2015 USA Physics Olympiad Exam AAPT UNITED STATES PHYSICS TEAM AIP 2015 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. 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 minutes 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 4 parts:the cover sheet (page 2), Part A (pages 3-7),Part B(pages 9-11),and several answer sheets for one of the questions in Part A(pages 13-13).Examinees should be provided parts A and B individually,although they may keep the cover sheet.The answer sheets should be printed single sided! 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 15,2015. 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. Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam 1 AAPT AIP 2015 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. • 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 minutes 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 4 parts: the cover sheet (page 2), Part A (pages 3-7), Part B (pages 9-11), and several answer sheets for one of the questions in Part A (pages 13-13). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided! • 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 15, 2015. • 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. Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP 2015 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 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 15, 2015. Possibly Useful Information.You may use this sheet for both parts of the exam. g=9.8 N/kg G=6.67×10-11N·m2/kg2 k=1/4r60=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)n≈1+nx for<1 sin0≈0-ae3 for0l<1 cos0≈1-02forl0l<1 Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Cover Sheet 2 AAPT AIP 2015 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 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 15, 2015. 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 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 2 Part A Question Al Consider a particle of mass m that elastically bounces off of an infinitely hard horizontal surface under the influence of gravity.The total mechanical energy of the particle is E and the acceleration of free fall is g.Treat the particle as a point mass and assume the motion is non-relativistic. a.An estimate for the regime where quantum effects become important can be found by simply considering when the deBroglie wavelength of the particle is on the same order as the height of a bounce.Assuming that the deBroglie wavelength is defined by the maximum momentum of the bouncing particle,determine the value of the energy Ea where quantum effects become important.Write your answer in terms of some or all of g,m,and Planck's constant h. Solution Full points will only be awarded if it is clear that the examinee knew the deBroglie wavelength expression. The deBroglie wavelength is p=h/ so if the height H of the bounce is given by E mgH 2m and入=H,then h2 mgH= 2mH2 or H3= h2 2m2g or 瓦=5mg22 An answer of mg2h2 is acceptable,but will not receive full points if it was derived by dimensional analysis alone. b.A second approach allows us to develop an estimate for the actual allowed energy levels of a bouncing particle.Assuming that the particle rises to a height H,we can write -(+ H where p is the momentum as a function of height r above the ground,n is a non-negative integer,and h is Planck's constant. i.Determine the allowed energies En as a function of the integer n,and some or all of g, m,and Planck's constant h. Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 3 Part A Question A1 Consider a particle of mass m that elastically bounces off of an infinitely hard horizontal surface under the influence of gravity. The total mechanical energy of the particle is E and the acceleration of free fall is g. Treat the particle as a point mass and assume the motion is non-relativistic. a. An estimate for the regime where quantum effects become important can be found by simply considering when the deBroglie wavelength of the particle is on the same order as the height of a bounce. Assuming that the deBroglie wavelength is defined by the maximum momentum of the bouncing particle, determine the value of the energy Eq where quantum effects become important. Write your answer in terms of some or all of g, m, and Planck’s constant h. Solution Full points will only be awarded if it is clear that the examinee knew the deBroglie wavelength expression. The deBroglie wavelength is p = h/λ so if the height H of the bounce is given by E = mgH = p 2 2m and λ = H, then mgH = h 2 2mH2 or H3 = h 2 2m2g or Eq = 3 r 1 2 mg2h 2 An answer of p3 mg2h 2 is acceptable, but will not receive full points if it was derived by dimensional analysis alone. b. A second approach allows us to develop an estimate for the actual allowed energy levels of a bouncing particle. Assuming that the particle rises to a height H, we can write 2 Z H 0 p dx = n + 1 2 h where p is the momentum as a function of height x above the ground, n is a non-negative integer, and h is Planck’s constant. i. Determine the allowed energies En as a function of the integer n, and some or all of g, m, and Planck’s constant h. Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 4 ii.Numerically determine the minimum energy of a bouncing neutron.The mass of a neutron is mn=1.675x 10-27 kg=940 MeV/c2;you may express your answer in either Joules or eV. iii.Determine the bounce height of one of these minimum energy neutrons. Solution The integral is not particular difficult to solve, H 2v2m VE-mgx dx; 2V2mE V1-mgx/E da, Jo 2V2mE E V1-u du, mg Jo E 2V2mE vu dv, mg Jo =2v2 E3/22 mg3 So 9mg2h2 1 2/3 En= 32 n+2) Solving for the minimum energy we get E0= 9mg2h2 9(mc2)g2h2 =1.1×10-12eV 128 128c2 The bounce height is given by H= E0=10μm mg This is a very measurable distance! c.Let Eo be the minimum energy of the bouncing neutron and f be the frequency of the bounce. Determine an order of magnitude estimate for the ratio E/f.It only needs to be accurate to within an order of magnitude or so,but you do need to show work! Solution Asf=l/△t and Eo≈△E,Heisenberg's uncertainty relation yeilds △E△t≈h Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 4 ii. Numerically determine the minimum energy of a bouncing neutron. The mass of a neutron is mn = 1.675×10−27 kg = 940 MeV/c 2 ; you may express your answer in either Joules or eV. iii. Determine the bounce height of one of these minimum energy neutrons. Solution The integral is not particular difficult to solve, n + 1 2 h = 2 Z H 0 p dx = 2√ 2m Z H 0 p E − mgx dx, = 2√ 2mE Z H 0 p 1 − mgx/E dx, = 2√ 2mE E mg Z 1 0 √ 1 − u du, = 2√ 2mE E mg Z 1 0 √ v dv, = 2√ 2 E3/2 √ mg 2 3 so En = 3 r 9mg2h 2 32 n + 1 2 2/3 Solving for the minimum energy we get E0 = 3 r 9mg2h 2 128 = 3 r 9(mc2)g 2h 2 128c 2 = 1.1 × 10−12eV The bounce height is given by H = E0 mg = 10 µm This is a very measurable distance! c. Let E0 be the minimum energy of the bouncing neutron and f be the frequency of the bounce. Determine an order of magnitude estimate for the ratio E/f. It only needs to be accurate to within an order of magnitude or so, but you do need to show work! Solution As f = 1/∆t and E0 ≈ ∆E, Heisenberg’s uncertainty relation yeilds ∆E∆t ≈ h Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 5 Question A2 Consider the circuit shown below.Is is a constant current source,meaning that no matter what device is connected between points A and B,the current provided by the constant current source is the same. 4R 2R 2R ● ● 2R3 A B 4R a.Connect an ideal voltmeter between A and B.Determine the voltage reading in terms of any or all of R and Is. Solution An ideal voltmeter has infinite resistance,so no current flows between A and B. Out of symmetry,the same current must flow down each leg,so the current in a leg isIs/2. Assume the potential at the bottom is zero. The potential at A is the same as the junction to the left of A,or,by simple application of Ohm's law VA-52R-LR. The potential at B is found the same way,so R=2I,R. VB =2 The difference is VA-VB=-Is R. The negative is not important for scoring purposes. b.Connect instead an ideal ammeter between A and B.Determine the current in terms of any or all of R and Is. Solution An ideal ammeter has zero resistance.So the problem is simply finding the current through the effective 6R resistor that connects the two vertical branches.This current will flow to the left. Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 5 Question A2 Consider the circuit shown below. Is is a constant current source, meaning that no matter what device is connected between points A and B, the current provided by the constant current source is the same. 2R 4R 4R 2R Is 4R 2R A B a. Connect an ideal voltmeter between A and B. Determine the voltage reading in terms of any or all of R and Is. Solution An ideal voltmeter has infinite resistance, so no current flows between A and B. Out of symmetry, the same current must flow down each leg, so the current in a leg is Is/2. Assume the potential at the bottom is zero. The potential at A is the same as the junction to the left of A, or, by simple application of Ohm’s law VA = Is 2 2R = IsR. The potential at B is found the same way, so VB = Is 2 4R = 2IsR. The difference is VA − VB = −IsR. The negative is not important for scoring purposes. b. Connect instead an ideal ammeter between A and B. Determine the current in terms of any or all of R and Is. Solution An ideal ammeter has zero resistance. So the problem is simply finding the current through the effective 6R resistor that connects the two vertical branches. This current will flow to the left. Copyright c 2015 American Association of Physics Teachers