2012 Semifinal Exam 1 AAPT UNITEDSTATES PHYSICS TEAM AIP 2012 Semifinal 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. The first page that follows is a cover sheet.Examinees may keep the cover sheet for both parts of the exam. The parts are then identified by the center header on each page.Examinees are only allowed to do one part at a time,and may not work on other parts,even if they have time remaining Allow 90 minutes to complete Part A.Do not let students look at Part B.Collect the answers to Part A before allowing the examinee to begin Part B.Examinees are allowed a 10 to 15 minute break between parts A and B. Allow 90 minutes to complete Part B.Do not let students go back to Part A. Ideally the test supervisor will divide the question paper into 3 parts:the cover sheet (page 2),Part A (pages 3-4),and Part B(pages 6-7).Examinees should be provided parts A and B individually,although they may keep the cover sheet. The supervisor must collect all examination questions,including the cover sheet,at the end of the exam,as well as any scratch paper used by the examinees.Examinees may not take the exam questions.The examination questions may be returned to the students after April 1.2012. Examinees are allowed calculators,but they may not use symbolic math,programming,or graphic features of these calculators.Calculators may not be shared and their memory must be cleared of data and programs.Cell phones,PDA's or cameras may not be used during the exam or while the exam papers are present.Examinees may not use any tables,books, or collections of formulas. Please provide the examinees with graph paper for Part A. Copyright C2012 American Association of Physics Teachers
2012 Semifinal Exam 1 AAPT UNITED STATES PHYSICS TEAM AIP 2012 Semifinal 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. • The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam. • The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining. • Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minute break between parts A and B. • Allow 90 minutes to complete Part B. Do not let students go back to Part A. • Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-4), and Part B (pages 6-7). Examinees should be provided parts A and B individually, although they may keep the cover sheet. • The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after April 1, 2012. • Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas. • Please provide the examinees with graph paper for Part A. Copyright c 2012 American Association of Physics Teachers
2012 Semifinal Exam Cover Sheet 2 AAPT UNITEDSTATES PHYSICS TEAM AIP 2012 Semifinal 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 AAPT ID number,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, AAPT ID# Doe,Jamie A1-1/3 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 1,2012. Possibly Useful Information.You may use this sheet for both parts of the exam. g=9.8 N/kg G=6.67×10-11Nm2/kg2 k=1/4re0=8.99×109N.m2/c2 km=4o/4r=10-7T.m/A c=3.00×108m/s k3=1.38×10-23J/K NA=6.02×1023(mol)-1 R=NA=8.31J/mol.K)） o=5.67×10-8J/(s·m2.K4) e=1.602×10-19C 1eV=1.602×10-19J h=6.63×10-34J.s=4.14×10-15eV.s me=9.109×10-31kg=0.511MeV/c2(1+x)m≈1+nx for<1 sin0≈0-03forl9l≤1 cos0≈1-02forl0l<1 Copyright C2012 American Association of Physics Teachers
2012 Semifinal Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP 2012 Semifinal 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 AAPT ID number, 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, AAPT ID # Doe, Jamie A1 - 1/3 • 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 1, 2012. 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 = 1/4π0 = 8.99 × 109 N · m2/C 2 km = µ0/4π = 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 = NAkB = 8.31 J/(mol · K) σ = 5.67 × 10−8 J/(s · m2 · K4 ) e = 1.602 × 10−19 C 1eV = 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/c 2 (1 + x) n ≈ 1 + nx for |x| 1 sin θ ≈ θ − 1 6 θ 3 for |θ| 1 cos θ ≈ 1 − 1 2 θ 2 for |θ| 1 Copyright c 2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 3 Part A Question Al A newly discovered subatomic particle,the S meson,has a mass M.When at rest,it lives for exactly T =3 x 10-8 seconds before decaying into two identical particles called P mesons(peons?) that each have a mass of aM. a.In a reference frame where the S meson is at rest,determine i.the kinetic energy, ii.the momentum,and iii.the velocity of each P meson particle in terms of M,a,the speed of light c,and any numerical constants. b.In a reference frame where the S meson travels 9 meters between creation and decay,determine i.the velocity and ii.kinetic energy of the S meson. Write the answers in terms of M,the speed of light c,and any numerical constants. Solution a.Apply conservation of four momentum.For the S meson,we have psc=(Es;0) and for the two P mesons we have ppc=(Ep,±p), where p is the magnitude of the (relativistic)three momentum of the P mesons. This yields Es =2EP We must also satisfy the relation E2 p2c2+m2c4 for each particle,so Es2=M2c4 and Ep2 p2c2+a2M2c4. Therefore,the kinetic energy of each P meson is Kp-Ep-oMe-iMle-aNe- 3-awe Copyright C2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 3 Part A Question A1 A newly discovered subatomic particle, the S meson, has a mass M. When at rest, it lives for exactly τ = 3 × 10−8 seconds before decaying into two identical particles called P mesons (peons?) that each have a mass of αM. a. In a reference frame where the S meson is at rest, determine i. the kinetic energy, ii. the momentum, and iii. the velocity of each P meson particle in terms of M, α, the speed of light c, and any numerical constants. b. In a reference frame where the S meson travels 9 meters between creation and decay, determine i. the velocity and ii. kinetic energy of the S meson. Write the answers in terms of M, the speed of light c, and any numerical constants. Solution a. Apply conservation of four momentum. For the S meson, we have pSc = (Es, 0) and for the two P mesons we have pP c = (Ep, ±p), where p is the magnitude of the (relativistic) three momentum of the P mesons. This yields ES = 2EP We must also satisfy the relation E 2 = p 2 c 2 + m2 c 4 for each particle, so ES 2 = M2 c 4 and EP 2 = p 2 c 2 + α 2M2 c 4 . Therefore, the kinetic energy of each P meson is KP = EP − αM c2 = 1 2 M c2 − αM c2 = 1 2 − α M c2 . Copyright c 2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 4 Square the energy conservation expression,and combine with the momentum/energy/mass relations: 2m2e4=p22+a2m2e4, a2Mc2 =pc. The velocity of each P meson will be found from the relativistic three momentum, p=myu and the relativistic energy, E=me2 so mcyu E Yme2 =. Putting in the values for the P meson, ev1-4a2 b.From relativistic kinematics, d=ut =vYT, So v d Call this k for now.Then k =By, k2=822, k2= 32 1-321 k21-32)=32, k2=(1+2)32, k2 1+k2 =32 Combine, d2 v=cV22+2 Copyright C2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 4 Square the energy conservation expression, and combine with the momentum/energy/mass relations: 1 4 M2 c 4 = p 2 c 2 + α 2M2 c 4 , 1 4 − α 2 M2 c 4 = p 2 c 2 r 1 4 − α2M c2 = pc. The velocity of each P meson will be found from the relativistic three momentum, p = mγv and the relativistic energy, E = γmc2 so pc E = mcγv γmc2 = β. Putting in the values for the P meson, v = c q 1 4 − α2M c2 1 2M c2 = c p 1 − 4α2 b. From relativistic kinematics, d = vt = vγτ, so v c γ = d cτ Call this k for now. Then k = βγ, k 2 = β 2 γ 2 , k 2 = β 2 1 − β 2 , k 2 (1 − β 2 ) = β 2 , k 2 = (1 + k 2 )β 2 , k 2 1 + k 2 = β 2 . Combine, v = c r d 2 c 2τ 2 + d 2 Copyright c 2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 5 SO 92 v=Vg+9=反 Then Y=1/V1-1/2=2 It isn't much work to find the kinetic energy, K=y-1)Mc2=(W2-1)Mc2. c.This is a velocity addition problem,so US +US w=1+s212 or,using the numbers from the first part of the problem, V1-4a2 U= 1-2a2 Copyright C2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 5 so v = c r 9 2 9 2 + 92 = c √ 2 Then γ = 1/ p 1 − 1/2 = √ 2 It isn’t much work to find the kinetic energy, K = (γ − 1)M c2 = (√ 2 − 1)M c2 . c. This is a velocity addition problem, so v = vS + vS 1 + vS 2/c2 or, using the numbers from the first part of the problem, v = c √ 1 − 4α2 1 − 2α2 Copyright c 2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 6 Question A2 An ideal(but not necessarily perfect monatomic)gas undergoes the following cycle. The gas starts at pressure Po,volume Vo and temperature To The gas is heated at constant volume to a pressure aPo,where a>1. The gas is then allowed to expand adiabatically (no heat is transferred to or from the gas)to pressure Po The gas is cooled at constant pressure back to the original state. The adiabatic constant y is defined in terms of the specific heat at constant pressure Cp and the specific heat at constant volume Co by the ratio y=Cp/Co. a.Determine the efficiency of this cycle in terms of a and the adiabatic constant y.As a reminder,efficiency is defined as the ratio of work out divided by heat in. b.A lab worker makes measurements of the temperature and pressure of the gas during the adiabatic process.The results,in terms of To and Po are Pressure units of Po1.211.411.591.732.14 Temperature units of T62.112.212.282.342.49 Plot an appropriate graph from this data that can be used to determine the adiabatic constant. c.What is y for this gas? Solution a.Label the end points as 0,1,and 2.A quick application of PV =nRT requires that Ti=aTo It takes more work to do the process 1-2;it is acceptable to simply state the adiabatic law of PVY=constant;if you don't know this,you will need to derive it. In the case that you know the adiabatic process law, PiVi=PV=aPiV, so that =(a)号 Another quick application of PV =nRT requires that T2=(a)To. Heat enters the gas during isochoric process 0-1,so Qin nCuAT=nCu(a-1)To Heat exits the system during process2一→0，so Qout nCpAT nCp(a/-1)To Copyright C2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 6 Question A2 An ideal (but not necessarily perfect monatomic) gas undergoes the following cycle. • The gas starts at pressure P0, volume V0 and temperature T0. • The gas is heated at constant volume to a pressure αP0, where α > 1. • The gas is then allowed to expand adiabatically (no heat is transferred to or from the gas) to pressure P0 • The gas is cooled at constant pressure back to the original state. The adiabatic constant γ is defined in terms of the specific heat at constant pressure Cp and the specific heat at constant volume Cv by the ratio γ = Cp/Cv. a. Determine the efficiency of this cycle in terms of α and the adiabatic constant γ. As a reminder, efficiency is defined as the ratio of work out divided by heat in. b. A lab worker makes measurements of the temperature and pressure of the gas during the adiabatic process. The results, in terms of T0 and P0 are Pressure units of P0 1.21 1.41 1.59 1.73 2.14 Temperature units of T0 2.11 2.21 2.28 2.34 2.49 Plot an appropriate graph from this data that can be used to determine the adiabatic constant. c. What is γ for this gas? Solution a. Label the end points as 0, 1, and 2. A quick application of P V = nRT requires that T1 = αT0. It takes more work to do the process 1 → 2; it is acceptable to simply state the adiabatic law of P V γ = constant; if you don’t know this, you will need to derive it. In the case that you know the adiabatic process law, P1V γ 1 = P2V γ 2 = αP1V γ 2 , so that V2 = V1 (α) 1 γ . Another quick application of P V = nRT requires that T2 = (α) 1 γ T0. Heat enters the gas during isochoric process 0 → 1, so Qin = nCv∆T = nCv(α − 1)T0 Heat exits the system during process 2 → 0, so Qout = nCp∆T = nCp(α 1/γ − 1)T0 Copyright c 2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 7 We only consider absolute values,and insert negative signs later as needed. The work done is the difference,so W Qin-Qout nCu(a-1)To -nCp(a1/y-1)To and the efficiency is then e= C.(a-1)-Cn(a/n-1) C(a-1) This can be greatly simplified to a2/h-1 e=1-ya-1 b.Along an adiabatic path,the relationship between pressure and temperature is given by PVY constant o P so PT=constant As such, Px T Note that,for an ideal gas, Cp/Cu Cp y-1=cC-1=f This means that we want to plot a log-log plot with log T horizontal and log P vertical.The slope of the graph will be Cp/R. For the data given,Cp=(7/2)R,so y=7/5. Copyright C2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 7 We only consider absolute values, and insert negative signs later as needed. The work done is the difference, so W = Qin − Qout = nCv(α − 1)T0 − nCp(α 1/γ − 1)T0 and the efficiency is then e = Cv(α − 1) − Cp(α 1/γ − 1) Cv(α − 1) This can be greatly simplified to e = 1 − γ α 1/γ − 1 α − 1 b. Along an adiabatic path, the relationship between pressure and temperature is given by P V γ = constant ∝ P T P γ so P T γ 1−γ = constant As such, P ∝ T γ γ−1 Note that, for an ideal gas, γ γ − 1 = Cp/Cv Cp/Cv − 1 = Cp R This means that we want to plot a log-log plot with log T horizontal and log P vertical. The slope of the graph will be Cp/R. For the data given, Cp = (7/2)R, so γ = 7/5. Copyright c 2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 8 Question A3 This problem inspired by the 2008 Guangdong Province Physics Olympiad Two infinitely long concentric hollow cylinders have radii a and 4a.Both cylinders are insulators; the inner cylinder has a uniformly distributed charge per length of +A;the outer cylinder has a uniformly distributed charge per length of-A. An infinitely long dielectric cylinder with permittivity e=Keo,where k is the dielectric constant, has a inner radius 2a and outer radius 3a is also concentric with the insulating cylinders.The dielectric cylinder is rotating about its axis with an angular velocity wc/a,where c is the speed of light.Assume that the permeability of the dielectric cylinder and the space between the cylinders is that of free space,Ho. a.Determine the electric field for all regions. b.Determine the magnetic field for all regions. Solution a.Consider a Gaussian cylinder of radius r and length I centered on the cylinder axis.Gauss's Law states that E dA=Qenc 2πrEL= 入encl E0 入encd E二2rr0 where Aenc is the enclosed linear charge density. The field due to the hollow cylinders alone is therefore 0 r4a The field within the dielectric is reduced by a factor K,so that in total 0 r4a Copyright C2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 8 Question A3 This problem inspired by the 2008 Guangdong Province Physics Olympiad Two infinitely long concentric hollow cylinders have radii a and 4a. Both cylinders are insulators; the inner cylinder has a uniformly distributed charge per length of +λ; the outer cylinder has a uniformly distributed charge per length of −λ. An infinitely long dielectric cylinder with permittivity = κ0, where κ is the dielectric constant, has a inner radius 2a and outer radius 3a is also concentric with the insulating cylinders. The dielectric cylinder is rotating about its axis with an angular velocity ω c/a, where c is the speed of light. Assume that the permeability of the dielectric cylinder and the space between the cylinders is that of free space, µ0. a. Determine the electric field for all regions. b. Determine the magnetic field for all regions. Solution a. Consider a Gaussian cylinder of radius r and length l centered on the cylinder axis. Gauss’s Law states that Z E dA = qencl 0 2πrEl = λencll 0 E = λencl 2πr0 where λencl is the enclosed linear charge density. The field due to the hollow cylinders alone is therefore Eapplied = 0 r 4a The field within the dielectric is reduced by a factor κ, so that in total E = 0 r 4a Copyright c 2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 9 b.We can apply the results of the previous section to obtain the enclosed charge density Aenc as a function of radius: (0 r4a Defining =0-) we conclude that a charge density-A;exists on the inner surface of the dielectric,a charge density Ai on the outer surface,and no charge on the interior. As with the case of a very long solenoid,we expect the magnetic field to be entirely parallel to the cylinder axis,and to go to zero for large r.Consider an Amperian loop of length l extending along a radius,the inner side of which is at radius r and the outer side of which is at a very large radius.We have on this loop B dl=Holenct Letting B now be the magnetic field at radius r, IB Holencl B=Holend For r>3a,Iend=0,since the charge on the hollow cylinders is not moving.For 2a3a or,using our expression forλ， 0 r3a Copyright C2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 9 b. We can apply the results of the previous section to obtain the enclosed charge density λencl as a function of radius: λencl = 0 r 4a Defining λi = 1 − 1 κ λ we conclude that a charge density −λi exists on the inner surface of the dielectric, a charge density λi on the outer surface, and no charge on the interior. As with the case of a very long solenoid, we expect the magnetic field to be entirely parallel to the cylinder axis, and to go to zero for large r. Consider an Amperian loop of length l extending along a radius, the inner side of which is at radius r and the outer side of which is at a very large radius. We have on this loop I B dl = µ0Iencl Letting B now be the magnetic field at radius r, lB = µ0Iencl B = µ0Iencl l For r > 3a, Iencl = 0, since the charge on the hollow cylinders is not moving. For 2a 3a or, using our expression for λi , B = 0 r 3a Copyright c 2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 10 Question A4 Two masses m separated by a distance l are given initial velocities vo as shown in the diagram. The masses interact only through universal gravitation. a.Under what conditions will the masses eventually collide? b.Under what conditions will the masses follow circular orbits of diameter 1? c.Under what conditions will the masses follow closed orbits? d.What is the minimum distance achieved between the masses along their path? Solution a.In order for the masses to collide,the total angular momentum of the system must be zero, which only occurs if vo =0. b.In this case,the masses undergo uniform circular motion with radius and speed vo,so that Gm2 mvo2 12 Gm 072 c.The masses follow closed orbits if they do not have enough energy to escape,i.e.if the total energy of the system is negative.The total energy of the system is 1 02、Gm2 2·2m so that the condition required is Gm2 1 Copyright C2012 American Association of Physics Teachers
2012 Semifinal Exam Part A 10 Question A4 Two masses m separated by a distance l are given initial velocities v0 as shown in the diagram. The masses interact only through universal gravitation. l v0 v0 a. Under what conditions will the masses eventually collide? b. Under what conditions will the masses follow circular orbits of diameter l? c. Under what conditions will the masses follow closed orbits? d. What is the minimum distance achieved between the masses along their path? Solution a. In order for the masses to collide, the total angular momentum of the system must be zero, which only occurs if v0 = 0. b. In this case, the masses undergo uniform circular motion with radius l 2 and speed v0, so that Gm2 l 2 = mv0 2 l 2 Gm v0 2l = 2 c. The masses follow closed orbits if they do not have enough energy to escape, i.e. if the total energy of the system is negative. The total energy of the system is 2 · 1 2 mv0 2 − Gm2 l so that the condition required is mv0 2 − Gm2 l 1 Copyright c 2012 American Association of Physics Teachers