Aapt United States Physics Team Aip 2017-Books Download

AAPT UNITED STATES PHYSICS TEAM AIP 2017

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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
c American Association of Physics Teachers
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
Copyright 2017
c American Association of Physics Teachers
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
Copyright 2017
c American Association of Physics Teachers
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
Copyright 2017
c American Association of Physics Teachers
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
Copyright 2017
c American Association of Physics Teachers
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
Copyright 2017
c American Association of Physics Teachers
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
Copyright 2017
c American Association of Physics Teachers
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
Copyright 2017
c American Association of Physics Teachers
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
Copyright 2017
c American Association of Physics Teachers
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
Copyright 2017
c American Association of Physics Teachers
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
Copyright 2017
c American Association of Physics Teachers
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
Copyright 2017
c American Association of Physics Teachers
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
Copyright 2017
c American Association of Physics Teachers
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
Copyright 2017
c American Association of Physics Teachers


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