[john p. cullerne, anton machacek] the language of(1)

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 , and average speed between t = 2 and t = 2 + t is 232 tts +=

tt

ttt

ststs =

=+=

1)2()2(2

.

Now for t = 0.1s, 0.01s, 0.001s, 1.1=

ts , 1.01, 1.001. Of course, 23 t

dtds = so 1=

dtds at t = 2.

Q2 (i) 25ts = (ii) ts 2= (iii) 13 += ts (iv) )1( += ttsQ3

v

(i) (ii) (iii)

t

v

a t

v

a t

v

a

v 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 is the angle between the initial velocity and the x-axis and the angle 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

Page 2


=

sincos

gg

g

x

y

x'

y' u

O

Q

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:

=

)sin()cos(

uu

u .

So, in the primed coordinate system:

2cos21)sin(' tgtuy = ,

which allows us to find the time the projectile arrives at Q since this will be one of the solutions of y = 0:

= tgut cos21)sin(0 ,

so,

t = 0 or t =

cos)sin(2

gu .

The position X of the point Q along the x-axis is just

X =

cos)sin(cos2cos

2

gutu = ,

and OQ = Xsec, so OQ =

2

2

cos)sin(cos2

gu .

Page 3


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

( ))2cos(cos20 2

2

= gu ,

or

22 = or =

2.

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

2

22

2

2

2

2

coscos2

cos2

sincos2

cos)sin(cos2

gu

g

u

gur =

== .

Now,

coscos2

cos)sin(2

gu

gut == ,

hence

2

2tgr = .

Q5

d h

A B

u

Time t to go from A to B is just

cos' udt = .

The vertical component of velocity at A is

ghuvy 2sin22 =

so another expression for t would be

ghug

t 2sin2' 22 = ,

which means that u and must satisfy:

Page 4


ghugu

d 2sin2cos

22 = .

Differentiating this expression with respect to and setting 0=ddu for a

minimum, we get:

sec2

gdu = .

Introducing this result back into the expression u and must satisfy we get: )2( hdgu += .

Q6 The acceleration and velocity vectors are as follows:

=

gkg

a ,

=

=

gtukgtu

tvtv

y

x

sincos

)()(

v

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

guT sin2= .

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

gukuu

gukuuTvTvR zzxxx )(2sin)sincos(2

2))0()(( ==+= ,

and

2

2

2sin)(

2cos)(

tgtutz

tkgtutx

=

=

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

Page 5


x

z

Q7 We need to calculate the displacement vectors of the two aircraft at 15:00.

400 km

60o

375 km r

r1

r2

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:

20230cos3754002375400 22 =+=r km. The bearing is just 90o + , and is given by:

sin30sin1rr = .

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

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x

y

r

R

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:

)sin1(cos

+==

ryrx

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

++

+=

+++=

vat

vat

at

tr

sin1

cos

)sin(1)cos(

R ,

for r = a and v the uniform speed of the rotating point (v = a). So that R = 0 at t = 0 we must have = /2.

Q9

30o

8 m/s

vR

V

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


9.1330tan

8 ==Rv m/s Q10 (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) Newtons second law for the 5 blocks would be written:

mFaamF

5)5( == .

(b) The resultant force on each cube must be 5F as maF =

5 and

mFa

5= .

(c) The resultant force F* on the 5th cube due to the 4th cube must accelerate 2m at a, therefore

52

52.2* F

mFmamF =

== .

Now that means that the force on the 4th cube due to the 5th cube is easily

obtained through Newtons 3 rd law as 5

2F , which of course is just enough

to make the resultant on cubes 1, 2 and 3 equal to 5

35

2 FFF = (just what is

needed to accelerate 3 blocks at mFa

5= ).

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. Lets 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:

WaRaWRaF +==== mmm 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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Q14 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:

110)600)(10(2)10(2 222 =+=+= vghuv 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:

0cos = mvMV

Barrel

Recoil velocity

Direction of shells initial velocity

V

v

Using the vector diagram:

sin)sin(vV = .

Using our first expression in conjunction with this we get:

tan1tan

sin)(costansincos)sincoscos(sin

+=+=

=

Mm

MmMmM

Q16 We of course have one of the standard triangles:

Page 9


pi = 5m

pf = 10m

60o

p = 5 3 m

So, if

=

=

mmm

fi 355

,0

5pp

then

=

m350

p .

If this impulse is applied to a mass of 5m at rest then the resulting motion will be the velocity vector:

==

30

5mpv .

Q17

m m

M

30o

u v

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


32

30coscoscos1

12

21

uv

uv

uuvve ==

== Conservation of momentum in the original direction of u gives:

xxu 330cos2 mvmvM == Therefore,

233

233 ===

uv

mM .

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:

2cos zyx

mvmvmvI ++= momentum in YX direction

sin2

30 xz mvmv +=

Also, the fact that the strings are inextensible means that the motions of Y and Z are restricted:

momentum YX direction

cos= )60cos( += xz

vv xyvv

30o

With our restriction on the velocities of Y and Z we immediately see that:

)sin3(cos2 = vv2

cos2 +=xz

zx

vvmI

So,

60o vx

vy

vz

X Y

Z

m

m

m

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mIvvvv

mI

zzz

z 152

215

27 ==+=

and

cos7

2233cos2 xzzxz

vvvvv == .

Now,

sin23

cos27

xz

xz

vv

vv

=

=

therefore 73tan = . Finally we can put our results for vz and back into

our expressions for vx and vy to obtain:

mIvy 15

7= m

Ivx 15132=

Q19

R2

m M C

R1

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:

vMm

mv

vMm

Mv

uMm

mu

uMm

Mu

+=+=

2

1

So, the impulse I on the smaller mass is given by

+=+=

2

1

)(11 uvMmmMmumvI +== ,

Page 12


but, , soeuv = ueMm

mMI )1( ++= . Q20

firs pact After the t im

m m m u

2)1(

2)1(

2

1

uev

uev

+=

=

2)1(

2ueu += , For the second impact

So, finally

ueueev

ueueev

final

final

4)1(

2)1(

2)1(

4)1(

2)1(

2)1(

2

3

2

2

+=++=

=+=

Page 13


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.

( )1

20

20

1144

+=+=== Q

qaxQq

xa

Qq

xxa

xaq

xQ

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

( )2020 44 xaq

xQE = , thus ( )3030 22 xa

qx

QdxdE

= .

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 Earths 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 / 9.81183Nkg1. Thus we may say that 72.00kg = W / 9.81183Nkg1. My weight in Madras is given by 9.78281Nkg1 m, where m is my true mass. Putting these two together gives

kg21.7278281.981183.9kg00.72kg00.72

Nkg81183.9Nkg78281.9

1

1

===

mm .

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.

Page 14


The gravitational field at the Earth caused by the Sun is 2RGM , while the

electrostatic field at the Earth caused by the Sun would be 204 R

Mq . Thus the

gravitational force of attraction is 2RGMm , while the electrostatic force of

repulsion is 204 R

Mqmq . These two are equal when

11100

2

20

2

2 Ckg1061.8444==== GqqG

RMmq

RGMm .

Accordingly, the charge on the Earth would have to be

mq = 5.981024 kg 8.611011 Ckg1 = 5.151014 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

.22100

kadxkxdxFWaa

x === Thus the potential energy function ( ) 221 kxxW = . b) If we write 22 yxr += 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 ( ) ( )2221221 yxkkrrW +== .

c) If the line joining the point to the origin makes angle with the x-axis, then the x-component of the force is Fx = F cos = kr x/r = kx, while the y-component of the force is Fy = F sin = kr y/r = ky. Thus as a vector

.

=

yx

kF

d) ( ) kxyxkxx

W =+=

2221

( ) kyyxkyy

W =+=

2221

Thus x

WFx = and

yWFy

= as we should expect.

Page 15


Q5

2

92

92

96

zxyy

zg

zxz

yg

zyx

xg

z

y

x

+==

==

==

, so the total force would be

+

==

2

92

92

96

zxyyzxzzyx

mmgF .

Q6 a) zy 265 = b) zxy 354 = c) Not possible.

d) Not possible.

e) ( )22223 zyx ++= Q7 a) 06.0=

xh , this means that for every metre you travel East, you rise by

0.06m.

b) 02.0=yh , this means that for every metre you travel North, you fall by

0.02m.

c) 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).

d) The gradient is steepest in the direction of the vector . In

this direction, you go 0.02m North for every 0.06m travelled East, and accordingly your bearing is given by tan

= 02.0

06.0

1(0.06/0.02) = tan1 3 = 108.4.

If you wish to check, suppose you travel a distance s in a on a bearing . This means that you move s sin East, and s cos North. Your new height will be different to your old one by h = 0.06s sin 0.02s cos . We now choose the value of to maximise the value of h for a given, fixed, value of s.

( ) sin02.0cos06.0 ssh +=

. The maximum value of h occurs when this partial derivative is zero, so 0.06 cos + 0.02 sin = 0, hence tan = 3 as before.

e) The maximum gradient is given by the magnitude of , and

is thus equal to

= 02.0

06.0

( ) K0632.002.006.0 22 =+ We can check this by feeding the bearing calculated in part (d) into the formula for h. = 108.4, so h = 0.0632s. The gradient equals h/s = 0.0632 as before.

Page 16


f) You need to walk perpendicular to the direction calculated in part (d) that is along a bearing of 18.4or 198.4. This is the direction in which h = 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 0.06200 = 12m. We have also walked 600m North, and thus expect to rise 0.02600 = 12m. When these are combined, we see that we have neither risen nor fallen.

Q8 a&b) Any positive scalar multiple of points in the same direction as r.

This vector has length

=

zyx

r

222 zyx ++ , and so to make a vector which has unit length but points in the same direction as r, we divide r by

222 zyx ++ . c) Using our information from part (a), where r = 222 zyx ++ ,

( )

++=

=

==

zyx

zyxGM

zyx

rGM

zyx

rrGM

rGM

23222322

1rg .

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

( ) ( )

( ) ( ) ( ) ( )zyCzyxGMxd

zyxGM

dxxzyx

GMdxzyx

GMx

,2

22

21222

223222

2322223222

+++=++=++=++=

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

Page 17


direction to our path. The magnitude of E is given by ArxAyAEEE yx =+=+= 222222 , and is the same for all points along

the circular path. The line integral is therefore equal to E length of path = 2r E = 2Ar2, where the minus sign comes from the fact that our path is in the opposite direction to E, and thus cos = 1, with defined as in Workshop 2.6.

Q10 ( ) 121126

20

NC 59913mFm108.8544

C1064

=

==r

QE

Q11 Using equation (2.27) we require d

2112 1mFm108.8541F =

, so

m108.8541F

1mFm108.854 122112 ==d . This is clearly impossible, 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...

Q12 Normally, using an inverse square law, we would expect that as our distance to the Earths centre has halved, the gravitational field should have quadrupled to 39.2 Nkg1. 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 doesnt), 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 Earths centre, a sphere drawn through your position, centred on the Earths centre will include 0.53 = 0.125 of the Earths mass. We reduce our estimate for the Earths field strength accordingly, and get a final answer of 4.9Nkg1, 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 9.8Nkg1 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

13 s rad10237.1 == EE Rg . The time period is 2/ = 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 Earths centre be L. The direct distance of our object to the Earths centre is given by Pythagoras theorem as 22 Ldr += . Accordingly, the local value of the 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


can affect the motion along it. The true gravitational field makes angle = cos1 (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 tunnels centre of magnitude gE d/RE, and so we have simple harmonic motion as before, with the same time period

13 s rad10237.1 == EE Rg .

r

d

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 irons 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

S

dSE = 0 over this surface.

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

Page 19


expect S

dSg = 0 on any spherical surface within this cavity, and thus g = 0 at

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 33 Rr , and thus the charge enclosed is 33 RQr . Accordingly, on our surface we have

3

3

RQr

S = dSE , 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

points on our surface (which has area S = 4r2). Therefore 33

24R

QrEr = and

34 RQrE = . Please note that this formula only applies when r < R. Once r > R,

then the normal Coulomb formula 24 rQE = applies. Given that a positive test

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 34 RQrE

r ==

, so

CR

Qr += 32

8 , where C is the arbitrary constant of integration. While we can choose C to be any value, it is usually taken as

RQC 8

3+= so that the function is continuous across r=R if we take the potential outside the material to be

RQ 4= in order that the potential falls to zero as we travel to infinity and

away from the influence of the charge.

Page 20


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: m or 1.9 1011105.1''60'60360 o 17 m. The radius of such a circle is then: 2

109.1 17 = 3.0 1016 m (this is in fact called a parsec of parallax-second).

Q2 The circumference of a wheel of 40 cm radius is: 2 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

12)(2

== rr

rd

rrdn .

The second expression may be expanded using a Taylor expansion:

11....,1)1(11....,1)1(

4321

4321


2v

v

x

y

x-axis

y-axis

P P

vP

vP'

A

Q

From the diagram we see that the lengths AP and AP are just

AP = AP yRyx 222 =+= Therefore the speed with respect to the ground (magnitudes of vP and vP) is just:

RyvyRyv 22)( == ;

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

Q5

AB

y O

r

v

x

z P

C

The position is momentarily at: , and at this point the velocity in m/s is:

03.000.005.0

Page 22


=

=

=

0.05.00.0

03.000.005.0

1000

03.000.005.0

v

After 5 s the vector will have rotated a further 50 radians or 250 = 7.96

revolutions.

Q6

y O

x

z

u v

w

r

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 are respectively , and , then may be written as:

,

=

coscoscos

where is the magnitude of the angular velocity. The velocity of the point due to this pure rotation is then just

v = r = ,

=

coscoscoscoscoscos

coscoscos

xyzxyz

zyx

that is, if the velocity were a consequence of a pure rotation, the

components of the vector v above would equal to u, v and w respectively. Notice that

wvu

0=++ wzvyux ,

Page 23


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

Q7 To differentiate (3.11) we need to replace all t with (t), so that is the first derivative of (t) with respect to time:

+

=

'''

1000cossin0sincos

'''

0000sincos0cossin

zyx

zyx

zyx

dtd

&&&

.

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

=

'''

0000sincos0cossin

2

2

zyx

zyx

dtd

&

+

+

'''

0000sincos0cossin

'''

0000cossin0sincos

zyx

zyx

&&&

+

+'''

1000cossin0sincos

'''

0000sincos0cossin

zyx

zyx

&&&&&&

&&&

and collecting these terms in , and we get:

'''

zyx

'''

zyx

&&&

'''

zyx

&&&&&&

+

'''

0000cossin0sincos

'''

0000sincos0cossin

2

zyx

zyx

&

'''

0000sincos0cossin

2zyx

&&&

'''

1000cossin0sincos

zyx

&&&&&&

The first of the terms in is recognisable as

'''

zyx

dtdr r, where

dtdr is the rate

of change of angular velocity represented as a vector of magnitude & and direction along the axis of rotation, which in this case is along the z-axis:

Page 24


==

=

zyxxy

zyx

&&& 00

0'''

0000sincos0cossin zyx

dtdr r

The second term in is immediately recognisable as =

'''

zyx

zyx

2 2r , but as

we see in Q9 this is just:

( r) = 2r,

r in these expressions represents the displacement vector in the primed coordinates rotated into the un-primed coordinates.

The term in also has the property of a vector cross product:

'''

zyx

&&&

==

=

zyxxy

zyx

&&&&&

&&&

00

20

2'''

0000sincos0cossin

2zyx

2 dtdr r

Where dtdr r here represents the velocity vector in the primed coordinates

rotated into the un-primed coordinates.

The final term in is obviously the acceleration vector in the primed

coordinates as seen from the un-primed coordinates:

'''

zyx

&&&&&&

'''

1000cossin0sincos

zyx

&&&&&&

= 22

dtdr r

Our 5 terms can therefore be summarised symbolically as follows:

dtda v = 2

2

dtdr r +

dtdr r + 2

dtdr r + ( r).

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:

dtda v = ( r) = 2r,

Page 25


as is r. So, the magnitude of dtda v = 82 0.6 = 35 m/s2 and the direction

is towards the centre of the millstone.

(ii) If there is an angular deceleration the acceleration expression collapses to:

dtda v =

dtdr r + ( r).

These two components are perpendicular to each other.

r

2r

dtdr

dtdr r

The resultant acceleration will have a magnitude:

222 )6.02.0()6.08( +=a = 38 m/s2,

with a direction that makes an angle

6.0646.02.0tan 1 = 18o with the radius

vector r. Q9 Using the expressions for vector triple products,

dtda v = ( r) = ( r) 2r = 2r as and r are perpendicular.

Q10 dtda v = 2

2

dtdr r +

dtdr r + 2

dtdr r + ( r). If

dtda v = 0 and

dtdr = 0,

then

0 = 22

dtdr r + 2

dtdr r + ( r).

Now,

dtda r =

dtdr r + r

dtdr r =

dtda r r,

so,

Page 26


0 = 22

dtdr r + 2 (

dtda r r )+ ( r) = 2

2

dtdr r + 2

dtda r ( r).

With ddt

a r = r = and = we have: v x , r x z2

2dtdr r = 2 2 ;

that is, a combination of a centrip

11& by the

r y r x

etal acceleration and a Coriolis effect.

Q 12 The force keeping the satellite in circular orbit is provided gravitational interaction:

22

rGmMrm = ,

where m and M are respectively the mass of the satellite and the mass of the planet, and is angular frequency of the orbit. This of course leads to:

2324 2 rTGMr == . 2 GMT

rRr +=

r

R

r

So,

( ) 23

23

122

343

34

+=+=Rr

GrR

RGT

,

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

11....,1)1( 382

81

212

1 ++=+ xxx 33


Q13

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

0RR =+ 2211 mm . This of course means that:

221

212

22

221

211

21

)(

)(

RRmGmRm

RRmGmRm

+=+=

Adding these two expressions together we get:

221

212 ()(mGRR +=+ 21 )(

)RR

m+ .

So,

3

21

22

)(4 a

mmGT +=

,

where a = (R1 + R2). speed, which is essentially made up of a radial bit and a

y

Q14 Now V is the orbitaltangential bit. The vector V would be:

'xV 'rr += & , where and are unit vectors in the ro coordinate system attached to

Using the expressions for r and h given in the text of the question we find:

'x 'y tating the planet. The square modulus of this vector is:

2222 rrV += & .

)cos1(

sin)sin( ehepr ==&)cos1( 2

ephr

pe

+=+

Therefore we can write V2 as:

22

222

22 ehV = 2 )cos1(sin ep

hp

++

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:

m1 m2 C

R1 R2

Page 28


[ ] +=+++= rpepheeephV 2)1(coscos21sin 222

22222

22

Giving:

=arp

hV 122

2 .

For e greater than 1:

[ ] +=+++= rpepheeephV 2)1(coscos21sin 222

22222

22

Giving:

+=arp

hV 122

2

For e = 1 the expression for V2 collapses to:

=rp

hV 22

2 .

Finally,

phmmG

mmGhp

2

2121

2

)()(

=++= .

Q15 Keplers 3rd law and the results of the last exercise give us:

32

2 4 aGM

T = and

=ar

GMV 122 .

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:

222 ~22

aaGM

aar

rGMVV

+= .

as the boost is very short and in that time r ~ 0. Differentiating the Kepler law gives:

aaGM

TT = 22

342 . Combining these two differentials with the 3rd law:

32

35

34

38

)2(

34

6222

2

GM

VVTTGMVTVMG

TT =

= .

Page 29


4

)

Oscillations & Waves Q1 = 2f = 8 rad/s. Given that we start with the displacement equal to the

amplitude, we should use a cosine wave. y = 4cm cos (8t), where we assume that t is in seconds.

Q2 = 2f = 20 rad/s. The amplitude is 5cm, so our equation will take the form y = 5cm cos (20t+). Using the initial condition y = 3cm, we have 0.6 = cos and so = 0.927 rad. We choose the negative value of to ensure that y gets bigger as t increases from zero. Our solution is therefore y = 5cm cos (20t 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 wouldnt work.

Q3 If we write our oscillation as ( +=+= tAtty cossin4cos3 , then it follows that sinsincoscossin4cos3 tAtAtty =+= , and so A cos = 3 and A sin = 4. From this we can tell that = tan1(1.3333) = 0.927 rad or 2.214 rad, and we choose the smaller so that cos is positive, while sin is negative. Meanwhile ( ) ( ) 25sincos 222 ==+ AAA , so A = 5.

Q4 The particle describes a figure of eight motion as in the diagram below.

Q5 Firstly, we remember that = 2f, and k = 2/. Therefore cffk

===

22 .

Q6 = 2f = 27 = 43.98 rad/s, and k = 2/ = 3.142 m1. 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

Page 30


( ) ( )

( ) ( )xtkxtA

kxtAkxtAkxtAkxtAkxtAkxtAy

142.3cos98.43coscm6coscos2

sinsincoscossinsincoscoscoscos

==

++=++=

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)

( ) cos2AAeeAAAeAAeT iiii +=++=++= , where we use the result from part (p) of Workshop 4.2.

Modulus square amplitude = ( )2* cos2 AATT += Full constructive interference occurs when cos = 1, so = 0, 2, 4... Full destructive interference occurs when cos = 0.5, so = 2/3, 4/3, 8/3,

10/3... Q10 We use the standard result that the sum of the series

a

aaaaap

p

=+++++

111 132 L . A proof of this in the case of an infinite

series is given as part (g) of Workshop 6.5.3, and the proof is readily adaptable to the finite series here.

The complex amplitude for the waves when they all meet is given by

( )( ) iiNNiiiNp

ip

eeAeeeAAeT

=++++==

= 111 12

1

0

L , where we use the

geometric series formula with a = ei. The modulus square amplitude is given by

( )( )

212

212

2222*

sinsin

cos22cos22

22

11

11 NANA

eeeeA

ee

eeATT ii

iNiN

i

iN

i

iN

==

=

=

where we use the trignometrical result that 1 cos 2 = 2 sin2 . It is immediately apparent that we shall have full destructive interference

whenever sin (N/2) = 0, providing sin(/2)0. Thus full destructive interference occurs when = 2/N, 4/N, 6/N... but not including = 2, 4, 6...

For the constructive interference, we need to be a bit more sneaky. When = 0, all the waves are in step with no phase difference, so there must be constructive interference. The same is expected whenever = 2, 4, 6... since this is in practice indistinguishable from the =0 case. We just need to check this from our formula. When is small, we may approximate sin , and so

( )( )

( )( ) 22221

221

2

212

212

2*

sinsin

NAN

AN

ATT ==

. Thus the intensity when we have

constructive interference is A2N2. Q11 N=3:

Page 31


N=5:

N=10

Page 32


When N=10 000, we expect bright constructive interference when = 0, 2, 4..., 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

Tfc == , so

Tf =22 and 22 fT= . The densities can then be

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 violins neck to one side.

Q13 For a transverse wave on a string, T = c2, as in section 4.8. Therefore

cTc

cTcZ == 2 .

Q14 ( ) 222122212221 ATckckAcTZAP ===

Q15 T

ZZTc

k === . As the question says, the tensions in the two strings must 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 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.

Q16 Using equation (4.37), the power incident on the join is 2221 iLi AZP = , while the power transmitted is 2221 tRt AZP = . It follows that the fraction of incident power transmitted is

( )222

2

2 42

RL

RL

RL

L

L

R

i

t

L

R

iL

tR

i

t

ZZZZ

ZZZ

ZZ

AA

ZZ

AZAZ

PP

+=

+=

== , where we have used

equation (4.48) for the ratio of the amplitudes. Similarly

22

2

2

+=

==

RL

RL

i

r

iL

rL

i

r

ZZZZ

AA

AZAZ

PP . For energy to be conserved, we require

that Pr + Pt = Pi, and so...

Page 33


( )( ) ( ) ( ) 1

22442

22

2

22

2

2

=+++=+

++=++=+

RL

RLRL

RL

RLRLRL

RL

RLRL

i

rt

ZZZZZZ

ZZZZZZZZ

ZZZZZZ

PPP

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

( ) ( )( )( ) ( )( ) ( ) ( )zkykxktEkkzkykxktE

xzkykxkt

zkykxktdzkykxktdE

zkykxktEx

tExx

E

zyxxxzyx

zyx

zyx

zyx

zyx

==

=

=

=

sinsin

cos

coscos rk

Q18 a) ( ) 221126

1380Wmm101.504

W103.9 ===

API

b) ( ) 22826

MWm1.46m106.964

W103.9 ===

API

Page 34


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 1.5106 m2.

143292619 ms105.42m101.0m101.5C101.613A

=== qAnIu

Q3 A = 0.02mm2 = 2108 m2.

143202819 ms103.13m101.0m102C101.60.01A

=== qAnIu .

Q4 Charge carried per second = 30mC. Charge on one Cu2+ ion is 21.61019C

So number of ions leaving anode each second = 1619 109.38C103.20.03C = .

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 = ( ) J101.34ms105.42kg109.1 372143121221 ==mv Potential energy = qV = 1.61019 C 230V = 3.681017 J. Ratio of potential to kinetic energy = 2.751020 : 1. Assumption justified.

In practice, actual velocities of the electrons are typically much higher than given in Q2, with many travelling at speeds over 106ms1 (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

Page 35


energy), and show that this is equal to 2max103 Nmv . Next, you can ask them to 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 Kirchhoffs 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:

55.11711

100600175

==

=

VV

VV

so the voltage is 1.55V as required. Using our branch equations and our knowledge of V, we find the currents as

mA45.15100

V55.1

mA18.8300

1.55V4V

mA27.7200

1.55V3V

3

2

1

==

==

==

I

I

I

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 k 4.5 V

1.0 k 4.5 V

1.5 V 3.0 V

IL1 IL1IL2 IL2

For the left hand circuit, the simultaneous equations are

( )

( ) 0300060000600010005.4

212

211

==

LLL

LLL

IIIIII

thus 21

21

90006000600070005.4

LL

LL

IIII

==

and these equations can be solved to give IL1 = 1.5 mA, IL2 = 1 mA. The currents in the resistors are therefore

1k: 1.5 mA to the right 6k: 0.5 mA downwards 3k: 1 mA downwards For the right hand circuit, the simultaneous equations are

Page 36


( )

( ) 03300060005.105.1600010005.4

212

211

==

LLL

LLL

IIIIII

thus 21

21

900060005.1600070003

LL

LL

IIII

==

and these equations can be solved to give IL1 = 0.667 mA, IL2 = 0.278 mA. The currents in the resistors are therefore

1k: 0.667 mA to the right 6k: 0.389 mA downwards 3k: 0.278 mA downwards. Q8 a) ( ) A

18100cos325 t

RVI == , so the amplitude is 18.1A.

b) The velocity is given (as in Q2) by

( )

( ) 143292619

mst100cos107.54m10m101.5C101.6

t100cos18.1A

=

==

qAn

Iv ,

and so the amplitude is 0.754mms1.

c) Equation (4.9) tells us that the amplitude of the velocity is equal to A where is the angular frequency (here 100 rads1) and A is the amplitude in distance terms. Thus here

m102.40rads100

ms107.54 61

14

=== amplitudevA .

Q9 325VV223020 === rmsVV Q10 18.4AA21320 === rmsII Q11 W12060W2222000 ===== rmsrmsrms PVIVIP Q12 (a) tVV 2202 cos= (b) Mean value of cos t equals one half. Accordingly the mean value of V2

is 2021 V .

(c) 0212

021 VV =

Q13 (a) Multiplication gives , the real part of which (using the methods of part (m) of Workshop 4.2) is equal to V

tieIV 2000I0 cos 2t. The average value

of this real part is zero.

(b) Using the method suggested:

( ) ( ) ( ) ( )

( )

iiitiiti

itiitititiititi

eeeeIV

eeIeeVeIeVVI+

++

+++=++==

22004

1

021

021

00 ReRe

(c) Using the approach suggested:

Page 37


( )

( )( )( )( )

cos2coscos22cos2

0021

0041

22004

1

++=++=

+++= +

tIVtIV

eeeeIVVI iiitiiti

.

Taking a time average, the cosine term containing t + averages to zero, leaving behind cos0021 IVVI = , and as we know from section 5.3.2

rmsrms IVIV =0021 , and so the reasoning is complete. (d) Here we go...

( )( )

sin2sincoscos

sinsincoscoscoscossinsincoscoscos

coscos

00212

00

0000

00

00

tIVtIVttIVttIV

tttIVtItVVI

===

+=,

where we have used the relationship sin 2t = 2 sin t cos t in the final line. Taking time averages, the cos2 averages to one half, while the second term averages to zero. This gives cos0021 IVVI = 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 Z1, which also must be real for the current and voltage to be in phase.

( )( )( )( )( )

( )222

222

222

2222

221

1

11

LRCLLCRiR

LRCLLiCRiRCLRCLR

LiRLiRLiRCLCRi

LiRCLCRiZ

+++=

+++=

++=+

+=

The imaginary part will be zero if

, and so we require that LCLCR =+ 222 222 LRLC += .

Q15 When R is set equal to zero, the impedance becomes

CL

LiZ 21

= , and therefore we see that this will be undefined (infinite) when LC1= . With small, non-zero R, we would also expect very large

impedances at this frequency.

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 . 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.

Page 38


Q16 Using the reasoning in and by figure 5.10 we require that ILR sin A = CV where 222 LRVI LR += , and tan A = L/R as in figure 5.9. It follows from the geometry of figure 5.9 that 222sin LRLA += , and therefore we can write our condition as

CLR

L

CVLR

LLR

VCVAI LR

=+

=++

=

222

222222

sin

as before.

Page 39


6 Thermal Physics Q1 %71

)273700()2737(11

1

2 =++==

TT .

Q2 so . 21 TTT = TTT = 12

11

1

1

2 11TT

TTT

TT === , which means to maximise we need to

maximise T. 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:

Entropy decrease of hot block = mcmcTdTmc

TdQ f

i

f

i

T

T

T

T

14.0373323ln =

==

Entropy increase of cold block = mcmcTdTmc

TdQ f

i

f

i

T

T

T

T

17.0273323ln =

==

Stotal = Shot + Scold = mcmcmc 03.017.014.0 =+ . 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:

=

+=+=

C

o

H

oT

T

T

T

T

T

T

T TT

TT

TdTmc

TdTmc

TdQ

TdQ o

C

o

H

o

C

o

H

lnln0 , which means

that:

CHo TTT = , 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 kTef= ,

TkT

ef kT

= and 1~2~ff

fff + ,

Page 40


For T ~ 300 K, T ~ 10 K 19223

102.110

3001038.1~

= J Q5 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 NkTH

ep , where H is the heat of vaporisation for a sample of N molecules of the substance. This means

constant =plnNkTH

2121 lnln NkT

HNkT

Hpp += .

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:

RTMgh

epp o= , 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:

kTmu

ef 22= .

u

Page 41


However, the volume of a small element* of the spiky ball volume is (see the diagram above):

sin2uV = du d d so the number of vectors of velocity with lengths between u and u + du is proportional to:

0

2

0

2

2

kTmu

e sin2u du d d ~ kTmueu 2 22 du.

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 kTmu

eu 22

2 .

Q9 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:

MRTu 32 = , where M is the molar mass of nitrogen.

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 222 udVd RMS = . 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 then we can define a mean free path as:

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

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

= 2222 1

dud

u = ,

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

V1 V2

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 we need to see what happens if we use the RMS value of this relative velocity rather than the

RMS value 2u :

(V1 V2)RMS = NVV + 21 VV22221 , where the sum is over all the

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 = 22 u , which changes our expression to =

21

2 d= .

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:

5151028

29831.8333

2 === M

RTu m/s,

and RT

pN

pnRTnN

VnN AAA === , so

Page 43


= 210100)102(

29831.822

1321022

== AA NpNd

RTd .

= 2.3 10-7 m. Q12 The volume V goes to V/100 adiabatically. The two expressions we need are:

2211 VpVp =

2

22

1

11

TVp

TVp =

Putting in what we know:

100

100 1221ppVpVp ==

so,

( )1

1

21

1

1 100100100

2

=

=

TT

T

Vp

TVp

,

which means that:

( ) 1112 1100 TTTT == Work done can be obtained from the first law of thermodynamics:

WQU += . Adiabatic compression means that Q = 0, so U = W = pV.

Throughout the process the pressure and volume are related on the adiabatic so:

kpV = , 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:

1001100100

)1(''

'1'

V

V

V

V

V

V

VkdVV

pdVU

===

( )1

11

1

)1(1100

)1( TTkVkVU

==

.

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.

Page 44

1 Linear Mechanics 2 Fields 3 Rotation 4 Oscillations & Waves 5 Circuits 6 Thermal Physics

[john p. cullerne, anton machacek] the language of(1)

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