2014 USA Physics Olympiad Exam AAPT UNITEDSTATES PHYSICS TEAM AIP 2014 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 3-16),Part B(pages 17-23),and several answer sheets for two of the questions in part A(pages 25-28).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 15,2014. 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 C2014 American Association of Physics Teachers
2014 USA Physics Olympiad Exam 1 AAPT UNITED STATES PHYSICS TEAM AIP 2014 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 3-16), Part B (pages 17-23), and several answer sheets for two of the questions in part A (pages 25-28). 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, 2014. • 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 2014 American Association of Physics Teachers
2014 USA Physics Olympiad Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP 2014 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, 2014. 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/2(1+x)n≈1+nz for<1 sin0≈0-ae3 for0l<1 cos0≈1-02forl0l<1 Copyright C2014 American Association of Physics Teachers
2014 USA Physics Olympiad Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP 2014 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, 2014. 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 2014 American Association of Physics Teachers
2014 USA Physics Olympiad Exam Part A 2 Part A Question Al Inspired by:http://www.wired.com/wiredscience/2012/04/a-leaning-motorcycle-on-a-vertical-wall/ A unicyclist of total height h goes around a circular track of radius R while leaning inward at an angle 0 to the vertical.The acceleration due to gravity is g. a.Suppose h<R.What angular velocity w must the unicyclist sustain? Solution Work in the rotating frame,where four forces act on the unicyclist:a normal and frictional force at the point of contact,gravity downwards at the center of mass,and a (fictitious) centrifugal force. IfhR,all parts of the unicyclist are at a distance of approximately R from the center of the circle,so the centripetal acceleration of every part of the unicyclist is w2R.The centrifugal force can then be taken to act at the center of mass for purposes of computing the torque. If the center of mass is a distance l from the point of contact,the torque about the point of contact is T mw2RI cos 0-mgl sin 0 Since the unicyclist is stationary in this frame,r=0: mw2RI cos0-mgl sin0=0 9 W= tan R b.Now model the unicyclist as a uniform rod of length h,where h is less than R but not negligible.This refined model introduces a correction to the previous result.What is the new expression for the angular velocity w?Assume that the rod remains in the plane formed by the vertical and radial directions,and that R is measured from the center of the circle to the point of contact at the ground. Solution The centripetal acceleration now varies meaningfully along the length of the unicyclist.In the rotating frame,the torque about the point of contact is given by Te=w2rz dm where r is the distance from the center of the circle,z is the height above the ground,and dm is a mass element.Because the mass of the unicyclist is uniformly distributed along a length h,the mass element dm can be written as mds for a length element ds,and we have Te= si)(d Copyright C2014 American Association of Physics Teachers
2014 USA Physics Olympiad Exam Part A 3 Part A Question A1 Inspired by: http://www.wired.com/wiredscience/2012/04/a-leaning-motorcycle-on-a-vertical-wall/ A unicyclist of total height h goes around a circular track of radius R while leaning inward at an angle θ to the vertical. The acceleration due to gravity is g. a. Suppose h R. What angular velocity ω must the unicyclist sustain? Solution Work in the rotating frame, where four forces act on the unicyclist: a normal and frictional force at the point of contact, gravity downwards at the center of mass, and a (fictitious) centrifugal force. If h R, all parts of the unicyclist are at a distance of approximately R from the center of the circle, so the centripetal acceleration of every part of the unicyclist is ω 2R. The centrifugal force can then be taken to act at the center of mass for purposes of computing the torque. If the center of mass is a distance l from the point of contact, the torque about the point of contact is τ = mω2Rl cos θ − mglsin θ Since the unicyclist is stationary in this frame, τ = 0: mω2Rl cos θ − mglsin θ = 0 ω = r g R tan θ b. Now model the unicyclist as a uniform rod of length h, where h is less than R but not negligible. This refined model introduces a correction to the previous result. What is the new expression for the angular velocity ω? Assume that the rod remains in the plane formed by the vertical and radial directions, and that R is measured from the center of the circle to the point of contact at the ground. Solution The centripetal acceleration now varies meaningfully along the length of the unicyclist. In the rotating frame, the torque about the point of contact is given by τc = Z ω 2 rz dm where r is the distance from the center of the circle, z is the height above the ground, and dm is a mass element. Because the mass of the unicyclist is uniformly distributed along a length h, the mass element dm can be written as m h ds for a length element ds, and we have τc = Z h 0 ω 2 (R − s sin θ)(s cos θ) m h ds Copyright c 2014 American Association of Physics Teachers
2014 USA Physics Olympiad Exam Part A 4 h Te mw2h cos 0 R 2-3sin Gravity continues to act at the center of mass,a distancefrom the point of contact,and in the opposite direction: Ta=-mg2 sin0 Again,the total torque is zero,so mw?h cos 0 R h h 、23sin0 mg sin0=0 tan 2h 3R sin0 Question A2 A room air conditioner is modeled as a heat engine run in reverse:an amount of heat QL is absorbed from the room at a temperature TL into cooling coils containing a working gas;this gas is compressed adiabatically to a temperature TH;the gas is compressed isothermally in a coil outside the house,giving off an amount of heat QH;the gas expands adiabatically back to a temperature TL;and the cycle repeats.An amount of energy W is input into the system every cycle through an electric pump.This model describes the air conditioner with the best possible efficiency. heating coil pump room valve cooling coil Assume that the outside air temperature is TH and the inside air temperature is TL.The air-conditioner unit consumes electric power P.Assume that the air is sufficiently dry so that no condensation of water occurs in the cooling coils of the air conditioner.Water boils at 373 K and freezes at 273 K at normal atmospheric pressure. a.Derive an expression for the maximum rate at which heat is removed from the room in terms of the air temperatures TH,TL,and the power consumed by the air conditioner P.Your derivation must refer to the entropy changes that occur in a Carnot cycle in order to receive full marks for this part. Copyright C2014 American Association of Physics Teachers
2014 USA Physics Olympiad Exam Part A 4 τc = mω2h cos θ R 2 − h 3 sin θ Gravity continues to act at the center of mass, a distance h 2 from the point of contact, and in the opposite direction: τg = −mg h 2 sin θ Again, the total torque is zero, so mω2h cos θ R 2 − h 3 sin θ − mg h 2 sin θ = 0 ω = s g R tan θ 1 − 2 3 h R sin θ −1 Question A2 A room air conditioner is modeled as a heat engine run in reverse: an amount of heat QL is absorbed from the room at a temperature TL into cooling coils containing a working gas; this gas is compressed adiabatically to a temperature TH; the gas is compressed isothermally in a coil outside the house, giving off an amount of heat QH; the gas expands adiabatically back to a temperature TL; and the cycle repeats. An amount of energy W is input into the system every cycle through an electric pump. This model describes the air conditioner with the best possible efficiency. room cooling coil pump valve heating coil Assume that the outside air temperature is TH and the inside air temperature is TL. The air-conditioner unit consumes electric power P. Assume that the air is sufficiently dry so that no condensation of water occurs in the cooling coils of the air conditioner. Water boils at 373 K and freezes at 273 K at normal atmospheric pressure. a. Derive an expression for the maximum rate at which heat is removed from the room in terms of the air temperatures TH, TL, and the power consumed by the air conditioner P. Your derivation must refer to the entropy changes that occur in a Carnot cycle in order to receive full marks for this part. Copyright c 2014 American Association of Physics Teachers
2014 USA Physics Olympiad Exam Part A 6 Solution From Carnot cycles,and by entropy conservation,we have 8肥-光 Also,by energy conservation, QH=QL+W So heat is removed at a rate QL/t.But T里-W Q=QH-W=QL or W=QL (院 Rearrange and divide by time, =P(） b.The room is insulated,but heat still passes into the room at a rate R=KAT,where AT is the temperature difference between the inside and the outside of the room and k is a constant. Find the coldest possible temperature of the room in terms of TH,k,and P. Solution Equate. kar=P器-pn △T or k(△T)2=PTH-P△T, which is a quadratic that can be solved as △T=-P±VPm+4PkH 2k but only the positive root has physical significance.Writing x=P/k, Ar=5(V+4a- That's the amount the room is colder than the outside,so 五=H-5(V1+4a- Copyright C2014 American Association of Physics Teachers
2014 USA Physics Olympiad Exam Part A 5 Solution From Carnot cycles, and by entropy conservation, we have QH QL = TH TL Also, by energy conservation, QH = QL + W So heat is removed at a rate QL/t. But QL = QH − W = QL TH TL − W or W = QL TH TL − 1 Rearrange and divide by time, QL t = P TL TH − TL b. The room is insulated, but heat still passes into the room at a rate R = k∆T, where ∆T is the temperature difference between the inside and the outside of the room and k is a constant. Find the coldest possible temperature of the room in terms of TH, k, and P. Solution Equate. k∆T = P TL ∆T = P TH − ∆T ∆T or k(∆T) 2 = P TH − P ∆T, which is a quadratic that can be solved as ∆T = −P ± p P2 + 4P kTH 2k , but only the positive root has physical significance. Writing x = P/k, ∆T = x 2 p 1 + 4TH/x − 1 That’s the amount the room is colder than the outside, so TL = TH − x 2 p 1 + 4TH/x − 1 Copyright c 2014 American Association of Physics Teachers