2016 USA Physics Olympiad Exam AAPT UNITED STATES PHYSICS TEAM AIP 2016 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 3-12),Part B(pages 14-19),and several answer sheets for one of the questions in Part A(pages 21-22).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 15.2016. 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 C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam 1 AAPT AIP 2016 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 3-12), Part B (pages 14-19), and several answer sheets for one of the questions in Part A (pages 21-22). 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, 2016. • 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 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP 2016 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, 2016. 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=1/4πe0=8.99×109N.m2/C2 km=40/4r=10-7T·m/A c=3.00×103m/s kB=1.38×10-23J/K NA=6.02×1023(mol)-1 R=NAkB =8.31 J/(mol.K) σ=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+na for≤1 sin0≈0-03forl0l<1 cos0≈1-号02forl0l<1 Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Cover Sheet 2 AAPT AIP 2016 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, 2016. 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 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A Part A Question Al The Doppler effect for a source moving relative to a stationary observer is described by fo f=1-(v/c)cos0 where f is the frequency measured by the observer,fo is the frequency emitted by the source,v is the speed of the source,c is the wave speed,and 6 is the angle between the source velocity and the line between the source and observer.(Thus 0=0 when the source is moving directly towards the observer and 6=when moving directly away.) A sound source of constant frequency travels at a constant velocity past an observer,and the observed frequency is plotted as a function of time: 450 000 448 00 00 oObserved Frequency 0 446 444 442 440 到 438 436 434 0 432 430 0 428 0 426 424 0 0 0 422 42 0 2 345 67 891011121314 Time (s) The experiment happens in room temperature air,so the speed of sound is 340 m/s. a.What is the speed of the source? Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 3 Part A Question A1 The Doppler effect for a source moving relative to a stationary observer is described by f = f0 1 − (v/c) cos θ where f is the frequency measured by the observer, f0 is the frequency emitted by the source, v is the speed of the source, c is the wave speed, and θ is the angle between the source velocity and the line between the source and observer. (Thus θ = 0 when the source is moving directly towards the observer and θ = π when moving directly away.) A sound source of constant frequency travels at a constant velocity past an observer, and the observed frequency is plotted as a function of time: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 420 422 424 426 428 430 432 434 436 438 440 442 444 446 448 450 Time (s) Frequency (Hz) Observed Frequency The experiment happens in room temperature air, so the speed of sound is 340 m/s. a. What is the speed of the source? Copyright c 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 4 Solution For 0=0 we have a≈fo/1-v/c) and for0=π， f6=fo/(1+v/c): Read fa and fo off the early and late time portions of the graph and use fa/f6=(1+v/c)/(1-v/c) giving an answer of v=10.7 m/s. Alternatively,we can see that v<c and approximate a/f6≈1+2v/c which makes the calculation of v slightly faster.This is acceptable because the error terms are of order (v/c)2~0.1%. b.What is the smallest distance between the source and the observer? Solution Let d be the (fixed)distance between the observer and the path of the source;let x be the displacement along the path,with x=0 at closest approach.Then for d, cos0≈cot0=x/d so we have f fo/(1-(v/c)(x/d))fo(1+(v/c)(z/d)). Taking the time derivative,and noting that z'is simply v, f'=fo(v2/c)d Therefore we can read f'off the center region of the graph.We still need to find fo,which we can do using our result from part (a)or simply by averaging fa and fo,since v<c,giving fo =435 Hz and an answer of d =17.8 m. There's also a nice trick to speed up this computation.Draw lines at the asymptotic values and through the central data points.The two horizontal lines are 2fo(v/c)apart in frequency, so the time between their intersections with the third line is simply 2d/v. Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 4 Solution For θ = 0 we have fa ≈ f0/(1 − v/c) and for θ = π, fb = f0/(1 + v/c). Read fa and fb off the early and late time portions of the graph and use fa/fb = (1 + v/c)/(1 − v/c) giving an answer of v = 10.7 m/s. Alternatively, we can see that v c and approximate fa/fb ≈ 1 + 2v/c which makes the calculation of v slightly faster. This is acceptable because the error terms are of order (v/c) 2 ∼ 0.1%. b. What is the smallest distance between the source and the observer? Solution Let d be the (fixed) distance between the observer and the path of the source; let x be the displacement along the path, with x = 0 at closest approach. Then for |x| d, cos θ ≈ cot θ = x/d so we have f = f0/(1 − (v/c)(x/d)) ≈ f0(1 + (v/c)(x/d)). Taking the time derivative, and noting that x 0 is simply v, f 0 = f0(v 2 /c)d Therefore we can read f 0 off the center region of the graph. We still need to find f0, which we can do using our result from part (a) or simply by averaging fa and fb, since v c, giving f0 = 435 Hz and an answer of d = 17.8 m. There’s also a nice trick to speed up this computation. Draw lines at the asymptotic values and through the central data points. The two horizontal lines are 2f0(v/c) apart in frequency, so the time between their intersections with the third line is simply 2d/v. Copyright c 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 5 Question A2 A student designs a simple integrated circuit device that has two inputs,Va and V,and two outputs, Vo and V.The inputs are effectively connected internally to a single resistor with effectively infinite resistance.The outputs are effectively connected internally to a perfect source of emf &The integrated circuit is configured so that &=G(Va-V),where G is a very large number somewhere between 107 and 109.The circuits below are chosen so that the precise value of G is unimportant. On the left is an internal schematic for the device;on the right is the symbol that is used in circuit diagrams. 0 Solution The key idea is that if is finite,then VaVo,since G is so large.If we work exactly,then the answers will contain terms like (V-Va)/G which are negligible.Thus we can find the same answers by just setting Va =Vo. a.Consider the following circuit.R1 =8.2 kn and R2 =560 are two resistors.Terminal g and the negative side of Vin are connected to ground,so both are at a potential of 0 volts. Determine the ratio Vout/Vin Vout Solution For this first part,we will not assume Va =V.Since terminal g is grounded,V=0 and Va =Vin,so Vout =G(Vin-V).No current runs between a and b,so any current through Ri also flows through R2.Then Ohm's law gives Vout Vout=G1 R2 → -Vout R1+R2 and solving for Vout gives Vou V But since G>RI/R2,we can neglect the 1/G term,giving Vout≈ 1+B2 a R2 This circuit is an amplifier with feedback. Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 5 Question A2 A student designs a simple integrated circuit device that has two inputs, Va and Vb, and two outputs, Vo and Vg. The inputs are effectively connected internally to a single resistor with effectively infinite resistance. The outputs are effectively connected internally to a perfect source of emf E. The integrated circuit is configured so that E = G(Va − Vb), where G is a very large number somewhere between 107 and 109 . The circuits below are chosen so that the precise value of G is unimportant. On the left is an internal schematic for the device; on the right is the symbol that is used in circuit diagrams. Va Vb Vo Vg a b o g Solution The key idea is that if E is finite, then Va ≈ Vb, since G is so large. If we work exactly, then the answers will contain terms like (Vb − Va)/G which are negligible. Thus we can find the same answers by just setting Va = Vb. a. Consider the following circuit. R1 = 8.2 kΩ and R2 = 560 Ω are two resistors. Terminal g and the negative side of Vin are connected to ground, so both are at a potential of 0 volts. Determine the ratio Vout/Vin. a b o g R1 R2 Vout Vin Solution For this first part, we will not assume Va = Vb. Since terminal g is grounded, Vg = 0 and Va = Vin, so Vout = G(Vin − Vb). No current runs between a and b, so any current through R1 also flows through R2. Then Ohm’s law gives Vb R2 = Vout R1 + R2 ⇒ Vout = G Vin − Vout R2 R1 + R2 and solving for Vout gives Vout = Vin 1 1 G + R2 R1+R2 . But since G R1/R2, we can neglect the 1/G term, giving Vout Vin ≈ R1 + R2 R2 . This circuit is an amplifier with feedback. Copyright c 2016 American Association of Physics Teachers