The Language of Physics: A Foundation for University Study - Solutions Manual

THE LANGUAGE OF PHYSICS FOUNDATIONS FOR UNIVERSITY STUDY

SOLUTIONS MANUAL

JOHN P. CULLERNE HEAD OF PHYSICS, WINCHESTER COLLEGE, WINCHESTER

ANTON MACHACEK HEAD OF PHYSICS, ROYAL GRAMMAR SCHOOL, HIGH WYCOMBE

J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

1 Linear Mechanics Q1

s = 2 + 3t − t 2 , and average speed between t = 2 and t = 2 + Δt is

Δs s (2 + Δt ) − s (2) − Δt − Δt 2 = = = −1 − Δt . Δt Δt Δt Now for Δt = 0.1s, 0.01s, 0.001s, Δs ds ds = −1.1 , −1.01, −1.001. Of course, = 3 − t 2 so = −1 at t = 2. Δt dt dt

Q2

(i) s = 5t 2 (ii) s = 2t (iii) s = 3t + 1 (iv) s = t (t + 1)

Q3 (i) v

(ii) v

t a

(iii) v

t a

v

v

t a

v

(i)

Constant acceleration.

(ii)

Acceleration is proportional to speed. At rest at t = 0, so v = 0 for all t.

(iii) Acceleration linear in v. a = 0 when v = g/k and this is the terminal speed. Q4 Let us begin with quite an important result in applied mathematics. The diagram below shows the projection of the projectile up the inclined plane. The angle α is the angle between the initial velocity and the x-axis and the angle β is the angle the inclined plane makes with the x-axis. Now let us rotate the axes of the problem so that the new x’-axis coincides with the surface of the inclined plane. In this new coordinate system the gravitational acceleration will be

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual ⎛ g cos β ⎞ ⎟⎟ g = ⎜⎜ ⎝ g sin β ⎠

y

x' Q y'

u

α

β x

O

Let Q be the point where the projectile lands up the slope. So OQ is the range. The vector u in the primed coordinate system is: ⎛ u cos(α − β ) ⎞ ⎟⎟ . u = ⎜⎜ ⎝ u sin(α − β ) ⎠ So, in the primed coordinate system: y ' = u sin(α − β )t −

1 g cos β t 2 , 2

which allows us to find the time the projectile arrives at Q since this will be one of the solutions of y’ = 0: 1 ⎛ ⎞ 0 = t ⎜ u sin(α − β ) − g cos β t ⎟ , 2 ⎝ ⎠

so, t = 0 or t =

2u sin(α − β ) . g cos β

The position X of the point Q along the x-axis is just X = u cos αt =

2u 2 cos α sin(α − β ) , g cos β

and OQ = Xsecβ, so OQ =

2u 2 cos α sin(α − β ) . g cos 2 β

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

To find the maximum of OQ (which we shall call r) we differentiate this expression with respect to α and find the stationary points. After some manipulation: 0=

2u 2 (cos(2α − β ) ) , g cos 2 β

or 2α − β =

π 2

or α − β =

π 2

−α .

This result means that the maximum range occurs when the direction of the vector u bisects the angle between the inclined plane and the y – axis. It follows that ⎛π ⎞ 2u 2 cos α sin ⎜ − α ⎟ 2 2 2u cos α sin(α − β ) ⎝2 ⎠ = 2u cos α . = r= g cos 2 β g cos 2 β g cos 2 β 2

Now, t=

2u sin(α − β ) 2u cos α = , g cos β g cos β

hence r=

g 2 t . 2

Q5

B

A u

h

α

d

Time t’ to go from A to B is just t' =

d . u cos θ

The vertical component of velocity at A is

v y = u 2 sin 2 θ − 2 gh so another expression for t’ would be t' =

2 u 2 sin 2 θ − 2 gh , g

which means that u and θ must satisfy:

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

d 2 = u 2 sin 2 θ − 2 gh . u cos θ g Differentiating this expression with respect to θ and setting

du = 0 for a dθ

minimum, we get:

u=

gd secθ . 2

Introducing this result back into the expression u and θ must satisfy we get:

u = g ( d + 2h) . Q6

The acceleration and velocity vectors are as follows: ⎛ − kg ⎞ ⎟⎟ , a = ⎜⎜ ⎝ g ⎠

⎛ v x (t ) ⎞ ⎛ u cos θ − kgt ⎞ ⎟ = ⎜⎜ ⎟⎟ v = ⎜⎜ ⎟ v t ( ) y ⎝ ⎠ ⎝ u sin θ − gt ⎠

The time of flight, T, is given by twice the time for the vertical component of v to reach zero; that is, T=

2u sin θ . g

Using the horizontal velocity in conjunction with this time we get the range R to be R=

(v x (T ) + v x (0)) 2(u cos θ − ku sin θ )u sin θ 2(u x − ku z )u z , T= = g g 2

and kg 2 t 2 g z (t ) = u sin θt − t 2 2 x(t ) = u cos θt −

so, the trajectory curve must intercept both the x and z axes twice – both x and z are quadratic in time.

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

z

x

Q7

We need to calculate the displacement vectors of the two aircraft at 15:00.

θ

375 km

r

r2 60o r1

ϕ

400 km This diagram is constructed by looking at the facts of the question. The first aircraft travels due east at 400 km/h for an hour so its displacement vector is r1 and the second travels 60o east of north at 450 km/h for 50 minutes so r2 is its displacement vector. The magnitude of r is then just:

r = 400 2 + 375 2 − 2 × 400 × 375 × cos 30 = 202 km. The bearing θ is just 90o + ϕ, and ϕ is given by: r r = 1 . sin 30 sin ϕ

This gives ϕ as 68.2o and hence θ as 158o. Q8

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

y

r

θ R

x

X

The question is asking for the components of the vector R in some coordinate system. We will choose the coordinates depicted in the diagram above. The end of the vector R has coordinates: x = r cos θ y = r (1 + sin θ )

If the rate at which angle is swept out by r is ω, then

⎛ ⎛ at ⎞ ⎞ ⎜ cos⎜ + ϕ ⎟ ⎟ + ω ϕ cos( t ) ⎛ ⎞ ⎝v ⎠ ⎟, ⎟⎟ = a⎜ R = r ⎜⎜ ⎜ ⎞⎟ ⎛ at ⎝1 + sin(ωt + ϕ ) ⎠ ⎜1 + sin ⎜ + ϕ ⎟ ⎟ ⎝v ⎠⎠ ⎝ for r = a and v the uniform speed of the rotating point (v = ωa). So that R = 0 at t = 0 we must have ϕ = − π/2. Q9

30o

V

8 m/s

vR The speed of the car will of course be the magnitude of vR, which is the relative horizontal velocity of the rain with respect to the car: Page 7

J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual vR =

Q10

8 = 13.9 m/s tan 30

(i) For components that are equal in magnitude then each component has a length of 10sin(45) = 7.1 m/s. (ii) When one component has a magnitude twice that of the other we have 5 v2 = 102 v = 4.5 m/s.

Q11

(a) Newton’s second law for the 5 blocks would be written: F = (5m)a ⇒ a =

F . 5m

(b) The resultant force on each cube must be

F F F as . = ma and a = 5m 5 5

(c) The resultant force F* on the 5th cube due to the 4th cube must accelerate 2m at a, therefore ⎛ F ⎞ 2F . F * = 2m.a = 2m⎜ ⎟= 5 ⎝ 5m ⎠

Now that means that the force on the 4th cube due to the 5th cube is easily 2F obtained through Newton’s 3 rd law as − , which of course is just enough 5 2 F 3F (just what is to make the resultant on cubes 1, 2 and 3 equal to F − = 5 5 F ). needed to accelerate 3 blocks at a = 5m Q12

Let u be the initial speed of the bullet and v be its final speed. Then v = u + at: 0 = 300 − a (0.01), So, a = 3 × 104 m/s2. This means that the resistive force has a magnitude of: F = ma = 20 × 103 × 3 × 104 = 600 N.

Q13

In this sort of problem it is always tempting to mix up acceleration and field strength, so be very careful. A common mistake is to think that the N/kg of a field strength is the same thing as the m/s2. To do this is to confuse gravitational and inertial masses. Let’s see how we can avoid this confusion: When the lift is accelerating upwards there must be a resultant force on the man to accelerate him upwards. This of course must be provided by the normal reaction R of the floor of the lift:

F = ma = R − W = ma ⇒ R = ma + W The reading on the scale is of course just going to have the magnitude of R, so (a) When the lift is accelerating upwards at 5 m/s the reading on the scales is: 1040 N. (b) When the lift is accelerating downwards at 5 m/s the reading on the scales is: 337 N.

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J.P. Cullerne & A.C. Machacek Q14

Language of Physics: Solutions Manual

As the balloon is rising at 10 m/s when the sand bag is released, the sand bag continues to rise a little with decreasing speed until it reaches 0 m/s. It will then begin to fall accelerating downwards at 10 m/s2. It will of course reach 10 m/s once it returns to the height at which it was initially released. Therefore the speed when it hits the ground must be given by: v 2 = u 2 + 2 gh = (10) 2 + 2(10)(600) ⇒ v = 110 m/s Assuming no re-bound we have Impulse = 10 kg × 110 m/s = 1100 kg m/s.

Q15

As the gun fires the truck (mass M) recoils and because the shell (mass m) is in a barrel for the first few split seconds of its journey, the recoil motion of the truck is also imparted to the shell. For such a problem the first thing to do is to translate the situation into a vector diagram (see diagrams below). Let the recoil velocity be V and the final velocity of the shell be v. Using the conservation of momentum in the horizontal plane gives:

MV − mv cosα = 0

V Recoil velocity Direction of shell’s initial velocity

v

Barrel

θ

α

α

Using the vector diagram: V sin(α − θ )

=

v . sin θ

Using our first expression in conjunction with this we get: M (sin α cos θ − cos α sin θ ) = m cos α sin θ M tan α cos θ = (m + M ) sin θ m⎞ ⎛ tan α = ⎜1 + ⎟ tan θ ⎝ M⎠

Q16

We of course have one of the standard triangles:

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θ

J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

pf = 10m Δp = 5 3 m

60o

pi = 5m

So, if ⎛ 5m ⎞ ⎛ 5m ⎞ ⎟⎟ p i = ⎜⎜ ⎟⎟, p f = ⎜⎜ ⎝ 0 ⎠ ⎝ 5 3m ⎠ then ⎛ 0 ⎞ ⎟⎟ . Δp = ⎜⎜ ⎝ 5 3m ⎠ If this impulse is applied to a mass of 5m at rest then the resulting motion will be the velocity vector: v=

Δp ⎛ 0 ⎞ = ⎜ ⎟. 5m ⎜⎝ 3 ⎟⎠

Q17 v

u

m

30o

m

M

With the balls being smooth, the subsequent motion of the two balls of mass m must be along the lines joining the centres (dotted lines). Using the expression of part (a) in 1.2.4 we get: Page 10

J.P. Cullerne & A.C. Machacek

e =1=

Language of Physics: Solutions Manual

v1 − v 2 v 2v = = u 2 cos φ − u1 cos θ u cos 30 u 3

Conservation of momentum in the original direction of u gives: Mu = 2mv cos 30xˆ = mv 3xˆ Therefore, M v 3 3 = 3= 3= . m u 2 2

Q18

It really pays in questions like this one to resolve everything along appropriate coordinate axes. The diagram below has the direction YX in the horizontal and we consider momentum equations in this direction or in the direction perpendicular to it:

I = mv x cos ϕ + mv y + 0=

mv z 2

momentum in YX direction

mv z 3 + mv x sin ϕ 2

momentum ⊥ YX direction

Also, the fact that the strings are inextensible means that the motions of Y and Z are restricted: v y = v x cos ϕ v z = v x cos(60 + ϕ )

Z m

vz 30o

m vy Y

o

60

m ϕ

vx

X

With our restriction on the velocities of Y and Z we immediately see that: v I = 2v x cos ϕ + z m 2 2v z = v x (cos ϕ − 3 sin ϕ )

So,

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

15v z v 2I I = 7v z + z = ⇒ vz = 2 2 15m m and 2v 3 v z ⇒ v z = x cos ϕ . 2 7

2v z = v x cos ϕ − 3

Now, 7 v z = v x cos ϕ 2 3 v z = v x sin ϕ 2 3 . Finally we can put our results for vz and ϕ back into 7 our expressions for vx and vy to obtain:

therefore tan ϕ =

2 13I 15m 7I vy = 15m

vx =

Q19

R1

R2 m

C

M

The point C is the centre of mass of this system of two masses. Let us stay in the 1-d line of the line joining the centres of these two masses. The position vectors as determined with respect to the centre of mass are shown. In terms of the relative initial speed u and the relative final speed v the initial and final speeds of the masses are: M u m+M m u2 = u m+M M v1 = v m+M m v v2 = m+M u1 =

So, the impulse I on the smaller mass is given by I = mv1 − mu1 =

mM (v − u ) , m+M

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J.P. Cullerne & A.C. Machacek

but, v = −eu , so I = −

Language of Physics: Solutions Manual

mM (1 + e)u . m+M

Q20

m

m

u

After the first impact (1 − e)u 2 (1 + e)u v2 = 2 v1 =

For the second impact u 2 =

(1 + e)u , 2

So, finally (1 − e) (1 + e) (1 − e 2 ) u= u 2 2 4 (1 + e) (1 + e) (1 + e) 2 u= u = 2 2 4

v 2final = v3final

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m

J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

2 Fields Q1

For there to be no acceleration, the fields due to the other charges must be equal and opposite. Let the position of the test charge be x. Let us denote the position of charge q as a = 0.5m. ⎛ q a−x q a q q⎞ ⎟ = ⇒ = ⇒ = 1+ ⇒ x = a⎜⎜1 + 2 2 ⎟ x Q x Q Q 4πε 0 x 4πε 0 (a − x ) ⎝ ⎠

−1

Q

If the test charge is moved off the x-axis, then the electric field will subsequently have a component pushing away from the x-axis. Thus if the charge is positive it will not return to equilibrium, but if it is negative, it will. However, if it remains on the x-axis, we can calculate the net field at position x as

E=

Q 4πε 0 x

2

−

q dE Q q =− − , thus . 2 3 3 dx 2πε 0 x 4πε 0 (a − x ) 2πε 0 (a − x )

Thus in the region 0 < x < a, we see that dE/dx < 0. Now, when the test charge is at the equilibrium position, E = 0. Given that dE/dx < 0, if we move it to larger x, E < 0; while if we move it to smaller x, E > 0. This means that a positive test charge will tend to return to equilibrium if displaced along the x-axis, but a negative charge will not. To sum up, positive test charges are in stable equilibrium in the x-direction, but unstable equilibrium in the y-direction. The converse is true for negative test charges. Q2

Firstly, the Earth is not a perfect sphere, so parts of the Earth are further from its centre, and thus experience a lower gravitational field strength. Given that the Earth is oblate (equator bulges outwards), we would expect weaker gravitational field at the equator, and stronger field at the poles. The matter is complicated because when you measure the weight of a mass on the rotating Earth, you automatically get the field strength with a component of centripetal acceleration subtracted from it. Given that you are moving faster as a result of the Earth’s rotation when you are on the Equator, this subtraction is greatest on the Equator, and so we expect the measured value of g to be lowest relative to the ‘true gravitational’ value of g at the Equator too. Scales calibrated in London measure apparent weight W, and give a reading on the dial in kilograms equal to W / 9.81183Nkg−1. Thus we may say that 72.00kg = W / 9.81183Nkg−1. My weight in Madras is given by 9.78281Nkg−1 m, where m is my true mass. Putting these two together gives 9.78281 Nkg −1 m 9.81183 = 72.00kg ⇒ m = 72.00kg × = 72.21kg . −1 9.78281 9.81183 Nkg

Q3

Let us denote the mass of Sun and Earth by M and m respectively, with R the distance between them. Let us now say that for each unit of mass, the objects are given charge q. Thus the charge on the Sun is now Mq, while the charge on the Earth is mq.

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

GM , while the R2 Mq . Thus the electrostatic field at the Earth caused by the Sun would be 4πε 0 R 2 GMm , while the electrostatic force of gravitational force of attraction is R2 Mqmq . These two are equal when repulsion is 4πε 0 R 2

The gravitational field at the Earth caused by the Sun is

GMm q 2 Mm q2 = ⇒G= ⇒ q = 4πε 0 G = 8.61 × 10 −11 Ckg −1 . 2 2 4πε 0 4πε 0 R R

Accordingly, the charge on the Earth would have to be mq = 5.98×1024 kg × 8.61×10−11 Ckg−1 = 5.15×1014 C.

Q4

a)

When the particle is on the x-axis, we can write the x-component of the force as Fx = −kx. Accordingly, the force needed to push the particle away from the origin is –Fx, and the work done by this force in moving the particle to position x=a is given by a

a

0

0

W = ∫ − Fx dx = ∫ kx dx = 12 ka 2 . Thus the potential energy function W ( x ) = 12 kx 2 . b)

If we write r = x 2 + y 2 as the distance from the point to the origin, we see that F = kr, and is pointed back towards the origin. Thus the integral in part (a) can be re-run in the r-direction to give a potential energy for any point in space of W (r ) = 12 kr 2 = 12 k (x 2 + y 2 ) .

c)

If the line joining the point to the origin makes angle θ with the x-axis, then the x-component of the force is Fx = −F cos θ = −kr × x/r = −kx, while the y-component of the force is Fy = −F sin θ = −kr × y/r = −ky. Thus as a vector ⎛ x⎞ F = −k ⎜⎜ ⎟⎟ . ⎝ y⎠

d)

∂ 1 ∂W = k x 2 + y 2 = kx ∂x 2 ∂x

(

)

∂W ∂ 1 2 2 ( ) = ky = 2k x + y ∂y ∂y

Thus Fx = −

∂W ∂W and Fy = − as we should expect. ∂y ∂x

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Language of Physics: Solutions Manual

Q5

∂φ 9y 9y ⎞ ⎛ = −6 x − ⎜ − 6x − ⎟ ∂x z z ⎟ ⎜ ∂φ 9x 9x ⎟ = −2 z − , so the total force would be F = mg = m⎜ − 2 z − . gy = − ⎜ ∂y z ⎟ z ⎜ 9 xy ⎟ ∂φ 9 xy − 2y + 2 ⎟ ⎜ = −2 y + 2 gz = − z ⎠ ⎝ ∂z z

Q6

a)

φ = −65 y − 2 z

b)

φ = −54 xy − 3 z

c)

Not possible.

d)

Not possible.

e)

φ=

gx = −

Q7

a)

b)

c)

d)

3 2

(x

2

+ y2 + z2

)

∂h = 0.06 , this means that for every metre you travel East, you rise by ∂x 0.06m.

∂h = −0.02 , this means that for every metre you travel North, you fall by ∂y 0.02m.

If the surface were curved, at least one of the gradients calculated in parts (a) or (b) would be dependent upon position (i.e. they would be functions of x or y). ⎛ 0.06 ⎞ ⎟⎟ . In The gradient is steepest in the direction of the vector ∇φ = ⎜⎜ ⎝ − 0.02 ⎠ this direction, you go −0.02m North for every 0.06m travelled East, and accordingly your bearing is given by tan−1(−0.06/0.02) = tan−1 −3 = 108.4°.

If you wish to check, suppose you travel a distance s in a on a bearing θ. This means that you move s sin θ East, and s cos θ North. Your new height will be different to your old one by Δh = 0.06s sin θ – 0.02s cos θ. We now choose the value of θ to maximise the value of Δh for a given, fixed, value of s. ∂ (Δh ) The maximum value of Δh occurs = 0.06 s cos θ + 0.02 s sin θ . ∂θ when this partial derivative is zero, so 0.06 cos θ + 0.02 sin θ = 0, hence tan θ = −3 as before.

e)

⎛ 0.06 ⎞ ⎟⎟ , and The maximum gradient is given by the magnitude of ∇φ = ⎜⎜ ⎝ − 0.02 ⎠

is thus equal to 0.06 2 + (− 0.02) = 0.0632K We can check this by feeding the bearing calculated in part (d) into the formula for Δh. θ = 108.4°, so Δh = 0.0632s. The gradient equals Δh/s = 0.0632 as before. 2

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Language of Physics: Solutions Manual

f)

You need to walk perpendicular to the direction calculated in part (d) – that is along a bearing of 18.4°or 198.4°. This is the direction in which Δh = 0, as can be checked using the equation for Dh given in part (d).

g)

Using the formula, the height of (100,200) is 102m, while (300,800) has a height of 102m. Thus there is no change of height. Using our answers to (a) and (b), we have walked 200m East, and thus expect to rise 0.06×200 = 12m. We have also walked 600m North, and thus expect to rise −0.02×600 = −12m. When these are combined, we see that we have neither risen nor fallen.

Q8

⎛ x⎞ ⎜ ⎟ a&b) Any positive scalar multiple of r = ⎜ y ⎟ points in the same direction as r. ⎜z⎟ ⎝ ⎠ This vector has length x 2 + y 2 + z 2 , and so to make a vector which has unit length but points in the same direction as r, we divide r by x2 + y2 + z2 .

c)

x2 + y2 + z2 ,

Using our information from part (a), where r =

⎛ x⎞ GM GM GM 1 ⎜ ⎟ g = − 2 rˆ = − 2 ⎜ y ⎟ = − 3 r r r r⎜ ⎟ ⎝z⎠

⎛ x⎞ ⎜ ⎟ GM ⎜ y⎟ = − 2 x + y2 + z2 ⎜z⎟ ⎝ ⎠

(

)

32

⎛ x⎞ ⎜ ⎟ ⎜ y⎟. ⎜z⎟ ⎝ ⎠

To find the potential function, we integrate the x-component of g with respect to x, and put a minus sign in front:

φ = −∫ − =∫

(

(x

GMx

+ y2 + z GM 2

2x +y +z 2

2

)

2 32

)

2 32

dx = ∫

( )

d x2 = −

GM

(

2 x2 + y2 + z2 GM

(x

2

+ y2 + z2

)

)

12

32

2 x dx

+ C ( y, z )

where the arbitrary constant can not be a function of x, but could be a function of y or z. We get identical answers (except for the arbitrary constants) by integrating the y-component of g with respect to y, or by integrating the z-component of g with respect to z, so this must be an acceptable solution for the potential. We usually take the convention of setting the arbitrary constant to zero so that the potential is zero at infinite distance from the mass M. Q9 a)

Integrating the x-component of E with respect to x gives –Axy + C(y,z). Integrating the y-component of E with respect to y gives Axy + C(x,z). There is no choice of the arbitrary constants which allows these to formulae to be consistent. Therefore E can not be written as the gradient of a scalar function in this case.

b)

We are going to use our first method of evaluating path integrals, as in part (a) of Workshop 2.6. If we take our path of integration as a circle in the xy plane of radius r, centred on the origin in the clockwise direction, then we find that at all points on the circle, E points in the opposite

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J.P. Cullerne & A.C. Machacek direction

to

our

Language of Physics: Solutions Manual path.

The

magnitude

of

E

is

given by

E = E x2 + E y2 = A 2 y 2 + A 2 x 2 = Ar , and is the same for all points along the circular path. The line integral is therefore equal to –E × length of path = −2πr E = −2Aπr2, where the minus sign comes from the fact that our path is in the opposite direction to E, and thus cos θ = −1, with θ defined as in Workshop 2.6. Q10 E = Q11

Q 4πε 0 r 2

=

6 × 10 −6 C = 5991 NC −1 2 −12 −1 4π × 8.854 × 10 Fm × (3m )

8.854 × 10 −12 Fm −1 × 1m 2 , so Using equation (2.27) we require 1F = d 8.854 × 10 −12 Fm −1 × 1m 2 = 8.854 × 10 −12 m . This is clearly impossible, 1F but if the area of the plates could be made bigger, then an insulating layer made one molecule thick by an electrochemical process at the interface between the plates might be possible. That said, using certain kinds of dielectric (insulating) materials between the plate can reduce the area needed because of their ability to polarize. A discussion of polarizability is beyond this text... d=

Q12

Normally, using an inverse square law, we would expect that as our distance to the Earth’s centre has halved, the gravitational field should have quadrupled to 39.2 Nkg−1. However, only the mass of the Earth closer to the centre than our position will cause field lines at our position – so only this mass should be counted. Assuming the Earth to have uniform density (which it doesn’t), the mass contained within radius r of material will be proportional to its volume and hence to r3. Thus when you are half way to the Earth’s centre, a sphere drawn through your position, centred on the Earth’s centre will include 0.53 = 0.125 of the Earth’s mass. We reduce our estimate for the Earth’s field strength accordingly, and get a final answer of 4.9Nkg−1, exactly half that measured at the surface.

Q13

Suppose the distance of the object from the centre is r. As shown in Q12, the gravitational field at this point is given by gE r/RE where gE is 9.8Nkg−1 and RE is the radius of the Earth. The object is therefore subject to a force equal to mgE r/RE directed towards the centre of the Earth, and experiences an acceleration of gE r/RE. Given that this is proportional to r, we have the conditions for simple harmonic motion (see section 4.1.1 for a workshop on this topic), with angular frequency −3 −1 ω = g E RE = 1.237 × 10 rad s . The time period is 2π/ω = 5077s or 1hr 24 mins 38s. The time for a journey from the U.K. to Australia is therefore half of this, namely about 42 minutes.

Q14

Let the distance of the object from the centre of the straight tunnel be d, and the distance of this centre from the Earth’s centre be L. The direct distance of our object to the Earth’s centre is given by Pythagoras’ Accordingly, the local value of the theorem as r = d 2 + L2 . gravitational field will be given by g = gE r/RE, and the force on the object will be mgE r/RE However, this force does not act in the direction of the tunnel, and only the component of the force which is parallel to the tunnel Page 18

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can affect the motion along it. The true gravitational field makes angle θ = cos−1 (d/r) to the direction of the tunnel pointing ‘inwards’ and thus the component of the force acting along the tunnel is given by mgE r/RE × d/r = mgE d/RE. Once more, we have an acceleration directed towards the tunnel’s centre of magnitude gE d/RE, and so we have simple harmonic motion as before, with the same time period −3 −1 ω = g E RE = 1.237 × 10 rad s .

d θ r L

Q15 Iron has a greater magnetic permeability than air, and so field lines will preferentially run in iron instead of air. This enables the pole pieces to funnel the field lines together before they reach the gap. Thus the field in the gap is stronger than it would be without the pole pieces, but is concentrated in a smaller area. Given that iron’s saturation magnetization is such that an iron bar can not be magnetized to produce much more than 1T (unaided), the use of pole pieces enables permanent magnets with fields in excess of this to be made.

N

S

Q16 First, please note that the surface is a spherical shell, not a sphere. Let us take our surface of integration as any spherical surface inside the shell, centred on the centre of the shell. The total charge enclosed by this surface is zero (since there is no charge inside the shell), and therefore ∫∫ E • dS = 0 over this surface. S

By symmetry, E must be the same at all points on the surface, and must either be directed inwards or outwards. The only way the integral can be true is if E = 0 at all points on the surface, and hence within the shell. If the very centre of the Earth were hollow, there would be no mass in this hollow, and so we must

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∫∫ g • dS = 0 on any spherical surface within this cavity, and thus g = 0 at S

all points within the cavity. This can be proved in a much more messy manner by showing that the gravitational attractions in all directions cancel out. Q17 Firstly, we set up a spherical surface centred on the centre of the sphere with radius r, where r < R. Given that the charge per volume of the material is uniform, the share of the volume enclosed by our surface is r 3 R 3 , and thus the charge enclosed is Qr 3 R 3 . Accordingly, on our surface we have Qr 3 ∫∫S E • dS = εR 3 , and given the symmetry of the situation, we can say that E will be directed either inwards or outwards and will have the same magnitude at all Qr 3 points on our surface (which has area S = 4πr2). Therefore 4πr 2 E = 3 and εR Qr . Please note that this formula only applies when r < R. Once r > R, E= 4πεR 3 Q then the normal Coulomb formula E = applies. Given that a positive test 4πεr 2 charge would be pushed outwards by this field, it is clear that E acts outwards at all points.

The potential function can then be written as

∂φ Qr , so = −E = − ∂r 4πεR 3

Qr 2 + C , where C is the arbitrary constant of integration. While we can 8πεR 3 3Q choose C to be any value, it is usually taken as C = + so that the function 8πεR φ is continuous across r=R if we take the potential outside the material to be Q in order that the potential falls to zero as we travel to infinity and φ= 4πεR away from the influence of the charge.

φ =−

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3 Rotation Q1

One astronomical unit (nominally, the distance from the Sun to the Earth) is about 1.5 × 1011 m. The circumference of a circle with an arc length of 1 AU subtended by 1 second or arc is: 360 o × 60'×60' '×1.5 × 1011 m or 1.9 × 1017 m. The radius of such a circle is then:

1.9 × 1017 = 3.0 × 1016 m (this is in fact called 2π

a parsec of ‘parallax-second’). Q2

The circumference of a wheel of 40 cm radius is: 2π × 0.4 m. As we can see, the answer to the question will now entirely depend on how well we know the measurement. Let us imagine we know (as the text suggests), d, the 140 km very well indeed (140.00000000000000 km). The number of revolutions n is given by: −1

n=

d d ⎛ δr ⎞ = ⎜1 ± ⎟ . r ⎠ 2π ( r ± δr ) 2πr ⎝

The second expression may be expanded using a Taylor expansion:

(1 + x) −1 = 1 − x + x 2 − x 3 + x 4 − ....,−1 < x < 1 (1 − x) −1 = 1 + x − x 2 + x 3 − x 4 − ....,−1 < x < 1 By putting

δr r

= x we have: n=

d dδr , m 2πr 2πr 2

if we neglect terms second order in

δr r

as they will be very small.

If we insist on knowing the revolutions to within one revolution then: dδr ~1 2πr 2

so,

δr ~ 7 × 10-6 m or less. Q3&4 The diagram below shows a wheel rotating at the instant the point A is in contact with the ground. The velocities of various points are shown for this instant in time. With no slip, the point A is instantaneously at rest with respect with the ground and the lines AP, AQ and AP’ instantaneously rotate about A with an angular speedω. Therefore the speed of the gum on the rim with respect to the axle is just v = ωr = 30 m/s. In the coordinate system shown, the coordinates of points on the rim are related by the following algebraic expression: x 2 + ( y − R) 2 = R 2

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2v

Q

v vP P

P’ vP'

y x

A

x-axis

From the diagram we see that the lengths AP and AP’ are just AP = AP’ = x 2 + y 2 = 2 yR Therefore the speed with respect to the ground (magnitudes of vP and vP’) is just:

2y ; R

v( y ) = ω 2 yR = v

that is, the instantaneous speed of points on the wheel is a function of the height of the points above the ground. Q5 ω A

B

z

v

P

θ

r y

O x C

⎛ 0.05 ⎞ ⎜ ⎟ The position is momentarily at: ⎜ 0.00 ⎟ , and at this point the velocity in m/s is: ⎜ 0.03 ⎟ ⎝ ⎠ Page 22

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⎛ 0.05 ⎞ ⎛ 0 ⎞ ⎛ 0.05 ⎞ ⎛ 0.0 ⎞ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ v = ω × ⎜ 0.00 ⎟ = ⎜ 0 ⎟ × ⎜ 0.00 ⎟ = ⎜ 0.5 ⎟ ⎜ 0.03 ⎟ ⎜10 ⎟ ⎜ 0.03 ⎟ ⎜ 0.0 ⎟ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ After 5 s the vector will have rotated a further 50 radians or

50 = 7.96 2π

revolutions. Q6

z ω

w u

θ

v

r

y

O x

The angular velocity vector may be represented in terms of its orientation in our coordinate system. If the angles made with the x, y and z axes by the vector ω are respectively α, β and γ , then ω may be written as: ⎛ cos α ⎞ ⎜ ⎟ ω = ω ⎜ cos β ⎟ , ⎜ cos γ ⎟ ⎝ ⎠ where ω is the magnitude of the angular velocity. The velocity of the point due to this pure rotation is then just ⎛ z cos β − y cos γ ⎞ ⎛ cos α ⎞ ⎛ x ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ v = ω × r = ω ⎜ cos β ⎟ × ⎜ y ⎟ = ω ⎜ x cos γ − z cos α ⎟ , ⎜ y cos α − x cos β ⎟ ⎜ cos γ ⎟ ⎜ z ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛u⎞ ⎜ ⎟ that is, if the velocity ⎜ v ⎟ were a consequence of a pure rotation, the ⎜ w⎟ ⎝ ⎠ components of the vector v above would equal to u, v and w respectively. Notice that ux + vy + wz = 0 ,

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for all t if the velocity were a consequence of a pure rotation. Q7

To differentiate (3.11) we need to replace all ωt with θ (t), so that ω is the first derivative of θ (t) with respect to time: ⎛ − sin θ ⎛ x⎞ ⎜ d ⎜ ⎟ ⎜ y ⎟ = ω ⎜ cos θ dt ⎜ ⎟ ⎜ 0 ⎝ ⎝z⎠

− cos θ − sin θ

0 ⎞⎛ x' ⎞ ⎛ cos θ ⎟⎜ ⎟ ⎜ 0 ⎟⎜ y ' ⎟ + ⎜ sin θ 0 ⎟⎠⎜⎝ z ' ⎟⎠ ⎜⎝ 0

0

− sin θ cos θ 0

0 ⎞⎛ x& ' ⎞ ⎟⎜ ⎟ 0 ⎟⎜ y& ' ⎟ . 1 ⎟⎠⎜⎝ z& ' ⎟⎠

Differentiating this expression with respect to t a second time leads to 5 terms:

⎛ x⎞ ⎛ − sin θ ⎜ d2 ⎜ ⎟ & ω = y ⎜ ⎟ ⎜ cos θ dt 2 ⎜ ⎟ ⎜ 0 ⎝z⎠ ⎝ ⎛ ⎛ − cos θ ⎜ ⎜ + ω ⎜ ω ⎜ − sin θ ⎜ ⎜ 0 ⎝ ⎝

⎛ − sin θ ⎜ + ω ⎜ cos θ ⎜ 0 ⎝

− cos θ − sin θ 0

sin θ − cos θ 0

− cos θ − sin θ 0

0 ⎞⎛ x' ⎞ ⎟⎜ ⎟ 0 ⎟⎜ y ' ⎟ 0 ⎟⎠⎜⎝ z ' ⎟⎠

0 ⎞⎛ x' ⎞ ⎛ − sin θ ⎟⎜ ⎟ ⎜ 0 ⎟⎜ y ' ⎟ + ⎜ cos θ 0 ⎟⎠⎜⎝ z ' ⎟⎠ ⎜⎝ 0

0 ⎞⎛ x& ' ⎞ ⎛ cos θ ⎟⎜ ⎟ ⎜ 0 ⎟⎜ y& ' ⎟ + ⎜ sin θ 0 ⎟⎠⎜⎝ z& ' ⎟⎠ ⎜⎝ 0

⎛ x' ⎞ ⎜ ⎟ and collecting these terms in ⎜ y ' ⎟ , ⎜ z' ⎟ ⎝ ⎠ •

⎛ − sin θ ⎜ ω& ⎜ cos θ ⎜ 0 ⎝

− sin θ cos θ

•

•

⎛ cos θ ⎜ ⎜ sin θ ⎜ 0 ⎝

0

0 ⎞⎛ x' ⎞ ⎛ − cos θ ⎟⎜ ⎟ ⎜ 2 0 ⎟⎜ y ' ⎟ + ω ⎜ − sin θ ⎜ 0 0 0 ⎟⎠⎜⎝ z ' ⎟⎠ ⎝ − cos θ 0 ⎞⎛ x& ' ⎞ ⎟⎜ ⎟ − sin θ 0 ⎟⎜ y& ' ⎟ 0 0 ⎟⎠⎜⎝ z& ' ⎟⎠

− sin θ cos θ 0

0 ⎞⎛ x& ' ⎞ ⎞ ⎟⎜ ⎟ ⎟ 0 ⎟⎜ y& ' ⎟ ⎟ 0 ⎟⎠⎜⎝ z& ' ⎟⎠ ⎟⎠

0 ⎞⎛ &x&' ⎞ ⎟⎜ ⎟ 0 ⎟⎜ &y&' ⎟ 1 ⎟⎠⎜⎝ &z&' ⎟⎠

⎛ &x&' ⎞ ⎛ x& ' ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ y& ' ⎟ and ⎜ &y&' ⎟ we get: ⎜ &z&' ⎟ ⎜ z& ' ⎟ ⎝ ⎠ ⎝ ⎠

− cos θ − sin θ

⎛ − sin θ ⎜ 2ω ⎜ cos θ ⎜ 0 ⎝

− cos θ − sin θ 0

sin θ − cos θ 0

0 ⎞⎛ x' ⎞ ⎟⎜ ⎟ 0 ⎟⎜ y ' ⎟ 0 ⎟⎠⎜⎝ z ' ⎟⎠

0 ⎞⎛ &x&' ⎞ ⎟⎜ ⎟ 0 ⎟⎜ &y&' ⎟ 1 ⎟⎠⎜⎝ &z&' ⎟⎠

⎛ x' ⎞ ⎜ ⎟ d d The first of the terms in ⎜ y ' ⎟ is recognisable as r ω × r, where r ω is the rate dt dt ⎜ z' ⎟ ⎝ ⎠ of change of angular velocity represented as a vector of magnitude ω& and direction along the axis of rotation, which in this case is along the z-axis:

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Language of Physics: Solutions Manual

− cos θ − sin θ

0 ⎞⎛ x' ⎞ ⎛ − y ⎞ xˆ yˆ zˆ ⎟⎜ ⎟ ⎜ ⎟ d 0 ⎟⎜ y ' ⎟ = ω& ⎜ x ⎟ = 0 0 ω& = r ω × r dt ⎜ 0 ⎟ x y z 0 ⎟⎠⎜⎝ z ' ⎟⎠ ⎝ ⎠

0

⎛ x' ⎞ ⎛ x⎞ ⎜ ⎟ ⎜ ⎟ The second term in ⎜ y ' ⎟ is immediately recognisable as − ω 2 ⎜ y ⎟ = −ω2r , but as ⎜ z' ⎟ ⎜z⎟ ⎝ ⎠ ⎝ ⎠ we see in Q9 this is just: ω × (ω × r) = −ω2r,

r in these expressions represents the displacement vector in the primed coordinates rotated into the un-primed coordinates. ⎛ x& ' ⎞ ⎜ ⎟ The term in ⎜ y& ' ⎟ also has the property of a vector cross product: ⎜ z& ' ⎟ ⎝ ⎠

⎛ − sin θ ⎜ 2ω ⎜ cos θ ⎜ 0 ⎝

− cos θ − sin θ 0

0 ⎞⎛ x& ' ⎞ xˆ yˆ zˆ ⎛ − y& ⎞ ⎟⎜ ⎟ ⎜ ⎟ d 0 ⎟⎜ y& ' ⎟ = 2ω ⎜ x& ⎟ = 2 0 0 ω = 2ω × r r dt ⎜ 0 ⎟ 0 ⎟⎠⎜⎝ z& ' ⎟⎠ x& y& z& ⎝ ⎠

dr r here represents the velocity vector in the primed coordinates dt rotated into the un-primed coordinates.

Where

⎛ &x&' ⎞ ⎜ ⎟ The final term in ⎜ &y&' ⎟ is obviously the acceleration vector in the primed ⎜ &z&' ⎟ ⎝ ⎠ coordinates as seen from the un-primed coordinates: ⎛ cos θ ⎜ ⎜ sin θ ⎜ 0 ⎝

− sin θ cos θ 0

0 ⎞⎛ &x&' ⎞ ⎟⎜ ⎟ d2 0 ⎟⎜ &y&' ⎟ = r2 r dt 1 ⎟⎠⎜⎝ &z&' ⎟⎠

Our 5 terms can therefore be summarised symbolically as follows: d2 da d d v = r2 r + r ω × r + 2 ω × r r + ω × (ω × r). dt dt dt dt Q8 (i)

The point on the millstone is stationary with respect to the millstone and there is no angular acceleration so the acceleration expression collapses to: da v = ω × (ω × r) = −ω2r, dt

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da v = − 82 × 0.6 = 35 m/s2 and the direction dt is towards the centre of the millstone.

as ω is ⊥ r. So, the magnitude of (ii)

If there is an angular deceleration the acceleration expression collapses to: da d v = r ω × r + ω × (ω × r). dt dt

These two components are perpendicular to each other. ω

dr ω dt

r −ω2r dr ω×r dt

The resultant acceleration will have a magnitude: a = (8 2 × 0.6) 2 + (0.2 × 0.6) 2 = 38 m/s2,

⎛ 0 .2 × 0 .6 ⎞ o with a direction that makes an angle tan −1 ⎜ ⎟ = 18 with the radius × 64 0 . 6 ⎝ ⎠ vector r.

Q9

Using the expressions for vector triple products, da v = ω × (ω × r) = (ω • r) ω −ω2r = −ω2r as ω and r are perpendicular. dt

Q10

d2 da d d d d v = r2 r + r ω × r + 2 ω × r r + ω × (ω × r). If a v = 0 and r ω = 0, dt dt dt dt dt dt then

0=

d r2 d r + 2 ω × r r + ω × (ω × r). 2 dt dt

Now,

da d d d r = r r + ω × r ⇒ r r = a r − ω × r, dt dt dt dt so,

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0=

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da d d r2 d r2 r + 2 ω × ( r − ω × r )+ ω × (ω × r ) = r + 2 ω × a r − ω × (ω × r). 2 2 dt dt dt dt With

da r = v xˆ , r = r xˆ and ω = ω zˆ we have: dt d r2 r = − 2ωr yˆ − ω2r xˆ ; 2 dt

that is, a combination of a centripetal acceleration and a Coriolis effect. Q11&12 The force keeping the satellite in circular orbit is provided by the gravitational interaction: mω 2 r =

GmM , r2

where m and M are respectively the mass of the satellite and the mass of the planet, and ω is angular frequency of the orbit. This of course leads to: 3 4π 2 2π r = GM ⇒ T = r2 . 2 T GM

r

R

δr r = R + δr So, T=

2π G 43 πR 3 ρ

(R + δr )

3 2

=

2π

3

⎛ δr ⎞ 2 ⎜1 + ⎟ , R⎠ G 43 πρ ⎝

where ρ is the density of the planet material. The expression in the bracket can be expanded using a Taylor expansion: 1

(1 + x) 2 = 1 + 12 x − 18 x 2 + 383 x 3 − ...., −1 < x ≤ 1 ,

and with x = δr/R << 1 we see that: T=

3π ; Gρ

that is, to a good approximation, T for this low orbit satellite is only a function of the density of the material that the planet is made of.

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Q13

R2

R1

m1

C

m2

If we choose our origin of coordinates to be C the centre of mass, then

m1 R 1 + m2 R 2 = 0 . This of course means that: m1ω 2 R1 =

Gm1 m2 ( R1 + R2 ) 2

m 2 ω 2 R2 =

Gm1 m2 ( R1 + R2 ) 2

Adding these two expressions together we get:

ω 2 ( R1 + R2 ) =

G (m1 + m2 ) . ( R1 + R2 ) 2

So, T2 =

4π 2 a3 , G (m1 + m2 )

where a = (R1 + R2). Q14 Now V is the orbital speed, which is essentially made up of a radial bit and a tangential bit. The vector V would be: V = r&xˆ '+ rωyˆ ' ,

where xˆ ' and yˆ ' are unit vectors in the rotating coordinate system attached to the planet. The square modulus of this vector is: V 2 = r& 2 + r 2ω 2 .

Using the expressions for r and h given in the text of the question we find: h −p (−e sin θ )ω = e sin θ 2 p (1 + e cos θ ) h rω = (1 + e cos θ ) p r& =

Therefore we can write V2 as: V2 =

h2 2 h2 2 e sin θ + (1 + e cos θ ) 2 2 2 p p

This expression may be factorised a number of different ways depending on the value one ascribes to e. If e is less that 1 we would try:

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J.P. Cullerne & A.C. Machacek V2 =

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[

]

2p⎤ h2 2 h2 ⎡ 2 2 2 + + + = − (1 − e 2 ) + θ θ θ sin 1 2 cos cos e e e 2 2 ⎢ r ⎥⎦ p p ⎣

Giving:

V2 =

h2 p

⎡2 1⎤ ⎢r − a⎥ . ⎣ ⎦

For e greater than 1:

[

]

h2 2 h2 2 2 2 V = 2 e sin θ + 1 + 2e cos θ + e cos θ = 2 p p 2

2p⎤ ⎡ 2 ⎢⎣(e − 1) + r ⎥⎦

Giving:

V2 =

h2 p

⎡2 1⎤ ⎢⎣ r + a ⎥⎦

For e = 1 the expression for V2 collapses to:

h2 p

V2 =

⎡2⎤ ⎢r ⎥ . ⎣ ⎦

Finally, h2 h2 ⇒ G (m1 + m2 ) = . p= G (m1 + m2 ) p

Q15 Kepler’s 3rd law and the results of the last exercise give us: T2 =

4π 2 3 a GM

and ⎡2 1⎤ V 2 = GM ⎢ − ⎥ . ⎣r a⎦

The mass of the satellite compared to that of the Sun means that the (m1 + m2) that would appear if their masses were comparable is replaced by M. A boost is applied momentarily so that it may be treated as an impulse. Differentiating the V2 expression we get: Δa Δa ⎞ ⎛ 2 2VΔV = GM ⎜ − 2 Δr + 2 ⎟ ~ GM 2 . a ⎠ a ⎝ r

as the boost is very short and in that time Δr ~ 0. Differentiating the Kepler law gives: 2TΔT =

4π 2 2 3a Δa . GM

Combining these two differentials with the 3rd law: 4

5

8 6π 2 3T 3 VΔV ⎛ GM ⎞ 3 . TΔT = 2 2 VΔVT 3 ⎜ 2 ⎟ ⇒ ΔT = 2 G M ⎝ 4π ⎠ (2πGM ) 3

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4 Oscillations & Waves Q1

ω = 2πf = 8π rad/s. Given that we start with the displacement equal to the amplitude, we should use a cosine wave. y = 4cm cos (8πt), where we assume that t is in seconds.

Q2

ω = 2πf = 20π rad/s. The amplitude is 5cm, so our equation will take the form y = 5cm cos (20πt+φ). Using the initial condition y = 3cm, we have 0.6 = cos φ and so φ = ±0.927 rad. We choose the negative value of φ to ensure that y gets bigger as t increases from zero. Our solution is therefore y = 5cm cos (20πt − 0.927). Please note that it is essential that we work in radians throughout – else our angular frequency will need to be different (if it were in degrees/s) and the calculus we are about to do later in the chapter wouldn’t work.

Q3

If we write our oscillation as y = 3 cos ωt + 4 sin ωt = A cos(ωt + φ ) , then it follows that y = 3 cos ωt + 4 sin ωt = A cos ωt cos φ − A sin ωt sin φ , and so A cos φ = 3 and A sin φ = −4. From this we can tell that φ = tan−1(−1.3333) = −0.927 rad or 2.214 rad, and we choose the smaller so that cos φ is positive, while sin φ 2 2 is negative. Meanwhile ( A cos φ ) + ( A sin φ ) = A 2 = 25 , so A = 5.

Q4

The particle describes a figure of eight motion as in the diagram below.

Q5

Firstly, we remember that ω = 2πf, and k = 2π/λ. Therefore

Q6

ω = 2πf = 2π×7 = 43.98 rad/s, and k = 2π/λ = 3.142 m−1.

ω k

=

2πfλ = fλ = c . 2π

Given the parameters, we want a peak at x=0 when t=0, so we choose a cosine wave. y = 3cm × cos(43.98t – 3.142x), where we assume that t is in seconds, and x in metres.

Q7

The only thing we need to change is the sign of the co-efficient of x. y = 3cm × cos(43.98t + 3.142x)

Q8

Adding the results of Q6 and Q7, we get

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y = A cos(ωt − kx ) + A cos(ωt + kx ) = A cos ωt cos kx + A sin ωt sin kx + A cos ωt cos kx − A sin ωt sin kx = 2 A cos ωt cos kx = 6cm cos(43.98t )cos(3.142 x )

The nodes occur whenever cos(3.142x) = 0, namely at x = 0.5m, 1.5m, 2.5m... The amplitude of the oscillation is given by everything that multiplies the cos (43.98t), namely 6cm cos 3.142x. Q9

Complex amplitude after superposition (summing)

(

)

T = Ae − iφ + A + Ae iφ = A + A e − iφ + e iφ = A + 2 A cos φ , where we use the result from part (p) of Workshop 4.2.

Modulus square amplitude = TT * = ( A + 2 A cos φ )

2

Full constructive interference occurs when cos φ = 1, so φ = 0, 2π, 4π... Full destructive interference occurs when cos φ = −0.5, so φ = 2π/3, 4π/3, 8π/3, 10π/3... Q10 We use the standard result that the sum of the series 1− a p . A proof of this in the case of an infinite 1− a series is given as part (g) of Workshop 6.5.3, and the proof is readily adaptable to the finite series here. 1 + a + a + a + L + a p −1 = 2

3

The complex amplitude for the waves when they all meet is given by 1 − e iNφ , where we use the 1 − e iφ p =0 geometric series formula with a = eiφ. The modulus square amplitude is given by N −1

(

)

T = ∑ Ae ipφ = A 1 + e iφ + e 2iφ + L + e i ( N −1)φ = A

2 1 iNφ − e − iNφ 1 − e iNφ 1 − e −iNφ 2 2−e 2 2 − 2 cos Nφ 2 sin ( 2 Nφ ) = = = A A A 2 − 2 cos φ sin 2 ( 12 φ ) 1 − e iφ 1 − e − iφ 2 − e iφ − e −iφ where we use the trignometrical result that 1 − cos 2θ = 2 sin2 θ.

TT * = A 2

It is immediately apparent that we shall have full destructive interference whenever sin (Nφ/2) = 0, providing sin(φ/2)≠0. Thus full destructive interference occurs when φ = 2π/N, 4π/N, 6π/N... but not including φ = 2π, 4π, 6π... For the constructive interference, we need to be a bit more sneaky. When φ = 0, all the waves are in step with no phase difference, so there must be constructive interference. The same is expected whenever φ = 2π, 4π, 6π... since this is in practice indistinguishable from the φ=0 case. We just need to check this from our formula. When φ is small, we may approximate sin φ ≈ φ, and so 1 sin 2 ( 12 Nφ ) 2 ( 2 Nφ ) ≈ A = A 2 N 2 . Thus the intensity when we have 2 2 1 1 sin ( 2 φ ) (2 φ ) constructive interference is A2N2. 2

TT * = A 2

Q11 N=3: Page 31

J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

N=5:

N=10

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

When N=10 000, we expect bright constructive interference when φ = 0, 2π, 4π..., and almost total destructive interference at all other places. Q12 The frequencies listed will be the fundamental notes for each string, and as such, the string will be half the wavelength of the sound involved. Thus the wavelength for all the waves considered here will be twice the string length – namely 1.00m. With the wavelength known, we have c = fλ =

T

, so f 2 λ2 =

T

T

and ρ =

. The densities can then be ρ ρ f 2 λ2 calculated as in the table. Please note that densities have been given in grams per metre to make the numbers nicer. String

Frequency (Hz)

Density (g/m)

G

196

1.30

D

293

0.58

A

440

0.26

E

660

0.11

If the tensions in the strings were different, the excess tension in the higher strings would cause a torsion which would tend to twist and warp the violin’s neck to one side. Q13 For a transverse wave on a string, T = c2ρ, as in section 4.8. Therefore Z ≡ ρc =

T T c= . 2 c c

Q14 P = 12 ZA 2ω 2 = =

ω

T 2 2 A (ck ) = 12 Tck 2 A 2 c

Zω . As the question says, the tensions in the two strings must c T Z T be equal. It is also true that when a wave moves from one medium to another, its frequency can not change, since the boundary can not make more waves each second than it is provided with. Given that ω and T are the same on either side of the boundary, it follows that k and Z must be in proportion, and thus the ratio of the impedances is the same as the ratio of the wave numbers.

Q15 k =

ω

1 2

=

Q16 Using equation (4.37), the power incident on the join is Pi = 12 Z L Ai2ω 2 , while the power transmitted is Pt = 12 Z R At2ω 2 . It follows that the fraction of incident power transmitted is 2

2

Pt Z R At2 Z R ⎛ At ⎞ 4Z L Z R Z ⎛ 2Z L ⎞ ⎜⎜ ⎟⎟ = R ⎜⎜ ⎟⎟ = , where we have used = = 2 Pi Z L Ai Z L ⎝ Ai ⎠ ZL ⎝ ZL + ZR ⎠ (Z L + Z R )2 equation (4.48) for the ratio of the amplitudes. Similarly 2

Pr Z L Ar2 ⎛ Ar ⎞ ⎛ Z L − Z R = =⎜ ⎟ =⎜ Pi Z L Ai2 ⎜⎝ Ai ⎟⎠ ⎜⎝ Z L + Z R that Pr + Pt = Pi, and so...

2

⎞ ⎟⎟ . For energy to be conserved, we require ⎠

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

Pt + Pr 4Z L Z R + (Z L − Z R ) 4Z L Z R + Z L2 + Z R2 − 2Z L Z R Z L2 + Z R2 + 2Z L Z R = = = =1 Pi (Z L + Z R )2 (Z L + Z R )2 (Z L + Z R )2 2

as it should be. Q17 Using the definition of the scalar product and its form in Cartesian co-ordinates (as shown in Workshop 2.3), we have ∂ ∂E ∂ = E cos(ωt − k • r ) = E cos(ωt − k x x − k y y − k z z ) ∂x ∂x ∂x dE cos(ωt − k x x − k y y − k z z ) ∂ (ωt − k x x − k y y − k z z ) × = d (ωt − k x x − k y y − k z z ) ∂x

= − E sin (ωt − k x x − k y y − k z z )× (− k x ) = Ek x sin (ωt − k x x − k y y − k z z )

Q18 a)

I=

P 3.9 × 10 26 W = A 4π 1.50 × 1011 m

b)

I=

P 3.9 × 10 26 W = = 64.1MWm −2 2 8 A 4π(6.96 × 10 m )

(

)

2

= 1380Wm −2

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

5 Circuits Q1

Using the suggestions in the hints at the bottom of the page a)

Number of free electrons per metre of wire = Number of free electrons per m3 × volume of 1m of wire = n A

b)

The number of electrons which occupy length u of wire = Number of electrons per metre of wire × u = n A u.

c)

Total charge on these electrons = no. of electrons × charge on 1 electron = n A u × q = quAn.

Q2

Note that area of 1.5mm2 is the same as 1.5×10−6 m2. u=

Q3

A = 0.02mm2 = 2×10−8 m2. u=

Q4

I 13A = = 5.42 × 10 − 4 ms −1 −19 −6 −3 2 29 qAn 1.6 × 10 C × 1.5 × 10 m × 1.0 × 10 m I 0.01A = = 3.13 × 10 4 ms −1 . −19 qAn 1.6 × 10 C × 2 × 10 −8 m 2 × 1.0 × 10 20 m −3

Charge carried per second = 30mC. Charge on one Cu2+ ion is 2×1.6×10−19C So number of ions leaving anode each second =

0.03C = 9.38 × 1016 . −19 3.2 × 10 C

Number of electrons leaving anode is twice this. If same current of 30mA were carried equally by sulphate and copper ions, number of copper ions leaving anode would be half the value above, while number of electrons leaving the anode would remain the same. Q5

Kinetic energy =

1 2

mv 2 = 12 × 9.1 × 10 −31 kg × (5.42 × 10 −4 ms −1 ) = 1.34 × 10 −37 J 2

Potential energy = qV = 1.6×10−19 C × 230V = 3.68×10−17 J. Ratio of potential to kinetic energy = 2.75×1020 : 1. Assumption justified. In practice, actual velocities of the electrons are typically much higher than given in Q2, with many travelling at speeds over 106ms−1 (as explained in section 5.1) and accordingly the kinetic energy of these electrons is greater than their potential energy. Some electrons move one way, and an almost identical number move the opposite way, and this is why their velocities virtually cancel out when the average is calculated. However the electrons remain close to these phenomenal speeds even when there is zero current – thus it is still fair to say that the increase in the kinetic energy of the electrons as a whole when a current starts flowing is still negligible in comparison to the potential energy, even when the calculation is done properly. In fact you can prove that the additional kinetic energy is equal to the value calculated above where it was assumed that all of the electrons were moving at the average (or drift) velocity. Note to tutor: as an extension exercise, you may wish to ask students to derive an expression for the total kinetic energy of N electrons (each of mass m) whose velocities are uniformly distributed in three dimensional v-space subject to the condition that none travel faster than vmax (the speed corresponding to the Fermi

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

2 . Next, you can ask them to energy), and show that this is equal to 103 Nmv max show that if a constant velocity u (say, parallel to the positive x-axis) is added to all of the original velocities, the new total kinetic energy is greater than the original value by Nmu2.

Q6

Firstly we use Kirchhoff’s First Law to write I3 = −I1 – I2. We can then use our Second Law equation for the lowest branch to write V = 100 (I1+I2). Adding three times our top branch equation to two times our middle branch equation gives 5V = 17 – 600 (I1+I2). Now eliminating I1+I2 between our two equations, we get:

5V = 17 − 600

V 100

11V = 17 V = 1.55 so the voltage is 1.55V as required. Using our branch equations and our knowledge of V, we find the currents as 3V − 1.55V = 7.27mA 200Ω 4V − 1.55V I2 = = 8.18mA 300Ω 1.55V I3 = − = −15.45mA 100Ω I1 =

and so we realize that currents the top and middle branches travel in the same direction as the arrows, but that the current in the bottom branch travels in the opposite direction to the arrow on the diagram. Q7

We use the loop current method, with IL1 and IL2 defined in the diagram below: 1.0 kΩ

4.5 V

IL1

1.0 kΩ IL2

4.5 V IL2

IL1 1.5 V

3.0 V

For the left hand circuit, the simultaneous equations are 4.5 − 1000 I L1 − 6000(I L1 − I L 2 ) = 0 − 6000(I L 2 − I L1 ) − 3000 I L 2 = 0

thus

4.5 = 7000 I L1 − 6000 I L 2 6000 I L1 = 9000 I L 2

and these equations can be solved to give IL1 = 1.5 mA, IL2 = 1 mA. The currents in the resistors are therefore 1kΩ: 1.5 mA to the right 6kΩ: 0.5 mA downwards 3kΩ: 1 mA downwards For the right hand circuit, the simultaneous equations are

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

4.5 − 1000 I L1 − 6000(I L1 − I L 2 ) − 1.5 = 0

thus

1.5 − 6000(I L 2 − I L1 ) − 3000 I L 2 − 3 = 0

3 = 7000 I L1 − 6000 I L 2 1.5 = 6000 I L1 − 9000 I L 2

and these equations can be solved to give IL1 = 0.667 mA, IL2 = 0.278 mA. The currents in the resistors are therefore 1kΩ: 0.667 mA to the right 6kΩ: 0.389 mA downwards 3kΩ: 0.278 mA downwards. Q8

V 325 cos(100πt ) = A , so the amplitude is 18.1A. 18 R

a)

I=

b)

The velocity is given (as in Q2) by v=

I 18.1A × cos(100πt ) = −19 qAn 1.6 × 10 C × 1.5 × 10 −6 m 2 × 10 29 m −3

= 7.54 × 10 cos(100πt ) ms −4

,

−1

and so the amplitude is 0.754mms−1. c)

Equation (4.9) tells us that the amplitude of the velocity is equal to ωA where ω is the angular frequency (here 100π rads−1) and A is the amplitude in distance terms. Thus here A=

Q9

v amplitude

ω

=

7.54 × 10 −4 ms −1 = 2.40 × 10 −6 m . −1 100π rads

V0 = Vrms 2 = 230 2 V = 325V

Q10 I 0 = I rms 2 = 13 2 A = 18.4A Q11 P0 = I 0V0 = I rms 2 × Vrms 2 = 2 Prms = 2 × 60W = 120W Q12 (a) (b)

(c) Q13 (a)

(b)

V 2 = V02 cos 2 ωt

Mean value of cos ωt equals one half. Accordingly the mean value of V2 is 12 V02 . 1 2

V02 =

1 2

V0

Multiplication gives V0 I 0 e 2iωt , the real part of which (using the methods of part (m) of Workshop 4.2) is equal to V0I0 cos 2ωt. The average value of this real part is zero. Using the method suggested:

(

) (

)

(

)

(

VI = Re V0 e iωt Re I 0 e iωt +iφ = 12 V0 e iωt + e − iωt × 12 I 0 e iωt +iφ + e − iωt −iφ

(

= 14 V0 I 0 e 2iωt +iφ + e − 2iωt −iφ + e iφ + e −iφ

(c)

Using the approach suggested:

Page 37

)

)

J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

VI = 14 V0 I 0 (e 2iωt +iφ + e −2iωt −iφ + e iφ + e − iφ ) = 14 V0 I 0 (2 cos(2ωt + φ ) + 2 cos φ )

.

= V0 I 0 (cos(2ωt + φ ) + cos φ ) 1 2

Taking a time average, the cosine term containing ωt + φ averages to zero, leaving behind VI = 12 V0 I 0 cos φ , and as we know from section 5.3.2 1 2 V0 I 0 = Vrms I rms , and so the reasoning is complete. (d)

Here we go... VI = V0 cos ωt × I 0 cos(ωt + φ )

= V0 I 0 cos ωt (cos ωt cos φ − sin ωt sin φ ) = V0 I 0 cos ωt cos ωt cos φ − V0 I 0 cos ωt sin ωt sin φ

,

= V0 I 0 cos 2 ωt cos φ − 12 V0 I 0 sin 2ωt sin φ where we have used the relationship sin 2ωt = 2 sin ωt cos ωt in the final line. Taking time averages, the cos2 averages to one half, while the second term averages to zero. This gives VI = 12 V0 I 0 cos φ as before. Q14 For current and voltage to be in phase, Z in equation (5.17) must be real. The denominator is nastier, and thus we would prefer to work with Z−1, which also must be real for the current and voltage to be in phase. Z −1 =

(

)

1 + iωCR − ω 2 CL 1 + iωCR − ω 2 CL (R − iωL ) = (R + iωL )(R − iωL ) R + iωL

(

R − ω 2 RCL + ω 2 RCL + iωCR 2 − iωL 1 − ω 2 CL R 2 + ω 2 L2 R + iω CR 2 − L + ω 2 CL2 = R 2 + ω 2 L2 =

(

)

)

The imaginary part will be zero if CR 2 + ω 2 CL2 = L , and so we require that C =

L . R + ω 2 L2 2

Q15 When R is set equal to zero, the impedance becomes iωL , and therefore we see that this will be undefined (infinite) when 1 − ω 2 CL ω = 1 LC . With small, non-zero R, we would also expect very large impedances at this frequency. Z=

If two resistors are put in parallel, the current will be shared between them in the inverse ratio of the resistances (the resistor with twice the size carries half the current). The same is true of impedances in general – here our circuit is in parallel with the detecting electronics. We choose our inductance L so that Z is quite small for all other values of ω. Thus the LCR circuit acts as a more or less perfect conductor for the unwanted frequencies, effectively short circuiting them, leaving virtually no current to pass through the rectifier and on to the audio amplifiers. The desired frequency can not pass through the LCR circuit, succeeds in driving a current through the rectifier and therefore is detected by the amplifier which is wired in parallel.

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

Q16 Using the reasoning in and by figure 5.10 we require that ILR sin A = ω CV where I LR = V

R 2 + ω 2 L2 , and tan A = ωL/R as in figure 5.9. It follows from

the geometry of figure 5.9 that sin A = ωL write our condition as

R 2 + ω 2 L2 , and therefore we can

I LR sin A = ωCV V R 2 + ω 2 L2

ωL

= ωCV R 2 + ω 2 L2 L =C 2 R + ω 2 L2

as before.

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

6 Thermal Physics T2 (7 + 273) = 1− = 71% . (700 + 273) T1

Q1

η = 1−

Q2

ΔT = T1 − T2 so T2 = T1 − ΔT . T2 T − ΔT ΔT = 1− 1 = , which means to maximise η we need to T1 T1 T1 maximise ΔT.

η = 1−

Q3

In this question it is crucial we understand how we obtain the final temperature of the two blocks in the two cases (a) and (b). Here we assume the steel has a specific heat capacity of c, mass m, and that the volume of the blocks remains constant. (a) For the simple conduction case we know that the conservation of energy leads to a final temperature of the two blocks of 50 oC = 323 K. Now, we cannot calculate the change in entropy directly for an irreversible process (we have heat passing across a temperature gradient), but we can come up with a process of infinitesimal reversible heat exchanges that happen over constant temperatures: Tf

Tf

dQ dT ⎛ 323 ⎞ Entropy decrease of hot block = − ∫ = − mc ∫ = mc ln⎜ ⎟ = 0.14mc T T ⎝ 373 ⎠ Ti Ti Tf

Tf

dQ dT ⎛ 323 ⎞ Entropy increase of cold block = ∫ = mc ∫ = mc ln⎜ ⎟ = 0.17 mc T T 273 ⎝ ⎠ Ti Ti

ΔStotal = ΔShot + ΔScold = − 0.14mc + 0.17mc = 0.03mc . Energy lost by hot block = 50mc Energy gained by cold block = 50mc. (b) Now, if we have a reversible heat engine between the two blocks (at temperatures TH and TC) then the total change in entropy around a cycle must be zero and so: o o ⎛T dQ o dQ dT dT +∫ = mc ∫ + mc ∫ ⇒ − ln⎜⎜ o 0= ∫ T TC T T T ⎝ TH TH TH TC

To

T

T

T

⎛T ⎞ ⎟⎟ = ln⎜⎜ o ⎠ ⎝ TC

⎞ ⎟⎟ , ⎠

which

means

that: To = TH TC , where To is the final temperature of the two blocks. This is the geometrical mean of the temperatures of the hot and cold blocks. The rest of the problem will be solved by applying the procedure above but with this geometrical mean as the final temperature. Q4

f =e

ε − kT

δf = e

ε − kT

,

δf f + δf ⎛ ε ⎞ ~2⇒ ~ 1, ⎜ ⎟δT and f f ⎝ kT ⎠ Page 40

J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

For T ~ 300 K, δT ~ 10 K ⇒ ε ~ Q5

1.38 × 10 −23 × 300 2 = 1.2 × 10 −19 J 10

H2O molar mass = 18 g 1 kg of H2O is 55.6 mol of H2O ⇒ 55.6 × 6.02 × 1023 = 3.34 × 1025 particles. So, Energy per molecule = 2.26 × 106 J ÷3.34 × 1025 = 6.76 × 10-20 J.

Q6

−

H

p ∝ e NkT , where H is the heat of vaporisation for a sample of N molecules of the substance. This means ln p = constant −

H H H ⇒ ln p1 − ln p2 = − . + NkT1 NkT2 NkT

With p1 = 100 kPa, H = 2.26 × 106 J, N = 3.34 × 1025, T1 = 373 K and T2 = 358 K, we get p2 = 57.6 kPa. Q7

Isothermal atmosphere of N2: p = po e

− Mgh RT

, with M as the molar mass (28 g) and po the pressure at h = 0.

With po = 100 kPa we get h = 4.47 km. Q8

If we were to gather the tails of all the velocity vectors of all the molecules in a gas to one point we would get a ‘spiky ball’. The spikes are the velocity vectors and they would have a certain distribution of lengths. The length of each velocity (spike) is of course a speed u and the distribution of speeds goes as a Boltzmann factor:

f =e

2

− mu 2 kT

.

u

θ ϕ

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

However, the volume of a small element * of the spiky ball volume is (see the diagram above): ΔV = u 2 sin θ du dθ dϕ

so the number of vectors of velocity with lengths between u and u + du is proportional to: π 2π

∫ ∫e

2

− mu 2 kT

u 2 sin θ du dθ dϕ ~ u 2 e

2

− mu 2 kT

du.

0 0

The integrals over the angles add up all the vectors of each length over all directions. So the fraction of molecules which have speed u is proportional to u 2 e Q9

2

− mu 2 kT

.

pV = nRT p = 100 × 103 Pa, T = 298 K, n = 1 mol, so V = 0.025 m3.

Q10

(10-10)3

Q11

The RMS speed of a nitrogen molecule is easily calculated from the Kinetic theory by:

u2 =

3RT , where M is the molar mass of nitrogen. M

VRMS d

The diagram above shows a molecule sweeping out a ‘tube of influence’ in 1 second. In the ideal gas theory the molecules themselves do not have interactions with each other except through collisions, so they can only influence each other if they collide. If the ‘position’ of a molecule is the position of its centre of mass then the diagram above shows that the molecule in the middle will just collide with others if the centres of the others lie on or in the tube. The diagram illustrates that the radius of this tube is the diameter of a molecule so the volume of the tube is just πd 2 V RMS = πd 2 u 2 . It is now a simple matter of multiplying this volume by the number density of the gas to obtain the number of collisions that a molecule is likely to have in 1 second. If the number density of the gas is ρ then we can define a mean free path λ as:

*

For more detail on the bizarre coordinate system see part c of Workshop 7.1.

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J.P. Cullerne & A.C. Machacek

Language of Physics: Solutions Manual

λ = (length of path in 1 second) ÷ (number of collisions in 1 second) λ=

u2

πd 2 u 2 ρ

1 , πd 2 ρ

=

which is the average distance a molecule will move before a collision occurs with another molecule. Actually, we have done the analysis without accounting for the relative velocities of the other molecules. To do this we need to just consider the vector diagram:

V1 − V2

V2 V1

If the vector V1 is the velocity of the molecule under consideration and V2 is the velocity of another molecule, then V1 − V2 is the relative velocity of the molecule under consideration with respect to the other molecule. Therefore, to see how this relative velocity modifies our expression for λ we need to see what happens if we use the RMS value of this relative velocity rather than the RMS value

u2 :

(V1 − V2)RMS =

∑V

2 1

+ V22 − 2V1 • V2

, where the sum is over all the N molecules in the gas. The scalar products will not contribute to the sum once it is performed over all molecules (there will be as many positive terms as negative). So (V1 − V2)RMS = λ= =

1

πd ρ 2 2

2 u 2 , which changes our expression to

.

Now we are in a position to answer the question. A good estimate of the speed of a nitrogen molecule at room temperature and pressure would be its RMS speed:

u2 = and ρ =

3RT 3 × 8.31 × 298 = = 515 m/s, M 28 × 10 −3

nN A nN A N A p = = , so nRT V RT p Page 43

J.P. Cullerne & A.C. Machacek λ=

1

πd ρ 2 2

=

Language of Physics: Solutions Manual

RT

πd pN A 2 2

=

8.31 × 298

π × (2 × 10

−10 2

) × 100 × 10 3 × N A × 2

.

λ = 2.3 × 10-7 m. Q12

The volume V goes to V/100 adiabatically. The two expressions we need are: γ

p1V1 = p 2V2

γ

p1V1 p 2V2 = T1 T2 Putting in what we know: p1V γ = p 2

Vγ ⇒ p 2 = p1100 γ γ 100

so, V ⎞ ( p 100 )⎛⎜ 100 ⎟ γ

p1V = T1 which means that:

1

⎝

T2

⎠ ⇒ T2 = 100 γ −1 , T1

ΔT = T2 − T1 = (100 γ −1 − 1)T1

Work done can be obtained from the first law of thermodynamics: ΔU = ΔQ + ΔW .

Adiabatic compression means that ΔQ = 0, so ΔU = ΔW = − pΔV. Throughout the process the pressure and volume are related on the adiabatic so: pV γ = k , where k is a constant. This means that we will need to do an integral since as the volume changes so will the pressure. This means that: V 100

V 100

V

V

ΔU = − ∫ pdV ' = − ∫ ΔU =

V

⎡ V 'γ −1 ⎤ 100 1 = − ' dV k ⎢ ⎥ V 'γ ⎣ (γ − 1) ⎦ V

kV 1−γ kV 1−γ ΔT 100 γ −1 − 1) = ( . (γ − 1) (γ − 1)T1

Notice this is positive – we have done work on the gas without allowing it to lose any heat so its temperature, and therefore, internal energy must rise.

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