Calculations
6. When a star collapses, becoming a Black Hole, there is a critical radius called the Schwarzschild Radius. When
the radius is below this value, the escape velocity exceeds the velocity of light and a Black Hole is formed. This
Mm
radius depends on G, the Gravitational Constant (defined by the formula F = G r2 ), the mass of the star and the
speed of light. Use the method of dimensions to find the form of the equation for the Schwarzschild Radius.
The Schwarzschild Radius can also be found by considering the radius at which the gravitational potential energy
1
for a body of mass m is equal to the non-relativistic kinetic energy at the speed of light, mc 2 . Use this method to
2
find an equation for the radius which now also includes the dimensionless constant.
Calculate the Schwarzschild radius for a sphere made of neutronium, which has a density of 10 17kgm-3, and for
water. The universe is approximately 10 billion light years in radius. Calculate the critical density, the density of
matter which would lead to the universe just failing to expand for ever.
7. Fermat explained the path taken by light as being the one of shortest time, not the shortest path length. A ray of
light passes from medium 1 into medium two, with refractive indices n1 and n2. Use Fermat’s assertion to find the
relation between the angles 1 and 2. You will need to use differential calculus and trigonometry.
Medium 1,
Refractive index n 1
y1

x
x1
x2

y2
Medium 2,
Refractive index n 2
8. I have two radioactive
substances, A and B.
There are 1010 atoms of
A which has a decay
constant A of 10-6 s-1
and 1011 atoms of B with
a decay constant B of 5
10-5 s-1.
After how long will there
be equal amounts of the
two materials? And how
long will it take before
the activities from the
two are equal?
9. It has been suggested that the reason there are two tides per day is that the effect of the pull of the moon on the
Earth is different at the near side, centre and back. If we look at the centre of mass frame (that is, essentially, we
subtract off the motion of the centre of the Earth) we get an effective pull on the near side and a push on the far
side. Draw a diagram to show this and calculate the difference in the forces at the front, middle and back. [Mass of
moon is about 7.5 1022Kg. Moon Earth distance is about 400 000km. Radius of Earth is about 6000km.]
10. As I walked past a fence made of separate posts I noticed that my dull, unmusical footsteps were reflected back
to me as a fairly pure tone. A friend with perfect pitch commented that the note was a high B - at about 1kHz. What
was the cause of the sound and the spacing of the posts?
11. As you walk under a street light your shadow stretches out ahead of you. If you walk at constant rate, does the
shadow accelerate, decelerate or stretch at a constant rate?
12.As you walk deeper into a wood, the amount of light coming from the evenly lit horizon will decrease. At what
rate? Assuming there are n trees per square metre and the tree are of mean diameter d, find an expression for the
rate at which the light intensity (the power per unit area) decreases with distance into the wood, x.
13. Grand Unification Theories are based around three fundamental scales called the Planck mass, Planck time and
 h 
Planck length. These are simply combinations of the constants c,   and G. There are no dimensionless
 2 
constants in these formulae. Find the form of the equation for each, and their size (c= 3.00  10 8 ms-1,
  1.05  10 34 Js and G= 6.67  10 11 Nm2kg-2). Is there a combination of the 3 constants which are dimensionless?
If there were, it would be possible to multiply your formulae by any power of that combination and still end up
with the same dimension, making the definitions ambiguous.
14. In an optical fibre, light can travel directly down the middle of the fibre. Alternatively a meridional ray is one
which bounces off the walls of the fibre yet stays in a single plane. The minimum angle a ray can bounce at is
controlled by the critical angle, which is the inverse sine of the ratio of the refractive indices of the inner and outer
parts of the fibre. (Remember that angles in optics are always measured from the normal). For a glass fibre with a
core index of 1.500 and an outer index of 1.450, and length 1km calculate the maximum path length for a
meridional ray. Hence calculate the time difference for this ray and a ray which passes straight through. If square
(in time) pulses of light are used to send information down the fibre, calculate the maximum rate at which
information can be sent.
15. Coriolis cat is standing on a large turntable which very slowly rotates at a constant angular velocity, . Coriolis
walks from the centre, directly out to the edge in a straight line (as she sees it) at a velocity v relative to the
turntable. Standing on the ground I see Coriolis walk towards me, but also off to my right as the turntable carries
her round. Find a formula for her acceleration. What shape path does she appear to follow?
16. Find a relationship between a runner’s speed, v, his stride length, l, and his number of paces per second, f.
17. The electron tube in a cathode ray oscilloscope (CRO) accelerates the electrons through a potential of 5kV. The
electrons are deflected by applying a potential to a pair of plates separated by 1cm. Describe the motion of the
electron as it passes between the plates, and then after it leaves as it flies on towards the screen. The plates are 5cm
long and the electrons travel a further 20cm to the screen. Calculate the voltage required across the plates to cause
a deflection of 1cm on the screen.
18. As part of the Millenium celebrations a string is wrapped along the Greenwich Mean Line, the line of 0°
longitude around the Earth, so it just sits on the Earth’s surface. A friend of a friend decides to have some fun with
the string and splices in an extra 1m long piece when no one is looking. When the string is pulled tight, how far off
the surface of the Earth does it now sit?
19. Today I can buy US dollars at the rate of 1.60 dollars to the pound. I can invest my pounds now for a fixed term
of 6 months which will return 5% annual interest. Alternatively I can buy dollars and invest them in the USA at a
fixed rate of 4% annually. If I made a contract today where I agreed to buy US dollars in six months time, what is
the price I would be offered? Is this the predicted price of dollars six months from now?
The present value of a deal is simply the amount of money I would have to invest now, at the current interest rate
offered for the period, to end up with the same amount of money as the deal offers. For example a $1000 bond
which pays back capital plus 4% after one year has a present value of $1000 when the one year interest rate is 4%.
However, if the one year interest rate is only 3% then I would have to invest 1.04/1.03 *1000 = $1009.71 at 3% to
end up with the same amount the bond is offering. So the present value of the bond is $1009.71. However, if the
interest rate increases to 5%, the present value of the bond will drop to less that $1000. This is a very simple way
of comparing two different financial instruments.
I am offered two bonds. One pays 5% plus my capital back after 12 months. The other pays 1% interest after 6
months and then 4% plus my capital back after 12 months. Work out the present value of a $1000 bond of each
type given the current 6 month interest rate is 3% and the 12 month rate is 4%. Which is the better deal?
20. A slightly different problem now. Traders love to gamble. One of them offers me a deal. He’ll toss a coin. If it
comes up heads I have to pay him a certain amount, say £x and we stop the game. If it come up tails he’ll pay me
£1 and he then tosses the coin again. If it comes up heads I pay him £x, and we stop. If it comes up tails again he
pays me £1 and we continue until it comes up heads, when I pay him £x. How much should I be willing to spend,
i.e. what is the largest value of x where I would still expect to gain in the long run?
Now we change the rules, so that each time tails comes up the payback doubles. So if it comes up first time he pays
me £1. Second time £2, third time £4 and so on. Now how much should I pay to enter the game?
21. To celebrate the Millenium I want to build a huge clock which will strike a loud gong at the stroke of midnight
on 31st December 1999 (or 2000, if you want to be a purist about when the Millenium starts). However, being a
physicist I want my gong to signal the right time locally not just Greenwich Mean Time. Local time means that
anywhere to the west of Greenwich is slightly later because it takes a little while for the Earth’s rotation to carry it
to where Greenwich time is defined. Fortunately for me, I live at 0° longitude so it is no problem. But it occurs to
me that I can ensure that my signal reaches everyone due West of me at exactly the right time too so long as I place
my clock at the right latitude. What latitude would I need to place the clock at? (The speed of sound in air is
330ms-1 and the radius of the earth is 6400km.)
22. Given a constant (velocity independent) coefficient of friction, which is the fastest route round a corner, the
tightest radius or the widest? In the light of that, explain why race cars take the line that they do.
23. I have a set of electronic scales which weigh to the nearest 5g. However, I do not know whether they register
5g if the weight on them is anything from 5g to 10g, or if they change at 7.5g. The scales have a ‘tare’ button
which resets the display to 0g. Making bread I use the following recipe:
175g white bread flour
175g brown bread flour
15g oatbran
15g wheatgerm
and a few other bits and pieces.
The two flours do much the same thing in the recipe and the oatbran and wheatgerm also are fairly interchangeable.
Is there an optimum method for adding the ingredients to minimise possible errors?
24. A rocket is of mass M0. It burns fuel at a steady rate until it is of mass Mf. What is the optimum change in
velocity it can acquire? (Assume that g remains constant throughout the flight and that there are no drag or steering
losses).
25. In a course for drivers to make them aware of the dangers of speeding, it was claimed that if a car travelling at
30mph can come to a complete stop in a certain distance, then in the same distance, the same car travelling at
35mph would be still going at 21mph at the end – making a serious impact of course. Discuss. It may help to know
that 1mph = 2.2m/s and the Highway code uses a reaction time of 0.7s in calculating stopping distances.
26. The earth’s orbit varies from 91.5 to 94.5 million miles from the sun, being at its closest in January. Assuming
the Earth is in thermal equilibrium due to the heat from the sun being balanced by the Earth radiating, calculate the
approximate change in temperature due to distance alone. Radiant power depends on the fourth power of absolute
temperature (Stefan’s Law).
Calculations - Answers
6) In this case the only dimensions involved in the problem will be length [L], mass [M] and time [T].
We are trying to form an equation for the radius which is a length so has dimension [L]. The radius depends on the
mass, of dimension [M], the speed of light, of dimension [L][T] -1, and the Gravitational Constant whose dimension
we must first find.
Mm
We have the equation F=G r2 which we can rearrange to give
Fr2
G = Mm
The dimensions of the two sides must be the same. The dimensions of M,m and r are trivial but what about force,
F? We know the equation F=ma and so we can see that the dimensions of F are [M][L][T]-2.
So now the dimensions of G must be [M][L][T]-2[L]2[M]-2 = [L]3[T]-2[M]-1
We now write the equation for the Schwarzschild Radius:
r = kGmc
Where r is the radius and k is a dimensionless constant. The dimensions have to match on both sides so let’s look at
the dimension form of the equation:
[L] = [L]3[T]-2[M]-[M][L][T]-
Equating the dimensions we get three equations:





(mass)
(time)
(length)
We can solve these equations to find =1, =1, =-2 so the equation for the radius is
GM
r = k c2
where k is a dimensionless constant which we cannot find by this method.
1
mc 2 in a non-relativistic calculation. At rest on a sphere of
2
GMm
mass M it will have gravitational potential energy 
where r is the radius of the sphere. To escape it must
r
have kinetic energy just equal to the potential energy so we put the two terms equal and solve for r which gives
2GM
r
.
c2
A body travelling at a speed c has a kinetic energy
It is important to understand this formula. If a solid body has a Schwarzschild radius which is less than its actual
radius then light can escape – even from inside it because there the gravitational potential energy is closer to zero.
If the Schwarzschild radius is larger than the size of the body then there will be a region in which nothing, not even
light, has sufficient escape velocity. It is surprising, but the non-relativistic equation works in this case!
4
3c 2
Now for a sphere of density  the mass will be M  r 3  . Putting this into our equation we get RS 
.
3
8G
For neutronium (the stuff neutron stars are made of) this gives RS =40km. For water it gives RS  4  1011 m . This
shows that any body sufficiently large will become a black hole (assuming constant density and a bunch of other
things!). Clearly the mass goes up as the cube of the radius while the potential energy goes down linearly with the
radius but up with mass. Hence eventually the potential energy increases to be greater than the maximum kinetic
energy.
Now for the universe. We use the same formula and rearrange it to find the density. This gives 1.8  10 26 kgm -3 .
This is about 10 hydrogen atoms per cubic meter.
7) In this kind of problem it helps to decide first of all what is fixed and what is a variable. In this case y1, y2 and
x=x1+x2 are all fixed. The variables are x1 and x2 although they are not independent of each other.
We want to find an expression for the time taken for the light journey so we need the length of the two paths. The
first path, in medium one, is of length l1 given by:
l1 = (y12 + x12)1/2
The second path is l2 which is given by
l2 = (y22 + x22)1/2
The time taken on each path is given by l1/c1 and l2/c2 where c1 = c/n1 and c2 = c/n2. So the total time is:
ln l n
t= 1 1  2 2
c
c
We want to minimise this time with respect to x1 or x2. So we differentiate and set the differential to zero.
dt
cdx = n1x1(y12 + x12)-1/2 - n2x2(y22 + x22)-1/2 = 0
1
To get the second term I have used the chain rule. Differentiating the function with respect to x 1 I differentiated
dx2
with respect to x2 and then multiplied by dx which is -1.
1
We can substitute back for l1 and l2 to get
x1
x2
n1 l = n2 l
1
2
x
Looking back at the original diagram we can see that l = sin so the equation reduces to
n1 sin1 = n2sin2
which is known as Snell’s Law.
8) The number of atoms at a time t is given by
N(t) = N(0)e-t
So we simply equate the numbers for each of the substances A and B:
1010exp(-10-6t) = 1011exp(-510-5t)
exp(4.910-5t) = 10 (dividing through on both sides)
4.910-5t = ln 10
t = 47000s
= 13h 3m
The activity is given by e-t so again we equate the expressions:
10-6  1010 exp(-10-6t) = 510-5  1011 exp(-5  10-5t)
exp(4.910-5t) = 500 (dividing through on both sides)
4.910-5t = ln 500
t = 127000s
= 35h 14m
9) First we must draw the diagram:
Earth
Moon
The arrows indicate the strength of the gravitational force on the Earth. It is strongest at the point nearest the moon
of course.
If we subtract off the acceleration of the centre of the earth, then we subtract two arrows from each of the forces
above, giving the following diagram :
Earth
Moon
So the effect of a force which is purely attractive can be partly attractive and partly repulsive when you look at a
scale large enough for the field to vary, and subtract off the centre of mass motion.
The formula for the field is
GM
g = r2
G = 6.710-11Nm2kg-2
M = 7.51022Kg
r=4108m for the nearside, 4.06108m for the middle and 4.12108m for the far side.
gnear = 3.1410-5 Nkg-1
gmiddle = 3.0510-5 Nkg-1
gfar = 2.9610-5 Nkg-1
The difference in the field strengths is 910-7Nkg-1 between each end and the middle. In fact the difference
between the near side and middle will be slightly greater than between the middle and the far side, but the distances
are sufficiently small for it not to matter.
10) What is happening in the musical echo is that the sounds waves sent out as my feet hit the ground are reflected
by the posts of the fence. If there is a regular spacing of the posts then there will be a staccato sound reflected back
at regular intervals - and a rapidly repeated click or tap becomes a musical note.
The note I hear is at 1kHz. That means that the sound travels 1ms extra between each post. So the posts are a
distance
v
d = 2f (v = velocity of sound, f = frequency).
apart. The factor of two comes in because the sound has to travel to the post and back.
Putting in the numbers given we get d=0.17m=17cm.
11) Look at the diagram below:
d
x2
h
h
l2
x1
l1
What we want to know is how the length of the shadows, l1 and l2, varies with the distance from the lamppost, x1
dx
dl
and x2. Walking speed, dt , is constant and we need to find a formula for dt .
h
d
For each shadow, l = x+l because of the two similar triangles. h and d are constants here. Rearrange the
expressions and we find:
d-h
x= h l
 
So we can immediately see that l is directly proportional to x and if x increases linearly so does the shadow length.
12) Consider a light ray of intensity I and width b at some distance x into the wood. In penetrating a further
distance dx it sweeps out an area of bdx. It will be blocked by a number of trees nbdx and the fraction of light
removed by each tree will be d/b if d is the width of each tree. So we can write an equation, putting the change in
intensity to be dI:
dI -nbd dx
= -dn dx
I =
b
We can rearrange and integrate to give
I = I0 e-ndx
So the light intensity decays exponentially. Actually, in integrating we have assumed that the trees can be
subdivided indefinitely into slices, which is not true as they have a finite diameter. So in this case we have used
integration to approximate summation rather than vice versa.
13) First find the dimensions of the three constants:
[c] = [L][T]-1
[  ] = [M][L]2[T]-1
[G] = [M]-1[L]3[T]-2
In each case the Planck length, time and mass will be given by c   G  . We just need to solve for in each
case.
You should find the Planck Mass is given by
c
 2  10 8 kg , the Planck Time by
G
G
 5  10 44 s and the
c5
G
 10 35 m . If you try to solve for no dimensions you should find all three powers are zero
3
c
is the only solution, so no combination of them is dimensionless.
Planck Length by
1.496
14) The critical angle for this system is sin-1 1.500 = 85.81. So in the diagram below the angle to the surface is
4.19. The path length of the meridional ray is just l /cos so for 1km of fibre we have a path length of 1.0027km.
2.7
This extra 2.7 meters will take a time
s where c is the speed of light in a vacuum – remember that the light
c
1.500
travels more slowly in the glass. This gives a time difference of 1.34  10 8 s. Square pulse will end up with a long
tail so no matter how short the pulse there needs to be a gap of 1.34  10 8 s between it and the next pulse. So at
most 75 million pulses can be sent per second if there is to be absolutely no overlap between them. In practice a
small amount of overlap may be permissible allowing a faster transfer rate. Fibres used for high speed
communication may be of the ‘single mode’ type which allows only a central ray to be sent, and no other path is
possible. Then a second problem arises. A pulse is necessarily made up of a spread of frequencies. Different
frequencies travel at different speeds in glass (which is why a prism creates a spectrum from white light) and so
again pulse will spread out. This effect is called dispersion.
15) Consider the situation from Coriolis’ point of view. She walks along a radius of the turntable so at some time, t,
she has reached a distance r=vt from the centre. The turntable has a constant angular velocity. From my point of
view this gives her a sideways velocity (recall the turntable is moving slowly so I shall not worry here about her
going round corners). The sideways velocity, vs, is given by vs=r. So the velocity components are
v (towards me) = v
vs (to the side) = vt
So the sideways velocity increases linearly with time, representing a constant acceleration of v. Notice that this
acceleration is dependent on the forwards velocity – when she reaches the edge she will be travelling at r 
regardless of how quickly she got there.
However, if she pounces from the centre to the edge, the story is somewhat different. She takes off with no
sideways velocity, so just before she lands she is still going in the same direction. As far as I am concerned, she
travels in a straight line, but from the point of view of an observer on the turntable (a small furry rodent for
example) the cat appears to move in a curve. If she pounced in the direction of the rodent she would miss, and the
further away the rodent is, the more she misses by. The same is true for a ballistic missile launched in a north-south
direction on the Earth, or for weather systems too.
16) In each second the runner covers a distance of number of paces  length of pace = fl. The distance covered per
second is the speed so we get v=fl, or the same equation as for a wave.
17) First of all we need to calculate the velocity of the electrons after they are accelerated. With a potential of 5kV,
1
they gain an energy qV which will equal their kinetic energy, mv 2 . That gives a velocity of 4.2  10 7 ms -1 .
2
As it passes between the plates the electron feels a constant force and hence undergoes a constant acceleration
(ignoring relativistic effects). So its path will be parabolic – exactly like a projectile. Once it leaves the plates,
assuming no fringe effects, the electron will carry on in a straight line in the direction is was travelling.
If a voltage V is applied across the plates there will be a field E=V/d = 100V for a separation of 0.01m. This leads
to a force Eq on the electron and an acceleration a=Eq/m.
The time of flight through the plates is t=l/v. During that time the electron will be at co-ordinates x=vt and
1
1 a 2
x . At the end, when x=0.05m, the direction of travel can be found from the gradient of the path,
y= at 2 
2
2 v2
axs
dy ax
 2 . In a distance s after the plates the electron will travel a further 2 . So the total displacement, which is
v
dx v
0.01m, is given by
1 a 2 als
l  2
2 v2
v
where l is the length of the plates, 0.05m. Substituting in the numbers and putting a=Eq/m = 100Vq/m I get
V=89V.
0.01 =
18. The circumference increases by a distance of 1m. Therefore the radius increases by
1
m, and the string will
2
sit about 16cm above the ground. It seems counter-intuitive doesn’t it?
19. The financial markets operate on a principle of no arbitrage, that is there should never exist deals which make
free money. Given that, the two different ways to make US dollars in 6 months time must produce the same
number of dollars. Let the exchange rate in 6 months (the forward rate as it is known) be R. Then if I start with x
x
1.02 = 0.6375x dollars. (I use 1.02 not 1.04 to
pounds, convert them now and invest in the USA I shall have
1.60
calculate the interest because I have invested for 6 months not a year). If I invest in the UK and then convert I shall
1.025 x
have
dollars. These two amounts are equal so we can solve for R, giving 1.608 dollars to the pound.
R
This is not a market prediction for the rate – in fact it has nothing whatsoever to do with future exchange rates. It is
a necessary rate to offer to avoid arbitrage in a forward agreement.
First look at the simple bond. It will pay me a total of $1050 at the end of the 12 months. I would have to invest
$1050/1.04 = $1009.62 (to the nearest cent) now to make the same amount of money. So the present value is
$1009.62.
Now the other bond – it pays me back $1040 at the end of the 12 months, the present value of which is $1000
because that is the same as the current interest rate. In addition the bond pays me $10 (1%) after 6 months. The
present value of that is $10/1.03 = $9.71. In other words I would have to invest $1000 for a year and $9.71 for 6
months in order to make back the same amount as this bond is offering me. So the present value of this bond is
$1009.71. The second bond is better value.
In general the present value is given by the sum of the discounted cash flows. The cash flows are the individual
returns such as the 1% and 4% in the second example. The discounted part means they have to be figured back in
time using the current interest rate so they can be added together.
You may ask “why don’t I then invest that $10 for a further 6 months and get more interest – surely that is going to
increase the present value still further?”. Well, it will, but that is a different deal again because I don’t get the $10
as cash in 6 months time. A shrewd investor might buy a stack of the second bond, reinvest the $10 on each one
(he could even get an agreed rate now, a forward rate, to invest for 6 months in 6 months) and offer the whole
package for sale to others. Clearly it would pay back a little more than the 5% bond in the first example so it would
be a better deal.
20. We know that the cost of this deal to me is going to be £x – I have to pay the trader eventually. Now we have to
calculate my expected return on this deal. Assuming the coin is fair then I expect to get £1 half the time on the first
throw. This is worth 50p. On the second throw (which I will only reach half the time) I can expect a further 50p,
but my expected return is 25p because I only get the chance 1 time in two. For the third round, I have a one in four
chance of getting there, and again I expect to win 50p on average. So my expected return is
1
1
1
50p + 2  50p + 4  50p + 8  50p + ...
You may recognise this series as having a finite sum of 250p = £1. (If you don’t recognise it, it is simply a
1
geometric progression of common ratio 2 .) So the most I should be willing to pay to play this game is 99p if I still
want to make some profit.
Now the second game. In this case the cost is again £x. The return however is going to be
1
1
1
50p + 2  £1 + 4  £2 + 8  £4 + ...
Each term in the series is 50p. The series is infinite so the sum of this series is also infinite. I should be willing to
spend any amount of money to join this game.
21. The gong’s signal will spread out at the speed of sound. We need to find the position on the earth such that the
surface is moving at 330ms-1, meaning the sound reaches the right longitude at the right time. At the equator, the
surface is moving (26400)km in 24 hours. This is 465ms-1. However, as you move north (or south) the radius of
the circle of the earth decreases as cos where  is the angle of latitude. The rotation time stays the same so the
surface is rotating at slower speed. We simply need to find  such that 465cos = 330. This gives a latitude of
44°50’ north or south. North puts me a few miles north east of Bordeaux, whereas south dumps me in the South
Atlantic Ocean.
22. A constant  means a centripetal force F=R where R is the weight of the car (plus any downforce). This must
equal mv2/r. The time taken for a corner is the distance over velocity. Assuming the corner turns through an angle
 radians, that time is r/v. So we have v = r/T and therefore
R = m2r2/rT2
 r  T2
or T  r
In other words the larger the radius the longer the time. Racing cars take the widest possible radius round a corner
which should be the longest in time. There are two reasons for this. The first, and more minor I suspect is that the
reaction force is enhanced by the downthrust produced by spoilers. This increases with speed, but is limited by
design rules. However the main reason is that although the time taken round the corner is greater, the corner is
taken at higher speed so less speed is lost on the straights either side during braking and accelerating. Hence the
average speed on the straight is higher and the time taken there must more than make up for the time lost on the
corners.
23. I zero the scales then add the two flours, one after another, aiming for a total of 350g. The possible 5g error
trading white for brown flour is unimportant and the possible 5g error in the overall amount is a smaller percentage
than if I reset the scales between each weighing. Similarly with the oatbran and wheatgerm – I add oatbran until the
scales read 15g, the wheatgerm until the scales read 30g.
24. This is known as Tsiolkovsky’s equation, after the Russian schoolteacher who worked it out in the 1920s.
From Newton’s second law, F = ma so
dV
dm
m dt = -Ve dt
where Ve is the exhaust velocity. The left hand term is the effect of the force on the rocket and m is a function of
time. The right hand term is the thrust produced by the exhaust fuel eject at a constant rate with a constant velocity
relative to the rocket. This can be rewritten as:
m dV = -Ve dm
Vf
Mf
dm
m
M0
 dV  V 
e
V0
Vf – V0 = V = Ve ln(M0/Mf)
25. First off let’s look at the distance involved. 30mph is 13.6m/s. The thinking distance will be 0.7*13.6 = 9.5m.
Let the deceleration be a so the braking distance is given by 13.62/(2a) = 93/a m.
At 35mph (15.9m/s) the thinking distance is 11.1m. The distance left for braking will then be (9.5+93/a-11.1) =
(93/a – 1.6). Putting this into v2=u2 + 2as gives us:
u2 = 15.92 –2a(93/a-1.6)
= 253.1 – 186 – 3.2a
= 67.1 – 3.2a
We were told the final speed was 21mph, or 9.5m/s which gives a=-7.5m/s2 or a deceleration of ¾ of that due to
gravity. That seems like a sensible value. We can check it with the other figures in the highway code. At 70mph it
gives a 75m braking distance. That’s a deceleration of 6.7m/s2 so not too far off. If we use the Highway Code
deceleration we would get a final speed of 9.4m/s or a little under 21mph. Quite a sobering thought.
26. The power falling on the earth due to the sun varies with the inverse square of distance. The power radiated out
equals that in if the earth is in equilibrium so we must assume that the earth’s temperature varies such that T4  1/r2
or T1/r. The easiest way to tackle this is by percentage change. R changes by about 3% (3 million miles in 93
million). So T varies by half that (the square root halves the percentage change) so 1.5%. As the mean temperature
of the earth is about 300K, 1.5% change is about 4.5C.