

−31 6 2π mf cyc 2π ( 9.11 × 10 kg )( 45 × 10 Hz ) qB ⇒ B= = = 1.61 × 10−3 T q 1.60 × 10−19 C 2π m

33.30. Solve: (a) From Equation 33.19, the cyclotron radius is relec =

−31 6 mv ( 9.11 × 10 kg )(1.0 × 10 m/s ) = = 0.114 m = 11.4 cm qB (1.60 ×10−19 C )(5.0 ×10−5 T )

(b) For the proton,

(1.67 ×10 (1.60 ×10

−27

rproton =

kg )( 5.0 × 104 m/s )

−19

C )( 5.0 × 10−5 T )

= 10.4 m

33.31. Model: Charged particles moving perpendicular to a uniform magnetic field undergo circular motion at constant speed. Solve: The magnetic force on a proton causes the centripetal acceleration of

evB =

mv 2 mv ⇒ B= er r

The maximum radius r is 0.25 m, and the desired speed v is 0.10c. So, the required field B is

(1.67 ×10 kg ) ( 0.10) ( 3 ×10 (1.6 ×10 C ) ( 0.25 m ) −27

B=

8

−19

m/s )

= 1.25 T

33.32. Model: Assume the magnetic field is uniform over the Hall probe. Solve:

The Hall voltage is given by Equation 33.24: IB ΔVH I ΔVH = ⇒ = tne B tne ΔVH In both uses, the quantities I, t, n, and e are unchanged, so the ratio is constant. Thus B 1.9 × 10−6 V 2.8 × 10−6 V = ⇒ B = 81 mT 55 × 10−3 T B

33.33. Model: Assume the magnetic field is uniform over the Hall probe.

Visualize: Please refer to Figure EX33.42(a). The thickness is t = 4.0 × 10−3 m. Solve: The Hall voltage is given by Equation 33.24: ΔVH =

IB IB (15 A )(1.0 T ) ⇒n= = = 2.9 × 1028 m −3 −3 teΔVH (1.0 × 10 m )(1.60 × 10−19 C )( 3.2 × 10−6 V ) tne

Assess: The conduction electron density in metals is of the order of ≈ 5 × 10+28 m−3 (Table 31.1). The value obtained for the charge carrier density is reasonable.

33.34. Model: Assume that the field is uniform. The wire will float in the magnetic field if the magnetic force on the wire points upward and has a magnitude mg, allowing it to balance the downward gravitational force. Visualize: Please refer to Figure EX33.34. Solve: We can use the right-hand rule to determine which current direction experiences an upward force. The G current being from right to left, the force will be up if the magnetic field B points out of the page. The forces will balance when

F = ILB = mg ⇒ B =

G Thus B = (0.131 T, out of page).

−3 2 mg ( 2.0 × 10 kg )( 9.8 m/s ) = = 0.131 T IL (1.5 A )( 0.10 m )

33.35. Model: Assume that the magnetic field is uniform over the 10 cm length of the wire. Force on top and bottom pieces will cancel. Visualize: Please refer to Figure EX33.35. The figure shows a 10-cm-segment of a circuit in a region where the magnetic field is directed into the page. Solve: The current through the 10-cm-segment is E 15 V = =5A R 3Ω and is flowing down. The force on this wire, given by the right-hand rule, is to the right and perpendicular to the current and the magnetic field. The magnitude of the force is I=

G Thus F = (0.025 N, right).

F = ILB = (5 A)(0.10 m)(50 mT) = 0.025 N

33.36. Model: Two parallel wires carrying currents in opposite directions exert repulsive magnetic forces on each other. Two parallel wires carrying currents in the same direction exert attractive magnetic forces on each other. Visualize: Please refer to Figure EX33.36. Solve: The magnitudes of the various forces between the parallel wires are

F2 on 1 =

−7 μ0 LI1I 2 ( 2 × 10 T m/A ) ( 0.50 m )(10 A )(10 A ) = = 5.0 × 10−4 N = F2 on 3 = F3 on 2 = F1 on 2 2π d 0.02 m

F3 on 1 =

−7 μ0 LI1I 3 ( 2 × 10 T m/A ) ( 0.50 m )(10 A )(10 A ) = = 2.5 × 10−4 N = F1 on 3 2π d 0.04 m

Now we can find the net force each wire exerts on the other as follows: G G G Fon 1 = F2 on 1 + F3 on 1 = 5.0 × 10 −4 ˆj N + −2.5 × 10 −4 ˆj N = 2.5 × 10 −4 ˆj N = ( 2.5 × 10−4 N, up ) G G G Fon 2 = F1 on 2 + F3 on 2 = −5.0 × 10 −4 ˆj N + +5.0 × 10 −4 ˆj N = 0 N G G G Fon 3 = F1 on 3 + F2 on 3 = 2.5 × 10 −4 ˆj N + −5.0 × 10 −4 ˆj N = −2.5 × 10 −4 ˆj N = ( 2.5 × 10−4 N, down )

(

)

(

(

(

)

)

)

(

(

)

)

33.37. Model: Two parallel wires carrying currents in the same direction exert attractive magnetic forces on each other. Visualize: Please refer to Figure EX33.37. The current in the circuit on the left is I1 and has a clockwise direction. The current in the circuit on the right is I2 and has a counterclockwise direction. Solve: Since I1 = 9 V 2 Ω = 4.5 A, the force between the two wires is

F = 5.4 × 10−5 N =

μ0 LI1I 2 = 2π d

( 2 ×10

⇒ I2 = 3.0 A ⇒ R =

−7

T m/A ) ( 0.10 m )( 4.5 A ) I 2 0.0050 m

9V = 3.0 Ω 3.0 A

33.38. Model: The torque on a current loop is due to the magnetic field. Visualize:

Solve: (a) We can use the right-hand rule to find the force direction on the G currents at the top, bottom, front, and G back segments of loop 1 and loop 2 in Figure P26.36. We see that Ftop and Fbottom are equal and opposite to each other. G G Ffront and Fback are not seen in the figure above because they are perpendicular to the page. But, they are also equal and opposite toGeach other. G G G G G Thus, Ftop + Fbottom = 0 and Ffront + Fback = 0 . Since the top-bottom or front-back forces act along the same line, they cause no torque. Thus, both the loops are in static equilibrium. G G (b) Now let us rotate each loop slightly and re-examine the forces. The two forces on loop 1 still give Fnet = 0 , but now there is a torque that tends to rotate the left loop back to its upright position. This is a restoring torque, so this loop position is stable. But for loop 2, the torque causes the loop to rotate even further. Any small angular displacement gets magnified into a large displacement until the loop gets flipped over. So, the position of loop 2 is unstable.

33.39. Solve: From Equation 33.28, the torque on the loop exerted by the magnetic field is G

G

G

τ = μ × B ⇒ τ = μ B sin θ = IAB sin θ = ( 0.500 A )( 0.050 m × 0.050 m )(1.2 T ) sin 30° = 7.5 × 10 −4 N m

33.40. Solve: From Equation 33.28, the torque on the bar magnet exerted by the magnetic field is τ 0.020 N m G G G = = 0.28 A m 2 τ =μ×B⇒μ = B sin θ ( 0.10 T ) sin 45°

33.41. Model: The torque on the current loop is due to the magnetic field produced by the current-carrying wire. Assume that the wire is very long. Visualize: Please refer to Figure EX33.41. Solve: (a) From Equation 33.27, the magnitude of the torque on the current loop is τ = μ B sin θ , where μ = Iloop A and B is the magnetic field produced by the current Iwire in the wire. The magnetic field of the wire is G tangent to a circle around the wire. At the position of the loop, B points up and is θ = 90° from the axis of the loop. Thus,

τ = ( I loop A )

( 0.20 A ) π ( 0.0010 m ) ( 2 × 10−7 T m/A ) ( 2.0 A ) sin 90° μ0 I wire sin θ = = 1.26 × 10−11 N m 2π d 2.0 × 10−2 m 2

G Note that the magnetic field produced by the wire on the current loop is up so that the angle θ between B and the normal to the loop is 90°. (b) The loop is in equilibrium when θ = 0° or 180°. That is, when the coil is rotated by ± 90°.

33.42. Model: The magnetic field is that of two long wires that carry current. Visualize:

Solve: (a) For x > +2 cm and for x < −2 cm, the magnetic fields due to the currents in the two wires add. The point where the two magnetic fields cancel lies on the x-axis in between the two wires. Let that point be a distance x away from the origin. Because the magnetic field of a long wire is B = μ0 I 2π r , we have

μ0 ( 5.0 A ) = μ0 ( 3.0 A ) ⇒ 5(0.020 m – x) = 3(0.020 m + x) ⇒ x = 0.0050 m = 0.50 cm 2π ( 0.020 m + x ) 2π ( 0.020 m − x ) (b) The magnetic fields due to the currents in the two wires add in the region −2.0 cm < x < 2.0 cm. For x < −2.0 cm, the magnetic fields subtract, but the field due to the 5.0 A current is always larger than the field due to the 3.0 A current. However, for x > 2.0 m, the two fields will cancel at a point on the x-axis. Let that point be a distance x away from the origin, so μ0 5.0 A μ 3.0 A = 0 ⇒ 5(x − 0.020 m) = 3(x + 0.020 m) ⇒ x = 8.0 cm 2π x + 0.020 m 2π x − 0.020 m

33.44. Model: The resistor and capacitor form an RC circuit. Assume that the length of the wire next to the dot is much larger than 1.0 cm.

Visualize: Please refer to Figure P33.44 Solve: As the capacitor discharges through the resistor, the current through the circuit will decrease as

I= ⇒B=

ε R

50 V −t ( 5 Ω )( 2 μ F) e = (10 A ) e− t 10 μ s 5Ω ( 2.0 × 10−7 T m/A ) (10 A ) e−t 10 μs = 2.0 × 10−4 T e−t 10 μs = ( ) 0.010 m

e −t RC =

μ0 I μ = 0 (10 A ) e− t 10 μs 2π r 2π r

A graph of this equation is shown above.

33.45. Model: Assume that the superconducting niobium wire is very long. Solve:

The magnetic field of a long wire carrying current I is

Bwire =

μ0 I 2π d

We’re interested in the magnetic field of the current right at the surface of the wire, where d = 1.5 mm. The maximum field is 0.10 T, so the maximum current is

I=

( 2π d ) Bwire μ0

=

2π (1.5 × 10−3 m ) ( 0.10 T ) 4π × 10−7 T m/A

= 750 A

Assess: The current density in this superconducting wire is of the order of 1 × 108 A/m2. This is a typical value for conventional superconducting materials.

33.46. Model: Use the Biot-Savart law for a current carrying segment. Visualize: Please refer to Figure P33.46. G Solve: (a) The Biot-Savart law (Equation 33.6) for the magnetic field of a current segment Δs is G μ I ΔsG × rˆ B= 0 4π r 2

where the unit vector rˆ points from current segment Δs to the point, a distance r away, at which we want to G G evaluate the field. For the two linear segments of the wire, Δs is in the same direction as rˆ, so Δs × rˆ = 0. For G G the curved segment, Δs and rˆ are always perpendicular, so Δs × rˆ = Δs. Thus

B=

μ 0 I Δs 4π r 2

Now we are ready to sum the magnetic field of all the segments at point P. For all segments on the arc, the distance to point P is r = R. The superposition of the fields is B=

μ0 I μ IL μ Iθ ds = 0 2 = 0 2 ∫ 4π R arc 4π R 4π R

where L = Rθ is the length of the arc. (b) Substituting θ = 2π in the above expression, Bloop center =

μ0 I 2π μ0 I = 4π R 2 R

This is Equation 33.7, which is the magnetic field at the center of a 1-turn coil.

33.47. Model: Use the Biot-Savart law for a current carrying segment. Visualize: Please refer to Figure P33.47. The distance from P to the inner arc is r1 and the distance from P to the outer arc is r2. G Solve: As given in Equation 33.6, the Biot-Savart law for a current carrying small segment Δs is G μ I ΔsG × rˆ B= 0 4π r 2 G G For the linear segments of the loop, BΔs = 0 T because Δs × rˆ = 0. Consider a segment Δs on length on the inner G arc. Because Δs is perpendicular to the rˆ vector, we have π 2

B=

μ0 I Δs μ0 Ir1Δθ μ0 I Δθ μ Idθ μ0 I μI π= 0 = = ⇒ Barc 1 = ∫ 0 = 2 2 4π r1 4π r1 4π r1 4π r1 4r1 − π 2 4π r1

A similar expression applies for Barc 2 . The right-hand rule indicates an out-of-page direction for Barc 2 and an intopage direction for Barc 1. Thus,

G ⎛μ I ⎤ ⎞ ⎛μ I ⎞ ⎡μ I ⎛ 1 1 ⎞ B = ⎜ 0 , into page ⎟ + ⎜ 0 , out of page ⎟ = ⎢ 0 ⎜ − ⎟ , into page ⎥ 4 r 4 r 4 r r ⎝ 1 ⎠ ⎝ 2 ⎠ ⎣⎢ ⎝ 1 2⎠ ⎦⎥ The field strength is B=

( 4π ×10

G Thus B = (7.9 × 10–5 T, into page).

−7

T m/A ) ( 5.0 A ) ⎛ 1 1 ⎞ −5 − ⎜ ⎟ = 7.9 × 10 T 4 ⎝ 0.010 m 0.020 m ⎠

33.48. Model: Assume that the wire is infinitely long. Visualize: Please refer to Figure P33.48. The wire, looped as it is, consists of a circular part and a linear part. Solve: Using Equation 33.7 and Example 33.3, the magnetic field at P is μI μI BP = Bloop center + Bwire = 0 + 0 2 R 2π R 4π (10−7 T m/A ) ( 5.0 A ) 4π (10−7 T m/A ) ( 5.0 A ) = + = 4.1 × 10−4 T 2 ( 0.010 m ) 2π ( 0.010 m )

33.49. Model: Assume that the solenoid is an ideal solenoid. Solve: The magnetic field of a solenoid is Bsol = μ0 NI l , where l is the length and N the total number of turns of wire. If the wires are wound as closely as possible, the spacing between one turn and the next is simply the diameter dwire of the wire. The number of turns that will fit into length L is N = l d wire . For #18 wire, N = (20 cm)/(0.102 cm) = 196 turns. The current required is I #18

( 0.20 m ) ( 5 × 10−3 T ) LBsol = = = 4.1 A μ0 N ( 4π × 10−7 T m/A ) (196 )

For #26 wire, dwire = 0.41 mm, leading to N = 488 turns and I#26 = 1.63 A. The current that would be needed with #26 wire exceeds the current limit of 1 A, but the current needed with #18 wire is within the current limit of 6 A. So use #18 wire with a current of 4.1 A.

33.50. Model: Assume that the solenoid is an ideal solenoid. Solve: A solenoid field is Bsol = μ0NI/l, so the necessary number of turns is N=

lBsol ( 0.08 m )( 0.10 T ) = 3180 = μ0 I ( 4π × 10−7 T m/A ) ( 2 A )

Problem 33.49 states that the rating is 6 A for a wire with dwire = 1.02 mm and the rating is 1 A for a wire having dwire = 0.41 mm. This means the rating of a wire can be increased 1 A by increasing the diameter by approximately 0.12 mm. We can assume that a wire that can carry a 2 A current will probably have a diameter larger than 0.41 mm + 0.12 mm. Let us take the wire’s diameter to be 0.6 mm. The number of turns that will fit in a length of 8 cm is

N1 layer =

8 cm = 133 0.06 cm

Thus 3180 turns would require 3180/133 = 24 layers. Assess: With this many layers, the thickness of the layers of wires becomes larger than the radius of the solenoid. Maybe not impossible, but it is probably not feasible.

33.51. Model: A 1000-km-diameter ring makes a loop of diameter 3000 km. Visualize:

Solve:

(a) The current loop has a diameter of 3000 km, so its nominal area, ignoring curvature effects, is Aloop = π r2 = π (1500 × 103 m)2 = 7.07 × 1012 m2

Because the magnetic dipole moment of the earth is modeled to be due to a current flowing in such a loop, μ = IAloop. The current in the loop is

I=

μ Aloop

=

8.0 × 1022 A m 2 = 1.13 × 1010 A 7.07 × 1012 m 2

(b) The current density J in the above loop is

J loop =

I 1.13 × 1010 A = = 0.014 A/m 2 A π ( 1 × 1000 × 103 m )2 2

(c) The current density in the wire is

J wire =

I 1.0 A = = 1.3 × 106 A/m 2 A π ( 1 × 1.0 × 10−3 m )2 2

You can see that Jloop << Jwire. The current in the earth’s core is large, but the current density is actually quite small.

33.52. Model: The coils are identical, parallel, and carry equal currents in the same direction. The magnetic field is that of the currents in these coils. Solve: (a) The on-axis magnetic field of an N-turn current loop at a distance z from the loop center is

Bloop =

μ0

NIR 2

2 ( z + R 2 )3 / 2 2

If the spacing between the loops is R, then the midpoint between them is z = R/2. Since the currents are in the same direction, the field of each loop is in the same direction and the net field is simply twice the field due to a single loop. Thus

B=

(R

μ0 NIR 2 2

/4+ R

)

2 3/ 2

= (1.25)3 / 2

μ0 NI R

(b) The magnetic field is

B = ( 0.716 )

4π (10−7 T m/A )10 (1.0 A ) 0.050 m

= 1.80 × 10−4 T

33.53. Model: The magnetic field is that of a current in the wire. Visualize: Please refer to Figure P33.53. G Solve: As given in Equation 33.6 for a current carrying small segment Δs , the Biot-Savart law is G μ I ΔsG × rˆ B= 0 4π r 2 G G For the straight sections, Δs × rˆ = 0 because both Δs and rˆ point along the same line. That is not the case with the G G curved section over which Δs and r are perpendicular. Thus, B=

μ0 I Δs μ0 IR dθ μ0 I dθ = = 4π r 2 4π R 2 4π R

where we used Δs = RΔθ ≈ R dθ for the small arc length Δs. Integrating to obtain the total magnetic field at the center of the semicircle, π /2

B= −

μ0 I dθ μ0 I μI = π= 0 4π R 4π R 4R /2

∫ π

33.54. Model: The toroid may be viewed as a solenoid that has been bent into a circle. Visualize: Please refer to Figure P33.54. Solve: (a) A long solenoid has a uniform magnetic field inside and it is roughly parallel to the axis. If we bend the solenoid to make it circular, we will have circular magnetic field lines around the inside of the toroid. However, as explained in part (c) the field is not uniform. (b)

A top view of the toroid is shown. Current is into the page for the inside windings and out of the page for the outside windings. The closed Ampere’s path of integration thus contains a current NI where I is the current flowing through the wire and N is the number of turns. Applied to the closed line path, the Ampere’s law is G G úB ⋅ d s = μ0 I through = μ0 NI G G Because B and d s are along the same direction and B is the same along the line integral, the above simplifies to Búds = B 2π r = μ0 NI ⇒ B =

μ0 NI 2π r

(c) The magnetic field for a toroid depends inversely on r which is the distance from the center of the toroid. As r increases from the inside of the toroid to the outside, Btoroid decreases. Thus the field is not uniform.

33.55.

Model: Visualize:

The magnetic field is that of the current which is distributed uniformly in the hollow wire.

Ampere’s integration paths are shown in the figure for the regions 0 m < r < R1, R1 < r < R2, and R2 < r. G G Solve: For the region 0 m < r < R1, úB ⋅ d s = μ0 I through . Because the current inside the integration path is zero,

B = 0 T. To find Ithrough in the region R1 < r < R2, we multiply the current density by the area inside the integration path that carries the current. Thus, I I through = π r 2 − R12 2 π R2 − R12

(

)

(

)

where the current density is the first term. Because the magnetic field has the same magnitude at every point on the circular path of integration, Ampere’s law simplifies to

G G I (r úB ⋅ d s = B ∫ ds = B ( 2π r ) = μ ( R 0

2

2 2

− R12 )

− R12 )

⇒B=

μ0 I ⎛ r 2 − R12 ⎞ ⎜ ⎟ 2π r ⎝ R22 − R12 ⎠

For the region R2 < r, Ithrough is simply I because the loop encompasses the entire current. Thus, G G μI úB ⋅ d s = B ∫ ds = B 2π r = μ0 I ⇒ B = 2π0 r Assess: The results obtained for the regions r > R2 and R1 < r < R2 yield the same result at r = R2. Also note that a hollow wire and a regular wire have the same magnetic field outside the wire.

33.56. Model: The electron is a point mass undergoing cyclotron motion. Solve:

The electron’s angular momentum is L=mvr. The cyclotron radius is r =

mv . Multiplying the latter qB

equation by r and combining it with the expression for L, we have

r2 =

L L ⇒ = qB qB

8.0 × 10−26 kg m 2 /s = 0.010 m = 1.0 cm (1.60 ×10−19 C )( 5.0 ×103 T )

33.57. Model: Magnetic fields exert a force on moving charges. Visualize:

G G The right-hand rule applied to the first proton requires B lie towards the +y-axis from v1 , while when G G G applied to the second proton requires B lie towards the +x-axis from v2 . Thus B lies in the first quadrant of the xy-plane. The force on each proton is F1 = qv1B sin α F2 = qv2 B sin ( 90° − α ) = qv1B cosα Solve:

⎛ 2.00 × 106 m/s 1.20 × 10−16 N ⎞ F1 v1 = tan α ⇒ α = tan −1 ⎜ ⋅ ⎟ = 30° −16 6 F2 v2 ⎝ 1.00 × 10 m/s 4.16 × 10 N ⎠ The magnetic field strength is thus F1 1.20 × 10−16 N B= = = 1.50 × 10−3 T −19 qv1 sin α (1.60 × 10 C )(1.00 × 106 m/s ) sin 30° G Thus B = (1.50 mT, 30o ccw from +x-axis). ⇒

33.58. Visualize: Please refer to Figure P33.58. Solve:

The electric field is

G ⎛ 200 V ⎞ E =⎜ , down ⎟ = ( 20,000 V/m, down ) ⎝ 1 cm ⎠ G G G The force this field exerts on the electron is Felec = qE = −eE = ( 3.2 × 10−15 N, up ) . The electron will pass through G G G without deflection if the magnetic field also exerts a force on the electron such that Fnet = Felec + Fmag = 0 N . That is, G Fmag = (3.2 × 10 −15 N, 3.2 × 10 −15 N, down). In this case, the electric and magnetic forces cancel each other. For a G G G negative charge with v to the right to have Fmag down requires, from the right-hand rule, that B point into the page. The magnitude of the magnetic force on a moving charge is Fmag = qvB, so the needed field strength is

B=

Fmag ev

=

3.2 × 10−15 N = 2.0 × 10−3 T = 2.0 mT (1.60 ×10−19 C )(1.0 × 107 m/s )

G Thus, the required magnetic field is B = (2.0 mT, into page).

33.59. Model: Energy is conserved as the electron moves between the two electrodes. Assume the electron starts from rest. Once in the magnetic field, the electron moves along a circular arc. Visualize:

The electron is deflected by 10° after moving along a circular arc of angular width 10°. Solve: Energy is conserved as the electron moves from the 0 V electrode to the 10,000 V electrode. The potential energy is U = qV with q = –e, so K f + U f = Ki + U i ⇒ 12 mv 2 − eV = 0 + 0

v=

2eV 2(1.60 × 10−19 C)(10,000 V) = = 5.93 × 107 m/s 9.11× 10−31 kg m

The radius of cyclotron motion in a magnetic field is r = mv/eB. From the figure we see that the radius of the circular arc is r = (2.0 cm)/sin10°. Thus mv (9.11 × 10−31 kg)(5.93 × 107 m/s) B= = = 2.9 × 10−3 T er (1.60 × 10−19 C)(0.020 m)/sin10°

33.60. Model: Charged particles moving perpendicular to a uniform magnetic field undergo uniform circular motion at a constant speed. Solve: (a) From Equation 33.20, the magnetic field is 9 −31 2π fm 2π ( 2.4 × 10 Hz )( 9.11 × 10 kg ) = = 0.086 T = 86 mT e 1.6 × 10−19 C (b) The maximum kinetic energy is for an orbit with radius 1.25 cm.

B=

2 K = 12 mv 2 = 12 m ( 2π rf ) = 12 ( 9.11 × 10−31 kg ) ⎡⎣ 2π ( 0.0125 m ) ( 2.4 × 109 Hz ) ⎤⎦ = 1.62 × 10−14 J 2

33.61. Model: Electric and magnetic fields exert forces on a moving charge. The fields are uniform throughout the region. Visualize: Please refer to Figure P33.61. Solve: (a) We will first find the net force on the antiproton, and then find the net acceleration using Newton’s second law. The magnitudes of the electric and magnetic forces are

FE = eE = (1.60 × 10−19 C ) (1000 V/m ) = 1.60 × 10−16 N FB = evB = (1.60 × 10−19 C ) ( 500 m/s )( 2.5 T ) = 2.00 × 10−16 N G The directions of these two forces on the antiproton are opposite. FE points up whereas, using the right-hand G rule, FB points down. Hence, G Fnet = ( 2.0 × 10−16 N − 1.60 × 10−16 N, down ) ⇒ Fnet = 0.40 × 10−16 N = ma

⎞ G ⎛ 0.40 × 10−16 N ⇒ a =⎜ = 2.4 × 1010 m/s 2 , down ⎟ −27 1.67 10 kg × ⎝ ⎠ G G G (b) If v were reversed, both FE and FB will point up. Thus, G ⎞ G ⎛ 3.6 × 10−16 N Fnet = (1.6 × 10−16 N + 2.0 × 10−16 N, up ) ⇒ a = ⎜ = 2.2 × 1011 m/s 2 , up ⎟ −27 ⎝ 1.67 × 10 kg ⎠

G

G

G

33.62. Model: A magnetic field exerts a magnetic force on a moving charge given by F = qv × B. Assume the magnetic field is uniform. Visualize: Please refer to Figure P33.62. The magnetic field points in the +z-direction. If the charged particle is G G G moving along B, F = 0 N. If v is perpendicular to B, the motion of the charged particle is a circle. However, G G when v makes an angle with B, the motion of the charged particle is like a helix or a spiral. The perpendicular component of the velocity is responsible for the circular motion, and the parallel component is responsible for the linear motion along the magnetic field direction. Solve: From the figure we see that v y = v cos30° and vz = v sin 30°. For the circular motion, the magnetic force

causes a centripetal acceleration. That is,

ev y B =

mv y2 r

⇒ r=

mv y eB

( 9.11×10 kg )( 5.0 ×10 m/s ) cos30° = 0.82 mm (1.60 ×10 C ) ( 0.030 T ) −31

=

6

−19

The time for one revolution is T=

2π ( 8.2 × 10−4 m ) 2π r = = 1.19 × 10−9 s vy ( 5.0 ×106 m/s ) cos30°

The pitch p is the vertical distance covered in time T. We have

p = vzT = ( 5.0 × 106 m/s ) sin 30° (1.19 × 10−9 s ) = 3.0 × 10−3 m = 3.0 mm

33.63. Model: Charged particles moving perpendicular to a uniform magnetic field undergo uniform circular motion at constant speed. Solve: (a) The magnetic force on a proton causes a centripetal acceleration:

evB =

mv 2 eBr ⇒v= r m

Maximum kinetic energy is achieved when the diameter of the proton’s orbit matches the diameter of the cyclotron: −19 e 2 B 2 r 2 (1.60 × 10 C ) ( 0.75 T ) ( 0.325 m ) K = mv = = = 4.6 × 10−13 J 2m 2 (1.67 × 10−27 kg ) 2

1 2

2

2

2

(b) The proton accelerates through a potential difference of 500 V twice during one revolution. The energy gained per cycle is

2 qΔV = 2e (500 V) = 1.60 × 10−16 J Using the maximum kinetic energy of the proton from part (a), the number of cycles before the proton attains this energy is 4.6 × 10−13 J = 2850 1.60 × 10−16 J

33.64. Model: Charged particles moving perpendicular to a uniform magnetic field undergo circular motion at constant speed. Visualize: Please refer to Figure P33.64. Solve: The potential difference causes an ion of mass m to accelerate from rest to a speed v. Upon entering the magnetic field, the ion follows a circular trajectory with cyclotron radius r = mv/eB. To be detected, an ion’s trajectory must have radius d = 2r = 8 cm. This means the ion needs the speed v=

eBr eBd = m 2m

This speed was acquired by accelerating from potential V to potential 0. We can use the conservation of energy equation to find the voltage that will accelerate the ion: K1 + U1 = K2 + U2 ⇒ 0 J + eΔV = 12 mv 2 + 0 J ⇒ ΔV =

mv 2 2e

Using the above expression for v, the voltage that causes an ion to be detected is 2

ΔV =

mv 2 m ⎛ eBd ⎞ eB 2 d 2 = ⎜ ⎟ = 2e 2e ⎝ 2m ⎠ 8m

An ion’s mass is the sum of the masses of the two atoms minus the mass of the missing electron. For example, the mass of N +2 is m = mN + mN – melec = 2(14.0031 u)(1.661 × 10−27 kg/u) – 9.11 × 10−31 kg = 4.65174 × 10−26 kg Note that we’re given the atomic masses very accurately in Exercise 28. We need to retain this accuracy to tell the difference between N +2 and CO+. The voltage for N +2 is

(1.6 ×10 C ) ( 0.200 T ) ( 0.08 m ) 8 ( 4.65174 × 10 kg ) 2

−19

ΔV =

= 110.07 V

Ion

Mass (kg)

N +2

4.65174 × 10−26

110.07

+ 2

−26

96.36 110.11

O

+

CO Assess:

−26

2

Accelerating voltage (V)

5.31341 × 10 4.64986 × 10−26

The difference between N +2 and CO+ is not large but is easily detectable.

33.65. Model: Assume that the magnetic field is uniform over the Hall probe. Solve: Equation 33.24 gives the Hall voltage and Equation 33.20 gives the cyclotron frequency in terms of the magnetic field. We have ΔVH =

⇒ tne =

2π mf cyc I eΔVH

=

2π (1.67 × 10

−27

IB tne

B=

2π mf cyl e

kg )(10.0 × 10 Hz )( 0.150 × 10−3 A )

(1.60 × 10

6

−19

C )( 0.543 × 10−3 V )

= 0.1812 T A/V

With this value of tne, we can once again use the Hall voltage equation to find the magnetic field:

⎛ 1.735 × 10−3 V ⎞ ⎛ ΔV ⎞ B = ⎜ H ⎟ tne = ⎜ ⎟ ( 0.1812 TA/V ) = 2.10 T −3 ⎝ I ⎠ ⎝ 0.150 × 10 A ⎠

33.66. Model: The loop will not rotate about the axle if the torque due to the magnetic force on the loop balances the torque of the weight. Visualize: Please refer to Figure P33.66. G Solve: The rotational equilibrium condition Στ net = 0 N m is about the axle and means that the torque from the weight is equal and opposite to the torque from the magnetic force. We have

( 50 ×10 kg ) g ( 0.025 m ) = μ B sin 90° = ( NIA) B ( 50 × 10 kg )( 9.8 m/s ) ( 0.025 m ) = 0.123 T ⇒B= −3

−3

2

(10 )( 2.0 A )( 0.050 m )( 0.100 m )

Assess:

The current in the loop must be clockwise for the two torques to be equal.

33.67. Model: A magnetic field exerts a magnetic force on a length of current-carrying wire. We ignore gravitational effects, and focus on the B effects. Visualize: Please refer to Figure P33.67. The figure shows a wire in a magnetic field that is directed out of the page. The magnetic force on the wire is therefore to the right and will stretch the springs. Solve: In static equilibrium, the sum of the forces on the wire is zero: FB + Fsp 1 + Fsp 2 = 0 N ⇒ ILB + (−kΔx) + (−kΔx) ⇒ I =

2k Δx 2 (10 N/m )( 0.01 m ) = = 2.0 A LB ( 0.20 m )( 0.5 T )

33.68. Model: The bar is a current-carrying wire in a perpendicular uniform magnetic field. The current is constant. Visualize: Please refer to figure P33.68. Solve: (a) The right-hand rule as described in section 33.8 requires the current to be into the page. (b) The net force on the bar is F = IlB, and is constant throughout the motion. The acceleration of the bar is F IlB = thus . Using constant acceleration kinematics, m m 2 IlBd 2 ⎛ IlB ⎞ v 2f = v02 + 2aΔs = ( 0 m/s ) + 2 ⎜ ⎟d ⇒ vf = m ⎝ m ⎠

33.69. Model: A magnetic field exerts a magnetic force on a length of current-carrying wire. Visualize:

Please refer to Figure P33.69. Solve: The above figure shows a side view of the wire, with the current moving into the page. From the rightG hand rule, the magnetic field B points down to give a leftward force on the current. The wire is hanging in static G G G G equilibrium, so Fnet = Fmag + FG + T = 0 N. Consider a segment of wire of length L. The wire’s linear mass

density is μ = 0.050 kg/m, so the mass of this segment is m = μL and its weight is FG = mg = μLg. The magnetic force on this length of wire is Fmag = ILB. In component form, Newton’s first law is

( Fnet ) x = T sin θ − Fmag = T sin θ − ILB = 0 N ⇒ T sin θ = ILB ( Fnet ) y = T cosθ − FG = T cosθ − μ Lg = 0 N

⇒ T cosθ = μ Lg

Dividing the first equation by the second, 2 μ g tan θ ( 0.050 kg/m ) ( 9.8 m/s ) tan10° ⎡ T sin θ ⎤ ⎡ ILB IB ⎤ = = 0.0086 T ⎢ T cosθ = tan θ ⎥ = ⎢ μ Lg = μ g ⎥ ⇒ B = 10 A I ⎣ ⎦ ⎣ ⎦ G The magnetic field is B = ( 8.6 mT, down ) .

33.70. Model: The wire will float in the magnetic field if the magnetic force on the wire points upward and has a magnitude mg, allowing it to balance the downward gravitational force. Visualize:

Solve: Each lower wire exerts a repulsive force on the upper wire because the currents are in opposite directions. The currents are of equal magnitude and the distances are equal, so F1 = F2 . Consider segments of the wires of length L. Then the forces are

F1 = F2 =

μ0 LI 2 2π d

The horizontal components of these two forces cancel, so the net magnetic force is upward and of magnitude Fmag = 2 F1 cos30° =

μ0 LI 2 cos30° πd

In equilibrium, this force must exactly balance the downward weight of the wire. The wire’s linear mass density is μ = 0.050 kg/m, so the mass of this segment is m = μL and its weight is w = mg = μLg. Equating these gives

μ0 LI 2 cos30° μ gπ d = μ Lg ⇒ I = = πd μ0 cos30°

( 0.050 kg/m ) ( 9.8 m/s 2 )π ( 0.040 m )

( 4π ×10

−7

T m/A ) cos30°

= 238 A

33.71. Model: The coils of wire are wrapped around the center of the cylinder, and are wrapped tightly enough so that the loops all have the same area. Visualize: Please refer to figure P33.71.

Solve: (a) For equilibrium, the gravitational torque on the cylinder is balanced by the torque on the current loops exerted by the magnetic field. The magnetic torque is τ B = μ B sin θ = 10 ( 2 R ) LIB sin θ . G G Note that θ is the angle between the magnetic dipole μ of the loops and B. This is balanced by the gravitational torque τ G . Using the figure above,

τ G = FG R sin θ = (π R 2 ) Lρ gR sin θ = π R 3 Lρ g sin θ .

Equilibrium requires

τ B = τG 20 RLIB sin θ = π R 3 Lρ g sin θ ⇒I =

(b) Evaluating,

π R2ρ g 20 B

π ( 0.025 m ) ( 2500 kg/m3 )( 9.8 m/s 2 ) 2

I=

20 ( 0.25 T )

= 9.6 A

Assess: This is a reasonable current. Note that the angle of the incline does not matter since both the gravitational and magnetic torques have the same dependence on θ.

33.72. Model: Assume that the 20 cm is large compared with the effective radius of the magnetic dipole (bar magnet). Solve: From Equation 33.9, the on-axis field of a magnetic dipole at a distance z from the dipole is Bdipole =

μ0 2 μ 4π z 3

The magnetic dipole moment can be obtained from the magnitude of the torque. We have G

G

G

τ =μ×B⇒μ =

τ B sin θ

=

0.075 Nm

( 0.50 T ) sin 90°

= 0.15 Am 2

Thus, the magnetic field produced by the magnet is

Bdipole = (10−7 T m/A )

2 ( 0.15 Am 2 )

( 0.20 m )

3

= 3.8 × 10−6 T

33.73. Model: A current loop produces a magnetic field. Visualize:

Solve: The field at the center of a current loop is Bloop = μ0I/2R. The electron orbiting an atomic nucleus is, on average, a small current loop. Current is defined as I = Δq/Δt. During one orbital period T, the charge Δq = e goes around the loop one time. Thus the average current is Iavg = e/T. For circular motion, the period is T=

−11 2π R 2π ( 5.3 × 10 m ) 1.60 × 10−19 C = = 1.514 × 10−16 s ⇒ I avg = = 1.057 × 10−3 A 6 v 2.2 × 10 m/s 1.514 × 10−16 s

Thus, the magnetic field at the center of the atom is

Bcenter =

μ0 I avg 2R

=

( 4π ×10

−7

T m/A )(1.057 × 10−3 A )

2 ( 5.3 × 10−11 m )

= 12.5 T

33.74. Model: The superposition principle holds for forces and torques. Visualize:

Solve: (a) A graph of B vs x is shown above. (b) The magnetic force on a segment of wire Δx in length located at xi is

ΔFi = IB ( xi ) Δx The total force on the wire is L

Fnet = ΣΔFi = ΣIB ( xi ) Δx → ∫ IBdx = ∫ IB0 0

x 1 dx = ILB0 L 2

G 1 The right-hand rule gives the direction of the force along the +y-axis, so Fnet = ILB0 ˆj. 2 (c) Each segment of wire experiences a torque about the origin of

τ i = ri ΔFi sin 90° = xi IB ( xi ) Δx. The total torque on the wire is thus L

τ = Στ i = Σxi IB ( xi ) Δx → ∫ IB0 0

x2 1 dx = IL2 B0 L 3

The direction of the torque is along the +z-axis by the right-hand rule, indicating that the magnetic field will exert a torque causing the wire to rotate counter clockwise in the xy-plane.

33.75. Model: All the domains representing a byte are the same size. Solve:

The total number of domains needed is Nd = 8(250×109) = 2.0×1012

The area available for each domain is

π ( 4.75 × 10−2 m ) − π (1.25 × 10−2 m ) 2

A=

2.0 × 1012

2

= 3.3 × 10−15 m 2

Since the domains are square, the edge-length of a square with this area is

L = 3.3 × 10−15 m2 = 5.7 × 10−8 m = 57 nm Assess: This is much smaller than the typical domain size in a ferromagnet of 0.1 mm. However, the cobalt alloy is able to maintain much smaller domains than iron.

33.76. Model: A magnetic field exerts a force on a segment of current. Visualize: The figure shows the forces on two small segments of current, one in which the current enters the plane of the page and one in which the current leaves the plane of the page.

Solve: (a) Consider a small segment of the loop of length Δs. The magnetic force on this segment is perpendicular both to the current and to the magnetic field. The figure shows two segments on opposite sides of the loop. The horizontal components of the forces cancel but the vertical components combine to give a force toward the bar magnet. The net force is the sum of the vertical components of the force on all segments around the loop. For a segment of length Δs, the magnetic force is F = IBΔs and the vertical component is Fy = (IBΔs)sinθ. Thus the net force on the current loop is

Fnet = ∑ Fy = ∑ ( IBΔs )sin θ = IB sin θ ∑ Δs We could take sinθ outside the summation because θ is the same for all segments. The sum of all the Δs is simply the circumference 2πR of the loop, so (b) The net force is

Fnet = 2πRIBsinθ Fnet = 2π (0.020 m)(sin 20°)(0.50 A)(200 × 10−3 T) = 4.3 × 10−3 N

33.77. Solve: From Equation 33.7, the magnetic field at the center (z = 0 m) of an N-turn coil is μ NI μ0 1.0 m I Bcoil center = 0 = 2 R 2 2π R R where we used the requirement that the entire length of wire be used. That is, 1.0 m = (2πR)N. Solving for R,

R=

2π (10−7 T m/A ) (1.0 m )(1.0 A ) μ0 (1.0 m ) I = = 0.010 m = 1.0 cm 2 2π Bcoil center 2π (1.0 × 10−3 T )

33.78. Model: The magnetic field is that of a finite length of current-carrying wire. Solve: (a) Let the wire be on the x-axis and the point of interest be at distance d on the y-axis. The wire extends from x = −L/2 to x = +L/2. As explained in Example 33.3 and Figure 33.13, each segment dx of the wire generates a field at the observation point of strength μ0 Id dx dB = 3/ 2 4π x 2 + d 2

(

)

These infinitesimal fields all point the same direction, so we can add them (superposition) to find the total field. The sum becomes an integral, giving L/2

μ Id L / 2 dx μ Id x = 0 B= 0 3/ 2 ∫ 2 2 2 2 4π − L / 2 ( x + d ) 4π d ( x + d 2 )1/ 2

= −L / 2

μ0 IL 4π d

( L / 2)

2

+ d2

(b)

A square is formed of four straight-wire segments, with each segment contributing the same field at the center. The length is L = 2R, and the observation point is distance d = R from the center of each segment. Using the formula from part (a),

Bsquare = 4 Bseg = 4 ×

μ0 I ( 2 R ) 4π R

( 2R / 2)

2

+R

2

= 4×

μ0 I 2μ0 I = πR 2 2π R

(c) The ratio of the field strengths of a square loop and a circular loop, each with “radius” R, is

Bsquare Bcircle

=

2 μ0 I π R 2 2 = = 0.900 = 90.0% μ0 I 2 R π

The formula for the magnetic field at the center of a circular loop is from Equation 33.7.

33.79. Model: The spinning charged disk forms a series of concentric current loops. Solve:

The magnetic field at the center of a current loop of radius r is taken from Example 33.5 with z = 0:

Bloop =

μ0 I 2r

Consider a ring of charge on the disk at radius ri with width Δr. The charge on the ring is ΔQ = time for one revolution of the ring is Δt =

2π

ω

Ii =

, so the ring can be considered a current loop with

ΔQ 2Qri Δr ⎛ ω = ⎜ Δt R 2 ⎝ 2π

⎞ ωQ r Δr ⎟= 2 i ⎠ πR

Each ring contributes a field (Bloop)i, so by superposition

Bcenter = ∑ ( Bloop ) = ∑ i

μ0 I i 2ri

=∑

μ0ωQ μ ωQ dr = 0 2 2π R 2π R 0

R

→∫

μ0ωQ Δr 2π R 2

2π ri Δr Q . The π R2

33.80. Model: The net force is the sum of the forces on individual segments of the wire. Visualize:

Solve:

Please refer to figure CP33.80.

The magnetic field due to the wire carrying the current I1 = 10 A at a distance r from the wire is

B(r) =

μ0 I1 2π r

The force on a segment of the horizontal wire located at xi of length Δx and a distance ri from the 10 A current is

ΔFi = I 2 ΔxB⊥ ( ri ) = I 2 ΔxB ( ri ) sin θi =

μ0 I1I 2 sin θi Δx 2π ri

Note that only the perpendicular component of the magnetic field contributes to the force on the wire. From the x 2 figure, we can identify sin θi = i and ri = ( 0.010 m ) + xi 2 . The total force is the sum of the forces on each ri segment:

F = ∑ ΔFi = =

μ0 I1I 2 xi μ I I 0.050 m xdx Δx → 0 1 2 ∫ ∑ 2 2π 2π 0 m ( 0.010 m )2 + x 2 ri

(

μ0 I1I 2 1 2 ln x 2 + ( 0.010 m ) 2π 2

= (10

−7

)

= 0m

μ0 I1I 2 ⎛ ( 0.050 m ) + ( 0.010 m ) ln ⎜ 2 ⎜ 2π ( 0.010 m ) ⎝ 2

0.050 m

Tm/A ) (10 A )( 5.0 A ) ln ( 26 ) = 1.63 × 10−5 N

2

⎞ ⎟ ⎟ ⎠

Assess: This is a fairly small force, but can be understood with the realization that at the closest points between the wires, where B is strongest, its direction is nearly parallel to the horizontal wire, so the force it exerts on the wire is nearly zero there.

33.81. Model: The magnetic field is that of a conducting wire that has a nonuniform current density. Visualize:

Solve: is

(a) Consider a small circular disk of width dr at a distance r from the center. The current through this disk 2π J 0 r 2 dr ⎛r⎞ di = JdA = J 0 ⎜ ⎟ ( 2π r ) dr = R ⎝R⎠

Integrating this expression, we get R

2π J 0 2 2π J 0 ⎡ r 3 ⎤ 2π J 0 R 2 3I r dr = ⇒ J0 = ⎢ ⎥ = 3 3 2 R R π R2 ⎣ ⎦0 0

R

I = ∫ di = ∫ (b) Applying

G

G

úB ⋅ d s = μ I

0 through

to the circular path of integration, we note that the wire has perfect cylindrical

symmetry with all the charges moving parallel to the wire. So, the magnetic field must be tangent to circles that are concentric with the wire. The enclosed current is the current within radius r. Thus, r

2π J 0 2 μ 2π J 0 r 3 ⎛ μ0 2π ⎞⎛ 3I ⎞ r 3 μ0 Ir 2 =⎜ ⇒ = B r dr ⇒ B ( 2π r ) = 0 ⎟⎜ ⎟ R R 3 ⎝ R ⎠⎝ 2π R 2 ⎠ 3 2π R 3 0 r

∫ Bds = μ0 ∫ di = μ0 ∫ 0

(c) At r = R,

μ0 IR 2 μ0 I = 2π R 3 2π R This is the same result as obtained in Example 33.3 for the magnetic field of a long straight wire. B=

33.82. Model: The magnetic field is that of a coaxial cable consisting of a solid inner conductor surrounded by a hollow outer conductor of essentially zero thickness. Visualize: Please refer to Figure CP33.82. The solid inner conductor and the hollow outer conductor carry equal currents but in opposite directions. The coaxial cable has perfect cylindrical symmetry. So the magnetic field must be tangent to circles that are concentric with the wire. Solve: (a) Ampere’s law is G G μ0 I through úB ⋅ d s = μ0 I through = B ∫ ds = B ( 2π r ) ⇒ B = 2π r For r < R1, ⎛ I ⎞ 2 Ir 2 μ ⎛ Ir ⎞ I through = ⎜ πr = 2 ⇒ B = 0 ⎜ 2 ⎟ 2 ⎟ R1 2π ⎝ R1 ⎠ ⎝ π R1 ⎠

For R1 < r < R2, Ithrough = I. Hence, B = μ0 I 2π r . For r > R2, Ithrough = 0 A and B = 0 T. (b) In the region r < R1, B is linearly proportional to r; in the region R1 < r < R2, B is inversely proportional to r; and in the region for r > 2R2, B = 0 T. A B versus r graph in the range r = 0 m to r = 2R2 is shown below.

33.83. Model: The current sheet extends infinitely far in both directions. Visualize: (a) The shape of the magnetic field must be consistent with the symmetry of the current sheet. Consider two current-carrying wires. The magnetic field above the midpoint of the wires is horizontal and to the left. The magnetic field below the midpoint of the wires is horizontal and to the right. If we imagine the current sheet to consist of many such pairs of wires, closely spaced, then we see that the magnetic field above the sheet is everywhere horizontal and to the left while the magnetic field below the sheet is everywhere horizontal and to the right. The figure shows a closed path for using Ampere’s law.

Solve: (b) The closed path of width L shown in the figure is parallel to the field along the top and bottom edges, perpendicular to the field along the left and right edges. Thus the line integral in Ampere’s law is G G G G G G G G G G úB ⋅ d s = B ⋅ d s + B ⋅ d s + B ⋅ d s + B ⋅ d s = BL + 0 + BL + 0 = 2BL

∫

top

∫

left

∫

bottom

∫

right

Because the current per unit width is Js, the amount of current in a length L of the current sheet (the current through the closed path) is Ithrough = JsL. Thus Ampere’s law gives G G úB ⋅ d s = μ0 I through ⇒ 2BL = μ0 J s L ⇒ B = 12 μ0 J s Assess: The magnetic field strength is independent of the distance from the sheet. This is not unexpected since the electric field of an infinite plane of charge is independent of the distance from the plane.

33-1

34.1.

Visualize:

To develop a motional emf the magnetic field needs to be perpendicular to both, so let’s say its direction is into the page. Solve: This is a straightforward use of Equation 34.3. We have

v= Assess:

E 1.0 V = = 2.0 × 104 m/s lB (1.0 m ) ( 5.0 × 10−5 T )

This is an unreasonable speed for a car. It’s unlikely you’ll ever develop a volt.

34.2.

Model: Assume the magnetic field is uniform. Visualize: Please refer to Figure Ex34.2. Since a motional emf was developed the field must be perpendicular G to v . The positive charges experienced a magnetic force to the left. By the right-hand rule the field must be out G G of the page so that v × B is to the left. Solve: This is a straightforward use of Equation 34.3. We have

B= Assess:

E 0.050 V = = 0.10 T vl ( 5.0 m/s )( 0.10 m )

This is reasonable. Laboratory fields are typically up to a few teslas in magnitude.

34.3.

Visualize:

The wire is pulled with a constant force in a magnetic field. This results in a motional emf and produces a current in the circuit. From energy conservation, the mechanical power provided by the puller must appear as electrical power in the circuit. Solve: (a) Using Equation 34.6,

P = Fpullv ⇒ v =

P 4.0 W = = 4.0 m/s Fpull 1.0 N

(b) Using Equation 34.6 again,

P= Assess:

v 2l 2 B 2 ⇒B= R

RFpull vl

2

=

( 0.20 Ω )(1.0 N ) 2 ( 4.0 m/s )( 0.10 m )

This is reasonable field for the circumstances given.

= 2.2 T

34.4.

Model: Assume the field changes abruptly at the boundary between the two sections. Visualize: Please refer to Figure Ex34.4. The directions of the fields are opposite, so some flux is positive and some negative. The total flux is the sum of the flux in the two regions. G Solve: The field is constant in each region so we will use Equation 34.10. Take A to be into the page. Then, it is parallel to the field in the left region so the flux is positive, and it is opposite to the field in the right region so the flux is negative. The total flux is G G Φ = A ⋅ B = AL BL cosθ L + AR BR cosθ R

= ( 0.20 m ) ( 2.0 T ) − ( 0.20 m ) (1.0 T ) = 0.040 Wb 2

2

Assess: The flux is positive because the areas are equal and the stronger field is parallel to the normal of the surface.

34.5.

Model: Consider the solenoid to be long so the field is constant inside and zero outside. Visualize: Please refer to Figure Ex34.5. The field of a solenoid is along the axis. The flux through the loop is only nonzero inside the solenoid. Since the loop completely surrounds the solenoid, the total flux through the loop will be the same in both the perpendicular and tilted cases. G Solve: The field is constant inside the solenoid so we will use Equation 34.10. Take A to be in the same direction as the field. The magnetic flux is G G G G 2 Φ = Aloop ⋅ Bloop = Asol ⋅ Bsol = π rsol2 Bsol cosθ = π ( 0.010 m ) ( 0.20 T ) = 6.3 × 10−5 Wb G G When the loop is tilted the component of B in the direction of A is less, but the effective area of the loop surface through which the magnetic field lines cross is increased by the same factor.

34.6.

Model: Assume the field is uniform inside the rectangular region and zero outside. Visualize: The flux measures how much of the field penetrates the chosen surface. We will break the surface up because the magnetic field has different strengths over different parts of the surface of the loop. Solve: For convenience we choose the normal to the loop to be into the page so it is in the same direction as the magnetic field. The total flux is G G G G G G Φ = ∫ B ⋅ dA = ∫ B ⋅ dA + ∫ B ⋅ dA = ∫ BdA cosθ = BAinside = Bab loop

Assess:

inside rectangle

outside rectangle

inside rectangle

rectangle

Only the region with nonzero magnetic field inside the loop contributes to the flux.

34.7.

Model: Assume the field strength is uniform over the loop. Visualize: Please refer to Figure Ex34.7. According to Lenz’s law, the induced current creates an induced field that opposes the change in flux. Solve: The original field is into the page within the loop and is changing strength. The induced, counterclockwise current produces a field out of the page within the loop that is opposing the change. This implies that the original field must be increasing in strength so the flux into the loop is increasing.

34.8.

Model: Assume that the magnet is a bar magnet with field lines pointing away from the north end. Visualize: As the magnets move, if they create a change in the flux through the solenoid, there will be an induced current and corresponding field. According to Lenz’s law, the induced current creates an induced field that opposes the change in flux. Solve: (a) When magnet 1 is close to the solenoid there is flux to the left through the solenoid. As magnet 1 moves away there is less flux to the left. The induced current will oppose this change and produce an induced current and a corresponding flux to the left. By the right-hand rule, this corresponds to a current in the wire into the page at the top of the solenoid and out of the page at the bottom of the solenoid. So, the current will be right to left in the resistor. (b) When magnet 2 is close to the solenoid the diverging field lines of the bar magnet produce a flux to the left in the left half of the solenoid and a flux to the right in the right half. Since the flux depends on the orientation of the loop, the flux on the two halves have opposite signs and the net flux is zero. Moving the magnet away changes the strength of the field and flux, but the total flux is still zero. Thus there is no induced current.

34.9.

Visualize: Please refer to Figure Ex34.9. The changing current in the solenoid produces a changing flux in the loop. By Lenz’s law there will be an induced current and field to oppose the change in flux. Solve: The current shown produces a field to the right inside the solenoid. So there is flux to the right through the surrounding loop. As the current in the solenoid increases there is more field and more flux to the right through the loop. There is an induced current in the loop that will oppose the change by creating an induced field and flux to the left. This requires a counterclockwise current.

34.10.

Model: Assume the plane of the loop is perpendicular to the field direction. Visualize: Please refer to Figure Ex34.10. The flux is due to the field through the area of the triangle. Only the left half gives a contribution as the field strength is zero on the right half. G G G Solve: (a) The flux is Φ = A ⋅ B . Take A to be into the page, perpendicular to the triangle, and thus parallel G to B . In this case Φ = AB where A is the area of half of the triangle. This smaller triangle has a base of 10 cm and height 20sin 60° cm = 17.32 cm. Thus, 1 Φ = AB = (0.10 m)(0.1732 m) × 0.10 T = 8.7 × 10−4 Wb 2

(b) The flux is directed into the loop. According to Lenz’s law, the induced current will try to prevent the decrease of flux. To do this, the field of the induced current will have to point into this loop. This requires the induced current to flow clockwise.

34.11.

Model: Assume the field is uniform. Visualize: Please refer to Figure Ex34.11. If the changing field produces a changing flux in the loop there will be a corresponding induced emf and current. Solve: (a) The induced emf is E = d Φ/dt and the induced current is I = E /R. The field B is changing, but the G G area A is not. Take A to be out of the page and parallel to B , so Φ = AB. Thus, E= A

dB dB 2 = π r2 = π ( 0.050 m ) ( 0.50 T/s ) = 3.9 × 10 −3 V = 3.9 mV dt dt

I=

E 3.93 × 10−3 V = = 2.0 × 10−2 A = 20 mA 0.2 Ω R

The field is increasing out of the page. To prevent the increase, the induced field needs to point into the page. Thus, the induced current must flow clockwise. (b) As in part (a), E = A(dB/dt ) = 3.9 mV and I = 20 mA. Here the field is into the page and decreasing. To prevent the decrease, the induced field needs to point into the page. Thus the induced current must flow clockwise. G G G G (c) Now A (left or right) is perpendicular to B and so A ⋅ B = 0 Wb. That is, the field does not penetrate the plane of the loop. If Φ = 0 Wb, then E = |dΦ/dt| = 0 V/m and I = 0 A. There is no induced current.

Assess:

Note that the induced field opposes the change.

34.12.

Model: Assume the field is uniform. Visualize: Please refer to Figure Ex34.12. The motion of the loop changes the flux through it. This results in an induced emf and current. Solve: The induced emf is E = |dΦ/dt| and the induced current is I = E/R. The area A is changing, but the field G G B is not. Take A as being out of the page and parallel to B , so Φ = AB and dΦ/dt = B(dA/dt). The flux is through that portion of the loop where there is a field, that is, A = lx. The emf and current are

E=B

d ( lx ) dA =B = Blv = ( 0.20 T )( 0.050 m )( 50 m/s ) = 0.50 V dt dt I=

E 0.50 V = = 5.0 A R 0.10 Ω

The field is out of the page. As the loop moves the flux increases because more of the loop area has field through it. To prevent the increase, the induced field needs to point into the page. Thus, the induced current flows clockwise. Assess: This seems reasonable since there is rapid motion of the loop.

34.13.

Model: Visualize:

Assume the field strength is changing at a constant rate.

The changing field produces a changing flux in the coil and there will be a corresponding induced emf and current. Solve: The induced emf of the coil is G G d A⋅ B dΦ dB dB 2 ⎛ 0.10 T ⎞ E=N =N = NA = Nπ r 2 = (103 ) π ( 0.010 m ) ⎜ ⎟ = 3.1 V −3 dt dt dt dt ⎝ 10 × 10 s ⎠

(

)

G G where we’ve used the fact that B is parallel to A . Assess: This seems to be a reasonable emf as there are many turns.

34.14.

Model: Assume the field is uniform across the loop. Visualize: Please refer to Figure Ex34.14. There is a current in the loop so there must be an emf that is due to a changing flux. With the loop fixed the area is constant so the change in flux must be due to a changing field strength. Solve: The induced emf is E = |dΦ/dt| and the induced current is I = E/R. The B field is changing, but the area G G A is not. Take A as being into the page and parallel to B , so Φ = AB and dΦ/dt = A(dB/dt). We have

E=

−3 dΦ dB dB IR (150 × 10 A ) ( 0.20 Ω ) =A ⇒ = = = 4.7 T/s 2 dt dt dt A ( 0.080 m )

The original field and flux is into the page. The induced counterclockwise current produces an induced field and flux that is out of the page. Since the induced field opposes the change, the field must be increasing.

34.15. Model: A changing magnetic field creates an electric field. Visualize: Please refer to Figure Ex34.15 in your textbook.

Solve:

The magnetic field produced by a current I in a solenoid is

Bsolenoid =

μ0 NI

L The magnitude of the induced electric field inside a solenoid is given by Equation 34.26. The direction can be found using Lenz’s law, but the problem only asks for field strength, which is positive. Only the current is changing, thus

( 4π ×10 E=

−7

)

T m/A ( 0.010 m )( 400 ) dI 2 ( 0.20 m ) dt

V s ⎞ dI ⎛ = ⎜1.26 × 10−5 A m ⎟⎠ dt ⎝

In the interval t = 0 s to t = 0.1 s, dI 5A = = 50 A/s dt 0.1 s In the interval t = 0.1 s to 0.2 s, the change in the current is zero. In the interval t = 0.2 s to 0.4 s, dI dt = 25 A/s. Thus in the interval t = 0 s to t = 0.1 s, E = 6.3 × 10–4 V/m. In the interval t = 0.1 s to 0.2 s, E is zero. In the interval t = 0.2 s to 0.4 s, E = 3.1 × 10–4 V/m. The E-versus-t graph is shown earlier, in the Visualize step.

34.16. Model: Visualize:

A changing magnetic field creates an electric field.

Solve: (a) Apply Equation 34.26. For a point on the axis, r = 0 m, so E = 0 V/m. (b) For a point 2.0 cm from the axis, E=

r dB ⎛ 0.020 m ⎞ = ⎟ ( 4.0 T/s ) = 0.040 V/m 2 dt ⎜⎝ 2 ⎠

34.17. Model: Assume the magnetic field inside the circular region is uniform. The region looks exactly like a cross section of a solenoid of radius 2.0 cm. The proton accelerates due to the force of the induced electric field. Visualize: Please refer to Figure Ex34.17. Solve: Equation 34.26 gives the strength of the induced electric field inside a solenoid with changing magnetic r dB , where r indicates the radial distance of each point to the field. The magnitude at each point is thus E = 2 dt center of the circular region. The direction of E can be determined with Lenz’s Law. As B decreases, a clockwise electric field is induced that would create a current that opposes the decrease if a conducting loop were present. Using Newton’s second law, F = eE = ma ⇒ a =

eE er dB = m 2m dt

The sign indicates the acceleration is in the same direction as the induced electric field. At point a the acceleration is

a=

(

)(

− 1.6 × 10−19 C 1.0 × 10−2 m

(

2 1.6 × 10

−27

kg

)

) ( 0.10 T/s ) ⇒ aG =

( 4.8 × 10

4

m/s 2 , up

)

G G At point b, a = 0 m/s2. At point c, a = 4.8 × 10 4 m/s2 , down . At point d, a = 9.6 × 10 4 m/s2 , down .

(

)

(

)

34.18.

Model: Visualize:

Assume that the current drops uniformly.

Assume that the current is flowing left to right as in the figure. The changing current produces a changing flux in the inductor and a corresponding emf is developed across the inductor. Solve: For the inductor the potential difference is

ΔV = − L

dI ΔI ( 50 A − 150 A ) × 10 = −L = − (10 × 10−3 H ) dt Δt 10.0 × 10−6 s

−3

= 100 V

As the current decreases, so does the flux. The induced current will oppose the change and try to maintain the original flux. This means that the induced current will be in the same direction as the original current. This induced current carries positive charge carriers to the right, so the potential will increase in the direction of the current.

34.19.

Model: Assume that the current changes uniformly. Visualize: We want to increase the current without exceeding a maximum potential difference. Solve: Since we want the minimum time, we will use the maximum potential difference:

ΔV = − L Assess:

dI ΔI ΔI 3.0 A − 1.0 A = −L ⇒ Δt = L = ( 200 × 10−3 H ) = 1.0 ms dt Δt ΔVmax 400 V

If we change the current in any shorter time the potential difference will exceed the limit.

34.20. Solve:

Visualize: The battery will cause a current through the inductor. Hence, there will be stored energy. The energy stored in the inductor depends on the inductance and the current. According to Ohm’s law,

I=

ΔV Ebat 12 V = = = 2.0 A R Rind + r 4.0 Ω+2.0 Ω

U L = 12 LI 2 = 12 (100 × 10−3 H ) ( 2.0 A ) = 0.20 J 2

34.21.

Visualize: The solenoid has inductance and when a current flows there is energy stored in the magnetic field. Solve: The inductance and energy of the solenoid are

Lsol =

μ0 N 2 A

( 4π × 10

−7

H/m ) ( 200 ) π ( 0.015 m ) 2

2

= 2.96 × 10−4 H l 0.12 m U L = 12 LI 2 = 12 (2.96 × 10−4 H)(0.80 A) 2 = 9.5 × 10−5 J =

34.22.

Visualize: A capacitor in combination with an inductor will have a resonant frequency determined by the inductance and capacitance. Solve: We know the frequency and the capacitance so we can find the inductance. We have 1 1 1 ω= ⇒L= 2 = = 2.5 × 10−7 H = 0.25 μ H 6 ω C ( 2π ×100 × 10 Hz )2 (10 × 10−12 F ) LC

34.23.

Visualize: Changing the variable capacitor in combination with a fixed inductor will change the resonant frequency of the LC circuit. Solve: Since the resonant frequency depends on the inverse square root of the capacitance a lower capacitance will produce a higher frequency and vice versa. The maximum frequency is

ωmax =

1 = LCmin

1 = 2.2 × 106 rad/s −3 −12 2.0 10 H 100 10 F × × ( )( )

The corresponding minimum is ωmin = 2.0 × 106 rad/s . These are angular frequencies so we can use f = ω 2π to find f min = 2.5 × 105 Hz and f max = 3.6 × 105 Hz, giving a range of 250 kHz to 360 kHz.

34.24.

Visualize: The rate at which the charges leave the capacitor will determine the current through the inductor. Solve: The charge on the capacitor is Q = Q0 cos ωt where Q0 = C ΔV is the maximum charge at t = 0 s. The

current is given by I = − dQ dt = ωQ0 sin ωt . Since the sine function oscillates between +1 and −1, the maximum current is

C 0.10 × 10−6 F ⎛ 1 ⎞ −2 I max = ωQ0 = ⎜ ⎟ C ΔV = ΔV L = ( 5.0 V ) 1.0 × 10−3 H = 5.0 × 10 A = 50 mA LC ⎝ ⎠

34.25.

Visualize: Please refer to Figure Ex34.25. This is a simple LR circuit if the resistors in parallel are treated as an equivalent resistor in series with the inductor. Solve: We can find the equivalent resistance from the time constant since we know the inductance. We have

τ=

L L 3.6 × 10−3 H ⇒ Req = = = 360 Ω Req τ 10 × 10−6 s

The equivalent resistance is the parallel addition of the unknown resistor R and 600 Ω. We have

( 600 Ω )( 360 Ω ) = 900 Ω 1 1 1 = + ⇒R= Req 600 Ω R 600 Ω − 360 Ω

34.26.

Visualize: Please refer to Figure Ex34.26. This is a simple LR circuit if the two resistors in series are treated as an equivalent resistor in series with the inductor. Solve: The time constant of the equivalent circuit is

τ=

L L 50 × 10−3 H = = = 1.00 × 10−4 s Req R1 + R2 500 Ω

If the current is initially at the maximum, then it will decay exponentially with time according to I (t ) = I 0e −t / τ .

The time t1/ 2 when the current drops to half the initial value is found as follows: 1 2

I 0 = I 0e−t1/ 2 / τ ⇒ ln ( 12 ) = −t1/ 2 /τ ⇒ t1/ 2 = τ ln 2 = (1.00 × 10−4 s ) 0.69 = 6.9 × 10−5 s

Assess: This is less than τ , as expected, since the time constant is the amount of time it takes to decay to 0.37 of the initial current and we are only going to 1/2. This result is quite general and is characteristic of all exponential decay phenomena.

34.27.

Visualize: Please refer to Figure P34.27. To calculate the flux we need to consider the orientation of the normal of the surface relative to the magnetic field direction. We will consider the flux through the surface in the two parts corresponding to the two different directions of the surface normals. Solve: The flux is G G G G Φ = Φ top + Φ left = Atop ⋅ B + Aleft ⋅ B = Atop B cos 45° + Aleft B cos 45°

= 2 × (0.050 m × 0.10 m)(0.050 T)cos 45° = 3.5 × 10−4 Wb

34.28. Model: Assume the field is uniform in space though it is changing in time. Visualize: The changing magnetic field strength produces a changing flux through the coil, and a corresponding induced emf and current. Solve: (a)

G G (b) Since the field is perpendicular to the plane of the coil, A is parallel to B and Φ = AB . The emf is E (t ) =

dΦ dB = NA = 20π (0.025 m) 2 ( 0.020 + 0.020t ) T/s = 7.9 × 10−4 (1 + t ) V dt dt

⇒ I (t ) =

E (t ) 7.9 × 10−4 (1 + t ) V = = 1.6 × 10−3 (1 + t ) A 0.50 Ω R

(c) Using the expression for I(t) in part (b), I ( 5 s ) = 1.6 × 10−3 (1 + 5 ) A = 9.6 × 10−3 A

I (10 s ) = 1.7 × 10−2 A

Assess: The field is always increasing and increasing more rapidly with time so the induced current is greater at 10 s than at 5 s.

34.29. Model: Assume the field is uniform in space though it is changing in time. Visualize: The changing magnetic field strength produces a changing flux through the loop, and a corresponding induced emf and current. G G Solve: (a) Since the field is perpendicular to the plane of the loop, A is parallel to B and Φ = AB . The emf is E=

E dΦ dB =A = (0.20 m) 2 ( 4 − 4t ) T/s = 0.16 (1 − t ) V ⇒ I = = 1.6 (1 − t ) A dt dt R

The magnetic field is increasing over the interval 0 s < t < 1 s and is decreasing over the interval 1 s < t < 2 s, so the induced emf and current must have opposite signs in the second half of the time interval. We arbitrarily choose the sign to be positive during the first half. Time (s) 0.0 0.5 1.0 1.5 2.0

B (T) 0.00 1.50 2.00 1.50 0.00

E (volts) 0.16 0.08 0.00 –0.08 –0.16

I (A) 1.6 0.8 0.0 –0.8 –1.6

(b) To plot the field and current we look at the form of the equations as a function of time. The magnetic field strength is quadratic with a maximum at t = 1 s and vanishing at t = 0 s and t = 2 s. The current equation is linear and decreasing, starting at 1.6 A at t = 0 s and going through zero at t = 1 s.

Assess: Notice in the graph how I = 0 A at t = 1 s, the instant in time when B is a maximum, that is, when dB/dt = 0. At this point the flux is (instantaneously) not changing so the corresponding induced emf and current are zero.

34.30.

Model: Assume the field is uniform in space though it is changing in time. Visualize: The changing magnetic field strength produces a changing flux through the coil, and a corresponding induced emf and current. Solve: (a)

G G (b) Since the field is perpendicular to the plane of the coil, A is parallel to B and Φ = AB . The emf and current are

E (t ) =

dΦ dB = NA = 100π (0.020 m) 2 (1 − 12 t ) T/s = ( 0.126 ) (1 − 12 t ) V dt dt

I (t ) =

E (t ) ( 0.126 ) (1 − 12 t ) V = = ( 0.126 ) (1 − 12 t ) A R 1.0 Ω

(c) Using the expression for I(t) from part (b),

I (1 s ) = ( 0.126 ) (1 − 12 × 1) A = 6.3 × 10−2 A = 63 mA Also, I ( 2 s ) = 0.0 A and I ( 3 s ) = −63 mA .

34.31.

Model: Visualize:

Assume that B changes uniformly with time.

The magnetic strength is changing so the flux is changing and this will create an induced emf. The magnetic field is at an angle θ = 45° to the normal to the plane of the coils. G G Solve: The flux for a single loop of the coil is Φ = A ⋅ B = BA cosθ . The radius and angle don’t change with time, but B does. According to Faraday’s law, the induced emf is

E=N

ΔB dΦ dB = NA cosθ = N π r 2 cosθ Δt dt dt

= 25π (0.050 m) 2 ( cos 45° )

0.20 T − 0.80 T = 4.2 × 10−2 V = 42 mV 2.0 s

34.32.

Model: Visualize:

Assume that B changes uniformly with time.

The magnetic strength is changing so the flux is changing and this will create an induced emf. The magnetic field is at an angle θ = 60° to the normal of the plane of the coils. G G Solve: The flux for a single loop of the coil is Φ = A ⋅ B = BA cosθ . The radius and angle don’t change with time, but B does. According to Faraday’s law, the induced emf is ΔB dΦ dB = NA cosθ = N π r 2 cosθ E=N Δt dt dt

= 100π (0.010 m) 2 ( cos60° )

1.50 T − 0.50 T = 2.6 × 10−2 V = 26 mV 0.60 s

34.33.

Model: Assume that the field is uniform in space over the coil. Visualize: We want an induced current so there must be an induced emf created by a changing flux. Solve: To relate the emf and the current we need to know the resistance. From Equation 32.2,

R=

ρCu lwire Awire

ρ Cu N 2π r 2 (100 ) (1.7 × 10 Ω m ) ( 0.040 m ) = = 2.2 Ω 2 2 π rwire ( 2.5 ×10−4 m ) −8

=

The magnetic field is perpendicular to the plane of the coil so the flux for a single loop of the coil is G G Φ = A ⋅ B = BA , if we take the normal to the coil to be in the same direction as the field. Using Faraday’s law,

E = IR = NA

dB dB dB IR ( 2.0 A )( 2.2 Ω ) = 8.7 T/s = Nπ r 2 ⇒ = = dt dt dt N π r 2 100π ( 0.040 m )2

34.34. Model: The loop remains perpendicular to the magnetic field as it shrinks. The magnetic field remains uniform and steady. Visualize: The changing area of the loop causes the magnetic flux through the loop to change, inducing an emf in the loop. G Solve: The area of the loop at a time t is A(t ) = π r 2 = π r02e−2 β t . Since the normal to the loop dA and the field G G dΦm , where direction are parallel, B ⋅ dA = BdA . The induced emf E = dt G G Φ m (t ) = ∫ B ⋅ dA = BA(t ) Thus E=

dΦm dA = B = π r02 (−2 β )e −2 β t = 2βπ r02 Be −2 β t dt dt

Assess: As the loop area shrinks to zero radius as t → ∞ the induced emf also approaches zero. The induced emf is proportional to the time rate of change of the area of the loop.

34.35. Model: The magnetic field changes with time, but not position. The loop does not move. Solve: (a) The changing magnetic field will induce a current in the loop. We can compute the magnetic flux G through the loop, then apply Faraday’s Law. The magnetic field B does not depend on the position variables x or G G G y, so the flux Φ m = B ⋅ A . The area vector A points along the kˆ direction. G G With B = (0.030t iˆ + 0.50t 2 kˆ) T and A = (10cm × 10cm) kˆ = 0.010m 2 kˆ , G G Φ = B ⋅ A = ( 0.50t 2 T )( 0.010 m 2 ) = ( 5.0 × 10−3 ) t 2 Tm 2 .

The induced emf is thus E=

dΦ = 2 ( 5.0 × 10−3 Tm 2 ) t = (10.0 × 10−3 Tm 2 ) t = (10.0 × 10−3 V/s ) t . dt

At t = 0.5 s, E = 5.0 × 10−3 V = 0.0050 V. (b) Similarly, at t = 10 s, E =10 .0 × 1.0−3 V = 0.0100 V. G G Assess: Only the component of B along A contributes to the magnetic flux.

34.36.

Visualize:

The bottom of the loop is on the x-axis. The field is changing with time so the changing flux will produce an induced emf and corresponding induced current in the loop. The field strength varies in space as well as time. Solve: Let’s take the normal of the surface of the loop to be in the z-direction so it is parallel to the magnetic field. The flux through the shaded strip is G G d Φ = B ⋅ dA = B dA = B(b dy ) = ( 0.80 y 2t ) b dy To find the total flux we integrate over the area of the loop. We obtain

Φ = ∫ ( 0.80 y 2t ) b dy = 0.80tb ∫ y 2 dy = b

b

0

0

0.80tb 4 3

To find the induced current we take the time derivative of the flux: I loop =

Eloop

R

1 d Φ 1 d ⎛ 0.80tb 4 ⎞ 0.80b 4 0.80 ( 0.20 m ) = = = 8.5 × 10−4 A = 0.85 mA ⎜ ⎟= R dt R dt ⎝ 3 ⎠ 3R 3 ( 0.50Ω ) 4

=

The induced current in the loop is a constant.

34.37.

Model: “infinite” wire. Visualize:

Assume the wire is long enough so we can use the formula for the magnetic field of an

The magnetic field in the vicinity of the loop is due to the current in the wire and is perpendicular to the loop. The current is changing so the field and the flux through the loop are changing. This will create an induced emf and induced current in the loop. Solve: The induced current depends on the induced emf and is

I loop =

Eloop R

=

1 dΦ R dt

The flux through a rectangular loop due to a wire was found in Example 34.5. The total flux is

Φ= ⇒ I loop =

μ0 Ib ⎛ c + a ⎞ ln ⎜ ⎟ 2π ⎝ c ⎠

−7 1 μ0b ⎛ c + a ⎞ dI ( 4π × 10 T m/A ) ( 0.020 m ) ⎛ 0.030 m ⎞ ln ⎜ ln ⎜ = ⎟ ⎟ (100 A/s ) = 44 μ A R 2π ⎝ c ⎠ dt ( 0.010 Ω ) 2π ⎝ 0.010 m ⎠

34.38.

Model: Assume the wire is infinitely long. Visualize: See Figure P34.38. The current through the wire produces a steady magnetic field, but the magnetic field strength depends on the distance from the wire. The loop moves away from the wire at constant speed. The flux through the loop varies due to its motion. Faraday’s law gives the magnitude of the induced emf, and Ohm’s law will yield the current strength. The direction of the magnetic field is into the paper at the location of the loop. G G The magnetic field is perpendicular to the plane of the loop, so B ⋅ dA = BdA . Solve: The flux through the loop is decreasing as it moves away from the wire. Lenz’s law implies that the induced current is clockwise in order to increase the flux. Using the results of Example 34.5, which treats the rectangular loop as a series of tall infinitesimally thin strips, μ I (4.0 cm) ⎛ x + 1.0 cm ⎞ ln ⎜ Φm = 0 ⎟ 2π x ⎝ ⎠

where x is the distance from the wire to the closer edge of the loop, and changes with time. Note d Φ m d Φ m dx d Φ m dx is the velocity of the loop. v , where v = = = dt dx dt dx dt The induced emf is thus

E=

d Φ m dx μ0 I (4.0 cm) 1.0 cm = v dx dt 2π x ( x + 1.0 cm )

At x = 2.0 cm, E=

⎛ ⎞ μ0 I 1.0 cm −6 ( 4.0 cm ) ⎜⎜ ⎟⎟ (10 m/s ) = 1.3 × 10 V 2π 2.0 cm 3.0 cm ( ) ⎝ ⎠

Since the loop resistance is R = 0.020 Ω, the induced current is thus E = 6.7 × 10−5 I R Assess: The induced emf is proportional to the current in the wire. For reasonable currents (~1 A) the induced emf is reasonable. I loop =

34.39.

Model: Since the solenoid is fairly long compared to its diameter and the loop is located near the center, assume the solenoid field is uniform inside and zero outside. Visualize: Please refer to Figure P34.39. The solenoid’s magnetic field is perpendicular to the loop and creates a flux through the loop. This flux changes as the solenoid’s current changes, causing an induced emf and corresponding induced current. Solve: (a) Using Faraday’s law, the induced current is

I loop =

Eloop R

=

1 d Φ Asol dBsol π rsol2 d ⎛ μ0 NI ⎞ = = ⎜ ⎟ R dt R dt R dt ⎝ l ⎠

=

π rsol2 μ0 N dI Rl

dt

We have used the fact that the field is approximately zero outside the solenoid, so the flux is confined to the area Asol = π rsol2 of the solenoid, not the larger area of the loop itself. The current is constant from 0 s ≤ t ≤ 1 s so dI/dt = 0 A/s and Iloop = 0 A during this interval, and I(0.5s) = 0. (b) During the interval 1 s ≤ t ≤ 2 s, the current is changing at the rate | dI/dt | = 40 A/s, so the current during this interval is

π ( 0.010 m ) ( 4π × 10−7 T m/A ) (100 ) 2

I loop =

( 0.10 Ω )( 0.10 m )

( 40 A/s ) =1.58 × 10−4A=158 μ A

Thus I(1.5 s) = 158 μA. (c) The current is constant from 2 s ≤ t ≤ 3 s, so again dI/dt = 0 A/s and Iloop = 0 A so I(2.5s) = 0. Assess: The induced current is proportional to the negative derivative of the solenoid current, and is zero when there is a constant current in the solenoid.

34.40.

Model: Since the solenoid is fairly long compared to its diameter and the coil is located near the center, assume the solenoid field is uniform inside and zero outside. Visualize: Please refer to Figure P34.40. The solenoid’s magnetic field is parallel to the coil’s axis and creates a flux through the coil. This flux changes as the solenoid’s current changes, causing an induced emf and corresponding induced current in the coil. Solve: The induced current from the induced emf is given by Faraday’s law. We have

I coil =

2 2 Ecoil 1 d Φ coil N coilπ rcoil d ⎛ μ0 NI ⎞ N coilπ rcoil μ0 N dI = N coil = ⎜ ⎟ = R R dt R dt ⎝ l ⎠ Rl dt

2 We used the fact that the coil flux is confined to the area Acoil = π rcoil of the coil, not the larger area of the

solenoid. The current is changing uniformly over the interval 0 s ≤ t ≤ 0.02 s at the rate |dI/dt| = 50 A/s so the induced current during this interval is 5π ( 0.0050 m ) ( 4π × 10−7 T m/A ) (120 ) 2

I coil =

( 0.10 Ω )( 0.080 m )

( 50 A/s ) = 3.7 × 10−4 A = 0.37 mA

The current in the solenoid is changing steadily so the induced current in the coil is constant. The current is initially positive so the field is initially to the right and decreasing. The induced current will oppose this change and will therefore produce an induced field to the right. This requires an induced current in the coil coming out of the page at the top, so it is also positive. That is, it is clockwise when seen from the left.

34.41. Model: Assume that the magnetic field of coil 1 passes through coil 2 and that we can use the magnetic field of a solenoid for coil 1. Visualize: Please refer to Figure P34.41. The field of coil 1 produces flux in coil 2. The changing current in coil 1 gives a changing flux in coil 2 and a corresponding induced emf and current in coil 2. Solve: (a) From 0 s to 0.1 s and 0.3 s to 0.4 s the current in coil 1 is constant so the current in coil 2 is zero. Thus I (0.05 s) = 0 A. (b) From 0.1 s to 0.3 s, the induced current from the induced emf is given by Faraday’s law. The current in coil 2 is

I2 =

N π r 2 μ N dI dΦ 2 dB E2 1 1 1 d ⎛μ NI ⎞ = N2 = N2 A2 1 = N2π r22 ⎜ 0 1 1 ⎟ = 2 2 0 1 1 R R dt R dt R dt ⎝ l1 ⎠ Rl1 dt

(

)

20π ( 0.010 m ) 4π × 10 −7 T m/A ( 20 ) 2

=

( 2Ω )( 0.020 m )

20 A/s = 7.95 × 10 −5 A = 79 μ A

We used the facts that the field of coil 1 is constant inside the loops of coil 2 and the flux is confined to the area A2 = π r22 of coil 2. Also, we used l1 = N1d = 20 (1.0 mm ) = 0.020 m and |dI/dt| = 20 A/s. From 0.1 s to 0.2 s the current in coil 1 is initially negative so the field is initially to the right and the flux is decreasing. The induced current will oppose this change and will therefore produce a field to the right. This requires an induced current in coil 2 that comes out of the page at the top of the loops so it is negative. From 0.2 s to 0.3 s the current in coil 1 is positive so the field is to the left and the flux is increasing. The induced current will oppose this change and will therefore produce a field to the right. Again, this is a negative current. Hence I(0.25 s) = 79 μA right to left through the resistor.

34.42.

Model: Assume the field due to the solenoid is uniform inside and vanishes outside. Visualize: The changing current in the solenoid produces a changing field and flux through the coil. This changing flux creates an induced emf in the coil. Solve: The flux is only nonzero within the area of the solenoid, not the entire area of the coil. The flux is G G Φ = A ⋅ B = BA and the induced current in the coil is

I coil = =

dΦ 1 dBsol 1 d ⎛μ N I ⎞ Ecoil 1 = N coil = N coil As = N coilπ rs2 ⎜ 0 s s ⎟ R R dt R dt R dt ⎝ ls ⎠ N coilπ rs2 μ0 N s dI s N coilπ rs2 μ0 N s = ( 2π f I 0 ) cos ( 2π ft ) Rls dt Rls

50π ( 0.010 m ) ( 4π × 10−7 T m/A ) ( 200 ) 2

= Assess:

( 0.50Ω )( 0.20 m )

( 2π )( 60 Hz )( 0.50 A ) cos ( 2π ft ) = ( 7.4 mA ) cos ( 2π ft )

The induced current oscillates because the current and field of the solenoid oscillate.

34.43.

Model: Assume the magnetic field is uniform over the plane of the loop. Visualize: The oscillating magnetic field strength produces a changing flux through the loop and an induced emf in the loop. Solve: (a) The normal to the surface of the loop is in the same direction as the magnetic field so that G G Φ = A ⋅ B = BA . The induced emf is E=

dΦ dB dB =π r 2ω B0 cos ωt =A = π r2 dt dt dt

The cosine will oscillate between +1 and −1 so the maximum emf is

Emax = π r 2ω B0 = π r 2 ( 2π f ) B0 = 2π 2 ( 0.125 ) (150 × 106 Hz )( 20 × 10−9 T ) = 0.93 V 2

(b) If the loop is rotated so that the plane is perpendicular to the electric field, then the normal to the surface will be parallel to the magnetic field. There is no magnetic flux through the loop and no induced emf.

34.44.

Model: Assume the field due to the solenoid is uniform inside and vanishes outside. Visualize: The changing current in the solenoid produces a changing field and flux through the coil. This changing flux creates an induced emf in the coil. Solve: The flux is only nonzero within the area of the solenoid, not the entire area of the coil. The induced current in the coil is

I coil = =

dΦ 1 dBs 1 d ⎛μ N I ⎞ Ecoil 1 = N coil = N coil As = N coilπ rs2 ⎜ 0 s s ⎟ R R dt R dt R dt ⎝ ls ⎠ N coilπ rs2 μ0 N s dI s N coilπ rs2 μ0 N s = ( 2π f I 0 ) cos ( 2π ft ) Rls dt Rls

The maximum coil current occurs when the cosine is +1. The maximum current in the solenoid is

I0 = Assess:

I coil,max Rls

N coilπ rs2 μ0 N s ( 2π f )

=

( 0.20 A )( 0.40Ω )( 0.20 m )

40π ( 0.015 m ) ( 4π × 10−7 T m/A ) ( 200 ) 2π ( 60 Hz ) 2

This is a large current, but the induced current is large as well.

= 6.0 A

34.45.

Model: Assume an ideal transformer. Visualize: An ideal transformer changes the voltage, but not the power (energy conservation). Solve: (a) The primary and secondary voltages are related by Equation 34.30. We have

V2 =

N2 V 15,000 V1 ⇒ N1 = 1 N 2 = 100 = 12,500 turns N1 V2 120

(b) The input power equals the output power and we recall that P = I ΔV , so

Pout = Pin ⇒ I1ΔV1 = I 2ΔV2 ⇒ I1 =

I 2 ΔV2 ( 250 A )120 V = = 2.0 A ΔV1 15,000 V

Assess: These values seem reasonable, because houses have low voltage and high current while transmission lines have high voltage and low current.

34.46.

Model: The inner loop is small enough that the field due to the outer loop is essentially uniform over its area. Also, assume the current in the outer loop changes uniformly. Visualize: The current in the outer loop creates a field at the location of the inner loop. Changing the current in the outer loop results in a change in flux and a corresponding induced emf and current in the inner loop. Solve: The magnetic field due to the outer loop is perpendicular to the plane of the inner loop, so Φ inner = G G Ainner ⋅ Bouter = Ainner Bouter , where Bouter = μ0 I /2router is the field of a current loop. The radius and angle don’t change

with time, but Bouter does. According to Faraday’s law, the induced emf is I inner = =

2 2 2 1 d Φ π rinner d ⎛ μ0 I outer ⎞ π rinner μ0 dI outer π rinner μ0 ΔI outer Einner = = = ⎜ ⎟ = 2 Rinner router Δt Rinner Rinner dt Rinner dt ⎝ 2router ⎠ 2 Rinner router dt

π (0.0010 m) 2 ( 4π × 10−7 T m/A ) −1.0 T − 1.0 T 2 ( 0.020 Ω )( 0.050 m )

0.10 s

= 4.0 × 10−8 A = 4.0 nA

34.47.

Model: Assume the field changes abruptly at the boundary and is uniform. Visualize: Please refer to Figure P34.47. As the loop enters the field region the amount of flux will change as more area has field penetrating it. This change in flux will create an induced emf and corresponding current. While the loop is moving at constant speed, the rate of change of the area is not constant because of the orientation of the loop. The loop is moving along the x-axis. Solve: (a) If the edge of the loop enters the field region at t = 0 s, then the leading corner has moved a distance x = v0t at time t. The area of the loop with flux through it is

A = 2 ( 12 ) yx = x 2 = ( v0t )

2

where we have used the fact that y = x since the sides of the loop are oriented at 45° to the horizontal. Take the G G surface normal of the loop to be into the page so that Φ = A ⋅ B = BA . The current in the loop is I=

⎛ 2 ( 0.80 T )(10 m/s )2 ⎞ 1 E 1 d Φ 1 dA 1 d 2 ⎟ t = (1.6 × 103 A ) t = = B = B ( v0t ) = B ( 2 ) v02t = ⎜ ⎜ ⎟ 0.10 R R dt R dt R dt R Ω ⎝ ⎠

The current is increasing at a constant rate. This expression is good until the loop is halfway into the field region. The time for the loop to be halfway is found as follows: 10 cm = v0t = (10 m/s ) t ⇒ t = 7.1 × 10−3 s = 7.1 ms 2 At this time the current is 11 A. While the second half of the loop is moving into the field, the flux continues to increase, but at a slower rate. Therefore, the current will decrease at the same rate as it increased before, until the loop is completely in the field at t = 14 ms. After that the flux will not change and the current will be zero.

(b) The maximum current of 11 A occurs when the flux is changing the fastest and this occurs when the loop is halfway into the region of the field.

34.48. Visualize: Please refer to Figure 34.27. The moving wire in a magnetic field results in a motional emf and a current in the loop. Solve: (a) The induced emf is Eloop = vlB = (100 m/s)(0.040 m)(1.0 T) = 4.0 V (b) The total resistance of the square loop is R = 4(0.010 Ω) = 0.040 Ω. The induced current is I = Eloop /R = (4.0 V)/(0.040 Ω) = 100 A (c) The slide wire generates the emf, Eloop. However, the slide wire also has an “internal resistance” due to the resistance of the slide wire itself—namely r = 0.01 Ω. Thus, the slide wire acts like a battery with an internal resistance. The potential difference between the ends is ΔV =Eloop – Ir = 4.0 V – (100 A)(0.01 Ω) = 3.0 V

It is this 3.0 V that drives the 100 A current through the remaining 0.03 Ω resistance of the “external circuit.” Assess: The slide wire system acts as an emf just like a battery does.

34.49.

Model: Visualize:

Assume that the field is uniform in space over the loop.

The moving slide wire in a magnetic field develops a motional emf and a corresponding current in the wire and the rails. The current through the resistor will cause energy dissipation and subsequent warming. Solve: (a) The induced emf depends on the changing flux, not on where the resistance in the loop is located. We have

Eloop = vlB = (10 m/s)(0.20 m)(0.10 T) = 0.20 V

The total resistance around the loop is due entirely to the carbon resistor, so the induced current is

I = Eloop /R = (0.20 V)/(1.0 Ω) = 0.20 A (b) To move with a constant velocity the acceleration must be zero, so the pulling force must balance the retarding magnetic force on the induced current in the slide wire. Thus, Fpull = Fmag = IlB = (0.20 A)(0.20 m)(0.10 T) = 4.0 × 10−3 N

(c) The current in the resistor will result in power being dissipated. The power is

P = I 2 R = (0.20 A) 2 (1.0 Ω) = 0.040 W = 0.040 J/s During a 10 second period Q = 0.40 J of energy is dissipated by the current, increasing the internal energy of the carbon resistor and raising its temperature. This is, effectively, a heat source. The heat is related to the temperature rise by Q = mcΔT, where c is the specific heat of carbon. Thus,

ΔT = Assess:

0.40 J Q = = 11°C −5 mc ( 5.0 × 10 kg ) ( 710 J/kg °C )

This is a significant change in the temperature of the resistor.

34.50.

Model: Assume there is no resistance in the rails. If there is any resistance, it is accounted for by the resistor. Visualize: Please refer to Figure P34.50.The moving wire will have a motional emf that produces a current in the loop. Solve: (a) At constant velocity the external pushing force is balanced by the magnetic force, so

E B 2l 2v ( 0.50 T ) ( 0.10 m ) ( 0.50 m/s ) ⎛ Blv ⎞ = = = 6.25 × 10−4 N ≈ 6.3 × 10−4 N lB = ⎜ lB ⎟ 2.0 Ω R R ⎝ R ⎠ 2

Fpush = Fmag = IlB =

2

(b) The power is

P = Fv = ( 6.25 × 10−4 N ) ( 0.50 m/s ) = 3.1 × 10−4 W (c) The flux is out of the page and decreasing and the induced current/field will oppose the change. The induced field must have a flux that is out of the page so the current will be counterclockwise. The magnitude of the current is

⎛ Blv ⎞ ( 0.50 T )( 0.10 m )( 0.50 m/s ) I =⎜ = 1.25 × 10−2 A ⎟= 2.0 Ω ⎝ R ⎠ (d) The power is

P = I 2 R = (1.25 × 10−2 A ) ( 2.0 Ω ) = 3.1 × 10−4 W 2

Assess: From energy conservation we see that the mechanical energy put in by the pushing force shows up as electrical energy in the resistor.

34.51.

Model: Assume that the magnetic field is uniform in the region of the loop. Visualize: Please refer to Figure P34.51. The rotating semicircle will change the area of the loop and therefore the flux through the loop. This changing flux will produce an induced emf and corresponding current in the bulb. Solve: (a) The spinning semicircle has a normal to the surface that changes in time, so while the magnetic field is constant, the area is changing. The flux through in the lower portion of the circuit does not change and will not contribute to the emf. Only the flux in the part of the loop containing the rotating semicircle will change. The flux associated with the semicircle is G G Φ = A ⋅ B = BA = BA cosθ = BA cos ( 2π ft )

where θ = 2π ft is the angle between the normal of the rotating semicircle and the magnetic field and A is the area of the semicircle. The induced current from the induced emf is given by Faraday’s law. We have

I=

E 1 dΦ 1 d B π r2 BA cos ( 2π ft ) = = = 2π f sin ( 2π ft ) R R dt R dt R 2 2 ( 0.20 T ) π 2 ( 0.050 m ) f sin ( 2π ft ) = 4.9 × 10−3 f sin ( 2π ft ) A 2 (1.0Ω ) 2

=

where the frequency f is in Hz. (b) We can now solve for the frequency necessary to achieve a certain current. From our study of DC circuits we know how power relates to resistance: P = I 2 R ⇒ I = P / R = 4.0 W /1.0Ω = 2.0 A

The maximum of the sine function is +1, so the maximum current is 2.0 A = 4.1 × 10 2 Hz 4.9 × 10−3 A s This is not a reasonable frequency to obtain by hand. I max = 4.9 × 10−3 f A s = 2.0 A ⇒ f =

Assess:

34.52.

Model: Assume the magnetic field is uniform over the region where the bar is sliding and that friction between the bar and the rails is zero. Visualize: Please refer to Figure P34.52. The battery will produce a current in the rails and bar and the bar will experience a force. With the battery connected as shown in the figure, the current in the bar will be down and by the right-hand rule the force on the bar will be to the right. The motion of the bar will change the flux through the loop and there will be an induced emf that opposes the change. Solve: (a) As the bar speeds up the induced emf will get larger until finally it equals the battery emf. At that point, the current will go to zero and the bar will continue to move at a constant velocity. We have E = Blvterm = E bat ⇒ vterm =

E bat Bl

(b) The terminal speed is vterm =

Assess:

This is pretty fast, about 70 mph.

1.0 V = 33 m/s ( 0.50 T )( 0.060 m )

34.53.

Model: Visualize:

Assume the magnetic field is uniform in the region of the loop.

The moving wire creates a changing area and corresponding change in flux. This produces an induced emf and induced current. The flux through the loop depends on the size and orientation of the loop. G G Solve: (a) The normal to the surface is perpendicular to the loop and the flux is Φ inner = A ⋅ B = AB cosθ . We

can get the current from Faraday’s law. Since the loop area is A = lx, We have E 1 dΦ 1 d Bl cosθ dx Blv cosθ = = lxB cosθ = = R R dt R dt R dt R

I=

(b) Using the free-body diagram shown in the figure, we can apply Newton’s second law. The magnetic force on a straight, current-carrying wire is Fm = IlB and is horizontal. Using the current I from part (a) gives

∑F

x

= − Fm cosθ + mg sin θ = −

B 2l 2v cos 2 θ + mg sin θ = max R

Terminal speed is reached when ax drops to zero. In this case, the two terms are equal and we have vterm =

mgR tan θ l 2 B 2 cosθ

34.54.

Model: Assume the magnetic field is uniform over the loop. Visualize: Please refer to Figure P34.54. As the wire falls, the flux into the page will increase. This will induce a current to oppose the increase, so the induced current will flow counterclockwise. As this current passes through the slide wire, it experiences an upward magnetic force. So there is an upward force—a retarding force—on the wire as it falls in the field. As the wire speeds up the retarding force will become larger until it balances the weight. Solve: (a) The force on the current-carrying slide wire is Fm = IlB. The induced current is I=

E 1 dΦ 1 d B d Blv = = AB = lx = R R dt R dt R dt R l 2 B 2v ⎛ l 2 B 2 ⎞ Fm = =⎜ ⎟v R ⎝ R ⎠

Consequently, the retarding magnetic force is

The important point is that Fm is proportional to the speed v. As the wire begins to fall and its speed increases, so does the retarding force. Within a very short time, Fm will increase in size to where it matches the weight w = mg. At that point, there is no net force on the loop, so it will continue to fall at a constant speed. The condition that the magnetic force equals the weight is

⎛ l 2B2 ⎞ mgR ⎜ ⎟ vterm = mg ⇒ vterm = 2 2 l B ⎝ R ⎠ (b) The terminal speed is

vterm =

( 0.010 kg ) ( 9.8 m/s2 ) ( 0.10Ω ) = 0.98 m/s 2 2 ( 0.20 m ) ( 0.50 T )

34.55.

Model: Assume the magnetic field is uniform over the loop. Visualize: The motion of the eye will change the orientation of the loop relative to the fixed field direction resulting in an induced emf. Solve: We are just interested in the emf due to the motion of the eye so we can ignore the details of the time dependence of the change. From Faraday’s law,

Ecoil = N

Nπ r 2 B cosθ f − cosθ i dΦ ΔΦ Δ cosθ =N = NAB = dt Δt Δt Δt

20π ( 3.0 × 10−3 m ) (1.0 T ) cos85° − cos90° 2

= Assess:

0.20 s

= 2.5 × 10−4 V

This is a reasonable emf to measure, although you might need some amplification.

34.56.

Model: Visualize:

Assume the coil is moved to a location where the magnetic field is zero.

The removal of the coil from the field will change the flux and produce an induced emf and corresponding induced current. The current will charge the capacitor. Solve: The induced current is N dΦ E I coil = coil = R R dt The definition of current is I = dq/dt. Consequently, the charge flow through the coil and onto the capacitor is given by Δq N ΔΦ dq N d Φ N N = ⇒ = ⇒ Δq = ΔΦ = ( Φ f − Φ i ) Δt R Δt dt R dt R R We are only interested in the total charge that flows due to the change in flux and not the details of the time dependence. In this case, the flux is changed by physically pulling the coil out of the field. Since the coil is oriented for maximum flux, the initial flux through the coil is

Φ i = π r 2 B = π (0.0050 m) 2 (0.0010 T) = 7.85 × 10 −8 Wb After being pulled from the field, the final flux is Φf = 0 Wb. The charge that flows onto the capacitor is

Δq =

(10 ) 0 Wb − 7.85 × 10−8 Wb 0.20 Ω

= 3.93 × 10−6 C ⇒ ΔVC =

Δq 3.93 × 10−6 C = = 3.9 V C 1.0 × 10−6 F

34.57.

Model: Visualize:

Assume the field is uniform in the region of the coil.

The rotation of the coil in the field will change the flux and produce an induced emf and a corresponding induced current. The current will charge the capacitor. Solve: The induced current is N dΦ E I coil = coil = R R dt The definition of current is I = dq/dt. Consequently, the charge flow through the coil and onto the capacitor is given by dq N d Φ Δq N ΔΦ N N = ⇒ = ⇒ Δq = ΔΦ = ( Φ f − Φ i ) dt R dt Δt R Δt R R

We are only interested in the total charge that flows due to the change in flux and not the details of the time dependence. In this case, the flux is changed by physically rotating the coil in the field. The flux is Φ = G G A ⋅ B = AB cosθ . The change in flux is

ΔΦ = AB ( cosθ f − cosθi ) = π ( 0.020 m ) ( 55 × 10−6 T ) ( cos30° − cos 210° ) = 1.2 × 10−7 Wb 2

Note that the field is 60° from the horizontal and the normal to the plane of the loop is vertical. The final angle, G when A points down, is θ f = 30°, so the initial angle is θi = θ f + 180° = 210°. The charge that flows onto the capacitor is Δq =

200 (1.2 × 10−7 Wb ) Δq 1.2 × 10−5 C N = = 12 V = 1.2 × 10−5 C ⇒ ΔVC = ( Φf − Φi ) = R 2.0 Ω C 1.0 × 10−6 F

34.58.

Model: uniform. Visualize:

Solve:

Use Faraday’s law of electric induction and assume that the magnetic field inside the solenoid is

(a) With B = 10.0 T + (2.0 T)sin[2π (10 Hz)t], Equation 34.26 gives

E=

r dB 0.015 m V = ( 2.0 T ) ⎡⎣2π (10 Hz )⎤⎦ cos ⎡⎣ 2π (10 Hz ) t ⎤⎦ = ⎛⎜ 0.94 ⎞⎟ cos ⎡⎣ 2π (10 Hz ) t ⎤⎦ 2 dt 2 m ⎝ ⎠

V . m (b) E is maximum when cos[2π (10 Hz)t] = 1 or –1 which means when sin[2π (10 Hz)t] = 0. Under this condition, B = (10.0 T) + (2.0 T) sin [2π (10 Hz)t] = 10.0 T That is, B = 10.0 T at the instant E has a maximum value of 0.94 V/m.

The field strength is maximum when the cosine function is equal to 1 or –1. Hence Emax = 0.94

34.59.

Model: Visualize:

Use Faraday’s law of electric induction.

G

G

dB , describes the relationship between an induced electric field and the flux dt through a fixed area A. To solve this equation, choose a clockwise direction around a circle of radius r as the closed curve. The electric field vectors, as can be seen from the figure, are everywhere tangent to the curve. The G line integral of E then is G G úE ⋅ d s = ( 2π r ) E Equation 34.23,

úE ⋅ d s = A

The direction of the induced electric field is chosen to agree with the right-hand rule applied to the flux calculation for magnetic field. Note that B = 0 outside the solenoid. Solve: Equation 34.23 now simplifies to dB R2 dB ⇒E= dt 2r dt At r = R, E = 21 R dB dt , which is the same result as Equation 34.26. E ( 2π r ) = π R2

Assess:

Model: Assume the solenoid is an ideal solenoid, with R = 0 Ω, so that the magnetic field at any instant in time is the same everywhere within the solenoid. Assume dI/dt is constant. Visualize: An electric field is induced by the changing magnetic field. Solve: Using Equation 34.26, the electric field strength and rate of change of the magnetic field in a solenoid r dB , where r is the distance from the solenoid’s central axis at which E is measured. Thus are related by E = 2 dt

34.60.

E=

r dB r d 1 dI = ( μ0nI ) = μ0 nr dt 2 dt 2 dt 2

dI = 1.6 × 102 A s . dt Assess: This current change is occurring in the solenoid coils. At first glance this seems like a rapid rate of change, but is really quite acceptable. For example, if a circuit carrying 1.0 A decreases to zero current in a millisecond (the time for a switch to disconnect) the rate of change of the current is 1000 A/s.

With E = 5.0 × 10– 4 V/m at r = 0.50 cm,

34.61. Solve:

Visualize: The changing current produces an emf across the inductor. When the current is changing we know the induced emf, so we can get the inductance. We have dI ΔVL 0.20 V ΔVL = − L ⇒ L = = = 20 mH dt dI dt 10.0 A/s

We know the flux per turn for a given current so we can connect the inductance to the magnetic field and the flux per turn. The flux through one turn of the solenoid is ⎛ μ NI ⎞ ⎛ μ NA ⎞ Φ (1) = Bsol A = ⎜ 0 ⎟ A = ⎜ 0 ⎟I ⎝ l ⎠ ⎝ l ⎠

But the inductance is Lsol =

μ0 N 2 A ⎛ μ0 NA ⎞ l

=⎜ ⎝

l

⎛ μ0 NA ⎞ Lsol ⎟N ⇒ ⎜ ⎟= ⎠ ⎝ l ⎠ N

Substituting into the previous expression we get

( 20 ×10−3 H ) ( 0.10 A ) = 400 turns L ⎛L ⎞ Φ (1) = ⎜ sol ⎟ I ⇒ N = sol I = 5.0 × 10−6 Wb/turn Φ (1) ⎝ N ⎠

34.62. Visualize: The current through the inductor has energy stored in the magnetic field. Solve: The energy stored in the inductor depends on the inductance and the current. We know the energy and the current so we can get the inductance. We have U L = 12 LI 2 ⇒ L =

−3 2U L 2 (1.0 × 10 J ) = = 5.0 × 10−2 H 2 I2 ( 0.20 A )

We know the flux per turn for a given current so we can relate the inductance to the magnetic field and the flux per turn. The flux through one turn of the solenoid is ⎛ μ NI ⎞ ⎛ μ NA ⎞ Φ (1) = Bsol A = ⎜ 0 ⎟ A = ⎜ 0 ⎟I ⎝ l ⎠ ⎝ l ⎠

But the inductance is Lsol =

μ0 N 2 A ⎛ μ0 NA ⎞ l

=⎜ ⎝

l

⎛ μ0 NA ⎞ Lsol ⎟N ⇒ ⎜ ⎟= ⎠ ⎝ l ⎠ N

Substituting into the previous expression,

( 5.0 ×10−2 H ) ( 0.20 A ) = 500 turns L ⎛L ⎞ Φ (1) = ⎜ sol ⎟ I ⇒ N = sol I = Φ (1) 20 × 10−6 Wb/turn ⎝ N ⎠

34.63.

Model: Assume the inductance is that of a solenoid. Visualize: The inductance of a solenoid depends on the number of turns, the length, and the cross-sectional area. Solve: The length in terms of the number of turns in a layer and the wire diameter is l = N l d wire . The total

number of turns is N = 4 N l . For a solenoid we have

Lsol =

μ0 N 2 A l

⇒N=

lLsol = μ0 A

⇒ N l = 14 N = 56 ⇒ d wire =

( 0.050 m ) (100 × 10−6 H ) ( 4π × 10−7 H/m )π ( 0.0050 m )2

= 225

l 0.050 m = = 8.9 × 10−4 m = 0.89 mm Nl 56

34.64.

Model: Assume the straight wire in part (b) is “infinite.” Visualize: The energy density depends on the field strength in a particular region of space. Solve: (a) We will use the formula for the field at the center of a current loop and then find the energy density. We have

B=

μ0 I 2 rloop

=

( 4π ×10

−7

T m/A ) (1.0 A )

2 ( 0.020 m )

= 3.14 × 10−5 T

( 3.14 × 10−5 T ) = 3.9 ×10−4 J/m3 1 2 uB = B = 2μ0 2 ( 4π × 10−7 T m/A ) 2

(b) We will use the standard formula for the magnetic field of a long straight wire. We have 2 8π 2 ( 0.020 m ) ( 3.9 × 10−4 J/m3 ) μ0 I 1 2 1 ⎛ μ0 I ⎞ 8π 2 r 2u B B = I ⇒ uB = ⇒ = = = 3.1 A ⎜ ⎟ μ0 2π r 2μ0 2μ0 ⎝ 2π r ⎠ ( 4π ×10−7 T m/A ) 2

B=

34.65.

Model: Assume the solenoid is long enough that we can approximate the field as being constant inside and zero outside. Visualize: The energy density depends on the field strength in a particular region of space. Solve: (a) We can use the given field strength to find the energy density and then determine the volume of a cylinder. We have

1 2 ( 5.0 T ) = 9.95 × 106 J/m3 B = 2μ0 2 ( 4π × 10−7 T m/A ) 2

uB =

Vsol = π rsol2 l = π ( 0.20 m ) (1.0 m ) = 0.126 m3 ⇒ U tot = uBVsol = ( 9.95 × 106 J/m 3 )( 0.126 m 3 ) = 1.25 × 106 J 2

(b) Using the magnetic field of the solenoid, the number of turns is calculated as follows:

B= Assess:

μ0 NI l

⇒N=

lB (1.0 m )( 5.0 T ) = = 4.0 × 104 μ0 I ( 4π × 10−7 T m/A ) (100 A )

This is a reasonable number for a solenoid that is a meter long.

34.66.

Model: Assume the field is constant over the cross section of the patient. Visualize: The changing field results in a changing flux and an induced emf. Solve: To determine the largest emf, assume the normal to the surface is parallel to the field . From Faraday’s law, the emf is

E=

dΦ d dB ΔB = AB = A =A dt dt dt Δt

The time interval acceptable for the change is

Δt = A Assess:

2 ΔB ( 0.060 m ) 5.0 T − 0 T = = 3.0 s E 0.10 V

This is a reasonable amount of time in which to change the field.

34.67.

Model: Assume we can ignore the sharp corners when the current changes abruptly. Visualize: The changing current produces a changing flux, an induced emf, and a corresponding potential difference. Solve: Break the current into time intervals over which the current is changing linearly or not at all. For the intervals 2 ms to 3 ms and 5 ms to 6 ms, the current does not change, so the potential difference is zero. For the interval 0 s to 2 ms, the current goes from 0 A to 2 A, so the potential difference is

ΔVL = − L

dI ΔI 2 A−0 A = −L ⇒ ΔVL = − (10 × 10−3 H ) = −10 V dt Δt ( 2 s − 0 s ) × 10−3

Similarly for the interval 3 ms to 5 ms, the potential difference is +20 V.

Assess:

The potential difference is proportional to the negative slope of the current versus time graph.

34.68.

Model: Assume we can ignore the edges when the potential difference changes abruptly. Visualize: The existence of a potential difference indicates there is a changing current. If the potential difference is zero, the current is constant. Solve: Break the potential difference into time intervals and determine the corresponding rate of change of the current for each time interval. Knowing the starting current for an interval, we can determine the current at the end of the interval by finding the change. For the intervals 10 to 20 ms and 30 to 40 ms, the potential difference is zero, so the current remains the same as at the start of the interval. For the interval 0 ms to 10 ms, the potential difference is −1 V. The rate of change of the current is calculated as follows:

dI ΔI −ΔVL Δt ( −1 V )(10 s − 0 s ) × 10−3 s = 0.20 A = −L ⇒ ΔI 0 →10 = =− dt Δt L 50 × 10−3 H To find the current at 10 ms we need the current at the start of the interval. We have I (10 ms) = I (0 ms) + ΔI 0 →10 = 0.20 A + ( 0.20 A ) = 0.40 A ΔVL = − L

A constant potential difference implies that the current is changing linearly. So, the current starts at 0.2 A and goes linearly to 0.40 A over the first 10 ms. For the interval 10 ms to 20 ms, the current remains at 0.40 A. For the interval 20 ms to 30 ms, the current goes linearly from 0.40 A to 0 A. For the interval 30 ms to 40 ms, the current remains at 0 A.

34.69. Visualize: The potential difference across the inductor depends on the rate of change of the current with respect to time. Solve: (a) We can find the potential difference by taking a derivative. We have ΔVL = − L

dI d = − L I 0 sin ω t = − Lω I 0 cos ω t dt dt

(b) The cosine function oscillates between +1 and −1 so we ignore it and the negative sign. The maximum potential difference is ΔV 0.20 V ΔVL,max = ω LI 0 ⇒ I 0 = L,max = = 1.27 × 10−3 A ωL 2π ( 500 × 103 Hz )( 50 × 10−6 H )

34.70. Visualize: The current through the inductor is changing with time, which leads to a changing potential difference across the inductor. Solve: (a) To find the potential difference we differentiate the current. We have ΔVL = − L

dI d ⎛ LI ⎞ = − L ( I 0e − t / τ ) = ⎜ 0 ⎟ e − t / τ dt dt ⎝ τ ⎠

(b) For t = 1 ms, the potential difference is

ΔVL

( 20 × 10 =

−3

H )( 50 × 10−3 A )

1.0 × 10−3 s

e −1.0×10

−3

s /1.0×10−3 s

= (1.0 V ) e −1 = 0.37 V

Similar calculations give 1.0 V, 0.13 V, and 0.05 V for t = 0 ms, 2 ms, and 3 ms. (c)

34.71.

Model: Assume any resistance is negligible. Visualize: The potential difference across the inductor and capacitor oscillate. Solve: (a) The current is I (t ) = I 0 cos ωt = ( 0.50 A ) cos ωt . Looking at Figure 34.46, we see that the capacitor is

fully charged one-quarter cycle after the current is a maximum (or minimum), so the time needed is one quarter of a cycle. We have

ω= ⇒T =

2π

ω

=

1 = LC

1

( 20 × 10

−3

H )( 8.0 × 10

−6

F)

= 2.5 × 103 rad/s

2π = 2.51 × 10−3 s = 2.51 ms ⇒ Δt = 14 T = 0.63 ms 2.5 × 103 rad/s

(b) The maximum inductor current and maximum capacitor charge are related by I 0 = ωQ0 . The potential across the capacitor is

ΔVC =

0.50 A Q0 I = 0 = = 25 V C ωC ( 2.5 × 103 rad/s )( 8.0 × 10−6 F )

34.72.

Model: Assume any resistance is negligible. Visualize: The potential difference across the inductor and capacitor oscillate. The period of oscillation depends on the resistance and capacitance and the potential difference across the capacitor depends on the charge. Solve: The current is I (t ) = I 0 cos ωt = ( 0.60 A ) cos ωt . We can relate the extreme values of the current and the

capacitor potential difference. Using ω = 1

LC and Q = C ΔVC , we find

( 0.60 A ) (10 × 10−3 H ) I 2L 1 C ΔVC,max ⇒ C = 02 = = 1.0 μ F 2 ΔVC,max LC ( 60 V ) 2

I 0 = ωQ0 = ωC ΔVC,max = Assess:

This is a reasonable capacitance.

34.73.

Model: Assume any resistance is negligible. Visualize: Energy in the capacitor and inductor oscillates as the charge and current oscillate, but the total energy is conserved. Solve: The current through the inductor is zero when the charge on the capacitor is maximum. Thus the total energy is stored in the capacitor: U total = U C =

1 Q0 2 C2

At a later time, when the capacitor’s energy equals the inductor’s energy, they each have half the total energy. Thus

1 1 Q 1 ⎛ 1 Q0 ⎞ Q U C = U L = U total ⇒ = ⎜ ⇒ Q = 0 = 0.707Q0 2 2 ⎟ 2 2C 2⎝ 2 C ⎠ 2

34.74. Visualize: The oscillation frequency depends on the capacitance and the inductance. The inductance will depend on the number of turns, the length, and the cross-sectional area of the solenoid. Solve: First we’ll find the inductance necessary to give the proper frequency, then we can design our inductor. For an LC circuit, 1 1 1 = 2π f ⇒ L = 2 2 = 2 = 2.53 × 10−2 H 2 2 4 f C π LC 4π (1.0 Hz ) (1.0 F )

ω=

The inductor length is l = Nd , where N is the number of turns and d is the diameter of the wire used. We find that L=

μ0 N 2 A l

=

μ0 N 2π r 2 Nd

⇒N=

( 0.25 × 10−3 m )( 2.53 × 10−2 H ) = 4000 dL = μ0π r 2 ( 4π × 10−7 T m/A )π ( 0.02 m )2

If we have one layer, the length will be l = Nd = 4000(0.25 × 10−3 m) = 1.0 m. Our inductor can be 1.0 m long with 4000 turns of wire. Alternatively, a 0.5-m-long coil with two layers of 2000 turns would also work.

34.75.

Model: Assume there is no resistance in the LC circuit. Visualize: The LC circuit oscillates at a constant rate. The energy in the circuit alternates between the inductor and capacitor. Solve: The maximum energy stored in the capacitor is the same as the total energy of the circuit, which is the same as the maximum energy stored in the inductor. Maximum energy in the inductor occurs when the circuit 1 2 has maximum current, I max = 0.10 A. The maximum energy in the inductor is U L = LI max = 1.0 × 10 −5 J. 2 1 2 = ( 2π ⋅ 10 kHz ) , thus Solving yields L = 2.0 mH. The circuit oscillates with angular frequency ω 2 = LC C = 0.13 μ F. Assess: These are reasonable values for L and C.

34.76. Model: Assume negligible resistance in the LC part of the circuit. Visualize: With the switch in position 1 for a long time the capacitor is fully charged. After moving the switch to position 2, there will be oscillations in the LC part of the circuit. Solve: (a) After a long time the potential across the capacitor will be that of the battery and Q0 = C ΔVbatt . When the switch is moved, the capacitor will discharge through the inductor and LC oscillations will begin. The maximum current is I 0 = ωQ0 = ωC ΔVbatt =

C ΔVbatt C = ΔVbatt = L LC

( 2.0 ×10 F) (12 V ) = 7.6 × 10 ( 50 × 10 H ) −6

−3

−2

A = 76 mA

(b) The current will be a maximum one-quarter cycle after the maximum charge. The period is T=

2π

ω

= 2π LC = 2π

( 50 × 10

So the current is first maximum at tmax = 14 T = 0.50 ms.

−3

H )( 2.0 × 10−6 F ) = 2.0 ms

34.77.

Model: Assume negligible resistance in the circuit. Visualize: Energy is conserved. The maximum voltage on the right capacitor will occur when all of the energy from the left capacitor is transferred to the right one. Solve: (a) The voltage is calculated as follows: 1 2

2 C1ΔVC12 = 12 C2 ΔVC2 ⇒ ΔVC2 =

300 C1 ΔVC1 = (100 V ) = 50 V 1200 C2

(b) Closing S1 causes the charge and current of the left LC circuit to oscillate with period TL. After one-quarter of a period, the 300 μF capacitor is completely discharged and the current through the inductor is maximum. At that instant we’ll open S1 and close S2. Then the right LC circuit will start to oscillate with period TR and the inductor current will charge the 1200 μF capacitor. The capacitor will be fully charged after one-quarter of a period, so we will open S2 at that time to keep the charge on the 1200 μF capacitor. The periods are 2π TL = = 2π LC1 = 2π (5.3 H)(300 × 10−6 F) = 0.25 s ⇒ 14 TL = 0.063 s

ωL 2π TR = = 2π LC2 = 2π (5.3 H)(1200 × 10−6 F) = 0.50 s ⇒ 14 TR = 0.125 s ωR

So the procedure is to close S1 at t = 0 s, open S1 and close S2 at t = 0.0625 s, then open S2 at t = 0.0625 s + 0.1250 s = 0.1875 s.

34.78.

Model: Assume an ideal inductor and an ideal (resistanceless) battery. Visualize: Please refer to Figure P34.78. The current through the battery is the sum of the currents through the left and right branches of the circuit. Solve: (a) Because the switch has been open a long time, no current is flowing the instant before the switch is closed. A basic property of an ideal inductor is that the current through it cannot change instantaneously. This is because the potential difference ΔVL = –L(dI/dt) would become infinite for an instantaneous change of current, and that is not physically possible. Because the current through the inductor was zero before the switch was closed, it must still be zero (or very close to it) immediately after the switch is closed. Conservation of current requires the current through the entire right branch to be the same as that through the inductor, so it is also zero immediately after the switch is closed. The only current is through the left 20 Ω resistor, which sees the full battery potential of the battery. Thus Ibat = Ileft = ΔVbat/R = (10 V)/(20 Ω) = 0.50 A. (b) The current through the inductor increases as time passes. Once the current Iright reaches a steady value and is no longer changing, the potential difference across the inductor is ΔVL = –L(dI/dt) = 0 V. An ideal inductor has no resistance, so the inductor simply acts like a wire and has no effect on the circuit. The circuit is that of two 20 Ω resistors in parallel. The equivalent resistance is 10 Ω and the battery current is Ibat = (10 V)/(10 Ω) = 1.0 A.

34.79.

Model: Assume an ideal inductor and an ideal (resistanceless) battery. Visualize: Please refer to Figure P34.79. Solve: (a) Because the switch has been open a long time, no current is flowing the instant before the switch is closed. A basic property of an ideal inductor is that the current through it cannot change instantaneously. This is because the potential difference ΔVL = –L(dI/dt) would become infinite for an instantaneous change of current, and that is not physically possible. Because the current through the inductor was zero before the switch was closed, it must still be zero (or very close to it) immediately after the switch is closed. Consequently, the inductor has no effect on the circuit. It is simply a 10 Ω resistor and 20 Ω resistor in series with the battery. The equivalent resistance is 30 Ω, so the current through the circuit (including through the 20 Ω resistor) is I = ΔVbat/Req = (30 V)/(30 Ω) = 1.0 A. (b) After a long time, the currents in the circuit will reach steady values and no longer change. With steady currents, the potential difference across the inductor is ΔVL = –L(dI/dt) = 0 V. An ideal inductor has no resistance (R = 0 Ω), so the inductor simply acts like a wire. In this case, the inductor “shorts out” the 20 Ω resistor. All current from the 10 Ω resistor flows through the resistanceless inductor, so the current through the 20 Ω resistor is 0 A. (c) When the switch has been closed a long time, and the inductor is shorting out the 20 Ω resistor, the current passing through the 10 Ω resistor and through the inductor is I = (30 V)/(10 Ω) = 3.0 A. Because the current through an inductor cannot change instantaneously, the current must remain 3.0 A immediately after the switch reopens. This current must go somewhere (conservation of current), but now the open switch prevents the current from going back to the battery. Instead, it must flow upward through the 20 Ω resistor. That is, the current flows around the LR circuit consisting of the 20 Ω resistor and the inductor. This current will decay with time, with time constant τ = L/R, but immediately after the switch reopens the current is 3.0 A.

34.80. Visualize: Please refer to Figure P34.80. Solve: (a) After a long time has passed the current will no longer be changing. With steady currents, the potential difference across the inductor is ΔVL = –L(dI/dt) = 0 V. An ideal inductor has no resistance (R = 0 Ω), so the inductor simply acts like a wire. The circuit is simply that of a battery and resistor R, so the current is I0 = ΔVbat/R. (b) In general, we need to apply Kirchhoff’s loop law to the circuit. Starting with the battery and going clockwise, the loop law is dI ΔVbat + ΔVR + ΔVL = ΔVbat − IR − L = 0 dt dI ΔVbat IR R ⎛ ΔV ⎞ R = − = ⎜ − I ⎟ = (I0 − I ) dt L L L⎝ R ⎠ L This is a differential equation that we can solve by direct integration. The current is I = 0 A at t = 0 s, so separate the current and time variables and then integrate from 0 A at 0 s to current I at time t:

∫

t 0

⎛I −I ⎞ dI R t R t t I dt ⇒ − ln( I 0 − I ) 0 = t 0 ⇒ − ln ⎜ 0 = ⎟= / I0 − I L ∫0 L I L R ⎝ 0 ⎠

Taking the exponential of both sides gives I0 − I = e− t /( L / R ) ⇒ I 0 − I = I 0e − t /( L / R ) I0

Finally, solving for I gives

I = I 0 (1 − e −t /( L / R ) )

The current is 0 A at t = 0 s, as expected, and exponentially approaches I0 as t → ∞. (c)

Assess: The current is zero at the start and approaches the steady final value. The behavior is similar to the charging of a capacitor.

34.81.

Model: Assume the straight wire is very long so we can use the formula for the magnetic field produced by an “infinite” wire. Visualize: Please refer to Figure CP34.81. As the short wire moves in the field of the long wire, the charges in the short wire will experience a magnetic force. They will move inside the wire until the electric field created by the charge separation has a corresponding force that balances the magnetic force. The potential difference between the ends of the wire is due to this electric field. Solve: To find the potential difference between the ends of the short wire we need to know the electric field along it. The magnetic field of the long wire is into the page near the short wire, so the magnetic force on positive charges will be to the left. Positive charges will move to the left, leaving negative charges on the right. This creates an electric field and corresponding force to the right on the positive charges. Consider the charges at a distance x from the long wire. In equilibrium, the magnetic and electric forces balance. Using the magnetic field for an “infinite” wire,

qvB = qE ⇒ E = vB = v

μ0 I 2π x

Integrating the electric field from the negative end of the wire (right end) to the positive end (left end), the potential difference is ΔV = − ∫

d d +l

G G E ⋅ dl =

∫

d +l d

⎛ vμ0 I ⎞ ⎛ vμ0 I ⎞ d + l dx ⎛ vμ0 I ⎞ ⎛ d + l ⎞ =⎜ ⎜ ⎟ xˆ ⋅ dx xˆ = ⎜ ⎟∫ ⎟ ln ⎜ ⎟ ⎝ 2π x ⎠ ⎝ 2π ⎠ d x ⎝ 2π ⎠ ⎝ d ⎠

G where we have used dl = − dx xˆ . Assess: This seems reasonable, because the potential difference increases with current (stronger magnetic field) and velocity (larger motional emf).

34.82.

Model: Assume the RC circuit has wires with zero resistance. Assume that only the bottom horizontal wire in the RC circuit closest to the metal loop contributes to the magnetic flux through the loop. Assume that the wire appears infinite in length compared to the metal loop. Visualize: See Figure CP34.82. The current in the RC circuit decreases as the capacitor is discharged. There is a changing magnetic flux through the metal loop due to the changing current through the RC circuit, inducing an emf through the loop. Solve: (a) Current I(t) = I0e-t/RC flows from right to left through the wire at the bottom of the RC circuit, causing a magnetic field out of the paper through the metal loop. Since the RC circuit current is decreasing, the induced current through the loop is counterclockwise in order to increase the magnetic flux. (b) Using the result of Example 34.5, the flux through the metal loop due to the wire is Φm =

μ0 I (t )(2.0 cm) ⎛ 0.5 cm + 1.0 cm ⎞ −t −9 RC ln ⎜ ⎟ = ( 4.4 × 10 ) I 0e 2π 0.5 cm ⎝ ⎠

The magnitude of the induced emf in the metal loop is

d Φ ( 4.4 × 10 ) I 0e E= = dt RC −9

−t

RC

= ( 4.4 × 10−4 ) I 0e

−t

RC

The initial current through the RC circuit I0 = 20 V/2.0 Ω = 10 A. At t = 5.0 μs, E = 0.267 mV. The current through the loop I loop = E /R = 0.267 mV/0.050 Ω = 5.3 mA. Assess: As the current through the RC circuit decreases to a constant value of zero at large times, the induced emf also approaches zero.

34.83. Visualize: Please refer to Figure CP34.83. The area within in the loop is changing so the flux will change. This will produce an induced emf and corresponding current. The area of the loop is the area of a square. Solve: (a) The field is out of the page, so the flux is increasing outward as the loop expands. According to Lenz’s law, the induced current tries to prevent the increase of outward flux. It does so by generating a field into the page. This G requires aG clockwise current flow. (b) Take A parallel to B , so Φ = AB and E = |dΦ/dt| = B(dA/dt). The field is constant but the loop area is changing. Let the length of each edge of the loop be x at time t. This length is increasing linearly with time as the corner of the loop moves outward with speed v = 10 m/s. We have x = x0 + vxt = 0 + [(10 m/s)cos 45°]t = 7.07t m The loop’s area at time t is A = x 2 = (7.07t m)2 = 50t 2 m 2 ⇒

dA = 100t m 2 /s dt

Consequently, the induced emf at time t is dA = (0.10 T)(100t m 2 /s) = 10t V dt The induced current is I = E/R, where R is the resistance of the loop. The wire has resistance 0.010 Ω per meter, so R = (0.010 Ω/m)l where l = 4x is the perimeter of the square at time t. That is, R = 4(0.010 Ω/m)(7.07t m) = 0.283t Ω E=B

Thus, the induced current is I=

E 10t V = = 35 A R 0.283t Ω

(c) At t = 0.1 s, E = 1.0 V and I = 35 A. Assess: The induced emf depends on time, but so does the resistance. This means the induced current is constant.

34.84.

Model: Assume that the sides remain straight during the collapse. Also, assume the magnetic field is constant in the region of interest. Visualize: As the loop collapses the area will change, there will be a corresponding change in flux and, therefore, an induced emf and current. The loop is made up of four right triangles of equal size with width x, height y, and hypotenuse b = 10 cm. The corner of the loop is moving in the x direction.

Solve: (a) For simplicity we will focus on the upper-right triangle. The width of the upper-right triangle is given by x = x0 + vt , with x0 = b 2 = ( 0.10 m ) 2 = 0.0707 m and v = 0.293 m/s. The height y of the triangle

is y = b 2 − x 2 . The loop will be a straight line when x = b. The time when this occurs is b − x0 0.10 m − 0.0707 m = = 0.100 s v 0.293 m/s (b) The induced current is due to the changing flux as a result of the changing area. We have tcollapse =

I=

E 1 d Φ B dA = = R R dt R dt

The total area at any time is four times the area of the upper-right triangle: A = 4 × 12 xy = 2 x b 2 − x 2 Using the chain rule,

I=

(

)

B dA dx B d 2 Bv ⎛ b 2 − 2 x 2 ⎞ 2 x b2 − x 2 = = v ⎜ ⎟ R dx dt R dx R ⎝ b2 − x 2 ⎠

2 2 ⎛ 2 ( 0.293 m/s )( 0.50 T ) ⎜ ( 0.10 m ) − 2 ⎣⎡0.0707 m + ( 0.293 m/s ) t ⎦⎤ ⎜ 2 0.10 Ω ⎜ ( 0.10 m )2 − ⎡⎣0.0707 m + ( 0.293 m/s ) t ⎤⎦ ⎝ 2 2 ⎛ ( 0.10 ) − 2 ⎡⎣0.0707 + ( 0.293) t ⎤⎦ ⎟⎞ = 2.93 ⎜⎜ A 2 ⎟ ⎜ ( 0.10 )2 − ⎡⎣0.0707 + ( 0.293) t ⎤⎦ ⎟ ⎝ ⎠

=

where t is in seconds. (c)

⎞ ⎟ ⎟ ⎟ ⎠

t (s) 0.000 0.020 0.040 0.060 0.080 0.090 0.100

I (A) 0.000 0.078 0.186 0.348 0.671 1.078 diverges

34.85.

Model: Visualize:

Assume the field changes abruptly at the boundary and is constant.

The loop moving through the field will have a changing flux. This will produce an induced emf and corresponding current. The induced current will oppose the change in flux and result in a retarding force on the loop. Solve: (a) The loop moving into the field will have an increasing outward flux. The induced current will oppose this change and create a field that is into the page. This requires that the induced current flow clockwise. The current-carrying wires of the loop will experience a force. The force on the leading edge is to the left, retarding the motion. The forces on the top and bottom edges cancel and the trailing edge experiences no force since the field is zero there. The induced current is I=

E 1 d Φ B dA Bl dx Blv = = = = R R dt R dt R dt R

With this current we can find the force on the leading edge. Remembering that it opposes the motion, we have

⎛ B 2l 2 ⎞ ⎛ Blv ⎞ F = IlB = ⎜ lB = ⎜ ⎟v ⎟ ⎝ R ⎠ ⎝ R ⎠ Newton’s second law is

ma = m

⎛ B 2l 2 ⎞ ⎛ B 2l 2 ⎞ dv dv − bt = −⎜ = −⎜ ⎟v ⇒ ⎟ v ⇒ v = v0e dt dt ⎝ R ⎠ ⎝ mR ⎠

where b = l 2 B 2 mR and the initial velocity is v0. (b) The value of b is

( 0.1 m ) ( 0.1 T ) = 100 s −1 ( 0.001 kg )( 0.001 Ω ) 2

b=

Assess:

2

This exponential decay of the velocity is called eddy-current braking.

34.86.

Model: Visualize:

Assume the field changes uniformly when turned on and that the loop is initially at rest.

The changing magnetic field produces a changing flux that will generate an induced emf and current. The current-carrying wire will experience a force that opposes the change. Solve: (a) The increasing field strength produces an increasing outward flux. The induced current will oppose this change and create a field that is into the page. This requires that the induced current flow clockwise. The G G G G current-carrying wires of the loop will experience a force. Take A parallel to B so the flux is Φ = A ⋅ B = AB . Because the field strength is changing and the loop is moving, the emf has two terms. We have E=

d Φ 1 ⎛ dB dA ⎞ = ⎜A + B⎟ dt R ⎝ dt dt ⎠

We will assume that the loop does not move much and ignore the dA/dt term (see below). The current is E 1 dB l 2 ΔB ( 0.080 m ) (100 T/s ) = A = = = 32 A R R dt 2 R Δt 2 ( 0.010 Ω ) 2

I=

(b) For a clockwise current, the force on the leading edge is to the left, so it is away from the magnetic field. The forces on the top and bottom edges cancel, and the trailing edge experiences no force since the field is zero there. The force on the loop is

F = IlB = ( 32 A )( 0.080 m )( 0.50 T ) = 1.28 N (c) Using Newton’s second law,

a=

F 1.28 N = = 128 m/s 2 m 0.010 kg

If the acceleration is constant, the distance the loop moves is Δs = 12 a ( Δt ) = 12 (128 m/s 2 ) ( 0.010 s ) = 6.4 × 10−3 m = 0.64 cm 2

2

(d) The distance traveled is about a factor of ten smaller than the size of the loop, so ignoring the motion of the loop is reasonable. (e) To find the speed after being “kicked” we calculate the impulse. Using the average force found in the part (b), Fave Δt = Δp = mvf − mvi ⇒ vf =

Assess:

Fave Δt (1.28 N )( 0.010 s ) = = 1.28 m/s m 0.010 kg

This shows that the loop would be shot out of the field region.

34.87.

Model: Assume the conductor is long enough to use the formula for the magnetic field of an “infinite” wire. Visualize: Please refer to Figure CP34.87.The current in the inner conductor creates a field that circles it in the space between the inner and outer conductor. Magnetic flux exists in the space. Solve: (a) We will consider a rectangle between the conductors of length l that is perpendicular to the magnetic field G G produced by the inner wire. Take A parallel to B so the flux through a small piece of the rectangle of width dr is G G d Φ = B ⋅ dA = Bldr . The total flux is r2 ⎛ μ I ⎞ μ Il r 2 dr μ0 Il ⎛ r2 ⎞ Φ = ∫ d Φ = ∫ ⎜ 0 ⎟ l dr = 0 ∫ = ln ⎜ ⎟ r1 2π r1 r 2π ⎝ r1 ⎠ ⎝ 2π r ⎠

where we have used the formula for the field of an infinite wire. We want the inductance per unit length, so we take the inductance and divide by the length. We have

L=

⎛r ⎞ Φ Φ μ ⇒ L = = 0 ln ⎜ 2 ⎟ I Il 2π ⎝ r1 ⎠

(b) The inductance per unit length is L ( 4π × 10 T m/A ) ⎛ 3.0 ⎞ −7 = ln ⎜ ⎟ = 3.6 × 10 H/m = 0.36 μ H/m l 2π 0.50 ⎝ ⎠ −7

34-1

35.1. Model: Apply the Galilean transformation of velocity. Solve:

(a) In the laboratory frame S, the speed of the proton is

v=

(1.41×10

6

m/s ) + (1.41 × 106 m/s ) = 2.0 × 106 m/s 2

2

The angle the velocity vector makes with the positive y-axis is

⎛ 1.41 × 106 m/s ⎞ ⎟ = 45° 6 ⎝ 1.41 × 10 m/s ⎠ G (b) In the rocket frame S′, we need to first determine the vector v′ . Equation 34.1 yields: G G G v′ = v − V = 1.41 × 106 iˆ + 1.41 × 106 ˆj m/s − 1.00 × 106 iˆ m/s = 0.41 × 106 iˆ + 1.41 × 106 ˆj m/s

θ = tan −1 ⎜

(

)

(

)

(

The speed of the proton is v′ =

( 0.41×10

6

m/s ) + (1.41 × 106 m/s ) = 1.47 × 106 m/s 2

2

The angle the velocity vector makes with the positive y′-axis is ⎛ 0.41 × 106 m/s ⎞ ⎟ = 16.2° 6 ⎝ 1.41× 10 m/s ⎠

θ ′ = tan −1 ⎜

)

35.2. Model: Apply the Galilean transformation of fields. Visualize: Please refer to Figure EX35.2. Solve: (a) Equation 35.11 gives the Galilean field transformation equation for magnetic fields: G G 1 G G B′ = B − 2 V × E c G G G G G B is in the positive kˆ direction, B = Bkˆ . For B′ > B, V × E must be in the negative kˆ direction. Since E = Ejˆ, G G G V must be in the negative iˆ direction, so that V × E = − Viˆ × Ejˆ = −VEkˆ. The rocket scientist will measure B′ >

( ) ( )

B, if the rocket moves along the –x-axis. G G (b) For B′ = B, V × E must be zero. The rocket scientist will measure B′ = B if the rocket moves along either the +y-axis or the –y-axis. G G (c) For B′ < B, V × E must be in the positive kˆ direction. The rocket scientist will measure B′ < B, if the rocket moves along the +x-axis.

35.3. Model: Use the Galilean transformation of fields. Solve: Equation 35.11 gives the Galilean transformation equations for the electric and magnetic fields in S and S′ frames: G G G G G G 1 G G E′ = E + V × B B′ = B − 2 V × E c G G G G In a region of space where B = 0 , E ′ = E = −1.0 × 106 ˆj V/m. The magnetic field is G G 1 1.0 × 1012 ˆ B′ = 0 − 2 1.0 × 106 iˆ m/s × −1.0 × 106 ˆj V/m = k T = 1.11× 10−5 kˆ T 8 2 c ( 3.0 ×10 )

(

) (

)

35.4. Model: Use the Galilean transformation of fields.

G B′ = 1.0 ˆj T, and Visualize: Please refer to Figure EX35.4. We are given V = 2.0 × 106 iˆ m/s, G 6 ˆ E ′ = 1.0 × 10 k V/m. Solve: Equation 35.11 gives the Galilean transformation equations for the electric and magnetic fields in S and S′ frames: G G G G E = E ′ − V × B′

G G 1 G G B = B′ + 2 V × E′ c

The electric and magnetic fields viewed from earth are G E = 1.0 × 106 kˆ V/m − 2.0 × 106 iˆ m/s × 1.0 ˆj T = − (1.0 × 106 V/m ) kˆ

(

) (

)

G 1 2.0 × 1012 V/m ˆ B = 1.0ˆj T + 2 2.0 × 106 iˆ m/s × 1.0 × 106 kˆ V/m = 1.0ˆj T − j = 0.99998 ˆj T 2 c (3.0 ×108 m/s )

(

Assess:

) (

)

Although B < B′, you need 5 significant figures of accuracy to tell the difference between them.

35.5. Model: Use the Galilean transformation of fields. Visualize: Please refer to G E = 12 iˆ + 12 ˆj × 106 V/m.

(

)

Figure

EX35.5.

We

are

G given V = 1.0 × 106 iˆ m/s,

G B = 0.50kˆ T,

and

Solve: Equation 35.11 gives the Galilean transformation equation for the electric field in the S and S′ frames: G G G G E ′ = E + V × B . The electric field from the moving rocket is G E ′ = iˆ + ˆj 0.707 × 106 V/m + 1.0 × 106 iˆ m/s × 0.50kˆ T = 0.707 × 106 iˆ + 0.207 × 106 ˆj V/m

(

)

(

) (

) (

⎛ 0.207 × 106 V/m ⎞ ⎟ = 16.3° above the x′-axis 6 ⎝ 0.707 × 10 V/m ⎠

θ = tan −1 ⎜

)

35.6. Model: The net magnetic flux over a closed surface is zero. Visualize: Solve:

Please refer to Figure EX35.6.

Because we can’t enclose a “net pole” within a surface, Φ m = úB ⋅ dA = 0 . Since the magnetic field is

uniform over each face of the box, the total magnetic flux around the box is

(1 cm × 2 cm )( 2 T − 2 T − 1 T + 3 T ) + (1 cm × 1 cm )( 2 T ) + Φ unknown = 0 ⇒ Φ unknown = ∫ Bunknown ⋅ dA = −0.0006 T m 2 ⇒ Bunknown cosθ = −6 T

The angle θ must be 180°. Because θ is the angle between B and the outward normal of d A , the field B is directed into the face.

35.7. Solve: The units of ε 0 ( d Φ e dt ) are

( N C) ( m2 ) C C2 × = =A N m2 s s

35.8. Solve: The displacement current is defined as Idisp = ε0(dΦe/dt). The electric flux inside a capacitor with plate area A is Φe = EA. The electric field inside a capacitor is E = η/ε0 = (Q/A)/ε0, and thus the electric flux is Φ e = EA =

Q

ε0

=

CVC

ε0

where VC is the capacitor voltage. The capacitance C is constant, hence the displacement current is

I disp = ε 0

d Φe d ⎛ CV ⎞ dV = ε0 ⎜ C ⎟ = C C dt dt ⎝ ε 0 ⎠ dt

35.9. Model: The displacement current is numerically equal to the current in the wires leading to and from the capacitor. Solve: During the charging process, a parallel-plate capacitor develops charge as a function of time. If the charge on a capacitor plate is Q at time t, then Q = CVC, where VC is the voltage across the capacitor plates. Taking the derivative,

dQ dV dV I 1.0 A = I =C C ⇒ C = = = 1.0 × 106 V/s dt dt dt C 1.0 μ F

35.10. Model: The electric field inside a parallel-plate capacitor is uniform. As the capacitor is charged, the changing electric field induces a magnetic field. Visualize: The induced magnetic field lines are circles concentric with the capacitor. Please refer to Figure 35.18. Solve: (a) The Ampere-Maxwell law is G G dΦ dE úB ⋅ d s = ε 0μ0 dt e = ε 0μ0 A dt where EA is the electric flux through G G the circle of radius r. The magnetic field is everywhere tangent to the circle of radius r, so the integral of B ⋅ d s around the circle is simply B(2πr). The Ampere-Maxwell law becomes

2π rB = ε 0 μ0 (π r 2 )

dE r dE ⇒ B = ε 0 μ0 dt 2 dt

On the axis, r = 0 m, so B = 0 T. (b) At r = 3.0 cm,

B=

⎛ 0.030 m ⎞ 6 −13 ⎜ ⎟ (1.0 × 10 V/m s ) = 1.67 × 10 T 2 ⎠ ( 3.0 ×10 m/s ) ⎝ 1

8

2

(c) For r > 5.0 cm, the electric flux Φe is the flux through a 10-cm-diameter circle because E = 0 V/m outside the capacitor plates. The Ampere-Maxwell law is dE B ( 2π r ) = ε 0 μ0π R 2 dt

1 ( 0.050 m ) 1.0 × 106 V/m s = 1.98 × 10−13 T R 2 dE = ( ) 2 2r dt ( 3.0 × 108 m/s ) 2 ( 0.07 m ) 2

⇒ B = ε 0 μ0

35.11. Model: The displacement current is numerically equal to the current in the wires leading to and from the capacitor. Solve: The process of charging increases the charge on the plates of a parallel-plate capacitor. The charge Q on a capacitor plate at time t is Q = CVC, where VC is the voltage across the capacitor plates. Taking the derivative, 2 2 −12 dQ dV ε A dVC ( 8.85 × 10 C /N m ) ( 0.050 m ) =C C = 0 = ( 500,000 V/s ) = 22 μ A dt dt d dt 0.50 × 10−3 m 2

I disp = I =

35.12. Model: The electric and magnetic field amplitudes of an electromagnetic wave are related. Solve:

Using Equation 35.30,

B0 =

E0 10 V/m = = 3.3 × 10−8 Vs/m2 = 3.3 × 10−8 T c 3 × 108 m/s

35.13. Model: The electric and magnetic field amplitudes of an electromagnetic wave are related. Solve:

Using Equation 35.30,

E0 = cB0 = ( 3.0 × 108 m/s )( 2.0 × 10 −3 T ) = 6.0 × 105 V/m

35.14. Model: Electromagnetic waves are sinusoidal. Solve: (a) The magnetic field is Bz = B0 sin ( kx − ωt ) , where B0 = 3.00 μΤ and k = 1.00 × 107 m–1. The wavelength is

λ=

2π 2π = = 6.28 × 10−7 m = 628 nm k 1.00 × 107 m −1

(b) The frequency is f =

c

λ

=

3.0 × 108 m/s = 4.77 × 1014 Hz 6.28 × 10−7 m

(c) The electric field amplitude is E0 = cB0 = (3 × 108 m/s)(3.00 × 10–6 T) = 900 V/m

35.15. Model: Electromagnetic waves are sinusoidal. Solve: (a) The electric field is Ey = E0 cos (kx – ωt), where E0 = 20.0 V/m and k = 6.28 × 108 m–1. The wavelength is

λ=

2π 2π = = 1.00 × 10−8 m = 10.0 nm k 6.28 × 108 m −1

(b) The frequency is f =

c

λ

=

3.0 × 108 m/s = 3.00 × 1016 Hz 1.00 × 10−8 m

(c) The magnetic field amplitude is

B0 =

E0 20.0 V/m = = 6.67 × 10−8 T vem 3.0 × 108 m/s

G G G G G E and B are perpendicular to each other and E × B is in the direction of vem . G G G Solve: At any point, the Poynting vector S ≡ μ0 −1E × B points in the direction that an electromagnetic wave is G G traveling. We have B = Biˆ and S = − Sjˆ . So,

35.16. Model:

G 1 G − Sjˆ = E × Biˆ ⇒ E = − Ekˆ

μ0

The electric field points in the negative z-direction.

35.17. Solve: (a) The units of cB are m m N N ×T = × = s s ( C )( m / s ) C

G G G These are the units for E. In the unit conversion, the units of tesla are taken from the equation F = qv × B . (b) The magnitude of the Poynting vector is EB E E E2 = = ( c 2ε 0 μ0 ) = cε 0 E 2 S= μ0 μ0 c μ0c The units of S are cε 0 E 2 ≡

m C2 N V C V AV W × × × = × = = s N m2 C m s m2 m2 m2

In the unit conversion, we have used both N/C and V/m for the units of the electric field. Since P = IV for circuits, A V = W.

35.18. Model: The electric and magnetic field amplitudes of an electromagnetic wave are related to each other. Solve:

(a) Using Equation 35.30,

1 100 V/m B0 = E0 = = 3.33 × 10−7 T c 3.0 × 108 m/s (b) From Equation 35.37, the intensity of an electromagnetic wave is I=

8 −12 2 2 cε 0 2 ( 3 × 10 m/s )( 8.85 × 10 C /N m ) 2 E0 = (100 V/m ) = 13.3 W/m2 2 2

35.19. Model: A radio signal is an electromagnetic wave. Solve:

From Equation 35.37, the intensity of an electromagnetic wave is ( 3 ×108 m/s )(8.85 ×10−12 C2 /N m2 ) 300 × 10−6 V/m 2 = 1.195 × 10−10 W/m2 cε I = 0 E02 = ( ) 2 2

35.20. Model: The laser beam is an electromagnetic plane wave. Assume that the energy is uniformly distributed over the diameter of the laser beam. Solve: (a) Using Equation 35.37, the light intensity is

I=

( 3 × 108 m/s )(8.85 × 10−12 C2/N m2 ) E 2 ⇒ E = 2.2 × 1011 V/m P cε 0 2 200 × 106 W = E0 ⇒ = 0 0 2 A 2 2 π (1.0 × 10−6 m )

(b) The electric field between the proton and the electron is

E=

9 2 2 −19 q ( 9 × 10 N m /C )(1.60 × 10 C ) = = 5.1 × 1011 V/m 2 2 4πε 0 r ( 0.053 ×10−9 m )

1

The ratio of the laser beam’s electric field to this field is 2.2 × 1011 V/m = 0.43 5.1 × 1011 V/m

The laser beam’s electric field is approximately half the electric field that keeps the electron in its orbit.

35.21. Model: A radio wave is an electromagnetic wave.

Solve: (a) The energy transported per second by the radio wave is 25 kW, or 25 × 103 J/s. This energy is carried uniformly in all directions. From Equation 35.37, the light intensity is

I= (b) Using Equation 35.37 again,

P P 25 × 103 W = = = 2.2 × 10−6 W/m 2 A 4π r 2 4π ( 30 × 103 m )2

3 × 108 m/s )( 8.85 × 10−12 C 2 /N m 2 ) 2 ( cε 0 2 −6 2 I= E0 ⇒ 2.2 × 10 W/m = E0 ⇒ E0 = 0.041 V/m 2 2

35.22. Solve: (a) The energy transported per second by the wave is 10 W, or 10 J/s. This energy is carried uniformly in all directions. From Equation 35.37, the light intensity is I=

2 (10 W ) P cε 2P 599.45 24.48 = 0 E02 ⇒ r = = = = 4π r 2 2 4π cε 0 E02 E02 E0 4π ( 3 × 108 m/s )( 8.85 × 10−12 C2 /N m 2 ) E02

For E0 = 100 V/m, r = 0.245 m. (b) For E0 = 0.010 V/m, r = 2.45 km.

35.23. Model: An object gains momentum when it absorbs electromagnetic waves. Solve:

The radiation force on an object that absorbs all the light is F=

Assess:

P 1000 W = = 3.3 × 10−6 N c 3.0 × 108 m/s

The force is independent of the size of the beam and the wavelength.

35.24. Model: A polarized radio wave is an electromagnetic wave. Solve:

(a) From Equation 35.30, B0 =

E0 1000 V/m = = 3.33 × 10−6 T c 3.0 × 108 m/s

(b) Likewise,

E 500 V/m = = 1.67 × 10−6 T c 3.0 × 108 m/s G The direction of the wave is into the page ( − kˆ direction) and E is down ( − ˆj direction). Using the right-hand G G G G rule and S = μ0 −1E × B , B is to the left ( −iˆ direction). (c) The wavelength of the radio wave is B=

λ=

3 × 108 m/s 3 × 108 m/s = = 300 m f 1.0 × 106 Hz

The magnetic field has half its maximum value at a point where 2π x ⎛ 2π x ⎞ 1 = 1.05 rad ⇒ x = 0.1667λ = 50 m cos ⎜ ⎟= ⇒ λ ⎝ λ ⎠ 2

35.25. Model: Use Malus’s law for the polarized light. Visualize:

Solve: From Equation 35.42, the relationship between the incident and the transmitted polarized light is = Itransmitted I0 cos2θ where θ is the angle between the electric field and the axis of the filter. Therefore, 0.25 I0 = I0 cos2θ ⇒ cosθ = 0.50 ⇒ θ = 60° Assess: Note that θ is the angle between the electric field and the axis of the filter.

35.26. Model: Use Malus’s law for the polarized light. Visualize:

Solve: We can use Equation 35.42 for transmitted power as well as intensity if the area of the beam does not change. The transmitted polarized light is Ptransmitted = P0 cos2θ = (200 mW)cos2(90° –25°) = 36 mW Assess: Note that θ is the angle between the electric field and the axis of the filter, which is 90° – 25° = 65°.

35.27. Model: Use Malus’s law for polarized light. Visualize:

Solve: For unpolarized light, the electric field vector varies randomly through all possible values of θ. Because the average value of cos2θ is 12 , the intensity transmitted by a polarizing filter is I transmitted = 12 I 0 . On the other hand, for polarized light Itransmitted = I0 cos2θ. Therefore,

I transmitted 2 = I transmitted 1 cos 2 θ = 12 I 0 cos 2 θ = 12 ( 350 W/m 2 ) cos 2 30° = 131 W/m 2

Assess: Note that any particular wave has a clear polarization. It is only in a “sea” of waves that the resultant wave has no polarization.

35.28. Model: Electric and magnetic fields exert forces on charged particles. Assume the fields are uniform. Visualize: Please refer to Figure P35.28. The electric field is in the direction of the positive y-axis and the magnetic field is in the direction of the negative z-axis. Solve: Substituting into the Lorentz force law, G G G G Fnet = q E + v × B = (1.6 × 10−19 C ) −1.0 × 106 iˆ V/m + 1.0 × 107 iˆ m/s × −0.10kˆ T

(

)

= 1.6 × 10

⇒ Fnet =

−13

(1.6 )

2

( −iˆ + ˆj ) N

(

+ (1.6 ) × 10−13 N = 2.3 × 10−13 N 2

The angle θ is measured counterclockwise from the +y-axis.

) (

(

))

⎛ 1.6 × 10−13 N ⎞ ⎟ = 45° −13 ⎝ 1.6 × 10 N ⎠

θ = tan −1 ⎜

35.29. Model: Assume the electric and magnetic fields are uniform. Visualize: Please refer to Figure P35.29. Solve: The force on the proton, which is the sum of the electric and magnetic forces, is G G G F = FE + FB = − F cos30°iˆ + F sin 30° ˆj = (−2.77iˆ + 1.60 ˆj ) × 10−13 N G G G G Since v points out of the page, the magnetic force is FB = ev × B = 1.60 × 10−13 ˆj N. Thus G G G G G G FE = eE = F − FB = −2.77 × 10−13 iˆ N ⇒ E = FE / e = −1.73 × 106 iˆ V/m G That is, the electric field is E = (1.73×106 V/m, left).

35.30. Model: Assume that the electric and magnetic fields are uniform fields. Visualize: Please refer to Figure P35.30. The magnetic force on the negative electron by the right-hand rule is directed downward. So that the electron is undeflected, we must apply an electric field to cause an electric force directed upward. That is, the electric field must point downward. Solve: For the electron to not deflect, G G FB = FE ⇒ e v × B = eE ⇒ E = vB sin 90° = ( 2.0 × 107 m/s ) ( 0.010 T ) = 2.0 × 105 V/m

35.31. Model: Use the Galilean transformation of fields. Assume that the electric and magnetic fields are uniform inside the capacitor. Visualize: Please refer to Figure P35.31. The laboratory frame is the S frame and the proton’s frame is the S′ frame. Solve: (a) The electric field is directed downward, and thus the electric force on the proton is downward. The G magnetic field B is oriented so that the force on the proton is directed upward. Use of the right-hand rule tells us that the magnetic field is directed into the page. The magnitude of the magnetic field is obtained from setting the magnetic force equal to the electric force, yielding the equation evB = eE. Solving for B,

B=

E 1.0 × 105 V/m = = 0.10 T v 1.0 × 106 m/s

G Thus B = (0.10 T, into page). (b) In the S′ frame, the magnetic and electric fields are

(

) (

)

1.0 × 106 ˆj m/s × 1.0 × 105 iˆ V/m G G 1 G G B′ = B − 2 V × E = −0.10kˆ T − ≈ −0.10kˆ T 2 8 c × 3.0 10 m/s ( ) G G G G V V E ′ = E + V × B = 1.0 × 105 iˆ + 1.0 × 106 ˆj m/s × −0.10kˆ T = 0 m m

(

) (

)

(c) There is no electric force in the proton’s frame because E′ = 0, and there is no magnetic force because the proton is at rest in the S′ frame.

35.32. Model: Electric and magnetic fields exert forces on charged particles. Assume the fields are uniform. Solve:

Substituting into the Lorentz force law, G G G G V ⎡ ⎤ Fnet = q E + v × B = ( −1.6 × 10−19 C ) ⎢ 2.0 × 105 iˆ − ˆj + 5.0 × 106 iˆ m/s × −0.10kˆ T ⎥ m ⎣ ⎦ −14 −14 ˆ −14 ˆ −14 ˆ ˆ ˆ = − ( 3.2 × 10 ) i − j N − 8.0 × 10 j N = −3.2 × 10 i − 4.8 × 10 j N

(

)

(

(

)

(

)

)

(

(

) (

)

)

35.33. Model: Use the Galilean transformation of fields. Visualize:

A current of 2.5 A flows to the right through the wire, and the plastic insulation has a charge of linear density λ= 2.5 n C/cm. Solve: The magnetic field B at a distance r from the wire is

G ⎛μI ⎞ B = ⎜ 0 , clockwise seen from left ⎟ r 2 π ⎝ ⎠

G On the other hand, E is radially out along rˆ , that is,

G E=

λ rˆ 2πε 0 r

As the mosquito is 1.0 cm from the center of the wire at the top of the wire, B=

E=

( 4π ×10

( 2.5 ×10

−7

T m/A ) ( 2.5 A )

2π ( 0.010 m )

−7

= 5.0 × 10−5 T

C/m ) ( 2 ) ( 9.0 × 109 N m 2 /C 2 ) 0.010 m

= 4.5 × 105

V m

G G where the direction of B is out of the page and the direction of E is radially outward. In the mosquito’s frame G G (let us call it S′), we want B′ = 0 T . Thus, G 1 G G G G 1 G G B′ = B − 2 V × E = 0 ⇒ B = 2 V × E c c G G G G G Because V × E must be in the direction of B and E is radially outward, according to the right-hand rule V must be along the direction of the current. The magnitude of the velocity is −5 8 c 2 B ( 3.0 × 10 m/s ) ( 5.0 × 10 T ) V= = = 1.0 × 107 m/s E 4.5 × 105 V/m 2

The mosquito must fly at 1.0 ×107 m/s parallel to the current. This is highly unlikely to happen unless the mosquito is from Planet Krypton, like Superman.

35.34. Model: Use the Galilean transformation of fields. Assume that the wire is infinite. Visualize: Please refer to Figure P35.34. The laboratory frame is frame S and the circular loop’s frame is frame S′. G Solve: (a) From Equation 27.15, the electric field E due to a wire at rest is

G E=

λ 2πε 0 r

⎛ λ ⎞ rˆ = ⎜ , away from wire ⎟ πε 2 r 0 ⎝ ⎠

G The magnetic field B at a point on the loop is zero because there are no moving charges. These are the fields measured in frame S. G G G G G G (b) In frame S′, E ′ = E + v × B = E because B = 0 T. The magnetic field in frame S′ is

G G 1 G G ⎛ 1 vλ ⎞ 1 G λ , into the page at the top ⎟ =⎜ 2 B′ = B − 2 v × E = 0 T − 2 v × rˆ 2πε 0 r ⎝ c ε 0 2π r c c ⎠ (c) Consider a segment of the wire of length Δx with charge ΔQ = λΔx. The experimenter in the loop’s frame sees a charge on this segment passing by him/her with a velocity v to the left. Thus,

ΔQ λΔx = = λv Δt Δt (d) The magnetic field in the loop’s frame is due to current I: I=

B′ =

μ0 I μ0λ v μ0λ v ε 0 1 λv = = × = 2π r 2π r 2π r ε 0 ε 0c 2 2π r

Since the current I is moving to the left, the right-hand rule states the direction of B′ is into the page on the top. G G The electric field E′ which is due to the charge on the wire would be equal to E. (e) As we see, the results in parts (b) and (d) are the same. (f) There will not be any induced current if the loop is made of a conducting material. This is because (i) the field is perpendicular to the area of the loop and (ii) the flux is not changing.

35.35. Model: Use Faraday’s law of electric induction and assume that the magnetic field inside the solenoid is uniform. Visualize:

Equation 35.17 for Faraday’s law is

G

G

úE ⋅ d s = −

G G dΦm d = − ⎡ ∫ B ⋅ dA⎤ ⎣ ⎦ dt dt

To solve this equation, choose a clockwise direction around a circle of radius r as the closed curve. The electric G field vectors, as the figure shows, are everywhere tangent to the curve. The line integral of E then is G G úE ⋅ d s = E ( 2π r ) G G To do the surface integral, we need to know the sign of the flux or the integral ∫ B ⋅ dA . Curl your right fingers

around the circle in the clockwise direction. Your thumb points to the right, which is along the same direction as G G G the magnetic field B . That is, ∫ B ⋅ dA = Bπ r 2 is positive. Solve:

(a) Since B = 10.0 T + (2.0 T)sin[2π (10 Hz)t], Faraday’s law simplifies to

E ( 2π r ) = −π r 2

dB V ⎞ ⎛ = −π r 2 ( 2.0 T ) ⎡⎣ 2π (10 Hz ) ⎤⎦ cos ⎡⎣ 2π (10 Hz ) t ⎤⎦ ⇒ E = − ⎜ 20.0 2 ⎟ π r cos ⎡⎣ 2π (10 Hz ) t ⎤⎦ dt m ⎝ ⎠

The field strength is maximum when the cosine function is equal to –1. Hence at r = 1.5 cm,

V ⎞ ⎛ Emax = ⎜ 20.0 2 ⎟ π ( 0.015 m ) = 0.94 V/m m ⎠ ⎝ (b) E is maximum when cos[2π (10 Hz)t] = –1 which means when sin[2π (10 Hz)t] = 0. Under this condition,

B = (10.0 T) + (2.0 T) sin [2π (10 Hz)t] = 10.0 T

That is, B = 10.0 T at the instant E has a maximum value of 0.94 V/m.

35.36.

Model: Assume a uniform electric field inside the capacitor. The displacement current Idisp between the capacitor plates is numerically equal to the current I in the wires leading to and from the capacitor. Solve: (a) From Equation 32.36, the voltage across the capacitor that develops during charging is VC = E (1 – e–t/τ), where τ = RC and E = 25 V is the battery emf. Since Q = CVC, we have

I disp = I =

dQ dV CE − t / τ CE CE E 25 V e ⇒ I disp max = =C C = = = = = 0.167 A dt dt RC R 150 Ω τ τ

The maximum electric flux through the capacitor plates is

Φ max = Emax A =

−12 Vmax E Cd EC ( 25 V ) ( 2.5 × 10 F ) A= = = = 7.1 V m ε 0 8.85 × 10−12 C 2 /N m 2 d d ε0

(b) As a function of time, the flux is

Φ=

VCC

ε0

=

C

ε0

E (1 − e −t / τ ) = Φ max (1 − e −t / τ )

⇒ Φ = ( 7.1 V m ) ⎡1 − e ⎢⎣

(

− 0.50×10−9 s

) ( 2.5×10−12 F × 150 Ω ) ⎤ = 5.2 V m ⎥⎦

Likewise, from part (a) the displacement current is

I disp =

CE

τ

e −t τ = ( 0.167 A ) e

(

− 0.50×10−9 s

) ( 2.5×10−12 F × 150 Ω )

= 0.044 A

35.37. Model: Use Equation 35.21 for the definition of the displacement current. Solve:

The current in a conductor arises from the electric field E in the conductor. From Equation 31.18, J=

I I dI d d d =σE ⇒ = (σ EA ) = σ ( EA ) = σ Φ e = σ disp ε0 A dt dt dt dt

where Φe = EA is the electric flux through the wire and, by definition, I disp = ε 0 d Φ e dt . Thus

I disp = ( ε 0 σ ) dI dt . (b) Using the value for the conductivity of copper wire from Table 31.2, I disp =

ε 0 dI 8.85 × 10−12 C 2 /N m 2 = (1.0 × 106 A/s ) = 1.48 × 10−13 A σ dt 6.0 × 107 Ω −1 m −1

35.38. Model: Assume the electric field inside the capacitor is uniform and use the Ampere-Maxwell law. Visualize:

Solve:

(a) For a current-carrying wire, Example 33.3 yields an equation for the magnetic field strength:

μ0 I wire ( 4π × 10 T m/A ) (10 A ) = = 1.0 × 10−3 T −3 2π r 2 π 2.0 × 10 m ( )( ) −7

B=

(b) Example 35.3 found that the induced magnetic field inside a charging capacitor is

B=

μ0 r dQ μ0 r = I wire 2π R 2 dt 2π R 2

where we used Iwire = dQ/dt as the actual current in the wire leading to the capacitor. At r = 2.0 mm, (4π × 10−7 T m/A)(2.0 × 10−3 m) B= (10 A) = 1.60 × 10−4 T 2π (5.0 × 10−3 m) 2

35.39. Model: The displacement current through a capacitor is the same as the current in the connecting wires. Solve:

We have

I disp = ε 0

d Φe d ⎛ CV ⎞ dV = ε0 ⎜ c ⎟ = C c dt dt ⎝ ε 0 ⎠ dt

The displacement current is the slope of the Vc vs t curve times C.

35.40. Model: Assume that the electric field inside the cylinder is uniform. Use the Ampere-Maxwell law to obtain the induced magnetic field. Visualize:

Solve:

G G (a) The electric flux through the entire cylinder is Φ = ∫ E ⋅ dA. For a closed curve of radius R, assuming

a clockwise direction around the ring, the right-hand rule places the thumb into the page, which is the same G direction as that of the electric field E . Thus, we will take the flux Φ to have a positive sign: 2 V ⎞⎡ ⎛ 3 2 −3 ⎤ Φ = E (π R 2 ) = ⎜1.0 × 108 t 2 ⎟ ⎣⎢π ( 3.0 × 10 m ) ⎥⎦ = ( 2.8 × 10 t ) Vm m ⎝ ⎠

(b) From the Ampere-Maxwell law, with Ithrough = 0, G

G

úB ⋅ d s = ε μ 0

0

d Φe d Φe = ε 0 μ0 dt dt

Because Φe is positive and the field strength increases with time, d Φ e dt is positive. The line integral is therefore positive, which means that B is in the clockwise direction. This reasoning applies to both the r < R and r > R regions. Thus the field lines are clockwise and they are shown in the figure. (c) We can now make use of the Ampere-Maxwell law to find the magnitude of B for r < R. Based upon parts (a) and (b), B ( 2π r ) = ε 0 μ0

r ⎡⎛ ⎤ dE 8 V⎞ −9 = (π r 2 ) ⇒ B = ε 0μ0 ⎛⎜⎝ 2r ⎞⎟⎠ dE ⎜1.0 × 10 ⎟ 2t = (1.11 × 10 rt ) T dt m ⎠ ⎥⎦ dt 2c 2 ⎢⎣⎝

where r is in m and t is in s. At r = 2 mm and t = 2.0 s,

B = (1.11 × 10−9 )( 2.0 × 10−3 ) ( 2.0 ) T = 4.4 × 10−12 T

(d) For r > R, the flux is confined to a circle of radius R. Thus

R 2 ⎡⎛ V⎞ ⎤ ⎛ ⎛ dE ⎞ 2 −14 t ⎞ = B 1.0 × 108 ⇒ B ( 2π r ) = ε 0 μ0 ⎜ π R ⎟ 2t ⎥ = ⎜1.00 × 10 ⎟T ⎟ 2 ⎢⎜ 2c r ⎣⎝ m⎠ ⎦ ⎝ r⎠ ⎝ dt ⎠ where r is in m and t is in s. At r = 4.0 mm and t = 2.0 s, 1.00 × 10−14 ( 2.0 ) B= T = 5.0 × 10−12 T ( 4.0 × 10−3 )

G

G

B are perpendicular in electromagnetic waves, and their magnitudes are related. 35.41. Model: GE and G

Solve:

(a) Since E ⊥ B , the dot product must be zero. G G E ⋅ B = 0 = ( 200 )( 7.3) + ( 300 )( −7.3) + ( −50 ) a

⇒ a = −14.6 G G G Since B0 multiplies the complete vector B it does not effect the calculation for a. Requiring E = c B and

squaring yields 2 2 2 2 2 2 2 2 2 ( 200 ) + ( 300 ) + ( −50 ) ( V/m ) = B0 2 ( 3.0 × 108 m/s ) ( 7.3) + ( −7.3) + ( −14.6 ) ( μ T )

(

)

(

⇒ B0 = 6.8 × 10

−2

(b) The Poynting vector G is G G S = μ0 −1E × B

Assess:

⎧ ⎡⎣( 300 )( −14.6 ) − ( −7.3)( −50 ) ⎤⎦ iˆ ⎫ ⎪⎪ ⎪⎪ = μ0 −1B0 (10−6 ) ⎨+ ⎣⎡( −50 )( 7.3) − ( −14.6 )( 200 ) ⎤⎦ ˆj ⎬ ⎪ ⎪ ⎪⎩+ ⎣⎡( 200 )( −7.3) − ( 7.3)( 300 ) ⎤⎦ kˆ ⎪⎭ = μ0 −1B0 (10−3 ) ⎡⎣ −4.75iˆ + 2.56 ˆj − 3.65kˆ ⎤⎦ = −260iˆ + 140 ˆj − 200kˆ W/m 2 G G G G A quick check yields E ⋅ S = 0 and B ⋅ S = 0.

)

1 ε0E2 and the magnetic field energy density is uB 2 = (1/2μ0)B2. In an electromagnetic wave, the fields are related by E = cB. Using this and the fact that c2 = 1/(ε0μ0), we find

35.42. Solve: (a) The electric field energy density is uE =

uE =

ε0 2

E2 =

ε0 2

(cB ) 2 =

ε0 1 2 c 2ε 0 2 B = B2 = B = uB 2 2ε 0 μ0 2 μ0

(b) Since the energy density is equally divided between the electric field and the magnetic field, the total energy 1 2 density in an electromagnetic wave is utot = uE + uB = 2uE. We also know that the wave intensity is I = cε0 E 0 . 2 Thus 2 2⎛ 1 ⎞ 2 I 2(1000 W/m ) utot = 2uE = ε 0 E02 = ⎜ cε 0 E02 ⎟ = = = 6.67 × 10−6 J/m3 3.0 × 108 m/s c⎝2 ⎠ c

35.43. Model: Light is an electromagnetic wave. A quick measurement of a light bulb shows that its radius is r ≈ 2.8 cm. Solve: 100 W is the energy transported per second by the electromagnetic light wave. This energy is carried in all directions. The light intensity is given by Equation 35.37: P P cε = 0 E02 I= = 2 A 4π r 2 8 ( 3 ×10 m/s )(8.85 ×10−12 C2/N m2 ) E 2 ⇒ E ≈ 2800 V 100 W ⇒ = 0 0 2 2 m 4π ( 2.8 × 10−2 m ) ⇒ B0 =

E0 ⎛ 2800 V/m ⎞ −6 =⎜ ⎟ ≈ 9.3 × 10 T c ⎝ 3.0 × 108 m/s ⎠

35.44. Model: Sunlight is an electromagnetic wave. Solve: (a) The sun’s energy is transported by the electromagnetic waves in all directions. From Equation 35.37, the light intensity is

I=

2 P 2 ⇒ P = IA = (1360 W/m 2 )( 4π Rsun-earth = (1360 W/m 2 ) 4π (1.50 × 1011 m ) = 3.85 × 1026 W ) A

(b) The intensity of sunlight at Mars is 2 2 11 P (1360 W/m )( 4π Rearth ) 2 ⎛ 1.50 × 10 m ⎞ 2 = = 1360 W/m ( ) ⎜ ⎟ = 589 W/m 2 11 4π RMars A ⎝ 2.28 × 10 m ⎠ 2

I=

35.45. Model:

The microwave beam is an electromagnetic wave. The water does not lose heat during the process. Solve: The rate of energy transfer from the beam to the cube is

Cε 0 2 E0 A 2 (3 ×108 m/s )(8.85 × 10−12 C2 /Nm2 )

P = ( 0.80 ) IA = ( 0.80 ) = ( 0.80 )

2

(11× 10

3

V/m ) ( 0.10 m ) = 1.29 kW 2

2

The amount of energy required to raise the temperature by 50 °C is ΔE = mcΔT = ( 0.10 m ) (1000 kg/m 3 ) ( 4186 J/kg/ °C )( 50 C ) = 2.09 × 105 J 3

The time required for the water to absorb this much energy from the microwave beam is Δt =

Assess:

ΔE 2.09 × 105 J = = 162 s P 1.29 × 103 W

Raising 1 kg of water by 50°C in a microwave oven takes around 2–3 minutes, so this is reasonable.

35.46. Model: The converging lens is an ideal thin lens, and the laser beam is pointed along its principal axis. Visualize:

Solve: (a) The lens will change the cross sectional area A of the laser beam. The power of the beam does not change. Since the intensity I ∝ A−1 , reducing the area by ¼ will increase I by a factor of 4. This occurs when the radius of the laser beam is halved, so that 2

2 A ⎛ r ⎞ πr = . A1 = π r12 = π ⎜ ⎟ = 2 4 4 ⎝ ⎠

The distance past the lens is halfway to the focal point at 5 cm. (b) Since I ∝ E0 2 , quadrupling E0 requires increasing I by a factor of 16, reducing the area by 16, requiring a reduction in radius of 4. The distance past the lens is ¾ of the way to the focal point, at 7.5 cm.

35.47. Model: Radio waves are electromagnetic waves. Solve: (a) The radio transmitter is radiating energy in all directions at the rate of 21 J per second. The signal intensity received at the earth is given by Equation 35.37: P P 21 W I= = = = 8.3 × 10−26 W/m 2 2 2 A 4π r 4π ( 4.5 × 1012 m ) (b) Using Equation 35.37 again,

I=

cε 0 2 2I E0 ⇒ E0 = = 2 cε 0

( 3.0 ×10

2 ( 8.3 × 10−26 W/m 2 )

8

m/s )( 8.85 × 10

−12

2

C /N m

2

)

= 7.9 × 10−12

V m

35.48. Model: Radio waves are electromagnetic waves. Assume that the transmitter unit radiates in all directions. Solve: The transmitting unit radiates energy in all directions at the rate of 250 mJ per second. From Equation 35.37, the signal intensity at a distance of 42 m is

I=

P P 250 × 10−3 W = = = 1.13 × 10−5 W/m 2 2 A 4π r 2 4π ( 42 m )

Using Equation 35.37 again, I=

cε 0 2 2I E0 ⇒ E0 = = cε 0 2

2 (1.13 × 10−5 W/m 2 )

( 3.0 × 108 m/s )(8.85 × 10−12 C2/N m2 )

= 0.092

V m

A few steps before 42 m, the field strength was 0.100 V/m and the door opened. The manufacturer’s claims are correct.

35.49. Model: The laser beam is an electromagnetic wave. Solve: The maximum intensity of the laser beam is determined by the maximum electric field strength in air. Thus the maximum power delivered by the beam is

cε 0 2 E0 A 2 ( 3 ×108 m/s )(8.85 × 10−12 C2/Nm2 )

P = IA = =

= 9.4 × 107 W/m 2

2

( 3.0 × 10

6

V/m ) π ( 0.050 m ) 2

2

35.50. Model: The radar beam is an electromagnetic wave. There is no absorption of the beam by the atmosphere. Solve: The intensity at the airplane is

(150 ×10 W ) = 1.33 ×10−5 W/m2 Psource = 2 2 4π r 4π ( 30 × 103 m ) 3

I=

The airplane then acts as a source of microwaves with power

P = IA = (1.33 × 10−5 W/m 2 )( 31 m 2 ) = 4.11 × 10−4 W The intensity of the reflected radar beam at the airport is thus I airport = =

P cε 2P = 0 E0 2 ⇒ E0 = 4π r 2 2 4πε 0cr 2 2 ( 4.11 × 10−4 W )

4π ( 8.85 × 10−12 C2 /Nm 2 )( 3 × 108 m/s )( 30 × 103 m )

= 5.2 μ V/m

2

35.51. Model: The earth is a complete absorber of sunlight. An object gains momentum when it absorbs electromagnetic waves. Solve: The radiation pressure on an object that absorbs all the light is

prad =

I 1360 W/m 2 = = 4.533 × 10−6 Pa c 3.0 × 108 m/s

2 Seen from the sun, the earth is a circle of radius Rearth and area A = π Rearth . The pressure exerts a force on this area 2 Frad = prad A = prad (π Rearth ) = ( 4.533 × 10−6 Pa )π ( 6.37 × 106 m ) = 5.78 × 108 N 2

The sun’s gravitational force on the earth is Fgrav =

GM sun M earth ( 6.67 × 10 = 2 Rsun-earth

⇒ That is, Frad is 1.64 × 10–12 % of Fgrav.

−11

N m 2 /kg 2 )(1.99 × 1030 kg )( 5.98 × 1024 kg )

(1.50 ×10

11

m)

2

Frad 5.78 × 108 N = = 1.64 × 10−14 Fgrav 3.53 × 1022 N

= 3.53 × 1022 N

35.52. Model: Assume that the black foil absorbs the laser light completely. Use the particle model for the foil. Visualize:

For the foil to levitate, the radiation-pressure force must be equal to the gravitational force on the foil. Solve: Using Equation 35.39,

Frad = prad A =

I P A = ⇒ P = cFrad = cFG = (3.0 × 108 m/s)(25 × 10–6 kg)(9.8 m/s2) = 73.5 W c c

35.53. Model: Assume that the black paper absorbs the light completely. Use the particle model for the paper. Visualize:

For the black paper to be suspended, the radiation-pressure force must be equal to the gravitational force on the paper. Solve: From Equation 35.39, Frad = prad A = IA c . Hence,

I=

( 3.0 ×108 m/s )(1.0 × 10−3 kg )( 9.8 m/s2 ) = 4.9 × 107 W/m2 c c Frad = FG = 2 A A ( 8.5 inch × 11 inch ) ( 2.54 × 10−2 m inch )

35.54. Model: Assume that the block absorbs the laser light completely. Use the particle model for the block. Solve:

From Equation 35.39,

Frad = prad A =

P 25 × 106 W = = 0.0833 N c 3.0 × 108 m/s

Applying Newton’s second law, Frad = ma = (100 kg)a ⇒ a =

Frad 0.0833 N = = 8.33 × 10−4 m/s 2 100 kg 100 kg

From kinematics,

vf2 = vi2 + 2a ( sf − si ) = 0 m 2 / s 2 + 2 ( 8.33 × 10−4 m/s 2 ) (100 m ) ⇒ vf = 0.408 m/s Assess:

This does not seem like a promising method for launching satellites.

35.55. Model: Use the particle model for the astronaut. Solve: According to Newton’s third law, the force of the radiation on the astronaut is equal to the momentum delivered by the radiation. For this force we have F = prad A =

P 1000 W = = 3.333 × 10−6 N c 3.0 × 108 m/s

Using Newton’s second law, the acceleration of the astronaut is

a=

3.333 × 10−6 N = 4.167 × 10−8 m/s 2 80 kg

Using vf = vi + a(tf – ti) and a time equal to the lifetime of the batteries, vf = 0 m/s + (4.167 × 10–8 m/s2)(3600 s) = 1.500 × 10–4 m/s

The distance traveled in the first hour is calculated as follows: vf2 − vi2 = 2a ( Δs )first hour

⇒ (1.500 × 10–4 m/s)2 – (0 m/s)2 = 2(4.167× 10–8 m/s2)(Δs)first hour ⇒ (Δs)first hour = 0.270 m This means the astronaut must cover a distance of 5.0 m – 0.27 m = 4.73 m in a time of 9 hours. The acceleration is zero during this time. The time it will take the astronaut to reach the space capsule is Δt =

4.73 m = 31,533 s = 8.76 hours 1.500 × 10−4 m/s

Because this time is less than 9 hours, the astronaut is able to make it safely to the space capsule.

35.56. Model: Use Malus’s law for the polarized light. Visualize:

Solve: For unpolarized light, the electric field vector varies randomly through all possible values of θ. Because the average value of cos2θ is 12 , the intensity transmitted by a polarizing filter when the incident light is unpolarized is I1 = 12 I0. For polarized light, Itransmitted = I0 cos2θ. Therefore, I2 = I1cos2 45° I3 = I2cos2 45° ⇒ I3 = (I1 cos2 45°)cos2 45° =

1 2

I0(cos4 45°) = 18 I 0

35.57. Model: Assume that the electric and magnetic fields are uniform fields. Solve:

Substituting into the Lorentz force law, G G G G G F = q E + v × B ⇒ ( 9.6 × 10−14 N ) iˆ − kˆ = − (1.60 × 10−19 C ) ⎡ E + 5.0 × 106 iˆ m/s × 0.10 ˆj T ⎤ ⎣ ⎦ G 5 5 5 ˆ ˆ ˆ ˆ ˆ ⇒ E = − ( 6.0 × 10 N/C ) i − k − ( 5.0 × 10 N/C ) k = −6.0i + 1.0k × 10 N/C

(

)

(

)

(

)

(

(

)

) (

)

35.58. Model: Assume that the electric field inside the capacitor is uniform. Use the Ampere-Maxwell law to find the magnetic field. Visualize:

Solve: (a) This is an RC circuit with capacitance C = ε0A/d = ε0πR2/d = 1.11 × 10–11 F = 11.1 pF. We know from Chapter 32 that the current through the wire decays exponentially as

I wire =

ΔVC − t / τ e = (5000 A)e−t / τ R

where the time constant is τ = RC = 2.22 × 10–12 s = 2.22 ps. Example 35.3 found that the induced magnetic field strength at radius r inside a charging or discharging capacitor is

B=

μ0 r dQ μ0 r = I wire 2π R 2 dt 2π R 2

where we used Iwire = dQ/dt as the actual current in the wire leading to the capacitor. Thus B=

(4π × 10−7 T m/A)(1.0 × 10−2 m) (5000 A)e − t / 2.22 ps = (0.025 A)e − t / 2.22 ps 2π (2.0 × 10−2 m) 2

(b) The graph is shown above.

35.59. Model: The displacement current through a capacitor is the same as the current in the wires. Solve:

From Chapter 32, a discharging capacitor has circuit current

I = I 0e

−t

τ

=

( ΔVc )0 R

e

−t

τ

Identifying τ = 2.0 μs = RC, we get R = 2.0 Ω. Thus

I0 =

( ΔVc )0 R

⇒ ( ΔVc )0 = I 0 R = (10 A )( 2.0 Ω ) = 20 V

35.60. Model: Dust particles absorb sunlight completely. Solve: Let R be the radius of a dust grain and ρ its density. The area of cross section of the dust particles is 4πR2, its mass is m = ρ ( 43π R 3 ) . Letting d be the distance of the dust grain from the sun, the gravitational force on the dust grain is

FG =

GM Sm d2

On the other hand, the pressure on the grain due to the sun’s electromagnetic radiation is prad =

I Psun F F R2 P = = rad = rad 2 ⇒ Frad = 2 sun 2 c ( 4π d ) c A 4π R d c

For the dust particles to remain in the solar system over long periods of time, Frad < FG. Hence, R 2 Psun GM Sm GM S ρ 4π R 3 3 Psun < = ⇒R> 2 2 2 d c d 3d 4π cGM S ρ

⇒R>

3( 3.9 × 1026 W )

4π ( 3.0 × 108 m/s )( 6.67 × 10−11 N m 2 /kg 2 )(1.99 × 1030 kg )( 2000 kg/m3 )

= 1.17 × 10−6 m

For R > 1.17 μm, radiation pressure force is less than the sun’s gravitational force, so the particle can orbit the sun. However, for R < 1.17 μm, radiation pressure force is greater than the sun’s gravitational force, so the particle is gradually pushed out of the solar system. Thus, the diameter of the dust particle that can remain in the solar system for a very long period of time is 2.34 μm.

35.61. Model: Use Ampere’s law and assume that the current through the resistor is uniform. Visualize:

Solve:

(a) The electric field E in the resistor is

E=

ΔVR IR = = Esurface L L

For the magnetic field, we use the Ampere-Maxwell law:

G

G

úB ⋅ d s = μ I

0 through

= μ0 I ⇒ B ( 2π r ) = μ0 I ⇒ B =

μ0 I = Bsurface 2π r

G G G (b) The Poynting vector S = μ0−1E × B is directed into the curved surface. That is, into the wire. The magnitude is S=

1

μ0

EB =

1 IR μ0 I I 2R = μ0 L 2π r 2π rL

(c) The flux over the surface of the resistor is

G G ∫ S ⋅ dA =

∫

faces

G G S ⋅ dA +

G G I 2R S ⋅ dA = 0 + SA = ( 2π rL ) = I 2 R wall ∫ 2π rL wall

The Poynting vector is the power per unit area carried by electric and magnetic fields. Thus the Poynting flux (SAwall) is electromagnetic power directed into the resistor. It matches the power dissipated by the resistor (I2R), which is what we expect from energy conservation.

35.62. Model: Use Malus’s law for the polarized light.

Solve: For unpolarized light, the electric field vector varies randomly through all possible values of θ. Because the average value of cos2θ is 12 , the intensity transmitted by the first polarizing filter is I1 = 12 I0. For polarized light,

Itransmitted I0 cos2θ. For the second filter the transmitted intensity is I2 = I1 cos2θ =

1 2

I0 cos2θ. Similarly, I3 = I2 cos2θ =

(cos2θ )2, and so on. Thus,

I 7 = I1 ( cos 2 θ )

7 −1

= 12 I 0 ( cos 2 θ ) = 12 I 0 ( cos 2 15° ) = 0.33I 0 6

6

1 2

= I0

36.1. Model: A phasor is a vector that rotates counterclockwise around the origin at angular frequency ω. Solve:

(a) Referring to the phasor in Figure EX36.1, the phase angle is

ωt = 180° + 30° = 210° ×

π rad 180°

= 3.665 rad ⇒ ω =

3.665 rad = 2.4 × 102 rad/s 15 × 10−3 s

(b) The instantaneous value of the emf is

E = E0 cos ωt = (12 V ) cos ( 3.665 rad ) = −10.4 V Assess:

Be careful to change your calculator to the radian mode to work with the trigonometric functions.

36.2. Model: A phasor is a vector that rotates counterclockwise around the origin at angular frequency ω. Solve:

(a) Referring to the phasor in Figure EX36.2, the phase angle is

ωt = 135° = 135° ×

π rad 180°

=

3π 3π 4 = 1178 rad/s rad ⇒ ω = 4 2.0 ms

(b) From Figure EX36.2,

E = E0 cos ωt ⇒ E0 =

50 V E = = −71 V cos ωt cos ( 3π /4 rad )

36.3. Model: A phasor is a vector that rotates counterclockwise around the origin at angular velocity ω. Solve:

The emf is E = E0 cosωt = (50 V)cos(2π × 110 rad/s × 3.0 × 10–3 s) = (50 V)cos(2.074 rad) = (50 V)cos119°

36.4. Model: A phasor is a vector that rotates counterclockwise around the origin at angular velocity ω. Solve:

The instantaneous emf is given by the equation E = (170 V) cos[(2π × 60 Hz)t]

At t = 60 ms, E = (170 V) cos (22.619 rad). An angle of 22.619 rad corresponds to 3.60 periods, which implies that the phasor makes an angle of (0.60)2π rad or (0.60)(360°) = 216° in its fourth cycle.

36.5. Visualize: Please refer to Figure EX36.4 for an AC resistor circuit. Solve:

(a) For a circuit with a single resistor, the peak current is

IR =

E0 10 V = = 0.050 A = 50 mA R 200 Ω

(b) The peak current is the same as in part (a) because the current is independent of frequency.

36.6. Model: Current and voltage phasors are vectors that rotate counterclockwise around the origin at angular frequency ω. Visualize: Solve:

Please refer to Figure EX36.6.

(a) The frequency is f =

ω 1 1 = = = 25 Hz 2π T 0.04 s

(b) From the figure we note that VR = 10 V and IR = 0.50 A. Using Ohm’s law, V 10 V R= R = = 20 Ω I R 0.50 A (c) The voltage and current are vR = VR cos ωt = (10 V ) cos ⎡⎣ 2π ( 25 Hz ) t ⎤⎦

iR = I R cos ωt = ( 0.50 A ) cos ⎡⎣ 2π ( 25 Hz ) t ⎤⎦

For both the voltage and the current at t = 15 ms, the phase angle is

ωt = 2π (25 Hz)(15 ms) = 2π(0.375) rad = 135° That is, the current and voltage phasors will make an angle of 135° with the starting t = 0 s position.

Assess: Ohm’s law applies to both the instantaneous and peak currents and voltages. For a resistor, the current and voltage are in phase.

36.7. Solve:

Visualize: Figure EX36.7 shows a simple one-capacitor circuit. (a) The capacitive reactance at ω = 2πf = 2π (100 Hz) = 628.3 rad/s is

XC =

⇒ IC =

1 1 = = 5305 Ω ωC ( 628.3 rad/s ) ( 0.30 × 10−6 F )

VC 10 V = = 1.88 × 10−3 A = 1.88 mA X C 5.305 × 103 Ω

(b) The capacitive reactance at ω = 2π(100 kHz) = 628,300 rad/s is

XC =

1 1 = = 5.305 Ω 5 ωC ( 6.283 × 10 rad/s )( 0.30 × 10−6 F ) ⇒ IC =

Assess:

VC 10 V = = 1.88 A X C 5.305 Ω

Using reactance is just like using resistance in Ohm’s law. Because X C ∝ ω −1 , XC decreases with an

increase in ω , as observed above.

36.8. Solve: (a) For a simple one-capacitor circuit, IC =

VC V = C = ωCVC X C 1/ω C

When the frequency is doubled, the new current is I C′ = ω ′CVC = ( 2ω ) CVC = 2 (ωCVC ) = 2 I C = 20 mA (b) Likewise, when the voltage is doubled, the current doubles to 20 mA. (c) When the frequency is halved and the emf is doubled, the current remains the same at 10 mA.

36.9. Visualize: Figure EX36.7 shows a simple one-capacitor circuit. Solve:

(a) From Equation 36.11,

IC =

VC V 50 × 10−3 A = C = ωCVC = 2π fCVC ⇒ f = = 80 kHz X C 1/ω C 2π ( 5.0 V ) ( 20 × 10−9 F )

(b) The AC current through a capacitor leads the capacitor voltage by 90° or π/2 rad. For a simple one-capacitor circuit iC = I C cos (ωt + 12 π ) . For iC = IC, (ωt + 12 π ) must be equal to 2nπ, where n = 1, 2, … . This means

ωt =

3π 7π 11π , , , … 2 2 2

At these values of ωt, v C = VC cos(ωt ) = 0 V. That is, iC is maximum when vC = 0 V.

36.10. Solve: From Equation 36.11, IC =

I 65 × 10−3 A VC V = 81 × 10−9 F = 81 nF = C = ωCVC ⇒ C = C = ωVC 2π (15,000 Hz ) ⎡ 2 ( 6.0 V ) ⎤ X C 1/ω C ⎣ ⎦

36.11. Solve: (a) From Equation 36.11, Ic = ⇒C =

Vc Vc = = ωCVc = 2π fCVc X c 1/ω C

(330 ×10−6 A ) Ic = = 9.5 × 10−11 C = 95 pC 2π fVc 2π ( 250 × 103 Hz ) ( 2.2 V )

(b) Doubling the frequency will halve the capacitive reactance Xc, doubling the current, so Ic = 2 (330 μA) = 660 μA.

36.12. Model: The current and voltage of a resistor are in phase, but the capacitor current leads the capacitor voltage by 90°. Solve: For an RC circuit, the peak voltages are related through Equation 36.12. We have E02 = VR2 + VC2 ⇒ VR = E02 − VC2 =

(10.0 V ) − ( 6.0 V ) 2

2

= 8.0 V

36.13. Model: The crossover frequency occurs for a series RC circuit when VR = VC. Visualize: Please refer to Figure EX36.13a for the low-pass filter circuit. Solve: From Equation 36.15,

ωc = 2π f c = Assess:

1 1 1 ⇒C = = = 1.59 × 10−6 F = 1.59 μ F RC 2π Rf c 2π (100 Ω )(1000 Hz )

The output for a low-pass filter is across the capacitor.

36.14. Model: The crossover frequency occurs for a series RC circuit when VR = VC. Visualize: Please refer to Figure EX36.13b for the high-pass filter circuit. Solve: From Equation 36.15,

ωc = 2π f c = Assess:

1 1 1 ⇒C = = = 1.59 × 10−6 F = 1.59 μ F RC 2π Rf c 2π (100 Ω )(1000 Hz )

The output for a high-pass filter is across the resistor.

36.15. Model: The current and voltage of a resistor are in phase, but the capacitor current leads the capacitor voltage by 90°. Visualize: Please refer to Figure EX36.15. Solve: From Equations 36.13 and 36.14, the peak voltages are VR = IR and VC = IXC, where XC =

1 1 1 = = = 199 Ω 4 ωC 2π fC 2π (1× 10 Hz)(80 × 10−4 F)

The peak current is

I=

E0 X +R 2 C

2

=

10 V

(199 Ω ) + (150 Ω ) 2

2

= 0.0401 A

Thus, VR = (0.0401 A)(150 Ω) = 6.0 V and VC = IXC = (0.0401 A)(199 Ω) = 8.0 V.

36.16. Visualize: Please refer to Figure EX36.13b for a high-pass RC filter. Solve:

(a) From Equation 36.15, the crossover frequency is

fc =

1 1 = = 1000 Hz 2π RC 2π (100 Ω ) (1.59 × 10−6 F )

(b) The resistor voltage in an RC circuit is VR =

E 0R R 2 + 1/ω 2C 2

=

E0 1 + 1/ω 2 R 2C 2

where, in the last step, we factored R2 from the square root. Using the definition ωc = 1/RC, we can write the capacitor voltage as E0 E0 E0 = = VR = 2 2 2 2 2 1 + 1/ω R C 1 + ωc /ω 1 + ( fc / f )2 Using E 0 = 5.00 V, we find

Assess:

f

VR (V)

1 2 c

f

2.24

fc

3.53

2fc

4.47

As expected of a high-pass filter, the voltage increases with increasing frequency.

36.17. Visualize: Please refer to Figure EX36.13a for a low-pass RC filter. Solve:

(a) From Equation 36.15, the crossover frequency is

fc =

1 1 = = 10.0 Hz 2π RC 2π (159 Ω ) (100 × 10−6 F )

(b) The capacitor voltage in an RC circuit is VC =

E 0 /ω C R + 1/ω C 2

2

2

=

E 0 /ω RC 1 + 1/ω 2 R 2C 2

where, in the last step, we factored R2 from the square root. Using the definition ωc = 1/RC, we can write the capacitor voltage as E 0 /ω RC E 0ωc /ω E 0( f c / f ) VC = = = 2 2 2 2 2 1 + 1/ω R C 1 + ωc /ω 1 + ( fc / f )2 Using E 0 = 5.00 V, we find

Assess:

f

VC (V)

1 2 c

f

4.47

fc

3.53

2fc

2.24

As expected of a low-pass filter, the voltage decreases with increasing frequency.

36.18. Solve: (a) For a simple one-inductor circuit, IL =

VL V = L X L 2π fL

If the frequency is doubled, the new current will be VL I = L = 5.0 mA I L′ = 2π ( 2 f ) L 2 (b) If the voltage is doubled, the current will double to 20 mA. (c) If the voltage is doubled and the frequency is halved, the current will quadruple to 40 mA.

36.19. Visualize: Figure EX36.14 shows a simple one-inductor circuit. Solve:

(a) The peak current through the inductor is

IL =

VL VL V 10 V = = L = = 0.80 A X L ω L 2π fL 2π (100 Hz ) ( 20 × 10−3 H )

(b) At a frequency of 100 kHz instead of 100 Hz as in part (a), the reactance will increase by a factor of 1000 and thus the current will decrease by a factor of 1000. Thus, IL = 0.80 mA.

36.20. Solve: For a simple one-inductor circuit, IL =

2 ( 6.0 V ) VL V VL 2Vrms = L ⇒L= = = = 1.39 × 10 −3 H = 1.39 mH X L 2π fL 2π f I L 2π f I L 2π (15 × 103 Hz )( 65 × 10−3 A )

36.21. Model: The AC current through an inductor lags the inductor voltage by 90°. Solve:

(a) From Equation 36.21, V V 5.0 V I L = 50 mA = L = L ⇒ f = = 3.2 × 104 Hz X L 2π fL 2π 50 × 10−3 A 500 × 10−6 H

(

)(

)

(b) The current and voltage for a simple one-inductor circuit are iL = I L cos (ωt − 12 π )

vL = VL cos ωt

For iL = IL, ωt − 12 π must be equal to 2nπ, where n = 0, 1, 2, … . This means ωt = ( 2nπ + 12 π ) . Thus, the instantaneous value of the emf at the instant when iL = IL is vL = 0 V.

36.22. Model: The AC current through an inductor lags the inductor voltage by 90o. Solve:

(a) From Equation 36.21,

I L = 330 μ A = ⇒L=

VL V = L X L 2π fL

2.2 V = 2.4 × 10−5 H = 24 μ H 2π ( 45 × 106 Hz )( 330 × 10−6 A )

(b) Doubling the frequency will double the inductive reactance and halve the current, so

IL =

330 μ A = 165μ A 2

36.23. Solve: (a) When the resistance is doubled, the resonance frequency stays the same because f is independent of R. Hence, f = 200 kHz. (b) From Equation 36.30,

f =

1 2π

1 LC

When the capacitor value is doubled,

f′=

1 2π

1 L ( 2C )

=

f 200 kHz = = 141 kHz 2 2

36.24. Solve: (a) When the resistor value is doubled, the frequency f is unchanged because f does not depend on R. Hence f = 200 kHz. (b) From Equation 36.30, 1 1 2π LC When the capacitor value is doubled and the inductor value is halved, then f =

f′=

1 2π

1

( L 2 )( 2C )

= f = 200 kHz

36.25. Model: At the resonance frequency, the current in a series RLC circuit is a maximum. The resistor does not affect the resonance frequency. Solve: From Equation 36.30, f =

1 2π

1 1 1 ⇒ C= 2 2 = 2 = 1.27 × 10−6 F = 1.27 μ F 4π f L 4π (1000 Hz )2 ( 20 × 10−3 H ) LC

36.26. Model: At the resonance frequency, the current in the series RLC circuit is a maximum. The resistor does not affect the resonance frequency. Solve: From Equation 36.30, f =

1 2π

1 1 1 ⇒ L= 2 2 = 2 = 10.1 mH 4π f C 4π (1000 Hz )2 ( 2.5 × 10−6 F ) LC

36.27. Visualize: The circuit looks like Figure 36.17. Solve:

(a) The impedance of the circuit for a frequency of 3000 Hz is

Z = R + ( XL − XC ) 2

=

( 50 Ω )

2

2

⎡ ⎤ 1 ⎥ = ( 50 Ω ) + ⎢ 2π ( 3 × 103 Hz )( 3.3 × 10−3 H ) − 3 −9 2π ( 3 × 10 Hz )( 480 × 10 F ) ⎥⎦ ⎢⎣ 2

+ ( 62.20 Ω − 110.52 Ω ) = 69.53 Ω ≈ 70 Ω 2

The peak current is

I=

E0 5.0 V = = 0.072 A = 72 mA Z 69.53 Ω

The phase angle is ⎡ XL − XC ⎤ −1 ⎛ −48.32 Ω ⎞ ⎥ = tan ⎜ 50 Ω ⎟ = −44° R ⎣ ⎦ ⎝ ⎠

φ = tan −1 ⎢

(b) For 4000 Hz, Z = 50.0 Ω I = 0.100 A, and φ = 0°. (c) For 5000 Hz, Z = 62.42 Ω ≈ 62 Ω, I = 0.080 A, and φ = 37° The following table summarizes the results. Z (Ω) I (A)

φ

f = 3000 Hz 70 0.072 –44°

f = 4000 Hz 50 0.100 0°

f = 5000 Hz 62 0.080 37°

2

36.28. Solve: When a capacitive reactance XC and an inductive reactance XL become equal, ωL =

1 ⇒ω = ωC

1 1 ⇒ f = 2π LC

1

(1.0 ×10

−6

H )(1.0 × 10−6 F )

The reactance at this frequency is XC = XL = ωL = 2π (159 kHz)(1.0 × 10−6 H) = 1.0 Ω

= 159 kHz

36.29. Visualize: Assume a simple one-resistor circuit. Solve: An electrical device labeled 1500 W is designed to dissipate an average 1500 W at Vrms = 120 V. We can use Equation 36.39 to find that

(120 V ) = 9.6 Ω V2 R = rms = Pavg 1500 W 2

Assess:

Calculations with rms values are just like the calculations for DC circuits.

36.30. Visualize: Assume a simple one-resistor circuit. Solve: An electrical power outlet in the United States is 120 V/60 Hz. This is Erms. We can use Equation 36.39 to find 2 Vrms (120 V ) = 144 W = R 100 Ω 2

Pavg =

36.31. Model: The motor has inductance, otherwise the power factor would be 1. Treat the circuit as a series RLC circuit without the capacitor (XC = 0 Ω). Solve: The average power is

Psource = I rmsErms cos φ = ( 3.5 A )(120 V ) cos 20° = 4.0 × 102 W

36.32. Visualize: Assume a simple one-resistor circuit. Solve:

From Equation 36.39, 2 (10.0 V ) = 50 Ω Vrms = Pavg 2.0 W 2

R= Using the equation again,

Vrms = Pavg R =

(10.0 W )( 50 Ω ) = 22 V

36.33. Model: The energy supplied by the emf source to the RLC circuit is dissipated by the resistor. Because of the phase difference between the current and the emf, the energy dissipated is PR = I rmsErms cos φ = I rmsVrms (Equation 36.47). Solve: From Equation 36.28, VR = E0 cosφ ⇒ Vrms = Erms cos φ

Using Ohm’s law, R=

Vrms Erms cos φ (120 V )( 0.87 ) = = = 44 Ω 2.4 A I rms I rms

36.34. Model: The energy supplied by the emf source to the RLC circuit is dissipated by the resistor. Because of the phase difference between the current and the emf, the energy dissipated is PR = I rmsErms cos φ = I rmsVrms (Equation 36.47). Solve: The power supplied by the source and dissipated by the resistor is Psource = PR = Pmaxcos2φ where Pmax = Imax E0 is the maximum possible power. For an RLC circuit, Imax = E0 R , so Pmax =

2 E02 Erms (120 V) 2 = = = 144 W 2R R 100 Ω

Thus the power factor is

cosφ =

PR 80 W = = 0.75 Pmax 144 W

36.35. Solve: (a) From Equation 36.14, E0 R

VR =

R + (ωcapC ) 2

(b) At this frequency,

VC = IX C =

−2

=

E0 1 1 ⇒ R 2 + 2 2 = 4 R 2 ⇒ ωres = ωresC 2 3RC

(

VR ⎛ 1 ⎞ E0 / 2 ⎜ ⎟= R ⎝ ωresC ⎠ R

)

(

3RC

) C1 =

3 E0 2

(c) The crossover frequency is

ωc =

1 = 6280 rad/s ⇒ ωres = ωc RC

3 = 3630 rad/s

36.36. Solve: (a) From Equation 36.14, E0 ωC VC =

R 2 + (ωcapC )

−2

=

1 4 3 E0 ⇒ R 2 + 2 2 = 2 2 ⇒ ωcap = ωcapC ωcapC RC 2

(b) At this frequency, VR = IR =

E0 R R 2 + (ωcapC )

−2

=

E0 R R + R /3 2

2

=

3 E0 2

(c) The crossover frequency is

ωc =

1 = 6280 rad/s ⇒ ωcap = 3 ωc = 10,877 rad/s RC

36.37. Visualize: Please refer to Figure P36.37. Solve:

(a) The voltage across the capacitor is E0

VC = IX C = =

R +X 2

2 C

XC =

(

E0 1 /ωC

(

10 V 4π f 2

2

(16 Ω )

2

(1.0 × 10

)

R + 1 / ωC 2

−6

F) + 1 2

=

)

2

=

E0

(ω RC )

2

+1

10 V 1 + (1.0106 × 10−8 s 2 ) f 2

The values of VC at a few frequencies are in the following table. f (kHz) 1 3 10 30 100

VC (V) 9.95 9.57 7.05 3.15 0.990

(b)

Assess:

For the voltage across the capacitor, the circuit is a low-pass filter.

36.38. Visualize: Please refer to Figure P36.38. Solve:

(a) The voltage across the resistor is

VR = IR =

E0 R R2 +

1 ω C2 2

10 V 1

= 1+

( 2π f ) (100 Ω ) 2

2

10 V

=

(1.6 ×10

−6

F)

2

1+

( 994.72 Hz )

2

f2

The values of VR at various values of f are in the following table. f (Hz) VR (V) 100 1.00 300 2.89 1000 7.09 3000 9.49 10,000 9.95 (b)

Assess:

The circuit acts like a high-pass filter when the output voltage is taken across the resistor.

36.39. Visualize: Please refer to Figure 36.13. Solve:

For an RC filter, I = E0

R 2 + X C2 and E0 R

VR = IR =

(

R + 1 /ω C 2

)

2

E0 X C

VC = IX C =

(

)

E0 R

=

R + 1 /ω C 2

2

At the crossover frequency ω = ωc = 1 RC and thus 1 ωcC = R. Hence,

VR =

E0 R R +R 2

2

=

E0 2

VC =

(

E0 1 /ωC

(

)

R + 1 /ωC 2

)

2

=

R2 + R2

E0 2

36.40.

Visualize:

Solve:

For a circuit with parallel capacitors, Cp = C1 + C2. The current in the circuit is

I rms =

Erms Erms I 545 × 10−3 A = = ωCpErms ⇒ Cp = C1 + C2 = rms = = 8.674 × 10−6 F XC ωErms 2π (1.00 × 103 Hz ) (10.0 V ) 1 /ωC

(

)

When these two capacitors are connected in series, Cs−1 = C1−1 + C2−1 . Using the same formula for the current, Cs =

C1C2 I 126 × 10−3 A = rms = = 2.005 × 10−6 F C1 + C2 ω Erms 2π (1.00 × 103 Hz ) (10.0 V )

We have two simultaneous equations for C1 and C2. The solutions are C1 = 3.15 μF and C2 = 5.53 μF.

36.41. Visualize: Figure 36.18 defines the phase angle φ. Solve:

The phase angle is

⎛ XL − XC ⎞ ⎟ R ⎝ ⎠

φ = tan −1 ⎜

A capacitor-only circuit has no resistor (hence R = 0 Ω) and no inductor (hence XL = 0 Ω). Thus the phase angle is

π ⎛ 0 − XC ⎞ −1 ⎟ = tan (−∞) = − rad 0 2 ⎝ ⎠

φ = tan −1 ⎜

That is, the current leads the voltage by π/2 rad or 90°. That is exactly the expected behavior for a capacitor circuit.

36.42.

Visualize:

Solve: (a) The figure shows that the equivalent capacitance of the three capacitors is 2.0 μF. The capacitive reactance and peak current are

XC =

1 1 = = 397.9 Ω 2π fC 2π ( 200 Hz ) ( 2.0 × 10−6 F )

I=

E0 10 V = = 0.02513 A = 25 mA X C 397.9 Ω

(b) All the current passes through the 3.0 μF capacitor. Thus the peak voltage across the 3.0 μF capacitor is

V3 μ F = IX C = ( 25.13 × 10−3 A )

1 = 6.7 V 2π ( 200 Hz ) ( 3.0 × 10−6 F )

36.43. Solve: From Equation 32.32, the voltage of a discharging capacitor is VC = V0e − t /RC ⇒ 12 V0 = V0e − ( 2.5 ms ) / RC ⇒ ln ( 12 ) = −

( 2.5 ms ) ⇒ RC = − ( 2.5 × 10−3 s ) = 3.61× 10−3 s RC

ln 0.5

The crossover frequency for a low-pass circuit is

ωc =

1 1 1 ⇒ fc = = = 44 Hz RC 2π RC 2π ( 3.61 × 10−3 s )

36.44. Model: The AC current through a capacitor leads the capacitor voltage by 90°. On the other hand, the current and the voltage are in phase for a resistor. Visualize: Parallel circuit components have the same potential difference. The voltage phasor of both the resistor and capacitor match the phasor of the emf.

Please refer to Figure P36.44. Solve: (a) Because we have a parallel RC circuit, the voltage across the resistor and the capacitor is the same. The current phasor IR is therefore along the same direction as the voltage phasor E0 , and the current phasor IC is ahead of the voltage phasor E0 by 90°. The peak currents in the resistor and capacitor are IR =

VR E0 = R R

IC =

VC E0 = XC 1 / ωC

(

)

(b) From the figure, we see that the phasors IR and IC are perpendicular to each other. We can combine them and find the phasor for the peak emf current I as follows: 2

2

⎛ E ⎞ ⎛E ⎞ I = I C2 + I R2 = ⎜ 0 ⎟ + ⎜ 0 ⎟ = E0 ⎝ 1 / ωC ⎠ ⎝ R ⎠

( ωC )

2

+

1 R2

36.45. Model: The AC current through a capacitor leads the capacitor voltage by 90o. A phasor is a vector rotating counterclockwise. Visualize: Please refer to Figure P36.45.

Solve:

(a) The emf frequency and that of the capacitor voltage are the same, although the waves are out of 1 = 50 Hz. phase. From the figure, the period is 0.02 s, so f = 0.02 s (b) From Figure P36.45, VC = 10 V and IC = 15 mA. We have

IC = ⇒C =

VC V = C = ωCVC X C 1 / ωC

15 × 10−3 A IC = = 4.8 × 10−6 F = 4.8μ F ωVC 2π ( 50 Hz )(10 V )

(c) From Figure P36.45,

vC = (10 V ) cos ( 2π ( 50 Hz ) t )

π⎞ ⎛ iC = (15 mA ) cos ⎜ 2π ( 50 Hz ) t + ⎟ 2⎠ ⎝ At t = 7.5 ms, ωt = 2π ( 50 Hz ) ( 7.5 × 10−3 s ) = 2.356 rad = 135°. The voltage and current phasors are shown in the figure above.

36.46. Model: The AC current through an inductor lags the inductor voltage by 90o. A phasor is a vector rotating counterclockwise. Visualize: Please refer to Figure P36.46.

Solve:

(a) The emf frequency and that of the inductor voltage are the same, although the waves are out of 1 = 50 Hz . phase. From the figure, the period is 0.02 s, so f = 0.02 s (b) From Figure P36.46, VL = 1 V and IL = 2 A. We have

IL =

VL VL V 1V = ⇒L= L = = 1.6 × 10−3 H = 1.6 mH . X L ωL ω I L 2π ( 50 Hz )( 2 A )

(c) From Figure P36.46,

vL = (1 V ) cos ( 2π ( 50 Hz ) t )

π⎞ ⎛ iL = ( 2 A ) cos ⎜ 2π ( 50 Hz ) t − ⎟ 2⎠ ⎝ At t = 7.5 ms, ωt = 2π ( 50 Hz ) ( 7.5 × 10−3 s ) = 2.356 rad = 135° . The voltage and current phasors are shown in the figure above.

36.47. Model: While the AC current through an inductor lags the inductor voltage by 90°, the current and the voltage are in phase for a resistor. Visualize: Series elements have the same current, so we start with a common current phasor I for the inductor and resistor,

Please refer to Figure P36.47. Solve: Because we have a series RL circuit, the current through the resistor and the inductor is the same. The voltage phasor VR is along the same direction as the current phasor I. The voltage phasor VL is ahead of the current phasor by 90°. (a) From the phasors in the figure, E0 = VL2 + VR2 . Noting that VL =IωL and VR = IR, E0 = I ω 2 L2 + R 2 and I=

E0 R +ω L 2

2 2

VR =

E0 R

VL =

R +ω L 2

2 2

E0ω L R 2 + ω 2 L2

(b) As ω → 0 rad/s, VR → E0 R R = E0 and as ω → ∞, VR → 0 V. (c) The LR circuit will be a low-pass filter, if the output is taken from the resistor. This is because VR is maximum when ω is low and goes to zero when ω becomes large. (d) At the cross-over frequency, VL = VR . Hence,

I ωc L = IR ⇒ ωc =

R L

36.48. Visualize: The circuit looks like the one in Figure 36.17. Solve:

(a) The impedance of the circuit is

Z = R2 + ( X L − X C ) = 2

=

(100 Ω )

2

⎡ ⎤ 1 2 ⎥ (100 Ω ) + ⎢ 2π ( 60 Hz )( 0.10 H ) − −6 2π ( 60 Hz ) (100 × 10 F ) ⎥⎦ ⎢⎣

+ ( 37.70 Ω − 26.53 Ω ) = 100.62 Ω 2

The peak current I is I=

E0 (120 V ) 2 = = 1.69 A Z 100.62 Ω

(b) The phase angle is ⎡ XL − XC ⎤ −1 ⎛ 11.173 Ω ⎞ ⎥ = tan ⎜ 100 Ω ⎟ = 6.4° R ⎣ ⎦ ⎝ ⎠

φ = tan −1 ⎢

This is an inductive circuit because φ is positive. (c) The average power loss is 2 E2 (120 V ) cos 6.4° = 143 W PR = I rmsErms cos φ = rms cos φ = ( ) R 100 Ω

2

36.49. Visualize: The circuit looks like the one in Figure 36.17. Solve:

(a) The impedance of the circuit is Z = R + ( XL − XC ) 2

=

(100 Ω )

2

2

⎛ ⎞ 1 ⎟ = (100 Ω ) + ⎜ 2π ( 60 Hz )( 0.15 H ) − −6 ⎜ 2π ( 60 Hz ) ( 30 × 10 F ) ⎟⎠ ⎝ 2

+ ( 56.55 Ω − 88.42 Ω ) = 104.96 Ω. 2

The peak current is I=

E0 (120 V ) 2 = = 1.617 A = 1.62 A Z 104.96

(b) The phase angle is ⎛ XL − XC ⎞ −1 ⎛ 56.55 Ω − 88.42 Ω ⎞ ⎟ = tan ⎜ ⎟ = −17.7° R 100 Ω ⎝ ⎠ ⎝ ⎠

φ = tan −1 ⎜

Because φ is negative, this is a capacitive circuit. (c) The average power loss is 2 Erms (120 V ) cos −17.7° = 137 W cos φ = ( ) R 100 Ω 2

PR = I rmsErms cos φ =

2

36.50. Visualize: Please refer to Figure P36.50. Solve:

(a) The resonance frequency of the circuit is

ω=

1 = LC

1

(10 × 10

−3

H )(10 × 10

−6

F)

= 3160 rad/s ⇒ f =

ω = 5.0 × 102 Hz 2π

(b) At resonance, XL = XC. So, the peak values are

I=

E0 10 V = = 1.0 A ⇒VR = IR = (1.0 A)(10 Ω) = 10.0 V R 10 Ω

⇒ VL = IXL = IωL = (1.0 A)(3160 rad/s)(10 × 10−3 Η) = 32 V (c) The instantaneous voltages must satisfy vR + vC + vL = E . But vC and vL are out of phase at resonance and cancel. Consequently, it is entirely possible for their peak values VC and VL to exceed E0 . VR + VC + VL = E0 is not a requirement of an AC circuit.

36.51. Visualize: Please refer to Figure P36.51. Solve:

(a) The resonance frequency of the circuit is

ω=

1 = LC

1

(1.0 × 10

−3

H )(1.0 × 10

−6

F)

= 3.2 × 10 4 rad/s ⇒ f =

ω = 5.0 × 103 Hz 2π

(b) At resonance XL = XC. So, the peak values are

I=

E0 10 V = = 1.0 A ⇒ VR = IR = (1.0 A)(10 Ω) = 10.0 V R 10 Ω

1.0 A ⎛ 1 ⎞ ⇒ VC = IX C = I ⎜ = 32 V ⎟= 4 −6 ⎝ ωC ⎠ ( 3.16 × 10 rad/s )(1.0 × 10 F ) (c) The instantaneous voltages must satisty vR + vC + vL = E . But vC and vL are out of phase at resonance and cancel. Consequently, it is entirely possible for their peak values VC and VL to exceed E0 . VR + VC + VL = E0 is not a requirement of an AC circuit.

36.52. Visualize: Please refer to Figure P36.52. When the frequency is very small, X C = 1 ωC becomes very large and XL = ωL becomes very small. So, the branch in the circuit with the capacitor has a very large

impedance and most of the current flows through the branch with the inductor. When the frequency is very large, the reverse is true and most of the current flows through the branch with the capacitor. Solve: (a) When the frequency is very small, the branch with the inductor has a very small XL, so Z ≅ R . The current supplied by the emf is I rms =

10 V = 0.10 A 100 Ω

(b) When the frequency is very large, the branch with the capacitor has a very small XC, so Z ≅ R . The current supplied by the emf is

I rms =

10 V = 0.20 A 50 Ω

36.53. Model: An RLC is driven above the resonance frequency when the circuit current lags the emf. Visualize: The circuit looks like the one in Figure 36.17. Solve: The phase angle is +30° since the circuit is above resonance, and X L > X C . Thus tan 30° =

XL − XC ⇒ X L − X C = R tan 30° R

The impedance of the circuit is Z = R 2 + ( X L − X C ) = R 2 + ( R tan 30° ) = R 1 + ( tan 30° ) 2

2

2

The peak current is I= Assess:

E0 10 V = = 0.173 A Z ( 50Ω ) 1 + ( tan 30° )2

Remember to put your calculator back into degrees mode when calculating tan 30°.

36.54. Visualize: The circuit looks like the one in Figure 36.17. Solve:

(a) The impedance and phase angle are

Z = R + ( XL − XC ) 2

=

( 50 Ω )

2

2

⎡ ⎤ 1 ⎥ = ( 50 Ω ) + ⎢ 2π ( 5.0 × 103 Hz )( 3.3 × 10−3 H ) − 3 −9 2π ( 5.0 × 10 Hz )( 480 × 10 F ) ⎥⎦ ⎢⎣ 2

+ ( 37.36 Ω ) = 62.42 Ω 2

⎛ XL − XC ⎞ −1 ⎛ 37.36 Ω ⎞ ⎟ = tan ⎜ ⎟ = 36.8° R ⎝ ⎠ ⎝ 50 Ω ⎠

φ = tan −1 ⎜

Since E = E0 cos ωt , E = E0 implies that ωt = 0. From Equation 36.22, the instantaneous current is i = I cos (ωt − φ ) =

5.0 V E0 cos (ωt − φ ) = cos ( 0° − 36.8° ) = 0.064 A Z 62.42 Ω

(b) E = 0 V and E decreasing implies that ωt = 12 π rad. As before,

i=

5.0 V cos ( 90° − 36.8° ) = 0.048 A 62.42 Ω

2

36.55. Visualize: The circuit looks like the one in Figure 36.17. (a) The instantaneous current is i = I cos (ωt − φ ) . The phase angle is

Solve:

⎡ XL − XC ⎤ ⎛ −48.32 Ω ⎞ = tan −1 ⎜ ⎟ = −44° ⎥ R ⎣ ⎦ ⎝ 50 Ω ⎠

φ = tan −1 ⎢

Since i = I cos (ωt − φ ) , i = I implies that ωt − φ = 0 rad. That is, ωt = φ = −44° . Thus,

E = E0 cos ωt = ( 5.0 V ) cos ( −44° ) = 3.6 V (b) i = 0 A and is decreasing implies that ωt − φ = 12 π rad. That is, ωt = 12 π + φ = 90° − 44° = 46° . Thus,

E = ( 5.0 V ) cos ( 46° ) = 3.5 V .

(c) i = −I implies ωt − φ = π rad. That is, ωt = π + φ = 180° − 44° = 136° . Thus, E = (5.0 V)cos136° = −3.6 V.

36.56. Model: The average power is the total energy dissipated per second. Visualize: The circuit looks like the one in Figure 36.17. Solve: (a) The resonance frequency is ω0 = 1 LC . At ω = 12 ω0 , the inductive and capacitive reactance are

X L = ω L = 12 ω0 L =

1 L 1 L = 2 LC 2 C

⇒ XL − XC = − ⇒ tan φ = ⇒Z =

XC =

1 2 2 LC L = = =2 ω C ω 0C C C

3 L 3 10 × 10−3 H =− = −4743 Ω 2 C 2 1.0 × 10−9 F

−4743 Ω = 47.43 rad ⇒ φ = −88.8° 100 Ω

( XL − XC )

2

+ R2 =

( 4743 Ω )

2

+ (100 Ω ) = 4744 Ω 2

From Equation 36.47, the average power supplied to the circuit is

⎛ 10 V ⎞ −4 Psource = I rmsErms cos φ = ⎜ ⎟ (10 V ) cos ( −88.8° ) = 4.4 × 10 W ⎝ 4744 Ω ⎠ (b) At ω = ω0, XL = XC. So, Z = R and φ = 0 . The average power supplied by the emf is

E2 (10 V ) = 1.0 W ⎛E ⎞ Psource = I rmsErms cos φ = ⎜ rms ⎟ Erms = rms = R 100 Ω ⎝ R ⎠ 2

(c) At ω = 2ω0, the inductive and capacitive reactance are

XL = 2

L C

⇒ XL − XC =

XC =

1 L 2 C

3 L = 4743 Ω 2 C

As in part (a), φ = +88.8° and Z = 4744 Ω . The average power is Psource = 4.4 × 10−4 W. Assess: The power dissipated by an RLC circuit falls off dramatically on either side of resonance.

36.57. Solve: From Equation 36.25, the impedance is Z = R 2 + ( X L − X C ) . The difference between the 2

inductive reactance and the capacitive reactance can be written

ωL −

⎛ ω02 ⎞ 1 1 ⎞ ω02 ⎞ ⎛ 2 2 2⎛ = ω L ⎜1 − 2 ⎟ = ω L ⎜1 − 2 ⎟ ⇒ Z = R + ω L ⎜1 − 2 ⎟ ωC ⎝ ω LC ⎠ ⎝ ω ⎠ ⎝ ω ⎠

2

36.58. Solve: (a) The peak current in a series RLC circuit is I=

E0 ⎛ E0 ⎞ ⎛ R ⎞ =⎜ ⎟ ⎜ ⎟ Z ⎝ R⎠ ⎝Z⎠

The maximum current is I max = E0 R and it occurs only at resonance because this is when the impedance

Z = R 2 + ( X L − X C ) is the smallest. Using this expression for Z, 2

R = Z

R R + ( XL − XC ) 2

2

=

1 ⎛ X − XC ⎞ 1+ ⎜ L ⎟ R ⎝ ⎠

2

=

1 1 + tan φ 2

=

1 sec2φ

= cos φ

Thus, I = I max cosφ . (b) From Equation 36.44, the average power is

Pavg = 12 I E0 cos φ = 12 ( I max cos φ ) E0 cos φ = ( 12 I maxE0 ) cos 2 φ = Pmax cos 2 φ where Pmax = 12 I max E0 is the maximum power the source can deliver to the circuit.

36.59. Solve: (a) The resonance frequency is ω0 =

1 1 1 = 2π f 0 ⇒ C = 2 2 = = 11.64 pF 4π f 0 L 4π 2 (104.3 × 106 Hz )2 ( 0.200 μ H ) LC

(b) The current produced by the out-of-tune radio station is 0.10% of the resonance current. Therefore,

E0 R + ( XL − XC ) 2

2

= (10−3 )

E0 ⇒ R2 + (XL – XC)2 = 106 R2 ≅ (XL – XC)2 ⇒ X L − X C = 103 R R

⎡ ⎤ 1 ⎞ 1 ⎛ −3 −3 ⇒ R = 10−3 ⎜ ω L − ⎥ = 1.49 × 10 Ω ⎟ = 10 ⎢ 2π (103.9 MHz )( 0.20 μ H ) − ωC ⎠ 2π (103.9 MHz )(11.6 pF ) ⎥⎦ ⎝ ⎢⎣ Assess:

The impedance at resonance is Z = R because XL = XC.

36.60. Solve: (a) The resonance frequency is f 0 = 57 MHz =

1 1 1 ⇒L= 2 2 = = 0.487 μ H ≈ 0.49 μ H 4π f 0 C 4π 2 ( 57 × 106 Hz )2 (16 × 10−12 F ) 2π LC

(b) We have I end frequency = 12 I resonance . Therefore, 2

E0 R2 + ( X L − X C )

2

=

1 E0 1 ⎞ 1 ⎛ 2 ⇒ R2 + ⎜ ω L − = 3R ⎟ = 4R ⇒ ω L − 2R ωC ⎠ ωC ⎝

For f = 60 MHz,

⎤ 1 ⎡ 1 ⎢ 2π ( 60 × 106 Hz ) ( 0.487 μ H ) − ⎥ 6 −12 2π ( 60 × 10 Hz )(16 × 10 F ) ⎥⎦ 3 ⎢⎣ 1 = [183.70 Ω − 165.79 Ω] = 10.3 Ω 3

R=

For f = 54 MHz, R = 10.9 Ω. The minimum possible value of the circuit resistance is thus 10.9 Ω.

36.61. Model: The filament in a light bulb acts as a resistor. Visualize: Please refer to Figure P36.61. Solve: A bulb labeled 40 W is designed to dissipate an average of 40 W of power at a voltage of Vrms = 120 V. From Equation 36.39, the resistance of the 40 W light bulb is

V2 (120 V ) = 360 Ω R40 = rms = 40 W P40 2

Likewise, R60 = 240 Ω and R100 = 144 Ω . The rms current through the 40 W and 60 W bulbs, which are in series, is I 60 = I 40 =

120 V E0 = = 0.20 A 600 Ω 600 Ω

The rms voltage across the light bulbs in series are V40 = I40R40 = (0.20 A)(360 Ω) = 72

V60 = I60R60 = (0.20 A)(240 Ω) = 48 V

The voltage across the 100 W light bulb is V100 = 120 V. The powers dissipated in the light bulbs in series are P40 = V40I40 = (72 V)(0.20 A) = 14.4 W

P60 = V60I60 = (48 V)(0.20 A) = 9.6 W

The power dissipated in the 100 W bulb is 2 V100 (120 V ) = 100 W = R100 144 Ω 2

P100 =

36.62. Model: The wires are resistors as well as conductors. Solve:

(a) From Equation 36.40, Psource = (Irms Erms)1 + (Irms Erms)2 + (Irms Erms)3 ⇒ 450 MW = 3Irms Erms = 3Irms(120 V)

⇒ I rms =

450 × 106 W = 1.25 × 106 A 3 (120 V )

(b) Using the same equation,

450 MW = 3Irms Erms = 3Irms(500 × 103 V) ⇒ I rms =

450 × 106 W = 300 A 3 ( 500 × 103 V )

(c) Copper may be a good conductor, but it still has some resistance. Even a thick copper cable cannot carry a 1.25 million amp current. It would melt! A step up to much higher voltages allows the power to be transmitted at much lower currents.

36.63. Model: Assume the motor is a series RLC circuit. Solve:

(a) From Equation 36.44, the average power is Psource = I rmsErms cos φ . Thus, the power factor is cos φ =

Psource 800 W = = 0.833 I rmsErms (120 V )( 8.0 A )

(b) The resistance voltage is Vrms = Erms cosφ = (120 V )( 0.833) = 100 V . (c) Using Ohm’s law, the resistance of the motor is V 100 V R = rms = = 12.5 Ω I rms 8.0 A (d) From the results of part (a) and from Equation 36.27, the phase angle is ⎛ X − XC ⎞ φ = cos −1 ( 0.833) = 33.6° = tan −1 ⎜ L ⎟ R ⎝ ⎠

⇒ X L − X C = (12.5 Ω ) tan 33.6° = 8.292 Ω

The phase angle becomes zero and the power factor becomes 1.0 by increasing the capacitive reactance by 8.292 Ω. The capacitor with this reactance is 1 = 3.20 × 10−4 F = 320 μ F C= 2π ( 60 Hz )( 8.292 Ω )

36.64. Model: Assume that the circuit is a series RLC circuit. Solve: (a) From Equation 36.44, Psource = I rmsErms cos φ . Hence, 6.0 × 106 W = Irms(15,000 V)(0.90) ⇒ Irms = 444 A (b)

The figure shows the nature of R, XC, XL, and Z as a phasor diagram. The power factor is cosφ = 0.9. The total circuit impedance is found from Z = Erms/Irms = (15,000 V)/(444 A) = 33.8 Ω. You can see from the right triangle in the diagram that X L − X C = Z sin φ = Z 1 − cos 2 φ = (33.8 Ω) 1 − (0.9) 2 = 14.7 Ω

You can bring the power factor up to 1.0 by reducing XL – XC to zero. To do this, you can increase XC by 14.7 Ω. Since XC = 1/ωC, the extra capacitance needed to raise the power factor is

C=

1 = 1.80 × 10−4 F = 180 μ F (2π × 60 Hz)(14.7 Ω)

(c) When the power factor is 1.0, the power delivered is P = (Erms)2/R. From the figure above, we see that the resistance is R = Zcosφ = (0.9)(33.8 Ω) = 30.4 Ω. Thus (15,000 V) 2 Psource = = 7.4 MW 30.4 Ω

36.65. Model: The current and voltage of a resistor are in phase. For a capacitor, the current leads the voltage by 90°. Visualize: This is the phasor diagram for an RC circuit in which the current leads the emf voltage by 45°.

Solve: We have two tasks. First, to get the phase right. Second, to get the output voltage right. In the RC circuit above, the resistor voltage is in phase with the current. If we chose a capacitor and resistor such that VR = VC, then the current and the resistor voltage (the output) lead the emf (the input) by exactly 45°. Since VR = IR and VC = IXC, the phase angle will be 45° if R = XC = 1/ωC. This gets the phase right, but is the output voltage right? Under these conditions, the rms voltage across the resistor is

(VR ) rms = I rms R =

Erms R R +X 2

2 C

=

Erms R R +R 2

2

=

Erms = 0.707Erms 2

This is too much voltage. We want a circuit in which (VR)rms = 0.5 Erms . But we can reduce the voltage by splitting the resistor R into two series resistors whose total resistance is R.

Because the current is the same through both resistors,

V2 V R2 R V 0.500Erms = 2 = = 2 ⇒ R2 = 2 R = R = 0.707 R 0.707Erms V1 + V2 VR R1 + R2 R VR That is, if we split resistor R into R1 = 0.293R and R2 = 0.707R, and take the output from R2, then the total resistance will cause the phase angle to be 45° and the voltage across R2 will be half the input voltage. There are no unique values for R and C. Any values that obey these relationships will work. But to be specific, suppose we choose C = 100 μF, a very typical capacitance value. Then

R=

1

ωC

=

1 = 26.5 Ω ⇒ R1 = 0.293R = 7.8 Ω and R2 = 0.707 R = 18.7 Ω (2π × 60 Hz)(100 × 10−6 F)

Hence our circuit is as shown above.

36.66.

Visualize:

Solve: (a) The three phasors are shown in the figure. The second phasor has a phase 120° ahead of the first and the third phasor has a phase that is 120° behind the first. These phasors are labeled as ( E )1, ( E0 )2, and ( E0 )3. (b) As shown in the figure, the three vectors (or phasors) add to zero. This means that the sum of the three phasors is zero. (c) In the figure above, we can use the law of cosines to find E0′ = E02 + E02 − 2E0E0 cos φ = E02 + E02 − 2E02 cos (120° ) = 3E0

Thus the rms value of the difference between two phases is

′ = 3 Erms = 3 (120 V ) = 208 V Erms

36.67. Model: The phase angle is positive in an RLC circuit in which the current lags the emf. Visualize: Please refer to Figure CP36.67. The circuit looks like Figure 36.17. Solve: (a) From Figure CP36.67, we identify E0 = 10 V, I = 2 A, and f = The current is half its value at t = 0, so

1 = 10.0 kHz. 100μ s

1 = cosϕ ⇒ ϕ = 60°. 2

Thus

tan 60° =

XL − XC ⇒ X L − X C = R tan 60°. R

Since

I= ⇒R=

E0 E0 = 2 2 Z R + ( XL − XC )

E0 I 1 + ( tan 60° )

(b) The resonance frequency ωres =

X L − X C = R tan 60° = ω L −

2

=

10 V

(2 A)

1 + ( tan 60° )

2

= 2.5 Ω

1 . Rearranging the expression in part (a), LC

⎛ ωres 2 ⎞ 1 1 ⎞ R tan 60° ⎞ ⎛ 2 2⎛ = ω L ⎜1 − 2 ⎟ = ω L ⎜1 − 2 ⎟ ⇒ ωres = ω ⎜1 − ωC ω ⎠ ω L ⎟⎠ ⎝ ω LC ⎠ ⎝ ⎝

With ω = 2π (10.0 × 103 Hz ) , and L = 200 μ H ,

ωres = 5.1× 104 rad/s ⇒ f res = Assess:

ωres = 8.1 kHz . 2π

Since ϕ > 0 , the circuit is being driven above the resonance frequency.

36.68. Solve: (a) From Equation 36.44, the average power is Pavg = Erms I rms cosφ , where I rms = Erms Z and cosφ = R Z . We also have 2

2

1 ⎞ 1 ⎞ L2 2 ⎛ 2 2⎛ 2 2 2 Z 2 = R2 + ⎜ ωL − ⎟ = R + L ⎜ω − ⎟ = R + 2 (ω − ω0 ) ω ω ω C LC ⎝ ⎠ ⎝ ⎠ Combining these results we obtain 2 Rω 2 Erms Pavg = 2 ω 2 R 2 + L2 (ω 2 − ω02 ) (b) Energy dissipation will be a maximum, when dPavg dω = 0 . Taking the derivative,

dPavg dω

=

2ω

ω 2 R 2 + L2 (ω 2 − ω02 )

2

−

ω 2 ⎡⎣ 2ω R 2 + 2 L2 (ω 2 − ω02 ) ( 2ω ) ⎤⎦ ⎡ω 2 R 2 + L2 (ω 2 − ω 2 )2 ⎤ 0 ⎢⎣ ⎥⎦

2

=0

⇒ ω 2 R 2 + L2 (ω 2 − ω02 ) = ω 2 R 2 + 2 L2ω 2 (ω 2 − ω02 ) ⇒ ω 4 = ω04 ⇒ ω = ω0 2

36.69. Solve: (a) The peak inductor voltage in a series RLC circuit is E0ω L ⎛E ⎞ VL = IX L = I (ω L ) = ⎜ 0 ⎟ ω L = ⎝Z⎠

1 ⎞ ⎛ R2 + ⎜ ωL − ωC ⎟⎠ ⎝

2

For VL to be maximum, dVL dω = 0 . Taking the derivative and dividing by the constant factors,

dVL = dω

1 1 ⎞ ⎛ R + ⎜ωL − ⎟ ω C⎠ ⎝ 2

2

−

⎛ ⎝

ω ( 2) ⎜ ω L −

1 ⎞⎛

ωC ⎟⎜ ⎠⎝

L+

1 ⎞

ω 2C ⎟⎠

2 ⎡ 1 ⎞ ⎤ ⎛ 2 ⎢ R2 + ⎜ ωL − ⎟ ⎥ ωC ⎠ ⎦⎥ ⎝ ⎣⎢

3/ 2

=0

2

⎡ 1 R 2C 2 ⎤ 1 ⎞ ⎛ 1 ⎞⎛ 1 ⎞ 2 2L ⎛ ⇒ R + ⎜ωL − − R 2 ⇒ ωL = ⎢ 2 − ⎥ ⎟ = ⎜ωL − ⎟⎜ ω L + ⎟⇒ 2 2 = C 2 ⎦ ωC ⎠ ⎝ ωC ⎠⎝ ωC ⎠ ω C ⎝ ⎣ ω0

−1/ 2

2

where ω0 = 1 LC and the maximizing frequency is ωL. (b) The voltage across the inductor is E E VL = IX L = 0 X L = 0 ω L Z Z At ω = ω0 = 1

LC , Z = R = 1.0 Ω. We get

VL =

E0 E L 10 V 1.0 μ H = = 10 V ω0 L = 0 Z Z C 1 Ω 1.0 μ F

The maximizing frequency ωL is ⎡ 1

ωL = ⎢

⎣ ω0

2

−

R 2C 2 ⎤ ⎥ 2 ⎦

−1/ 2

⎡ R 2C 2 ⎤ = ⎢ LC − ⎥ 2 ⎦ ⎣

−1/ 2

2 2 ⎡ (1.0 Ω ) (1.0 μ F ) ⎤ = ⎢(1.0 μ H )(1.0 μ F ) − ⎥ 2 ⎢⎣ ⎥⎦ The impedance at this frequency is

Z = R 2 + ( ωL L − 1 ω L C ) =

(1.0 Ω )

2

−1/ 2

= 1.414 × 106 rad/s

2

⎡ ⎤ 1 ⎥ = 1.225 Ω + ⎢(1.414 × 10+6 rad/s ) (1.0 μ H ) − ⎢⎣ (1.414 ×10+6 rad/s ) (1.0 μ F) ⎥⎦

The voltage is VL =

E0 10.0 V ωL L = (1.414 ×106 rad/s ) (1.0 μ H ) = 11.55 V Z 1.225 Ω

36.70. Model: The capacitor current leads the voltage by 90°. The inductor current lags the voltage by 90°. Visualize: (a) The capacitor and inductor are in parallel. Parallel circuit elements have the same voltage, so VL = VC = VLC. Because this is a shared voltage, we’ve started the phasor diagram by drawing the VLC phasor. Then the capacitor phasor IC leads by 90° and the inductor phasor IL lags by 90°.

The capacitor and inductor currents add to give the phasor ILC = IL – IC. The resistor is in series with the capacitor/inductor combination, and series elements share the same current, hence IR = ILC. For a resistor, the voltage VR is in phase with the current IR, and this has been shown on the diagram. Finally, the voltages VR and add as vectors to give the VLC emf voltage E0 . We see from the right triangle that 2 E02 = V R2 + VLC

For the resistor, VR = IRR = IR. The capacitor/inductor combination requires a little more care. The peak inductor current and peak capacitor current are

IL =

VL VLC = XL XL

IC =

VC VLC = XC XC

Thus

I LC = I L − I C =

⎛ 1 ⎛ 1 VLC VLC 1 ⎞ 1 ⎞ − = VLC ⎜ − − ⎟ ⇒ VLC = I LC ⎜ ⎟ XL XC ⎝ XL XC ⎠ ⎝ XL XC ⎠

−1

But ILC = IR = I, thus our expression for E0 becomes −2

⎛ 1 1 ⎞ E02 = VR2 + VLC2 = I 2 R 2 + I 2 ⎜ − ⎟ ⇒I= ⎝ XL XC ⎠

E0 ⎛ 1 1 ⎞ R2 + ⎜ − ⎟ X X C ⎠ ⎝ L

−2

(b) As ω → 0 rad/s, XL → 0 Ω and XC → ∞. This means −2

⎛ 1 ⎛ 1 1 ⎞ 1 ⎞ 2 − − ⎟ →0 Ω ⎜ ⎟→∞ ⇒ ⎜ X X X X C ⎠ C ⎠ ⎝ L ⎝ L Thus, I = E0 R . That is, the inductor L becomes a short and E0 drives only the resistor R. As ω → ∞, XL → ∞ and XC → 0 Ω. This means −2

⎛ 1 ⎛ 1 1 ⎞ 1 ⎞ 2 − − ⎜ ⎟ → −∞ ⇒ ⎜ ⎟ →0 Ω ⎝ XL XC ⎠ ⎝ XL XC ⎠ Thus, I = E0 R . That is, the capacitor C is the short and E0 drives only the resistor R. (c) At resonance IL = IC . This means XL = XC or ω0 = 1 LC . Therefore,

−2

⎛ 1 ⎛ 1 1 ⎞ 1 ⎞ − − ⎜ ⎟=0 ⇒ ⎜ ⎟ →∞ X X X X C ⎠ C ⎠ ⎝ L ⎝ L Thus, I = 0 A.

36.71. Visualize: Voltage is the same for circuit elements in parallel, so start with the E0 phasor. Please refer to Figure CP36.71.

Solve:

(a) We see from the above phasor diagram that 2

2 2 2 ⎡1 ⎛ 1 E ⎞ 2 ⎛E ⎞ ⎛E ⎞ ⎛ E ⎞ ⎤ − ωC ⎟ ⎥ I 2 = I R2 + ( I L − I C ) ⇒ ⎜ 0 ⎟ = ⎜ 0 ⎟ + ⎜ 0 − 0 ⎟ ⇒ Z = ⎢ 2 + ⎜ ⎝ Z ⎠ ⎝ R ⎠ ⎝ ω L 1 ωC ⎠ ⎠ ⎦⎥ ⎢⎣ R ⎝ ω L

−1/ 2

Since I = E0 Z , the current is

I = E0

1 ⎛ 1 ⎞ +⎜ − ωC ⎟ 2 R ⎝ ωL ⎠

2

(b) As ω → 0 rad/s, ωL → 0 Ω, so I → ∞. That is, the inductor becomes a short for E0 . On the other hand, as ω → ∞, I → ∞ because now the capacitor becomes a short for E0 . (c) To find the frequency for which I is minimum, we set dI dω = 0 . We get

dI =0= dω

⎛1⎞ ⎛ 1 ⎞⎛ 1 ⎞ E0 ⎜ ⎟ 2 ⎜ − ωC ⎟ ⎜ − 2 − C ⎟ ω ω 2 L L ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⇒ ⎛ 1 − ωC ⎞ ⎛ − 1 − C ⎞ = 0 ⎜ ⎟⎜ ⎟ 2 2 ⎝ ωL ⎠⎝ ω L ⎠ 1 ⎛ 1 ⎞ − ωC ⎟ 2 2 +⎜ R ⎝ ωL ⎠ ⇒

1

ωL

− ωC = 0 ⇒ ω 2 =

1 = ω02 ⇒ ω0 = LC

1 LC

The resonance frequency is the same as a series RLC circuit. (d) We know the current is a minimum at ω = ω0 and diverges as ω → 0 or ω → ∞. Thus the current graph must look more or less as sketched above.

37.1. Model: S and S′ are inertial frames that overlap at t = 0. Frame S′ moves with a speed v = 5.0 m/s along x-direction relative to frame S. Visualize:

the

The figure shows a pictorial representation of the S and S′ frames at t = 1.0 s and 5.0 s. Solve: From the figure, the observer in S′ finds the position of the first explosion at x1′ = 5.0 m at t = 1.0 s. The

position of the second explosion is x2′ = −5.0 m at t = 5.0 s. We can get the same answers using the Galilean transformations of position:

x1′ = x1 − vt = 10 m − ( 5.0 m/s )(1.0 s ) = 5.0 m at 1.0 s x′2 = x2 − vt = 20 m − ( 5.0 m/s )( 5.0 s ) = −5.0 m at 5.0 s

37.2. Model: S and S′ are inertial frames. S′ moves relative to S with speed v. Solve:

(a) Using the Galilean transformations of position, x1′ = x1 − vt 1 ⇒ 4.0 m = x1 – v (1.0 s) ⇒ x1 = 4.0 m + v (1.0 s) x2′ = x 2 − vt 2 ⇒ −4.0 m = x2 − v (3.0 s) ⇒ x2 = −4.0 m + v (3.0 s)

Because x1 = x2, 4.0 m + v (1.0 s) = −4.0 m + v (3.0 s) ⇒ v = 4.0 m/s (b) The positions of the two explosions in the S frame are x1 = 4.0 m + (4.0 m/s)(1.0 s) = 8.0 m

x2 = −4.0 m + (4.0 m/s)(3.0 s) = 8.0 m

37.3. Model: S is the ground’s frame of reference and S′ is the sprinter’s frame of reference. Frame S′ moves relative to frame S with speed v. Visualize:

Solve: The speed of a sound wave is measured relative to its medium. The medium is still air on the ground, which is our frame S. The sprinter travels to the right with reference frame S′ at velocity v. Using the Galilean transformations of velocity,

u1′ = −360 m/s = u1 − v = −vsound − v

u′2 = 330 m/s = u2 − v = vsound − v

Adding the two above equations, −30 m/s = −2v ⇒ vsprinter = 15 m/s From the first equation, −360 m/s = −vsound − (15 m/s) ⇒ vsound = 345 m/s

Assess: Notice that the Galilean transformations use velocities and not speeds. It is for that reason u1′ = −360 m/s.

37.4. Model: You are on the ground in frame S and the baseball pitcher is in the pickup in frame S′. S′ moves relative to S with velocity v. Visualize:

The figure shows a pictorial representation of the two frames. The Galilean transformation uses velocities, not speeds, so u and u′ are negative. Solve: The speed of the baseball in the two frames is u′ = −40 m/s and u = −10 m/s. From Equation 37.2, u′ = u − v ⇒ v = u − u′= (−10 m/s) − (−40 m/s) = 30 m/s

37.5. Model: The boy on a bicycle is frame S′ and the ground is frame S. S′ moves relative to S with a speed v = 5.0 m/s. The frames S and S′ overlap at t = 0. Visualize:

The figure shows a pictorial representation of the two frames. Solve: (a) When the newspaper is thrown forward, u ′x = 8.0 m/s. The Galilean transformation of velocity is ux = u x′ + v = 8.0 m/s + 5.0 m/s = 13 m/s (b) When the newspaper is thrown backward, u ′x = −8.0 m/s. In this case ux = u x′ + v = −8.0 m/s + 5.0 m/s = −3.0 m/s Thus the speed is 3.0 m/s. (c) When the newspaper is thrown to the side, u y = u′y = 8.0 m/s. Also,

u x = u′x + v = 0 m/s + 5.0 m/s = 5.0 m/s Thus the newspaper’s speed is u = u x2 + u y2 =

( 5.0 m/s ) + ( 8.0 m/s ) 2

2

= 9.4 m/s

37.6. Model: Assume the spacecraft is an inertial reference frame. Solve: Light travels at speed c in all inertial reference frames, regardless of how the reference frames are moving with respect to the light source. Relative to the spacecraft, the starlight is approaching at the speed of light c = 3.00 × 108 m/s.

37.7.

Model: Assume the starship and the earth are inertial reference frames. Solve: It has been found that light travels at 3.00 × 108 m/s in every inertial frame, regardless of how the reference frames are moving with respect to each other. An observer on the earth will measure the laser beam’s speed as 3.00 × 108 m/s.

37.8. Model: Assume the earth is an inertial reference frame. Solve: Light travels at speed c in all inertial reference frames, regardless of their motion with respect to the light source. The speed of each photon will be c in any such inertial reference frame.

37.9. Model: The clocks are in the same reference frame. Visualize:

Solve:

The speed of light is c = 300 m/μs = 0.30 m/ns. The distance from the origin to the point (x, y, z) = (30 m, 40

m, 0 m) is

( 30 m )

2

+ ( 40 m ) = 50 m. So, the time taken by the light to travel 50 m is 2

50 m = 167 ns 0.30 m/ns

The clock should be preset to 167 ns.

37.10. Model: Bjorn and firecrackers 1 and 2 are in the same reference frame. Light from both firecrackers travels towards Bjorn at 300 m/μs. Visualize:

Solve: Bjorn is 600 m from the origin. Light with a speed of 300 m/μs takes 2.0 μs to reach Bjorn. Since this flash reaches Bjorn at t = 3.0 μs, it left firecracker 1 at t1 = 1.0 μs. The flash from firecracker 2 takes 1.0 μs to reach Bjorn. So, the light left firecracker 2 at t2 = 2.0 μs. Note that the two events are not simultaneous although Bjorn sees the events as occurring at the same time.

37.11. Model: Bianca and firecrackers 1 and 2 are in the same reference frame. Light from both firecrackers travels toward Bianca at 300 m/μs. Visualize:

Solve:

The flash from firecracker 1 takes 2.0 μs to reach Bianca ( 600 m ÷ 300 m/μ s ) . The firecracker

exploded at t1 = 1.0 μs because it reached Bianca’s eye at 3.0 μs. The flash from the firecracker 2 takes 1.0 μs to reach Bianca. Since firecrackers 1 and 2 exploded simultaneously, the explosion occurs at t2 = 1.0 μs. So, the light from firecracker 2 reaches Bianca’s eye at 2.0 μs. Although the events are simultaneous, Bianca sees them occurring at different times.

37.12. Model: You and your assistant are in the same reference frame. Light from the two lightning bolts travels toward you and your assistant at 300 m/μs. You and your assistant have synchronized clocks. Visualize:

Solve:

Bolt 1 is 9.0 km away, so it takes 30 μs for the light to reach you ( 9000 m ÷ 300 m/μs ) . Bolt 2 is 3.0 km

away from you, so it takes 10 μ s to reach you. Since both flashes reach your eye at the same time, event 1 happened 20 μs before event 2. If event 1 happened at time t1 = 0 then event 2 happened at time t2 = 20 μs. For your assistant, it takes light from bolt 1 10 μ s to reach her and light from bolt 2 30 μs to reach her. She sees the flash from bolt 1 at t = 10 μs and the flash from bolt 2 at t = 50 μs. That is, your assistant sees flash 2 40 μs after she sees flash 1.

37.13. Model: You and your assistant are in the same reference frame. Light from the two lightning bolts travels toward you and your assistant at 300 m/μs. You and your assistant have synchronized clocks. Visualize:

Solve: Bolt 1 hits 9.0 km away, so the light takes 30 µs to reach you (9000 m ÷ 300 m/μs). You see this flash at t = 50 μs, so the lightning hit at t1 = 20 μs. Light from bolt 2, which hits 3.0 km away, takes 10 μs to reach you. You see it at 10 μs, so the lightning hit at t2 = 0 μs. The strikes are not simultaneous. Bolt 2 hits first, 20 μs before bolt 1. Your assistant is in your inertial reference frame, so your assistant agrees that bolt 2 hits first, 20 μs before bolt 1. Assess: A simple calculation would show that your assistant sees the flashes at the same time. When the flashes are seen is not the same as when the events happened.

37.14. Model: Light from the two lightning bolts travels toward Jose at 300 m/μs. Visualize:

Solve:

Bolt 1 light takes 1 μs to reach Jose ( 300 m ÷ 300 m/μs ) . On the other hand, the flash from bolt 2

takes 3 μ s to reach him. The times when the flashes are seen are

t1 seen = t1 happened + 1.0 μs

t2 seen = t2 happened + 3.0 μs

Because Jose sees the tree hit 1.0 μs before he sees the barn hit, t2 seen − t1 seen = 1.0 μs. Subtracting the two equations, t2 seen − t1 seen = ( t2 happened − t1 happened ) + ( 3.0 μ s − 1.0 μ s ) ⇒ 1.0 μs = ( t2 happened − t1 happened ) + 2.0 μs ⇒ t1 happened = t2 happened

+ 1.0 μs

Thus, the barn was struck by lightning 1.0 μs before the tree.

37.15. Model: Your personal rocket craft is an inertial frame moving at 0.9c relative to stars A and B. Solve: In your frame, star A is moving away from you and star B is moving toward you. When you are exactly halfway between them, both the stars explode simultaneously. The flashes from the two stars travel toward you with speed c. Because (i) you are at rest in your frame, (ii) the explosions are equally distant, and (iii) the light speed is c, independent of the fact that the stars are moving in your frame, the light will arrive simultaneously.

37.16.

Model: The earth is in reference frame S and the cosmic ray is in reference frame S´. Frame S´ travels with velocity v relative to frame S. Solve: Two events are “cosmic ray enters the atmosphere” and “cosmic ray hits the ground.” These can both be measured with a single clock in the cosmic ray’s frame, frame S´, so the time interval between them in S´ is the proper time interval: ∆t´ = ∆τ. The time interval measured in the earth’s frame, frame S, is ∆t = 400 μs. The timedilation result is Δτ = 1 − β 2 Δ t

The cosmic ray’s speed in frame S is simply

v=

ΔL 60,000 m v = = 1.5 × 108 m/s ⇒ β = = 0.50 −6 Δt 400 × 10 s c

Thus the time interval measured by the cosmic ray is

Δτ = 1 − (0.50) 2 (400 μs) = 346 μs

37.17. Model: Let the moving clock be in frame S´ and an identical at-rest clock be in frame S. Solve: The ticks being measured are those of the moving clock. The interval between 2 ticks is measured by the same clock in S´—namely, the clock that is ticking—so this is the proper time: Δt ′ = Δτ . The rest clock measures a longer interval ∆t between two ticks of the moving clock. These are related by Δτ = 1 − β 2 Δ t

The moving clock ticks at half the rate of the rest clock when ∆τ =

1 2

∆t. Thus

1 − β 2 = 1 − v 2 /c 2 = 1/2 ⇒ v = c 1 − (1/2) 2 = 0.866c

37.18. Solve: (a) The starting event is the astronaut leaving earth. The finishing event is the astronaut arriving at the star system. The time between these events as measured on earth is Δt =

4.5 ly 4.5 ly = = 5.0 y 0.9c 0.9 ly/y

(b) For the astronaut, the two events occur at the same position and can be measured with just one clock. Thus, the time interval in the astronaut’s frame is the proper time interval. 2

Δτ = 1 −

v2 ⎛ 0.9c ⎞ Δt = 1 − ⎜ ⎟ 5.0 y = 0.19 5.0 y = 2.2 y c2 ⎝ c ⎠

(c) The total elapsed time is the time for the astronaut to reach the star system plus the time for light to travel from the star system to the earth. The time is Δt + 4.5 y = 5.0 y + 4.5 y = 9.5 y

37.19. Model: Let S be the earth’s reference frame and S′ the rocket’s reference frame. Solve: (a) The astronauts measure proper time i Δt ′ = Δτ . Thus Δt =

Δτ 1 − (v c)

2

⇒ 120 y =

10 y 1 − (v c)

2

⇒ v = 0.9965c

(b) In frame S, the distance of the distant star is

Δ x = vΔt = ( 0.9965c )( 60 y ) = ( 0.9965 ly y )( 60 y ) = 59.8 ly

37.20. Model: The earth’s frame is S and the airliner’s frame is S′. S′ moves relative to S with velocity v. Also, assume zero acceleration/ deceleration times. The first event is when the airliner takes off and almost instantly attains a speed of v = 250 m/s. The second event is when the airliner returns to its original position after 2 days. It is clear that the two events occur at the same position in frame S′ and can be measured with just one clock. This is however not the case for an observer in frame S. Solve: (a) You have aged less because your proper time is less than the time in the earth frame. (b) In the S frame (earth), 2 × 5 × 106 m Δt = = 4.0 × 104 s 250 m/s In the S′ frame (airliner), 1 ⎛ v2 ⎞ Δτ = Δt (1 − v 2 c 2 ) 2 ≈ Δt ⎜1 − 2 ⎟ ⎝ 2c ⎠ 2

v2 1 ⎛ 250 m/s ⎞ −8 = ( 4.0 × 104 s ) ⎜ ⎟ = 1.4 × 10 s = 14 ns 2 2c 2 ⎝ 3.0 × 108 m ⎠ You age 14 ns less than your stay-at-home friends. ⇒ Δt − Δτ ≈ Δt

37.21. Model: The ground’s frame is S and the moving clock’s frame is S′.

Visualize: Δ t = 1.0 d + 1.0 ns = 86400.000000001 s and Δt′ = 1.0 d = 86,400 s. Solve: We want to solve for v in Δ t = γ Δ t′

Using the binomial approximation for γ : Δt 1v 2 ≈1+ 2 Δ t′ 2c

⎛ Δt ⎞ 1 v2 Δt −1 ≈ ⇒ 2c 2 ⎜ − 1⎟ = v 2 2 Δ t′ 2c ⎝ Δ t′ ⎠

⎛ Δt ⎞ ⎛ 1.0 d + 1.0 ns ⎞ ⎛ 1.0 ns ⎞ v = c 2⎜ − 1⎟ = c 2 ⎜ − 1⎟ = c 2 ⎜ ⎟ ′ Δ t 1.0d ⎝ ⎠ ⎝ ⎠ ⎝ 1.0d ⎠ ⎛ 1.0 × 10−9 s ⎞ v = (3.0× 108 m s) 2 ⎜ ⎟ = 46 m s ⎝ 86,400 s ⎠ Assess: This speed is about 100 mph, which is certainly doable. The calculation would be difficult without the binomial approximation due to limited calculator precision; fortunately, the approximation is excellent in this case.

37.22. Model: The ground is frame S and the moving rod is frame S´. The length of the rod in S´ is the proper length

because the rod is at rest in S´. The rod is length contracted in S to L = 1− β 2A

The length is contracted to 60% when L = 0.60 . Thus

1 − β 2 = 1 − v 2 / c 2 = 0.60 ⇒ v = c 1 − (0.60) 2 = 0.80c

37.23. Model: S′ is the rocket’s frame (Jill’s frame) and S is the ground’s frame (your frame). In the S′ frame, which moves with a velocity v relative to S, the length of the rocket is the proper length because it is at rest in this frame. So, L′ = A. Solve: You measure a length-contracted rocket L = 1 − β 2 A ⇒ 80 m =

She did exceed the 0.5c speed limit.

1 − β 2 (100 m ) ⇒ β = 0.6

37.24. Model: S′ is the muon’s frame and S is the ground’s frame. S′ moves relative to S with a speed of 0.9997c. Solve: For an experimenter in the ground’s frame, a distance of 60 km (or L) is always there for measurements. That is, L is the atmosphere’s proper length A. The muon measures the thickness of the atmosphere to be length contracted to L′ = 1 − β 2 A = 1 − ( 0.9997 ) ( 60 km ) = 1.47 km 2

37.25. Model: The length of an object is contracted when it is measured in any reference frame moving relative to the object. The contraction occurs only along the direction of motion of the object. Visualize:

Solve: The cube at rest has a density of 2000 kg/m3. That is, a cube of 1 m × 1 m × 1 m dimensions has a mass of 2000 kg. As the cube moves with a speed of 0.9c, its dimension along the direction of motion is contracted according to Equation 37.14:

L′ = 1 − ( v c ) L = 1 − ( 0.9 ) (1 m ) = 0.436 m 2

2

Because the other dimensions of the cube are not affected by the cube’s motion, the new density will be 2000 kg ρ= = 4600 kg/m3 0.436 m × 1 m × 1 m

37.26. Model: S is the galaxy’s reference frame and S′ is the spaceship’s reference frame. S′ moves relative to S with a speed v. Solve: (a) For an experimenter in the galaxy’s reference frame, the diameter (or length) of the galaxy is 105 ly. This is the proper length A = L because it is at rest and is always there for measurements. However, in the spaceship’s reference frame S′, the galaxy moves toward him/her with speed v. S′ measures the galaxy to be length contracted to L′ = 1.0 ly. Thus

L′ = A 1 − β 2 ⇒ 1.0 ly = (105 ly) 1 − β 2 ⇒ β = 1 − 10−10 ≈ 1 − 12 × 10−10 ⇒ v = (1 − 5.0 × 10−11 )c = 0.99999999995c

(b) In S, the spaceship travels 100,000 ly at speed v = 0.99999999995c, taking

Δt =

L ≈ 100,000 y v

37.27. Model: S is the ground’s reference frame and S′ is the meter stick’s reference frame. In the S′ frame, which moves with a velocity v relative to S, the length of the meter stick is the proper length because the meter stick is at rest in this frame. So L′ = A. Solve: An experimenter on the ground measures the length to be contracted to 1 ⎛ 1 ⎞ L = 1 − β 2 A ≈ ⎜ 1 − β 2 ⎟ A ⇒ shrinking = A − L = β 2A 2 ⎝ 2 ⎠ Thus the speed is

β=

2 ( 50 × 10−6 m ) 2(A − L) = = 0.01 ⇒ v = β c = 3.0 × 106 m/s 1.00 m A

37.28. Visualize: At t = t′ = t″ = 0 s, the origins of the S, S′, and S″ reference frames coincide. Solve:

We have γ = ⎡⎣ 1 − (v / c) 2 ⎤⎦

− 12

= ⎡⎣ 1 − (0.80) 2 ⎤⎦

− 12

= 1.667. (a) Using the Lorentz transformations,

x′ = γ ( x − vt ) = 1.667 ⎡⎣1200 m − (0.80)(3 × 108 m/s)(2.0 × 10−6 s) ⎤⎦ = 1200 m ⎡ ⎤ (0.80)(3 × 108 m/s)(1200 m) ⎥ ⎛ vx ⎞ t ′ = γ ⎜ t − 2 ⎟ = 1.667 ⎢ 2.0 × 10−6 s − = −2.0 μs 2 ⎢ ⎥ ⎝ c ⎠ ( 3 ×108 m/s ) ⎣ ⎦ (b) Using v = −0.80c, the above equations yield x ′′ = 2800 m and t ′′ = 8.67 μs.

37.29. Solve: We have γ = ⎡⎣ 1 − ( v c ) ⎤⎦ 2

− 12

2 = ⎡ 1 − ( 0.60 ) ⎤ ⎣ ⎦

− 12

= 1.25. In the earth’s reference frame, the

Lorentz transformations yield x = γ ( x′ + vt ′) = 1.25 ⎡⎣3.0 × 1010 m + ( 0.60 ) ( 3.0 × 108 m/s ) ( 200 ) ⎤⎦ = 8.25 × 1010 m ≈ 8.3 × 1010 m ⎡ ( 0.60 ) ( 3.0 × 108 m/s )( 3.0 × 1010 m ) ⎤⎥ v x′ ⎞ ⎛ = 325 s ≈ 330 s t = γ ⎜ t ′ + 2 ⎟ = 1.25 ⎢ 200 + 2 8 ⎢ ⎥ c ⎠ ⎝ × 3.0 10 m/s ( ) ⎣ ⎦

37.30. Model: S is the ground’s frame and S′ is the rocket’s frame. S′ moves with velocity v = 0.5c relative to S. −1

2 2 2 (a) We have γ = ⎡ 1 − ( v / c ) ⎤ = ⎡ 1 − ( 0.50 ) ⎤ ⎣ ⎦ ⎣ ⎦ lightning strike at x = 0 m and t = 10 μs,

Solve:

− 12

= 1.155. Applying the Lorentz transformations to the

x′ = γ ( x − vt ) = (1.155) ⎡⎣ 0 m − ( 0.5 ) ( 3.0 × 108 m/s )(1 × 10−5 s ) ⎤⎦ = −1732 m ≈ −1700 m ⎛ vx ⎞ t ′ = γ ⎜ t − 2 ⎟ (1.155)(1 × 10−5 s − 0 s) = 11.55 μs ≈ 12 μs ⎝ c ⎠

For the lightning strike at x = 30 km and t = 10 μs, x′ = (1.155 ) ⎡⎣3.0 × 104 m − ( 0.50 ) ( 3.0 × 108 m/s )(1 × 10−5 s ) ⎤⎦ = 32.91 m ≈ 33 m

⎡ ( 0.50 ) ( 3.0 × 108 m/s )( 3.0 × 104 m ) ⎤⎥ −5 ⎢ ′ = −46.2 μs ≈ −46 μs t = (1.155 ) 1 × 10 s − 2 ⎢ ⎥ ( 3.0 ×108 m/s ) ⎣ ⎦ (b) The events in the rocket’s frame are not simultaneous. The lightning is observed to strike the pole before the tree by 46 + 12 = 58 μs.

37.31. Model: The rocket and the earth are inertial frames. Let the earth be frame S and the rocket be frame S′. S′ moves with v = 0.8c relative to S. The bullet’s velocity in reference frame S′ is u′ = −0.9c. Solve:

Using the Lorentz velocity transformation equation, u′ + v −0.9c + 0.8c u= = = −0.36c 2 ′ 1 + u v / c 1 + ( −0.9c )( 0.8c ) / c 2

The bullet’s speed is 0.36c along the −x-direction. Note that the velocity transformations use velocity, which can be negative, and not speed.

37.32. Model: The proton and the earth are inertial frames. Let the earth be frame S and the proton be frame S′. S′ moves with v = 0.9c. The electron’s velocity in the laboratory frame is −0.9c. Solve: Using the Lorentz velocity transformation equation, u−v −0.9c − 0.9c u′ = = = −0.994c 1 − uv / c 2 1 − ( −0.9c )( 0.9c ) / c 2

The electron’s speed is 0.994c.

37.33. Model: The earth and the other galaxy are inertial reference frames. Let the earth be frame S and the other galaxy be frame S′. S′ moves with v = +0.2c. The quasar’s speed in frame S is u = +0.8c. Solve: Using the Lorentz velocity transformation equation,

u′ = Assess:

u −v 0.8c − 0.2c = = 0.71c 1 − uv / c 2 1 − ( 0.8c )( 0.2c ) / c 2

In Newtonian mechanics, the Galilean transformation of velocity would give u′ = 0.6c.

37.34. Solve: (a) The relativistic momentum is p=

mu 1 − u 2 / c2

=

(1.67 × 10

−27

kg ) ( 0.999 ) ( 3.0 × 108 m/s ) 1 − ( 0.999 )

2

= 1.12 × 10−17 kg m/s

(b) The ratio of the relativistic momentum and the Newtonian momentum is prelativistic mu 1 1 = = = 22.4 2 2 pclassical 1 − u / c mu 1 − u 2 / c2

37.35. Solve: The relativistic momentum is p= ⇒ 1−

mu 1 − u 2 c2

⇒ 400,000 kg m/s =

(1.0 × 10

−3

kg ) u

1 − u 2 c2

u2 u2 9 u2 1.0 × 10−3 kg ⎛ u ⎞ 3 ⎛ u ⎞ = c = ⇒ − = ⇒ u = 0.80c 1 ⎜ ⎟ ⎜ ⎟ c 2 400,000 kg m/s ⎝ c ⎠ 4 ⎝ c ⎠ c 2 16 c 2

Assess: In Newtonian mechanics, the momentum would be p = mu = (1.0 × 10−3 kg)(0.80)(3.0 × 108 m/s) = 240,000 kg m/s.

37.36. Solve: The Newtonian momentum is pNewton = mu. We have p=

mu 1 − u 2 / c2

= 2mu ⇒ 1 − u 2 / c 2 =

1 3 ⇒u = c = 0.866c 4 2

37.37. Solve: We have p=

mu 1 − u 2 c2

= mc ⇒ 1 − u 2 c 2 =

u 2u 2 c ⇒1= 2 ⇒ u = = 0.707c c c 2

Assess: The particle’s momentum being equal to mc does not mean that the particle is moving with the speed of light. We must use the relativistic formula for the momentum as the particle speeds become high.

37.38. Model: The particle is highly relativistic since u = 0.8c. Solve:

We have

γp =

1 1− u c 2

2

=

The kinetic energy is K = ( γ p − 1) E0 , where E0 is

1 1 − ( 0.80 )

2

=

5 5 2 ⇒ γ p −1 = −1 = 3 3 3

E0 = mc 2 = (1.0 × 10−3 kg )( 3.0 × 108 m/s ) = 9.0 × 1013 J 2

⎛2⎞ ⇒ K = ⎜ ⎟ ( 9.0 × 1013 J ) = 6.0 × 1013 J ⎝3⎠ The rest energy is E0 = mc2 = 9.0 × 1013 J. The total energy is E = E0 + K = 9.0 × 1013 J + 6.0 × 1013 J = 1.5 × 1014 J.

37.39. Model: The hamburger is a classical particle whose rest energy is E0 = mc2.

Solve:

(a) We have

E0 = mc 2 = ( 200 × 10−3 kg )( 3.0 × 108 m/s ) = 1.8 × 1016 J 2

(b) The ratio of the energy equivalent to the food energy is 1.8 × 1016 J = 9.0 × 109 2 × 106 J

37.40. Solve: The rest energy and the total energy are given by Equations 37.43 and 37.42. We have mpc2 = γmec2 ⇒ γ =

mp me

= 1833 =

1 1 − u 2 / c2

⇒ u = 0.99999985c

37.41. Solve: Equation 37.42 is E = γpmc2 = E0 + K. For K = 2E0, γpmc2 = E0 + 2E0 = 3mc2 ⇒ γ p =

1 1 − u 2 / c2

= 3 ⇒1−

u2 1 8 c = 0.943c = ⇒u = 2 c 9 3

37.42. Solve: The total energy is E = γpmc2. For E = 2E0, 2E0 = γpmc2 = γp E0 and γp = 2. Hence, 1 1 − u 2 / c2

=2⇒

u2 3 = ⇒ u = 0.866c c2 4

37.43. Model: Let S be the laboratory frame and S′ be the reference frame of the 100 g ball. S′ moves to the

right with a speed of v = 2.0 m/s relative to frame S. The 50 g ball’s speed in frame S is u50 = 4.0 m/s. Because these speeds are much smaller than the speed of light, we can use the Galilean transformations of velocity. Visualize:

′ initial = 0 m/s . Using the Galilean velocity Solve: Transform the collision from frame S into frame S′, where u100 transformation, ′ initial = u50 initial − v = 4.0 m/s − 2.0 m/s = 2.0 m/s u50

Using Equation 10.43,

50 g − 100 g ⎛1⎞ ⎛ 2.0 ⎞ ′ initial = − ⎜ ⎟ ( 2.0 m/s ) = − ⎜ u50 ⎟ m/s 50 g + 100 g ⎝ 3⎠ ⎝ 3 ⎠ 2 ( 50 g ) ⎛2⎞ ⎛ 4.0 ⎞ ′ initial = ⎜ ⎟ ( 2.0 m/s ) = + ⎜ = u50 ⎟ m/s 50 g + 100 g ⎝3⎠ ⎝ 3 ⎠

′ final = u50 ′ final u100

Using the Galilean transformations of velocity again to go back to the S frame,

⎛ 2.0 ⎞ ′ final + v = − ⎜ u50 final = u50 ⎟ m/s + 2.0 m/s = +1.33 m/s ⎝ 3 ⎠ ⎛ 4.0 ⎞ ′ final + v = ⎜ u100 final = u100 ⎟ m/s + 2.0 m/s = + 3.33 m/s ⎝ 3 ⎠ Because of plus signs with u50 final and u100 final, both masses are moving to the right.

37.44. Model: Let S be the laboratory frame and S′ be the reference frame of the 100 g ball. S′ moves to the

left with a speed of v = −8.0 m/s relative to frame S. In frame S, the 300 g ball has a velocity u300 initial = 2.0 m/s. Because these speeds are much smaller than the speed of light, we can use the Galilean transformation of velocity. Visualize:

′ initial = 0 m/s . Using the Galilean Solve: Transform the collision from frame S into the frame S′, where u100 transformation of velocity ′ initial = u300 initial − v = ( 2.0 m/s ) − ( −8.0 m/s ) = 10.0 m/s u300

From Equation 10.43,

⎛ 300 g − 100 g ⎞ 1 ′ final = ⎜ ′ initial = ⎛⎜ ⎞⎟ (10.0 m/s ) = 5.0 m/s u300 ⎟ u300 + 300 g 100 g ⎝2⎠ ⎝ ⎠ 2 ( 300 g ) ⎛3⎞ ′ initial = ⎜ ⎟ (10.0 m/s ) = 15.0 m/s u300 300 g + 100 g ⎝2⎠ Using the Galilean transformation of velocity again to go back to frame S, ′ final = u100

′ final + v = 5.0 m/s + ( −8.0 m/s ) = −3.0 m/s u300 final = u300

′ final + v = 15.0 m/s + ( −8.0 m/s ) = 7.0 m/s u100 final = u100 The 300 g ball is moving to the left with a speed of 3.0 m/s and the 100 g ball is moving to the right with a speed of 7.0 m/s.

37.45. Model: Let S be the laboratory frame and S′ be the reference frame of ball 2 after the collision. S′ moves to the right with a speed of v = 4.0 m/s relative to the frame S. Because these speeds are much smaller than the speed of light, we can use the Galilean transformation of velocity. Visualize:

Solve: Transform the collision from frame S into frame S´, where u′2 final = 0 m/s. Using the Galilean velocity transformation,

u′1 final = u1 final − v = −2.0 m/s − 4.0 m/s = −6.0 m/s In Chapter 10, we found that an elastic collision between two balls of equal mass, where ball 1 was initially at rest (v1i = 0), results in v1f = v2i and v2f = 0. That is, the balls simply “exchange” velocity. Thus in the S′ frame, the precollision velocities must have been u′2 initial = u1′ final = – 6.0 m/s and u1′ initial = u ′2 final = 0 m/s. We can now use the Galilean transformation again to transform the initial velocities in S′ to frame S: u1 initial = u′1 initial + v = 0 m/s + 4.0 m/s = 4.0 m/s u2 initial = u′2 initial + v = –6.0 m/s + 4.0 m/s = −2.0 m/s Before the collision, ball 1 was moving to the right at 4.0 m/s and ball 2 was moving to the left at 2.0 m/s.

37.46. Model: Let S be the reference frame of the ground and S′ be the reference frame of the shell. Visualize:

The masses of the two fragments are m1 = 3 kg and m2 = 6 kg. Solve: (a) The conservation of momentum equation pafter = pbefore in the ground frame S is m1u1f + m2u2f = (m1 + m2)(100 m/s) ⇒ (3 kg)u1f + (6 kg)u2f = 900 kg m/s ⇒ u1f + 2u2f = 300 m/s ⇒ u1f = 300 m/s – 2u2f Because 900 J of energy is released in the explosion, Kafter = Kbefore + 900 J. Note that the rest energy stays the same during the collision process. The conservation of energy equation is ⎡⎣ 12 ( 3 kg ) u1f2 + 12 ( 6 kg ) u2f2 ⎤⎦ =

1 2

( 9 kg )(100 m/s )

2

+ 900 J ⇒

3 2

( 300 m/s − 2u2f )

2

+ 3u2f2 = 45,900 m 2 /s 2

⇒ 6u2f2 − (1200 m/s ) u2f + 90,000 m 2 /s 2 = 30,600 m 2 /s 2 ⇒ u2f2 − ( 200 m/s ) u2f + 9900 m 2 /s 2 = 0 The solution of this quadratic equation gives u2f = 110 m/s and 90 m/s. Substituting these values of u2f into the equation for the conservation of momentum, we have u1f = 80 m/s and 120 m/s. That is, (u1f, u2f) equals either (80 m/s, 110 m/s) or (120 m/s, 90 m/s). Since the heavier fragment was found to be in front of the lighter fragment, u2f = 110 m/s and u1f =80 m/s. (b) The conservation of momentum equation p after ′ = pbefore ′ in the shell’s reference frame is

( 6 kg ) u2f′ + ( 3 kg ) u1f′ = ( 9 kg )( 0 m/s ) = 0 ⇒ u1f′

= −2 u 2f′

Since Kbefore = 0 J, the energy conservation equation is

K after = 900 J =

1 2

( 3 kg ) u1f′2 + 12 ( 6 kg ) u′2f2 ⇒ 1800 J = ( 3 kg )( −2u′2f )2 + ( 6 kg ) u2f′2 2 ′ ) = 100 m 2 /s 2 ⇒ u′2f = ±10 m/s ⇒ ( u2f

Once again, because the heavier fragment was found to be ahead of the lighter fragment (in frame S), in frame S′ the velocities of the two fragments are u′2f = 10 m/s and u1f′ = −20 m/s. We can now transform these velocities back to the S frame by using the Galilean transformations of velocity

u1f = u1f′ + v = −20 m/s + 100 m/s = 80 m/s u2f = u2f′ + v = 10 m/s + 100 m/s = 110 m/s

(c) Yes, the problem is much easier to solve in the shell’s frame.

37.47. Model: Let the earth be reference frame S and let the spaceship be the reference frame S′. S′ moves relative to S with speed v. Solve: For an observer in the earth’s frame S, the length of the solar system is 10 lh. The time interval for the spaceship to cross is Δt = 15 hours. The time interval measured in S′ is the proper time because this can be measured with one clock at both positions (i.e., both edges of the solar system). The velocity v is

v=

10 lh 2 2 = lh/h = c 15 h 3 3

Because Δt′ = Δτ, from Equation 36.9 we have

Δτ = Δt 1 − β 2 = (15 h ) 1 − ( 32 ) = 11.2 h 2

37.48. Model: Let the earth be frame S and the train be frame S´. S´ moves with velocity v = 0.5c relative to S. The 30 m length is measured in the train’s reference frame, frame S´. Visualize: The light flashes at t = t´ = 0 s as the origins of S and S´ coincide.

Solve: (a) For passengers on the train, light travels 15 m in both directions at speed c. The fact that the train is moving relative to the earth doesn’t affect the speed of light. Thus the light flash arrives at both ends of the train simultaneously, causing the bell and siren to be simultaneous. Since the light flashed at t´ = 0 s, the time of these two simultaneous events is tB′ = tS′ = (15 m)/(300 m/μs) = 0.050 μs. (b) The spacetime coordinates of the event “bell rings” are ( xB′ , tB′ ) = (15 m, 0.050 μs). The coordinates of the event “siren sounds” are ( xS′, t S′) = (–15 m, 0.050 μs). We can use the Lorentz time transformation to find the times of these events in frame S. To do so, we first need to calculate

γ=

1 1 − ( v/c )

2

=

1 1 − (0.50) 2

= 1.1547

Consequently, the times are tB = γ ( tB′ + vx′B / c 2 ) = γ ( tB′ + (v / c)( x′B / c) ) = 1.1547 ( 0.050 μs + (0.50)((15 m) / (300 m/μs)) ) = 0.087 μ s

tS = γ ( tS′ + vxS′ / c 2 ) = γ ( tS′ + (v / c)( xS′ / c) )

= 1.1547 ( 0.050 μs − (0.50)((15 m)/(300 m/μ s)) ) = 0.029 μs Thus the siren sounds before the bell rings. The time interval between the two is ∆t = 0.087 μs – 0.029 μs = 0.058 μs.

37.49. Model: Let S be the galaxy’s frame and S′ the alien spacecraft’s frame. The spacetime interval s between the two events is invariant in all frames. Solve: (a) The light from Alpha’s explosion will travel 10 ly in 10 years. Since neither light nor any other signal from Alpha can travel 100 ly in 10 years to reach Beta, the explosion of Alpha could not cause the explosion of Beta. (b) Because the spacetime interval s between the two events is invariant,

s = c ( Δt ) − ( Δx ) 2

2

2

2

2

2

⎛ 1 ly ⎞ ⎛ 1 ly ⎞ 2 2 2 2 = c ( Δt ′ ) − ( Δx′ ) ⇒ ⎜ ⎟ (10 y ) − (100 ly ) = ⎜ ⎟ ( Δt ′ ) − (120 ly ) ⎝ y ⎠ ⎝ y ⎠ 2

2

2

⇒ (10 y ) − (100 y ) = ( Δt ′ ) − (120 y ) ⇒ Δt ′ = 67.1 years 2

2

2

2

37.50. Model: The spacetime interval s between the two events is invariant in all frames. Solve:

(a) Equating the two spacetime intervals, 2

2

⎛ 300 m ⎞ ⎛ 300 m ⎞ 2 2 2 2 2 2 2 2 c 2 ( Δt ) − ( Δx ) = c 2 ( Δt ′ ) − ( Δx′ ) ⇒ ⎜ ⎟ (10 μs ) − ( 0 m ) = ⎜ ⎟ ( Δt′ ) − ( 2400 m ) s s μ μ ⎝ ⎠ ⎝ ⎠ ⇒ (10 μs ) + ( 8 μs ) = ( Δt ′ ) ⇒ Δt ′ = 12.8 μs 2

2

2

(b) Note that Δt = Δτ because the event occurs at the same point in space. Hence,

Δt ′ =

Δτ 1 − (v / c)

2

⇒ 12.8 μs =

10 μs 1 − v2 / c2

⇒ v = 0.625c

37.51. Model: The earth is frame S and the starship is frame S′. S′ moves relative to S with a speed v. Solve:

(a) The speed of the starship is v=

20 ly ( 20 y ) c = = 0.80c 25 y 25 y

(b) The astronauts measure the proper time while they are traveling. This is

Δτ = 1 −

v2 2 Δt = 1 − ( 0.8 ) ( 25 y ) = 15 y c2

Because the explorers stay on the planet for one year, the time elapsed on their chronometer is 16 years.

37.52. Model: Let S be the reference frame of the ground and S′ the reference frame of the muons. S′ travels with a speed of v relative to S. Solve: From Section 37.6, we note that the time to travel a distance of 60 km in the earth’s frame S, the muon takes a time of approximately 60,000 m/(3 × 108 m/s) = 200 μs. However, stationary muons decay with a half−40 life of 1.5 μs. So, the fraction of muons reaching the ground should be ≈ 10 . Let us take the speed of the muon to be 0.9997c. In the muon’s frame, the 60 km ‘‘length” of the atmosphere is contracted to

L′ = L 1 − ( v / c ) = 60 km 1 − (0.9997)2 = 1.5 km. 2

That is, in the muon’s frame the atmosphere is only 1.5 km thick, not 60 km. Thus, the time taken by the muon to 1500 m 1500 m ⎛ 1 μs ⎞ = travel a distance of 1.5 km is ⎜ ⎟ ≈ 5 μs. This shows that the fraction of muons reaching 0.9997c 0.9997 ⎝ 300 m ⎠ ⎛ 5 μs ⎞ ⎜ ⎟

⎛ 1 ⎞⎝ 1.5 μs ⎠ = 0.1, or 1 out of 10. the earth is ⎜ ⎟ ⎝2⎠

37.53. Model: S′ is the electron’s frame and S is the ground’s frame. S′ moves relative to S with a speed v = 0.99999997c. Solve: For an experimenter in the S frame, the length of the accelerator tube is 3.2 km. This is the proper length = L because it is at rest and is always there for measurements. The electron measures the tube to be length contracted to L′ = 1 − β 2 = 1 − ( 0.99999997 ) ( 3200 m ) = 0.78 m 2

37.54. Model: Let the earth be frame S and the rocket be frame S´. S´ moves with speed v relative to S. Solve: (a) The round-trip distance is 860 ly. If the rocket takes time ∆t to make the round trip, as measured on earth, its speed (as a fraction of c) is v 860 ly 860 yr = = c c Δt Δt where we used c = 1 ly/yr (1 light year per year). The astronaut’s elapsed time ∆t´ is the proper time, so ∆τ = 20 yr. The time dilation equation is Δτ 20 yr Δt = = ⇒ 1 − (860 yr/ Δt ) 2 = ( 20 yr/ Δt ) 2 2 1 − ( v/c ) 1 − (860 yr/Δt ) 2 Solving for ∆t gives ∆t = 860.2325 y, and thus v 860 y = = 0.99973 ⇒ v = 0.99973c c 860.2325 y (b) The rocket starts with rest energy Ei = mc2 and accelerates to have energy Ef = γpmc2. Thus the energy needed to accelerate the rocket is ∆E = Ef – E1 = (γp – 1)mc2 This is just the kinetic energy K gained by the rocket. We know the rocket’s speed, so

⎛ ⎞ 1 ΔE = ⎜ − 1⎟ (20,000 kg)(3.0 × 108 m/s )2 = 7.6 × 1022 J 2 ⎜ 1 − (0.99973) ⎟ ⎝ ⎠ (c) The total energy used by the United States in 2000 was ≈1.0 × 1020 J. To accelerate the rocket would require roughly 760 times the total energy used by the United States.

37.55. Model: Let S be the earth’s reference frame and S′ be the rocket’s reference frame. S′ travels at 0.5c relative to S. Solve: (a) For the earthlings, the total distance traveled by the rocket is 2 × 4.25 ly = 8.5 ly. The time taken by the rocket for the round trip is

8.50 ly 8.50 ly = = 17 y 0.5c 0.5 ly/y (b) The time interval measured in the rocket’s frame S′ is the proper time because it can be measured with a single clock at the same position. So,

Δt =

Δτ 1 − (v c)

2

⇒ 17 y =

Δτ 1 − ( 0.5 )

2

⇒ Δτ = 14.7 y ≈ 15 y

The distance traveled by the rocket crew is length contracted to L′ = L 1 − ( v c ) = ( 8.50 ly ) 1 − ( 0.5 ) = 7.36 ly ≈ 7.4 ly 2

2

Note that the speed is still the same:

v=

L′ 7.36 ly = = 0.5 c Δτ 14.7 y

(c) Both are correct in their own frame of reference.

37.56. Model: Let S be the earth’s reference frame. Let S′, S″, and S″′ be the reference frames of the three

spaceships cruising through the galaxy in the direction from Delta to Epsilon at velocities v1 = 0.3c, v2 = 0.5c, and v3 = 0.7c relative to the earth’s frame. Solve: (a) In frame S, xD = 0 ly and tD = 0 y. Also, xE = 2 ly and tE = 1 y. In the moving frames, tD′ = tD′′ = tD′′′ = 0 y. We can use the Lorentz transformation to find the time at which Epsilon explodes. In frame S′,

tE′ =

tE − xE v1 c 2 1− v c 2 1

2

=

1 y − ( 2 ly )( 0.3 c ) c 2 1 − ( 0.3)

2

=

1 y − ( 2 y )( 0.30 ) 1 − ( 0.3)

2

= 0.42 y

For t E′′ the terms in the numerator cancel and tE′′ = 0 y. Lastly, tE′′′ = −0.56 y. (b) Spaceship 2 finds that the explosions are simultaneous. (c) Spaceship 3 finds that Epsilon explodes before Delta. (d) No. The explosions are far enough apart that Delta can have no causal influence on Epsilon. Thus there’s no difficulty if Epsilon explodes before Delta in some reference frames.

37.57. Model: Let S be the earth’s reference frame and S′ be the reference frame of one rocket. S′ moves relative to S with v = −0.75c. The speed of the second rocket in the frame S is u = +0.75c. Visualize:

Solve:

Using the Lorentz velocity transformation equation,

u′ =

0.75c − ( −0.75c ) u −v = = 0.96c 2 1 − uv c 1 − ( 0.75c )( −0.75c ) c 2

Assess: In Newtonian mechanics, the Galilean transformation of velocity will give u′ = 0.75c − (−0.75c) = 1.50c. This is not permissible according to the theory of relativity.

37.58. Model: The earth’s frame is S and the rocket’s frame is S ′. S ′ (the rocket) moves relative to S (the

earth) with velocity v, which we want to know. We are given u = 0.90 c and u′ = 0.95 c. Visualize:

Solve:

We want to solve for v in the Lorentz velocity transformation equation: u′ =

v= Assess:

u −v u − u′ ⇒v= 2 1 − uv / c 1 − uu′ / c 2

u − u′ 0.90 c − 0.95 c = = −0.34 c 1 − uu′ / c 2 1 − (0.90 c)(0.95 c) / c 2

We expected the rocket to be moving to the left.

37.59. Model: Use the relativistic expression for kinetic energy in Equation 37.44: K = (γ p − 1) E0 . The electric potential energy of the electron is transformed into its kinetic energy. Visualize: Use the conservation of energy equation U f − U i + kf − K i = 0 J. Solve: (−e)(Vf − Vi ) + K f − 0 J = 0 J ⇒

K f (γ p − 1) E0 (γ p − 1)mc = = = e e e 2

ΔV = Assess:

(

Three million volts is easily obtainable.

1 1− 0.992

K f = eΔV

)

− 1 (9.1 × 10−31 kg)(3.0 × 108 m / s)2 1.6 × 10−19 C

= 3.1× 106 V

37.60. Model: Use the relativistic expression for the kinetic energy in Equation 37.44. The electric potential energy of the proton is transformed into its kinetic energy. Solve: The conservation of energy equation Uf – Ui + Kf – Ki = 0 J is (e)(Vf – Vi) + Kf – 0 J = 0 J ⇒ Kf = eΔV ⇒ (γp – 1)mc2 = eΔV

⇒ γ p =1+

(1.6 ×10−19 C )( 50 ×106 V ) = 1.05323 = 1 eΔV = 1 + 2 mc 2 1 − v2 / c2 (1.67 ×10−27 kg )( 3.0 × 108 m/s ) ⇒1 – v2/c2 = 0.90148 ⇒ v = 0.314c

37.61. Model: Let S be the ground’s reference frame and S′ the muon’s reference frame. S′ travels with a speed of v relative to S. Solve: (a) The half-life of a muon at rest is 1.5 μs. That is, the half-life in the muon’s rest frame S′ is 1.5 μs. So, Δt′ = Δτ = 1.5 μs. The half-life of 7.5 μs, when muons have been accelerated to very high speed, means that Δt = 7.5 μs. Thus

Δτ

Δt = 7.5 μs =

1 − (v c)

2

=

1.5 μs 1 − v2 c2

⇒ 1 − v 2 c 2 = 0.20 ⇒ v = 0.98c

(b) The muon’s total energy is

E = γ p mc 2 =

2 ⎛ 1 ⎞ 8 −31 −11 mc 2 = ⎜ ⎟ ( 207 ) ( 9.11 × 10 kg )( 3.0 × 10 m/s ) = 8.5 × 10 J 0.20 ⎝ ⎠ 1− v c

1

2

2

37.62. Model: Let S be the sun’s reference frame and S′ be the rocket’s reference frame. S′ moves with

speed v = 0.8c relative to S. The flare’s speed in the frame S is u = 0.9c. Visualize:

Solve:

Using the Lorentz velocity transformation equation,

u′ =

0.9c − 0.8c u −v = = 0.36c uv 0.9c )( 0.8c ) 1− 2 1− ( c c2

That is, the flare is approaching the rocket at a speed of 0.36c.

37.63. Model: The principle of relativity demands that all laws of physics be the same in all inertial frames. Solve: (a) If you are in the S frame, you see the blue paint nozzle approaching at high speed. If the perpendicular lengths contract, then the blue nozzle will be less than 1 meter up from the x-axis and the blue nozzle will paint a line of blue under the red nozzle. If you are in the S′ frame, you see the red nozzle approaching at a high speed. Then, the red bar will shorten and you will see a red line under the blue nozzle. (b) Physically you can’t have both of these happen. So, we conclude that lengths perpendicular to the motion are not affected.

37.64. Solve: The Lorentz transformation equation for the x-direction are x′ = γ ( x − vt )

x = γ ( x′ + vt ′ )

Substituting the expression for x′ into the expression for x,

⎛ v2 ⎞ ⎛ vt ′ ⎞ x vt ′ x = γ ⎡⎣γ ( x − vt ) + vt ′⎤⎦ = γ 2 ⎜ x − vt + ⎟ ⇒ 2 = x ⎜1 − 2 ⎟ = x − vt + γ ⎠ γ γ ⎝ ⎝ c ⎠ 2 xv vt ′ ⎛ vx ⎞ ⇒ − 2 + vt − = 0 ⇒ t′ = γ ⎜ t − 2 ⎟ c γ ⎝ c ⎠ The equation for t is found in exactly the same way by substituting the expression for x into the expression for x´.

37.65. Model: Let S be the earth’s frame and S′ the rocket’s frame. S′ moves at speed 0.8c relative to S. Also, u ′y = 0.6c and u′x = 0.

(a) Using Equation 36.23 and making note of the relationship y = y′, uy 1 dy / dt 1 dy′ dy′ dy = = = = u′y = 2 dt ′ d ⎡γ ( t − vx / c ) ⎤ γ dt − v γ dx γ 1 − v dx γ ⎛ u x v ⎞ ⎣ ⎦ ⎜1 − 2 ⎟ c2 c 2 dt c ⎠ ⎝ Similarly u′y dy dy′ uy = = = dt ⎛ ′ vdx′ ⎞ ⎛ u′x v ⎞ γ ⎜ dt + 2 ⎟ γ ⎜ 1 + 2 ⎟ c ⎠ c ⎠ ⎝ ⎝

Solve:

(b) The rocket travels past the earth at v = 0.8c. It launches the projectile with velocity components ux′ = 0 and uy′ = 0.6c. In the earth’s frame, the x- and y-components of velocity are

u′x + v 0+v = = v = 0.8c 2 1 + u′x v / c 1+ 0 u′y u′y 0.6c = = = 0.36c uy = 2 γ (1 + u′x v / c ) γ (1 + 0 ) 1/ 1 − (0.80) 2 ux =

Thus the projectile’s speed in the earth’s frame is u = u x2 + u y2 = (0.80c) 2 + (0.36c) 2 = 0.877c

37.66. Solve: The relationship between energy, momentum, and velocity is pc = βp E. Thus β p E (0.95)(2.0 × 10−10 J) −19 p=

c

=

3.0 × 108 m/s

= 6.33 × 10

kg m/s

37.67. Solve: The relationship between energy, momentum, and rest energy is E2 – (pc)2 = (mc2)2. With E =

4mc2, this becomes

(4mc 2 ) 2 − ( pc ) 2 = ( mc 2 ) 2 ⇒ ( pc ) 2 = 15( mc 2 ) 2 ⇒ p = 15mc = 3.87 mc

37.68. Model: Use Equations 37.35 and 37.42 for the momentum and total energy. Also, the quantity E0 = mc2 is an invariant in all inertial reference frames. Solve: (a) The momentum and energy are 1 1 p = γ mu = mu = 1.67 × 10−27 kg ⎡⎣( 0.99 ) 3 × 108 m/s ⎤⎦ = 3.5 × 10−18 kg m/s 2 2 2 1− u c 1 − ( 0.99 )

(

E = γ mc 2 =

1 1− u c 2

2

mc 2 =

)

1 1 − ( 0.99 )

2

(1.67 ×10

(

−27

)

kg )( 3.0 × 108 m/s ) = 1.1× 10−9 J 2

(b) From Equation 36.45, E ′2 − p′2c 2 = E02 = m 2c 4 . Thus,

( 5.0 ×10−10 J ) − 1.67 × 10−27 kg 2 3.0 × 108 m/s 2 E ′2 p′ = 2 − m 2c 2 = ( )( ) 2 c ( 3.0 ×108 m/s ) 2

2

= ( 2.778 × 10−36 − 0.2510 × 10−36 ) kg 2 m 2 /s 2 ⇒ p′ = 1.6 × 10−18 kg m/s

37.69. Solve: Using Equation 37.45 for the relativistic kinetic energy and K = 12 mu 2 for the Newtonian kinetic energy, we have

(γ

2 2 2 1 p − 1) mc = 2 ( 2 mu ) = mu ⇒ γ p − 1 =

⇒ u4 + c2u2 – c4 = 0 ⇒ u 2 =

2 u2 1 ⇒ = 1 + u 2 / c 2 ⇒ 1 = (1 − u 2 / c 2 )(1 + u 2 / c 2 ) 2 2 2 c 1− u /c

−c 2 ± c 4 + 4c 4 −c 2 ± 5c 2 = ⇒ u2 = 2 2

(

)

5 − 1 c2 2

⇒ u = 0.786 c

37.70. Solve: The rest energy and the total energy are given by Equations 37.43 and 37.42. We have mpc2 = γmec2 ⇒ γ =

mp me

= 1833 =

1 1 − u 2 / c2

⇒ u = 0.99999985c

37.71. Solve:

Model: Mass and energy are equivalent and given by Equation 37.43. (a) The power plant running at full capacity for 80% of the year runs for (0.80)(365 × 24 × 3600) s = 2.52 × 107 s

The amount of thermal energy generated per year is 3 × (1000 × 106 J/s) × (2.52 × 107 s) = 7.56 × 1016 J ≈ 7.6 × 1016 J (b) Since E0 = mc2, the mass of uranium transformed into thermal energy is

m=

E0 7.56 × 1016 J = = 0.84 kg 2 2 c ( 3.0 ×108 m/s )

37.72. Model: Mass and energy are equivalent and given by Equation 37.43.

(a) The sun radiates energy for 3.154 × 107 s per year. The amount of energy radiated per year is (3.8 × 1026 J/s)(3.154 × 107 s) = 1.198 × 1034 J/y Since E0 = mc2, the amount of mass lost is Solve:

m=

E0 1.198 × 1034 J = = 1.33 × 1017 kg ≈ 1.3 × 1017 kg 2 2 8 c ( 3.0 ×10 m/s )

(b) Since the mass of the sun is 2.0 × 1030 kg, the sun loses 6.7 × 10 (c) The lifetime of the sun can be estimated to be

T=

−12

% of its mass every year.

2.0 × 1030 kg = 1.5 × 1013 years 1.33 × 1017 kg/y

The sun will not really last this long in its current state because fusion only takes place in the core and it will become a red giant when the core hydrogen is all fused.

37.73. Model: Mass and energy are equivalent and given by Equation 37.43. Solve:

The mass “lost” during each disintegration is 222.017 u + 4.003 u – 226.015 u = 0.005 u = (0.005)(1.661 × 10−27 kg/u) = 8.30 × 10−30 kg

Since E0 =mc2, the energy produced with this mass loss is E = mc2 = (8.30 × 10−30 kg)(3.0 × 108 m/s)2 = 7.5 × 10−13 J

37.74. Model: Mass and energy are equivalent and given by Equation 37.43. Solve:

(a) The mass “lost” in each reaction is 4mp – mHe = 4(1.67 × 10−27 kg) – 6.64 × 10−27 kg = 0.04 × 10−27 kg

In each fusion reaction, the energy released is E = mc2 = (0.04 × 10−27 kg) (3.0 × 108 m/s)2 = 3.6 × 10−12 J (b) The initial mass of the 4 protons is 6.68 × 10−27 kg. The percent of mass lost during each fusion reaction is

0.04 × 10−27 kg × 100 = 0.60% 6.68 × 10−27 kg Since m = E / c 2 , the energy released in one fusion reaction is also 0.60% of the initial rest energy.

37.75. Model: Particles can be created from energy, and particles can return to energy. When a particle and its antiparticle meet, they annihilate each other and create two gamma ray photons. Solve: The energy of the electron is

Eelectron = γ p mec 2 =

1 1 − ( 0.9 )

2

( 9.11×10

−31

kg )( 3.0 × 108 m/s ) = 1.88 × 10−13 J 2

The energy of the positron is the same, so the total energy is Etotal = Eelectron + Epositron = 3.76 × 10−13 J. The energy is converted to two equal-energy photons. Thus, Etotal = 2hf = 2hc λ . The wavelength is

λ= Assess:

−34 8 2hc 2 ( 6.62 × 10 J s )( 3 × 10 m/s ) = = 1.06 × 10−12 m ≈ 1 pm Etotal 3.76 × 10−13 J

This wavelength is typical of γ-ray photons.

37.76.

Model: Particles can be created from energy. Solve: (a) The decrease in kinetic energy is exactly equal to the energy-equivalent of the mass of the two particles that have been created. The threshold kinetic energy is when the electron and positron are created at rest, thus K = 2mec2 = 2(9.11 × 10−31 kg)(3.0 × 108 m/s)2 = 1.64 × 10−13 J

(b) The kinetic energy of the fast electron is K = 2mec2 = (γp – 1)mec2 ⇒ γ p = 3 =

1 1 − u2 / c2

⇒ u = 0.943c

37.77. Model: Let S be the earth’s reference frame, and S′ be the frame of the rocket B. S′ moves with a velocity v = −0.8c relative to S. In S, rocket A has a speed u = 0.8c. Visualize:

Solve:

Using the Lorentz velocity transformation equation, the speed of rocket A as observed from rocket B is

u′ =

0.8c − ( −0.8c ) u −v = = 0.976c uv 0.8c )( −0.8c ) 1− 2 1− ( c c2

The length of rocket A as observed from rocket B is thus L = (100 m ) 1 − ( 0.976 ) = 22 m. 2

37.78. Model: Let S be the earth’s reference frame, and S′ be the rocket Sirius’s reference frame. S′ moves

relative to S with a speed of v = 0.6c. In S, rocket Orion’s speed is u = 0.8c. Visualize:

Solve:

Orion’s speed relative to Sirius is 0.8c − 0.6c u −v = = 0.385c u′ = 1 − uv c 2 1 − ( 0.8c )( 0.6c ) c 2

Thus, the length of the Orion rocket as measured in S′ (Sirius rocket) is L′ = (1000 m ) 1 − ( 0.385 ) = 923 m 2

Now, Orion is moving at a speed of 0.385c and has to move 1923 m to completely pass Sirius. Thus, the time taken is Δt =

1923 m = 17 μ s 0.385 ( 300 m 1 μ s )

37.79. Model: Particles can be created from energy and particles can return to energy. When a particle and its antiparticle meet, they annihilate each other and can create two gamma ray photons and a particle-antiparticle pair. Solve: The energy of the proton-antiproton pair is equal to the sum of the energies of two γ-ray photons and the electron-positron pair. Because the electron and positron are ejected with a speed of u = 0.9999995c, their energies are equal. The relativistic energy equation is Eelectron = Epositron = γpmc2 with

γp =

1 1− u c 2

2

=

1 1 − ( 0.9999995 )

2

= 1000

⇒Eelectron = 1000(9.11 × 10−31 kg)(3.0 × 108 m/s)2 = 8.20 × 10−11 J The energy of one of the photons is

Egamma hf =

hc

λ

=

( 6.63 ×10

−34

J s )( 3.0 × 108 m/s )

1.0 × 10−6 × 10−9 m

= 1.99 × 10−10 J

Because a pair of proton-antiproton produces a pair of electron-positron and a pair of γ-ray photons, the energy of a proton (or antiproton) is Eproton = Eelectron + Egamma = 0.82 × 10−10 J + 1.99 × 10−10 J = 2.81 × 10−10 J

⇒ 2.81 × 10

−10

J = γ p mp c

2

(1.67 ×10 =

−27

kg )( 3.0 × 108 m/s )

1 − ( up c )

2

2

⇒ 1 − ( up c ) = 0.535 ⇒ u = 0.85c 2

37.80. Model: Let the Huns’ rocket be frame S and the Goths’ rocket be frame S´. S´ moves with velocity v

= 0.8c relative to S. Visualize: Begin by considering the situation from the Huns’ reference frame.

Solve: The key to resolving the paradox is the fact that two events simultaneous in one reference frame are not simultaneous in a second reference frame. The Huns do, indeed, see the Goths’ rocket length contracted to LGoths = (1 – (0.8)2)1/2(1000 m) = 600 m. Let event 1 be the tail of the Goths’ rocket passing the nose of the Huns’ rocket. Since we’re free to define the origin of our coordinate system, define this event to be at time t1 = 0 μs and at position x1 = 0 m. Then the spacetime coordinates of event 2, the firing of the laser cannon, are (x2, t2) = (1000 m, 0 μs). The nose of the Goths’ rocket is at x = 600 m at t = 0 μs, thus the laser cannon misses the Goths by 400 m. This distance is a length because both ends of the interval are measured simultaneously at t = 0. Now use the Lorentz transformations to find the spacetime coordinates of the events in the Goths’ reference frame. The nose of the Huns’ rocket passes the tail of the Goths’ rocket at ( x ′1 , t′1 ) = (0 m, 0 μs). The Huns fire their laser cannon at

5 x′2 = γ ( x2 − vt2 ) = (1000 m – 0 m) = 1667 m 3 vx2 5⎛ 1000 m ⎞ t ′2 = γ (t2 − 2 ) = ⎜ 0 μ s − (0.8) ⎟ = −4.444 μs c 3⎝ 300 m/μs ⎠ Events 1 and 2 are not simultaneous in S´. In fact, the Huns fire the laser cannon 4.444 μs before the nose of their rocket reaches the tail of the Goths’ rocket. The laser is fired at x′2 = 1667 m, which misses the Goths’ nose by 667 m. In fact, since the Huns’ rocket is length contracted to 600 m, the nose of the Huns’ rocket is at x´ = 1067 m at the instant when they fire the laser cannon.

Traveling at a speed of v = 0.8c = 240 m/μs, in 4.444 μs the nose of the Huns’ rocket travels ∆x´ = (240 m/μs)(4.444 μs) = 1067 m—exactly the right distance to be at the tail of the Goths’ rocket at t′1 = 0 μs. We could also note that the 667 “miss distance” in the Goths’ frame is length contracted to (1 – (0.8)2)1/2(667 m) = 400 m in the Huns’ frame—exactly the amount by which the Huns think they miss the Goths’ rocket. Thus we end up with a consistent explanation. The Huns miss the Goths’ rocket because, to them, the Goths’ rocket is length contracted. The Goths find that the Huns miss because event 2 (the firing of the laser cannon) occurs before event 1 (the nose of one rocket passing the tail of the other). The 400 m distance of the miss in the Huns’ reference frame is the length-contracted miss distance of 667 m in the Goths’ reference frame.

37.81. Model: Let the farmer be frame S and the pole vaulter be frame S′. S′ moves with velocity v = 0.888c relative to S. The value of γ is γ = 2. Visualize: Begin by considering the situation from the farmer’s reference frame.

Solve: The key to resolving the paradox is the fact that two events simultaneous in one reference frame are not simultaneous in a second reference frame. The farmer does, indeed, see the pole length contracted to L = (1 – (0.888)2)1/2(16 m) = 8 m. Let event 1 be the farmer closing the left door just as the back edge of the pole enters. Since we’re free to define the origin of our coordinate system, define this event to be at time t1 = 0 μs and at position x1 = 0 m. Then the spacetime coordinates of event 2, the simultaneous closing of the right door, are (x2, t2) = (10 m, 0 μs). The front of the pole is at x = 8 m at t = 0 μs, thus the right door misses the pole by 2 m. This distance is a length because both ends of the interval are measured simultaneously at t = 0. To fully understand the situation, it is also useful to define a third event—the front of the pole crashing through the right door. The pole has to move 2 m at 0.866c, taking (2 m)/(0.866 × 300 m/μs) = 0.0077 μs, so the spacetime coordinates of event 3 are (x3, t3) = (10 m, 0.0077 μs). Now use the Lorentz transformations to find the spacetime coordinates of the events in the pole vaulter’s reference frame. The left door of the barn is closed as it reaches the back of the pole at ( x ′1 , t′1 ) = (0 m, 0 μs). The farmer closes the right barn door at

x′2 = γ ( x2 − vt2 ) = 2(10 m – 0 m) = 20 m t ′2 = γ (t2 −

⎛ vx2 10 m ⎞ ) = 2 ⎜ 0 μs − (0.866) ⎟ = −0.0577 μs c2 300 m/ μs ⎠ ⎝

Events 1 and 2 are not simultaneous in S′. In fact, the farmer closes the right door 0.0577 μs before closing the left door. At the instant it closes, the right door is at x2′ = 20 m, missing the front of the pole by 4 m. Since the barn is length contracted to 5 m, the left door of the barn is at x′ = 15 m at the instant the right door closes.

Traveling at a speed of v = 0.866c = 260 m/μs, in 0.0577 μs the left door travels ∆x´ = (260 m/μs)(0.0577 μs) = 15 m—exactly the correct distance to be at the back end of the pole at t′1 = 0 μs. We could also note that the 4 m “miss distance” in the pole vaulter’s frame is length contracted to (1 – (0.866)2)1/2(4 m) = 2 m in the farmer’s frame—exactly the amount by which the farmer thinks the right door misses the front end of the pole. According to the pole vaulter, the on-rushing right door of the barn—now closed—will crash into the tip of his pole at x3′ = 16 m after traveling 4 m at 0.866c, taking (4 m)/(0.866 × 300 m/μs) = 0.0154 μs. The door closed at t2′ = –0.0577 μs, so the door reaches his pole at t3′ = –0.0577 μs + 0.0154 μs = –0.0423 μs. Thus, in S′, the spacetime coordinates of event 3 appear to be ( x3′ , t3′ ) = (16 m, –0.0423 μs). We can check this by transforming the frame S spacetime coordinates to S´. We find that x3′ = γ ( x3 − vt3 ) = 2(10 m – (0.866 × 300 m/ μs)(0.0077 μ s)) = 16 m t3′ = γ (t3 − The agreement is perfect.

⎛ 10 m ⎞ vx3 ) = 2 ⎜ 0.0077 μs − (0.866) ⎟ = −0.0423 μs 300 m/ μ s ⎠ c2 ⎝

Thus we end up with a consistent explanation. The farmer is able to close both doors without breaking the pole because, to the farmer, the pole is length contracted to less than the length of the barn. The pole vaulter also finds that the farmer can close both doors without breaking the pole because event 2 (the closing of the right door) occurs before event 1 (the closing of the left door). The 2 m distance by which the door misses the front of the pole in the farmer’s reference frame is the length-contracted miss distance of 4 m in the pole vaulter’s frame. According to the farmer, the left door is closed before the pole breaks through the right door. The order of these two events is reversed in the pole vaulter’s frame, where the pole breaks through the right door before the left door is closed. These two events can occur in a different order in the two reference frames because the two events are not causally related.

38.1. Model: Current is defined as the rate at which charge flows across an area of cross section. Solve:

Since the current is ΔQ / Δt and ΔQ = N / e , the number of electrons per second is

N 10 nA 1.0 × 10−8 C/s = = = 6.25 × 1010 s −1 ≈ 6.3 × 1010 s −1 Δt e 1.60 × 10−19 C

38.2. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Visualize: Please refer to Figures 38.7 and 38.8. Solve: (a) The speed with which a particle can pass without deflection is

E ΔV/d 600 V/5.0 × 10−3 m = = = 6.0 × 107 m/s B B 2.0 × 10−3 T (b) The radius of cyclotron motion in a magnetic field is v=

−31 7 ⎛ m ⎞ v ⎛ 9.11 × 10 kg ⎞⎛ 6.0 × 10 m ⎞ r =⎜ ⎟ =⎜ ⎟⎜ ⎟ = 0.17 m = 17 cm −19 −3 ⎝ e ⎠ B ⎝ 1.60 × 10 C ⎠⎝ 2.0 × 10 T ⎠

38.3. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Visualize: Without the external magnetic field B, the electrons will be deflected up toward the positive electrode. The magnetic field must therefore be directed out of the page to exert a balancing downward force on the negative electron. Solve: In a crossed-field experiment, the magnitudes of the electric and magnetic forces on the electron are given by Equation 38.4. The magnitude of the magnetic field is

B=

E ΔV/d 200 V/8.0 × 10−3 m = = = 5.0 × 10−3 T v v 5.0 × 106 m

Thus B = (5.0 × 10–3 T, out of page).

38.4. Model: Assume the electric field (E = ΔV/d) between the plates is uniform. Visualize: Please refer to Figure 38.9. Solve: (a) The mass of the droplet is 3 4π ⎛ 4π 3 ⎞ mdrop = ρV = ρ ⎜ R ⎟ = ( 885 kg/m3 ) ( 0.4 × 10−6 m ) = 2.37 × 10−16 kg ≈ 2.4 × 10−16 kg 3 ⎝ 3 ⎠

(b) In order for the upward electric force to balance the gravitational force, the charge on the droplet must be

mdrop g

( 2.37 × 10

−16

kg )( 9.8 m/s 2 )

= 1.28 × 10−18 C ≈ 1.3 × 10−18 C E 20 V 11 × 10−3 m (c) Because the electric force is directed toward the electrode at the higher potential (or more positive plate), the charge on the droplet is negative. The number of surplus electrons is qdrop =

=

N=

qdroplet e

=

1.28 × 10−18 C =8 1.60 × 10−19 C

38.5. Model: Assume the electric field (E = ΔV/d) between the plates is uniform. Visualize: Please refer to Figure 38.9. To balance the weight, the electric force must be directed toward the upper electrode, which is more positive than the lower electrode. Solve: Since mdrop = ρV = ρ ( 43π ) R 3 , the equation mdrop g = qdrop E is R3 =

( ΔV d )(15e ) = ( 25 V 0.012 m )(15) (1.60 × 10−19 C ) = 1.4163 × 10−19 m3 ⇒ R = 0.52 μ m 4π 3

ρg

4π 3

(860 kg/m )( 9.8 m/s ) 3

2

38.6. Model: The charge on an object is an integral multiple of a certain minimum charge value.

Solve: From smallest charge to the largest charge, the measured charges on the drops are 2.66 × 10−19 C, 3.99 × 10−19 C, 6.65 × 10−19 C, 9.31 × 10−19 C, and 10.64 × 10−19 C. The differences between these values are 1.33 × 10−19 C, 2.66 × 10−19 C, 2.66 × 10−19 C, and 1.33 × 10−19 C. These differences are a multiple of 1.33 × 10−19 C. Thus, the largest value of the fundamental unit of charge that is consistent with the charge measurements is 1.33 × 10−19 C.

38.7. Model: The electron volt is a unit of energy and is defined as the kinetic energy gained by an electron or proton if it accelerates through a potential difference of 1 volt. Solve: (a) Converting electron volts to joules,

100 eV = 100 eV ×

1.60 × 10−19 J = 1.60 × 10−17 J 1 eV

Using the definition of kinetic energy K = 12 mv 2 , v=

2 (1.60 × 10−17 J ) 2K = = 5.9 × 106 m/s m 9.11 × 10−31 kg

(b) Likewise, the speed of the neutron is v=

( 2 )( 5.0 MeV ) =

1.67 × 10

−27

kg

2 ( 8.0 × 10−13 J ) 1.67 × 10−27 kg

= 3.1 × 107 m/s

(c) The mass of the particle is

m=

2 ( 3.34 × 1013 J ) 2 ( 2.09 MeV ) 2K = = = 6.69 × 10−27 kg v 2 (1.0 × 107 m/s )2 (1.0 × 107 m/s )2

The mass of the particle is the same as the mass of four protons (or two protons and two neutrons). It is an alpha particle.

38.8. Model: The electron volt is a unit of energy and is defined as the kinetic energy gained by an electron or proton if it accelerates through a potential difference of 1 volt. Solve: (a) Converting electron volts to joules,

6 MeV = 6 × 106 eV ×

1.6 × 10−19 J = 9.6 × 10−13 J 1 eV

Using the definition of kinetic energy K = 12 mv 2 , v=

2 ( 9.6 × 10−13 J ) 2K = = 3.4 × 107 m/s m 1.67 × 10−27 kg

(b) Likewise, the speed of the helium atom is v=

2 ( 20 MeV )

4 (1.67 × 10−27 kg )

=

2 ( 3.20 × 10−12 J )

4 (1.67 × 10−27 kg )

= 3.1 × 107 m/s

(c) The mass of the particle is

m= The particle is an electron.

2 (1.82 × 10−16 J ) 2 (1.14 keV ) 2K = = = 9.12 × 10−31 kg 2 2 2 7 7 v ( 2.0 ×10 m/s ) ( 2.0 × 10 m/s )

38.9. Model: The electron volt is a unit of energy. It is defined as the energy gained by an electron if it accelerates through a potential difference of 1 volt. Solve: (a) The kinetic energy is

K = 12 mv 2 = 12 ( 9.11× 10−31 kg )( 5.0 × 106 m/s ) = 1.139 × 10−17 J × 2

1 eV = 71 eV 1.60 × 10−19 J

(b) The potential energy is

U=

1

( e )( −e )

4πε 0 0.10 nm

=

− ( 9.0 × 109 N m 2 /C 2 )(1.60 × 10−19 ) 0.10 × 10−9 m

2

= −2.30 × 10−18 J ×

1 eV = −14 eV 1.60 × 10−19 J

(c)

The figure shows a proton accelerating from rest across a parallel-plate capacitor with a potential difference of ΔV = 5000 V. The energy conservation equation Kf + qVf = Ki + qVi is Kf = Ki + q(Vi – Vf) = 0 J + eΔV = e(5000 V) = 5000 eV = 5.0 keV

38.10. Model: The electron volt is a unit of energy. It is defined as the energy gained by an electron if it accelerates through a potential difference of 1 volt. Visualize:

Solve: (a) The figure for part (a) shows a Li++ ion accelerating from rest across a parallel-plate capacitor. The energy conservation equation Kf + qVf = Ki + qVi is Kf = Ki + q(Vi – Vf) = 0 + (2e)ΔV = 2e(5000 V) = 10,000 eV = 10 keV (b) The potential energy is

U=

1

( e )( e ) = ( 9.0 × 109 N m 2 /C2 )(1.60 × 10−19 C )

4πε 0 10 fm

10.0 × 10−15 m

2

= 2.30 × 10−14 J ×

1 eV = 144 keV 1.60 × 10−19 J

(c) The energy-conservation equation for the figure for part (c) Kf + Uf = Ki + Ui is

K f = 0 J + (U i − U f ) = mg ( yi − yf ) = ( 0.200 kg ) ( 9.8 m/s 2 ) (1.0 m − 0 m ) = 1.96 J ×

1 eV = 1.2 × 1019 eV 1.60 × 10−19 J

38.11. Visualize: The work-kinetic energy theorem is ΔK = Wnet . Work done on the proton in slowing it down will be W = − qΔV . K f − K i = −qΔV Solve:

But since K f = 0 , then

K i = qΔV = (1e)(75V) = 75eV Assess:

The plate separation doesn’t matter.

38.12. Model: The potential of a sphere of charge is V = (1/ 4πε 0 )( q/r ) if r ≥ R . Visualize:

We are given R = 0.50 mm, and q = 0.2 nC. Use conservation of mechanical energy.

K f + U f = Ki + U i Solve:

Solve for Kf . K f = Ki + U i − U f = 520 eV + (9.0 × 109 N ⋅ m 2 /C 2 ) 0.2 nC − (9.0 × 109 N ⋅ m 2 /C2 ) 0.2 nC 0.50 mm 2.50 mm 9 2 2 −9 1 1 = 520 eV + (9.0 × 10 N ⋅ m /C )(0.2 × 10 C) − 0.50 mm 2.50 mm = 3400 eV

(

Assess: sphere.

)

We expected the proton to have more kinetic energy as it was pushed away by the positively charged

38.13. Model: For a neutral atom, the number of electrons is the same as the number of protons.

Solve: (a) Since Z = 4, 9Be has 4 electrons, 4 protons, and 9 – 4 = 5 neutrons. (b) Since Z = 7, the 14N+ ion has 6 electrons, 7 protons, and 14 – 7 = 7 neutrons. (c) Since Z = 6, the doubly charged 13C++ ion has 4 electrons, 6 protons, and 13 – 6 = 7 neutrons.

38.14. Model: For a neutral atom, the number of electrons is the same as the number of protons.

Solve: (a) Since Z = 5, the 10B atom has 5 electrons, 5 protons, and 10 – 5 = 5 neutrons. (b) Since Z = 7, 13N+ has 6 electrons, 7 protons, and 13 – 7 = 6 neutrons. (c) Since Z = 8, the triply charged 17O+++ ion has 5 electrons, 8 protons, and 17 – 8 = 9 neutrons.

38.15. Model: For a neutral atom, the number of electrons is the same as the number of protons, which is the

atomic number Z. An atom’s mass number A is A = Z + N, where N is the number of neutrons. Solve: (a) This is a neutral atom with the symbol 10Be. (b) This is a doubly positively charged ion with A = 6 + 5 = 11. We have 6 protons, but there are only 4 electrons. The symbol is 11C++.

38.16. Model: For a neutral atom, the number of electrons is the same as the number of protons, which is the

atomic number Z. An atom’s mass number is A = Z + N, where N is the number of neutrons. Solve: (a) This is a neutral atom with A = 1 + 2 = 3. The symbol is 3H. This is also known as tritium. (b) This is a positively charged ion with A = 8 + 10 = 18. The symbol is 18O+.

38.17. Model: For a neutral atom, the number of electrons is the same as the number of protons, which is the

atomic number Z. An atom’s mass number is A = Z + N, where N is the number of neutrons. Solve: (a) For a 197Au atom, Z = 79. So, N = 197 – 79 = 118. A neutral 197Au atom contains 79 electrons, 79 protons, and 118 neutrons. (b) Assuming that the neutron rest mass is the same as the proton rest mass, the density of the gold nucleus is

ρ nucleus =

197mproton

4π 3

( 7.0 ×10

−15

m)

3

=

197 (1.67 × 10−27 kg ) 4π 3

( 7.0 ×10

−15

m)

3

= 2.29 × 1017 kg/m3

(c) The nuclear density in part (b) is 2.0 × 1013 times the density of lead. Assess: The mass of the matter is primarily in the nuclei and the volume of the matter is essentially due to the electrons around the nuclei.

38.18. Model: For a neutral atom, the number of electrons is the same as the number of protons, which is the atomic number Z. An atom’s mass number is A = Z + N, where N is the number of neutrons. Solve: (a) For a 207Pb atom, Z = 82. So, N = 207 – 82 = 125. The neutral protons, and 125 neutrons. (b) The electric potential at the surface of a lead nucleus is

V=

207

Pb atom contains 82 electrons, 82

9 2 2 −19 1 Q ( 9.0 × 10 N m /C ) ( 82 ) (1.60 × 10 C ) = = 1.66 × 107 V −15 4πε 0 r 7.1 × 10 m

The electric field strength at the surface of a lead nucleus is

E=

9 2 2 −19 Q ( 9.0 × 10 N m /C ) ( 82 ) (1.60 × 10 C ) = = 2.34 × 1021 V/m 2 2 4πε 0 r ( 7.1×10−15 m )

1

38.19. Model: Use Equation 38.11 which is the Balmer formula. Visualize: Please refer to Figure 38.22. Solve: (a) The wavelengths in the hydrogen emission spectrum are 656.6 nm, 486.3 nm, 434.2 nm, and 410.3 nm. The formula for the Balmer series can be written

1 1 91.18 nm − = λ m2 n2 where m = 1, 2, 3, . . . and n = m + 1, m + 2, . . . For the first wavelength,

1 1 91.18 nm n2 − m2 − 2= = 0.1389 ⇒ 2 2 = 0.1389 2 m n 656.5 nm nm This equation is satisfied when m = 2 and n = 3. For the second wavelength (486.3 nm) the equation is satisfied for m = 2 and n = 4. Likewise, for the next two wavelengths m = 2 and n = 5 and 6. (b) The fifth line in the spectrum will correspond to m = 2 and n = 7. Its wavelength is

λ=

91.18 nm

( )

1 2 2

− ( 17 )

2

⎛ 196 ⎞ =⎜ ⎟ ( 91.18 nm ) = 397.1 nm ⎝ 45 ⎠

38.20. Model: Use Equation 38.11 which is the Balmer formula. Visualize: Please refer to Figure 38.22. Solve: (a) The wavelengths in the hydrogen emission spectrum are 656.6 nm, 486.3 nm, 434.2 nm, and 410.3 nm. The formula for the Balmer series can be written

1 1 91.18 nm − = λ m2 n2 where m = 1, 2, 3, . . . and n = m + 1, m + 2, . . . For the first wavelength,

1 1 91.18 nm n2 − m2 − 2= = 0.1389 ⇒ 2 2 = 0.1389 2 m n 656.5 nm nm This equation is satisfied when m = 2 and n = 3. For the second wavelength (486.3 nm) the equation is satisfied for m = 2 and n = 4. Likewise, for the next two wavelengths m = 2 and n = 5 and 6. (b) From part (a), the lines in the spectrum in Figure 38.22 (b) are represented by the Balmer formula when m = 2. The limiting wavelength of the series occurs when n → ∞, which means 1 n 2 → 0. In this limit,

λ=

91.18 nm

( ) −( ) 1 2 2

1 2 n

→

91.18 nm

( 12 )

2

= 4 ( 91.18 nm ) = 364.7 nm

38.21. Model: Use Equation 38.11 with m = 1 and n = 2, 3, 4, and 5. This series of spectral lines is called the Lyman series. Solve: The formula for the Lyman series simplifies to λ=

91.18 nm 1 − (1 n )

2

where n = 2, 3, 4 and 5

For n = 2, λ = ( 91.18 nm ) (1 − 14 ) = 121.6 nm. For n = 3, λ = ( 91.18 nm ) (1 − 19 ) = 102.6 nm. Likewise, for n = 4 −1

and n = 5, λ = 97.3 nm and λ = 95.0 nm, respectively.

−1

38.22. Model: Use Equation 38.9 which is the Balmer formula. Solve:

(a) Equation 38.9 can be written as

1 1 n 2 − m 2 91.18 nm − = 2 2 = λ m2 n2 nm where m = 1, 2, 3, . . . and n = m + 1, m + 2, . . . For λ = 102.6 nm, n 2 − m 2 91.18 nm = = 0.8887 102.6 nm n2m2

This equation is satisfied by m = 1 and n = 3. For λ = 1876 nm, n 2 − m 2 91.18 nm = = 0.0486 1876 nm n2m2

This equation is satisfied by m = 3 and n = 4. (b) The ultraviolet wavelengths are <400 nm. The infrared wavelengths are >700 nm. So, λ = 102.6 nm is in the ultraviolet region and λ = 1876 nm is in the infrared region.

38.23. Model: Assume the blackbodies obey Wein’s law in Equation 38.9: λpeak = (2.90 × 106 nm ⋅ K)/T . Visualize:

We want to solve for T in the equation above. We are given λpeak = 300 nm, 3000 nm.

Solve:

T=

2.90 × 106 nm ⋅ K

λpeak

(a) T=

2.90 × 106 nm ⋅ K = 9667 K = 9394°C 300 nm

T=

2.90 × 106 nm ⋅ K = 966.7 K = 694°C 3000nm

(b)

38.24. Model: Assume the metal sphere is a blackbody (so the emissivity e =1 ). Visualize: First use Wein’s law (Equation 38.9) to find the temperature, then use Stefan’s law (Equation 38.8) to determine the power radiated. We are given R = 1.0 cm and λpeak = 2000 nm. Solve:

T=

2.90 × 106 nm ⋅ K = 1450 K 2000 nm

Q = eσ AT 4 = (1)(5.67 × 10−8 W/m 2 ⋅ K 4 )[4π (1.0 cm) 2 ](1450 K) 4 = 315 W Δt

Assess: The sphere radiates more than 3100W light bulbs, but it has a larger surface area than the filaments, so the answer is reasonable.

38.25. Model: Assume the ceramic cube is a blackbody (so the emissivity e =1 ). Visualize: First use Stefan’s law (Equation 38.8), ΔQt = eσ AT 4 , to find the temperature, then use Wein’s law (Equation 38.9) to get the peak wavelength. We are given A = 6(3.0 cm × 3.0 cm) = 0.0054 m 2 and Q/Δt = 630 W. Solve: Solve Stefan’s law for T.

T=4

Q/Δt 630 W =4 = 1198 K −8 eσ A (1)(5.67 × 10 W/m 2 ⋅ K 4 )(0.0054 m 2 )

Now plug this temperature into Wein’s law.

λpeak =

2.90 × 106 nm ⋅ K 2.90 × 106 nm ⋅ K = = 2420 nm = 2.42 μ m T 1198 K

38.26. Model: Use the relativistic expression for the total energy. Solve:

(a) The energy of the proton is

E = γ p mc 2 =

1 1− v c 2

= 1.065 × 10−9 J ×

2

mc 2 =

1 1 − ( 0.99 )

2

mc 2 = ( 7.089 ) (1.67 × 10−27 kg )( 3.0 × 108 m/s )

2

1 eV = 6.66 × 109 eV = 6660 MeV 1.6 × 10−19 J

(b) Likewise, the energy of the electron is

E= Assess:

1 1 − ( 0.99 )

2

( 9.11×10

−31

kg )( 3.0 × 108 m/s ) = 5.812 × 10−13 J × 2

1 eV = 3.63 × 106 eV = 3.63 MeV 1.6 × 10−19 J

The total energy E for the proton is larger than that for the electron by the factor mproton/melectron = 1833.

38.27. Model: Use the relativistic expression for the total energy. Solve:

(a) The energy of the proton is

E = γ p mc 2 = 500 GeV =

(1.67 × 10

−27

kg )( 3.0 × 108 m/s ) 1− v c 2

2

2

= 500 × 109 eV ×

⇒ 1 − v c = 1.879 × 10−3 ⇒ v = 0.999998c 2

2

(b) Likewise for the electron,

( 9.11×10 E = 2.0 GeV =

−31

kg )( 3.0 × 108 m/s )

2

1 − v2 c2

⇒ 1 − v 2 c 2 = 2.562 × 10−4 ⇒ v = 0.99999997c

1.60 × 10−19 J 1 eV

38.28. Solve: The rest energy of an electron is E0 = mc 2 = ( 9.11× 10−31 kg )( 3.0 × 108 m/s ) = 8.199 × 10−14 J × 2

1 eV = 0.512 MeV 1.6 × 10−19 J

The rest energy of a proton is

E0 = mc 2 = (1.67 × 10−27 kg )( 3.0 × 108 m/s ) = 1.503 × 10−10 J × 2

1 eV = 939 MeV 1.6 × 10−19 J

38.29. Solve: (a) The kinetic energy of the electron is K = ( γ − 1) mc 2 = (1.01 − 1) ( 9.11× 10−31 kg )( 3.0 × 108 m/s ) = 0.8199 × 10−15 J × 2

1 eV = 0.00512 MeV 1.6 × 10−19 J

(b) Likewise for the proton,

K = (1.01 − 1) (1.67 × 10−27 kg )( 3.0 × 108 m/s ) = 9.39 MeV 2

(c) Likewise for the alpha particle,

K = (1.01 − 1)( 4 ) (1.67 × 10−27 kg )( 3.0 × 108 m/s ) = 37.6 MeV 2

38.30. Model: Mass and energy are equivalent. Solve:

The energy released as kinetic energy is

K = ΔE = ( Δm ) c 2 = ( 0.185 ) (1.66 × 10−27 kg )( 3.0 × 108 m/s ) = 2.76 × 10−11 J × 2

1 eV = 173 MeV 1.6 × 10−19 J

38.31. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Visualize: Please refer to Figures 38.7 and 38.8. Solve: In a crossed-field experiment, the deflection is zero when the magnetic and electric forces exactly balance each other. From Equation 38.4, the speed of the electrons is v=

E ΔV/d 150 V/5.0 × 10−3 m = = = 3.0 × 107 m/s B B 1.0 × 10−3 T

When the potential difference across the plates is set equal to zero, the magnetic field deflects the electron into a circular path with radius

mv ( 9.11 × 10 kg )( 3.0 × 10 m/s ) = = 0.1708 m eB (1.6 ×10−19 C )(1.0 × 10−3 T ) −31

r=

7

From Figure 38.8, the angle through which the electron is deflected as it passes through the magnetic field is the angle subtended by the arc from where the electron enters the magnetic field to where the electron leaves the magnetic field. Calling this angle θ, we have L 2.5 cm sin θ = = = 0.1463 ⇒ θ = 8.4° r 17.08 cm Assess: In view of a rather small magnetic field strength, the small deflection angle is reasonable.

38.32. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Solve: In a crossed-field experiment, the deflection is zero when the magnetic and electric forces exactly balance each other. From Equation 38.4, B=

E ΔV d 500 V 1.0 × 10 −2 m 5.0 × 104 V/m = = = v v v v

We need to obtain v before we can find B. We know that (i) a potential difference of 500 V across the plates causes the proton to deflect vertically down by 0.5 cm and (ii) the proton travels a horizontal distance of 5.0 cm in the same time (t) as it travels vertically down by 0.5 cm. From kinematics,

xf − xi = vxi ( tf − ti ) + 12 ax ( tf − ti ) = v ( t − 0 s ) ⇒ v = 2

xf − xi 5.0 cm = t t

The time t can be found from the vertical motion as follows:

yf − yi = vyi ( tf − ti ) + 12 a y ( tf − ti ) ⇒ 0.50 cm = 2

1 2

a yt 2

The force on the electron is

FE = eE = ⇒ 0.50 × 10

−2

eΔV eΔV = ma y ⇒ a y = d md

−19 2 1 eΔV 2 1 (1.6 × 10 C ) ( 500 V ) t m= ⇒ t = 4.569 × 10−8 s t = 2 md 2 (1.67 × 10−27 kg )(1.0 × 10−2 m )

The speed of the electron is v=

5.0 cm 5.0 × 10 −2 m 5.0 × 104 V/m = = 1.094 × 106 m/s ⇒ B = = 46 mT −8 t 4.569 × 10 s 1.094 × 106 m/s

Using the right hand rule we determine that this magnetic field must be directed into the page in order for the G magnetic force to point opposite of the electric force. Thus B = (0.046 T, into page).

38.33. Model: Assume the fields are uniform. Visualize:

Solve: Between the plates, the fields are crossed so that the particle passes through the plates without deflection. From Equation 38.4, the speed of the particle is

v=

E 187,500 V/m = = 1.50 × 106 m/s B 0.125 T

After leaving the plates, the particle experiences only a magnetic force and thus follows a circular path of radius r. Thus the charge-to-mass ratio is

r=

mv q v 1.50 × 106 m/s ⇒ = = = 9.58 × 107 C/kg qB m Br ( 0.125 T ) ( 12 × 0.2505 m )

This must be a proton because the proton’s charge-to-mass ratio is e 1.6 × 10−19 C = = 9.58 × 107 C/kg m 1.67 × 10−27 kg

38.34. Model: Assume that the electrons transfer all their energy to the foil. Solve:

(a) From Chapter 17, the rise in thermal energy of the foil is

ΔEth = mcΔT = (10 × 10−6 kg ) ( 385 J/kg K )( 6.0°C ) = 0.0231 J This energy is provided by the electron impacts. Each electron that strikes the foil has been accelerated through a potential difference of 2000 V. Its kinetic energy when it strikes the foil is K = eΔV = 2000 eV = 3.2 × 10–16 J. The number of electrons that strike the foil in 10 seconds is

ΔEth = 7.2 × 1013 3.2 × 10−16 J (b) The current of the electron beam is 13 −19 Δq ( 7.2 × 10 )(1.6 × 10 C ) = = 1.2 μ A Δt 10 s

38.35. Model: Model the Li atom as a single valence electron orbiting a sphere with net charge q = +1e due to the 3 protons and 2 inner electrons. A sphere of charge acts like a point charge with the total charge concentrated at the center of the sphere. Visualize:

Solve: The electron’s energy is both kinetic and potential: E = 12 mv2 + kq(–e)/r. To say that the energy needed to ionize the atom is 5.14 eV means that you would need to increase the electron’s energy by 5.14 eV to remove it from the atom, taking it to r ≈ ∞. Since E = 0 for charged particles that are infinitely separated, the energy of the atom must be E = –5.14 eV = –8.224 × 10–19 J. Negative energy indicates that the system is bound, and the absolute value is the binding energy. Thus from energy considerations we learn that

E = 12 mv 2 −

Ke2 = −8.224 × 10−19 J r

The Coulomb force on the electron provides the centripetal acceleration of circular motion. The force equation is Fr =

Ke 2 mv 2 Ke2 = mar = ⇒ 12 mv 2 = 2 2r r r

Substitute this expression for the kinetic energy into the energy equation: Ke2 Ke 2 Ke2 Ke 2 = − =− = −8.224 × 10−19 J r 2r r 2r (8.99 × 109 N m 2 /C2 )(1.60 × 10−19 C) 2 ⇒r= = 1.40 × 10−10 m = 0.140 nm 2(8.224 × 10−19 J) 1 2

mv 2 −

With the radius now known, we can use the result of the force equation to find that v=

Ke2 = 1.34 × 106 m/s mr

38.36. Model: The mass of an atom is concentrated almost entirely in the nucleus. Solve:

The volumes of the nucleus and the atom are 3

Vnucleus =

4π ⎛ 1.0 × 10−14 m ⎞ 3 −42 ⎜ ⎟ = 0.524 × 10 m 3 ⎝ 2 ⎠ ⇒

3

Vatom =

4π ⎛ 1.2 × 10−10 m ⎞ 3 −31 ⎜ ⎟ = 9.05 × 10 m 3 ⎝ 2 ⎠

Vnucleus 0.524 × 10−42 m3 = ≈ 5.8 × 10−13 Vatom 9.05 × 10−31 m3

Thus 5.8 × 10–11% = 0.000000000058% of the atom’s volume contains all the mass. The percent of empty space in an atom is 99.999999999942%.

38.37. Solve: (a) Try relating the wavelengths to the number 125.00. We get 125.00 =

125.00 1

31.25 =

125.00 22

13.89 =

125.00 32

7.81 =

125.00 42

5.00 =

125.00 52

Thus, the series can be represented by the formula

λ=

(125.00 nm ) , n2

n = 1, 2, 3, . . .

(b) Notice that the common denominator of each of the wavelengths is 75. The six wavelengths are 5 ⋅ 75, 12 ⋅ 75, 21 ⋅ 75, 32 ⋅ 75, 45 ⋅ 75, and 60 ⋅ 75 nm. The series 5, 12, 21, 32, 45, 60 is increasing faster than linear, but these aren’t squares of integers. However, a little thought finds that each of these is 4 less than the square of = 32 – 4, 12 = an integer. That is, 5 2 2 4 – 4, 21 = 5 – 4, and so on. That is, the series of multipliers is (n2 – 4) with n = 3, 4, 5, . . . Thus the wavelengths in this series can be represented by the formula

λ = (75.00 nm)(n 2 − 4), n = 3, 4, 5, . . . An alternative formula that starts from n = 1 is

λ = n (n + 4)(75.00 nm), n = 1, 2, 3, . . .

38.38. Solve: (a) Vatom =

4π 3 ratom = 9.0 × 10−31 m3. Since the atomic mass of aluminum is 27 u, the mass per 3

atom is m = 27 × (1.661 × 10

−27

kg ) = 4.5 × 10

−26

kg

Therefore, the average density of an aluminum atom is

ρ atom =

4.5 × 10−26 kg = 50,000 kg/m 3 9.0 × 10−31 m3

(b) The average density of an aluminum atom is found in part (a) to be larger than the density of solid aluminum (ρAl = 2700 kg/m3). The volume per atom in the solid is 4.5 × 10−26 kg = 1.67 × 10 −29 m 3 ≈ 1.7 × 10−29 m3 2700 kg/m 3 ⎛ 4π ⎞ 3 −10 −29 3 So, using ⎜ ⎟ rsphere = 1.67 × 10 m , we get rsphere = 1.6 × 10 m . That is, the average spacing between atoms is ⎝ 3 ⎠ −10 2rsphere = 3.2 × 10 m, which is larger than the diameter of the aluminum atom by a factor of 2.7. (c) The atomic mass is almost entirely in the nucleus, so the density of the nucleus is

ρ nucleus =

4.5 × 10−26 kg 4π 3

( 4 × 10

−15

m)

3

= 1.7 × 1017 kg/m 3

Compared to the density of solid aluminum, which is 2700 kg/m3, the nuclear density is approximately ≈1014 times larger.

38.39. Model: The nucleus of an atom contains Z protons and A – Z neutrons. Solve:

(a) The values of Z, A, and Z/A for nuclei with Z = 1, 5, 10, 15, … 90 are given in the following table.

Z

1

A

1 10.8

Z/A

1 0.46 0.5 0.48 0.5 0.45 0.46 0.44 0.44 0.44 0.42 0.41 0.42 0.41 0.40 0.40 0.40 0.40 0.39

5

10

15

20

25

30

35

40

45

50

55

60

65

70

75

80

85

20

31

40

55

65

80

91

103

119

133

144

159

173

186

201

210 232

90

A graph of Z/A as a function of Z is shown below.

(b) The Z/A values are in the range 0.45–0.50 for values of Z up to approximately 45. For Z in the range 65 to 90, Z/A drops down to approximately 0.40. In short, Z/A decreases from around 0.50 to 0.40 with an increase in Z. (c) What we are plotting is Z/Z + N as a function of Z. With increasing Z, Z/A decreases because the number of neutrons in the nuclei increases more rapidly than Z.

38.40. Model: The nucleus of an atom is very small and it contains protons and neutrons. Solve:

(a) The repulsive electric force between two protons in the nucleus is FE =

e2

1

4πε 0 ( 2.0 fm )2

=

(8.99 × 10

9

N m 2 /C 2 )(1.60 × 10−19 C )

( 2.0 × 10

−15

m)

2

2

= 58 N

(b) The attractive gravitational force between two protons in the nucleus is

FG =

Gm 2

( 2.0 fm )

2

( 6.67 ×10 =

−11

N m 2 /kg 2 )(1.67 × 10−27 kg )

( 2.0 ×10

−15

m)

2

2

= 4.7 × 10−35 N

Because FG << FE, gravitational force could not be the force to hold two protons together. (c) The nuclear force must be very strong to overcome FE and it must be independent of charge because both protons and neutrons are held in the nucleus very tightly. Furthermore, nuclear force is a very short range force since it is not felt outside the nucleus.

38.41.

Model: Visualize:

Solve:

Assume the nucleus is at rest. Use the conservation of energy equation.

The energy conservation equation Kf + Uf = Ki + Ui is

0 J+

(9.0 ×10 ⇒

9

1 ( 2e )( Ze ) = 6.24 MeV + 0 J 4πε 0 ( 6 fm )

N m 2 /kg 2 ) ( 2 ) (1.60 × 10−19 C ) Z 2

6.0 × 10−15 m

Solving for Z gives Z = 13. The element is aluminum.

= 6.24 × 106 eV ×

1.60 × 10−19 J 1 eV

38.42. Model: Assume the 238U nucleus is at rest. Energy is conserved. Visualize:

Solve:

The energy conservation equation Kf + Uf = Ki + Ui is

0 J+ ⇒ Ki =

1 (2e)(92e) = Ki + 0 4πε 0 7.5 × 10−15 m

(9.0 × 109 N m 2 /C 2 )184(1.60 × 10−19 C ) 2 1 eV = 5.65 × 10−12 J × = 35.33 MeV 7.5 × 10−15 m 1.60 × 10−19 J

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2e gains kinetic energy K eV by accelerating through a potential difference ∆V = 12 K V. Thus the alpha particle will just reach the 238U nucleus after being accelerated through a potential difference 35.33 ΔV = MV = 1.8 × 107 V 2

38.43. Model: Assume the 16O nucleus is at rest. Energy is conserved. Visualize:

A proton with initial velocity vi and zero electric potential energy (due to its infinite separation from an 16O nucleus) moves toward 16O and turns around when it is 1.0 fm from the surface. Since r is measured from the center of the nucleus, rf = 4 fm. Solve:

(a) The energy conservation equation Kf + Uf = Ki + Ui is

0J +

⇒v = 2 i

(b) K = 12 mvi2 =

1 2

e ( 8e ) = 1 mvi2 + 0 J 4πε 0 ( 3.0 fm + 1.0 fm ) 2 1

( 2 ) ( 9.0 × 109 N m 2 /C2 ) ( 8) (1.6 × 10−19 C )

( 4.0 ×10−15 m )(1.67 × 10−27 kg )

(1.67 × 10

−27

2

= 5.52 × 1014 m 2 /s 2 ⇒ vi = 2.3 × 107 m/s

kg )( 5.52 × 1014 m 2 /s 2 ) = 4.60 × 10−13 J = 2.9 MeV

38.44. Model: Assume the 12C nucleus is at rest. Energy is conserved. Visualize:

Solve: (a) A proton with an initial velocity vi and zero electric potential energy (due to its infinite separation from a 12 C nucleus) moves toward the 12C nucleus. It must impact the 12C nucleus, which has a radius of 2.75 fm, with an energy of 3.0 MeV. The energy conservation equation Kf + Uf = Ki + Ui is

3.0 MeV +

e ( 6e ) 1 1 = mproton vi2 + 0 J 4πε 0 ( 2.75 × 10−15 m ) 2

−19 9 2 2 1.6 × 10−19 J ( 9.0 × 10 N m /C ) ( 6 ) (1.60 × 10 C ) 1 ⇒ 3.0 × 10 eV × + = (1.67 × 10−27 kg ) vi2 −15 1 eV 2.75 × 10 m 2 2

6

⇒ vi = 3.43 × 107 m/s 1 (b) The initial kinetic energy is K i = mvi2 = eΔV . Hence, 2 −27 7 1 (1.67 × 10 kg )( 3.43 × 10 m/s ) = 6.14 × 106 V 2 1.6 × 10−19 C 2

ΔV =

38.45. Model: Assume the 137Cs nucleus is at rest. Energy is conserved.

Visualize: The initial kinetic energy of the beta particle as it is ejected from the 137Cs nucleus is Ki. Because the beta particle is ejected from the surface, it also has electric potential energy U i from the coulombic attraction to the nucleus. As the beta particle moves away from the nucleus and is practically at an infinite distance, Uf = 0 J and Kf = 300 keV. Solve: The conservation of energy equation Kf + Uf = Ki + Ui is e ( 55e ) 1 300 × 103 eV + 0 J = K i − 4πε 0 ( 6.2 × 10−15 m )

⇒ 3.0 × 10

5

( 9.0 ×10 eV = K − i

9

N m 2 /C 2 ) ( 55 ) (1.6 × 10−19 C ) 6.2 × 10−15 m ⇒ K i = 13.1 MeV

2

×

1 eV = K i − 127.6 × 105 eV 1.6 × 10−19 J

38.46. Model: Momentum is conserved during the interaction. Assume the nucleus is initially at rest. Visualize:

Solve:

Momentum is a vector, so both the x- and y-components of p must be conserved. The x-equation is

mα viα = 4(1.50 × 107 m/s) = mα vfα cosθ + mn vfn cos φ = 4(1.49 × 107 m/s) cos 49° + 197vfn cos φ ⇒ vfn cos φ = (vfn ) x = 1.0609 × 105 m/s The y-equation is 0 = mα vfα sin θ − mn vfn sin φ = 4(1.49 × 107 m/s) sin 49° − 197vfn sin φ ⇒ vfn sin φ = (vfn ) y = 2.283 × 105 m/s

We can work with relative masses (4 and 192) since the conversion to kg occurs in all terms of the equation and cancels out. Dividing the y-equation by the x-equation gives tan φ =

vfn sin φ 2.283 = = 2.152 ⇒ φ = tan −1 (2.152) = 65° vfn cos φ 1.0609

The speed of the recoiling nucleus is vfn = (vfn ) x2 + (vfn ) y2 = (1.0609) 2 + (2.203) 2 × 105 m/s = 2.52 × 105 m/s

Thus the nucleus recoils at 2.52 × 105 m/s in a direction 65° below the x-axis.

38.47. Solve: (a) The Coulomb force between the electron and the proton causes the centripetal acceleration for the electron’s motion in a circular orbit. That is, Fr =

mv 2 1 e(e) 1 e2 = ⇒ mv 2 = 2 r 4πε 0 r 4πε 0 r

The total energy of the electron is a sum of its kinetic energy and electric potential energy: 2 1 1 ( −e ) e 1 ⎛ 1 e 2 ⎞ 1 ( −e ) e2 E = mv 2 + = ⎜ + =− ⎟ 2 4πε 0 r 2 ⎝ 4πε 0 r ⎠ 4πε 0 r 8πε 0 r

(b) The ratio of the potential energy to the kinetic energy is U −e 2 4πε 0 r − e 2 4πε 0 r = 1 2 =1 2 = −2 K mv e 4πε 0 r ) 2 2(

(c) The part (a) energy is negative because the electron is bound. To remove the electron to infinity, where E = 0, you need to give it energy Eionization = E . Thus

Eionization = 13.6 eV ×

e2 1.60 × 10−19 J = 1 eV 8πε 0 r

−19 9 2 2 1 ( 9.0 × 10 N m /C )(1.60 × 10 C ) (1 eV ) = 5.29 × 10−11 m −19 2 13.6 eV 1.60 10 J × ( )( ) 2

⇒r=

To calculate the speed, we have Eionization = − E = − (U + K ) = − ( −2 K + K ) = K = 12 mv 2 . Therefore, the speed is v=

2 (13.6 eV ) 1.60 × 10−19 J 2 Eionization = 2.19 × 106 m/s. = × m 9.11 × 10−31 kg 1 eV

38.48. Model: Assume that the electric field between the plates is uniform. Visualize: Please refer to Figure 38.9. Solve: (a) The electric force on a positive charge q in a field E is FE = qE. If a droplet is charged with charge q and is suspended between the plates by applying a field E0, then the electric force must exactly cancel the weight force: Fnet = FE + w = 0 N ⇒ qE0 – mg = 0 N ⇒ q =

mg E0

(b) A droplet falling at the terminal speed has no net force acting on it. The velocity is negative for a falling drop, or v = −vterm. Thus, Fnet = Fdrag + w = 0 N ⇒ −bv − mg = + bvterm − mg = 0 N ⇒ vterm =

mg b

(c) Because Fdrag = −6πηrv = −bv, b = 6πηr. The result in part (b) for vterm thus simplifies to v term = mg 6 πη r . Noting that m = ρV = ρ 43π r 3 , the expression for the terminal velocity is

vterm =

ρ ( 43π r 3 ) g 6πη r

=

2ρ r 2 g 9η vterm ⇒r= 9η 2ρ g

(d) We first need to find the mass m of the droplet before we can obtain q from the expression in part (a). Using the expression for r in part (c), the mass of the oil drop is

⎛ 4π 3 ⎞ ⎛ 4π m = ρ oilV = ρ oil ⎜ r ⎟ = ρ oil ⎜ ⎝ 3 ⎠ ⎝ 3

⎞ ⎡ 9ηair vterm ⎤ ⎥ ⎟⎢ ⎠ ⎣ 2 ρoil g ⎦

3/ 2

The terminal velocity is vterm = 3.00 × 10−3 m 7.33 s. The mass of the oil drop is ⎡ 9 (1.83 × 10−5 kg m/s )( 3.00 × 10−3 m 7.33 s ) ⎤ 4π 3 ⎥ m= 860 kg/m ( )⎢ 3 2 ( 860 kg/m3 )( 9.8 m/s 2 ) ⎢⎣ ⎥⎦ ⇒q=

( 2.881× 10 mg = ( ΔV / d )

−14

kg )( 9.8 m/s 2 )(1.0 × 10−2 m ) 1177 V

(e) The number of fundamental electric charges on the oil drop is

q 2.40 × 10−18 C = = 15 e 1.60 × 10−19 C

3/ 2

= 2.881 × 10−14 kg

= 2.40 × 10−18 C

38.49. Model: Assume the classical electric force is the centripetal force. F=

Ke 2 v2 = me ⇒ v = 2 r r

Ke 2 me r

Visualize: We are given λ = 600 nm. We also know that for electromagnetic waves c = λ f . Solve: (a) The speed is the circumference of the orbit divided by the period.

v=

2π r = 2π rf T

λ vλ mr r= = = e 2π f 2π c 2π c Ke 2

v

Square both sides. r = 2

Ke2 me r

λ2

4π 2c 2

Multiply both sides by r.

r3 =

Ke 2λ 2 4π 2c 2 me

Take cube roots. 2 2 r = 3 Ke2 λ2 4π c me

=3

(8.99 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C) 2 (600 × 10−9 m) 2 4π 2 (3.00 × 108 m/s) 2 (9.11× 10−31 kg)

= 2.95 × 10−10 m = 0.295 nm (b) The total mechanical energy of the classical atom was found in Problem 38.47. 2 (1.60 × 10−19 C) 2 E=− e =− = −3.90 × 10−19 J = −2.44eV −12 8πε 0 r 8π (8.85 × 10 C 2 /N ⋅ m 2 )(2.95 × 10−10 m)

Assess:

While we don’t expect the classical analysis to be perfect, the result is in a reasonable range.

39.1. Solve: A steady photoelectric current of 10 μA is indicated in the graph. The number of electrons per second is 10 μ A = 10

μC s

= 1.0 × 10 −5

C 1 electron × = 6.25 × 1013 electrons/s s 1.6 × 10 −19 C

39.2. Model: Light of frequency f consists of discrete quanta, each of energy E = hf. Solve:

(a) The energy of the light quantum is

E = hf = h

c

λ

=

( 6.63 × 10

−34

J s )( 3.0 × 108 m/s )

400 × 10−9 m

×

1 eV = 3.11 eV 1.6 × 10−19 J

From Table 39.1, the work functions for sodium and potassium are smaller than 3.11 eV. That is, light of wavelength 400 nm has enough energy to eject photoelectrons from sodium and potassium. (b) The energy of the light quantum is

E = hf = h

c

λ

=

( 6.63 × 10

−34

J s )( 3.0 × 108 m/s )

250 × 10−9 m

×

1 eV = 4.97 eV 1.6 × 10−19 J

Light of wavelength 250 nm has enough energy to eject photoelectrons from all of the metals on the table except gold.

39.3. Model: Light of frequency f consists of discrete quanta, each of energy E = hf. Solve:

The lowest photon energy that creates photoelectrons from the metal is

E=

hc

λ

=

( 6.63 ×10

−34

J s )( 3.0 × 108 m/s ) −9

388 × 10 m

The work function of the metal is E0 = 3.20 eV.

×

1 eV = 3.20 eV 1.6 × 10−19 J

39.4. Solve: From Equation 39.7, the maximum kinetic energy is K max = hf − E0 = h ⇒λ = Assess:

c

λ

− E0

( 6.63 × 10−34 J s )( 3.0 × 108 m/s ) × 1 eV = 209 nm hc = 4.65 eV + 1.30 eV 1.6 × 10−19 J E0 + K max

λ = 209 nm is the wavelength of light in the ultraviolet region of the spectrum.

39.5. Model: The threshold frequency for the ejection of photoelectrons is f0 = E0 /h where E0 is the work function. Solve: The visible region of light extends from 400 nm to 700 nm. For λ0 = 400 nm, the work function is

E0 = f 0 h = For λ0 = 700 nm,

hc

λ0

=

( 6.63 × 10

( 6.63 × 10

−34

J s )( 3.0 × 108 m/s )

400 × 10 −34

−9

m

J s )( 3.0 × 108 m/s )

×

1 eV = 3.11 eV 1.6 × 10−19 J

1 eV = 1.78 eV 700 × 10 m 1.6 × 10−19 J The cathode that will work in the entire visible range must have a work function of 1.78 eV or less. E0 =

−9

×

39.6. Visualize: Please refer to Figure 39.10.

Solve: (a) The threshold frequency is seen to be f0 = 4.39 × 1014 Hz. According to Einstein’s theory of the photoelectric effect, the work function is E0 = hf0. Using the modern value of h, the work function of cesium is E0 = hf 0 = ( 4.14 × 10 −15 eV s )( 4.39 × 1014 Hz ) = 1.82 eV

(b) According to Einstein’s theory, a graph of Vstop versus f should be linear with a slope of h/e. Millikan’s data is seen to be linear with an experimental slope of 4.124 × 10−15 V/Hz. So, an experimental value of h is h = e × slope = (1.60 × 10−19 C)(4.124 × 10−15 V/Hz) = 6.60 × 10−34 J s

39.7. Solve: (a) A metal can be identified by its work function. From Equation 39.8, the stopping potential is Vstop =

hf − E0 ⇒ E0 = hf – eVstop e

The frequency and energy of the photons are

f =

c

λ

=

3.00 × 108 m/s = 1.500 × 1015 Hz ⇒ hf = ( 4.14 × 10−15 eV s )(1.500 × 1015 Hz ) = 6.21 eV 200 × 10−9 m

If the stopping potential is Vstop = 1.93 V, then eVstop = 1.93 eV. Thus,

E0 = hf − eVstop = 6.21 eV − 1.93 eV = 4.28 eV Using Table 39.1, we can identify the metal as aluminum. (b) The kinetic energy of the electrons and thus the stopping potential are independent of the light intensity. A more intense light generates more electrons, but the electrons still have the same kinetic energy. The stopping potential is Vstop = 1.93 V after the intensity is doubled.

39.8. Solve: (a) The frequency of the photon is f =

c

λ

=

3.00 × 108 m/s = 4.29 × 1014 Hz 700 × 10−9 m

From Equation 39.4, the energy is

E = hf = ( 4.14 × 10−15 eV s )( 4.29 × 1014 Hz ) = 1.77 eV (b) The frequency of the photon is f =

E 5000 eV = = 1.208 × 1018 Hz h 4.14 × 10 −15 eV s

Thus, the wavelength is

λ= Assess:

c 3.00 × 108 m/s = = 2.5 × 10−10 m = 0.25 nm f 1.208 × 1018 Hz

Because x-ray photons are very energetic, their wavelength is small.

39.9. Solve: (a) Using Equation 39.4, the wavelength of the photon is λ=

−15 8 c c hc ( 4.14 × 10 eV s )( 3 × 10 m/s ) = = = = 4.14 × 10−6 m = 4140 nm f E h E 0.30 eV

This is infrared light. (b) Likewise, for an energy of 3.0 eV, the wavelength is λ = 414 nm and is in the visible region. (c) For an energy of 30.0 eV, the wavelength is λ = 41.4 nm and is in the ultraviolet region. −1 Assess: Since E ∝ λ , the higher the energy of the photon, the smaller its wavelength.

39.10. Solve: (a) From Equation 39.4, the energy of the radio-frequency photon is E = hf = ( 4.14 × 10−15 eV s )(100 × 106 Hz ) = 4.14 × 10−7 eV (b) The energy of the visible light photon

E=

hc

λ

=

( 4.14 ×10

−15

eV s )( 3.0 × 108 m/s )

500 × 10−9 m

= 2.49 eV

(c) The energy of the x-ray photon is

E=

hc

λ

=

( 4.14 ×10

−15

eV s )( 3.0 × 108 m/s )

0.10 × 10−9 m

= 12.4 keV

39.11. Solve: (a) From Equation 39.4, the energy of each photon is Ephoton = hf = (6.63 × 10−34 J s)(101 × 106 Hz) = 6.696 × 10−28 J The number of photons in 104 J is

N=

Etotal 104 J = = 1.5 × 1029 Ephoton 6.696 × 10−28 J

The antenna emits 1.5 × 1029 photons per second. (b) The number of photons emitted per second is so enormous that we couldn’t possibly recognize the effects of single photons. It’s safe to treat the broadcast as an electromagnetic wave.

39.12. Solve: The rate of photon emission by a laser is the number of photons emitted per second. If P is the power of the beam, the rate is Rphoton =

P P Pλ = = hf hc λ hc

For the same power P,

Rred ⎛ Pλred ⎞ ⎛ hc ⎞ λred 650 nm =⎜ = = 1.44 ⎟= ⎟⎜ Rblue ⎝ hc ⎠ ⎝ Pλblue ⎠ λblue 450 nm

39.13. Model: All the light emitted by the light bulb is assumed to have a wavelength of 600 nm. Solve:

(a) The frequency of the emitted light is

f =

c

λ

=

3.0 × 108 m/s = 5.0 × 1014 Hz 600 × 10−9 m

(b) The rate of photon emission is

Rphoton =

P 5W 5W = = 1.5 × 1019 photons/s ≈ 1× 1019 photons/s = −34 hf hf ( 6.63 ×10 J s )( 5.0 ×1014 Hz )

39.14. Solve: (a) The de Broglie wavelength is λ=

h h 6.63 × 10−34 J s = = = 1.0 × 10−12 m ⇒ v = 7.3 × 108 m/s p mv ( 9.11 × 10−31 kg ) v

This speed is larger than c, indicating a breakdown of de Broglie’s equation. This is an acceptable answer if you haven’t studied relativity. However, a better approach would be to use the relativistic form for the momentum, p = γ mv. Hence,

λ=

6.63 × 10−34 J s 1 − v 2 c 2 h = = 1.0 × 10−12 m γ mv ( 9.11× 10−31 kg ) v

8 −12 −31 ⎛ v ⎞ ( 3.0 × 10 m/s )(1.0 × 10 m )( 9.11 × 10 kg ) ⎛v⎞ ⇒ 1− v c = ⎜ ⎟ = 0.4122 ⎜ ⎟ −34 6.63 × 10 J s c ⎝ ⎠ ⎝c⎠ 2 2 v v ⇒ 1 − 2 = 0.170 2 ⇒ v = 0.925c = 2.8 × 108 m/s c c 2

2

(b) For λ = 1.0 × 10−9 m,

λ = 1.0 × 10−9 m =

h 6.63 × 10−34 J s = ⇒ v = 7.3 × 105 m/s mv ( 9.11 × 10−31 kg ) v

(c) Likewise, for λ = 1.0 × 10−6 m, v = 7.3 × 102 m/s. (d) For λ = 1.0 × 10−3 m, v = 0.73 m/s.

39.15. Solve: The de Broglie wavelength is λ = h/mv. Thus, v=

h 6.63 × 10−34 J s = = 1456 m/s mλ ( 9.11× 10−31 kg )( 500 × 10−9 m )

A potential difference of ΔV will raise the kinetic energy of a rest electron by

1 2

mv 2 . Thus,

−31 1 mv 2 ( 9.11 × 10 kg ) (1456 m/s ) eΔV = mv 2 ⇒ ΔV = = = 6.0 × 10−6 V 2 2e 2 (1.6 × 10−19 C ) 2

Assess:

A mere 6.0 × 10−6 V is able to increase an electron’s speed to 1456 m/s.

39.16. Solve: The de Broglie wavelength is λ = h/mv. Thus v=

h 6.63 × 10−34 J s = = 3.970 × 107 m/s mλ (1.67 × 10−27 kg )(10 × 10−15 m )

So, the kinetic energy of the proton is 2 1 1 1 eV = 8.2 MeV K = mv 2 = (1.67 × 10−27 kg )( 3.970 × 107 m/s ) × 2 2 1.6 × 10−19 J

39.17. Visualize: Equation 39.12 gives the de Broglie wavelength of a particle in a box. λn =

2L n

n = 1, 2, 3, 4, . . .

We are given λn = 1. 0 nm and L = 3. 0 nm. Solve: Solve the above equation for n.

n= Assess:

2L

λn

=

2(3.0 nm) =6 1.0 nm

6 is an unsurprising value for a quantum number.

39.18. Model: The energy of a confined particle in a one-dimensional box is quantized. Model a nucleus as a one-dimensional box of length L = 10 fm = 1.0 × 10−14 m. Solve: Protons and neutrons are particles of mass m confined in a box. From Equation 39.14, the allowed energies of the protons are

( 6.63 × 10−34 J s ) h2 2 = = n 2 ( 3.29 × 10−13 J ) = n 2 ( 2.06 MeV ) n 2 −27 −14 8mL2 8 (1.67 × 10 kg )(1.0 × 10 m ) 2

En = n 2

The first three energy levels are to two significant figures E1 = 2.1 MeV, E2 = 4E1 = 8.2 MeV, and E3 = 9E1 = 19 MeV.

39.19. Model: For a “particle in a box,” the energy is quantized. Solve:

The energy of the n = 1 state is

E1 = (1)

2

( 6.63 ×10−34 J s )( 600 × 10−9 m ) = 0.427 nm h2 hc hλ = E = ⇒ L = = photon 8mL2 8mc λ 8 ( 9.11 × 10−31 kg )( 3.0 × 108 m/s )

39.20. Model: To conserve energy, the emission and the absorption photons must have exactly the energy lost or gained by the atom in the appropriate quantum jumps. Visualize: The energy of a light quantum is E = hf = hc/λ. Solve: (a) The wavelength of the emission photon from the n = 2 to n = 1 transition is

λ=

( 4.14 × 10 hc = E2 − E1

−15

eV s )( 3.0 × 108 m/s ) 1.5 eV

= 828 nm

Likewise, λ = 497 nm for the 3 → 2 transition with ΔE = 2.5 eV, and λ = 311 nm for the 3 → 1 transition with ΔE = 4.0 eV. (b) Because the atom in the ground state is in the n = 1 state, the absorption lines correspond to the 1 → 2 and 1 → 3 transitions. The absorption wavelengths are 828 nm and 311 nm. The 2 → 3 transition is not seen in absorption.

39.21. Model: To conserve energy, the absorption spectrum must have exactly the energy gained by the atom in the quantum jumps. Visualize: Please refer to Figure EX39.20. Solve: (a) An electron with a kinetic energy of 2.00 eV can collide with an atom in the n = 1 state and raise its energy to the n = 2 state. This is possible because E2 – E1 = 1.50 eV is less than 2.00 eV. On the other hand, the atom cannot be excited to the n = 3 state. (b) The atom will absorb 1.50 eV of energy from the incoming electron, leaving the electron with 0.50 eV of kinetic energy.

39.22. Model: To conserve energy, the emission and the absorption spectra must have exactly the energy lost or gained by the atom in the appropriate quantum jumps. Solve: (a)

(b) From Equation 39.4, the energy of a light quantum is E = hf = hc/λ. We can use this equation to find the emission and absorption wavelengths. The emission energies from the above energy-level diagram are: E2→1 = 4.00 eV, E3→1 = 6.00 eV, and E3→2 = 2.00 eV. The wavelength corresponding to the 2→1 transition is

λ 2 →1 =

hc ( 4.14 × 10 = E21

−15

eV s )( 3.0 × 108 m/s ) 4.00 eV

= 311 nm

Likewise, λ 3→1 = hc E3→1 = 207 nm, and λ 3→2 = 622 nm. (c) Absorption transitions start from the n = 1 ground state. The energies in the atom’s absorption spectrum are E1→2 = 4.00 eV and E1→3 = 6.00 eV. The corresponding wavelengths are λ 1→ 2 = hc E1→ 2 = 311 nm and

λ1→3 = hc E1→3 = 207 nm.

39.23. Model: The electron must have k ≥ ΔEatom to cause collisional excitation. The atom is initially in the n = 1 ground state. Visualize:

Solve:

The kinetic energy of the incoming electron is E = 12 mv 2 =

1 2

( 9.11×10

−31

kg )(1.30 × 106 m/s ) = 7.698 × 10−19 J = 4.81 eV 2

The electron has enough energy to excite the atom to the n = 2 stationary state (E2 – E1 = 4.00 eV). However, it does not have enough energy to excite the atom into the n = 3 state which requires a total energy of 6.00 eV.

39.24. Solve: The Bohr radius is defined as 2 2 −12 −34 4πε 0= 2 4π ( 8.85 × 10 C /N m )(1.05 × 10 J s ) = = 5.26 × 10−11 m = 0.0526 nm 2 −31 −19 me 2 ( 9.11×10 kg )(1.60 × 10 C ) 2

aB =

This differs slightly from the accepted value of 0.0529 nm because of rounding error due to using constants accurate to only 3 significant figures. From Equation 39.29, the ground state energy level of hydrogen is − ( 9 × 109 N 2 m/C2 )(1.60 × 10−19 C ) −e 2 = = −2.18 × 10−18 J = − 13.61 eV 4πε 0 ( 2aB ) 2 ( 5.29 × 10−11 m ) 2

E1 =

The slight difference from the accepted −13.60 eV is due to rounding error.

39.25. Solve: (a) From Equation 39.25, the radius in state n of a hydrogen atom is rn = n2aB. A 100 nm diameter atom has rn = 50 nm. Thus, the quantum state is

rn 50 nm = = 30.74 aB 0.0529 nm

n=

Since n has to be an integer, we obtain n = 31. (b) From Equation 39.27, the electron’s speed is v31 =

v1 2.19 × 106 m/s = = 7.06 × 104 m/s 31 31

From Equation 39.30, the electron’s energy is

E31 = −

E1

( 31)

2

=−

13.60 eV

( 31)

2

= −0.0142 eV

39.26. Solve: (a) Using the data in Table 39.2, the wavelength of the electron in the n = 1 state is λ1 =

( 6.63 × 10−34 J s ) h = = 0.332 nm mv ( 9.11 × 10−31 kg )( 2.19 × 106 m/s )

Likewise, λ2 = 0.665 nm, and λ3 = 0.997 nm. (b) For n = 1, the circumference of the orbit is 0.0529 nm × 2π =0.332 nm, which is exactly equal to λ1. For n = 2, the circumference of the orbit is

0.212 nm × 2π =1.332 nm=2λ 2 = 2 ( 0.665 nm ) Likewise, the data from part (a) and Table 39.2 shows 3λ 3 = 2π r3 . (c)

39.27. Solve: From Equation 39.30, the energy of the hydrogen atom in its first excited state is E2 =

−13.60 eV

( 2)

2

= −3.40 eV

The ionization energy of the hydrogen atom in its first excited state (n = 2) is thus 3.40 eV.

39.28. Model: The angular momentum of a particle in circular motion is L = mvr. Solve:

For the ground state, the angular momentum in terms of Plank’s constant is

mv1r1 ( 9.11× 10 = =

−31

kg )( 2.19 × 106 m/s ) ( 0.053 nm )

( 6.63 ×10

−34

J s ) 2π

= 1.00

Likewise, mv2 r2 = = 1.99 and mv3r3 = = 3.00 . Thus, L1 = mv1r1 = =, L2 = 2=, and L3 = 3=.

39.29. Solve: The units of = are the units of angular momentum, L = mvr. These units are J ⋅ s = kg ⋅

m2 m ⋅ s = kg ⋅ ⋅ m s2 s

39.30. Solve: Photons emitted from the n = 4 state start in energy level n = 4 and undergo a quantum jump to a lower energy level with m < 4. The possibilities are 4 → 1, 4 → 2, and 4 → 3. According to Equation 39.36, the transition 4 → m emits a photon of wavelength. λ0 91.18 nm λ= = 1 ⎞ ⎛ 1 1⎞ ⎛ 1 ⎜ 2 − 2⎟ ⎜ 2− ⎟ m n m 16 ⎝ ⎠ ⎝ ⎠ These values are given in the table below. Transition 4→1 4→2 4→3

Wavelength 97.3 nm 486 nm 1876 nm

39.31. Model: Equation 39.36 predicts the absorption spectrum of hydrogen if we let m = 1. λ 1→ n =

91.18 nm 1− 1 12 n 2

(

n = 2, 3, 4, . . .

)

Visualize: The 1 → 2 transition will produce the longest wavelength; the 1 → 3 transition will produce the second-longest wavelength; and 1 → 4 transition will produce the third-longest wavelength. Solve:

λ 1→ 4 = Assess:

91. 18 nm = 97. 25 nm 1 − 1 12 42

(

)

This result is in the ultraviolet region, as expected for m = 1.

39.32. Model: Photons are absorbed in a quantum jump from a lower energy level to a higher energy level. Solve: Because absorption wavelengths are a subset of the wavelengths in the emission spectrum, we can use Equation 39.36 to analyze the hydrogen absorption spectrum. Because most of the atoms in a gas are in the ground state (n = 1), the only quantum jumps seen in the absorption spectrum start from this state. In other words, only the Lyman series (m = 1 and n = m + 1, m + 2, . . . . ) is observed in the absorption spectrum. Using m = 1 and n = 2, 3, 4, . . . in Equation 39.36, we do not find any spectral line with λ = 656.5 nm. However, a transition 2 → 3 does lead to a line with λ = 656.5 nm. However, this line is in the emission spectrum, not the absorption spectrum.

39.33. Solve: For hydrogen-like ions, Equation 39.37 is rn = n 2

aB ( rn )H = Z Z

vn = Z

v1 = Z ⋅ ( vn )H n

⎛ 13.60 eV ⎞ 2 En = − Z 2 ⎜ ⎟ = Z ( En ) H n2 ⎝ ⎠

Where (rn)H, (vn)H, and (En)H are the values of ordinary hydrogen. He+ has Z = 2. Using Table 39.2 for the values of hydrogen, we get vn (m/s) En (eV) n rn (nm) 1 2 3

0.026 0.106 0.238

4.38 × 106 2.19 × 106 1.46 × 106

−54.4 −13.6 −6.0

39.34. Solve: The laser light delivers 2.50 × 1017 photons per second and 100 × 10−3 J of energy per second. Thus, the energy of each photon is

100 × 10−3 J/s = 4.00 × 10−19 J 2.50 × 1017 s −1 From Equation 39.4, the wavelength of the photons is −34 8 c hc ( 6.63 × 10 J s )( 3.00 × 10 m/s ) λ= = = = 4.97 × 10−7 m = 497 nm f E 4.00 × 10−19 J

Assess:

The wavelength is in the visible region.

39.35. Solve: (a) The energy of each photon is Ephoton = hf =

hc

λ

=

( 6.63 × 10

−34

J s )( 3.00 × 108 m/s )

690 × 10−9 m

= 2.88 × 10−19 J

The number of photons in each pulse of light is N=

Etotal 0.500 J = = 1.7 × 1018 Ephoton 2.88 × 10−19 J

(b) The rate of emission, that is, the number of photons per second is R=

1.7 × 1018 photons = 1.7 × 1026 photons/s 10 × 10−9 s

39.36. Visualize: Equation 39.8 gives Vstop in terms of f ; we replace f with c/λ. Vstop =

hc/λ − E0 e

The value of E0 for aluminum is found in Table 39.1: E0 = 4. 28 eV = 6. 848 × 10−19 J. Solve:

For λ = 250 nm: Vstop =

hc/λ − E0 (6.63× 10−34 J ⋅ s)(3.00 × 108 m/s) / (250 × 10−9 m) − 6.848 × 10−19 J = = 0.693 V 1.60 × 10−19 C e

For λ = 200 nm: Vstop =

hc/λ − E0 (6.63× 10−34 J ⋅ s)(3.00 × 108 m/s) / (200 × 10−9 m) − 6.848 × 10−19 J = = 1.94 V 1.60 × 10−19 C e

So the change in stopping potential is 1. 94 V − 0. 693 V = 1. 24 V. Assess: This answer looks reasonable in light of Figure 39.10.

39.37. Solve: (a) The threshold frequency is f0 = E0/h. The threshold frequency for potassium and gold are given in the table in part (b). (b) The corresponding threshold wavelength is λ 0 = c/f0. The results of the calculations are in the table below. Metal E0 (eV) f0 (Hz) λ 0 (nm) Potassium Gold

5.56 × 1014 1.23 × 1015

2.30 5.10

540 244

(c) When light of wavelength λ is incident on the metal, the maximum kinetic energy of the photoelectrons is 2 K max = 12 mvmax = hf − E0 =

hc

λ

2 ⎛ hc ⎞ ⎜ − E0 ⎟ m⎝ λ ⎠

− E0 ⇒ vmax =

E0 must be converted to SI units of joules before this formula can be used. vmax for potassium and gold are given in the table in part (d). (d) The stopping potential is Vstop =

K max 1 ⎛ hc ⎞ = ⎜ − E0 ⎟ e e⎝ λ ⎠

where again E0 has to be joules. The results of the calculations are in the table below. Metal Potassium Gold

E0 (J)

vmax (m/s) −19

3.68 × 10 8.16 × 10−19

5

10.8 × 10 4.4 × 105

Vstop (V) 3.35 0.55

39.38. Solve: (a) The maximum kinetic energy of photoelectrons is Kmax = hf – E0. Substituting the given

values,

2.8 eV =

hc

λ

− E0

1.1 eV =

hc − E0 1.5λ

Multiplying the second equation by 1.5 and subtracting the second equation from the first, 1.15 eV = 0.5 E0 ⇒ E0 = 2.3 eV (b) Substituting E0 = 2.3 eV into the first equation,

2.8 eV =

( 4.14 × 10

−15

eV s )( 3.0 × 108 m/s )

λ

− 2.3 eV ⇒ λ = 244 nm

39.39. Solve: The stopping potential is eVstop = hf – E0. Substituting the given values, eVstop =

hc − E0 300 nm

0.257 ( eVstop ) =

hc − E0 400 nm

Substituting the first equation into the second,

hc 0.257 ⎞ ⎛ hc ⎞ ⎛ 1 0.257 ⎜ − E0 ⎟ = − E0 ⇒ E0 (1 − 0.257 ) = hc ⎜ − ⎟ ⎝ 300 nm ⎠ 400 nm ⎝ 400 nm 300 nm ⎠ ⇒ E0 =

( 4.14 × 10

−15

eV s )( 3.0 × 108 m/s )(1.643 × 10−3 /nm ) 0.743

From Table 39.1, we identify the cathode metal as sodium.

= 2.75 eV

39.40. Solve: (a) The stopping potential is Vstop =

h h f − f0 e e

A graph of Vstop versus frequency f should be linear with x-intercept f 0 and slope h e . Since the x-intercept is f0 = 1.0 × 1015 Hz, the work function is

E0 = hf0 = (4.14 × 10−15 eV s)(1.0 × 1015 Hz) = 4.14 eV (b) The slope of the graph is

ΔVstop Δf

=

8.0 V − 0 V = 4.0 × 10−15 V s 3.0 × 1015 Hz − 1.0 × 1015 Hz

Because the slope of the Vstop versus f graph is h e , an experimental value of Planck’s constant is h = e ( 4.0 × 10 −15 V s ) = (1.6 × 10−19 C )( 4.0 × 10−15 V s ) = 6.4 × 10−34 J s

Assess:

This value of the Planck’s constant is about 3.5% lower than the accepted value.

39.41. Solve: (a) The cathode is illuminated with light of wavelength 300 nm. The frequency of the light is f =

c

λ

=

3.0 × 108 m/s = 1.0 × 1015 Hz 300 × 10−9 m

To plot a graph of electron current I as a function of the potential difference ΔV, we need to know (i) the stopping potential and (ii) the saturation current. The equation for the stopping potential is

Vstop =

h h f − f0 e e

According to the data shown in Figure P39.41, the slope of the graph is h 2.5 V slope = = = 4.167 × 10−15 V s e 6.0 × 1014 Hz Also, f0 is the threshold frequency and its value is 6.0 × 1014 Hz. The stopping potential at f = 1.0 × 1015 Hz is Vstop = (4.167 × 10−15 Vs)(1.0 × 1015 Hz) – (4.167 × 10−15 Vs)(6.0 × 1014 Hz) = 1.67 V The rate of photons arriving at the cathode is P 10 × 10−6 W = = 1.508 × 1013 photons/s R= hf (6.63 × 10−34 J s)(1.0 × 1015 Hz) If 10% of the photons eject an electron, the electron current is i = 1.508 × 1012 electrons/s. Thus the actual current between the cathode and anode is I = ei = (1.60 × 10–19 C/electron)(1.508 × 1012 electrons/s) = 2.4 × 10–7 A = 0.24 μA For positive anode voltages, the current will level off at this value when all photoelectrons are collected at the anode. For negative voltages, the current will decrease until stopping at –1.67 V. Thus the current-versus-voltage graph looks as follows. (b)

39.42.

Visualize:

The frequency of the photons was obtained using the equation f = c/λ and is tabulated below. The figure is a graph of Vstop versus f.

λ (nm) f (Hz) Vstop (V)

500 6 × 1014 0.19

450 6.67 × 1014 0.48

400 7.50 × 1014 0.83

350 8.57 × 1014 1.28

300 1.0 × 1015 1.89

250 1.20 × 1015 2.74

Solve: (a) According to Equation 39.8, a graph of Vstop versus frequency f should be linear. The x-intercept is the threshold frequency f0. The slope of the graph is h/e. We can see from the graph that the x-intercept is f0 = 5.5 × 1014 Hz ⇒ E0 = hf0 = 2.3 eV

The work function E0 identifies the metal as potassium. (b) The graph rises ΔVstop = 3.0 V over a run of Δ f = (12.5 – 5.5) × 1014 Hz = 7.0 × 1014 Hz. The experimental slope is ΔVstop Δf

= 4.28 × 10−15 V/Hz

Equating the slope to h/e, we find that the experimental value for h is h = (slope)e = 6.85 × 10−34 J s = 4.28 × 10−15 eV s Assess:

This value of h is about 3% higher than the accepted value.

39.43. Model: Photons have both particle-like and wave-like properties. Solve:

(a) Because E0 = mc2 = 0 J for a photon,

E2 – p2c2 = 0 ⇒ p =

E c

(b) Using E = hf in the momentum equation in part (a),

p=

E hf h h = = ⇒λ = c c λ p

(c) Classically p = mv, so

λ= This is the expression for de Broglie wavelength.

h mv

39.44.

Model: Visualize:

Electrons have both particle-like and wave-like properties.

Please refer to Figure 39.13. Solve: (a) The kinetic energy is K = 12 mv 2 = 50 keV = 8.0 × 10−15 J . Using this formula, the electron’s speed is

v=

( 2 ) (8.0 × 10−15 J )

2K = m

9.11 × 10−31 kg

= 1.32 × 108 m/s ≈ 1.3 × 108 m/s

(b) From Equation 22.7, the fringe spacing in a double-slit interference experiment is Δy =λL/d, where d is the slit separation and L is the distance to the viewing screen. The wavelength of the electrons is their de Broglie wavelength. We have

λ=

h h 6.63 × 10−34 J s = = = 5.5 × 10−12 m p mv ( 9.11 × 10−31 kg )(1.32 × 108 m/s )

⇒ Δy =

λL d

=

( 5.5 × 10

−12

m ) (1.0 m )

1.0 × 10−6 m

= 5.5 × 10−6 m = 5.5 μ m

39.45. Model: Neutrons have both particle-like and wave-like properties. Visualize:

Solve:

(a) The kinetic energy of the neutron is K = 12 mv 2 =

1 2

(1.67 × 10

−27

kg ) ( 200 m/s ) = 3.34 × 10 −23 J = 2.1 × 10 −4 eV 2

(b) The de Broglie wavelength at this speed is

λ=

h 6.63 × 10−34 J s = = 2.0 × 10−9 m = 2.0 nm mv (1.67 × 10−27 kg ) ( 200 m/s )

(c) From Equation 22.7, the fringe spacing in a double-slit interference experiment is Δy =λL/d, where d is the slit separation and L is the distance to the detector. From Figure P39.45, the spacing between the two peaks with m = ±1 (on either side of the central maximum) is 1.4 times as long as the length of the reference bar, which gives Δy = 70 μm . Thus, the distance from the slits to the detector was

L=

d Δy

λ

=

(1.0 ×10

−4

m )( 7.0 × 10−5 m )

2.0 × 10−9 m

= 3.5 m

39.46. Model: Electrons have both particle-like and wave-like particles. Visualize:

Solve:

The kinetic energy of the electrons is

K f = 12 mv 2 = K i + eΔV = 0 J + e ( 250 V ) = (1.60 × 10−19 C ) ( 250 V ) = 4.00 × 10−17 J ⇒ v=

2 ( 4.00 × 10−17 J ) 9.11× 10−31 kg

= 9.37 × 106 m/s

The de Broglie wavelength at this speed is

λ=

h 6.63 × 10−34 J s = = 7.77 × 10−11 m mv ( 9.11 × 10−31 kg )( 9.37 × 106 m/s )

Circular-aperture diffraction produces the first minimum, defining the edge of the central maximum, at θ = 1.22λ D for small angles, as is the case here. The diameter of the opening is

D= Assess:

1.22 λ

θ

=

1.22 ( 7.77 × 10−11 m )

( 0.5° 180° )π

rad

= 1.09 × 10−8 m

The diameter of the hole that diffracts the electron beam is small indeed, and understandably so.

39.47. Model: The energy of a confined particle in a one-dimensional box is quantized.

Solve: The energy of the nth quantum state of a particle in a box is En = n2E1, where E1 is the lowest energy level. The energies 12 eV, 27 eV, and 48 eV have the ratios 4:9:16. Thus, they are the n = 2, n = 3, and n = 4 states of an electron that has E1 = 3 eV = 4.8 × 10−19 J. The lowest energy level is

( 6.63 ×10−34 J s ) h2 h2 E1 = ⇒L= = = 3.5 × 10−10 m = 0.35 nm 2 8mL 8mE1 8 ( 9.11 × 10−31 kg )( 4.8 × 10−19 J ) 2

39.48. Model: The energy of the emitted photon is exactly equal to the energy between the energy levels 1 and 2. Solve:

From Equation 39.14, the energy levels of the electron are

En = n 2

h2 8 mL2

n = 1, 2, 3, …

The energy of the emitted photon is E2 − E1 = hf = h ⇒L=

c

λ

=

4h 2 h2 3h 2 − = 8mL2 8mL2 8mL2

3 ( 6.63 × 10−34 J s )( 200 × 10−9 m ) 3h 2 ⎛ λ ⎞ 3hλ = = = 4.27 × 10−10 m = 0.427 nm ⎜ ⎟ 8m ⎝ hc ⎠ 8mc 8 ( 9.11 × 10−31 kg )( 3.0 × 108 m/s )

39.49. Model: The energy of the emitted gamma-ray photon is exactly equal to the energy between levels 1 and 2. Solve:

From Equation 39.14, the energy levels of the proton are

En = n 2

h2 8 mL2

n = 1, 2, 3, …

The energy of the emitted photon is

E2 − E1 =

4h 2 3h 2 h2 − = 2 2 8mL 8mL 8mL2

3 ( 6.63 × 10−34 J s ) 3h 2 1 eV ⇒L= = × = 1.8 × 10−14 m = 18 fm 6 −27 × 10−19 J 8m ( E2 − E1 ) 1.60 8 (1.67 × 10 kg )( 2.0 × 10 eV ) 2

Assess:

This is roughly the size of a typical nucleus.

39.50. Model: Photons are emitted when an atom undergoes a quantum jump from a higher energy level to a lower energy level. On the other hand, photons are absorbed in a quantum jump from a lower energy level to a higher energy level. Because most of the atoms are in the n = 1 ground state, the only quantum jumps in the absorption spectrum start from the n = 1 state. hc Solve: (a) Using Ephoton = = ΔEatom , the 3 wavelengths in the absorption spectrum give 2.49 eV, 4.14 eV,

λ

and 6.21 eV as the energies of the n = 2, 3, and 4 energy levels.

(b) The emission spectrum of the atom will contain the following wavelengths: hc hc hc hc λ 31 = = = 300 nm λ 41 = = = 200 nm E3 − E1 4.14 eV E4 − E1 6.21 eV

λ 21 =

hc hc = = 500 nm E2 − E1 2.49 eV

λ 42 =

hc hc = = 334 nm E4 − E2 3.72 eV

λ 43 =

hc hc = = 601 nm E4 − E3 2.07 eV

λ 32 =

hc hc = = 753 nm E3 − E2 1.65 eV

39.51. Model: Photons are emitted when an atom undergoes a quantum jump from a higher energy level to a lower energy level. On the other hand, photons are absorbed in a quantum jump from a lower energy level to a higher energy level. Because most of the atoms are in the n = 1 ground state, the only quantum jumps in the absorption spectrum start from the n = 1 state. Solve: (a) The ionization energy is E1 = 6.5 eV. (b) The absorption spectrum consists of the transitions 1 → 2 and 1 → 3 from the ground state to excited states. According to the Bohr model, the required photon frequency and wavelength are

f =

ΔE c hc ⇒ λ= = h f ΔE

where ΔE = Ef – Ei is the energy change of the atom. Using the energies given in the figure, we calculated the values in the table below. Ef (eV) Ei (eV) Δ E (eV) λ (nm) Transition 1→2 1→3

−3.0 –2.0

−6.5 −6.5

3.5 4.5

355 276

(c) Both wavelengths are ultraviolet (λ < 400 nm). (d) A photon with wavelength λ = 1240 nm has an energy Ephoton = hf = hc/λ = 1.0 eV. Because Ephoton must exactly match Δ E of the atom, a 1240 nm photon can be emitted only in a 3 → 2 transition. So, after the collision the atom was in the n = 3 state. Before the collision, the atom was in its ground state (n = 1). Thus, an electron with vi = 1.4 × 106 m/s collided with the atom in the n = 1 state. The atom gained 4.5 eV in the collision as it is was excited from the n = 1 to n = 3, so the electron lost 4.5 eV = 7.20 × 10−19 J of kinetic energy. Initially, the kinetic energy of the electron was

Ki = 12 melecvi2 = 12 ( 9.11× 10−31 kg )(1.40 × 106 m/s ) = 8.93 × 10−19 J 2

After losing 7.20 × 10−19 J in the collision, the kinetic energy is

K f = K i − 7.20 × 10−19 J = 1.73 × 10−19 J = 12 melecvf2 ⇒ vf =

2 (1.73 × 10−19 J ) 2Kf = = 6.16 × 105 m/s melec 9.11 × 10−31 kg

39.52. Solve: Equation 98.28 is 1 ⎛ =2 ⎞ e2 1 =2 ⎛ 1 ⎞ e2 En = m ⎜ 2 2 2 ⎟ − = ⎜ 2 ⎟− 2 2 ⎝ m aBn ⎠ 4πε 0 n aB 2 maB ⎝ n aB ⎠ 4πε 0 n 2 aB The Bohr radius is aB =

=2 e2 1 ⎛ e2 ⎞ ⎛ 1 ⎞ ⎛ e2 ⎞ ⎛ 1 ⎞ 1 ⎛ e2 ⎞ 1 1 ⎛ 1 e2 ⎞ 4πε 0= 2 = ⇒ = − = − = − E ⇒ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ n maB 4πε 0 2 ⎝ 4πε 0 ⎠ ⎝ n 2 aB ⎠ ⎝ 4πε 0 ⎠ ⎝ n 2aB ⎠ 2aB ⎝ 4πε 0 ⎠ n 2 n 2 ⎝ 4πε 0 2aB ⎠ me 2

39.53. Solve: The radius of the orbit in state n is rn = n2aB. The quantum number is found as follows: n2 =

rn 12 ( 5.18 nm ) = = 48.96 ≈ 49 ⇒ n = 7 aB 0.0529 nm

Thus the energy is

En = −

13.60 eV 13.60 eV ⇒ E7 = − = −0.278 eV 2 49 n

39.54. Solve: The wavelengths in the hydrogen spectrum are given by λn → m =

91.18 nm 1/ m 2 − 1/ n 2

For each m, the wavelengths range from a maximum value when n = m + 1 to a minimum value (the series limit) when n → ∞. A few calculations reveal the following behavior: m 1 2 3 4

λ max for n = m + 1

λ min for n = ∞

122 nm 656 nm 1876 nm 4050 nm

91 nm 365 nm 822 nm 1459 nm

We see that the only visible wavelengths (400–700 nm) in the hydrogen spectrum occur for m = 2, starting with n = 3. We can calculate that λ3→2 = 656.5 nm, λ4→2 = 486.3 nm, λ5→2 = 434.2 nm, and λ6→2 = 410.3 nm. The next transition, 6 → 2, has λ < 400 nm. Thus these are the only four visible wavelengths in the hydrogen spectrum.

39.55. Model: Photons are absorbed in a quantum jump from a lower energy level to a higher energy level. The energy of the emitted photon is exactly equal to the energy between the starting and the ending levels. Solve: The energy levels of the stationary states of the hydrogen atom are

En = −

13.60 eV n2

n = 1, 2, 3, …

In the ground state (n = 1), E1 = −13.60 eV. The change in energy when the hydrogen atom absorbs a 13.06 eV photon is 1 13.06 ⎛ 1 ⎞ En – E1 = 13.06 eV ⇒ ( −13.60 eV ) ⎜ 2 − 1⎟ = 13.06 eV ⇒ 2 = 1 − = 0.0397 ⇒ n = 5 n 13.60 ⎝n ⎠

When the atom, having been excited to n = 5, undergoes a quantum jump to the energy level two lower (corresponding to n = 3), the emitted wavelength is given by Equation 39.36:

λ4→3 =

91.18 nm = 1282 nm 1 32 − 1 52

39.56. Solve: (a) From Equation 39.36, the wavelengths of the emission spectrum are λ n→m =

91.18 nm m−2 − n −2

m = 1, 2, 3, … n = m + 1, m + 2, …

For the 200 → 199 transition,

λ 200→199 =

91.18 nm

(199 )

−2

− ( 200 )

−2

= 0.362 m

(b) For the 2 → 199 transition,

λ 2→199 =

91.18 nm

( 2)

−2

− (199 )

−2

= 4.000404071 (91.18 nm)

Likewise, λ 2 → 200 = 4.00040004(91.18 nm). The difference in the wavelengths of these two transitions is 0.0000041 (91.18 nm) = 0.000368 nm.

×

39.57. Solve: (a) The orbital radius and velocity for n = 99 are, r99 = 992aB = 518 nm

v99 =

v1 = 2.21 × 104 m/s 99

For n = 100, r100 = 1002aB = 529 nm

v100 =

v1 = 2.19 × 104 m/s 100

(b) For circular motion, the orbital frequency is f = v/2πr. Using r and v from part (a),

f99 = 6.79 × 109 Hz

f100 = 6.59 × 109 Hz

(c) Using Equation 38.36, the wavelength of a photon emitted in a 100 → 99 transition is

λ 100 →99 =

91.18 nm

( 99 )

−2

− (100 )

−2

= 4.49 × 107 nm = 0.0449 m

The frequency of this photon is f photon = c λ = 6.68 × 109 Hz. (d) The average of f99 and f100 from part (b) is favg = 6.69 × 109 Hz. This differs from the photon frequency fphoton by only 0.01 × 109 Hz, or 0.15%.

39.58.

Visualize:

Solve:

(a) He+ is a hydrogen-like ion with Z = 2. From Equation 39.37, its energy levels are En = − Z 2

13.60 eV 54.4 eV =− n2 n2

It is straightforward to compute E1 = −54.4 eV, E2 = −13.6 eV, E3 = −6.04 eV, E4 = −3.40 eV, and E5 = −2.18 eV. (b) The ionization energy is E1 = 54.4 eV. The ionization limit is shown as n = ∞. (c) The possible transitions from n = 4 are 4 → 3, 4 → 2, and 4 → 1. (d) The wavelengths of these transitions are λ = hc/ΔΕ: The computed values are shown in the table below. Ei (eV) ΔΕ (eV) λ (nm) Transition Ef (eV)

4→3 4→2 4→1

−6.04 −13.6 −54.4

−3.40 −3.40 −3.40

2.64 10.2 51.0

470 122 24

39.59. Visualize: A neutral oxygen atom that has lost 7 electrons is a hydrogen-like atom because it has one electron going around a nucleus with Z = 8. Solve: The energy levels for the O+7 ion are 13.60 ( 8 ) eV 13.60 Z 2 eV 870.4 eV =− =− n2 n2 n2 2

En = − The energy of the 3 → 2 transition is

hc ⎛1 1⎞ E3 − E2 = −870.4 eV ⎜ 2 − 2 ⎟ = 120.89 eV = λ ⎝3 2 ⎠ ⇒λ =

( 4.14 ×10

−15

eV )( 3.0 × 108 m/s )

120.89 eV

= 10.28 nm

Likewise for the 4 → 2 transition, ⎛ 1 1 ⎞ E4 − E2 = −870.4 eV ⎜ 2 − 2 ⎟ = 163.2 eV ⎝4 2 ⎠ The wavelength for this transition λ = 7.62 nm. For the 5 → 2 transition, E5 – E2 = 182.78 eV and λ = 6.80 nm. All the wavelengths are in the ultraviolet range.

39.60. Model: An inelastic collision between the atoms causes the electrons to be excited.

Solve: From Example 39.12, a wavelength of 121.6 nm corresponds to a 2 → 1 transition. Because the atoms were in their ground states before colliding, each atom lost kinetic energy ∆K = E2 – E1 and, in the process, excited the electron from n = 1 to n = 2, from where it subsequently emitted the photon. Because the atoms each end at rest, ∆K = Ki = 12 mv2. Thus 1.60 × 10−19 J 1 = 1.632 × 10−18 J matom v 2 = E2 − E1 = 10.2 eV × 2 1 eV ⇒ v=

2( E2 − E1 ) 2(1.632 × 10−18 J) = = 44,200 m/s matom 1.67 × 10−27 kg

39.61. Solve: (a) Nearly all atoms spend nearly all of their time in the ground state (n = 1). To cause an emission from the n = 3 state, the electrons must excite hydrogen atoms from the n = 1 state to the n = 3 state. The energy gained by each atom is E3 – E1 = −1.51 eV – (−13.60 eV) = 12.09 eV This means the electrons must each lose 12.09 eV of kinetic energy. Thus the electrons must have at least Kmin = 12.09 eV of kinetic energy to cause the emission of 656 nm light. The minimum speed of the electrons is

vmin =

2 (12.09 eV ) 1.6 × 10−19 J 2 K min = × = 2.06 × 106 m/s melec 9.11 × 10−31 kg 1 eV

(b) An electron gains 12.09 eV of kinetic energy by being accelerated through a potential difference of 12.09 V. This is simply the definition of electron volt.

39.62. Visualize: Please refer to Figure 39.12. Solve: (a) A single photon striking the cathode ejects 1 electron. This electron strikes the first dynode, causing the emission of ε electrons. Each of these electrons strikes the second dynode and ejects ε 2 electrons, so there are ε 2 electrons that strike the third dynode, and so on. There are ε N electrons after the N th dynode. The gain of the PMT is ε N. (b) The pulse “height” is 6 divisions. At 20 mV per division this is 120 mV = 0.120 V. It is a negative pulse because the current is due to electrons whereas positive current is defined as the flow of positive charges. From Ohm’s law, the maximum current is ΔV 0.120 V I= = = 0.0024 A = 2.4 mA R 50 Ω (c) The legend on the graph states that the width of the pulse is 300 ps. Thus, the charge is Q = ∫ I dt ≈ Area under I -versus-t curve ≈ height × width ≈ ( 0.0024 A ) ( 3.0 × 10−10 s ) = 7.20 × 10−13 C The number of electrons in the pulse is N elec =

Q 7.20 × 10−13 C = = 4.5 × 106 electrons e 1.60 × 10−19 C

(d) We found in part (a) that the number of electrons reaching the anode is ε N . These are the Nelec electrons that make the 120 mV pulse. So, Nelec = ε N ⇒ ε = (Nelec)1/N = (4.5 × 106)1/14 = 3.0

39.63. Solve: The energy of the ultraviolet light photon is E=

hc

λ

=

( 6.63 ×10

−34

J s )( 3.0 × 108 m/s )

70 × 10−9 m

×

1 eV = 17.76 eV 1.60 × 10−19 J

To ionize a hydrogen atom, a minimum energy of 13.60 eV is required. Thus, the kinetic energy of the freed electrons is K = 12 mv 2 = E − 13.60 eV = 17.76 eV − 13.60 eV = 4.16 eV

39.64. Visualize: Please refer to Figure 39.14. Solve:

(a) From chapter 18, the rms speed of the sodium atom is vrms =

3 (1.38 × 10−23 J/K ) ( 0.001 K ) 3kBT = = 1.04 m/s ≈ 1.0 m/s m 23 u × 1.67 × 10−27 kg/u

(b) The diffraction of a wave by a grating obeys the equation dsinθm = mθ, where m is the order of the diffraction. The de Broglie wavelength of the atoms is

λ=

h 6.63 × 10−34 J s = = 1.66 × 10−8 m mv ( 23 u × 1.67 × 10−27 kg/u ) (1.04 m/s )

The antinodes in the standing-wave laser beam form the grating. The spacing between adjacent antinodes of a standing wave is

d=

λ 2

= 295 nm = 2.95 × 10−7 m

Thus, the first order diffraction angle is 1.66 × 10−8 m ⎛λ⎞ θ = sin −1 ⎜ ⎟ = sin −1 = 3.2° d 2.95 × 10−7 m ⎝

⎠

(c) The waves traveling to points B and C are each diffracted an angle of 3.2°. From the geometry, the distance between B and C is 2 (10 cm ) tan 3.2° = 1.1 cm. (d) The atoms exhibit interference when recombined at point D. For interference to occur, each atom must have somehow traveled along both paths through the interferometer. So halfway through, each atom must have been present both at point B and at point C, even though these points are more than 1 cm apart. We’re forced to conclude that matter, at the atomic level, does not consist of localized, point-like “particles.” Instead, the atoms appear to have the nonlocalized behavior that we previously associated with waves.

39.65. Solve: (a) The orbital period of an electron in the n state is Tn =

2π ( 0.0529 nm ) 3 2π rn 2π r1n 2 2π r1 3 n = T1n3 = n = (1.52 × 10−16 s ) n3 = = 2.19 × 106 m/s vn v1 n v1

where we obtained the expression for the radius and speed from Equations 39.26 and 39.27. Thus, T1 = 1.52 × 10−16 s. (b) In the n = 2 state, the orbital period of the electron is T2 = (1.52 × 10−16 s)(2)3 = 1.216 × 10−15 s. The number of revolutions before the electron makes a quantum jump to the n = 1 state is

1.6 × 10−9 s = 1.32 × 106 revolutions 1.216 × 10−15 s

39.66. Solve: (a) The Bohr condition on the electron’s angular momentum is mvnrn = n=. For cyclotron motion in a magnetic field, the radial component of the force is

mvn2 = evn B ⇒ mvn = eBrn rn The Bohr condition becomes (eBrn)rn = n= ⇒ rn =

n= eB

(b) The first allowed radius is

r1 =

(1) = eB

=

6.63 × 10−34 J s = 25.7 nm 2π (1.6 × 10−19 C ) (1.0 T )

Likewise, r2 = 36.3 nm, r3 = 44.5 nm, and r4 = 51.4 nm. (c) For an electron undergoing cyclotron motion in a magnetic field, the kinetic energy is En = 12 mvn2 . Using the Bohr condition in part (a) first for momentum and then for the radius,

En =

1 m 2v 2 1 e2 B 2 rn2 eB = = = ( eBrn2 ) = eBn 2 m 2 m 2m 2m

From chapter 33, the cyclotron frequency is f cyc =

eB ⎛ eB ⎞ ⇒ En = ⎜ ⎟ (π n= ) = n (π =f cyc ) 2π m ⎝ 2π m ⎠

39.67. Solve: (a) The muon acts like an electron in a hydrogen-like ion of charge Z except that the muon’s mass is mμ = 207me. The different mass changes the Bohr radius to

( aB )μ =

4πε 0= 2 me 4πε 0= 2 1 1 aB = = = ( 0.0529 nm ) = 2.56 × 10−4 nm mμ e 2 mμ mee2 207 207

From Equation 39.37, the ground state radius of the muon in the carbon is r1 =

( aB ) μ

=

Z

2.56 × 10−4 nm = 4.26 × 10−5 nm 6

The speed in the ground state is v1 =

= 6.63 × 10−34 J s = = 1.31 × 107 m/s mμ r1 2π ( 207 × 9.11 × 10−31 kg )( 4.26 × 10−5 nm )

(b) From Equations 39.30 amd 39.37, the energy levels are En = −Z2(E1)μ /n2. Because of the different mass,

( E1 )μ =

e2 a 1 e2 = B = 207 (13.60 eV ) = 2815 eV 4πε 0 2 ( aB ) μ ( aB ) μ 4πε 0 2aB 1

For Z = 6, the n = 1 and n = 2 energy levels are

E1 = −

( 36 ) 2815 eV = −101,300 eV 12

E2 = −

( 36 ) 2815 eV = −25,300 eV 22

The photon emitted in a 2 → 1 transition has an energy Ephoton = E2 – E1 = 76,000 eV. The wavelength is

λ=

hc Ephoton

=

( 6.63 × 10

−34

J s )( 3.0 × 108 m/s )

76,000 eV

= 1.64 × 10−11 m = 0.0164 nm

(c) This is an x-ray photon. (d) For circular motion, the orbital period is

T=

−5 2π r1 2π ( 4.26 × 10 nm ) = = 2.04 × 10−20 s v1 1.31 × 107 m/s

The number of orbits completed during the muon’s half-life is N=

Δt 1.5 × 10−6 s = = 7.3 × 1013 T 2.04 × 10−20 s

This is such an incredibly large number of orbits that it “makes sense” to use the Bohr-model.

40.1. Model: The sum of the probabilities of all possible outcomes must equal 1 (100%). Solve:

The sum of the probabilities is PA + PB + PC + PD = 1. Hence,

0.40 + 0.30 + PC + PD = 1 ⇒ PC + PD = 0.30 Because PC = 2PD, 2PD + PD = 0.30. This means PD = 0.10 and PC = 0.20. Thus, the probabilities of outcomes C and D are 20% and 10%, respectively.

40.2. Model: The probability that the outcome will be A or B is the sum of PA and PB.

Solve:

(a)

Coin A

Coin B

Coin C

H H H H T T T T

H H T T H H T T

H T H T H T H T

(b) From the above table, we see that 2 heads and 1 tail occur three times (HHT, HTH, THH). Out of the possible eight outcomes, each outcome is equally probable and has a probability of occurrence of 1/8. So, the probability of getting 2 heads and 1 tail is 3/8 = 37.5%. (c) From the table, we see that at least two heads occur 4 times (HHH, HHT, HTH, and THH). So, the probability of getting at least two heads is 4/8 = 50%.

40.3. Model: The probability that the outcome will be A or B is the sum of PA and PB. Solve: (a) A regular deck of cards has 52 cards. Drawing a card from this deck has a probability of 1/52. Because there are 4 aces in the deck, the probability of drawing an ace is 4/52 = 0.077 = 7.7%. (b) Because there are 13 spades, the probability of drawing a spade is 13/52 = 0.25 = 25%.

40.4. Model: The probability that the outcome will be A or B is the sum of PA and PB. The expected value is

your best possible prediction of the outcome of an experiment. Solve: For each deck, there are 12 picture cards (4 Jacks, 4 Queens, and 4 Kings). Because the probability of drawing one card out of 52 cards is 1/52, the probability of drawing a card that is a picture card is 12/52 = 23.1%. The number of picture cards that will be drawn is 0.231 × 1000 = 231.

40.5. Model: The probability that the outcome will be A or B is the sum of PA and PB. Solve: (a) Each die has six faces and the faces have dots numbering from 1 to 6. We have two dice A and B. The various possible outcomes of rolling two dice are given in the following table. A B A B A B 1 3 1 5 1 1 1 2 3 2 5 2 1 3 3 3 5 3 1 4 3 4 5 4 1 5 3 5 5 5 6 3 6 5 6 1 1 4 1 6 1 2 2 2 4 2 6 2 2 3 4 3 6 3 4 4 4 6 4 2 2 5 4 5 6 5 2 6 4 6 6 6 There are 36 possible outcomes. From the table, we find that there are six ways of rolling a 7 (1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, 6 and 1). The probability is (1/36) × 6 = 1/6. (b) Likewise, the probability of rolling a double is 1/6. (c) There are 10 ways of rolling a 6 or an 8. The probability is (1/36) × 10 = 5/18.

40.6. Model: The probability density of finding a photon is directly proportional to the square of the lightwave amplitude A ( x ) . 2

Solve:

The probability of finding a photon within a narrow region of width δ x at position x is Prob(in δ x at x) ∝ A ( x ) δ x ⇒ 2

Prob ( in δ x1 at x1 )

Prob ( in δ x2 at x2 )

A ( x1 ) δ x 2

=

A ( x2 ) δ x 2

Let N be the total number of photons and N2 the number of photons detected at x2 in a width δ x. The above equation simplifies to

( 2000 )( 30 V/m ) = 18,000 2000 / N (10 V/m ) ( 0.10 mm ) = ⇒ N2 = 2 2 N2 N ( 30 V/m ) ( 0.10 mm ) (10 V/m ) 2

2

40.7. Visualize: Combine Equations 40.10 and 40.11 to show that N is proportional to | A( x) |2 δ x. N (in δ x2 at x2 ) N tot = 2 N (in δ x1 at x1 ) A( x1 ) δ x1 N tot 2

A( x2 ) δ x2

We are given N1 = 6000, δ x1 = 0.10 mm, A( x1 ) = 200 V/m, N 2 = 3000, and δ x2 = 0.20 mm. We are not given

N tot but it cancels anyway. Solve: Solve the above equation for A( x2 ) .

A( x2 ) = A( x1 )

δ x1 N (in δ x2 at x2 ) (0.10 mm)(3000) = (200 V/m) = 100 V/m δ x2 N (in δ x1 at x1 ) (0.20 mm)(6000)

Assess: The answer is half of the wave amplitude at the other strip, which seems reasonable.

40.8. Solve: The probability that a photon arrives at this 0.10-mm-wide strip is Prob(in 0.10 mm at x) =

N = P ( x )δ x 1.0 × 1010

where N is the number of photons detected in the strip and the total number of photons is 1.0 × 1010. We have

N = (1.0 × 1010 )( 20 m −1 )( 0.10 × 10−3 m ) ⇒ N = 2.0×107

40.9. Model: See Example 40.1. Visualize: We are given N tot = 5.0 × 1012 , N (in δ x at x) = 2.0 × 109 , and δ x = 0.10 mm. The probability that a

photon goes through the slit is Prob (in δ x at x) = 2.0 × 109 /5.0 × 1012 = 0.00040. Solve: Solve for the probability density P(x) in Equation 40.12.

P( x) =

Prob (in δ x at x) 0.00040 = = 4.0 m −1 1.0 × 10−4 m δx

Assess: This result is similar to that in Example 40.1.

40.10. Solve:

ψ ( x ) δ x is a probability, which is dimensionless. The units of δ x are m, so the units of 2

ψ ( x ) are m−1 and thus the units of ψ are m−1/2. 2

40.11. Model: The probability of finding a particle at position x is determined by ψ ( x ) . 2 Solve: (a) The probability of detecting an electron is Prob(in δx at x) = ψ ( x ) δ x. At x = 0 mm, the number of 2

electrons landing is calculated as follows: 2 N = ψ ( 0 mm ) δ x ⇒ N = ( 13 mm −1 ) ( 0.010 mm ) (1.0 × 106 ) = 3333 1.0 × 106

(b) Likewise, the number of electrons landing at x = 2.0 mm is N = ψ ( 2.0 mm ) δ xN total = ( 0.111 mm −1 ) ( 0.010 mm ) (1.0 × 106 ) = 1111 2

40.12.

Visualize:

Solve:

(a) We assume that ψ ( x ) = 0 at some point in between each of the peaks. 2

(b) Two factors are important for drawing ψ (x). First, the value at each point is the square root of the value on 2 the ψ ( x ) graph. Second, and especially important, is that ψ (x) is a wave function and so it oscillates between

positive and negative values each time ψ ( x ) reaches zero. 2

(c) Multiplying ψ (x) by −1 does not change ψ ( x ) . So another possible graph for ψ (x) is the “upside down” 2

version of the graph of part (b).

40.13. Model: The probability of finding a particle at position x is determined by ψ ( x ) . 2

Solve:

(a) The probability of detecting an electron is Prob ( in δ x at x ) = ψ ( x ) δ x . Hence the probability the 2

electron will land at x = 0.000 mm is Prob(in 0.010 mm at x = 0.000 mm) = (0.50 mm−1)(0.010 mm) = 5.0 × 10−3 (b) Since ψ ( 0.500 mm ) = 0.25 mm , the probability is 2.5 × 10−3. 2

(c) Since ψ (1.000 mm ) = 0, the probability is 0. 2

(d) Since ψ ( 2.000 mm ) = 0.25 mm −1 , the probability is 2.5 × 10−3. 2

40.14. Model: The probability of finding a particle is determined by the probability density P ( x ) = ψ ( x ) . 2

∞

Solve:

(a) The normalization condition for a wave function:

∫ ψ ( x)

2

dx = area under the curve = 1. In the

−∞

present case, the area under the ψ ( x ) -versus-x graph is 2a nm. Hence, a = 2

1 2

nm −1.

(b) Each point on the ψ (x) graph is the square root of the corresponding point on the |ψ (x)|2 graph. Where the |ψ (x)|2 graph has dropped to 1/2 its maximum value at x = 1 nm, the ψ (x) graph will have dropped only to

1/ 2 = 0.707 of its maximum value. Thus the graph shape is convex upward. Since a = 12 nm–1, the peak value of ψ (x) is a = 1/ 2 nm −1/ 2 = 0.707 nm–1/2. The graph is shown below. The negative of this graph, curving downward, would also be an acceptable wave function.

(c) The probability of the electron being located in the interval 1.0 ≤ x ≤ 2.0 nm is Prob(1.0 nm ≤ x ≤ 2.0 nm) = area under the curve between 1.0 nm and 2.0 nm −1

1⎛ a ⎞ (0.50 nm )(1.0 nm) = ⎜ ⎟ (2.0 nm – 1.0 nm) = = 0.125 2⎝ 2 ⎠ 4

40.15. Model: The probability of finding the particle is determined by the probability density 2 P ( x) = ψ ( x) . ∞

Solve:

(a) According to Equation 40.18,

∫ ψ ( x)

2

dx = area under the curve = 1. The area of the ψ ( x ) 2

−∞

versus-x graph is 4a mm. Hence, a = 14 mm −1. (b) Each point on the ψ (x) graph is the square root of the corresponding point on the |ψ (x)|2 graph. Where the |ψ (x)|2 graph has reached 1/2 its maximum value at x = 2 mm, the ψ (x) graph will have reached to 1/ 2 = 0.707 of its maximum value. Thus the graph shape is convex. Since a = 14 mm–1, the maximum value of ψ (x) is

a = 1/ 4 nm −1/ 2 = 0.50 mm–1/2. The graph is shown below. The negative of this graph, curving from negative to positive, would also be an acceptable wave function.

(c) The probability of the neutron being located in the interval |x| ≥ 2.0 fm is

Prob( x ≥ 2.0 fm) = 1 − area under the curve between –2.0 fm and 2.0 fmω0 z −1

1⎛ a ⎞ (0.25 fm )(2.0 fm) = 1 − 2 × ⎜ ⎟ (2.0 fm – 0 fm) = 1 − 2 × = 0.75 2⎝ 2 ⎠ 4

40.16. Model: The probability of finding the particle is determined by the probability density 2 P ( x) = ψ ( x) . ∞

Solve:

(a) According to Equation 40.18,

∫ ψ ( x)

2

dx = 1. We calculate this integral by drawing the graph of

−∞

ψ ( x ) and finding the area under the curve. 2

The area of the ψ ( x ) versus x graph is 2

−1 nm

∫

−2nm

ψ ( x ) dx + 2

1 nm

∫

ψ ( x ) dx + 2

−1 nm

2 nm

∫ ψ ( x)

2

dx = ( 0.25c 2 ) (1 nm ) + ( c 2 ) ( 2 nm ) + ( 0.25c 2 ) (1 nm ) = 2.5c 2

1 nm

⇒ 2.5c 2 nm = 1 ⇒ c =

1 nm −1/ 2 = 0.632 nm −1/ 2 2.5

(b) The graph is shown in part (a). (c) The probability is

⎛ 1 ⎞ nm −1 ⎟ ( 2 nm ) = 0.80 Prob(−1.0 nm ≤ x ≤ 1.0 nm) = area = ( c 2 ) ( 2 nm ) = ⎜ 2.5 ⎝ ⎠

40.17. Model: The probability of finding the particle is determined by the probability density 2 P ( x) = ψ ( x) . ∞

Solve:

(a) According to the normalization condition,

∫ ψ ( x)

2

dx = 1. From the given ψ (x)-versus-x graph, we

−∞

first generate a ψ ( x ) -versus-x graph and then find the area under the curve. 2

The area under the ψ ( x ) -versus-x graph is 2

4.0 mm

∫

−4.0 mm

c 2 dx = ( 8.0 mm ) c 2 = 1 ⇒ c =

1 = 0.354 mm −1/ 2 8.0 mm

(b) The graph is shown above. (c) The probability is

⎛ 1 ⎞ Prob(1.0 mm ≤ x ≤ 1.0 mm) = area = c 2 ( 2.0 mm ) = ⎜ mm −1 ⎟ ( 2.0 mm ) = 0.25 ⎝ 8.0 ⎠

40.18. Model: A radio-frequency pulse is an electromagnetic wave packet, hence it must satisfy the relationship

Δ f Δ t ≈ 1. Solve: A 1.00 MHz oscillation has a time period of T = 1.00 μs. A pulse consisting of 100 cycles of the 1.00 MHz oscillation will have a duration of 100(1.00 μs) = 0.10 ms. That is, Δt = 0.10 ms. Using Equation 40.20,

1 1 = = 10 kHz Δt 0.10 ms Thus the bandwidth of the wave centered at 1.00 MHz is 10 kHz. Δf ≈

40.19. Model: A radio-frequency pulse is an electromagnetic wave packet, hence it must satisfy the relationship

Δ f Δ t ≈ 1. Solve: The waves that must be superimposed to create the pulse of smallest duration span the frequency range f – Δ f / 2 ≤ f ≤ f + Δ f / 2 . Because Δ f = 120 MHz – 80 MHz = 40 MHz, f = 100 MHz. Using Equation 40.20, 1 1 Δt ≈ = = 2.5 × 10−8 s = 25 ns Δf 40 MHz

Thus, a radio wave centered at 100 MHz and having a frequency span 80 MHz to 120 MHz can be used to create a wave of duration 25 ns.

40.20. Model: The beating of two waves of different frequencies produces a series of wave packets. Solve:

The beat frequency is fbeat = f1 – f2 = 502 Hz – 498 Hz = 4 Hz. The period of one beat is

Tbeat =

1 f beat

=

1 = 0.25 s 4 Hz

During 0.25 s, the wave moves forward ∆ x = vsoundTbeat = (340 m/s)(0.25 s) = 85 m. Thus the length of each wave packet is 85 m.

40.21. Model: A laser pulse is an electromagnetic wave packet, hence it must satisfy the relationship Δ f Δt ≈ 1. Solve:

Because c = λ f , the frequency and period are f =

1 1 3.0 × 108 m/s = 5.0 × 10−15 s = 2.0 × 1014 Hz ⇒ T = = −6 f 2.0 × 1014 Hz 1.5 × 10 m

Since Δ f = 2.0 GHz, the minimum pulse duration is

Δt ≈

1 1 = = 5.0 × 10−10 s Δf 2.0 × 109 Hz

The number of oscillations in this laser pulse is 5.0 × 10−10 s 5.0 × 10−10 s = = 1.0 × 105 oscillations T 5.0 × 10−15 s

40.22. Visualize: The uncertainty in velocity is Δvx = 3.58 × 105 m/s − 3.48 × 105 m/s = 1.0 × 104 m/s. We recall (or look up) the mass of an electron: me = 9.11× 10−31 kg. Solve: Solve for the position uncertainty Δ x in Equation 40.28. Δx ≈

h h 6.63 × 10−34 J ⋅ s = = = 3.6 × 10−8 m = 36 nm 2Δpx 2me Δvx 2(9.11 × 10−31 kg)(1.0 × 104 m/s)

Assess: The answer is a few dozen atomic diameters.

40.23. Model: Andrea is subject to the Heisenberg uncertainty principle. She cannot be absolutely at rest

(∆vx = 0) without violating the uncertainty principle. Andrea’s mass is 50 kg. Solve: Because Andrea is inside her room, her position uncertainty could be as large as ∆ x = L = 5.0 m. According to the uncertainty principle, her velocity uncertainty is thus Δvx =

h 6.63 × 10−34 J s = = 1.3 × 10−36 m 2m Δ x 2(50 kg)(5.0 m)

Because her average velocity is zero, her velocity is likely to be in the range –0.65 × 10–36 m/s to +0.65 × 10–36 m/s. Assess: At a speed of 1 × 10–36 m/s, Andrea would have moved less than 1% the diameter of the nucleus of an atom in the entire age of the universe. It makes perfect sense to think that macroscopic objects can be at rest.

40.24. Model: Electrons are subject to the Heisenberg uncertainty principle. Solve:

Uncertainty in our knowledge of the position of the electron as it passes through the hole is Δ x = 10

μm. With a finite Δ x , the uncertainty Δpx cannot be zero. Using the uncertainty principle, h h 6.63 × 10−34 J s Δpx = mΔvx = ⇒ Δvx = = = 36 m/s 2Δx 2m Δ x 2 ( 9.11 × 10−31 kg )(10 × 10−6 m )

Because the average velocity is zero, the best we can say is that the electron’s velocity is somewhere in the interval −18 m/s ≤ vx ≤ 18 m/s.

40.25. Model: Protons are subject to the Heisenberg uncertainty principle. Solve: We know the proton is somewhere within the nucleus, so the uncertainty in our knowledge of its position is at most Δ x = L = 4.0 fm. With a finite Δ x , the uncertainty Δpx is given by the uncertainty principle:

Δpx = mΔvx =

h/2 h 6.63 × 10−34 J s ⇒ Δvx = = = 5.0 × 107 m/s Δx 2mL 2 (1.67 × 10−27 kg )( 4.0 × 10−15 m )

Because the average velocity is zero, the best we can say is that the proton’s velocity is somewhere in the range −2.5 × 107 m/s to 2.5 × 107 m/s. Thus the smallest range of speeds is 0 to 2.5 × 107 m/s.

40.26. Model: The probability of finding the particle is determined by the probability density P(x). Solve: (a) Because δ x = 0, the probability of finding the particle at exactly x = 50.0 mm is zero. (b) The particle is equally likely to be found anywhere in the region 0 mm ≤ x ≤ 100 mm. The region 49.0 mm ≤ x ≤ 51.0 mm is 2% of the 0-to-100 mm interval, so the probability is 2.0%. (c) One-quarter of the 0-to-100 mm interval has x ≥ 75 mm, so the probability is 25%.

40.27. Model: A radar pulse comprised of electromagnetic waves is a wave packet, hence it must satisfy the relationship Δ f Δt ≈ 1. Solve: The period 0.100 ns for a wave corresponds to a frequency of

f =

1 1 = = 1.0 × 1010 Hz T 0.100 ns

A pulse of 1.0 ns in duration is 10 oscillations of the wave. Although the station broadcasts at a frequency of 1.0 × 1010 Hz, this pulse is not a pure 1.0 × 1010 Hz oscillation. Instead, this pulse has been created by the superposition of many waves whose frequencies span 1 1 Δf ≈ = = 109 Hz Δt 1.0 × 10−9 s This range of frequencies is centered at the 1.0 × 1010 Hz broadcast frequency, so the waves that must be superimposed to create this pulse span the frequency range:

f − Δ f 2 ≤ f ≤ f + Δ f 2 ⇒ 9.5 GHz ≤ f ≤ 10.5 GHz

40.28. Model: The ultrasound pulse is a wave packet, so it must satisfy the relationship Δ f Δt ≈ 1. Solve:

(a) The frequency of the ultrasound pulse is 1.000 MHz, so its wavelength is v 1500 m/s λ= = = 1.5 × 10−3 m = 1.5 mm f 1.000 × 106 Hz

Since each pulse is 12 mm long, one pulse contains 8 complete cycles (12 mm 1.5 mm ) . (b) Because T = f −1 = 1.000 × 10−6 s and there are 8 cycles, the pulse length is Δt = 8.000 × 10−6 s. Using Δ f Δt ≈ 1,

Δ f ≈ 1 Δt = 1.25 × 105 Hz. The range of frequencies that must be superimposed to create this pulse is from ( f – Δ f / 2) ( f + Δ f / 2). That is, from 0.938 MHz to 1.063 MHz.

to

40.29. Model: The radio-wave pulses are wave packets, so each packet satisfies the relationship Δ f Δt ≈ 1. Visualize: Please refer to Figure P40.29. Solve: Because the frequency bandwidth is Δ f = 200 kHz, the shortest possible pulse width is 1 1 Δt ≈ = = 5 × 10−6 s Δ f 200 kHz This means the time period of the pulse train is T = 2Δt = 2(5 × 10−6 s) = 10 × 10−6 s So, the frequency of the pulse train is f = 1 T = 1.0 × 105 Hz. That is, the maximum transmission rate is

1.0 × 105 pulses/s.

40.30. Model: The probability of finding a particle at position x is determined by ψ ( x ) . 2

Visualize:

Solve: (a) Electrons are most likely to arrive at the points of maximum intensity. No electrons will arrive at points of zero intensity. (b) The graph of ψ ( x ) looks just like the classical intensity pattern of single-slit diffraction. 2

(c) The wave function ψ (x) is square root of ψ ( x ) . It oscillates because it alternates between the positive and 2

negative roots.

40.31. Model: The probability of finding a particle at position x is determined by P ( x ) = ψ ( x ) . 2

Visualize:

Solve:

(a) Since the electrons are uniformly distributed over the interval 0 ≤ x ≤ 2 cm, the probability density

P ( x ) = ψ ( x ) is constant over this interval. P(x) = 0 outside this interval because no electrons are detected. 2

Thus ψ ( x ) is a square function, as shown in the figure. To be normalized, the area under the probability curves 2

must be 1. Hence, the peak value of ψ ( x ) must be 0.5 cm−1. 2

(b) The interval is δ x = 0.02 cm. The probability is

Prob(in δ x at x = 0.80 cm) = ψ ( x = 0.80 cm ) δ x = (0.5 cm−1)(0.02 cm) = 0.01 = 1% 2

(c) From Equation 39.7, the number of electrons is N(in δ x at x = 0.80 cm) = NtotalProb(in δ x at x = 0.80 cm) = 106 × (0.01) = 104 (d) The probability density is P(x = 0.80 cm) = ψ ( x = 0.80 cm ) = 0.5 cm−1. 2

40.32. Model: Probability is the area under the probability density curve.

Solve: 5000 of 10,000 electrons are found in the region –1.0 cm ≤ x ≤ 1.0 cm, which spans ∆ x = 2.0 cm, so the probability of being in the region is 50%. Thus

area under the curve = P( x ) ⋅Δ x = P( x )(2.0 cm) = 0.50 ⇒

P( x ) = 0.25 cm −1

2500 electrons are found equally distributed in the two regions –2.0 cm ≤ x ≤ –1.0 cm and 1.0 cm ≤ x ≤ 2.0 cm, so the probability is 0.25. These two regions together span ∆ x = 2.0 cm, thus

area under the curve = P ( x) ⋅Δ x = P ( x)(2.0 cm) = 0.25 ⇒ P( x) = 0.125 cm −1 Similarly, 2500 electrons in the regions –3.0 cm ≤ x ≤ –2.0 cm and 2.0 cm ≤ x ≤ 3.0 cm, which also span ∆ x = 2.0 cm, gives a probability of 0.25 and P(x) = 0.125 cm–1. This information is shown on the probability density graph below.

40.33. Model: The probability of finding a particle at position x is determined by P ( x ) = ψ ( x ) . 2

Visualize:

Solve:

(a) Yes, because the area under the ψ ( x ) curve is equal to 1. 2

(b) There are two things to consider when drawing ψ (x). First ψ (x) is an oscillatory function that changes sign every time it reaches zero. Second, ψ (x) must have the right shape. Each point on the ψ (x) curve is the square root of the

corresponding point on the ψ ( x )

2

curve. The values ψ ( x ) = 1 cm −1 and ψ ( x ) = 0 cm −1 clearly give 2

2

ψ (x) = ±1 cm −1/2 and ψ (x) = 0 cm−1/2 , respectively. But consider x = 0.5 cm, where ψ ( x ) = 0.5 cm −1. Because 2

0.5 = 0.707 , ψ (x = 0.5 cm) = 0.707 cm−1/2. This tells us that the ψ (x) curve is not linear but bows upward over the interval 0 ≤ x ≤ 1 cm. Thus, ψ (x) has the shape shown in the above figure. (c) δ x = 0.0010 cm is a very small interval, so we can use Prob(in δ x at x) = ψ ( x ) δ x. The values of ψ ( x ) 2

can be read from Figure P40.33. Thus, Prob(in δ x at x = 0.0 cm) = ψ ( x = 0.0 cm ) δ x = ( 0.0 cm −1 ) ( 0.0010 cm ) = 0.000 2

Prob(in δ x at x = 0.5 cm) = ψ ( x = 0.5 cm ) δ x = ( 0.5 cm −1 ) ( 0.0010 cm ) = 0.0005 2

Prob(in δ x at x = 0.999 cm) = ψ ( x = 1.0 cm ) δ x = (1.0 cm −1 ) ( 0.0010 cm ) = 0.0010 2

(d) The number of electrons in the interval −0.3 cm ≤ x ≤ 0.3 cm is N(in −0.3 cm ≤ x 0.3 cm) = Ntotal × Prob(in −0.3 cm ≤ x ≤ 0.3 cm) The probability is the area under the probability density curve. We have 0.3 cm

Prob(in −0.3 cm ≤ x ≤ 0.3 cm) =

∫

ψ ( x ) dx = 2 × ( 12 × 0.3 cm × 0.3 cm −1 ) = 0.090 2

−0.3 cm

Thus, the number of electrons expected to land in the interval −0.3 cm ≤ x ≤ 0.3 cm is 10,000 × 0.090 = 900.

2

40.34. Model: The probability of finding a particle at position x is determined by P ( x ) = ψ ( x ) . 2

Visualize:

∞

Solve:

(a) According to Equation 40.18,

∫ ψ ( x)

2

dx = 1. In the interval 0 ≤ x ≤ 0.25 nm, ψ (x) = 4cx, where x is

−∞

in nm, and in the interval 0.25 nm ≤ x ≤ 1 nm, ψ ( x ) = 34 c (1 nm − x ) . Thus, 0.25

∫ 0

0.25

1.0

⎡ x3 ⎤ 16 ⎡ 2 x 2 x3 ⎤ + ⎥ =1 ( 4cx ) dx + ∫ ⎡⎣ c (1 − x )⎤⎦ dx = 1 ⇒ 16c ⎢ ⎥ + c 2 ⎢ x − 9 ⎣ 2 3 ⎦ 0.25 ⎣ 3 ⎦0 0.25 2

1.0

4 3

2

2

After some manipulations, 1 − ⎛ 64 ⎞ 2 2 ⎜ ⎟ c = 1 ⇒ c = 3 nm ⎝ 192 ⎠

We’ve not carried units through the integration, but we know that c has the units of the wave function ψ (ξ ), namely nm−1/2. (b) The particle wave function is ψ ( x) = (4 3) x for the interval 0 nm ≤ x ≤ 0.25 nm and ψ ( x) = ⎡⎣(4/ 3) ⎤⎦ (1 − x) for 2

2

the interval 0.25 nm ≤ x ≤ 1.0 nm. Thus, ψ ( x) = (48) x 2 for 0 nm ≤ x ≤ 0.25 nm and ψ ( x) = (16/3)(1 − x)2 for 0.25 nm ≤ x ≤ 1.0 nm. ψ ( x)

2

has units of nm–1. The graph is shown above.

(c) The particle is most likely to be found at the points where ψ ( x ) is a maximum. The dot picture is shown 2

above. (d) The probability is 0.25 nm

Prob(0.0 nm ≤ x ≤ 0.3 nm) =

∫

0.0 nm 0.25 nm

ψ ( x ) dx + 2

0.3 nm

∫

ψ ( x ) dx 2

0.25 nm 0.3 nm

⎡x ⎤ x3 ⎤ 2 ⎛ 16 ⎞⎡ = 0.25 + 0.14 = 0.39 = ( 48 nm −3 ) ⎢ ⎥ +⎜ nm −3 ⎟ ⎢(1.0 nm ) x − (1.0 nm ) x 2 + ⎥ 3 ⎦ 0.25 nm ⎠⎣ ⎣ 3 ⎦ 0.0 nm ⎝ 3 3

40.35. Model: The probability of finding a particle at position x is determined by the probability density 2 P( x) = ψ ( x) . Solve: (a) The wave function is a straight line passing through the origin such that it is +c at x = + 4 mm and –c at x = – 4 mm. That is, the wave function is

ψ ( x ) = cx/4, where x is in mm and c is in mm −1 2 . Note that the units of c must be that of ψ (x). Thus ∞

∫ ψ ( x ) dx = 2

−∞

∞

∫ ψ ( x)

Because

2

dx = 1,

−∞

4

∫ (c x

2 2

−4

4

16 ) dx = 2∫ ( c 2 x 2 16 ) dx = 0

4

2c 2 ⎡ x 3 ⎤ 8 2 ⎢ ⎥ = c 16 ⎣ 3 ⎦ 0 3

8 2 3 c =1⇒ c = mm −1/ 2 3 8

(b) From part (a), we have

ψ ( x ) = c 2 x 2 16 = 3 x 2 128 = 2

3x 2 ( mm−1 ) 128

A ψ ( x ) -versus-x graph is shown in the figure. 2

(c) The particle is most likely to be found at the positions where ψ ( x ) is a maximum. The graph above gives a 2

dot picture of the first few particles. 2.0 mm

(d) Prob( − 2.0 mm ≤ x ≤ 2.0 mm) = 2

∫

−2.0 mm

2.0

ψ ( x ) dx = 2 ∫ 2

0

2.0

3 3x 2 ⎛ 3 ⎞⎡ x ⎤ dx = 2 ⎜ ⎟ ⎢ ⎥ = 0.125 128 ⎝ 128 ⎠ ⎣ 3 ⎦ 0

40.36. Model: The probability of finding a particle at position x is determined by the probability density P ( x) = ψ ( x) . 2

Solve:

(a) The probability density extends from −4 cm to +4 cm. The area under the P(x)-versus-x graph is 4 cm

∫

ψ ( x ) dx = (4 cm)a = 1 ⇒ a = 0.25 cm−1 2

−4 cm

(b) The particle is most likely to be found at a position where ψ ( x ) is a maximum. This will occur at x = 0 cm 2

because P(x) has its greatest value at x = 0 cm. (c) 75% of the area under the curve occurs between x = −2 cm and 2 cm, so the range is −2 cm ≤ x ≤ 2 cm.

(d)

40.37. Model: The probability of finding a particle at position x is determined by the probability density P ( x) = ψ ( x) . 2

Solve: (a) The wave function ψ (x) = (1.414 nm−1/2)e−x/1 nm changes over a length scale of ≈ 1 nm. The distance δ x = 0.01 nm is very small compared to 1 nm. So we can use Prob(in δ x = 0.01 nm at x = 1 nm) = ψ ( x = 1 nm ) δ x 2

= ⎡⎣(1.414 nm −1/ 2 ) e −1 ⎤⎦ ( 0.01 nm ) = 2e −2 ( 0.01) = 0.0027 = 0.27% 2

(b) The interval 0.5 nm ≤ x ≤ 1.5 nm is not small compared to 1 nm, so we’ll need to integrate 1.5

Prob(0.5 nm ≤ x ≤ 1.5 nm) =

∫ ψ ( x)

0.5

2

1.5

dx = 2 ∫ e−2 x dx = −e−2 x 0.5

1.5 0.5

= 0.318 = 31.8%

40.38. Model: The probability of finding a particle at position x is determined by the probability density 2 P ( x) = ψ ( x) . (a) ψ ( x) = ce x / L for x ≤ 0 nm and ψ ( x) = ce− x / L for x ≥ 0 mm. The probability density will thus be

Solve:

ψ ( x ) = c 2e 2 x / L for x ≤ 0 mm and ψ ( x ) = c 2e−2 x / L for x ≥ 0 mm. With L = 2.0 mm, ψ and ψ 2

2

values of x are displayed in the table below. x (mm) 0 0.5 1.0 −x/L

ce c2e−2x/L

0.78c

c c2

0.61c 0.37c2

0.61c2

2

at various

1.5

2.0

3.0

4.0

5.0

0.47c 0.22c2

0.37c 0.14c2

0.22c 0.05c2

0.14c 0.022

0.08c 0.01c2

(b) Normalization of the wave function requires that ∞

∞

∞

⎛ L⎞

1

∞

1

2 −2 x / L 2 −2 x L −1/ 2 ∫ ψ ( x ) dx = 1 = 2∫0 ψ ( x ) dx = 2∫0 c e dx = 2c ⎜⎝ − 2 ⎟⎠ ⎣⎡e ⎤⎦ 0 ⇒ c = L = 2.0 mm = 0.707 mm −∞ 2

2

(c) The probability is 1.0 mm

Prob(−1.0 mm ≤ x ≤ 1.0 mm) =

∫

ψ ( x ) dx 2

−1.0 mm

1.0 mm

=2

1.0 mm ⎛ L⎞ ⎛ −2.0 mm ⎞ −1 c 2e −2 x / L dx = 2c 2 ⎜ − ⎟ ⎡⎣e−2 x / L ⎤⎦ = 2c 2 ⎜ ⎟ ⎡⎣e − 1⎤⎦ = 0.632 = 63.2% 0 mm 2 2 ⎝ ⎠ ⎝ ⎠ 0 mm

∫

(d) The region –1 mm ≤ x ≤ 1 mm is shaded on the probability density graph.

40.39. Model: The probability of finding a particle at position x is determined by the probability density 2 P ( x) = ψ ( x) . ∞

Solve:

(a) Normalization of the wave function requires that

∫ ψ ( x)

2

dx = 1. Therefore,

−∞ 1

1 ⎡ 3 4 x3 ⎤ = 0.866 cm −1/ 2 1 = c 2 ∫ (1 − x 2 ) dx = c 2 ⎢ x − ⎥ = c 2 ⇒ c = 4 3 ⎦ −1 3 ⎣ −1

(b) The value of ψ (x) decreases from 0.866 at x = 0 cm to 0.75 at x = 0.5 cm and then to 0 cm at x = 1 cm. Thus, the graph is bowed upward over the interval 0 ≤ x ≤ 1 cm. ψ (x) = 0 for x > 1 cm. The graph is also symmetrical about x = 0 cm.

(c) The probability density is ψ ( x ) = c 2 (1 − x 2 ) = 2

3 4

(1 − x ). The value of ψ ( x ) 2

2

decreases from 0.75 at x = 0

cm to 0.56 at x = 0.5 cm and then to 0 at x = 1 cm. This graph is also bowed upward, although not as sharply as

ψ (x). The graph of ψ ( x ) is shown in the above figure. 2

(d) The number of electrons is

N(in 0 cm ≤ x ≤ 0.5 cm) = Ntotal × Prob(in 0 cm ≤ x ≤ 0.5 cm) 0.5 cm

Prob(in 0 cm ≤ x ≤ 0.5 cm) =

∫

0 cm

ψ ( x ) dx = 2

3 4

0.5

2 ∫ (1 − x dx ) = 0

0.5

3⎡ x3 ⎤ ⎢ x − ⎥ = 0.344 4⎣ 3 ⎦0

Thus, the number of electrons detected in the interval 0 cm ≤ x ≤ 0.5 cm is 10,000 × 0.344 = 3440.

40.40. Model: The probability of finding a particle at position x is determined by the probability density 2 P( x) = ψ ( x) . ∞

Solve:

(a) Normalization of the wave function requires that

∫ ψ ( x)

2

dx = 1. By changing variables to θ =

−∞

2πx/L, we find ⎛ 2π x ⎞ 2 L 1 = c 2 ∫ sin 2 ⎜ ⎟ dx = c L 2π ⎝ ⎠ 0 L

2π

∫ sin

2

⎛ L ⎞⎛ 2π ⎞ ⎟⎜ ⎟ ⇒ c= ⎝ 2π ⎠⎝ 2 ⎠

θ dθ = c 2 ⎜

0

2 L

(b) The wave function is ψ ( x ) = 2 L sin ( 2π x L ) for 0 ≤ x ≤ L and ψ (x) = 0, x < 0 m or x > L. Note that the

wave function is nonzero only in the range 0 m ≤ x ≤ L.

(c) The probability density ψ ( x ) = ( 2 L ) sin 2 ( 2π x L ) is zero at x = 0 m, x = 12 L and x = L. The graph of 2

ψ ( x ) is shown in the figure above. 2

(d) The probability is

Prob ( 0 ≤ x ≤ L /3) = c 2

2π / 3

L/3

L L ∫ sin ( 2π x L ) dx = 2 2π ∫ 2

0

0

2π / 3

=

sin 2 θ dθ

1 ⎡θ 1 ⎤ − sin 2θ ⎥ π ⎢⎣ 2 4 ⎦0

⎛ 1 ⎞⎛ π ⎞ = ⎜ ⎟⎜ + 0.2165 ⎟ = 0.402 = 40.2% ⎝ π ⎠⎝ 3 ⎠

40.41. Model: The probability of finding a particle at position x is determined by the probability density 2 P ( x) = ψ ( x) . Solve:

(a) The given probability density means that ψ 1 ( x ) = a (1 − x ) in the range −1 mm ≤ x ≤ 0 mm and

ψ 2 ( x ) = b (1 − x ) in the range 0 mm ≤ x ≤ 1 mm. Because ψ1(x = 0 mm) = ψ2(x = 0 mm),

a = b and thus a =

b. (b) At x = −1 mm, P ( x ) = 12 a; at x = 0 mm, P(x) = a; and at x = 1 mm, P(x) = 0. Furthermore, P(x) is a linear

function of x for 0 ≤ x ≤ 1 mm.

(c) Normalization of the wave function requires integrating over the entire range. We have ∞

1

0 1 ⎡ 0 x2 ⎤ b a ⎛ a ⎞ ∫−∞ P ( x ) dx = −∫1⎝⎜ 1 − x ⎠⎟ dx + ∫0 b (1 − x ) dx = −a ⎣⎡ln (1 + x )⎦⎤ −1 + b ⎣⎢ x − 2 ⎦⎥ = a ln 2 + 2 = a ln 2 + 2 = a ( ln 2 + 12 ) 0

where we have used a = b from part (a). Because

∫ P ( x ) dx = 1 , we have a = b = 1 ( ln 2 + ) = 1 1.193 = 0.838.

(d) The probability is 0 0 0 a 1 ∫−1 P ( x ) dx = −∫11 − x dx = −a ⎡⎣ln (1 − x )⎤⎦ −1 = a ln 2 = 1.193 ( 0.693) = 0.581 = 58.1%

1 2

40.42. Model: A pulse is a wave packet, hence it must satisfy the relation Δ f Δt ≈ 1. Solve:

(a) The wavelength of 600 nm corresponds to a center frequency of f0 =

c

λ0

=

3 × 108 m/s = 5.0 × 1014 Hz 600 × 10−9 m

(b) The pulse duration is 6.0 fs, that is, Δt = 6.0 × 10−15 s. Because the time period of the center frequency is T = f 0 −1 = 2.0 × 10−15 s , the number of cycles in the pulse is

Δt 6.0 × 10−15 s = =3 T 2.0 × 10−15 s (c) The frequency bandwidth for a 6.0-fs-long pulse is Δf =

1 1 = = 1.67 × 1014 Hz Δt 6.0 × 10−15 s

This bandwidth is centered on f0 = 5.00 × 1014 Hz, so the necessary range of frequencies from f0 – 1 2

14

1 2

∆f to f0 +

14

∆f is from 4.17 × 10 Hz to 5.83 × 10 Hz.

(d) The pulse travels at speed c, so the length is ∆ x = c∆t = (3.0 × 108 m/s)(6.0 × 10–15 s) = 1.8 × 10–6 m = 1.8 μm. This is 3λ, in agreement with the finding that there are 3 cycles in the pulse. (e) The graph has three oscillations spanning 1.8 µm = 3λ.

40.43. Model: Electrons are subject to the Heisenberg uncertainty principle. Solve: When we confine a particle in a box, we know the particle is somewhere in the box. So the uncertainty in our knowledge of its position is at most Δ x = L. With a finite Δ x , the uncertainty Δpx cannot be zero. If speed is less than 10 m/s, then the range of velocities is –10 m/s to +10 m/s. Thus Δv x is 20 m/s. The uncertainty principle is Δpx = mΔvx =

h h2 h 6.63 × 10−34 J s = ⇒L= = = 18 μ m 2Δ x L 2m Δvx 2 ( 9.11 × 10−31 kg ) ( 20 m/s )

The smallest box length is 18 μm.

40.44. Model: A dust speck is a particle and is thus subject to the Heisenberg uncertainty principle. Solve: The uncertainty in our knowledge of the position of the dust speck is Δ x = 10 μ m. The uncertainty in the dust speck’s momentum is h / 2 6.63 × 10−34 J s = = 3.32 × 10−29 kg m/s Δ x 2 (10 × 10−6 m )

Δp x =

Equivalently, the uncertainty in the dust particle velocity is

Δv x =

Δpx 3.32 × 10−29 kg m/s = = 3.32 × 10−13 m/s m 1.0 × 10−16 kg

The average velocity is 0 m/s, so the range of possible velocities is –1.66 × 10–13 m/s to +1.66 × 10–13 m/s. The particle could have a top speed of up to 1.66 × 10–13 m/s. The maximum kinetic energy the speck has is 2 1 1 mv 2 = (1.0 × 10−16 kg )(1.66 × 10−13 m/s ) 2 2 = 1.4 × 10−42 J To get out of the hole, the particle would have to acquire potential energy

K=

U = mgh = (1.0 × 10 −16 kg )( 9.8 m/s 2 )(1.0 × 10−6 m ) = 9.8 × 10−22 J

The energy gain needed to get out of the hole is much larger than the available kinetic energy. The particle does not have anywhere near enough kinetic energy that it could, by any process, transform into potential energy and escape. Using K = mgh, the deepest hole from which the dust speck could have a good chance of escaping is h=

Assess:

K 1.4 × 10−42 J = = 1.4 × 10−27 m mg (1.0 × 10−16 kg )( 9.8 m/s 2 )

This is not a very deep hole.

40.45. Model: An atom is a particle and is thus subject to the Heisenberg uncertainty principle. Solve: (a) Since the atom is confined within a box 1 mm in length, the uncertainty in our knowledge of its position is Δx = 1 mm. The uncertainty in the atom’s momentum and velocity are

Δp x =

h 2 6.63 × 10−34 J s = = 3.32 × 10−31 kg m/s Δx 2 (1 × 10−3 m )

Δv x =

Δpx 3.315 × 10−31 kg m/s = = 8.6 × 10−6 m/s m 23 × 1.67 × 10−27 kg

This range of possible velocities will be centered on vx = 0 m/s, so all we can know is that the atom’s velocity is somewhere in the range − 4.3 × 10− 6 m/s ≤ vx ≤ 4.3 × 10− 6 m/s and thus its speed is in the range 0 m/s ≤ v ≤ 4.3 × 10– 6 m/s. 1 (b) With vrms = umax = 2.15 × 10−6 m/s , the lowest possible temperature is 2

( 23 ×1.67 ×10−27 kg )( 2.15 × 10−6 m/s ) = 4 × 10−15 K mv 2 T = rms = 3kB 3 (1.38 × 10−23 J/K ) 2

Assess:

The limit set on temperature by the uncertainty principle is much lower than 1 nK.

40.46. Model: The electron is subject to the Heisenberg uncertainty principle. Model the nucleus as a onedimensional box. Solve: (a) The uncertainty in our knowledge of the position of the electron is Δ x = 10 fm. The uncertainty in the electron’s momentum and velocity are Δp x =

h 2 6.63 × 10−34 J s = = 3.32 × 10−20 kg m/s Δ x 2 (10 × 10−15 m )

Δv x =

Δpx 3.32 × 10−20 kg m/s = = 3.64 × 1010 m/s m 9.11 × 10−31 kg

The range of possible velocities is –1.82 × 1010 m/s to +1.82 × 1010 m/s, so the range of speeds is from 0 m/s to 1.8 × 1010 m/s. (b) The minimum range of speeds for an electron confined to a nucleus exceeds the speed of light, so it is not possible.

40.47. Solve: (a) For a photon, E = hf which means ΔE = hΔ f . Assuming the photon is a wave packet, the relationship that is applicable to a wave packet Δ f Δt ≈ 1 becomes ⎛ ΔE ⎞ ⎜ ⎟ Δt ≈ 1 ⇒ ΔEΔt ≈ h ⎝ h ⎠ (b) The energy of a photon cannot be exactly known. The uncertainty in our knowledge of a photon’s energy depends on the length of time Δ t that is available to measure it. (c) The uncertainty in the energy is

ΔE ≅

h 6.63 × 10−34 J s ≅ = 6.63 × 10−26 J = 4.14 × 10−7 eV Δt 10 × 10−9 s

(d) The energy of the photon is

E=

hc

λ

=

( 6.63 × 10

−34

J s )( 3.0 × 108 m/s )

500 × 10−9 m ⇒

= 3.978 × 10−19 J ×

ΔE 4.14 × 10−7 eV = = 1.7 × 10−7 E 2.49 eV

1 eV = 2.49 eV 1.6 × 10−19 J

40.48. Model: The dust particle is subject to the Heisenberg uncertainty principle. Ignore air-resistance effects. Solve: (a) The uncertainty in the momentum and velocity of the dust particle are

Δp x =

h 2 6.63 × 10−34 J s = = 3.315 × 10−28 kg m/s Δ x 2 (1.0 × 10−6 m )

Δv x =

Δpx 3.13 × 10−28 kg m/s = = 3.32 × 10−13 m/s m 1.0 × 10−15 kg

Thus the range of velocities is –1.66 × 10–13 m/s to +1.66 × 10–13 m/s. Or the range of speeds is 0 m/s to 1.66 × 10–13 m/s. Assuming the dust particle falls freely with the acceleration due to gravity, the time taken by the dust particle to travel a vertical distance of d = 1.0 m is calculated as follows:

yf − yi = −1.0 m = vit − 12 gt 2 = ( 0 m/s ) t − 12 ( 9.8 m/s 2 ) t 2 ⇒ t = 0.45 s With a horizontal velocity of 1.66 × 10−13 m/s, the horizontal distance traveled in this time is dx = Δvxt = 7.5 × 10−14 m. Half the particles move to the right and half to the left, so the diameter of the circle in which they land exceeds 1.0 μm by 1.5 × 10–13 m. This certainly cannot be detected. (b) To get a circle with a 1.1 μm diameter, we need the dust particle to move 0.05 μm in either direction to increase radius from 0.50 μm to 0.55 μm. With the horizontal velocity obtained in part (a), to get this distance under the force of gravity will take t=

0.05 μ m = 3.0 × 105 s 1.66 × 10−13 m/s

We can now find the free-fall distance:

yf − yi = vit + 12 at 2 = ( 0 m/s ) t + 12 ( 9.8 m/s ) ( 3.0 × 105 s ) = 4.4 × 1011 m 2

Assess: This is larger than the diameter of the earth’s orbit around the sun.

40.49. Model: The probability of finding a particle at position x is determined by the probability density 2 P ( x) = ψ ( x) . Solve:

We first need to see if the wave function is normalized: ∞

∫ ψ ( x ) dx =

−∞

2

∞

∞

b b ⎡ 1 −1 x ⎤ 1 ⎡π ⎛ π ⎞⎤ ∫−∞ π ( x 2 + b2 ) dx = π ⎢⎣ b tan b ⎥⎦ −∞ = π ⎢⎣ 2 − ⎝⎜ − 2 ⎠⎟ ⎥⎦ = 1

The function was already normalized. The probability is Prob ( −b ≤ x ≤ b ) =

b

∫ ψ ( x ) dx = 2

−b

=

1

⎡ tan π⎣

−1

b

b

dx b 1 ⎡ −1 x ⎤ tan = ∫ 2 2 b ⎥⎦ − b π −b ( x + b ) π b ⎢⎣ b

(1) − tan −1 ( −1)⎤⎦ =

1 = 50% 2

40.50. Model: The probability of finding a particle at position x is determined by the probability density 2 P ( x) = ψ ( x) . Solve:

(a)

ψ ( x) =

b (1 + x 2 )

= c (1 + x )

− 1 mm ≤ x ≤ 0 mm 2

0 mm ≤ x ≤ 1 mm

=0

elsewhere

Because the value of the wave function must be the same at x = 0 (note that the x = 0 mm point is covered in both the left ( −1 mm ≤ x ≤ 0 mm ) and the right ( 0 mm ≤ x ≤ 1 mm ) parts of the wave function), b 1 + ( 0 mm )

2

= c (1 + 0 mm )

2

⇒b=c (b) So the wave function is

ψ ( x) =

c 1 + ( x2 )

= c (1 + x )

− 1 mm ≤ x ≤ 0 mm 2

0 mm ≤ x ≤ 1 mm

=0

elsewhere

and the probability density is

ψ ( x) = 2

c2 (1 + x 2 )

= c 2 (1 + x )

− 1 mm ≤ x ≤ 0 mm 4

0 mm ≤ x ≤ 1 mm

=0

elsewhere

ψ ( x ) and ψ ( x ) are shown below. 2

(c) Prob ( x ≥ 0 ) =

1 mm

∫

ψ ( x ) dx = 2

0 mm

1 mm

∫

c 2 (1 + x ) dx, when the wave function is normalized to unity. Let us first 4

0 mm

find c to normalize the function: 0

1

c2 4 2 ∫−1 (1 + x2 ) dx + ∫0 c (1 + x ) dx = 1 The first integral is

0

⎡ dx x tan −1 x ⎤ ⎛1 π ⎞ 2 2 = c2 ⎢ + ⎥ = ⎜ + ⎟ c = 0.643c 2 2 2 (1 x ) 2(1 x ) 2 4 8 + + ⎝ ⎠ ⎣ ⎦ −1 −1 0

c2 ∫ The second integral gives

1

⎡ x5 4 x 4 6 x3 4 x 2 ⎤ 31 c ∫ ( x + 4 x + 6 x + 4 x + 1) dx = c ⎢ + + + + x ⎥ = c 2 = 6.20c 2 5 4 3 2 ⎣ ⎦0 5 0 1

2

4

3

2

2

Thus the sum of the two integrals is c 2 ( 6.843) = 1 ⇒ c = 0.382

Finally, Prob ( x ≥ 0 ) =

1 mm

∫

c 2 (1 + x ) dx 4

0 mm 1 mm

= c2

∫ (x

4

+ 4 x 3 + 6 x 2 + 4 x + 1) dx

0 mm

1 mm

5 ⎤ 4 x 4 6 x3 4 x 2 2⎡x = ( 0.38 ) ⎢ + + + + x⎥ 4 3 2 ⎣5 ⎦ 0 mm 2 ⎡1 ⎤ = ( 0.38 ) ⎢ + 1 + 2 + 2 + 1⎥ ⎣5 ⎦ 31 2 ⎛ ⎞ = ⎜ ⎟ ( 0.382 ) = 0.91 = 91% ⎝ 5⎠

40.51. Model: The probability of finding a particle at position x is determined by the probability density P ( x ) = ψ ( x ) . 2

∞

Solve:

∫ ψ ( x)

(a) The normalization condition demands

2

dx = 1. Hence,

−∞ −1

0

∞

1

∞

1

c2 c2 c2 2 2 2 2 2 2 ∫−∞ x2 dx + −∫1 c x dx + ∫0 c x dx + ∫1 x2 dx = 2∫0 c x dx + 2∫1 x 2 dx = 1 1

∞

⎡ x3 ⎤ 2 3 ⎡ 1⎤ ⇒ 2c 2 ⎢ ⎥ + 2c 2 ⎢ − ⎥ = c 2 + 2c 2 = 1 ⇒ c = 8 ⎣ x ⎦1 3 ⎣ 3 ⎦0 (b)

(c) The probability density is

⎛3⎞

ψ ( x ) = ⎜ ⎟ x2 ⎝8⎠ 2

( x ≤ 1 nm )

⎛ 3 ⎞

ψ ( x) = ⎜ 2 ⎟ ⎝ 8x ⎠ 2

( x ≥ 1 nm )

The figure above shows the ψ ( x ) -versus-x graph. 2

(d) The probability is 1.0 nm

Prob(−1.0 nm ≤ x ≤ 1.0 nm) =

∫

1.0 nm

c 2 x 2 dx = 2c 2

−1.0 nm

6

∫

0 nm

2 2⎛ 3⎞ x 2 dx = c 2 = ⎜ ⎟ = 0.25 = 25% 3 3⎝8⎠

5

The number of electrons detected is (0.25)(10 ) = 2.5 × 10 .

40-1

41.1. Model: Model the electron as a particle in a rigid one-dimensional box of length L.

Solve: Absorption occurs from the ground state n = 1. It’s reasonable to assume that the transition is from n = 1 to n = 2. The energy levels of an electron in a rigid box are

En = n 2

h2 8mL2

The absorbed photons must have just the right energy, so hc 3h 2 = ΔEelec = E2 − E1 = Eph = hf = λ 8mL2

⇒L=

3( 6.63 × 10−34 J s )( 6.00 × 10−7 m ) 3hλ = = 7.39 × 10−10 m = 0.739 nm 8mc 8 ( 9.11 × 10−31 kg )( 3.0 × 108 m/s )

41.2. Model: Model the electron as a particle in a rigid one-dimensional box of length L. Solve: (a) The wavelength 1484 nm is in the infrared range. (b) The energy levels of an electron in a rigid box are

En = n 2

h2 8mL2

The emitted photons must have just the right energy, so Eph = hf = ⇒L=

hc

λ

= ΔEelec = E3 − E2 =

5h 2 8mL2

5 ( 6.63 × 10−34 J s )(1484 × 10−9 m ) 5hλ = = 1.5 × 10−9 m = 1.5 nm 8mc 8 ( 9.11 × 10−31 kg )( 3.0 × 108 m/s )

41.3. Model: Model the electron as a particle in a rigid one-dimensional box of length L. Solve:

The energy levels for a particle in a rigid box are

En = n 2

π 2= 2 2 L2

The wave function shown in Figure Ex41.3 corresponds to n = 3. This is also shown in Figure 41.7. Thus, L=

3 ( 6.63 × 10−34 J s ) 3π = 3h = = = 0.75 nm 2mE3 2 2mE3 2 2 ( 9.11 × 10−31 kg )( 6.0 eV × 1.6 × 10−19 J/eV )

41.4. Model: Model the electron as a particle in a rigid one-dimensional box of length L.

Solve: From Equation 41.23, the energies of the stationary states for a particle in a box are En = n2E1, where En is the energy of the stationary state with quantum number n. It can be seen either from Figure 41.7 or from the wave function equation ψ n ( x ) = A sin ( nπ x L ) that the wave function given in Figure Ex41.4 corresponds to n =

4. Thus, E4 = 16 E ⇒ E1 =

E4 12.0 eV = = 0.75 eV 16 16

41.5. Solve: From Equation 41.41, the units of the penetration distance are η=

= 2m (U 0 − E )

⇒

( kg × m2/s2 ) × s = kg × m2/s = kg × m2/s = m J×s = kg × m/s kg × J kg × kg m 2 /s 2 kg 2 × m 2 /s 2

41.6. Solve: (a)

(b) For n = 2, the probability of finding the particle at the center of the well is zero. This is because the wave function is zero at that point. (c) This is consistent with standing waves. The n = 2 standing wave on a string has a node at the center of the string.

41.7. Model: The wave function decreases exponentially in the classically forbidden region. Solve: The probability of finding a particle in the small interval δ x at position x is Prob(in δ x at x) = |ψ ( x) |2 δ x. Thus the ratio

Prob(in δ x at x = L + η ) |ψ ( L + η ) |2 δ x |ψ ( L + η ) |2 = = Prob(in δ x at x = L) | ψ ( L ) |2 δ x |ψ ( L) |2 The wave function in the classically forbidden region x ≥ L is

ψ ( x) = ψ edgee− ( x − L ) /η At the edge of the forbidden region, at x = L, ψ (L) = ψ edge. At x = L + η, ψ (L + η) = ψ edgee –1. Thus −1 2 Prob(in δ x at x = L + η ) |ψ ( L + η ) |2 (ψ edgee ) = = = e−2 = 0.135 Prob(in δ x at x = L) | ψ ( L ) |2 (ψ edge ) 2

41.8. Solve: (a) According to Equation 41.41, the penetration distance is η=

= 2m (U 0 − E )

=

1.05 ×10−34 J s

2 ( 9.11×10−31 kg ) ( 2.0 eV − 0.5 eV ) ×1.60 ×10−19 J/eV

= 0.159 nm

(b) Likewise for E = 1.00 eV, η = 0.195 nm. (c) For E = 1.50 eV, η = 0.275 nm. Assess: These values are of the correct order of magnitude as you can see by referring to Figure 41.14 (a).

41.9. Solve: According to Equation 41.41, the penetration depth is η = = U0 − E =

2m (U 0 − E ) . Hence,

(1.05 ×10−34 J s ) =2 1 eV = = 6.05 ×10−21 J × = 0.038 eV 2 2 − − 31 9 2mη 1.6 ×10−19 J 2 ( 9.11×10 kg )(1.0 ×10 m ) 2

The electron’s energy is 0.038 eV below U 0 .

41.10. Solve: According to Equation 41.41, the penetration depth is η=

= 2m (U 0 − E )

=

2 ( 4 ×1.661×10

1.05 ×10−34 J s −27

kg )(1.0 eV ×1.60 × 10

−19

J/eV )

= 2.28 × 10−12 m = 2.28 pm

41.11. Visualize:

Solve: There are three factors to consider. First, the de Broglie wavelength increases as the particle’s speed and kinetic energy decreases. Thus, the spacing between the nodes of ψ (x) increases in regions where U is larger. Second, a particle is more likely to be found where it is moving the slowest. Thus, the amplitude of ψ (x) increases in regions where U is larger. Third, for n = 6 there will be six antinodes to place.

41.12. Visualize:

Solve: There are three factors to consider. First, the de Broglie wavelength increases as the particle’s speed and kinetic energy decreases. Thus, the spacing between the nodes of ψ (x) increases in regions where U is larger. Second, a particle is more likely to be found where it is moving the slowest. Thus, the amplitude of ψ (x) increases in regions where U is larger. Third, for n = 8 there will be eight antinodes to place.

41.13. Visualize:

Solve: (a) The energy diagram is shown above. (b) There are three factors to consider. First, the de Broglie wavelength increases as the particle’s speed and kinetic energy decreases. Thus, the spacing between the nodes of ψ (x) increases in regions where U is larger. Second, a particle is more likely to be found where it is moving the slowest. Thus, the amplitude of ψ (x) increases in regions where U is larger. Third, n = 3 has 3 antinodes and n = 6 has six antinodes.

41.14. Visualize:

The steps of Tactics Box 41.1 have been followed to sketch the wave functions shown in the figure.

41.15. Model: The electron is a quantum harmonic oscillator.

Solve: Using Equation 41.48 for the energy levels of the electron, the energy of the photon emitted in the 3 → 2 and 3 → 1 quantum jumps are 1⎞ 1⎞ hc ⎛ ⎛ E3 − E2 = ⎜ 3 − ⎟ =ωe − ⎜ 2 − ⎟ =ωe = =ωe = 2⎠ 2⎠ λ32 ⎝ ⎝ 1⎞ hc ⎛ ⎛ 1⎞ E3 − E1 = ⎜ 3 − ⎟ =ωe − ⎜ 1 − ⎟ =ωe = 2=ωe = λ 2 2 ⎝ ⎠ ⎝ ⎠ 31

⇒ λ31 =

λ32 2

=

300 nm = 150 nm 2

41.16. Model: The electron is a quantum harmonic oscillator. Solve:

(a) The energy levels of a harmonic oscillator are E = ( n − 12 ) ω =

1 2

ω , 32 ω , 52 ω , … The classical

angular frequency of a mass on a spring is

ω=

2.0 N/m k = = 1.48 × 1015 rad/s m 9.11 × 10−31 kg

⇒ ω = (1.05 × 10−34 J s )(1.48 × 1015 rad/s ) = 1.56 × 10−19 J = 0.972 eV The first three energy levels are E1 = 0.49 eV, E2 = 1.46 eV, and E3 = 2.43 eV. (b) The photon energy equals the energy lost by the electron: Ephoton = ΔEelec = E3 – E1 = 1.94 eV = 3.10 × 10−19 J. The wavelength is

λ=

( 6.63 ×10−34 J s )( 3.0 ×108 m/s ) = 640 nm c hc = = f ΔEelec 3.10 ×10−19 J

41.17. Model: See Example 41.9. Visualize:

For a harmonic oscillator the energy difference between adjacent energy levels is ΔE = =ωe . We also

know the emitted photon has energy Ephoton = hf photon = ΔE. Combine these to get ωe = 2π λc . Solve:

We also recall that ωe =

k m

. Solve this for k and substitute for ωe . 2

2

⎛ c⎞ 3.0 × 108 m/s ⎞ ⎛ k = mωe2 = m⎜ 2π ⎟ = (9.11 × 10−31 kg)⎜ 2π ⎟ = 2.25 N/m −9 ⎝ λ⎠ ⎝ 1200 × 10 m ⎠ Assess: This answer is one-fourth of the answer in Example 41.9, which makes sense. An electron would emit a 1200 nm photon in any n → n − 1 jump in this quantum harmonic oscillator; not just the 3 → 2 jump.

41.18. Model: The electron is a quantum harmonic oscillator. Solve:

The energy levels of the quantum harmonic oscillator are En = (n –

1 2

) =ω. The longest wavelength of

light that can be absorbed is in a transition from level n to the next level n + 1. Thus Ephoton = hf =

⇒ λ = 2π c

hc

λ

= ΔEelectron = En +1 − En = =ω =

h 2π

k m

m 9.11 × 10−31 kg = 2π (3.0 × 108 m/s) = 5.19 × 10−7 m = 519 nm k 12.0 N/m

41.19. Model: The electron is a quantum harmonic oscillator. The given levels are adjacent levels. Solve:

Let the two adjacent levels be n and n + 1. From Equation 41.48,

En = ( n − 12 ) =ω = 2.0 eV

⇒

En +1 = ( n + 1 − 12 ) =ω = 2.8 eV

2.0 eV 2.8 eV = ⇒ 2.0n + 1.0 = 2.8n – 1.4 ⇒ n = 3 n − 12 n + 12

Thus, E3 = ( 3 − 12 ) =ω = 2.5=ω = 2.0 eV. Using Equation 41.43 for the angular frequency, 2 ⎡ ( 2.0 eV ) (1.6 × 10−19 J/eV ) ⎤ k ⎛ 2.0 eV ⎞ −31 ⎥ = 1.4 N/m = 2.0 eV ⇒ k = m ⎜ 2.5= ⎟ = ( 9.11 × 10 kg ) ⎢ −34 m ⎝ 2.5= ⎠ ⎢⎣ ( 2.5 ) (1.05 × 10 J s ) ⎥⎦ 2

41.20. Solve: Electrons are bound inside metals by an amount of energy called the work function E0. This is

the energy that must be supplied to lift an electron out of the metal. In our case, E0 is the amount of energy (U0 – E) appearing in Equation 41.41 for the penetration distance. Thus,

η=

(1.05 ×10 J s ) kg )( 4.0 eV ×1.60 ×10 −34

= 2m (U 0 − E )

=

2 ( 9.11×10−31

−19

J/eV )

= 9.72 ×10−11 m

The probability that an electron will tunnel through a w = 4.54 nm gap (from a metal to an STM probe) is

Ptunnel = e −2 w η = e

(

−2 0.45×10−9 m

) ( 9.72×10−11 m )

= 9.5 ×10−5 = 0.0095%

41.21. Solve: (a) The probability of an electron tunneling through a barrier is Ptunnel = e −2 w η

η=

⇒ ln Ptunnel = −2 w η ⇒ [ ln Ptunnel ] = 4 w2 2

−31

2m (U 0 − E )

2m (U 0 − E ) =2 2 ⇒ E = U0 − [ln Ptunnel ] 2 = 8mw2

(1.05 ×10 J s ) ⇒ E = 5.0 eV − 8 ( 9.11 × 10 kg )(1.0 × 10 −34

=

2

−9

m)

2

1 eV 2 [ln Ptunnel ] −19 (1.60 ×10 J )

= 5.0 eV − ( 0.009455 eV ) [ ln Ptunnel ]

2

For Ptunnel = 0.10, E = 5.0 eV – 0.050 eV = 4.95 eV. (b) For Ptunnel = 0.010, E = 5.0 eV – 0.20 eV = 4.80 eV. (c) For Ptunnel = 0.0010, E = 5.0 eV – 0.45 eV = 4.55 eV.

41.22. Model: Model the droplet as a particle in a one-dimensional rigid box of length L. Solve:

(a) The droplet’s mass is

m = ρ waterV = ρ water ( 34 π r 3 ) = (1000 kg/m3 ) ( 34 π ) (1.0 ×10−6 m ) = 4.2 × 10−15 kg 3

The droplet’s energy, which is entirely kinetic energy, is E = K = 12 mv 2 =

1 2

( 4.2 ×10

−15

kg )(1.0 ×10−6 m/s ) = 2.1× 10−27 J 2

From Equation 41.22, 8 ( 4.2 × 10−15 kg ) ( 20 μ m ) ( 2.1× 10−27 J ) n2h2 8mL2 E n ⇒ = = ≈ 2.5 ×108 = 250,000,000 2 −34 h2 8mL2 ( 6.63 ×10 J s ) 2

En =

(b) The correspondence principle says that the behavior of a system approaches the classical limit as the quantum number n → ∞. Even for a droplet this small and slow, n is so large (>> 1) that we can safely use classical physics to describe its motion.

41.23. Solve: A function ψ (x) is a solution to the Schrödinger equation if d 2ψ 2m = − 2 [ E − U ( x) ]ψ ( x) dx 2 = Let ψ (x) = Aψ 1(x) + Bψ 2(x), where ψ 1(x) and ψ 2(x) are both known to be solutions of the Schrödinger equation. The second derivative of ψ (x) is d 2ψ d2 d 2ψ d 2ψ 2 = 2 ( Aψ 1 ( x) + Bψ 2 ( x) ) = A 21 + B 2 dx dx dx dx 2 Since ψ 1(x) and ψ 2(x) are solutions, it must be the case that d 2ψ 1 2m = − 2 [ E − U ( x) ]ψ 1 ( x) and = dx 2

d 2ψ 2 2m = − 2 [ E − U ( x) ]ψ 2 ( x) = dx 2

Using these results, the second derivative of ψ (x) becomes d 2ψ d 2ψ 1 d 2ψ 2 ⎛ 2m ⎞ ⎛ 2m ⎞ A B = + = A ⎜ − 2 [ E − U ( x) ]ψ 1 ( x) ⎟ + B ⎜ − 2 [ E − U ( x)]ψ 2 ( x) ⎟ 2 2 2 dx dx dx ⎝ = ⎠ ⎝ = ⎠ =−

2m [ E − U ( x)]( Aψ 1 ( x) + Bψ 2 ( x) ) =2

=−

2m [ E − U ( x)]ψ ( x) =2

Thus ψ (x) is a solution to the Schrödinger equation.

41.24. Model: Model the hydrogen atom as a one-dimensional Coulomb potential energy. Solve:

41.25. Model: Model the particle as a particle in a rigid one-dimensional box of length L. Solve:

(a) From Equation 41.22, the particle’s energies are

En =

n2h2 3h 2 h2 ⇒ E2 − E1 = 22 − 12 ) = 2 2 ( 8mL 8mL 8mL2

Since E2 − E1 = hf = hc λ2 →1 , we have λ2 →1 = 8mcL2 3h . (b) The length of the box is L=

3 ( 6.63 ×10−34 J s )( 694 ×10−9 m ) 3hλ2→1 = = 0.795 nm 8mc 8 ( 9.11×10−31 kg )( 3.0 ×108 m/s )

41.26. Solve: (a) The energy levels of a particle in a rigid box are

( 6.63 ×10−34 J s ) ⎛ h2 ⎞ 1 eV = n2 × = ( 37.70 eV ) n 2 En = n ⎜ 2 2 ⎟ −31 −10 8 1.60 ×10−19 J mL ⎝ ⎠ 8 ( 9.11×10 kg )(1.0 ×10 m ) 2

2

So, E1 = 37.7 eV, E2 = 4E1 = 151 eV, E3 = 9E1 = 339 eV, and E4 = 16E1 = 603 eV. (b) The emission spectrum will contain the following transitions λ2→1, λ3→1, λ3→2, λ4→1, λ4→2, and λ4→3. Since En − Em = hc λn → m , −31 −10 8 ⎛ 8mL2 ⎞ ( 3.0 ×10 m/s ) ( 8 ) ( 9.11×10 kg )(1.0 ×10 m ) hc 3.298 ×10−8 m = 2 = = ⎜ ⎟ 2 n2 − m2 ( n − m2 ) ⎝ h2 ⎠ ( n2 − m2 )( 6.63 ×10−34 J s ) 2

λn→m

So, λ2→1 = 11.0 nm, λ3→1 = 4.12 nm, λ3→2 = 6.59 nm, λ4→1 = 2.20 nm, λ4→2 = 2.75 nm, and λ4→3 = 4.71 nm. (c) All are in the ultraviolet range. (d) Although the stationary states of this model have positive energies, unlike the Bohr atom which has negative energies, both models have E1 < E2 < E3 < E4. Thus, the transition energies, which involve differences of two energy values, are not changed at all. All energy levels are relative to an arbitrary zero of energy so the difference is not significant. (e) The two models are similar in the sense that both lead to quantized energy levels. However, whereas the electron in the Bohr model undergoes orbital motion around the nucleus, the electron in a rigid box undergoes translational motion. The energy levels in the Bohr model get closer together as n increases, whereas the energy levels of the box model get further apart as n increases.

41.27. Solve: From Equation 41.20, the wave functions for a particle in a box of length L are ψ n = An sin ( nπ x L ) . The wave function is nonzero only for 0 ≤ x < L. The normalization requirement is

∫

∞ −∞

L 2 ⎛ nπ x ⎞ ψ ( x ) dx = An2 ∫ sin 2 ⎜ ⎟ dx = 1 0 ⎝ L ⎠

Change the variable to u = nπ x/L. Then, dx = (L/nπ )du. The integration limits become u = 0 at x = 0 and u = nπ at x = L. The normalization integral, with the use of sin 2 u = ( 12 ) (1 − cos 2u ) , becomes 1 = An2

L nπ

∫

nπ 0

sin 2 u du = An2

L nπ

∫

nπ 0

1 2

(1 − cos 2u ) du = An2

nπ L 1 LA2 ⎡⎣ 2 u − 14 sin 2u ⎤⎦ 0 = n ⇒ An = nπ 2

2 L

41.28. Model: Model the particle as being confined in a rigid one-dimensional box of length L. Visualize:

Solve: (a) From Equation 41.22, the energy levels are En = n2h2/8mL2. Two adjacent energy levels have the energy ratio

En +1 ( n + 1) n +1 = ⇒ = 2 En n n 2

En +1 51.4 MeV 5 = = 1.25 = ⇒ n = 4 and n + 1 = 5 En 32.9 MeV 4

(b) We have En = n2E1, so E1 = 32.9 MeV/16 = 2.06 MeV. We can then find E2 = 8.2 MeV and E3 = 18.5 MeV. (c) ψ 5 has five antinodes and is zero at x = 0 fm and x = L. (d) The photon energy is Ephoton = hf = hc λ = ΔEnm . Hence,

λ=

−15 8 hc ( 4.14 ×10 eV s )( 3.00 ×10 m/s ) = = 6.71×10−5 nm ΔE 51.4 ×106 eV − 32.9 × 106 eV

This is a factor 107 smaller than typical visible-light wavelengths. (e) Using E4 = 32.9 MeV = 5.26 × 10−12 J, 2 ( 6.63 ×10−34 J s ) 42 h 2 = = 1.67 × 10−27 kg 8 E4 L2 ( 5.26 ×10−12 J )(10 ×10−15 m )2 2

m=

This is either a proton or a neutron.

41.29. Model: Model the particle as being confined in a rigid one-dimensional box of length L. Visualize:

(a) The probability density is ψ n ( x ) = ( 2 L ) sin 2 ( nπ x L ) . Graphs of ψ 1 ( x ) , 2

Solve:

2

ψ 2 ( x ) , and 2

ψ 2 ( x ) are shown above. 2

(b) The particle is most likely to be found at x where ψ ( x ) is a maximum. See table in part (d). 2

(c) The particle is least likely to be found at x where ψ ( x ) = 0. See table in part (d). 2

(d) The probability of finding the particle in the left one-third of the box is the area under the ψ ( x ) curve 2

between x = 0 and x = 13 L. From examining the graphs, we can determine whether this is more than, less than, or equal to one-third of the total area. The results are shown in the table below. n 1

Most likely 1 2L 1 4

2 3

1 6

L,

L and 3 6

3 4

Least likely 0 and L 0, 12 L, and L

L

L, and

5 6

L

0,

1 3

L,

2 3

Probability in left one-third < 13 > 13

L, and L

= 13

(e) The probability of finding the particle in the range 0 ≤ x ≤ 13 L is

Prob ( 0 ≤ x ≤ 13 L ) = ∫

L/3 0

ψ n ( x ) dx = 2

2 L/3 2 ⎛ nπ x ⎞ sin ⎜ ⎟ dx L ∫0 ⎝ L ⎠

Change the variable to u = nπ x/L. Then, dx = (L/nπ )du. The integration limits become u = 0 at x = 0 m and u = nπ /3 at x = 13 L. Then, Prob ( 0 ≤ x ≤ 13 L ) =

2 nπ

∫

nπ /3 0

sin 2 udu =

nπ /3 2 1 1 1 ⎛ 2nπ ⎞ sin ⎜ ⎡ u − 14 sin 2u ⎦⎤ 0 = − ⎟ nπ ⎣ 2 3 2nπ ⎝ 3 ⎠

The probability is 0.195 for n = 1, 0.402 for n = 2, and 0.333 for n = 3. Assess: The results agree with the earlier estimates of the probability.

41.30. Visualize: Please refer to Figure 41.16. Solve:

Finding the electron in the GaAlAs is equivalent to finding the electron in the classically-forbidden

region outside the GaAs potential well. The probability of this is the area under the ψ ( x ) curve that is outside 2

the potential well. Simply by inspecting the graph, we can estimate that it is ≈10% of the total area. So, the probability that the electron will be in one of the GaAlAs layers is ≈10%.

41.31. Model: The nucleus can be modeled as a potential well. Visualize:

Please refer to Figure 41.17. Solve: The gamma ray wavelength λ = 1.73 × 10−4 nm corresponds to a photon energy of Ephoton = hc/λ = 7.2 MeV. From Fig. 41.17, we can see that a photon of this energy is emitted in a transition from the n = 2 to n = 1 energy level. This can happen after a proton-nucleus collision if the proton’s impact excites the nucleus from the n = 1 ground state to the n = 2 excited state. To cause such an excitation, the proton’s kinetic energy at the instant of impact must be K ≥ 7.2 MeV. Let v1 be the proton’s initial speed at the distance r1 ≈ ∞. If v1 is the minimum speed that can excite the n = 2 state in the nucleus, then the proton has K2 = 7.2 MeV at the distance r2 equal to the radius of nucleus (4 fm). Its potential energy at this point is the electrostatic potential energy between the proton of charge +e and the nucleus of charge +Ze, with Z = 13. The conservation of energy equation K1 + U1 = K2 + U2 is 1 2

mv12 + 0 J = K 2 +

Ze 2 4πε 0 r

⎛ −19 Ze 2 ⎞ 2⎛ 2 ⎜ ( 7.2 MeV ) ⎛⎜ 1.60 × 10 ⇒ v1 = ⎜ K2 + ⎟= m⎝ 4πε 0 r ⎠ 1.67 × 10−27 kg ⎜ 1 eV ⎝ ⎝

2 −19 9 2 2 ⎞ J ⎞ ( 9.0 × 10 Nm /C )13 (1.6 × 10 C ) ⎟ + ⎟ ⎟ 4.0 × 10−15 m ⎠ ⎠

= 4.77 × 107 m/s This is the minimum speed, so any v1 ≥ 4.77 × 107 m/s can cause the emission of a gamma ray.

41.32. Solve: From Fig. 41.24, we see that absorption due to the C = O bond is at λ = 5.8 μm. The room-

temperature oscillating atoms spend most of their time in the n = 1 ground state, so the absorption occurs due to a 1 → 2 quantum jump. Because E1 = 12 =ω and E2 = 32 =ω ,

ΔE = E2 − E1 = =ω = Ephoton = hf = =

( 6.63 ×10

−34

hc

λ

J s )( 3.0 × 10 m/s ) 8

5.8 ×10

−6

m

×

1 eV = 0.214 eV 1.6 × 10−19 J

Thus the first three energy levels are E1 = 12 =ω = 0.107 eV

E2 = 32 =ω = 0.321 eV

E3 = 52 =ω = 0.535 eV

41.33. Solve: From Equations 41.45 and 41.46, ψ 1 ( x ) = A1e − x

2 2b 2

, with b2 = h/mω and ω = k m . This is a

solution to the Schrödinger equation if it’s true that

d 2ψ 1 2m ⎡ 1 ⎤ = − 2 ⎢ E − kx 2 ⎥ψ 1 ( x ) 2 dx 2 = ⎣ ⎦ where we’ve used U ( x ) = 12 kx 2 . This equality is a hypothesis to be tested. To do so, we need the second derivative of ψ 1(x):

dψ 1 ( x ) d 2ψ 1 ( x ) ⎛ 1 x2 ⎞ 2 2 A A1 − x 2 2b 2 A1 2 − x 2 2b 2 e x e = − 21 xe − x 2b ⇒ = − + = − ⎜ 2 − 4 ⎟ψ 1 ( x ) dx b dx 2 b2 b4 ⎝b b ⎠ In the last step we used the definition of ψ 1(x). Using the definition of b2 and ω, this equation is d 2ψ 1 ( x ) ⎛ mω m 2ω 2 x 2 ⎞ ⎛ mω mkx 2 ⎞ 2m x = − − = − − 2 ⎟ψ 1 ( x ) = − 2 ( 12 =ω − 12 kx 2 )ψ 1 ( x ) ψ ( ) ⎜ ⎟ 1 ⎜ dx 2 =2 ⎠ = ⎠ = ⎝ = ⎝ = Does this result for d2ψ1(x)/dx2 equal − ( 2m = 2 )( E − 12 kx 2 )ψ 1 ( x ) , as required by the Schrödinger equation? We can see by inspection that they are equal if E = 12 =ω. But this is the correct ground-state energy for the quantum harmonic oscillator. So we’ve shown that ψ1(x) is a solution of the Schrödinger equation.

41.34. Solve: (a) From Equation 41.46, b = = mω . The units of b are J×s kg × m 2 s 2 = =m kg × s −1 s 2 kg

(b) The classical turning point is the point where U = E and K = 0 J. Since U = 12 kx 2 and E = E1 = 12 =ω , the

classical turning point for an oscillator in the n = 1 state is =ω =ω = 1 2 1 kxtp = =ω ⇒ xtp = ± =± =± = ±b 2 ωm ωm k 2 2

41.35. Solve: (a) The ground-state wave function of the quantum harmonic oscillator is ψ 1 ( x ) = A1e − x

2

2 b2

.

Normalization requires

∫

∞ −∞

∞

ψ 1 ( x ) dx = A12 ∫ e − x 2

2/ 2 b 2

−∞

dx = 1

Change the variable to u = x/b. Then, dx = bdu. The integration limits don’t change, so ∞

1 = bA12 ∫ e− u du 2

−∞

The definite integral can be looked up in a table of integrals. The result is

1 = bA12 π = A12 π b 2 ⇒ A1 =

π . Hence,

1 (π b 2 )1/ 4

(b) The forbidden region is both x < −b and x > b. ψ 1 ( x ) is symmetrical about x = 0 m, so 2

∞

2

Prob(x < −b or x > b) = (2)Prob(x > b) = 2 ∫ ψ 1 ( x) dx = 2

πb

b

2

∫

∞ b

e− x

2/b 2

dx

(c) The integral of part (b) cannot be evaluated in closed form, but the answer can be found with a numerical integration. First, change the variable to u = x/b, making dx = bdu. But unlike the variable change in part (c), this does change the lower limit of integration. Thus,

Prob(x < −b or x > b) =

2

πb

∞

2

b ∫ e − u du = 1

2

2

π ∫

∞ 1

2

e− u du

The definite integral can be evaluated numerically with a calculator or computer, giving

∫

∞ 1

2

e − u du = 0.139

The probability of finding the harmonic oscillator in the forbidden region is 2 (π )

− 12

( 0.139 ) = 0.157 = 15.7% .

41.36. Solve: (a) The classical probability density of finding the oscillator of mass m at position x is given by Equation 41.32: 2 1 2 ω = = Pclass ( x ) = Tv ( x ) 2π ω v ( x ) π v ( x ) where v(x) is the velocity as a function of x, and T is the period of oscillation. From Chapter 14, the energy of a harmonic oscillator is

E = 12 kA2 = 12 mω 2 A2 = 12 mv 2 + 12 kx 2 = 12 mv 2 + 12 mω 2 x 2 ⇒ v ( x ) = ω 2 A2 − ω 2 x 2 = ω A2 − x 2 ⇒ Pclass ( x ) =

ω πω A − x 2

2

=

1

π A2 − x 2

(b)

(c) The oscillator is most likely to be found near the turning points at x = ± A. This is because v → 0 m/s at A and the oscillator spends more time at these points. For the same reason, the oscillator spends the least amount of time around x = 0 m and thus the probability density is the least in the center.

41.37. Model: The collisions with the ground are perfectly elastic. Solve: (a) The classical probability density at position y of finding a ball that bounces between the ground and height h is given by Equation 41.32: 2 Pclass ( y ) = Tv( y )

where v(y) is the ball’s velocity as a function of y and T is the period of oscillation. For a freely falling object, energy conservation gives

mgh = 12 mv 2 + mgy ⇒ v ( y ) = 2 g ( h − y ) The time t = 12 T to reach a height h after a collision with the ground can be found from kinematics:

Δy = h = 12 gt 2 ⇒ t =

⇒ Pclass ( y ) =

2 2 2h g

1 2g ( h − y )

=

g 2h

2h g

1 2g ( h − y )

=

1 1 ⎛ 1 ⎞ =⎜ ⎟ 2 h h − y ⎝ 2h ⎠ 1 − ( y h )

(b)

(c) The ball is most likely to be found near the upper turning point at y = h. This is because v → 0 m/s at y = h so the ball spends more time at this point. For the same reason, the ball spends the least time near the ground, where it is moving fastest, and thus the probability density is the least at y = 0 m.

41.38. Model: We model the situation as a finite square well as shown in Figure 43.9. Visualize: As shown in Figure 43.9, the potential well has a depth of about 50 MeV, but the neutron in this problem is only 20 MeV below the top of the potential well; that is, U 0 − E = 0 MeV − (−20 MeV) = 20 MeV = 3.2 × 10−12 J. We look up the mass of a neutron: m = 1.67 ×10−27 kg. Solve: The penetration distance is given in Equation 41.41

η=

= = 2m(U 0 − E )

1 . 05 × 10−34 J ⋅ s 2(1 . 67 × 10−27 kg)(3 . 2 × 10−12 J)

= 1 . 0 fm

This is 1/4 of the nuclear radius. Assess: This shows that a nucleus is fairly fuzzy since the penetration distance is 1/4 of the radius.

41.39. Solve: Electrons are bound inside metals by an amount of energy called the work function E0. This is

the energy that must be supplied to lift an electron out of the metal. In this case, the expression U0 – E that appears in Equation 41.41 is E0. Thus, the penetration distance is

η=

= 2m (U 0 − E )

=

2 ( 9.11×10

1.05 ×10−34 J s −31

kg ) ( 4.3 eV ) (1.6 × 10

−19

J/eV )

= 9.38 × 10−11 m

From Equation 41.53, the probability that an electron will tunnel between the two aluminum pieces is

Ptunnel = e −2 w η = e

(

−2 50×10−9 m

) ( 9.38×10−11 m )

= e −1066 = 10−463 ≈ 0

41.40. Solve: From Equation 41.41, the penetration distance is η=

= 2m (U 0 − E )

(1.05 ×10 kg )(1.0 × 10

−34

=

2 (1.67 ×10

−27

6

J s)

eV )(1.60 × 10

−19

J/eV )

The probability the proton will tunnel through the barrier is

Ptunnel = e −2 w η = e

(

−2 10×10−15 m

) ( 4.54×10−15 m )

= 0.012 = 1.2%

= 4.54 ×10−15 m

41.41. Model: A particle is confined in a rigid one-dimensional box of length L. Visualize:

Solve: (a) ψ (x) is zero because it is physically impossible for the particle to be there because the box is rigid. (b) The potential energy within the region −L/2 ≤ x ≤ L/2 is U(x) = 0 J. The Schrödinger equation in this region is

d 2ψ ( x ) 2m = − 2 Eψ ( x ) = − β 2ψ ( x ) 2 dx where β = 2mE

2

.

(c) Two functions ψ (x) that satisfy the above equation are sinβ x and cosβ x . A general solution to the Schrödinger equation in this region is ψ (x) = Asinβ x + Bcosβ x

where A and B are constants to be determined by the boundary conditions and normalization. (d) The wave function must be continuous at all points. ψ = 0 just outside the edges of the box. Continuity requires that ψ also be zero at the edges. The boundary conditions are ψ (x = −L/2) = 0 and ψ (x = L/2) = 0. (e) The two boundary conditions are ⎛ βL ⎞ ⎛ βL ⎞ ⎛ βL ⎞ ⎛ βL⎞ ⎟ + B cos ⎜ − ⎟ = − A sin ⎜ ⎟ + B cos ⎜ ⎟=0 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠

ψ ( − L 2 ) = A sin ⎜ −

⎛ βL ⎞ ⎛ βL ⎞ ⎟ + B cos ⎜ ⎟=0 2 ⎝ ⎠ ⎝ 2 ⎠

ψ ( L 2 ) = A sin ⎜

These are two simultaneous equations. Unlike the boundary conditions in the particle in a box problem of Section 41.4, there are two distinct ways to satisfy these equations. The first way is to add the equations. This gives ⎛ βL ⎞ 2 B cos ⎜ ⎟ = 0 ⇒ B = 0 ⇒ ψ (x) = Asinβ x ⎝ 2 ⎠ To finish satisfying the boundary conditions, ⎛ βL ⎞ sin ⎜ ⎟ = 0 ⇒ β L = 2π , 4π , 6π , … = 2nπ with n = 1, 2, 3, … ⎝ 2 ⎠

With this restriction on the values of β , the wave function becomes ψ ( x ) = A sin ( 2nπ x L ) . Using the definition of β from part (b), the energy is En =

( 2nπ )

2

2

2

2mL

= ( 2n )

2

h2 8mL2

n = 1, 2, 3, …

The second way is to subtract the second equation from the first. This gives ⎛ βL ⎞ −2 A cos ⎜ ⎟ = 0 ⇒ A = 0 ⇒ ψ (x) = Bcosβ x ⎝ 2 ⎠ To finish satisfying the boundary conditions,

⎛ βL ⎞ cos ⎜ ⎟ = 0 ⇒ β L = π , 3π , 5π , … = (2n – 1)π ⎝ 2 ⎠

n = 1, 2, 3, …

With this restriction on the values of β , the wave function becomes ψ ( x ) = B cos ( 2 ( n − 1) π x L ) . Using the definition of β , the energy is En =

2 ( 2n − 1) π 2 2

2mL

2

= ( 2n − 1)

2

h2 8mL2

n = 1, 2, 3, …

Summarizing this information, the allowed energies and the corresponding wave functions are 2 ⎧ ⎛ ( 2n − 1) π x ⎞ 2 h = E1 , 9 E1 , 25E1 , … ⎪ B cos ⎜ ⎟ En = ( 2n − 1) L 8mL2 ⎪ ⎝ ⎠ ψ ( x) = ⎨ 2 2 h ⎪ A sin ⎛ ( 2n ) π x ⎞ = = 4 E1 , 16 E1 , 36 E1 , … E n 2 ( ) ⎜ ⎟ n ⎪ 8mL2 ⎝ L ⎠ ⎩

where E1 = h2/8mL2. (f) The results are actually the same as the results for a particle located at 0 ≤ x ≤ L. That is, the energy levels are the same and the shapes of the wave functions are the same. This has to be, because neither the particle nor the potential well have changed. All that is different is our choice of coordinate system, and physically meaningful results can’t depend on the choice of a coordinate system. The new coordinate system forces us to use both sines and cosines, whereas before we could use just sines, but the shapes of the wave functions in the box haven’t changed.

41.42. Model: The electron is in the finite potential well whose energies and wave functions are shown in Figure 41.14a. Solve: (a) From Equations 41.40 and 41.41, the wave function as it penetrates into a barrier (with edge at x = L) is ψ ( x ) = ψ edgee −( x − L ) η , where η is the penetration distance. The ratio of the probability densities at x = d + L and x = L is 2

ψ edge e−2 d η P (d + L) ψ (d + L) = = = e−2 d η 2 2 P ( L) ψ ( L) ψ edge e0 2

To have energy −2.7 eV compared to a free electron means that an electron in the sodium metal is bound inside the metal by U0 – E = 2.7 eV, where U0 is the potential energy “barrier” of the surface of the metal. The penetration distance of the electron is

η=

= 2m (U 0 − E )

=

1.05 × 10−34 J s 2 ( 9.11 × 10

−31

kg )( 2.7 eV × 1.60 × 10

−19

J/eV )

= 1.184 × 10−10 m = 0.1184 nm

If L is the position of the surface, the distance d beyond the surface where the probability is 10% is found as follows:

P (d + L) = e −2 d η = 10% = 0.10 ⇒ d = − 12 η ln 0.1 = 0.136 nm P( L) (b) A typical atomic diameter is ≈ 2aB ≈ 0.1 nm. So, the penetration of an electron beyond the surface is roughly one atomic diameter.

41.43. Visualize:

Solve:

(a) A graph of ψ ( x ) = Axe− x

2

a2

is shown in the figure. Values of ψ (x) at some selected values of x are

in the table below. x/a

ψ (x)/(aA)

0 0

±0.25 ±.235

±0.50 ±0.389

±0.75 ±0.427

±1.0 ±0.368

±1.50 ±0.158

±2.0 ±0.037

±2.5 ±0.005

(b) The particle is most likely to be found at the place or places where ψ ( x ) is a maximum. This condition is 2

(

2 2 d d A2 x 2e −2 x ψ ( x) = dx dx

a2

) = 0 ⇒ A ⎡⎢⎣2xe 2

The particle will most likely be found at x = ± a (c) The Schrödinger equation is

−2 x 2 a 2

−

4 x3 −2 x2 e a2

a2

⎤ a ⎥ =0⇒ x =± 2 ⎦

2.

d 2ψ 2m 2m = − 2 [ E − U ( x) ]ψ ( x) = 2 [U ( x) − E ]ψ ( x) 2 dx = = = 2 d 2ψ / dx 2 ⇒ U ( x) = E + 2m ψ ( x ) where ψ (x) is the wave function for a particle with energy E. E = 0 for the given wave function, so U ( x) =

= 2 d 2ψ / dx 2 2m ψ ( x )

The first derivative of ψ (x) is 2 2 ⎛ 2x2 ⎞ 2 2 dψ d Axe − x /a = A ⎜1 − 2 ⎟ e − x /a = dx dx a ⎠ ⎝

(

)

The second derivative takes a bit of algebra to simplify, but it becomes d 2ψ d ⎛ ⎛ 2x2 ⎞ 2 2 = ⎜⎜ A ⎜1 − 2 ⎟ e − x /a 2 dx dx ⎝ ⎝ a ⎠

2 ⎞ 4 A ⎛ x 2 3 ⎞ − x2 /a 2 4 ⎛⎛ x ⎞ 3 ⎞ = 2 ⎜ ⎜ ⎟ − ⎟ψ ( x) ⎟⎟ = 2 ⎜ 2 − ⎟ xe a ⎜⎝ ⎝ a ⎠ 2 ⎟⎠ ⎠ a ⎝a 2⎠

Thus the potential energy is U ( x) =

2 2 = 2 d 2ψ / dx 2 = 2 4 ⎛ ⎛ x ⎞ 3 ⎞ 2= 2 ⎛ ⎛ x ⎞ 3 ⎞ = × 2 ⎜⎜ ⎟ − ⎟ = − ⎟ ⎜ ⎜ ⎟ 2m ψ ( x ) 2m a ⎜⎝ ⎝ a ⎠ 2 ⎟⎠ ma 2 ⎜⎝ ⎝ a ⎠ 2 ⎟⎠

This is a harmonic oscillator potential energy that has been shifted downward, starting at U(0) = −3= 2 /ma 2 instead of at 0. The potential energy is zero at (x/a) = (3/2)1/2 = 1.22. A graph of the potential energy is shown above.

41.44. Model: Ions in the crystal lattice behave like simple harmonic oscillators. Solve: (a) Suppose the middle charge is displaced slightly (x << b) to the right from its equilibrium position. The net force on the middle charge due to the two adjacent charges is −2 −2 G 1 e2 ˆ 1 e2 ˆ e2 ⎛ x⎞ ˆ e2 ⎛ x ⎞ ˆ Fnet = i i 1 i 1 − = + − − ⎜ ⎟ ⎜ ⎟ i 4πε 0 ( b + x )2 4πε 0 ( b − x )2 4πε 0b 2 ⎝ b ⎠ 4πε 0b 2 ⎝ b ⎠

Using the binomial approximation,

G Fnet =

⎛ e2 ⎡⎛ 2 x ⎞ ⎛ 2 x ⎞ ⎤ ˆ e 2 ⎛ −4 x ⎞ ˆ ⎜1 − ⎟ − ⎜1 + ⎟ ⎥ i = ⎟i = − ⎜ 3 ⎢ 2⎜ 4πε 0b ⎣⎝ 4πε 0b ⎝ b ⎠ b ⎠ ⎝ b ⎠⎦ ⎝ b πε 0 e2

2

⎞ x ⎟ iˆ ⎠

(b) The force in part (a) is a linear restoring force of the form F = –kx. This is Hooke’s law with, in this case, “spring constant” k = e2 b3 πε 0 . The potential energy of Hooke’s law is

1 e2 U = kx 2 = 3 x2 2 2b πε 0 Thus the angular frequency of vibration is

1 4e 2 k e2 ω= = = = 3 πε 0b m m 4πε 0 b3m

( 4 ) ( 9.0 ×109 N m 2 /C2 )(1.6 ×10−19 C )

( 0.30 ×10

−9

2

m ) ( 27 ×1.66 ×10−27 kg ) 3

= 2.76 ×1013 s −1

The ground-state energy is E1 = 12 =ω =

1 2

(1.05 ×10

−34

J s )( 2.76 ×1013 s −1 ) ×

1 eV = 0.0091 eV 1.60 ×10−19 J

Likewise, E2 = 32 =ω = 0.0272 eV, E3 = 52 =ω = 0.0453 eV, and E4 = 72 =ω = 0.0634 eV. (c) ∆E between two adjacent levels from part (b) is E2 – E1 = E3 – E2 = E4 – E3 = 0.0181 eV. The wavelength of a photon with this energy is

λ=

−34 8 hc ( 6.63 ×10 J s )( 3.0 ×10 m/s ) = = 69 μ m ΔE 0.0181 eV ×1.60 × 10−19 J/eV

The wavelength is in the far infrared region.

41.45. Visualize:

Solve: (a) The work function E0 = 4.0 eV is the energy barrier U0 – E that an electron must either go over (photoelectric effect) or tunnel through. From Equation 41.40, the electron’s penetration distance is

η=

= 2m (U 0 − E )

=

1.05 ×10−34 J s

2 ( 9.11×10−31 kg )( 4.0 eV ×1.6 ×10−19 J/eV )

= 9.72 ×10−11 m = 0.0972 nm

The tunneling probability through the barrier is Ptunnel = e −2 w η = e −2( 0.50 nm ) ( 0.0972 nm ) = 3.40 × 10−5

(b) In passing over an atom 0.05 nm high, the barrier width decreases to w = 0.45 nm. This changes the tunneling probability to

Ptunnel = e −2 w η = e −2( 0.45 nm ) ( 0.0972 nm ) = 9.52 ×10−5 The tunneling probability increases by a factor of 2.8 compared to part (a). If we assume that the tunneling current is proportional to the tunneling probability, then the current increases by a factor of 2.8, or 280%, as the probe passes over an atom. This is a huge change! (c) Detecting a 10% change means detecting a change in the tunneling probability from 3.40 × 10 −5 to Ptunnel = 1.10 × 3.40 × 10−5 = 3.74 × 10−5 (Note that we’re changing Ptunnel by 10%, not looking for a Ptunnel of 10%.) The barrier width that gives this probability is

Ptunnel = 3.74 ×10−5 = e−2 w η

⇒ w = − 12 η ln ( 3.74 × 10−5 ) = − 12 ( 0.0972 nm ) ln ( 3.74 × 10−5 ) = 0.495 nm This is a change in w of 0.005 nm from its original value of 0.500 nm. So the STM can detect height changes of 0.005 nm, or ≈ 201 of an atomic diameter.

41.46. Visualize:

Solve:

If a vmax = 200 mph ball is just sufficient to break the strings, then 2 U 0 = 12 mvmax = 12 ( 0.100 kg )( 89 m/s ) = 396 J , 2

where we converted vmax = 200 mph = 89 m/s. A tennis ball with v = 120 mph = 54 m/s has kinetic energy E = K = 146 J. From Equation 41.40, the penetration distance is

η=

=

2m (U 0 − E )

=

1.05 ×10−34 J s 2 ( 0.100 kg )( 396 J − 146 J )

= 1.485 ×10−35 m

From Equation 41.53, the tunneling probability is Ptunnel = e −2 w η = e

(

−2 ( 0.002 m ) 1.485×10 −35 m

) = e−2.69×1032

We can’t evaluate this number directly with a calculator. To express the result as a power of 10, take the base-10 logarithm:

log ( Ptunnel ) = −2.69 × 1032 log ( e ) = −1.17 × 1032 ⇒ Ptunnel = 10log ( Ptunnel ) = 10−1.17×10 It seems rather unlikely.

32

42.1.

Solve: (a) A 4p state corresponds to n = 4 and l = 1. From Equation 42.3, the orbital angular

momentum is L = 1(1 + 1)= = 2=. (b) In the case of a 5f state, n = 5 and l = 3. So, L = 3 ( 3 + 1)= = 12=.

42.2. Solve: (a) Excluding spin, a state is described by three quantum numbers: n, l, and m. 3p states correspond to n = 3 and l = 1. The quantum number m takes values from −l to l. The quantum numbers of the various 3p states are displayed in the table below. n 3 3 3 l 1 1 1 m 0 −1 +1 (b) A 3d state is described by n = 3 and l = 2. Including the quantum number m, the quantum numbers of the various 3d states are displayed in the table below. n l m

3 2 −2

3 2 −1

3 2 0

3 2 +1

3 2 +2

42.3. Solve: (a) The orbital angular momentum is L = l ( l + 1)=. Thus, 2

2

−34 ⎛ L ⎞ ⎛ 3.65 × 10 J s ⎞ l ( l + 1) = ⎜ ⎟ = ⎜ ⎟ = 12 ⇒ l = 3 −34 ⎝ = ⎠ ⎝ 1.05 × 10 J s ⎠

This is an f electron. (b) The l quantum number is required to be less than n. Thus, the minimum possible value of n for an electron in the f state is nmin = 4. The corresponding minimum possible energy is Emin = E4 = −

13.60 eV = −0.85 eV 42

42.4. Solve: From Equation 42.2, the hydrogen atom’s energy is En = −

13.60 eV = −0.544 eV ⇒ n = 5 n2

The largest l value for an n = 5 state is 4. Thus, the magnitude of the maximum possible angular momentum L is L = l ( l + 1)= = 4 ( 4 + 1)= = 20=.

42.5. Solve: A 6f state for a hydrogen atom corresponds to n = 6 and l = 3. Using Equation 42.2, E6 =

−13.6 eV = −0.378 eV 62

The magnitude of the angular momentum is L = l ( l + 1)= = 3 ( 3 + 1)= = 12=.

42.6. Model: No two electrons can have exactly the same set of quantum numbers (n, l, m, ms). Solve:

For n = 1, there are a total of two states with the quantum numbers given by (1, 0, 0, ± 12 ) . For n = 2,

there are a total of eight states:

( 2, 0, 0,

± 12 )

( 2, 1,

− 1, ± 12 )

( 2, 1, 0,

± 12 )

( 2, 1, 1,

± 12 )

( 3, 1,

− 1, ± 12 )

( 3, 1, 0,

± 12 )

( 3, 1, 1,

± 12 )

For n = 3, there are a total of 18 states:

( 3, 0, 0, ( 3, 2, 2,

± 12 )

± 12 )

( 3, 2, 1,

± 12 )

( 3, 2, 0,

± 12 )

( 3, 2,

− 1, ± 12 )

( 3, 2,

− 2, ± 12 )

42.7. Solve: (a) A lithium atom has three electrons, two are in the 1s shell and one is in the 2s shell. The electron in the 2s shell has the following quantum numbers: n = 2, l = 0, m = 0, and ms. ms could be either + 12 or

− 12 . Thus, lithium atoms should behave like hydrogen atoms because lithium atoms could exist in the following

two states: ( 2, 0, 0, + 12 ) and ( 2, 0, 0, − 12 ) . Thus there are two lines.

(b) For a beryllium atom, we have two electrons in the 1s shell and two electrons in the 2s shell. The electrons in both the 1s and 2s states are filled. Because the two electron magnetic moments point in opposite directions, beryllium has no net magnetic moment and is not deflected in a Stern-Gerlach experiment. Thus there is only one line.

42.8. Solve: Mg, Sr, and Ba are all in the second column of the periodic table. The electron configuration of Mg (Z = 12) is 1s22s22p63s2. The electron configuration of Sr (Z = 38) is 1s22s22p63s23p64s23d104p65s2 The electron configuration of Ba (Z = 56) is 1s22s22p63s23p64s23d104p65s24d105p66s2 Assess: It is necessary to recall that the 4s subshell fills before the 3d subshell, the 5s before the 4d, and the 6s before the 4f or 5d.

42.9. Solve: P, As, and Sb are all in the same column of the periodic table as nitrogen. The electron configuration P (Z = 15) is 1s22s22p63s23p3. The electron configuration of As (Z = 33) is 1s22s22p63s23p64s23d104p3 The electron configuration of Sb (Z = 51) is 1s22s22p63s23p64s23d104p65s24d105p3 Assess:

It is necessary to recall that the 4s subshell fills before the 3d subshell.

of

42.10. Solve: (a) Nine electrons (Z = 9) make the element fluorine (F). These are the nine lowest energy states, so this is the ground state of F. (b) Thirty-one electrons (Z = 31) make the element gallium (Ga). States fill in the order 4s – 3d – 4p. So 3d104s24p, with filled 3d and 4s shells, has the 31 electrons in the lowest possible energy states. This is the ground state of Ga.

42.11. Solve: (a) Ten electrons (Z = 10) make the element neon (Ne). These are not the ten lowest energy

states because 1s22s22p6 would be lower in energy than 1s22s22p53d. This is an excited state of Ne. (b) Twenty-six electrons (Z = 26) make the element iron (Fe). These are not the 26 lowest energy states because the 3d shell is not filled. This is an excited state of Fe.

42.12. Solve: Accurate to three significant figures, the factor is hc = ( 6.63 × 10−34 J s )( 3.00 × 108 m/s ) = 19.89 × 10−26 J m ×

1 eV = 1243 eV nm ≅ 1240 eV nm 1.60 × 10−19 J

42.13. Visualize: Please refer to Figure 42.15. Solve: The diagram shows the energy levels for electrons in a multielectron atom. A lithium atom in the ground state has two electrons in the 1s shell and 1 electron in the 2s shell. In other words, the ground-state electron configuration of lithium is 1s22s1. The electron configuration for the first excited state is 1s22p1 and for the second excited state is 1s23s1.

42.14.

Model: Assume the hydrogen atom starts in the ground state. Visualize: The electron gains 12.5 eV of energy in the acceleration. This is enough to excite the hydrogen atom to n = 3 (with the electron left with some energy) but no higher. (See Figure 42.4.) Solve: The atom could emit a photon in the two possible quantum-jump transitions: 3 → 2 and 3 → 1 . The −1.51 eV − (−3.40 eV) = 1.89 eV and corresponding energies of the emitted photons are −1.51 eV − (−13.60 eV) = 12.09 eV. The corre-sponding wavelengths are given by λ = hc/Ephoton .

transition 3→2 3 →1

Ephoton (eV) λ (nm) 1.89 656 12.09 102

Assess: The wavelength for the 3 → 2 transition is in the visible Balmer series; while the one for the 3 → 1 transition is in the ultraviolet region.

42.15. Solve: (a) A 4p → 4s transition is allowed because ∆l = 1. Using the sodium energy levels from Figure 42.25, the wavelength is

λ=

hc 1240 eV nm = = 2210 nm = 2.21 μ m ΔE 3.75 eV − 3.19 eV

(b) A 3d → 4s transition is not allowed because ∆l = 2 violates the selection rule that requires ∆l = 1.

42.16. Solve: Because the probability is very small, we can write the decay rate r=

0.010 P = = 0.10 ns −1 Δt 0.10 ns

The lifetime of the excited state is 1 r

τ= =

1 = 10 ns 0.10 ns −1

42.17. Solve: The interval ∆t = 0.50 ns is very small in comparison with the lifetime τ = 25 ns, so we can write Prob(decay in ∆t) = r ∆t where r is the decay rate. The decay rate is related to the lifetime by r = 1/τ = 1/(25 ns) = 0.040 ns–1. Thus Prob(decay in ∆t) = (0.040 ns–1)(0.50 ns) = 0.020 = 2.0%

42.18. Solve: (a) If there are initially N0 atoms in a state with lifetime τ, then the number remaining at time t is Nexc = N0e−t/τ. From Table 42.3, the 3p state has a lifetime τ = 17 ns. Hence, Nexc = 1.0 × 106e−10/17 = 555,000 (b) Likewise at t = 30 ns, Nexc = 1.0 × 106e−30/17 = 171,000. (c) For t = 100 ns, Nexc = 1.0 × 106e−100/17 = 3000.

42.19. Solve: (a) The number of atoms that have undergone a quantum jump to the ground state is 0.90N0 = 0.90(1.0 × 106) = 9.0 × 105 Because each transition is accompanied by a photon, the number of emitted photons is also 9.0 × 105. (b) 10% of the atoms remain excited at t = 20 ns. Thus

N exc = N 0e− t / τ ⇒ 0.10 N 0 = N 0e−20 ns/τ ⇒ ln ( 0.10 ) =

−20 ns

τ

⇒ τ = 8.7 ns

42.20. Visualize: We’ll first use Equation 42.18 to find the time t when 80% of the excited sodium atoms have decayed (that is, when 20% of the sodium atoms are still excited). Then we’ll take that t and use it again in Equa-tion 42.18 to find how many potassium atoms are still excited; t is the same in both equations. We look up the two lifetimes in Table 42.3: τ Na = 17 ns and τ P = 26 ns. We are given N 0 = 1.0 × 108 for both kinds of atoms. Solve: Solve Equation 42.18 for t and plug in the values for sodium. ⎛N ⎞ t = −τ ln ⎜ exc ⎟ = −(17 ns)ln(0.20) = 27.4 ns ⎝ N0 ⎠ Now use this t to find N exc for potassium. N exc = N 0e −t /τ = (1.0 × 108 )e −27.4 ns/ 26 ns = 3.5 × 107 Assess: 35% of the potassium atoms are still excited when only 20% of the sodium ones are; this seems reasonable.

42.21. Solve: Since 1 mW = 0.001 W, the laser emits Elight = 0.001 J of light energy per second. This energy consists of N photons. The energy of each photon is Ephoton = hf =

hc

λ

= 3.14 × 10−19 J

Because Elight = NEphoton, the number of photons is

N=

Elight Ephoton

So, photons are emitted at the rate 3.2 × 1015 s−1.

=

0.001 J = 3.2 × 1015 3.13 × 10−19 J

42.22. Solve: The energy of each photon is Ephoton = hf =

hc

λ

= 1.876 × 10−20 J

Because Elight = NEphoton, Elight = (5.0 × 1022 )(1.876 × 10−20 J) = 940 J So, the energy output is 940 J every second, or 940 W.

42.23. Solve: (a) The wavelength is λ=

1240 eV nm hc = = 1060 nm = 1.06 μ m ΔE 1.17 eV − 0 eV

(b) The energy per photon is Eph = 1.17 eV = 1.87 × 10−19 J. The power output of the laser is the number of photons per second times the energy per photon: P = (1.0 × 1019 s−1)(1.87 × 10−19 J) = 1.9 J/s = 1.9 W

42.24. Solve: (a) For l = 3, the magnitude of the angular momentum vector is L = l ( l + 1)= = 3 ( 3 + 1)= = 12= The angular momentum vector has a z-component Lz = Lcosθ along the z-axis. Lz = m= where m is an integer between −l and l, that is, between −3 and 3. The seven possible orientations of the angular momentum vector for l = 3 are shown in figure below.

G (b) From Equation 42.5, the minimum angle between L and the z-axis is ⎛ 3= ⎞ ⎟ = 30° ⎝ 12= ⎠

θ33 = cos −1 ⎜

42.25. Solve: (a) For s = 1, S = s ( s + 1)= = 2= = 1.48 × 10−34 J s. (b) The spin quantum number is ms = −1, 0, or 1. G (c) The figure below shows the three possible orientations of S .

42.26. Solve: (a) Since L2x + L2y + L2z = L2 , L2x + L2y = L2 − L2z . For a hydrogen atom with l = 2, the magnitude of L2 is always l(l + 1)=2 or 6=2. The value of Lz is m=, where m is an integer between −l and l. Hence, the maximum value of Lz is 2= and the minimum value of ( L2x + L2y )

1/ 2

is

l ( l + 1) = 2 − ( 2= ) = 6= 2 − 4= 2 = 2= 2

(b) Because the minimum value of Lz is 0 J s, the maximum value of

L2x + L2y is

L2 − L2z = 6= 2 − 0 J s = 6=

42.27. Solve: A hydrogen atom in its fourth excited state is in the n = 5 state. The energy of the wavelength of the emitted photon is hc 1240 eV nm ΔE = = = 0.967 eV 1282 nm λ From Equation 42.2, the hydrogen atom’s energy must be one of the following values 13.60 eV En = − n2 The transition 5 → n corresponds to the energy

⎛1 1⎞ ΔE = −13.60 eV ⎜ 2 − 2 ⎟ = 0.967 eV ⇒ n = 3 ⎝5 n ⎠ This means the angular momentum quantum number is l = 2, 1, or 0. The atom’s maximum possible orbital angular momentum after the emission is L = l ( l + 1)= = 2 ( 2 + 1)= = 6=

42.28. Solve: (a) From Equation 42.7, the radial wave function of hydrogen in the 1s state is R1s ( r ) =

1

πa

3 B

e − r aB ⇒ R1s ( 12 aB ) =

1

πa

3 B

e

− 12

=

0.342 aB3

(b) From Equation 42.10, the probability density is 2 ⎛a ⎞ Pr ( r ) = 4π r 2 Rnl ( r ) ⇒ P1s ( 12 aB ) = 4π ⎜ B ⎟ ⎝ 2 ⎠

2

2

0.368 ⎛ 1 −1 ⎞ e 2⎟ = ⎜⎜ 3 ⎟ aB ⎝ π aB ⎠

42.29. Solve: (a) From Equations 42.7 and 42.20, the radial wave function and radial probability density are R1s ( 12 aB ) =

1

π aB3

e

− 12

=

⎛a ⎞ ⇒ Pr ( 12 aB ) = 4π ⎜ B ⎟ ⎝ 2 ⎠ π aB3

0.607

2

2

⎛ 0.607 ⎞ 0.368 = ⎜⎜ 3 ⎟ ⎟ aB ⎝ π aB ⎠

From Equation 42.9, the probability is

⎛ 0.368 ⎞ −3 Prob(in δ r at r) = Pr ( r ) δ r = Pr ( 12 aB ) ⎣⎡( 0.01) aB ⎦⎤ = ⎜ ⎟ ( 0.01aB ) = 3.7 × 10 a ⎝ B ⎠ (b) Likewise,

1

R1s ( aB ) =

πa

3 B

e −1 =

0.368

πa

3 B

⇒ Pr ( aB ) = 4π aB2

( 0.368) πa

2

3 B

=

0.541 aB

The probability is Pr ( r ) δ r = Pr ( aB )( 0.01) aB = 5.4 × 10−3 . (c) For r = 2aB, R1s ( 2aB ) =

1

πa

3 B

e −2 =

0.135

πa

3 B

⇒ Pr ( 2aB ) = 4π ( 2aB )

The probability is Pr ( r ) δ r = Pr ( 2aB )( 0.01) aB = 2.9 × 10−3.

2

( 0.135) πa

3 B

2

=

0.293 aB

42.30. Solve: The normalization condition for the three-dimensional hydrogen atom is

∫

∞ 0

∞

Pr ( r ) dr = 4π ∫ r 2 Rnl ( r ) dr = 1 . 2

0

From Equation 42.7, the 1s radial wave function can be written R1s ( r ) = A1s e − r aB . Hence, ∞

1 = 4π A12s ∫ r 2e−2 r aB dr = 4π A12s 0

2!

( 2 aB )

3

= π a B3 A12s ⇒ A1s =

1

π aB3

42.31. Solve: From Equation 42.7, the 2p radial wave function of the hydrogen atom can be written ⎛ r ⎞ − r 2 aB R2 p ( r ) = A2 p ⎜ ⎟e ⎝ 2aB ⎠ The normalization condition for the three-dimensional hydrogen atom is

∫

∞ 0

2r

∞ ∞ ⎛ r2 ⎞ − 2 Pr ( r ) dr = 4π ∫ r 2 Rnl ( r ) dr = 4π ∫ A22p ⎜ 2 ⎟ e 2 aB r 2 dr 0 0 ⎝ 4aB ⎠

=π

A22p aB2

∫

∞

0

r 4e − r aB dr =

A22pπ ⎡ 4! ⎤ 1 = 24π aB3 A22p = 1 ⇒ A2 p = ⎢ 5⎥ 2 24π aB3 aB ⎢⎣ (1 aB ) ⎥⎦

42.32. Solve: Using Equations 42.10 and 42.7, the radial probability density for the 1s state is Pr ( r ) = 4π r 2 R1s ( r ) = 4π A12s r 2e −2 r aB 2

This radial probability density peaks at the point where dPr /dr = 0. The derivative is

⎡ dPr 2r 2 −2 r aB ⎤ r ⎤ −2 r aB 2 ⎡ = 4π A12s ⎢ 2re −2 r aB − e ⎥ = 8π A1s ⎢1 − ⎥ re dr aB ⎣ ⎦ ⎣ aB ⎦ The derivative is zero at r = aB, so Pr(r) is a maximum at this value of r.

42.33. Solve: (a) From Equation 42.7, the 2p radial wave function is R2 p ( r ) =

A2 p 2 aB

re − r 2 aB

The graph of R2p(r) is seen to have a single maximum.

(b) R2p(r) is a maximum at the point where dR2p/dr = 0. The derivative is

dR2 p dr

=

A2 p ⎡ − r 2 aB r − r 2 aB ⎤ A2 p ⎡ r ⎤ − r 2 aB − e ⎢e ⎥= ⎢1 − ⎥e a a 2aB ⎣ 2aB 2 2 ⎦ B ⎣ B⎦

The derivative is zero at r = 2aB, so R2p(r) is a maximum at r = 2aB . (c) The radial wave function R2p and the probability density P2p(r) is smaller at r = 4aB than it is at r = 2aB. We are less likely to find the electron at a point with r = 4aB than at a point with r = 2aB. However, there are many more points with r = 4aB than with r = 2aB. The increased number of points more than compensates for the decreased probability per point. As a result, it is more probable to find the electron at distance r = 4aB than it is at distance r = 2aB.

42.34. Solve: (a) For a hydrogen atom in the p-state, l = 1. Because Lz = m= and S z = ± 12 =, the three possible values of Lz are =, 0, and −=, and the two possible values of Sz are

1 2

= and − 12 =. We can now compute

Jz using Jz = Lz + Sz. The results are presented in the following table. Lz =

Sz + 12 =

3 2

Jz =

mj − 32

= 0

− 12 =

1 2

=

− 12

+ 12 =

1 2

=

− 12

0

1 2

− =

1 2

− =

− 12

−=

+ 12 =

− 12 =

− 12

−=

− 12 =

− 32 =

− 32

(b) The values of Jz found above can be divided into two groups:

Jz = mj= and mj = −j to j, the j values of the above two groups are

( 32 =, 3 2

and

1 2 1 2

=, − 12 =, − 23 = ) and

.

( 12 =, − 12 = ) . Because

42.35. Solve: The electron configuration in the ground state of K (Z = 19) is 1s22s22p63s23p64s1 This means that all the states except the 4s state are completely filled. For Ti (Z = 22), all the states up to the 4s state are completely filled. The electron configuration is 1s22s22p63s23p64s23d2 The d subshell has only two electrons. In the case of Fe (Z = 26), the 3d subshell has six electrons and all the lower level states are completely filled. The electron configuration is 1s22s22p63s23p64s23d6 The ground-state electron configurations of Ge (Z = 32) and Br (Z = 35) are 1s22s22p63s23p64s23d104p2

1s22s22p63s23p64s23d104p5

42.36. Solve: The electron configuration in the ground state of Ca (Z = 20) is 1s22s22p63s23p64s2. The

configuration for V (Z = 23) is 1s22s22p63s23p64s23d3. Ni (Z = 28), As (Z = 33), and Kr (Z = 36) are 1s22s22p63s23p64s23d8

1s22s22p63s23p64s23d104p3

1s22s22p63s23p64s23d104p6

42.37. Visualize: Please refer to Figure 42.25. Solve: (a) The allowed transitions are those with ∆l = ±1. Starting from an s-state, the only allowed transitions are to p-states. (b) Once the states are identified, the wavelength is λ = hc/∆E = 1240 eV nm/∆E. Transition 6s → 5p 6s → 4p 6s → 3p

∆E 0.17 eV 0.76 eV 2.41 eV

λ 7290 nm 1630 nm 515 nm

42.38. Visualize: Please refer to Figure 42.25. Solve: For sodium, the energy of the 3p state (which is the first excited state) is E3p = 2.104 eV, compared to the ground state at E = 0 eV. We need to add the transition energy E5d →3p to E3p to obtain the energy of the 5d state. The transition energy is E5 d →3 p =

hc

λ

=

1240 eV nm = 2.485 eV ⇒ E5 d = E3 p + E5 d →3 p = 2.104 eV + 2.485 eV = 4.59 eV 499 nm

Assess: E5d is larger than E6s, E5p, and E4d and less than the ionization energy limit of 5.14 eV. This is reasonable.

42.39. Visualize: Please refer to Figure 42.25. Solve:

A photon wavelength of 818 nm corresponds to an energy of

E=

hc

λ

=

1240 eV nm = 1.516 eV 818 nm

From Figure 42.25, the transition that obeys the selection rule ∆l = 1 and has a magnitude around 1.5 eV is 3p → 3d. Note that E3d – E3p = 3.620 eV – 2.104 eV = 1.516 eV, which is exactly equal to the photon energy. The atom was excited to the 3d state from the ground state. Thus the minimum kinetic energy of the electron was

2 ( 3.62 eV ) (1.60 × 10−19 J/eV ) 1 2 mv = 3.62 eV ⇒ v = = 1.13 × 106 m/s 2 9.11 × 10−31 kg

Model: Allowed transitions are those with ∆l = ±1. Visualize: For the fictitious atom, our task is to draw an energy level diagram that will have four emission wavelengths 310.0 nm, 354.3 nm, 826.7 nm, and 1240.0 nm. These wavelengths correspond to energies of 4.00 eV, 3.50 eV, 1.50 eV, and 1.00 eV. A ground p-state (l = 1), an excited s-state (l = 0) at 3.50 eV, an excited d-state (l = 2) at 4.0 eV, and an excited p-state (l = 1) at 5.0 eV will lead to the four energies or transitions. The excited p-state can decay only to the s and d states, and s and d states can decay only to the ground state. So, the four transitions make a diamond pattern on the diagram.

42.40.

42.41. Visualize: Please refer to Figure P42.41. Solve: We need to use the condition ∆l = 1 to determine the allowed transitions of the emission spectrum, and the equation E=

hc

λ

=

1240 eV nm

λ

to find the wavelengths. Visible wavelengths are in the range 400–700 nm, with infrared being greater than 700 nm and ultraviolet being less than 400 nm. The absorption spectrum has only transitions from the ground state, so the wavelengths of the emission spectrum that are in the absorption spectrum belong to transitions that end on the 2s ground state. Transition 2p → 2s 3s → 2p 3p → 2s 3p → 3s 3d → 2p 3d → 3p 4s → 2p 4s → 3p

(a) Wavelength 670 nm 816 nm 324 nm 2696 nm 611 nm 24,800 nm 498 nm 2430 nm

(b) Type VIS IR UV IR VIS IR VIS IR

(c) Absorption Yes No Yes No No No No No

42.42. Visualize: Please refer to Figure P42.42. Solve:

(a) We need to use the condition ∆l = 1 to determine the allowed transitions, and the equation

ΔE =

hc

λ

=

1240 eV nm

λ

to find the wavelengths. Transition 6p → 6s 6d → 6p 7s → 6p 7p → 6s 7p → 7s 8s → 6p 8s → 7p 8p → 6s 8p → 7s 8p → 8s 8p → 6d

∆E (eV) 6.70 2.14 1.22 8.84 0.92 2.52 0.38 9.53 1.61 0.31 0.69

λ (nm) 185 579 1016 140 1350 492 3260 130 770 4000 1800

(b) From part (a), the 492-nm-wavelength blue emission line in the Hg spectrum corresponds to the transition 8s → 6p. The energy of the 8s state is 9.22 eV. The atom is excited to the 8s state from the ground state. Thus the minimum kinetic energy of the electron must be 2 ( 9.22 eV ) (1.60 × 10−19 J/eV ) 1 2 mv = 9.22 eV ⇒ v = = 1.80 × 106 m/s 2 9.11 × 10−31 kg

42.43. Model: We have a one-dimensional rigid box with infinite potential walls and a length 0.50 nm. Solve:

(a) From Equation 41.22, the lowest energy level is

( 6.63 × 10−34 J s ) h2 1 eV E1 = = = 1.51 V 2 2 31 10 − − 8mL 8 ( 9.11 × 10 kg )( 5.0 × 10 m ) 1.60 × 10−19 J 2

The next two levels are E2 = 4E1 = 6.04 eV and E3 = 9E1 = 13.6 eV. The Pauli principle allows only two electrons in each of these energy levels, one with spin up and one with spin down. So five electrons fill the n = 1 and n = 2 levels, with the fifth electron going to n = 3.

(b) The ground-state energy of these five electrons is E = 2E1 + 2E2 + E 3 = 28.7 eV

42.44. Model: The electrons are in a one-dimensional rigid box.

Solve: (a) The energy levels in a rigid box are En = n2E1 where E1 = h2/8mL2 = 1.51 eV for L = 0.50 nm. The figure below shows the energy-level diagram. The electron in the n = 6 state can decay to lower states, but the transitions must obey the selection rule ∆n = odd. In principle, this allows transitions to n = 5, 3, and 1. However, the n = 1 state is full since the Pauli principle allows only two electrons, one with spin up and the other with spin down. Consequently, the only possible emission transitions are 6 → 5 and 6 → 3.

(b) The wavelengths of the photon are calculated from

Ephoton =

hc

λ

= ΔEelectron = E6 − En = (36 − n 2 ) E1 ⇒ λ =

hc (36 − n 2 ) E1

The two allowed transitions are 6 → 5 with λ65 = 75 nm and 6 → 3 with λ63 = 30 nm.

42.45. Solve: (a) From Table 42.3, the lifetime of the 2p state of hydrogen is τ = 1.6 ns. The decay rate is r=

1

τ

=

1 = 6.25 × 108 s −1 1.6 × 10−9 s

(b) From Equation 42.25, the number of excited atoms left at time t is Nexc = N0e−t/τ. If 10% of a sample decays, then 90% of the atoms in the sample are still excited. That is, Nexc = 0.90N0. The time for this to occur is calculated as follows:

Nexc = 0.90N0 = N0e−t/τ ⇒ e−t/τ = 0.90 ⇒ t = −τ ln0.90 = 0.17 ns

42.46. Solve: For very small time intervals ∆t, the probability that an atom decays is P = r∆t. The time interval for a 1% probability of decay is Δt =

P = Pτ = ( 0.01)(1.6 ns ) = 0.016 ns r

where we used τ = 1/r and took the value of τ from Table 42.3.

42.47. Solve: The number of excited atoms left at time t is given by Equation 42.25: Next = N0e−t/τ. If 1% of

the atoms in the excited state decay in t = 0.20 ns, then 99% of the atoms remain in the excited state. So, 0.99N0 = N0e−0.20 ns/τ ⇒ 0.99 = e−0.20 ns/τ ⇒ ln(0.99) = −0.20 ns/τ ⇒ τ = 19.90 ns

Having determined τ, we can now find the time during which 25% of the sample of excited atoms would decay, leaving 75% still excited. Applying Equation 42.25 once again, 0.75N0 = N0e−t/19.90 ns ⇒ t = 5.7 ns

42.48. Solve: (a) The half-life t1/2 is the time needed for half of the initially excited atoms to decay. This means that half of the excited atoms are still present at t = t1/2. Using the exponential decay law, N exc ( t1/ 2 ) = 12 N 0 = N 0e − t1/ 2 τ ⇒ e − t1/ 2 τ =

1 2

⇒ t1/2 = τ ln 2 = 0.693τ

(b) The half-life is 69.3% of the lifetime τ. From Table 42.3, we find τ = 17 ns for the 3p state of sodium. The half life is t1/2 = (0.693)(17 ns) = 12 ns.

42.49. Solve: If there are Nexc atoms in the excited state, the rate of decay is rNexc, where r = 1/τ is the decay rate. Each decay generates one photon, so the rate of photon emission (photons per second) is the same as the rate of decay (decays per second). Hence, the number of photons emitted per second is rN exc =

N exc

τ

=

1.0 × 109 = 5.0 × 1016 20 × 10−9 s

42.50. Solve: 100 MW is the peak power (P = ∆E/∆t). The energy in the laser pulse is ∆E = P∆t = (1.0 × 108 W)(1.0 × 10–8 s) = 1.0 J This light energy is due to N photons, each of energy Eph = hf = hc/λ. That is,

ΔE = NEph ⇒ N =

ΔE ΔE λ ΔE ( 690 nm )(1.0 J ) = = = = 3.5 × 1018 Eph hc λ hc ( 6.63 ×10−34 J s )(3.0 × 108 m/s )

Each photon is emitted when a chromium atom undergoes stimulated emission. So a total of 3.5 × 1018 atoms undergo stimulated emission to generate this laser pulse.

42.51. Solve: (a) To derive the formula, consider the change in the energy of the photon as the wavelength changes. The energy of an emitted photon is E = hf =

hc

λ

⇒

dE hc λ2 = − 2 ⇒ Δλ = Δ E λ dλ hc

(b) The energy difference between the two states is 1 ⎞ ⎛ 1 ΔE = E21 − E20 = 13.6 eV ⎜ − 2 + 2 ⎟ = 3.16 × 10−3 eV ⎝ 21 20 ⎠

To obtain Δλ from the formula in part (a), we need λ. The wavelength is calculated as follows:

hc 1240 eV nm 1 ⎞ 1240 eV nm ⎛1 E20 →1 = (13.6 eV ) ⎜ 2 − 2 ⎟ = 13.56 eV = = ⇒λ = = 91.4 nm 13.56 eV λ λ ⎝ 1 20 ⎠ The wavelength of the 21 → 1 transition is almost identical. The two wavelengths differ by Δλ =

λ2 hc

ΔE =

( 91.4 nm )

2

(1240 eV nm )

3.16 × 10−3 eV = 0.021 nm

42.52. Solve: The probability of finding a 1s electron at r > aB is ∞

∞

aB

aB

Prob ( r > aB ) = ∫ Pr ( r ) dr = 4π ∫ r 2 R1s ( r ) dr = 2

4 ∞ 2 −2 r aB re dr aB3 ∫ aB

where we used the expression for R1s ( r ) found in Equation 42.7. Change the variable to u = 2r/aB. This means r 2 dr =

( a 8) u du 3 B

2

and u = 2 when r = aB. The probability is

Prob(r > aB) =

∞ 1 ∞ 2 −u 1 1 u e du = ⎡⎣ − ( u 2 + 2u + 2 ) e− u ⎤⎦ = ⎡⎣( 4 + 4 + 2 ) e −2 ⎤⎦ = 5e−2 = 0.677 = 67.7% ∫ 2 2 2 2 2

42.53. Solve: The probability of finding a 1s electron at r < 12 aB is 1a

1a

Prob ( r < 12 aB ) = ∫ 2 Pr ( r ) dr = 4π ∫ 2 r 2 R1s ( r ) dr = B

0

B

0

2

4 12 aB 2 −2 r aB re dr aB3 ∫ 0

where we used the expression for R1s ( r ) found in Equation 42.7. Change the variable to u = 2r/aB. This means r 2 dr =

( a 8) u du. Also, the upper limit of the integral changes to u = 1 when r = 3 B

2

Prob ( r < 12 aB ) =

1 2

aB . The probability is

4 1 ⎛ aB3 ⎞ 2 − u 1 1 2 −u ⎜ ⎟ u e du = ∫ 0 u e du aB3 ∫ 0 ⎝ 8 ⎠ 2

1 1 1 5 = ⎡⎣ − ( u 2 + 2u + 2 ) e − u ⎤⎦ = ⎡⎣ −5e −1 + 2 ⎤⎦ = 1 − e−1 = 0.0803 = 8.03% 0 2 2 2

42.54. Solve: The radial probability density for the 2s state is 2

⎛ ⎛ 2 r ⎞ − r aB r3 r4 ⎞ Pr ( r ) = 4π r 2 R2 s ( r ) = 4π A22s r 2 ⎜1 − = C ⎜ r 2 − + 2 ⎟ e − r aB ⎟ e aB 4aB ⎠ ⎝ 2aB ⎠ ⎝ where C = 4π A22s is a constant. This is a maximum where dPr /dr = 0. The derivative is

⎡⎛ ⎤ dPr 3r 2 r 3 ⎞ − r aB 1 ⎛ 2 r 3 r4 ⎞ = C ⎢⎜ 2 r − + 2 ⎟e − ⎜r − + 2 ⎟ e − r aB ⎥ dr a a a a 4 a B B ⎠ B⎝ B B ⎠ ⎣⎢⎝ ⎦⎥ ⎡ 4 r 2 r 2 r 3 ⎤ − r aB = C ⎢2 − + − ⎥ re aB aB2 4aB3 ⎦ ⎣ The derivative is zero, giving the maxima and minima of Pr(r), when 2

3

⎛ r ⎞ ⎛ r ⎞ 1⎛ r ⎞ 2 − 4⎜ ⎟ + 2⎜ ⎟ − ⎜ ⎟ = 0 ⎝ aB ⎠ ⎝ aB ⎠ 4 ⎝ aB ⎠ This is a cubic equation, which will have three roots. This is expected because we know from the graph in Figure 42.8 that Pr(r) has not only two maxima but also, in between, a minimum. Although the cubic equation can be solved numerically, here we just need to verify that r = 5.236 aB is a solution. Substituting into the equation, 2 − 4(5.236) + 2(5.236) 2 − 14 = (5.236)3 = 0. Thus r = 5.236aB is the most probable distance of the electron in the 2s state.

42.55. Solve: The radial probability density for the 2s state is 2

⎛ ⎛ 2 r ⎞ − r aB r3 r4 ⎞ Pr ( r ) = 4π r 2 R2 s ( r ) = 4π A22s r 2 ⎜1 − = C ⎜ r 2 − + 2 ⎟ e − r aB ⎟ e aB 4aB ⎠ ⎝ 2aB ⎠ ⎝ where C = 4π A22s is a constant. This is a maximum where dPr /dr = 0. The derivative is

⎡⎛ ⎤ dPr 3r 2 r 3 ⎞ − r aB 1 ⎛ 2 r 3 r4 ⎞ = C ⎢⎜ 2 r − + 2 ⎟e − ⎜r − + 2 ⎟ e − r aB ⎥ dr a a a a 4 a B B ⎠ B⎝ B B ⎠ ⎣⎢⎝ ⎦⎥ ⎡ 4 r 2 r 2 r 3 ⎤ − r aB = C ⎢2 − + − ⎥ re aB aB2 4aB3 ⎦ ⎣ The derivative is zero, giving the maxima and minima of Pr(r), when 2

3

⎛ r ⎞ ⎛ r ⎞ 1⎛ r ⎞ 2 − 4⎜ ⎟ + 2⎜ ⎟ − ⎜ ⎟ = 0 ⎝ aB ⎠ ⎝ aB ⎠ 4 ⎝ aB ⎠ This is a cubic equation, which will have three roots. This is expected because we know from the graph of Figure 42.8 that Pr(r) has not only two maxima but also, in between, a minimum. The cubic equation can be solved numerically to give r = 0.764aB, 2.00aB, and 5.236aB. By referring to the graph, we see that the outer two roots correspond to the maxima and the middle root is the minimum. The separation between the peaks is Δr = ( 5.236aB − 0.764aB ) = 4.472aB

42.56. Solve: For an electron in the 1s and 2p states of hydrogen, the radial wave functions are given by Equation 42.7 and the radial probability density for a state nl is given by Equation 42.10. For the 1s state, ∞

∞

0

0

ravg = ∫ rPr ( r ) dr = ∫

r −2 r aB 4 ∞ 4 3! 24 a 4 = 3 B = 1.5aB e 4π r 2 ) dr = 3 ∫ r 3e−2 r aB dr = 3 ( 4 3 π aB aB 0 aB ( 2 aB ) aB 16

where we used formula for the definite integral given in Problem 42.33. For the 2p state, ravg = ∫

∞ 0

∞ ⎛ r 2 ⎞ −2 r 2 aB 1 1 5! = 5 aB 4π r 2 dr = r 5e − r aB dr = ⎜ ⎟e 24π a ⎝ 4aB2 ⎠ 24aB5 ∫ 0 24aB5 (1 aB )6

r

3 B

42.57. Solve: (a) The atoms leave the oven at 500 m/s, so their momentum is patom = mv = (1.4 × 10–25 kg)(500 m/s) = 7.0 × 10–23 kg m/s The momentum of a photon traveling in the –x-direction is

Ephoton

pphoton = −

c

=−

hf h 6.63 × 10−34 J s =− =− = −8.5 × 10−28 kg m/s λ c 780 × 10−9 m

The negative sign is due to the direction of the photon. (b) As the photon approaches the atom, the total momentum of the atom+photon system is (patom)i + pphoton. The photon is absorbed and disappears, so the final momentum is (patom)f. Momentum is conserved in the absorption process, so

( patom )f = ( patom )i + pphoton ⇒ Δpatom = pphoton ∆patom is negative because the atom loses momentum. Since ∆patom is the same for every absorption, independent of the atom’s velocity, the number of absorptions that will bring the atom to rest is N absorb =

( patom )i ( patom )i 7.0 × 10−23 kg m/s = = = 82,400 photons | Δpatom | | pphoton | 8.5 × 10−28 kg m/s

(c) An atom in the ground state instantly absorbs a photon then, on average, spends 15 ns in the excited state before it drops back to the ground state where it can absorb another photon. That is, the average time between photon absorptions is 15 ns. The time to absorb 82,400 photons is

∆t = (82,400 photons)(15 × 10–9 s/photon) = 1.24 × 10–3 s = 1.24 ms (d) The atom’s momentum changes by ∆patom = pphoton = –8.5 × 10–28 kg m/s during each ∆t = 15 ns. The force exerted on the atom and the consequent acceleration are

F=

Δpatom −8.5 × 10−28 kg m/s = = −5.67 × 10−20 N Δt 15 × 10−9 s

⇒a=

F −5.67 × 10−20 N = = −4.05 × 105 m/s 2 1.4 × 10−25 kg m

(e) From kinematics, vf 2 = 0 = vi 2 + 2aΔx ⇒ Δx = −

vi 2 (500 m/s) 2 =− = 0.31 m = 31 cm 2a 2(− 4.05 × 105 m/s 2 )

42-1

43.1. Model: The nucleus is composed of Z protons and Α − Ζ neutrons. Solve: (a) 3H has Z = 1 proton and 3 − 1 = 2 neutrons. (b) 40Ar has Z = 18 protons and 40 − 18 = 22 neutrons. (c) 40Ca has Z = 20 protons and 40 − 20 = 20 neutrons. (d) 239Pu has Z = 94 protons and 239 − 94 = 145 neutrons.

43.2. Model: The nucleus is composed of Z protons and Α − Ζ neutrons. Solve: (a) 6Li has Z = 3 protons and 3 − 3 = 3 neutrons. (b) 54Cr has Z = 24 protons and 54 − 24 = 30 neutrons. (c) 54Fe has Z = 26 protons and 54 − 26 = 28 neutrons. (d) 220Rn has Z = 86 protons and 220 − 86 = 134 neutrons.

43.3. Solve: (a) The radius and diameter of the nucleus of 4He are r = r0A1/3 = (1.2 fm)(4)1/3 = 1.90 fm ⇒ d = 3.8 fm (b) For 40Ar, r = (1.2 fm)(40)1/3 = 4.10 fm and d = 8.2 fm. (c) For 220Rn, r = (1.2 fm)(220)1/3 = 7.24 fm and d = 14.5 fm.

43.4. Solve: Using Equation 43.2, 3

⎛ 7.46 fm ⎞ r = r0 A1/ 3 ⇒ 7.46 fm = 2(1.2 fm)A1/ 3 ⇒ A = ⎜ ⎟ = 30 ⎝ 2.4 fm ⎠ Only silicon has a stable isotope of A = 30.

43.5. Visualize: The masses of the nuclei are found by subtracting the mass of Z electrons from the atomic masses in Appendix C, and then converting to kg.

m(6 Li) = 6.015121 u − 3(0.000548) u = 6.0135 u m( 207 Pb) = 206.975871 u − 82(0.000548) u = 206.9309 u The radii are computed from r = r0 A1/ 3 where r0 = 1.2 fm. The densities are computed from

ρ= Solve:

m 4 3

π r3

(a) For 6 Li:

⎛ 1.661 × 10−27 kg ⎞ −27 m = 6.0135 u ⎜ ⎟ = 9.988 × 10 kg 1 u ⎝ ⎠ r = r0 A1/ 3 = (1.2 × 10−15 m)(6)1/ 3 = 2.18 × 10−15 m ≈ 2.2 × 10−15 m

ρ= (b) For

207

4 3

m 9.988 × 10−27 kg =4 = 2.3 × 1017 kg/m3 −15 3 3 (2 18 10 m) πr π . × 3

Pb:

⎛ 1.661 × 10−27 kg ⎞ −25 m = 206.9309 u ⎜ ⎟ = 3.437 × 10 kg 1 u ⎝ ⎠ r = r0 A1/ 3 = (1.2 × 10−15 m)(207)1/ 3 = 7.10 × 10−15 m ≈ 7.1× 10−15 m

ρ=

4 3

m 3.437 × 10−25 kg =4 = 2.3 × 1017 kg/m3 −15 3 3 (7 10 10 m) πr π . × 3

Assess: The densities are very similar because the nucleons are tightly packed in both cases. All nuclei have similar densities. This is also a typical density for a neutron star.

43.6. Model: Assume that air is 21% O2 and 79% N2, and that its density is 1.2 kg/m3.

Solve:

The masses of oxygen and nitrogen molecules are

Moxygen = (0.21)ρairV = (0.21)(1.2 kg/m3)(1 m3) = 0.252 kg Mnitrogen = (0.79)ρairV = 0.79(1.2 kg/m3)(1 m3) = 0.948 kg The number of oxygen molecules is

N oxygen =

M oxygen MA

NA =

0.252 kg × 6.02 × 1023 mol−1 = 4.74 × 1024 0.032 kg/mol

The number of protons from oxygen molecules is N oxygen protons = 4.743 × 1024 × 16 = 7.59 × 1025 Likewise, the number of protons from nitrogen molecules is

N nitrogen protons =

0.948 kg × 6.02 × 1023 mol −1 × 14 = 2.85 × 1026 0.028 kg/mol

The total number of protons is 7.59 × 1025 + 2.85 × 1026 = 3.6 × 1026. Because nitrogen and oxygen each have an equal number of protons and neutrons, this is also the total number of neutrons.

43.7. Solve: The nuclear density was found to be 2.3 × 1017 kg/m3. Thus 3 ⎛ 4π ⎞ 11 −2 M = ρ nuclearV = ( 2.3 × 1017 kg/m3 ) ⎜ ⎟ ( 0.5 × 10 m ) = 1.2 × 10 kg ⎝ 3 ⎠

Assess:

The nuclear density is tremendously large compared to the density of familiar liquids and solids.

43.8. Solve: The chart shows stable and unstable nuclei for all nuclei with Z ≤ 8. A black dot represents stable isotopes, a dark gray dot represents isotopes that undergo beta-minus decay, and a light gray dot represents isotopes that undergo beta-plus decay or electron-capture decay.

43.9. Solve: (a) A = 36, for which 36S and 36Ar are stable.

(b) Nuclei with A = 8 and A = 5 have no stable nuclei.

43.10. Solve: From Equation 43.6, the binding energy for 3 H is B = ( ZmH + Nmn − matom ) = 1(1.00783 u ) + 2 (1.00866 u ) − 3.01605 u = 0.00910 u × 931.49 MeV/u = 8.48 MeV The binding energy per nucleon is

1 3

( 8.48 MeV ) = 2.83 MeV.

3

For He, the binding energy is B = ZmH + Nmn − matom = 2 (1.00783 u ) + 1(1.00866 u ) − 3.01603 u = 0.00829 u × 931.49 MeV/u = 7.72 MeV

The binding energy per nucleon is 13 ( 7.72 MeV ) = 2.57 MeV.

43.11. Solve: From Equation 43.6, the binding energy for 54Fe is B = ZmH + Nmn − matom = 26 (1.00783 u ) + 32 (1.00866 u ) − 57.933278 u = 0.54742 u × 931.49 MeV/u = 509.92 MeV The binding energy per nucleon is

1 58

( 509.92 MeV ) = 8.7917 MeV.

58

For Ni, the binding energy is B = ZmH + Nmn − matom = 28 (1.00783 u ) + 30 (1.00866 u ) − 57.935346 u

= 0.54369 u × 931.49 MeV/u = 506.45 MeV The binding energy per nucleon is

1 58

( 506.45 MeV ) = 8.7318 MeV.

Assess: The binding energy per nucleon is higher for iron but the binding energy per nucleon for nickel isn’t much different; we expect both of these results from Figure 43.6

43.12. Solve: From Equation 43.6, the binding energy for 3He is B = ZmH + Nmn − matom = 2 (1.00783 u ) + 1(1.00866 u ) − 3.01603 u = 0.00829 u × 931.49 MeV/u = 7.722 MeV The binding energy per nucleon is

1 3

( 7.722 MeV )

= 2.57 MeV. For 4He, the binding energy is

B = ZmH + Nmn − matom = 2 (1.00783 u ) + 2 (1.00866 u ) − 4.00260 u

= 0.03038 u × 931.49 MeV/u = 28.30 MeV The binding energy per nucleon is 4

3

4

1 4

( 28.30 MeV ) = 7.07 MeV.

for He than for He, He is more tightly bound.

Because the binding energy per nucleon is more

43.13. Solve: From Equation 43.6, the binding energy for 12C is B = ZmH + Nmn − matom = 6 (1.00783 u ) + 6 (1.00866 u ) − 12.00000 u = 0.09894 u × 931.49 MeV/u = 92.162 MeV The binding energy per nucleon is

1 12

( 92.162 MeV ) = 7.68 MeV.

13

For C, the binding energy is B = ZmH + Nmn − matom = 6 (1.00783 u ) + 7 (1.00866 u ) − 13.00336 u

= 0.10424 u × 931.49 MeV/u = 97.099 MeV The binding energy per nucleon is

1 13

( 97.099 MeV ) = 7.47 MeV.

Thus, 12C is slightly more tightly bound.

43.14. Solve: (a) From Equation 43.6, the binding energy for 12C is B = ZmH + Nmn − matom = 6 (1.00783 u ) + 6 (1.00866 u ) − 12.00000 u = 0.09894 u × 931.49 MeV/u = 92.162 MeV The binding energy per nucleon is

1 12

( 92.162 MeV ) = 7.6801 MeV.

58

(b) For Ni, the binding energy is

B = ZmH + Nmn − matom = 28 (1.00783 u ) + 30 (1.00866 u ) − 57.935346 u

= 0.54369 u × 931.49 MeV/u = 506.45 MeV The binding energy per nucleon is (c) For

1 58

( 506.45 MeV ) = 8.7318 MeV.

226

Ra,

B = ZmH + Nmn − matom = 88 (1.00783 u ) + 138 (1.00866 u ) − 226.025402 u

= 1.85872 u × 931.49 MeV/u = 1731.38 MeV The binding energy per nucleon is

1 226

(1731.38 MeV ) = 7.6610 MeV.

43.15. Solve: 90.51% of neon is 20Ne, 0.27% of neon is 21Ne, and the remaining 9.22% of neon is 22Ne. The chemical atomic mass of neon is 0.9051(19.992439 u) + 0.0027(20.993845 u) + 0.0922(21.991384 u) = 20.17945 u ≈ 20.179 u

43.16. Solve: 78.99% of magnesium is 24Mg, 10.00% of magnesium is 25Mg, and the remaining 11.01% of magnesium is 26Mg. The chemical atomic mass of magnesium is

0.7899(23.985045 u) + 0.1000(24.985839 u) + 0.1101(25.982596 u) = 24.30505 u ≈ 24.305 u

43.17. Model: The force is the negative of the slope of a potential energy graph. Visualize: Please refer to Figure 43.8. Solve: For the two nucleons that are separated by 1.5 fm, the potential energy is U ≈ −38 MeV. A line drawn tangent to the curve at this point goes from ≈ −70 MeV at 1 fm to ≈ −20 MeV at 2 fm. Its slope is

ΔU 50 MeV 50 MeV ≈ = × 1.6 × 10−19 J/eV = 8000 J/m Δr 1.0 fm 1 × 10−15 m Thus F = −ΔU Δr = −8000 N. The strength of the force is 8000 N.

43.18. Model: The force is the negative of the slope of a potential energy graph. Visualize: Please refer to Figure 43.8. Solve: The force is zero when U is at its maximum negative value at r ≈ 1.0 fm. The maximum force occurs at r ≈ 1.5 fm where the slope is a maximum. The slope is approximately 8000 N, so F = −8000 N. For r ≤ 1.0 fm, the slope is negative and hence the force is positive. For r ≥ 1.0 fm, the slope is positive and hence the force is negative.

43.19. Solve: From Figure 43.8, the nuclear potential energy at r = 1.0 fm is Unuclear = −50 MeV = −(50 MeV)(1.6 × 10−19 J/eV) = −8.0 × 10−12 J The gravitational potential energy is

U grav

( 6.67 ×10 Gm 2 =− =− r

−11

Nm 2 /kg 2 )(1.67 × 10−27 kg )

2

= −1.86 × 10−49 J 1.0 × 10−15 m U 1.86 × 10−49 J ⇒ grav = = 2.3 × 10−38 U nuclear 8.0 × 10−12 J

43.20. Solve: (a) The A = 10 nuclei listed in Appendix C are 10Be, 10B, and 10C. Their nuclear energy level

diagrams are shown in the figure below. 10Be has Z = 4, so N = 6. 10B has Z = 5, N = 5, and 10C has Z = 6, N = 4. These three nuclei are called the A = 10 isobars.

(b) 10B is a stable nucleus, but 10Be and 10C are radioactive. 10Be undergoes beta-minus decay and 10C undergoes beta-plus decay. In beta-minus decay, a neutron within the nucleus changes into a proton and an electron. In betaplus decay, a proton changes into a neutron and a positron. Assess: The above decays for 10Be and 10C are consistent with the fact that the line of stability follows the N = Z line for Z < 16.

43.21. Solve: (a) The A = 14 nuclei listed in Appendix C are 14C, 14N, and 14O. 14C has Z = 6, so N = 8. 14N

has Z = 7, N = 7, and 14O has Z = 8, N = 6. These 3 nuclei are the A = 14 isobars.

(b) 14N is a stable nucleus, but 14C and 14O are radioactive. 14C undergoes beta-minus decay and 14O undergoes beta-plus decay. In beta-minus decay, a neutron within the nucleus changes into a proton and an electron. In betaplus decay, a proton changes into a neutron and a positron. Assess: The above decays for 14C and 14O are consistent with the fact that the line of stability follows the N = Z line for Z < 16.

43.22. Model: The number of radioactive atoms decreases exponentially with time. t/t

⎛ 1 ⎞ 1/2 The number of remaining 226Ra nuclei at time t is N = N 0 ⎜ ⎟ . ⎝ 2⎠ (a) After 200 years, N = (1.0 × 1010)(1/2)200 years/1600 years = 9.17 × 109. (b) After 2000 years, N = (1.0 × 1010)(1/2)2000 years/1600 years = 4.20 × 109. (c) After 20,000 years, N = (1.0 × 1010)(1/2)20,000 years/1600 years = 1.73 × 106. Assess: Do not think that if half the nuclei decay during one half-life, then all will decay in two half-lives. Solve:

43.23. Model: The number of radioactive atoms decreases exponentially with time. Solve:

The number of remaining 109Cd nuclei at time t is

⎛1⎞ N = N0 ⎜ ⎟ ⎝ 2⎠

t/t1/2

(a) After 50 days, N = (1.00 × 1012)(1/2)50 days/462 days = 9.28 × 1011. (b) After 500 days, N = (1.00 × 1012)(1/2)500 days/462 days = 4.72 × 1011. (c) After 5000 days, N = (1.00 × 1012)(1/2)5000 days/462 days = 5.52 × 108. Assess: Do not think that if half the nuclei decay during one half-life, then all will decay in two half-lives.

43.24. Model: The number of radioactive atoms decreases exponentially with time.

Solve: (a) 3H is called tritium and has Z = 1 and N = 2. 3H undergoes beta-minus decay and gives rise to the daughter nucleus 3He, that is, 3

H →3He + β −

(b) From Equation 43.18, the decay rate is

r=

1

τ

=

0.693 0.693 1 year = × = 1.78 × 10−9 s −1 t1 2 12.33 years 3.15 × 107 s

The time constant is τ = 5.62 × 108 s. The half-life in the above calculation was obtained from Appendix C.

43.25. Model: The number of atoms decays exponentially with time. Solve:

The number of remaining radioactive atoms at t = 50 min and t = 200 min is

⎛1⎞ N 50 = N 0 ⎜ ⎟ ⎝2⎠

t / t1/ 2

⎛1⎞ N 200 = N 0 ⎜ ⎟ ⎝2⎠

t / t1/ 2

⎛1⎞ = (1.0 × 1010 ) ⎜ ⎟ ⎝ 2⎠

(50 min) /(100 min)

⎛1⎞ = (1.0 × 1010 ) ⎜ ⎟ ⎝ 2⎠

= 7.07 × 109 atoms (200 min) /(100 min)

= 2.50 × 109 atoms

Thus the number that decay between 50 min and 200 min is Ndecay = N50 – N200 = 4.57 × 109 atoms Each decay emits an alpha particle, so there are 4.57 × 109 alphas emitted.

43.26. Solve: The activity of a radioactive sample is R = rN. From Appendix C, the half-life of 60Co is t1/2 = 5.27 yr = 1.663 × 108 s. Thus the decay rate is

r=

1

τ

=

ln 2 0.693 = = 4.17 × 10−9 s −1 t1/ 2 1.663 × 108 s

We know that the sample’s activity is R = 1.25 × 109 Bq = 1.25 × 109 decays/s, so the number of atoms in the sample is N=

R 1.25 × 109 decays/s = = 3.00 × 1017 atoms r 4.17 × 109 s −1

The mass of the sample is M = mN = (60 × 1.661 × 10–27 kg)(3.00 × 1017) = 2.99 × 10–8 kg = 299 μg

43.27. Model: The activity R of a radioactive sample is the number of decays per second. Solve:

The rate of decay is

R=

⇒ t1 2 =

dN N ( 0.693) N = rN = = dt t1 2 τ

( 0.693) N = ( 0.693) ( 5.0 × 1015 atoms ) = 6.93 × 106 s × R

5.0 × 108 Bq

1 day = 80.2 days ≈ 80 days 86,400 s

43.28. Solve: (a) The decay is (b) The decay is

35

−

230

Th → X + 4 He, so X =

226

S → X + e + v , so X = Cl. 35

(c) The decay is X → 40 K + e + + v, so X = 40 Ca. (d) The decay is

24

Na →24 Mg + e− + v → X + γ , so X =

24

Mg.

Ra.

43.29. Solve: (a) The decay is X → 224 Ra + 4 He, so X = 228 Th. (b) The decay is X →

207

Pb + e − + v , so X =

−

207

(c) The decay is Be + e → X + v, so X = Li. 7

(d) The decay is X → 60 Ni + γ , so X =

7

60

Ni.

Tl.

Pu → 235 U + 4 He. Solve: Using Appendix C to find the masses, the energy released is

43.30. Model: The decay is

239

E = Δmc 2 = ⎡⎣ m ( 239 Pu ) − m ( 235 U ) − m ( 4 He ) ⎤⎦ c 2 = [ 239.052157 u − 235.043924 u − 4.002602 u ] c 2 = ( 0.00563 u ) c 2 × 931.49 MeV/c 2 u = 5.24 MeV Assess:

Essentially all of this energy goes into the alpha particle’s kinetic energy.

43.31. Model: Assume the energy released goes into the alpha particle’s kinetic energy so that Kα = 5.52 MeV. Visualize: Use Equation 43.22: Kα = (mX − mY − mHe )c 2 = 5.52 MeV. We are also given mX + mY = 452 u. Solve:

(mX − mY − mHe )c 2 = 5.52 MeV 1u ⎛ ⎞ (mX − mY − mHe ) = 5.52 MeV/c 2 ⎜ = 0.00593 u 2 ⎟ ⎝ 931.49 MeV/c ⎠ mX − mY = 0.00593 u + mHe = 0.00593 u + 4.00260 u = 4.00853 u We now have a system of two equations in two unknowns which we will add together to solve for mX .

⎧mX − mY = 4.00853 u ⎨ ⎩mX + mY = 452 u 2mX = 456 u mX = 228 u Looking in Appendix C for a nucleus with a mass of 228 u which decays by α decay, we find 228Th. Assess: 228Th is a step in the 232Th decay series. 238Th has a half-life of 1.9 yr and decays by α radiation releasing 5.52 MeV of energy.

43.32. Model: The decay is 3 H → 3 He + e − + v . Solve: Beta-minus decay leaves the daughter atom as a positive ion. However, the mass of the ion plus the mass of the escaping electron is the mass of a neutral atom, which is what is tabulated in Appendix C. Thus the mass loss is the mass difference between the two neutral atoms. In Appendix C we find that m ( 3 H ) = 3.016049 u, and m ( 3 He ) = 3.016029 u. The energy released in the beta-minus decay corresponds to a

mass change of 0.000020 u. The energy released is E = ∆mc2 = (0.000020 u)(931.49 MeV/u) = 0.0186 MeV Assess:

The energy E is shared between the electron and the antineutrino.

43.33. Model: The decay is 19 O → 19 F + e − + v . Solve: Beta-minus decay leaves the daughter atom as a positive ion. However, the mass of the ion plus the mass of the escaping electron are the mass of a neutral atom, which is what is tabulated in Appendix C. Thus the mass loss is the mass difference between the two neutral atoms. In Appendix C we find that m ( 19 O ) = 19.003577 u and m 19 F = 18.998404 u. The energy released in the beta-minus decay corresponds to a

mass change of 0.005173 u. The energy released is E = Δmc 2 = ( 0.005173 u ) c 2 × Assess:

931.49 MeV/c 2 = 4.82 MeV 1u

This energy is shared between the electron and the antineutrino.

43.34. Model: The decay is n → p+ + e– + v . Solve:

The mass lost in this decay is

∆m = m(n) – m(p+) – m(e–) = 1.00866 u – 1.00728 u – 0.00055 u = 0.00083 u The energy released is the energy equivalent of 0.00083 u, or Assess:

E = (0.00083 u) × (931.49 MeV/u) = 0.773 MeV This energy is shared between the electron and the neutrino.

43.35. Model: The rad (radiation absorbed dose) measures the energy deposited in an irradiated material. Solve: Because 1 Gy = 1.00 J/kg of absorbed energy, 1.5 Gy corresponds to a dose of 1.5 J/kg. This means a 150 g tumor absorbs an amount of energy equal to (1.50 J/kg)(0.150 kg) = 0.225 J

43.36. Model: The gray measures the energy deposited in an irradiated material. Solve: The energy to be absorbed by the 100 g tumor is 0.20 J. Because 1 Gy is defined as 1.00 J/kg of absorbed energy, the dose given is 0.20 J 1 Gy × = 2.0 Gy 100 g 1.00 J/kg

43.37. Model: The radiation dose in units of rems is a combination of deposited energy and biological effectiveness. The RBE for beta radiation is 1.5. Solve: 1 Gy is defined as 1.00 J/kg of absorbed energy. In the case of the 50 kg worker, the energy absorbed per kg is

20 × 10−3 J = 4.0 × 10−4 J/kg 50 kg This corresponds to a dose of

4.0 × 10−4 J/kg ×

1 Gy = 4.0 × 10−4 Gy 1.00 J/kg

The dose equivalent in rems is (4.0 × 10−4 Gy)(1.5) = 0.060 Sv = 60 mrem.

43.38. Solve: For α-particles, the RBE = 15. The dose in Sv of the alpha radiation is 30 Gy × 15 = 450 Sv. For γ - rays, the RBE = 1. The dose in Gy of 450 Sv of γ - rays is 450 Gy.

43.39. Model: Assume the 197Au nucleus remains at rest. Visualize:

The radii of the alpha particle and the gold nucleus are rα = (1.2 fm)(4)1/3 = 1.90 fm

rAu = (1.2 fm)(197)1/3 = 6.98 fm

If the alpha just touches the surface of the gold nucleus, the distance between their centers is rf = 1.90 fm + 6.98 fm = 8.88 fm.

Solve:

(a) The energy conservation equation Kf + Uf = Ki + Ui is

0 J+ ⇒ vi =

1

(2e)(79e) = 1 mvi2 + 0 J 4π ε 0 8.88 × 10−15 fm 2

2(9.0 × 109 N m 2 /C2 )(158)(1.60 × 10−19 C) 2 = 3.51 × 107 m/s (4 × 1.661 × 10−27 kg)(8.88 × 10 −15 fm)

(b) The energy of the alpha particle is

K = 12 mvi2 = 12 (4 × 1.661× 10−27 kg)(3.51 × 107 m/s) 2 = 4.09 × 10 −12 J × (1 MeV/1.60 × 10 −13 J) = 25.6 MeV

43.40. Model: Assume the 207Pb nucleus remains at rest. Visualize:

The radii of the proton and the lead nucleus are rp = (1.2 fm)(1)1/3 = 1.20 fm

rPb = (1.2 fm)(207)1/3 = 7.10 fm

At the momentum of impact on the lead nucleus, the distance between their centers is rf = 1.20 fm + 7.10 fm = 8.30 fm.

Solve:

The energy conservation equation Kf + Uf = Ki + Ui is

20 MeV + ⇒ K i = 20 MeV +

1

(e)(82e) = Ki + 0 J 4π ε 0 8.30 × 10−15 fm

(9.0 × 109 N m 2 /C 2 ) (82)(1.60 × 10−19 C) 2 1 MeV × = 34.2 MeV −15 8.30 × 10 fm 1.60 × 10−13 J

Solve: (a) The sun’s mass of 1.99 × 1030 kg is unchanged, but it assumes the density of nuclear matter, which we found to be ρnuc = 2.3 × 1017 kg/m3. The volume of the collapsed sun is

43.41.

4 M 1.99 × 1030 kg V = π r3 = S = = 8.65 × 1012 m3 ρ nuc 2.3 × 1017 kg/m3 3

Thus its radius is 1/ 3

⎛ 3(8.65 × 1012 m3 ) ⎞ r =⎜ ⎟ 4π ⎝ ⎠

= 12,700 m = 12.7 km

(b) We can use the conservation of angular momentum to find the new rotational period: 2 M R2 (Iω)after = (Iω)before ⇒ ωafter = I before ωbefore = 5 S S ωbefore 2 I after M Sr 2 5 2

2

⎛ 1.27 × 104 m ⎞ 2π ⎛ RS ⎞ 2π −4 ⇒ =⎜ ⎟ ⇒ Tafter = ⎜ ⎟ ( 27 days ) = 7.8 × 10 s = 780 μs 8 Tafter ⎝ r ⎠ Tbefore ⎝ 6.96 × 10 m ⎠

43.42. Visualize: Let x1 be the fraction of 69Ga with atomic mass 68.92 u and x2 be the fraction of 71Ga with atomic mass 70.92 u. We know that x1 + x2 = 1 or x2 = 1 − x1. We seek x1. Solve: x1 (68.92 u) + x2 (70.92 u) = 69.72

x1 (68.92 u) + (1 − x1 )(70.92 u) = 69.72 ⇒ x1 =

69.72 u − 70.92 u = 0.60 68.92 u − 70.92 u

The percent abundance of 69Ga is 60% Assess: While Ga does not appear in Appendix C, this result can be verified in other tables of isotopes of the elements.

43.43. Solve: (a) The binding energy of the electron in a hydrogen atom is B = 13.6 eV. That is, the mass decreases by the equivalent of 13.6 eV when an electron and proton form a hydrogen atom. Since B = ∆mc2, Δm =

13.6 eV 13.6 eV 1u = × = 1.46 × 10−8 u c2 c2 931.49 MeV/c 2

As a percentage of the hydrogen mass, the mass decrease is

Δm 1.46 × 10−8 u = = 1.45 × 10−6% 1.007825 u 1.007825 u (b) The mass decrease is ∆m = 2mp + 2mn – mHe nuc. These are nuclear masses, but Appendix C tabulates atomic masses. Add and subtract the mass of two electrons:

∆m = 2(mp + me) + 2mn – (mHe nuc – 2me) = 2mH + 2mn – mHe where mH is the mass of a hydrogen atom and mHe is the mass of a helium atom. Using Appendix C, ∆m = 2(1.007825 u) + 2(1.008665) – 4.002602 = 0.0304 u As a percentage of the helium mass, the mass decrease is 0.0304 u = 0.0076 = 0.76% 4.0026 u (c) Although mass does change in chemical reactions, the change is an incredibly small fraction of the mass of the atoms. No experiment will be sensitive to changes of ≈1 × 10–6%, so this small change in mass is easily neglected. Not so in nuclear reactions, where the mass change can be ≈1% of the particle masses. Not only is this mass change easily detectable, it is essential for understanding nuclear reactions.

43.44. Visualize: Please refer to Figure 43.6 for the graph of the binding energy versus mass number. Solve: For a nucleus of mass number of 240, the binding energy per nucleon is ≈7.8 MeV. Hence, the total binding is (7.8 MeV)(240) = 1870 MeV. For a nucleus of mass number of 120, the binding energy per nucleon is ≈8.5 MeV and the total binding energy is (8.5 MeV)(120) = 1020 MeV. When a nucleus with mass number 240 fissions into two nuclei with mass number 120, the total energy released is 2(1020) MeV – 1870 MeV = 170 MeV

43.45. Visualize: Please refer to Figure 43.6 for the graph of the binding energy versus mass number. Solve: For a 4He nucleus, the binding energy per nucleon is 7.2 MeV. Because A = 4, the binding energy of the 4 He nuclei is 7.2 MeV × 4 = 28.8 MeV. Three 4He nuclei thus have a total energy of 28.8 MeV × 3 = 86.4 MeV. For the 12C nucleus, the binding energy per nucleon is 7.7 MeV. Because A = 12, the binding energy of 12C is 7.7 MeV × 12 = 92.4 MeV. When three 4He nuclei fuse together to form a 12C nucleus, the total energy released is 92.4 MeV – 86.4 MeV = 6.0 MeV.

43.46. Visualize: Please refer to Figure 43.6 for the graph of the binding energy versus mass number. Solve: For a 56Fe nucleus, the binding energy per nucleon is 8.8 MeV. Because A = 56, the total binding energy is 8.8 MeV × 56 = 493 MeV. For a 28Al nucleus, the binding energy per nucleon is 8.4 MeV and the total binding energy is 28 × 8.4 MeV = 235 MeV. The total binding energy for two 28Al nuclei is 2 × 235 MeV = 470 MeV. Because the 56Fe is more tightly bound than two 28Al nuclei, 56Fe cannot fission spontaneously. It could perhaps be forced to fission with the input of 493 MeV – 470 MeV = 23 MeV.

43.47. Solve: (a) From Appendix C, the A = 17 isobars are 17N, 17O, and 17F. (b) The isotope 17O is stable but rare. (c) The daughter nucleus for both of the unstable isotopes is 17O. 17N decays by beta-minus decay. 17F decays by electron capture.

43.48. Solve: (a) From Appendix C, the A = 19 isobars are 19O, 19F, and 19Ne.

(b) The 19F isotope is stable. (c) The daughter nucleus for both of the unstable isotopes is 19F. 19O decays by beta-minus. 19Ne decays by betaplus.

43.49. Solve: Equation 43.15 is N = N0e–t/τ. The number e = 2.718 is some power of 12 . Define p such that ⎛1⎞ e=⎜ ⎟ ⎝2⎠

p

To determine p, take the logarithm of both sides: ln(e) = 1 = p ln(1/2) = − p ln 2 ⇒ p = −

1 ⎛1⎞ ⇒e=⎜ ⎟ ln 2 ⎝2⎠

−1/ ln 2

With this, the decay equation can be written ⎡⎛ 1 ⎞ −1/ ln 2 ⎤ N = N 0 e − t / τ = N 0 ⎢⎜ ⎟ ⎥ ⎣⎢⎝ 2 ⎠ ⎦⎥

−t / τ

Recall that (xa)b = xab. Since t1/2 = τ ln2, the decay equation is ⎡⎛ 1 ⎞ −1/ ln 2 ⎤ N = N 0 ⎢⎜ ⎟ ⎥ ⎢⎣⎝ 2 ⎠ ⎥⎦ This is Equation 43.19.

−t / τ

⎛1⎞ = N0 ⎜ ⎟ ⎝ 2⎠

( −1/ ln 2)( − t / τ )

⎛1⎞ = N0 ⎜ ⎟ ⎝2⎠

t / τ ln 2

⎛1⎞ = N0 ⎜ ⎟ ⎝2⎠

t / t1/ 2

43.50. Solve: The radius of a 238U nucleus is r = r0A1/3 = (1.2 fm)(238)1/3 = 7.436 fm For the de Broglie wavelength to be equal to the diameter of the 238U nucleus, p = h λ = h 2r. Hence, 2

K=

2

1 1 eV p 2 ⎛ h ⎞ 1 ⎛ 6.63 × 10−34 J s ⎞ =⎜ ⎟ =⎜ × = 0.93 MeV ⎟ −15 −27 2m ⎝ 2r ⎠ 2m ⎝ 2 × 7.436 × 10 m ⎠ 2 ( 4 × 1.67 × 10 kg ) 1.6 × 10−19 J

43.51. Solve: From Appendix C, the half-life of 3H is t1/ 2 = 12.33 years. The activity of a radioactive sample is defined as R = rN. The decay rate is r=

1

τ

=

0.693 0.693 1 year = × = 1.784 × 10−9 s −1 t1 2 12.33 years 3.15 × 107 s

The number of atoms in the sample is N =

2.0 × 10−3 g × 6.02 × 1023 mol −1 = 4.01 × 1020 3 g/mol

⇒ R = rN = (1.784 × 10−9 s−1)(4.01 × 1020) = 7.16 × 1011 Bq = 7.16 × 1011 Bq ×

1 Ci = 19.4 Ci 3.7 × 1010 Bq

43.52. Model: The 14C/12C ratio in living matter is 1.3 × 10–12. After death, this ratio decays exponentially with time. Solve: The decay equation in terms of N, the number of radioactive atoms, can equally well be written in terms of the ratio R = N/Nref where Nref is a stable reference. In dating experiments, the number of 12C atoms is a reference to which the number of 14C atoms can be compared. Thus ⎛1⎞ R = R0 ⎜ ⎟ ⎝2⎠

t / t1/ 2

⎛1⎞ = (1.3 × 10−12 ) ⎜ ⎟ ⎝2⎠

t / 5730 yr

where R0 = 1.3×10–12 is the ratio at the time of death and the half-life of 14C is known to be 5730 yr. We can find the time at which the ratio has decreased to R by taking the logarithm of both sides: ⎛1⎞ ⎜ ⎟ ⎝2⎠

t / 5730 yr

=

R t R ln(1.3 × 10−12 /R ) ⎛1⎞ ⎛ ⎞ ⇒ ln ⎜ ⎟ = ln ⎜ ⇒t = ⋅ 5730 yr −12 −12 ⎟ 1.3 × 10 5730 yr ⎝ 2 ⎠ ln(2) ⎝ 1.3 × 10 ⎠

This sample has R = 1.65 × 10–13, so its age is t=

ln(1.3 × 10−12 /1.65 × 10−13 ) ⋅ 5730 yr = 17,100 yr ln(2)

43.53. Solve: From Appendix C, the half-life of a 133Cs sample is t1 2 = 30 years ×

3.15 × 107 s = 9.45 × 108 s 1 year

The activity of a radioactive sample is R = rN. The decay rate is

r=

1

τ

=

⇒N =

0.693 0.693 = = 7.33 × 10−10 s −1 t1 2 9.45 × 108 s

R0 2.0 × 108 Bq = = 2.73 × 1017 atoms r 7.33 × 10−10 s −1

That is, at t = 0 s, we have 2.73 × 1017 137Cs atoms and after a very long time all will have decayed. Thus 2.73 × 1017 beta particles will have been emitted.

43.54. Model: The number of radioactive tracers decays exponentially. Solve:

(a) The activity of a radioactive sample decays as

⎛1⎞ R = R0 ⎜ ⎟ ⎝ 2⎠

t t1 2

16 hours t1 2

⎛1⎞ ⇒ 95 mCi = 115 mCi ⎜ ⎟ ⎝ 2⎠

⎛ 95 ⎞ ⎛ 16 hours ⎞ ⎛ 1 ⎞ ⇒ ln ⎜ ⎟ ln ⎜ ⎟ ⇒ t1 2 = 58.0 hours ⎟=⎜ ⎝ 115 ⎠ ⎜⎝ t1 2 ⎟⎠ ⎝ 2 ⎠

(b) Let t be the time it takes the activity to drop from 95 mCi to 10 mCi. Using the results of part (a),

()

10 mCi = 95 mCi 1 2

t 58.0 hours

t ⎛ 10 ⎞ ⎛ ⎞ 1 ⇒ ln ⎜ ⎟ = ⎜ ⎟ ln ( 2 ) ⇒ t = 188 hours ⎝ 95 ⎠ ⎝ 58.0 hours ⎠

Ra → 219 Rn + 4 He. Solve: (a) The energy released in the above decay is

43.55. Model: The decay is

223

E = ⎡⎣ m ( 223 Ra ) − m ( 219 Rn ) − m ( 4 He ) ⎤⎦ × 931.49 MeV/u = [ 223.018499 u − 219.009477 u − 4.002602 u ] × 931.49 MeV/u = 5.98 MeV That is, each α-particle is released with an energy of 5.98 MeV = 5.98 × 106 eV × 1.6 × 10−19 J/eV = 9.57 × 10−13 J The amount of energy needed to raise the temperature of 100 mL of water at 18°C to 100°C is ⎛ 1 kg ⎞ Q = mcΔt = ( 0.10 L ) ⎜ ⎟ ( 4190 J/kg K )(100 K − 18 K ) = 34,400 J ⎝1L ⎠

The number of decays we need to generate this amount of energy is

34,400 J = 3.59 × 1016 9.57 × 10−13 J The total number of radium atoms in the cube at t = 0 s is

N0 =

1g × 6.02 × 1023 atoms/mol = 2.70 × 1021 atoms 223 g/mol

The number of needed decays is very small compared to N0, so we can write dN ΔN ≈ = − rN dt Δt

⇒ Δt = −

t ΔN ΔN ΔN = −τ = − 1/ 2 rN N ln 2 N

where we used r = 1/τ. ∆N = –3.59 × 1016, with the negative sign due to the fact that the number of radium atoms decreases. Since t1/2 = 11.43 days = 9.876 × 105 s, the time for 3.59 × 1016 decays is Δt = −

9.876 × 105 s (−3.59 × 1016 ) = 18.9 s ≈ 19 s ln 2 2.70 × 1021

(b) It’s possible that an alpha-particle collision will break the molecular bond in a very small number of H2O molecules, causing H2 gas and O2 gas to bubble out of the water. The water that remains in the container has not changed or been altered.

43.56. Model: The number of radioactive atoms decreases exponentially with time. Solve:

(a) If 90% of the sample has decayed, 10% is still present. This happens at time t such that

⎛1⎞ N = 0.10 N 0 = N 0 ⎜ ⎟ ⎝2⎠

t / t1/ 2

t=

⎛1⎞ ⇒⎜ ⎟ ⎝2⎠

t / t1/ 2

= 0.10 ⇒

⎛1⎞ ln ⎜ ⎟ = ln(0.10) t1/ 2 ⎝ 2 ⎠ t

ln(0.10) t1/ 2 = 3.32t1/ 2 ln(0.50)

That is, 90% decay in 3.32 half-lives. (b) When 99% of the sample has decayed, 1% is left. The analysis is the same as in part (a), giving

t= That is, 99% decay in 6.64 half-lives.

ln(0.010) t1/ 2 = 6.64t1/ 2 ln(0.50)

43.57. Model: The number of radioactive atoms decays exponentially with time. Solve:

At t = 2 hours, the number of A and B atoms is equal. Thus

⎛1⎞ N A = ( N A )0 ⎜ ⎟ ⎝2⎠

(2.0 h)/(0.5 h)

=

( N A )0 ⎛1⎞ = N B = ( N B )0 ⎜ ⎟ 16 ⎝2⎠

(2.0 h)/t1/ 2

Initially, (NA)0 = 5(NB)0. With this information, 5 ⎛1⎞ =⎜ ⎟ 16 ⎝ 2 ⎠

(2.0 h)/t1/ 2

⇒ ln(5/16) =

2.0 h ln(1/ 2) ln(1/ 2) ⇒ t1/ 2 = ⋅ 2 h = 1.19 h ≈ 1.2 h ln(5 /16) t1/ 2

43.58. Model: The number of 131Ba and 47Ca atoms decays exponentially.

Solve: From Equation 43.15, the number of radioactive atoms is N = N0e−t/τ. At time t = 0 s, (N0)Ca = 2(N0)Ba. The ratio of atoms at time t is

N Ca ( N 0 )Ca e − t τ Ca = N Ba ( N 0 )Ba e − t τ Ba Relating the half-life to the time constant,

τ Ca =

(t )

1 2 Ca

ln 2

=

4.5 days = 6.492 days ln 2 ⇒

τ Ba =

(t )

1 2 Ba

ln 2

=

12 days = 17.31 days ln 2

2 ( 0.0675 ) N Ca e = ( 2 ) − ( 2.5) 7 days 17.31 days = = 0.371 N Ba 0.3639 e − ( 2.5 ) 7 days 6.492 days

43.59. Model: The number of 40K atoms decays exponentially with time.

Solve: Suppose the number of 40K atoms is N0 at the time the lava solidifies. There are no 40Ar atoms at that time. As time passes, some 40K decay into 40Ar, but the total number of atoms locked inside the lava is unchanged. That is, NAr + NK = N0, where NK is the remaining number of 40K atoms. Dividing by NK, we have

N Ar N N 1 + 1 = 0.12 + 1 = 0 ⇒ K = = 0.893 NK NK N 0 1.12 That is, 89.3% of the 40K remains at a time when the Ar/K ratio is 0.12. The 40K decays as t /t

1/ 2 t ⎛1⎞ N K = N0 ⎜ ⎟ ln(1/2 ) = ln( N K / N 0 ) ⇒ t1/ 2 ⎝2⎠ ln(0.893) ⇒ t = (1.28 billion years) = 0.21 billion years = 210 million years ln(0.50)

43.60. Model: The number of 235U atoms decays exponentially with time. Solve:

The 235U decays as

⎛1⎞ N U = ( N U )0 ⎜ ⎟ ⎝ 2⎠

t / t1/ 2

The ratio of 235U atoms when the earth formed to the number now present is

( N U )0 1 = = (2)t / t1/ 2 = (2)(4500 my)/(700 my) = 86.1 ≈ 86 (1/2 )t / t1/ 2 NU That is, the abundance of 235U was 86 times larger then than it is now.

43.61. Model: Assume that the tracer stays in the body for the life of the tracer. Solve:

The initial activity of the source is

R0 = 30 μ Ci = 30 × 10−6 Ci ×

3.7 × 1010 Bq = 1.11× 106 Bq 1 Ci

The decay rate is r=

1

τ

=

ln 2 ln 2 = = 1.605 × 10−6 s −1 t1 2 5.0 days × 24 hours/day × 3600 s/hour

The number of atoms that will decay by beta emission is N0 =

R0 1.11 × 106 Bq = = 6.916 × 1011 r 1.605 × 10−6 s −1

Since 10% of the beta particles escape, there are 0.90 N0 disintegrations in the body. Since the average energy of the beta particle is 0.35 MeV, the energy absorbed by the body is E = 0.35 MeV × (1.6 × 10−19 J/eV) × 0.90 × (6.926 × 1011) = 0.03485 J

Thus, the radiation dose is

0.03485 J 1 Gy = 4.647 × 10−4 J/kg × = 4.65 × 10−4 Gy 75 kg 1.00 J/kg The radiation dose in rems is 4.65 × 10−4 Gy × 1.5 = 6.97 × 10−4 Sv = 69.7 mrem ≈ 70 mrem.

43.62. Solve: The heat energy required to transform a block of ice at 0°C to liquid water at 0°C is Q = mLf = m(3.33 × 105 J/kg) ⇒

Q absorbed energy = = 3.33 × 105 J/kg m mass

By definition, 1.00 J/kg of absorbed energy is 1 Gy. Therefore, the dose of gamma radiation needed is

3.33 × 105 J/kg ×

1 rad = 3.33 × 105 Gy 1.00 J/kg

43.63. Solve: Since the dose equivalent in Sv is the dose in Gy times the RBE, the dose is 0.30 × 10−3 Sv 1.00 J/kg = 3.53 × 10−4 Gy × = 3.53 × 10−4 J/kg 0.85 1 Gy Since only one-fourth of the body received x-ray exposure, the total amount of energy received is 1 3.53 × 10−4 J/kg × × 60 kg = 5.29 × 10−3 J 4 The energy of each photon is

10 keV ×

1.6 × 10−19 J = 1.6 × 10−15 J 1 eV

Thus, the number of x-ray photons absorbed by the body is 5.29 × 10−3 J = 3.3 × 1012 1.6 × 10−15 J

43.64. Model: The rate of removal of a tracer is the rate of decay plus the rate of excretion. Solve:

The rate at which the number of tracer atoms in the body decays is

dN ⎛ dN ⎞ ⎛ dN ⎞ =⎜ +⎜ = − rd N − re N = −( rd + re ) N = − reff N ⎟ ⎟ dt ⎝ dt ⎠decay ⎝ dt ⎠excrete where rd and re are the decay rate and the excretion rate. The effective loss rate is the sum of these two rates. The effective half-life is t1/ 2 =

⎡ 1 ln 2 ln 2 ln 2 1 ⎤ = = =⎢ + ⎥ reff rd + re ln 2/(t1/ 2 )d + ln 2/(t1/ 2 )e ⎣(t1/ 2 )d (t1/ 2 ) e ⎦

−1

The effective half-life is the inverse of the sum of the inverses of the half-lives of the individual processes. Note that this is exactly like finding the equivalent resistance of two parallel resistors. Here the two removal processes run “in parallel.” Using the known values, −1

⎡ ⎤ 1 1 t1/ 2 = ⎢ + ⎥ = 3.6 days ⎣ 9.0 days 6.0 days ⎦

43.65. Model: The number of 239Pu atoms decays exponentially. Solve: (a) The number of 239Pu atoms in a 1.0-μm-diameter particle is ⎛ m ⎞ ( ρV ) N = ρ ( 43π ) r 3 N A N =⎜ ⎟ NA = A MA MA ⎝ MA ⎠

(19,800 kg/m ) ( ) ( 0.5 ×10 = 3

4π 3

−6

m ) ( 6.02 × 1023 mol −1 ) 3

239 × 10−3 kg/mol

= 2.61 × 1010 ≈ 2.6 × 1010

(b) The activity of the particle is

R=

ln 2 0.6931 dN = rN = N= × 2.61× 1010 = 0.0239 Bq ≈ 0.024 Bq 24,000 years × 3.15 × 107 s/year dt t1 2

(c) The volume of the 50-μm diameter sphere of tissue around the particle is 3 4π 25 × 10−6 m ) = 6.545 × 10−14 m3 ( 3

This volume of the tissue has a mass of

m = ρV = (1000 kg/m3 )( 6.545 × 10−14 m3 ) = 6.545 × 10−11 kg In one year, the activity changes insignificantly. Thus the number of decays per year is dN Δt = ( 0.0239 Bq ) ( 3.15 × 107 s ) = 7.529 × 105 dt Since each decay creates an α-particle with energy 5.2 MeV, the total energy received per year by the tissue is (7.529 × 105 × 5.2 MeV) × (1.6 × 10−19 J/eV) = 6.264 × 10−7 J Dividing this energy by the tissue’s mass, the dose received by the tissue is 6.264 × 10−7 J/6.545 × 10−11 kg = 9.57 × 103 J/kg = 9.57 × 103 Gy The dose per year in rem is (9.57 × 103 Gy)(RBE) = (9.57 × 103 Gy)(15) = 1.436 × 105 Sv = 1.436 × 107 rem ≈ 1.4 × 107 rem (d) This is a very high dose to a very small volume of body mass. Table 43.5 gives a typical exposure (in units of rem/year) from various radiation sources. The background radiation from various natural occurring sources is about 300 mrem. The exposure to this tissue is roughly 50 million times the background level.

43.66. Model: Although individual radon atoms decay, the radon is always being replaced such that the radon concentration remains constant with a constant activity of 4 pCi/L. Solve: (a) Activity is R = rN, so the number of atoms giving rise to a known activity is N = R/r. The half-life of 222 Rn is 3.82 days = 330,000 s, so its decay rate is

r=

ln 2 ln 2 = = 2.1 × 10−6 s −1 t1/ 2 330,000

One liter is 1 L = 1000 cm3 = 0.0010 m3, so 1 m3 of air is 1000 L. At 4 pCi/L, the activity in 1 m3 is R = 4000 pCi = 4 × 10−9 Ci ×

3.7 × 1010 Bq = 148 Bq 1 Ci

Thus the number of 222Rn atoms in 1 m3 of air is R 148 Bq = = 7.05 × 107 atoms ≈ 7 × 107 atoms r 2.10 × 10−6 s −1

N=

(b) We first need to find how many 222Rn atoms decay within 3 cm of the person in 1 year. The person is a 25cm-diameter cylinder 180 cm tall. The decays must be inside a 31-cm-diameter cylinder (i.e., 25 cm + 3 cm + 3 cm) 183 cm tall. The volume of air in this shell is

V = π (0.155 m) 2 (1.83 m) − π (0.125 m) 2 (1.80 m) = 0.0498 m3 The concentration of 222Rn is 7.05 × 107 m–3, so the number of 222Rn atoms in this shell is N = (0.0498 m3) × (7.05 × 107 m–3) = 3.51 × 106. The total number of decays per second is

R = rN = (2.10 × 10−6 s −1 )(3.51 × 106 ) = 7.4 decays per second Only half of these decays direct the alpha particle toward the person, so the number of alphas striking the person in 1 year is Nα =

1 × (7.4 decays/s) × (3.15 × 107 s/yr) = 1.2 × 108 alphas/yr 2

Each alpha deposits 5.50 MeV = 8.8 × 10–13 J of energy, so the energy absorbed in 1 year is E = (1.2 × 108 alphas) × (8.8 × 10–13 J/alpha) = 1.06 × 10–4 J. The dose of alpha radiation (with RBE = 15) received by a 65 kg person in a year is dose =

1.06 × 10−4 J 1 Gy × × 15 = 2.4 × 10−5 Sv = 2.4 mrem 65 kg 1.00 J/kg

(c) This dose is roughly 1% of the natural background of radioactivity. Although an exposure increase of 1% may seem too small to worry about, we assumed in using a 65 kg mass that the radiation is uniformly received. But alpha radiation is not very penetrating, so the skin and outer extremities would receive a dose considerably higher than 2.4 mrem. Since some people are housebound and really would receive this yearly dose, an upper limit of 4 pCi/L seems prudent.

43.67. Model: Assume the alpha particle is a point particle. Solve:

From Equation 18.3, the mean free path of a molecule moving through a gas is

λ=

1 4 2π ( N / V )r 2

Treating the alpha particle as a point particle, the “collision cylinder” of radius 2r in the derivation of Equation 18.3 becomes a cylinder of radius r. Since the radius is squared, the 4 in the denominator of the expression vanishes. In Example 18.1, the mean free path of nitrogen molecules in nitrogen gas is calculated to be 225 nm. A point particle will have a mean free path 4 times greater, that is, 900 nm. The stopping distance of the alpha particle is the number of collisions times the mean free path. The number of collisions is the energy of the alpha particle divided by the energy lost per collision. The stopping distance is estimated to be 5.0 MeV × 900 nm = 1.667 × 105 × 900 × 10−9 m = 15 cm 30 eV

43.68. Model: The decay is AXZ → AYZ – 1 + e+ + ν. Solve: (a) This decay is energetically possible only if the mass of the X nucleus exceeds the mass of the Y nucleus plus the mass of the positron. The X nucleus contains Z protons and N neutrons while the Y nucleus has (Z – 1) protons and (N + 1) neutrons. Thus the mass requirement is Zmp + Nmn > (Z – 1)mp + (N + 1)mn + me Tabulated masses are atomic masses, which include the electron masses. Atom X has Z electrons and atom Y has (Z – 1) electrons. Add the term Zme to both sides, obtaining

Zmp + Zme + Nmn > ( Z − 1)mp + Zme + ( N + 1)mn + me ⇒ Z (mp + me ) + Nmn > ⎡⎣( Z − 1)(mp + me ) + ( N + 1)mn ⎤⎦ + 2me The term on the left is the mass of an AXZ atom and the term in square brackets on the right is the mass of an AYZ – 1 atom. Thus the threshold condition for beta-plus decay is m(AXZ) > m(AYZ – 1) + 2me (b) From Appendix C, m(13N) = 13.005738 u and m(13C) = 13.003355 u. Thus m(13C) + 2me = 13.004455 u. This is less than m(13N), so beta-plus decay is allowed. The energy released is

E = Δmc 2 = (0.001283 u)(931.49 MeV/u) = 1.20 MeV

43.69. Model: The number of 235U and 238U atoms decays exponentially with time.

Solve: From Appendix C, the half-life of 235U is (t1/2)235 = 7.04 × 108 yr and that of 238U is (t1/2)238 = 4.47 × 109 yr. Today, N238/N235 = 0.9928/0.0072 = 137.9, but it is thought that (N238)0 /(N235)0 ≈ 1 immediately after the supernova that created these elements. The decays are such that

N 238 ( N 238 )0 (1/ 2)t /(t1/ 2 )238 (1/ 2)t /(t1/ 2 )238 ⎛ 1 ⎞ = ⇒ =⎜ ⎟ t /( t1/ 2 )235 N 235 ( N 235 )0 (1/ 2) (1/ 2)t /(t1/ 2 )235 ⎝ 2 ⎠

t /( t1/ 2 )238 − t /( t1/ 2 ) 235

= 137.9

To solve for t, take the logarithm of both sides:

⎡⎛ t ⎞ ⎡⎛ 1 ⎞ ⎛ t ⎞ ⎤ ⎛ 1 ⎞ ⎤ ⎢⎜ ⎟ −⎜ ⎟ ⎥ ln(1/2) = t × ⎢⎜ ⎟ −⎜ ⎟ ⎥ ln(1/2 ) = ln137.9 ⎣⎢⎝ t1/ 2 ⎠ 238 ⎝ t1/ 2 ⎠ 235 ⎦⎥ ⎣⎢⎝ t1/ 2 ⎠ 238 ⎝ t1/ 2 ⎠ 235 ⎦⎥ ⇒t =

ln137.9 = 5.9 × 109 years ≈ 6 billion years [(1/t1/ 2 )238 − (1/t1/ 2 )235 ] ln(1/2)

The supernova occurred approximately 6 billion years ago. Assess: Our sun and solar system are ≈4.5 billion years old, so the supernova occurred ≈1.5 billion years before the debris coalesced to form the solar system.

43.70 Model: If we ignore the antineutrino, we can model beta decay as n → p+ + e–. Solve:

(a) The mass difference is ∆m = mn – mp – me = 1.00866 u – 1.00728 u – 0.00055 u = 0.00083 u

This “mass loss” is transformed into the kinetic energy of the proton and electron, so K = (0.00083 u) × (931.40 MeV/u) = 0.773 MeV (b) The neutron is initially at rest, with γn = 1, so the energy conservation equation is mnc2 = γpmpc2 + γemec2 The c2 cancels, so we can also write the equation as

Δm = mn − γ p mp − γ e me (c) The neutron momentum is zero, so the sum of the proton and electron momenta must be zero: 0 = γpmpvp – γemeve or γpmpvp = γemeve (d) Using the definition of γ,

γ=

1 1 − v 2 /c 2

γ 2 −1 c v2 1 γ 2 −1 v = 1 − = ⇒ = ⇒ γ v = (γ 2 − 1)1/ 2 c c2 γ2 γ2 γ

⇒

(e) Each side of the momentum equation has a γv term. Using the result of part (d), the momentum equation becomes

(γ p2 − 1)1/ 2 mpc = (γ e2 − 1)1/ 2 mec ⇒ mp2(γ p2 − 1) = me2(γ e2 − 1) ⇒ mp2γ p2 − mp2 = me2γ e2 − me2 where, in the second step, we squared both sides of the equation. We can use the energy equation to find meγe in terms of γp:

meγ e = mn − mpγ p ⇒ me2γ e2 = ( mn − mpγ p ) 2 = mn2 − 2mn mpγ p + mp2γ p2 ⇒ me2γ e2 − me2 = mn2 − me2 − 2mn mpγ p + mp2γ p2 Substitute this expression for the right side of the momentum equation, giving mp2γ p2 − mp2 = mn2 − me2 − 2mn mpγ p + mp2γ p2 The term involving γ p2 cancels. Solving for γp then gives

γp =

mn2 + mp2 − me2 2mn mp

=

(1.00866 u) 2 + (1.00728 u) 2 − (0.00055 u) 2 = 1.000000788 2(1.00866 u)(1.00728 u)

With γp known, γe is found from the energy equation:

γe =

mn − γ p mp me

=

(1.00866 u) − (1.000000788)(1.00728 u) = 2.508 (0.00055 u)

Finally, using the next-to-last step of part (d) to find v from γ, we obtain

vp =

γ p2 − 1 c γp

= 0.00126c and ve =

γ e2 − 1 c = 0.917c γe

The electron is ejected at nearly the speed of light, but the proton’s speed is sufficiently small that we can treat the proton non-relativistically. (f) The kinetic energies are

K p = (γ p − 1)mp = (0.000000788)(1.00728 u) = 7.94 × 10−7 u × 931.59 MeV/u = 0.0007 MeV K e = (γ e − 1)me = (1.508)(0.00055 u) = 8.29 × 10−4 u × 931.59 MeV/u = 0.7722 MeV The total kinetic energy is Kp + Ke = 0.773 MeV, the same as we found in part (a). (g) The proton has almost no recoil energy. Since nucleons are bound inside the nucleus by many MeV, a recoil energy of 0.0007 MeV is far too small to eject the proton from the nucleus. To say that the electron is ejected because it is moving so fast—0.917c—is close to being correct, but not quite. Although the electron is ejected with ≈0.8 MeV of kinetic energy, the electric potential energy of an electron at the surface of a nucleus is typically –5 MeV. A classical electron, even one moving at 0.917c, does not have enough kinetic energy to escape the attractive electric force of the nucleus. However, the electron is not a classical particle. A simple calculation finds that the de Broglie wavelength of an electron with v = 0.917c is λ ≈ 1000 fm. This is ≈100 times the diameter of a nucleus. The uncertainty principle prevents a particle from being confined in a region much smaller than its de Broglie wavelength. Consequently, the electron cannot be confined within the nucleus.

43.71. Model: A quantum particle can tunnel through a classically forbidden region.

Solve: (a) Kinetic energy is K = E – U. The alpha particle has E = 5 MeV whether it is inside or outside the nucleus. Inside, where U = –60 MeV, the kinetic energy is Kin = 65.0 MeV. Outside, where U = 0 MeV, the kinetic energy is Kout = 5.0 MeV. (b) The particle’s speed is found from 1 2

mv 2 = K in ⇒ v =

2(65 × 106 eV)(1.60 × 10−19 J/eV) = 5.60 × 107 m/s 4 × (1.661 × 10−27 kg)

(Although this is >0.1c, the spirit of this simple model of alpha decay is best served with a classical calculation rather than a more complex relativistic calculation of the speed.) The time needed to move from one side of the potential well to the other, a distance of 15 fm, is

Δt =

15 × 10−15 m = 2.7 × 10−22 s 5.60 × 107 m/s

The particle collides with one wall or the other of the potential-energy barrier each time it moves across the potential well, so the rate of collisions is

R=

1 1 = = 3.7 × 1021 collisions/s Δt 2.7 × 10−22 s per collision

(c) From Chapter 41, the tunneling probability is Ptunnel = e–2w/η, where w is the width of the barrier and η is the penetration distance into the classically forbidden region:

η=

1.05 × 10−34 J s = = 2m(U 0 − E ) 2(4 × 1.661 × 10−27 kg)(25 × 106 eV × 1.60 × 10−19 J/eV)

= 4.55 × 10−16 m = 0.455 fm The probability of tunneling through the 20-fm-wide barrier is

Ptunnel = e −2(20 fm)/(0.455 fm) = 6.6 × 10−39 (d) Although the probability of tunneling is extremely small, the alpha particle collides with the barrier a very large number of times per second. The probability that the particle is still inside the nucleus after N collisions is Pin ≈ 1 – NPtunnel. The number of collisions needed to reduce Pin to 0.50, at which time half of a sample of nuclei would have decayed, is

N=

1 − Pin 0.50 = = 7.6 × 1037 collisions Ptunnel 6.6 × 10−39

At a collision rate of 3.7×1021 collisions/s, we can estimate the half-live to be t1/ 2 =

7.6 × 1037 collisions 1 year = 2.05 × 1016 s × 21 3.7 × 10 collisions/s 3.15 × 107 s

= 6.5 × 108 years = 650 million years Assess: The tunneling probability is very sensitive to the value of η. If you round your results slightly differently and get a slightly different value of η, your answer for t1/2 could differ by more than 100 million years. Even so, heavy nuclei that decay by the emission of alpha particles with K ≈ 5 MeV usually do have t1/2 of several hundred million years.

43-1