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2017 USA Physics Olympiad Exam 1 AAPT AIP 2017 UNITED STATES PHYSICS TEAM USA Physics Olympiad Exam DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor This examination consists of two parts: Part A has four questions and is allowed 90 minutes; Part B has two questions and is allowed 90 minutes.

2017 USA Physics Olympiad Exam Cover Sheet 2

AAPT UNITED STATES PHYSICS TEAM

USA Physics Olympiad Exam

INSTRUCTIONS

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Work Part A first You have 90 minutes to complete all four problems Each question is

worth 25 points Do not look at Part B during this time

After you have completed Part A you may take a break

Then work Part B You have 90 minutes to complete both problems Each question is worth

50 points Do not look at Part A during this time

Show all your work Partial credit will be given Do not write on the back of any page Do

not write anything that you wish graded on the question sheets

Start each question on a new sheet of paper Put your proctor s AAPT ID your AAPT ID

your name the question number and the page number total pages for this problem in the

upper right hand corner of each page For example

student AAPT ID

proctor AAPT ID

A hand held calculator may be used Its memory must be cleared of data and programs You

may use only the basic functions found on a simple scientific calculator Calculators may not

be shared Cell phones PDA s or cameras may not be used during the exam or while the

exam papers are present You may not use any tables books or collections of formulas

Questions with the same point value are not necessarily of the same difficulty

In order to maintain exam security do not communicate any information about

the questions or their answers solutions on this contest until after April 8 2017

Possibly Useful Information You may use this sheet for both parts of the exam

g 9 8 N kg G 6 67 10 11 N m2 kg2

k 1 4 0 8 99 109 N m2 C2 km 0 4 10 7 T m A

c 3 00 108 m s kB 1 38 10 23 J K

NA 6 02 1023 mol 1 R NA kB 8 31 J mol K

5 67 10 8 J s m2 K4 e 1 602 10 19 C

1 eV 1 602 10 19 J h 6 63 10 34 J s 4 14 10 15 eV s

me 9 109 10 31 kg 0 511 MeV c2 1 x n 1 nx for x 1

mp 1 673 10 27 kg 938 MeV c2 ln 1 x x for x 1

sin 16 3 for 1 cos 1 21 2 for 1

Copyright 2017

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2017 USA Physics Olympiad Exam Part A 3

Question A1

A pair of wedges are located on a horizontal surface The coefficient of friction both sliding and

static between the wedges is the coefficient of friction between the bottom wedge B and the

horizontal surface is and the angle of the wedge is The mass of the top wedge A is m and

the mass of the bottom wedge B is M 2m A horizontal force F directed to the left is applied to

the bottom wedge as shown in the figure

Determine the range of values for F so that the the top wedge does not slip on the bottom

wedge Express your answer s in terms of any or all of m g and

Wedge General Solution with mass ratio

Note that the problem was changed so that 2

Solution 1

For the entire system

F 1 mg 1 ma ma mg

Critical condition 1 When F and subsequently a is small the top wedge tends down and

static friction is UP the ramp

For the top wedge

to ramp N cos f sin mg N

f tan A1 1

f f tan A1 2

k to ramp N sin f cos ma mg A1 4

Rearrange the last equation one arrives at

sin 1 2 1 2 tan

F 1 mg 1 mg Fmin

cos sin 1 tan

Critical condition 2 When F is large the top wedge tends up and static friction is DOWN

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2017 USA Physics Olympiad Exam Part A 4

to ramp N cos f sin mg N

f tan A1 5

k to ramp N sin f cos ma mg A1 7

Again some rearranging late

F 1 mg Fmax

Carefully examining the results one should note that

When there is no relative motion between the wedges and the floor a 0 and F fstatic

1 mg In this case the condition that the top does not slip is

fstatic mg cos mg sin

Therefore the range of F is 0 1 mg

Examining the express of Fmax above one also notes that if cot F can be of any value

and the top wedge still does not slip Hence Fmax if cot

Solution 2

The no slip condition is given by

where is the angle between the plane and the vertical and the coefficient of friction

We can move into a virtual vertical however by recognizing that down will make an effective

angle of with the vertical according to

where a is the horizontal acceleration of the blocks which must be given by

So the top block will not slip so long as

Consider the minimum amin which happens when

Take the tangent of both sides and

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2017 USA Physics Olympiad Exam Part A 5

This is only meaningful for tan otherwise the answer is simply amin 0

Consider the minimum amax which happens when

Take the tangent of both sides and

This is only meaningful for cot otherwise the answer is simply amax

In short we toss out the negative answers for a

Question A2

Consider two objects with equal heat capacities C and initial temperatures T1 and T2 A Carnot

engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures

Assume that the temperature changes of both the hot and cold reservoirs is very small compared

to the temperature during any one cycle of the Carnot engine

a Find the final temperature Tf of the two objects and the total work W done by the engine

Since a Carnot engine is reversible it produces no entropy

By the definition of heat capacity dQi CdTi so

Integrating this equation shows that T1 T2 is constant so the final temperature is

The change in thermal energy of the objects is

C Tf T1 C Tf T2 C 2 T1 T2 T1 T2

By the First Law of Thermodynamics the missing energy has been used to do work so

W C T1 T2 2 T1 T2

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2017 USA Physics Olympiad Exam Part A 6

Now consider three objects with equal and constant heat capacity at initial temperatures

T1 100 K T2 300 K and T3 300 K Suppose we wish to raise the temperature of the

third object

To do this we could run a Carnot engine between the first and second objects extracting work

W This work can then be dissipated as heat to raise the temperature of the third object Even

better it can be stored and used to run a Carnot engine between the first and third object in

reverse which pumps heat into the third object

Assume that all work produced by running engines can be stored and used without dissipation

b Find the minimum temperature TL to which the first object can be lowered

By the Second Law of Thermodynamics we must have TL 100K Indeed if TL 100K then

it would be possible to extract work by restoring the objects to their original temperatures

c Find the maximum temperature TH to which the third object can be raised

The entropy of an object with constant heat capacity is

S C C ln T

Since the total entropy remains constant T1 T2 T3 is constant this is a direct generalization

of the result for Tf found in part a Energy is also conserved as it makes no sense to leave

stored energy unused so T1 T2 T3 is constant

When one object is at temperature TH the other two must be at the same lower temperature

T0 or else further work could be extracted from their temperature difference so

T1 T2 T3 TH 2T0 T1 T2 T3 TH T02

Plugging in temperatures with values divided by 100 for convenience and eliminating T0 gives

TH 7 TH 2 36

We know that TH 1 is one spurious solution since this is the minimum possible final

temperature as found in part b The other roots are TH 4 and TH 9 by the quadratic

formula The solution TH 9 is impossible by energy conservation so

It is also possible to solve the problem more explicitly For example one can run a Carnot

cycle between the first two objects until they are at the same temperature then run a Carnot

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2017 USA Physics Olympiad Exam Part A 7

cycle in reverse between the last two objects using the stored work At this point the first

two objects will no longer be at the same temperature so we can repeat the procedure this

yields an infinite series for TH Taking only the first term of this series yields a slightly worse

result TH 395K

Another explicit method is to continuously switch between running one Carnot engine forward

and another Carnot engine in reverse this yields three differential equations for T1 T2 and

T3 Solving the equations and setting T1 T2 yields T3 TH

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2017 USA Physics Olympiad Exam Part A 8

Question A3

A ship can be thought of as a symmetric arrangement of soft iron In the presence of an external

magnetic field the soft iron will become magnetized creating a second weaker magnetic field We

want to examine the effect of the ship s field on the ship s compass which will be located in the

middle of the ship

Let the strength of the Earth s magnetic field near the ship be Be and the orientation of the

field be horizontal pointing directly toward true north

The Earth s magnetic field Be will magnetize the ship which will then create a second magnetic

field Bs in the vicinity of the ship s compass given by

s Be Kb cos b Ks sin s

where Kb and Ks are positive constants is the angle between the heading of the ship and magnetic

north measured clockwise b and s are unit vectors pointing in the forward direction of the ship

bow and directly right of the forward direction starboard respectively

Because of the ship s magnetic field the ship s compass will no longer necessarily point North

a Derive an expression for the deviation of the compass from north as a function of Kb

Add the fields to get the local field The northward component is

Bnorth Be Be Kb cos cos Be Ks sin sin

while the eastward component is

Beast Be Kb sin cos Be Ks cos sin

The deviation is given by

1 Kb cos2 Ks sin2

In general Kb and Ks are small enough to ignore in the denominator

b Assuming that Kb and Ks are both much smaller than one at what heading s will the

deviation be largest

By inspection 45 will yield the largest deviation

A pair of iron balls placed in the same horizontal plane as the compass but a distance d away

can be used to help correct for the error caused by the induced magnetism of the ship

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2017 USA Physics Olympiad Exam Part A 9

A binnacle protecting the ship s compass in the center with two soft iron spheres to help correct for errors in the

compass heading The use of the spheres was suggested by Lord Kelvin

Just like the ship the iron balls will become magnetic because of the Earth s field Be As

spheres the balls will individually act like dipoles A dipole can be thought of as the field produced

by two magnetic monopoles of strength m at two different points

The magnetic field of a single pole is

where the positive sign is for a north pole and the negative for a south pole The dipole magnetic

field is the sum of the two fields a north pole at y a 2 and a south pole at y a 2 where

the y axis is horizontal and pointing north a is a small distance much smaller than the radius of

the iron balls in general a Ki Be where Ki is a constant that depends on the size of the iron

i from the iron a distance d a from the center

c Derive an expression for the magnetic field B

of the ball Note that there will be a component directed radially away from the ball and a

component directed tangent to a circle of radius d around the ball so using polar coordinates

is recommended

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2017 USA Physics Olympiad Exam Part A 10

This problem is not near as difficult as it looks

Consider first the diagram above with a focus on the black red green triangle The black

side has length a The angle between the green and black sides is so the length of the red

side is a sin and the length of the green side is a cos

The magnetic field strength from one magnetic pole a distance d away is given by

The sum of the two fields has two components The angular component is a measure of the

opening of the triangle formed by the two vectors and since the two vectors basically have

the same length we can use similar triangles to conclude

sin Be 3 sin

As expected this component vanishes for 0

The radial component is given by the difference in the lengths of the two field vectors or

1 1 1 1 1 2x

d d x d 1 x d d d

where x a cos is the green length so

Br Be 2 cos

That wasn t so bad was it

d If placed directly to the right and left of the ship compass the iron balls can be located at a

distance d to cancel out the error in the magnetic heading for any angle s where is largest

Assuming that this is done find the resulting expression for the combined deviation due

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2017 USA Physics Olympiad Exam Part A 11

to the ship and the balls for the magnetic heading for all angles

Note that the two iron balls created a magnetic field near the compass that behaves in a

similar fashion to the ship as a whole There is a component directed toward the bow given

Bb 2B sin cos

and a component directed toward the starboard given by

Bs 2Br cos sin

where the factors of 2 are because there are two iron balls Note that is the ship heading

while is the angle between north and the location of the compass relative to one of the

If it is corrected for the maximum angles it will necessarily cancel out the induced ship field

for all of the angles so that

for all Effectively this means placing the balls to make Kb Ks

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2017 USA Physics Olympiad Exam Part A 12

Question A4

Relativistic particles obey the mass energy relation

E 2 pc 2 mc2 2

where E is the relativistic energy of the particle p is the relativistic momentum m is the mass

and c is the speed of light

A proton with mass mp and energy Ep collides head on with a photon which is massless and

has energy Eb The two combine and form a new particle with mass m called or delta It

is a one dimensional collision that conserves both relativistic energy and relativistic momentum

a Determine Ep in terms of mp m and Eb You may assume that Eb is small

This problem will be solved in units of c 1 in order to make the math easier

First the exact solution which you didn t need to do

The easiest approach is to transform to an inertial frame where the proton particle is at rest

before the collision Then before the collision we have for the proton

and for the photon

and for the

E 2 p 2 m 2

By energy conservation

By momentum conservation we have

Combine the above and

mp E 2 E 2 m 2

But this is really E the energy of the photon in our special frame where the proton is at

rest So we want to boost to a frame where the energy of the photon is Eb We can use the

Lorentz transformation to do this but it is a little neater to realize that

for photons and apply the doppler shift In this case

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2017 USA Physics Olympiad Exam Part A 13

where is the velocity parameter of the proton in the inertial frame where the photon has

If the energy ratio is we have

m 2 mp 2 2 4mp 2 Eb 2

m 2 mp 2 2 4mp 2 Eb 2

The energy of the proton in this frame is given by knowing the Lorentz factor

If you substitute in the right order this isn t so bad

and then the proton energy in the correct rest frame is

2 m 2 mp 2 2mp Eb

After the problem was written it was decide to let Eb be a small quantity mainly because it

is This makes the math easier but before I do it notice that the second term is much larger

than the first so

Note that the following reuses symbols from above but does everything in the

lab frame as such the symbols are not necessarily the same as above

In the lab frame using the approximation that Eb is small conserve momentum

and energy

square both expressions and drop terms with Eb 2

pp 2 2pp pb p2

Ep 2 2Ep Eb E 2

subtract momentum squared expression from energy squared expression

mp 2 2Ep Eb 2pp Eb m 2

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2017 USA Physics Olympiad Exam Part A 14

Since Eb is small this quantity must be large But this means ultrarelativistic protons so

b In this case the photon energy Eb is that of the cosmic background radiation which is an EM

wave with wavelength 1 06 mm Determine the energy of the photons writing your answer

in electron volts

E 0 00112 eV

c Assuming this value for Eb what is the energy of the proton in electron volts that will allow

the above reaction This sets an upper limit on the energy of cosmic rays The mass of the

proton is given by mp c2 938 MeV and the mass of the is given by m c2 1232 MeV

Put the numbers into the adventure above

Ep 1 4 1020 eV

The following relationships may be useful in solving this problem

velocity parameter c

Lorentz factor

relativistic momentum p mc

relativistic energy E mc2

relativistic doppler shift f0 1

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2017 USA Physics Olympiad Exam Part A 15

STOP Do Not Continue to Part B

If there is still time remaining for Part A you should review your work for

Part A but do not continue to Part B until instructed by your exam

supervisor

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