2018 USA Physics Olympiad Exam AAPT UNITED STATES PHYSICS TEAM AIP 2018 USA Physics Olympiad Exam DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor This examination consists of two parts:Part A has three questions and is allowed 90 minutes; Part B also has three 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 3 parts:the cover sheet (page 2), Part A(pages 3-10),Part B(pages 12-18),and several answer sheets for one of the questions in Part A(pages 20-22).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 21.2018. 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 C2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam 1 AAPT AIP 2018 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 three questions and is allowed 90 minutes; Part B also has three 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 3 parts: the cover sheet (page 2), Part A (pages 3-10), Part B (pages 12-18), and several answer sheets for one of the questions in Part A (pages 20-22). 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 21, 2018. • 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 2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP 2018 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 three 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 three problems.Each question is worth 25 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 proctor's AAPT ID number,the question number and the page number/total pages for this problem, in the upper right hand corner of each page.For example, Student AAPT ID# Proctor AAPT ID# 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 13, 2018. 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) 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 x 10-31 kg=0.511 MeV/c2 (1+x)m1+nx for <1 sin0≈0-03forl0l<1 cos0≈1-号02forl0l<1 Copyright C2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Cover Sheet 2 AAPT AIP 2018 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 three 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 three problems. Each question is worth 25 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 proctor’s AAPT ID number, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example, Student AAPT ID # Proctor AAPT ID # 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 13, 2018. 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 1 eV = 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 2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 2 Part A Question Al a.Suppose you drop a block of mass m vertically onto a fixed ramp with angle 0 with coefficient of static and kinetic friction u.The block is dropped in such a way that it does not rotate after colliding with the ramp.Throughout this problem,assume the time of the collision is negligible. i.Suppose the block's speed just before it hits the ramp is v and the block slides down the ramp immediately after impact.What is the speed of the block right after the collision? Solution During the collision,the block receives impulses from the normal force,friction,and gravity.Since the collision is very short,the impulse due to gravity is negligible.Let p and pr be the magnitudes of the impulses from the normal force and friction force. Since the block stays on the ramp after the collision,its final momentum is parallel to the ramp.Then the normal force must completely eliminate the block's initial momentum perpendicular to the ramp,so PN =mu cos 0. Since the block still moves after the collision, PF upN. The block's initial momentum parallel to the ramp is mu sin0,so mu sin0-pF mu, where u is the final speed of the block.Solving for u gives u=v(sin0-μcos0) ii.What is the minimum u such that the speed of the block right after the collision is 0? Solution We set u=0 to obtain u=tan0. Note that this is simply the no-slip condition for a block resting on an inclined plane! This is because in both cases,equality is achieved when the normal force and maximal friction force sum to a purely vertical force. b.Now suppose you drop a sphere with mass m,radius R and moment of inertia BmR2 vertically onto the same fixed ramp such that it reaches the ramp with speed v. Copyright C2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 3 Part A Question A1 a. Suppose you drop a block of mass m vertically onto a fixed ramp with angle θ with coefficient of static and kinetic friction µ. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible. i. Suppose the block’s speed just before it hits the ramp is v and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision? Solution During the collision, the block receives impulses from the normal force, friction, and gravity. Since the collision is very short, the impulse due to gravity is negligible. Let pN and pF be the magnitudes of the impulses from the normal force and friction force. Since the block stays on the ramp after the collision, its final momentum is parallel to the ramp. Then the normal force must completely eliminate the block’s initial momentum perpendicular to the ramp, so pN = mv cos θ. Since the block still moves after the collision, pF = µpN . The block’s initial momentum parallel to the ramp is mv sin θ, so mv sin θ − pF = mu, where u is the final speed of the block. Solving for u gives u = v(sin θ − µ cos θ). ii. What is the minimum µ such that the speed of the block right after the collision is 0? Solution We set u = 0 to obtain µ = tan θ. Note that this is simply the no-slip condition for a block resting on an inclined plane! This is because in both cases, equality is achieved when the normal force and maximal friction force sum to a purely vertical force. b. Now suppose you drop a sphere with mass m, radius R and moment of inertia βmR2 vertically onto the same fixed ramp such that it reaches the ramp with speed v. Copyright c 2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 4 i.Suppose the sphere immediately begins to roll without slipping.What is the new speed of the sphere in this case? Solution If the sphere immediately begins to roll without slipping,we can calculate the frictional impulse independently of the normal impulse.We have nvsinθ-pF=mu. The frictional impulse is responsible for the sphere's rotation,so its angular momentum about its center of mass is L=pFR.But we also know that L BmR2w BmRu. Then PF Bmu. Substituting into the previous expression gives vsin0 mv sin0=(1+β)mu→ u=1+B ii.What is the minimum coefficient of friction such that the sphere rolls without slipping immediately after the collision? Solution As in part (a),the normal impulse is pN=mu cos0 and the maximal frictional impulse is pr =upN.From the previous part,we need Bmu sin 0 PF= 1+B and equating these expressions gives Btan0 μ= 1+B Copyright C2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 4 i. Suppose the sphere immediately begins to roll without slipping. What is the new speed of the sphere in this case? Solution If the sphere immediately begins to roll without slipping, we can calculate the frictional impulse independently of the normal impulse. We have mv sin θ − pF = mu. The frictional impulse is responsible for the sphere’s rotation, so its angular momentum about its center of mass is L = pF R. But we also know that L = βmR2ω = βmRu. Then pF = βmu. Substituting into the previous expression gives mv sin θ = (1 + β)mu ⇒ u = v sin θ 1 + β . ii. What is the minimum coefficient of friction such that the sphere rolls without slipping immediately after the collision? Solution As in part (a), the normal impulse is pN = mv cos θ and the maximal frictional impulse is pF = µpN . From the previous part, we need pF = βmv sin θ 1 + β and equating these expressions gives µ = β tan θ 1 + β . Copyright c 2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 5 Question A2 For this problem,graphical answers should be drawn on the answer sheets graphs provided.Supporting work is to be written on blank answer sheets.Incorrect graphs without supporting work will receive no partial credit. The current I as a function of voltage V for a certain electrical device is I=l0e-qVo/kBT (eaV/KBT-1) where g is the magnitude of the charge on an electron,kB is Boltzmann's constant,and T is the absolute temperature.To and Vo are non-zero positive constants.Throughout this problem assume low temperature values kBTVo.Then 0V<, as shown below.Accounting for finite temperature,which is not necessary for full credit, simply rounds the corners in all of the graphs. Copyright C2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 5 Question A2 For this problem, graphical answers should be drawn on the answer sheets graphs provided. Supporting work is to be written on blank answer sheets. Incorrect graphs without supporting work will receive no partial credit. The current I as a function of voltage V for a certain electrical device is I = I0e −qV0/kBT e qV /kBT − 1 where q is the magnitude of the charge on an electron, kB is Boltzmann’s constant, and T is the absolute temperature. I0 and V0 are non-zero positive constants. Throughout this problem assume low temperature values kBT qV0. a. On the answer sheets, sketch a graph of the current versus voltage for low temperature values kBT qV0, clearly indicating any asymptotic behavior. Solution The current is simply proportional to e qV /kBT − 1, which is a shifted exponential. Then I always has the same sign as V , and vanishes when V vanishes. The current grows quickly for high V and approaches a constant for low V . This answer is acceptable, but we can use the condition kBT qV0 to simplify the graph. For negative V , we have I/I0 ≈ e −qV0/kBT which is extremely small. For positive V , we have I/I0 ≈ e q(V −V0)/kBT which is extremely small when V V0. Then I I0 ≈ ( 0 V V0 as shown below. Accounting for finite temperature, which is not necessary for full credit, simply rounds the corners in all of the graphs. Copyright c 2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 6 10 2 (1/1) -2 -4 -6 -8 -10L -10 -8 -6 -4 -2 0 2 8 10 Voltage (V/Vo) Shown is a schematic for the device.Positive voltage means that the electric potential of the left hand side of the device is higher than the right hand side.For this device,Io=25pA and %=1.0V. ViVR Below is a circuit made up of these elements.The voltage supplied the circuit is sinusoidal, VAB =VA-VB Vs sinwt,and is also shown on answer sheets.The resistance is R=5.00 and V=5.0V. VA VD VB Copyright C2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 6 −10 −8 −6 −4 −2 0 2 4 6 8 10 −10 −8 −6 −4 −2 0 2 4 6 8 10 Voltage (V /V0) Current (I/I0) Shown is a schematic for the device. Positive voltage means that the electric potential of the left hand side of the device is higher than the right hand side. For this device, I0 = 25 µA and V0 = 1.0 V. VL VR Below is a circuit made up of these elements. The voltage supplied the circuit is sinusoidal, VAB = VA − VB = Vs sin ωt, and is also shown on answer sheets. The resistance is R = 5.0 Ω and Vs = 5.0 V. VB VA VC VD R Copyright c 2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 7 b.Sketch the potential difference Vcp =Vc-Vp as a function of time on the answer sheet.For your convenience,VAB is shown in light gray.Assume that VAB has been running for a long time. Solution When VAB 2Vo,no current flows.When IVABl>2Vo,current begins to fow,with each diode subtracting a potential difference of Vo.Note that the current flows in the same direction for both positive and negative VAB.This device is a rectifier. 2 (A) 0 -1 -2 -3 -4 -5 -6 010 2030405060708090100110120 Time (ms) A capacitor is connected to the circuit as shown below.The capacitance is C=50mF. VA- VB Copyright C2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 7 b. Sketch the potential difference VCD = VC −VD as a function of time on the answer sheet. For your convenience, VAB is shown in light gray. Assume that VAB has been running for a long time. Solution When |VAB| 2V0, current begins to flow, with each diode subtracting a potential difference of V0. Note that the current flows in the same direction for both positive and negative VAB. This device is a rectifier. 0 10 20 30 40 50 60 70 80 90 100 110 120 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 Time (ms) Voltage (V) A capacitor is connected to the circuit as shown below. The capacitance is C = 50 mF. VB VA VC VD C R Copyright c 2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 8 c.Sketch the new potential difference VCp Vc-Vp as a function of time on the answer sheet. For your convenience,VAB is shown in light gray.Assume that VAB has been running for a long time. Solution Let the voltage on the capacitor be Vs.Whenever VAB>Vs+2Vo,current flows through the diodes,charging the capacitor up to voltage VABI-2Vo.Whenever IVABI<V+2Vo,no current flows through the diodes,and the capacitor and resistor simply discharge as an RC circuit with time constant RC=250 ms. Since RC is much longer than the timescale on the answer sheets,the discharge is approxi- mately linear,with dVs 1.0V dt-83.3 ms Thus the capacitor charges completely to 3.0 V every cycle,then discharges approximately linearly by about a half volt during the reminder of the cycle.On the graph we show charging as red and discharging as blue. 6 4 3 9 0 -1 -2 -3 -4 -5 - 0 10 20 3040 506070 8090 100110 120 Time (ms) Copyright C2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 8 c. Sketch the new potential difference VCD = VC −VD as a function of time on the answer sheet. For your convenience, VAB is shown in light gray. Assume that VAB has been running for a long time. Solution Let the voltage on the capacitor be Vs. Whenever |VAB| ≥ Vs + 2V0, current flows through the diodes, charging the capacitor up to voltage |VAB| − 2V0. Whenever |VAB| < Vs + 2V0, no current flows through the diodes, and the capacitor and resistor simply discharge as an RC circuit with time constant RC = 250 ms. Since RC is much longer than the timescale on the answer sheets, the discharge is approximately linear, with dVs dt = 1.0 V 83.3 ms . Thus the capacitor charges completely to 3.0 V every cycle, then discharges approximately linearly by about a half volt during the reminder of the cycle. On the graph we show charging as red and discharging as blue. 0 10 20 30 40 50 60 70 80 90 100 110 120 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 Time (ms) Voltage (V) Copyright c 2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 9 Question A3 A vacuum system consists of a chamber of volume V connected to a vacuum pump that is a cylinder with a piston that moves left and right.The minimum volume in the pump cylinder is Vo,and the maximum volume in the cylinder is Vo+AV.You should assume that Av<V. chamber motor cylinder outlet inlet piston The cylinder has two valves.The inlet valve opens when the pressure inside the cylinder is lower than the pressure in the chamber,but closes when the piston moves to the right.The outlet valve opens when the pressure inside the cylinder is greater than atmospheric pressure Pa,and closes when the piston moves to the left.A motor drives the piston to move back and forth.The piston moves at such a rate that heat is not conducted in or out of the gas contained in the cylinder during the pumping cycle.One complete cycle takes a time At.You should assume that At is a very small quantity,but AV/At=R is finite.The gas in the chamber is ideal monatomic and remains at a fixed temperature of Ta. Start with assumption that Vo =0 and there are no leaks in the system. a.At t=0 the pressure inside the chamber is Pa.Find an equation for the pressure at a later time t. Solution During each cycle,the system sucks gas out of the chamber and pushes it into the atmosphere. Since Vo =0,the inlet valve opens the moment the piston starts moving to the left.When the piston is all the way to the left,a fraction Av/V+AV)of the gas is in the cylinder.As the piston moves to the right,all of this gas is pushed out,so after a single cycle, (a) and in general, Pd)=卫 (a) While this is technically correct,it can be simplified significantly.Write P0=n(+）=(+ -t/△t where x=AV/V<1.Then using the definition of e, e=lim (1+2)1/ x→0 we have P(t)=Pae-Ri/V. Copyright C2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 9 Question A3 A vacuum system consists of a chamber of volume V connected to a vacuum pump that is a cylinder with a piston that moves left and right. The minimum volume in the pump cylinder is V0, and the maximum volume in the cylinder is V0 + ∆V . You should assume that ∆V V . V chamber cylinder outlet inlet piston motor The cylinder has two valves. The inlet valve opens when the pressure inside the cylinder is lower than the pressure in the chamber, but closes when the piston moves to the right. The outlet valve opens when the pressure inside the cylinder is greater than atmospheric pressure Pa, and closes when the piston moves to the left. A motor drives the piston to move back and forth. The piston moves at such a rate that heat is not conducted in or out of the gas contained in the cylinder during the pumping cycle. One complete cycle takes a time ∆t. You should assume that ∆t is a very small quantity, but ∆V /∆t = R is finite. The gas in the chamber is ideal monatomic and remains at a fixed temperature of Ta. Start with assumption that V0 = 0 and there are no leaks in the system. a. At t = 0 the pressure inside the chamber is Pa. Find an equation for the pressure at a later time t. Solution During each cycle, the system sucks gas out of the chamber and pushes it into the atmosphere. Since V0 = 0, the inlet valve opens the moment the piston starts moving to the left. When the piston is all the way to the left, a fraction ∆V /(V + ∆V ) of the gas is in the cylinder. As the piston moves to the right, all of this gas is pushed out, so after a single cycle, Pf = Pi V V + ∆V and in general, P(t) = Pa V V + ∆V t/∆t . While this is technically correct, it can be simplified significantly. Write P(t) = Pa 1 + ∆V V −t/∆t = Pa (1 + x) 1/x−Rt/V where x = ∆V /V 1. Then using the definition of e, e = lim x→0 (1 + x) 1/x we have P(t) = Pae −Rt/V . Copyright c 2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 10 b.Find an expression for the temperature of the gas as it is emitted from the pump cylinder into the atmosphere.Your answer may depend on time. Solution When the piston is all the way to the left,the pressure is P(t)and the temperature is Ta.As the piston moves to the right,the gas is adiabatically compressed until its pressure reaches Pa and the outlet valve opens.Since PVy is constant during adiabatic compression and PV/T is constant by the ideal gas law, (品) 1-1/ Tout(t)=T -Ta Pa 2/5 P(t) =Tae2Rt/5V where we used y=5/3 for a monatomic ideal gas. For the remainder of this problem 00,the inlet valve will not open immediately when the piston begins moving to the left;instead it will open once the pressure in the cylinder equals the pressure in the chamber. Since the expansion of the cylinder is adiabatic,PVy is constant,so Pmin =Pa Vo Copyright C2018 American Association of Physics Teachers
2018 USA Physics Olympiad Exam Part A 10 b. Find an expression for the temperature of the gas as it is emitted from the pump cylinder into the atmosphere. Your answer may depend on time. Solution When the piston is all the way to the left, the pressure is P(t) and the temperature is Ta. As the piston moves to the right, the gas is adiabatically compressed until its pressure reaches Pa and the outlet valve opens. Since P V γ is constant during adiabatic compression and P V /T is constant by the ideal gas law, Tout(t) = Ta Pa P(t) 1−1/γ = Ta Pa P(t) 2/5 = Tae 2Rt/5V where we used γ = 5/3 for a monatomic ideal gas. For the remainder of this problem 0 0, the inlet valve will not open immediately when the piston begins moving to the left; instead it will open once the pressure in the cylinder equals the pressure in the chamber. Since the expansion of the cylinder is adiabatic, P V γ is constant, so Pmin = Pa V0 V0 + ∆V γ = Pa 1 + ∆V V0 −γ . Copyright c 2018 American Association of Physics Teachers
