Download Exam 2013 with Answers File - QMplus

BSc/MSci Examination by course unit.
???day ??th ??? 2013
10:00 – 12:30
PHY4116 From Newton to Einstein
Duration: 2 hours 30 minutes
YOU ARE NOT PERMITTED TO READ THE CONTENTS OF THIS QUESTION PAPER UNTIL
INSTRUCTED TO DO SO BY AN INVIGILATOR.
Instructions:
Answer ALL questions in section A. Answer ONLY TWO questions from section B.
Section A carries 50 marks, each question in section B carries 25 marks.
If you answer more questions than specified, only the first answers (up to the specified
number) will be marked. Cross out any answers that you do not wish to be marked.
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answer book the name and type of machine used.
Complete all rough workings in the answer book and cross through any work that is not to be
assessed.
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EXAM PAPERS MUST NOT BE REMOVED FROM THE EXAM ROOM
Examiners:
D.J. Dunstan
M. Baxendale
© Queen Mary, University of London, 2013
Page 2
PHY4116 (2013)
SECTION A. Answers all questions in Section A
Question A1
Given the vectors u = (2, 0, 1) and v = (1, 1, 0), calculate
i)
u+v
(2+1, 0+1, 1+0) = (3, 1, 1)
[1]
ii) u.v
21 + 01 + 10 = 2
[1]
iii) uv
(00 – 11, 11 – 20, 00 – 11) = (–1, 1, –1) [1]
State giving reasons which of the following quantities are vectors,
iv) iˆ  2 ˆj  3kˆ
Yes: hat means unit vector – and adding vectors gives a vector [1]
v) Stress
No – has magnitude and two directions
[1]
[5 marks]
Question A2
i)
ii)
State Newton’s Third Law of Motion and explain what it says about momentum.
Action = Reaction [1]; Momentum is conserved [1].
A falling rock in a quarry gains momentum of 100 kg m s–1 during its descent.
What other body must be considered in the application of Newton’s Third Law to
this problem? The Earth [1]. What momentum does it gain (state magnitude and
direction)? 100 kg m s–1 upwards OR –100 kg m s–1 downwards [2].
[5 marks 10]
Question A3
A pendulum has a bob of mass 10 kg; at rest the bob is at x = 0. We will approximate its
potential curve when it is displaced as E = 5x2 J m–2.
Sketch this curve for E and label the axes.
Potential energy, E J
i)
20
15
10
5
2
1
1
2
Position, x m
ii)
iii)
[2]
Calculate the force as a function of position. F = –dE/dx = –10x N m–1
Sketch this curve for F and label the axes.
[1]
PHY4116 (2013)
Page 3
Force, F N
10
5
2
1
1
2
5
10
Position, x m
iv)
What is the acceleration at x = 0? F = 0 so a = 0
[1]
[1]
[5 marks 15]
Question A4
A space station consists of a cylindrical shell of inner diameter 20 m. It rotates around
its axis to give artificial gravity of 1g at the “floor” (the inner surface of the shell).
i)
Calculate the speed v of the floor: a = v2/r so v = Sqrt[ar] = 10 m s–1. [1]
ii) Calculate the angular velocity : v = r so  = v / r = 1 rad s–1
[1]
–1
iii) If 0.5g were required, what would be the speed? v / Sqrt[2] = 7.07 m s [1]
iv) If the diameter were 40 m, what would be the speed for 1g? v  2½ = 14.14 m s–
1[1]
v) How does the artificial gravity vary from the axis to the floor? a = v2/r = 2r so it
increases linearly. [1]
[5 marks 20]
Question A5
A bicycle wheel weighs 1 kg, all of which is at the rim (i.e. neglect the hub). Its diameter
2r is 0.7 m. If the bicycle is ridden at the speed of 3.5 m s–1, then in the frame of
reference of the rider,
i)
Calculate the angular speed  of the wheel.  = v/r = 10 rad s–1.
[1]
ii) Calculate the momentum of the rim, p: mv = 13.5 = 3.5 kg ms–1 rad s–1 [1]
iii) Calculate the angular momentum of the wheel: pr = 3.50.35 = 1.225 kg m2 s–1
[1]
iv) Calculate the kinetic energy of the wheel ½mv = ½13.52 = 6.125 J
[1]
[4 marks 24]
Question A6
A planet orbits its star in a circular orbit at a speed v at a radius R.
i)
Give an expression for the centripetal force FC on the planet. FC = mv2/r [1]
ii) Give an expression for the gravitational force FG of the star on the planet, in
terms of the mass M of the star and the mass m of the planet: FG = GMm/r2
[1]
iii) Hence find an expression for the speed v of the planet in its orbit: FG = FC, so v =
Sqrt[GM/r]
[1]
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PHY4116 (2013)
G = 6.67  10-11 m3 kg–1 s–2, mass of Sun is 2.0  1030 kg and the distance of the Earth
from the Sun is 1.50  1011 m.
iv) Putting in the numerical values to your answer to (iii), find the speed of the Earth
in its orbit. v = Sqrt[6.67  10–11  2  1030 / 150  109] = 30 km s–1[1]
v) What will escape velocity be at the orbital distance of the Earth from the Sun?
vesc = Sqrt[2] vorb ~ 42 km s–1
[1]
[5 marks 29]
Question A7
A fast sprinter visiting the Antarctic research base crosses the South Pole, at 10 m s –1.
Assuming he weighs 80kg, and taking the Coriolis force as F = 2mv  ,
i)
Calculate the Coriolis force he experiences and give its direction. FC = 2mv =
2  80  10  7.29  10-5 = 0.12 N. For the direction, I would accept “North”, but
“to his right” would be better.
[2]
ii) He now goes to a tropical country and runs across the Equator. What is the value
of the Coriolis force now? v is now parallel to  so FC = 0
[1]
iii) Still at the Equator, he now climbs a tree, at 1 m s–1. What is the value of the
Coriolis force now? v is now perpendicular to  so FC = 0.012 N.
[1]
[4 marks 33]
Question A8
A diatomic molecule consists of two atoms separated by a distance x.
i)
Sketch the Lennard-Jones potential energy E(x) for two atoms separated by a
distance x, and mark the equilibrium separation x0.
Equilibrium separation
ii)
iii)
[3]
On your sketch add a dashed curve showing the force F(x) between the atoms.
See diagram
[1]
Approximating the minimum of the potential as the parabolic well E = a + b(x –
x0)2, give rough estimates in any convenient units of the constants a, b and x0 for
real atoms. The depth a should be a few eV; the equilibrium separation x0 should
be a few Angstroms. The minimum is quite sharp, so b should be a few eV Å–2.
[3]
[7 marks 40]
Question A9
A body has a rest mass of m0. It is accelerated to the speed corresponding to  = 5.
i)
What is now its mass? m = m0 = 5m0
[1]
ii) Give expressions for its total energy at rest and at speed. E = mc2, so at rest E =
m0c2 and at speed E = m0c2 = 5m0c2.
[2]
PHY4116 (2013)
iii)
What is its kinetic energy at speed? KE = E - m0c2 = 4m0c2
Page 5
[1]
[4 marks 44]
Question A10
i)
ii)
iii)
Special Relativity rests upon the Principles of Relativity, and the constancy of the
speed of light. Give the name of the third Principle required for General Relativity
and say what it asserts. The Principle of Equivalence; you cannot tell the
difference (in a small closed lab) between acceleration and gravity. [2]
Give a qualitative description and explanation of the path required by the third
Principle for a beam of light emitted by a horizontal torch or laser in a laboratory
that is accelerating upwards. Curved downwards (parabola) as if the light was
subject to gravity. [2]
Describe some experimental evidence for General Relativity and this Principle.
Eotvos experiments, OR the corrections needed for GPS OR etc. [2]
[6 marks 50]
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PHY4116 (2013)
SECTION B. Attempt two of the four questions in this section
Question B1
a)
An unstable sub-atomic particle is created in a collision in the beamline in the
Large Hadron Collider. It has a lifetime of 0.1 ns. It travels at almost the speed of light,
and is detected decaying 3 m away in a detector element in the ATLAS experiment.
i)
From its point of view, how long does the particle live? It is stationary in its own
inertial rest-frame and it can carry a clock from the event that created it to the
event that destroys it, at the same place in its frame. So it lives the lifetime of
0.1ns.
[1]
ii) From the point of view of the scientists running ATLAS, how long does it live? At
almost the speed of light, it has taken 10ns to travel 3m. Therefore they see it
existing for 10ns before its decay.
[1]
iii) Explain your answer to (ii), giving the value of . Time dilation – for the scientists,
the particle’s clock is running slow by the factor  = 100.
[1]
2
iv) What is the speed of the particle?  = 100 so Sqrt[1 – v2/c2] = 1/100 so v /c2 = 1
– 10–4 so v = 0.99995c
[1]
v) What is the Proper Time between the creation and decay of the particle? 0.1ns,
see (i), the particle may be considered to be a clock present at both events. [1]
vi) What is the distance between the beamline and the detector in the particle’s
frame of reference? 0.1ns  0.99995c ~ 3cm.
[1]
vii) What is the Proper Length between the beamline and the detector element? 3 m,
for the stationary scientists measure this stationary length. [1]
viii) What is the spacetime Lorentz-invariant interval between the creation and decay
of the particle? Is this spacelike or timelike? The proper length and the proper
time are Lorentz-invariant; he we have the proper time = 0.1 ns, hence timelike.
[1]
[8 marks]
b)
A roller-coaster includes a complete vertical circle as part of the track, of radius 5
m. A car coasting down the track weighs 200 kg. What is the minimum speed of the
car,
i)
At the top of the circle? Equate centrifugal and gravitational force, i.e. mv2/r =
mg, so v  gr  50 = 7.07 ms–1
[1]
ii)
At the bottom of the circle? KE at top is then ½ mv2 = 5 kJ. PE in going to bottom
is mgh = 2mgr = 2 kJ. So 5 + 2 = 7 kJ = ½ mv2, so v  70 = 8.34 ms–1
[2]
iii)
What is the force of the car on the track at the top and at the bottom of the circle?
In both cases, mv2/r + mg with due regard to signs. At top, 0. At bottom, 200 
70 / 5 + 200  10 = 4.8 kN
[2]
[5 marks]
c)
[This question is unseen – they have done problems on a fixed beam, but not a
moving one] The figure shows a beam of negligible weight pivoted at O with weights
attached as indicated.
PHY4116 (2013)
i)
Page 7
Calculate the mass m at A, and the normal reaction N of the system if it is to
satisfy these equilibrium conditions.
1.1 m
2.0 m
1.2 m
A
O
15 kg
m
3 kg
Considering translational equilibrium, N=(15+3+m)g
Considering rotational equilibrium, 15g × 1.1=(3g×3.2)+mg(1.2)
So m=(16.5-9.6)/1.2=5.75 kg, and hence N=(15+3+5.75)g=237.5 Newtons [4]
ii)
iii)
iv)
The mass m is now removed. Calculate the angular acceleration of the beam. M
of I about O is IO = Sum[Mi ri2] = 48.87 kg m2. Torque is  = m  1.2 = 6.9 N m.
So  =  / IO = 0.141 rad s–2.
[2]
After two seconds, a second pivot OCG is inserted at the position of the centre of
gravity of the beam. Where is this, relative to O? Let it be a distance x to the left
of O. Then we solve 15 (1.1 – x) = 3 (3.2 + x), i.e. x = 0.383 m.
[2]
What will be the angular momentum about OCG? L = I,  is unchanged. I is
given by theorem of parallel axes as the previous I – total mass  x2 = 46.2. So L
= 46.2  2  = 13.1kg m2 s–1.
[4]
[12 marks]
Question B2
a)
A pair of atoms separated by the distance r has the potential energy given by the
Lennard-Jones U(r). Find the radius at which the force is at a maximum. What
happens if an external tensile force is applied to pull the atoms to this separation? What
does this say about the strength of materials?
Solve dF(r)/dr = 0: (– 84 r06/r8 + 156 r012/r14 = 0, so r = (13/7)^(1/6) = 1.11r0. Once
reaching this, the atoms will fly apart. This says that the maximum theoretical strength
of any material corresponds to an elastic strain of about 10%, hence modulus / 10.
[5 marks]
b)
Describe the aim and methodology of the Michelson-Morley experiment.
The experiment uses a Michelson Interferometer:
Mirror M1 (movable)
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Page 8
PHY4116 (2013)
R1
Beam-splitter (halfsilvered mirror)
Monochromatic
light beam
Mirror M2 (fixed)
R2
Detector
The two arms are of lengths R1 and R2. If R1 = R2, the two paths interfere constructively
and the detector detects light. This is called a bright fringe. If the path lengths are
different, then if
R1  R2 
n
2
interference is still constructive and a bright fringe is observed. If
R1  R2 
n ½

2
then interference is destructive and no light falls on the detector (a dark fringe). If M1 is
moved, a succession of light and dark fringes is observed. Use as a Speedometer for
the Earth .
[5 marks]
c)
[This question is unseen – they have done problems on a fixed ramp, but not a
moving one] A hollow cylindrical roller (all its weight is at the rim) weighing 10 kg is
initially held stationary on the inclined surface of a ramp, a wedge also weighing 10 kg,
which itself is initially stationary but free to slide without friction on a horizontal surface
(air-bed). The wedge angle is 45°. The roller is released at time t = 0, and rolls without
slipping down the wedge.
i)
State what Conservation of Momentum requires of the horizontal speeds u of the
wedge and vH of the wedge when the roller has descended a height h. mW u = mR
vH so u = vH where u is the speed to the left and v is the speed to the right. [2]
ii) State what Conservation of Energy requires of the speeds u of the wedge and v
of the roller at subsequent times. mRgh = ½ mWu2 + ½ mR vH2 + ½ mR vV2
(translational terms) + ½ I2 (rotational term) so u2 + ½ vV2 + ½ vR2 = gh where
vR is the speed of rolling.
[3]
PHY4116 (2013)
Page 9
iii) Hence or otherwise, find the speed of the wedge when the roller has descended
through the distance of 110 cm. We need vV and vR. vV is tan  the relative
horizontal speed of roller and wedge, which is 2u tan = 2u. [4] But vV is also sin
 the rolling speed vR, so vR2 = 8u2. [4] Putting these into the expression from (ii),
we get 11u2 = gh, or u = Sqrt[g/10] = 1m s–1. [2]
[15 marks]
Question B3
a)
For a body in the gravitational field of a mass M, calculate the escape velocity at
a radius r. Compare it with the orbital velocity for a circular orbit at the same radius. PE
= –GMm/r and KE = ½ mv2. For escape velocity, TE = KE + PE = 0, so vesc2 = 2GM/r
and vesc = Sqrt[2GM/r]. For the orbital velocity, v2/r = GM/r^2, so vorb = Sqrt[GM/r]. Thus
vesc = Sqrt[2] vorb.
[5 marks]
b)
A crystalline solid may be represented as in the diagram (but continued
indefinitely), with atoms at cube corners connected by bonds along the cube edges. We
treat the bonds as Hooke’s Law springs with a spring constant k and an unstrained
length a0. Calculate the bulk modulus B = – V dP / dV for small volume changes.
[8 marks]
If a0 is reduced by an amount da, then the force in each bond is k da. [2] The pressure
is then dP = k da / a02. [2] The volume a03 becomes V – dV = (a0 – da)3 = a0 – 3da, so
V/dV is – a03/(3da). [2] So B = – k da / a02  – a03/(3da) = k / (3 a0). [2]
c)
[This question is unseen – they have seen a Minkowski diagram but not used it
for the analysis below.] In a frame of reference S, an event O occurs at the origin (x =
0, t = 0). Two other events are A (x = 5, t = 13) and B (x = 13, t = 5). Units of x are
lightyears and units of t are years. Using a Minkowski spacetime diagram, or otherwise,
i)
Calculate the three spacetime intervals OA, OB and AB. Using s2 = x2 – t2,
OA = 12, OB = 12, AB = 0
[1]
ii)
Identify a frame S’ in which events O and A occur at the same place. What is the
time between events O and A in this frame? Give the value of  for the relative
speed between S and S’ and state how the time interval of 13 in S is related to
the time interval in S’. [On a Minkowski diagram, if they draw one, this may be
the frame whose t-axis joins O to A. Otherwise . . . ] S’ is a frame travelling at the
speed 5/13 in S (for convenience, we may have O happen at its origin too). The
value of  is then 13/12. The interval in S’ is the Proper Time, 12, and the 13
observed in S is the dilated time.
[3]
iii)
Identify the frame S’’ in which events O and B occur at the same time. What is
the distance between events O and B in this frame? [On a Minkowski diagram, if
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PHY4116 (2013)
they draw one, this is the frame whose x-axis joins O to A. Otherwise . . . ] S’’ is a
frame travelling in the x-direction relative to S fast enough that the event B which
is later than O in S becomes simultaneous with O in S’. The spacetime interval is
Lorentz-invariant, so still 12, and as t’’ = 0 the length x’’ = s = 12.
[1]
iv)
Give reasons why frames S’ and S’’ are the same frame. (E.g.) the x’ and t’ axes
make the same angle to the light-line.
[1]
v)
Explain why the reasoning in (ii) and (iii) is not symmetrical, i.e. why we do not
deduce a dilated length in the same way as we deduce a dilated time. How does
it come about that the length between O and B is contracted in S from the Proper
Length of 12 determined in S’? The distance between events O and B in S is
certainly 13, but the observer in S would not consider this to be the length in S of
an object whose ends were present at O and B. That is because the object if
stationary in S’ would be moving in S [a Minkowski diagram can help here], and if
the measurement at the back (event O) is recorded first and the measurement at
the front (event B) later, a greater (dilated) length would be recorded but would
be known to be incorrect.
[6]
[12 marks]
Question B4
a)
A ladder stands on a floor with a coefficient of friction of  = 0.25 and leans
against a frictionless wall. The angle the ladder makes to the floor is .
i)
Find the minimum value of . Let the top be A and the bottom be B, and let the
ladder be of length L. The reaction at A is the horizontal force FA. Equating
moments about B,
FA L sin   ½ mgL cos 
tan  
mg
2 FA
mg
2 tan 
Resolving vertical forces on the ladder and summing to zero (the ladder isn’t
accelerating)
mg  FB
The maximum horizontal frictional force is FB =0.25 mg and, resolving horizontal
forces and equating to zero, this is equal to FA. So the minimum value of  is
given by tan = 2, i.e.  = 63½
[3]
FA 
ii)
Neglecting the weight of the ladder, what does the minimum value of  become if
a man stands on the top rung (approximate this to the very top) of the ladder?
The first equation becomes
FA L sin   mgL cos 
mg
4
FA
So the limiting value of  is tan–14 = 76.
tan  
[2]
PHY4116 (2013)
Page 11
iii) An assistant of equal weight now stands on the bottom rung (approximate this to
the very bottom), as recommended by Safety Officers. Using your previous
results or otherwise, find the minimum value of  in this situation. The centre of
gravity is now back at the centre of the ladder, so the original calculation and
answer to (i) applies, i.e.  = 63½.
[2]
[7 marks]
b)
The Coriolis force may be expressed as F = 2mv  .
i)
Derive this expression. Only radial velocity matters, so consider vR for an object
moving in a straight line from the centre to the rim of a turntable rotating at 
clockwise. Then  points straight down. During the time t = r / vR, the rim moves
a distance t = r / vR. Using s = ½ a t2, a = 2 r / vR t2 = 2 vR. So the force is
F = 2 vR. Note that if v is away from the centre, the rim has moved to the right,
so the object arrives at a point to the left, so the Coriolis force is to the left, at
right angles to v and to , in accordance with the cross-product v  , so F =
2mv  . [5]
ii)
A naval gun fires a shell at 900 km per hour (ignore the vertical component of the
motion). The target is 10 km away to the north. By how much will it miss the
target if no correction is made for the Coriolis effect, if the gun is at the South
Pole, at the latitude 45° South, and at the Equator? At the South Pole, vR = 900
kph = 250 m s–1. So the Coriolis acceleration is 2  250  7.29  10-5 = 0.0365 m
s–2. The time of flight t is 40 s, so the deflection s = ½ a t2 = 29 m. At 45° South,
the radial component of v is v cos 45° so the deflection is 29 / Sqrt[2] = 20.6 m.
At the Equator, vR = 0 so the deflection is 0.
[4]
[9 marks]
c)
[This question is unseen – they have not encountered a circular pendulum nor
addressed the question implicit in (iv)] A circular pendulum consists of a light
string attached to the ceiling, by which is suspended a 1 kg steel ball. The ball
follows a horizontal circular path, radius 2.5 m with a period of  seconds.
i)
Calculate the speed of the ball and the length of the string. Radius of orbit = r;
speed = v. Period = 2  r / v = . So v = 5 m s–1. Centrifugal acceleration = v2/r
= 10 = g, so string is at 45°. Then its length is Sqrt[2]  r = 3.5 m. [2]
ii)
Give the angular momentum of the ball and its kinetic energy. L = mvr = 12.5
kg m2 s–1, E = ½ mv2 = 12.5 J
[2]
iii)
While the ball continues to rotate, the string is shortened until the radius of the
path of the ball is 1.25 m. What are now the angular momentum, the speed of
the ball, the kinetic energy, and the length of the string? L = mvr = 12.5 kg m2
s–1, so v = 10 m s–1, E = ½ mv2 = 50 J
[2]
iv)
How much work has been done shortening the string against the tension in the
string? Conservation of Energy says that this work has gone into the increased
KE, so it is 50 – 12.5 = 37.5 J.
[3]
[9 marks]
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PHY4116 (2013)
End of Paper –
An appendix of 1 page with formulae and values of physical constants follows
PHY4116 (2013)
Page 13
FORMULA AND DATA SHEET
You may wish to use some of the following formulae.
Vector dot product
u.v = uxvx + uyvy + uzvz
Vector cross product
uv = (uyvz – uzvy, uzvx – uxvz, uxvy – uyvx)
Lennard-Jones potential
 r 12
r6 
U (r )   012  2 06 
r 
r
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