AAPT UNITED STATES PHYSICS TEAM AIP 2015
AIP 2015 UNITED STATES PHYSICS TEAM ... on this contest until after April 15, 2015. Possibly Useful Information. You may use this sheet for both parts of the exam.
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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 2AAPT UNITED STATES PHYSICS TEAMUSA Physics Olympiad ExamINSTRUCTIONSDO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGINWork Part A first You have 90 minutes to complete all four problems Each question isworth 25 points Do not look at Part B during this timeAfter you have completed Part A you may take a breakThen work Part B You have 90 minutes to complete both problems Each question is worth50 points Do not look at Part A during this timeShow all your work Partial credit will be given Do not write on the back of any page Donot write anything that you wish graded on the question sheetsStart each question on a new sheet of paper Put your proctor s AAPT ID your AAPT IDyour name the question number and the page number total pages for this problem in theupper right hand corner of each page For examplestudent AAPT IDproctor AAPT IDA hand held calculator may be used Its memory must be cleared of data and programs Youmay use only the basic functions found on a simple scientific calculator Calculators may notbe shared Cell phones PDA s or cameras may not be used during the exam or while theexam papers are present You may not use any tables books or collections of formulasQuestions with the same point value are not necessarily of the same difficultyIn order to maintain exam security do not communicate any information aboutthe questions or their answers solutions on this contest until after April 8 2017Possibly Useful Information You may use this sheet for both parts of the examg 9 8 N kg G 6 67 10 11 N m2 kg2k 1 4 0 8 99 109 N m2 C2 km 0 4 10 7 T m Ac 3 00 108 m s kB 1 38 10 23 J KNA 6 02 1023 mol 1 R NA kB 8 31 J mol K5 67 10 8 J s m2 K4 e 1 602 10 19 C1 eV 1 602 10 19 J h 6 63 10 34 J s 4 14 10 15 eV sme 9 109 10 31 kg 0 511 MeV c2 1 x n 1 nx for x 1mp 1 673 10 27 kg 938 MeV c2 ln 1 x x for x 1sin 16 3 for 1 cos 1 21 2 for 1Copyright 2017c American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 3Question A1A pair of wedges are located on a horizontal surface The coefficient of friction both sliding andstatic between the wedges is the coefficient of friction between the bottom wedge B and thehorizontal surface is and the angle of the wedge is The mass of the top wedge A is m andthe mass of the bottom wedge B is M 2m A horizontal force F directed to the left is applied tothe bottom wedge as shown in the figureDetermine the range of values for F so that the the top wedge does not slip on the bottomwedge Express your answer s in terms of any or all of m g andWedge General Solution with mass ratioNote that the problem was changed so that 2Solution 1For the entire systemF 1 mg 1 ma ma mgCritical condition 1 When F and subsequently a is small the top wedge tends down andstatic friction is UP the rampFor the top wedgeto ramp N cos f sin mg Nf tan A1 1f f tan A1 2k to ramp N sin f cos ma mg A1 4Rearrange the last equation one arrives atsin 1 2 1 2 tanF 1 mg 1 mg Fmincos sin 1 tanCritical condition 2 When F is large the top wedge tends up and static friction is DOWNCopyright 2017c American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 4to ramp N cos f sin mg Nf tan A1 5k to ramp N sin f cos ma mg A1 7Again some rearranging lateF 1 mg FmaxCarefully examining the results one should note thatWhen there is no relative motion between the wedges and the floor a 0 and F fstatic1 mg In this case the condition that the top does not slip isfstatic mg cos mg sinTherefore the range of F is 0 1 mgExamining the express of Fmax above one also notes that if cot F can be of any valueand the top wedge still does not slip Hence Fmax if cotSolution 2The no slip condition is given bywhere is the angle between the plane and the vertical and the coefficient of frictionWe can move into a virtual vertical however by recognizing that down will make an effectiveangle of with the vertical according towhere a is the horizontal acceleration of the blocks which must be given bySo the top block will not slip so long asConsider the minimum amin which happens whenTake the tangent of both sides andCopyright 2017c American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 5This is only meaningful for tan otherwise the answer is simply amin 0Consider the minimum amax which happens whenTake the tangent of both sides andThis is only meaningful for cot otherwise the answer is simply amaxIn short we toss out the negative answers for aQuestion A2Consider two objects with equal heat capacities C and initial temperatures T1 and T2 A Carnotengine is run using these objects as its hot and cold reservoirs until they are at equal temperaturesAssume that the temperature changes of both the hot and cold reservoirs is very small comparedto the temperature during any one cycle of the Carnot enginea Find the final temperature Tf of the two objects and the total work W done by the engineSince a Carnot engine is reversible it produces no entropyBy the definition of heat capacity dQi CdTi soIntegrating this equation shows that T1 T2 is constant so the final temperature isThe change in thermal energy of the objects isC Tf T1 C Tf T2 C 2 T1 T2 T1 T2By the First Law of Thermodynamics the missing energy has been used to do work soW C T1 T2 2 T1 T2Copyright 2017c American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 6Now consider three objects with equal and constant heat capacity at initial temperaturesT1 100 K T2 300 K and T3 300 K Suppose we wish to raise the temperature of thethird objectTo do this we could run a Carnot engine between the first and second objects extracting workW This work can then be dissipated as heat to raise the temperature of the third object Evenbetter it can be stored and used to run a Carnot engine between the first and third object inreverse which pumps heat into the third objectAssume that all work produced by running engines can be stored and used without dissipationb Find the minimum temperature TL to which the first object can be loweredBy the Second Law of Thermodynamics we must have TL 100K Indeed if TL 100K thenit would be possible to extract work by restoring the objects to their original temperaturesc Find the maximum temperature TH to which the third object can be raisedThe entropy of an object with constant heat capacity isS C C ln TSince the total entropy remains constant T1 T2 T3 is constant this is a direct generalizationof the result for Tf found in part a Energy is also conserved as it makes no sense to leavestored energy unused so T1 T2 T3 is constantWhen one object is at temperature TH the other two must be at the same lower temperatureT0 or else further work could be extracted from their temperature difference soT1 T2 T3 TH 2T0 T1 T2 T3 TH T02Plugging in temperatures with values divided by 100 for convenience and eliminating T0 givesTH 7 TH 2 36We know that TH 1 is one spurious solution since this is the minimum possible finaltemperature as found in part b The other roots are TH 4 and TH 9 by the quadraticformula The solution TH 9 is impossible by energy conservation soIt is also possible to solve the problem more explicitly For example one can run a Carnotcycle between the first two objects until they are at the same temperature then run a CarnotCopyright 2017c American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 7cycle in reverse between the last two objects using the stored work At this point the firsttwo objects will no longer be at the same temperature so we can repeat the procedure thisyields an infinite series for TH Taking only the first term of this series yields a slightly worseresult TH 395KAnother explicit method is to continuously switch between running one Carnot engine forwardand another Carnot engine in reverse this yields three differential equations for T1 T2 andT3 Solving the equations and setting T1 T2 yields T3 THCopyright 2017c American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 8Question A3A ship can be thought of as a symmetric arrangement of soft iron In the presence of an externalmagnetic field the soft iron will become magnetized creating a second weaker magnetic field Wewant to examine the effect of the ship s field on the ship s compass which will be located in themiddle of the shipLet the strength of the Earth s magnetic field near the ship be Be and the orientation of thefield be horizontal pointing directly toward true northThe Earth s magnetic field Be will magnetize the ship which will then create a second magneticfield Bs in the vicinity of the ship s compass given bys Be Kb cos b Ks sin swhere Kb and Ks are positive constants is the angle between the heading of the ship and magneticnorth measured clockwise b and s are unit vectors pointing in the forward direction of the shipbow and directly right of the forward direction starboard respectivelyBecause of the ship s magnetic field the ship s compass will no longer necessarily point Northa Derive an expression for the deviation of the compass from north as a function of KbAdd the fields to get the local field The northward component isBnorth Be Be Kb cos cos Be Ks sin sinwhile the eastward component isBeast Be Kb sin cos Be Ks cos sinThe deviation is given by1 Kb cos2 Ks sin2In general Kb and Ks are small enough to ignore in the denominatorb Assuming that Kb and Ks are both much smaller than one at what heading s will thedeviation be largestBy inspection 45 will yield the largest deviationA pair of iron balls placed in the same horizontal plane as the compass but a distance d awaycan be used to help correct for the error caused by the induced magnetism of the shipCopyright 2017c American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 9A binnacle protecting the ship s compass in the center with two soft iron spheres to help correct for errors in thecompass heading The use of the spheres was suggested by Lord KelvinJust like the ship the iron balls will become magnetic because of the Earth s field Be Asspheres the balls will individually act like dipoles A dipole can be thought of as the field producedby two magnetic monopoles of strength m at two different pointsThe magnetic field of a single pole iswhere the positive sign is for a north pole and the negative for a south pole The dipole magneticfield is the sum of the two fields a north pole at y a 2 and a south pole at y a 2 wherethe y axis is horizontal and pointing north a is a small distance much smaller than the radius ofthe iron balls in general a Ki Be where Ki is a constant that depends on the size of the ironi from the iron a distance d a from the centerc Derive an expression for the magnetic field Bof the ball Note that there will be a component directed radially away from the ball and acomponent directed tangent to a circle of radius d around the ball so using polar coordinatesis recommendedCopyright 2017c American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 10This problem is not near as difficult as it looksConsider first the diagram above with a focus on the black red green triangle The blackside has length a The angle between the green and black sides is so the length of the redside is a sin and the length of the green side is a cosThe magnetic field strength from one magnetic pole a distance d away is given byThe sum of the two fields has two components The angular component is a measure of theopening of the triangle formed by the two vectors and since the two vectors basically havethe same length we can use similar triangles to concludesin Be 3 sinAs expected this component vanishes for 0The radial component is given by the difference in the lengths of the two field vectors or1 1 1 1 1 2xd d x d 1 x d d dwhere x a cos is the green length soBr Be 2 cosThat wasn t so bad was itd If placed directly to the right and left of the ship compass the iron balls can be located at adistance d to cancel out the error in the magnetic heading for any angle s where is largestAssuming that this is done find the resulting expression for the combined deviation dueCopyright 2017c American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 11to the ship and the balls for the magnetic heading for all anglesNote that the two iron balls created a magnetic field near the compass that behaves in asimilar fashion to the ship as a whole There is a component directed toward the bow givenBb 2B sin cosand a component directed toward the starboard given byBs 2Br cos sinwhere the factors of 2 are because there are two iron balls Note that is the ship headingwhile is the angle between north and the location of the compass relative to one of theIf it is corrected for the maximum angles it will necessarily cancel out the induced ship fieldfor all of the angles so thatfor all Effectively this means placing the balls to make Kb KsCopyright 2017c American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 12Question A4Relativistic particles obey the mass energy relationE 2 pc 2 mc2 2where E is the relativistic energy of the particle p is the relativistic momentum m is the massand c is the speed of lightA proton with mass mp and energy Ep collides head on with a photon which is massless andhas energy Eb The two combine and form a new particle with mass m called or delta Itis a one dimensional collision that conserves both relativistic energy and relativistic momentuma Determine Ep in terms of mp m and Eb You may assume that Eb is smallThis problem will be solved in units of c 1 in order to make the math easierFirst the exact solution which you didn t need to doThe easiest approach is to transform to an inertial frame where the proton particle is at restbefore the collision Then before the collision we have for the protonand for the photonand for theE 2 p 2 m 2By energy conservationBy momentum conservation we haveCombine the above andmp E 2 E 2 m 2But this is really E the energy of the photon in our special frame where the proton is atrest So we want to boost to a frame where the energy of the photon is Eb We can use theLorentz transformation to do this but it is a little neater to realize thatfor photons and apply the doppler shift In this caseCopyright 2017c American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 13where is the velocity parameter of the proton in the inertial frame where the photon hasIf the energy ratio is we havem 2 mp 2 2 4mp 2 Eb 2m 2 mp 2 2 4mp 2 Eb 2The energy of the proton in this frame is given by knowing the Lorentz factorIf you substitute in the right order this isn t so badand then the proton energy in the correct rest frame is2 m 2 mp 2 2mp EbAfter the problem was written it was decide to let Eb be a small quantity mainly because itis This makes the math easier but before I do it notice that the second term is much largerthan the first soNote that the following reuses symbols from above but does everything in thelab frame as such the symbols are not necessarily the same as aboveIn the lab frame using the approximation that Eb is small conserve momentumand energysquare both expressions and drop terms with Eb 2pp 2 2pp pb p2Ep 2 2Ep Eb E 2subtract momentum squared expression from energy squared expressionmp 2 2Ep Eb 2pp Eb m 2Copyright 2017c American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 14Since Eb is small this quantity must be large But this means ultrarelativistic protons sob In this case the photon energy Eb is that of the cosmic background radiation which is an EMwave with wavelength 1 06 mm Determine the energy of the photons writing your answerin electron voltsE 0 00112 eVc Assuming this value for Eb what is the energy of the proton in electron volts that will allowthe above reaction This sets an upper limit on the energy of cosmic rays The mass of theproton is given by mp c2 938 MeV and the mass of the is given by m c2 1232 MeVPut the numbers into the adventure aboveEp 1 4 1020 eVThe following relationships may be useful in solving this problemvelocity parameter cLorentz factorrelativistic momentum p mcrelativistic energy E mc2relativistic doppler shift f0 1Copyright 2017c American Association of Physics Teachers2017 USA Physics Olympiad Exam Part A 15STOP Do Not Continue to Part BIf there is still time remaining for Part A you should review your work forPart A but do not continue to Part B until instructed by your examsupervisorCopyright 2017c American Association of Physics Teachers
AIP 2015 UNITED STATES PHYSICS TEAM ... on this contest until after April 15, 2015. Possibly Useful Information. You may use this sheet for both parts of the exam.
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