2017 USA Physics Olympiad Exam AAPT UNITED STATES PHYSICS TEAM AIP 2017 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 3 parts:the cover sheet (page 2), Part A (pages 3-13),Part B(pages 15-23).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 15.2017. 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 C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam 1 AAPT AIP 2017 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 3 parts: the cover sheet (page 2), Part A (pages 3-13), Part B (pages 15-23). 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, 2017. • 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 2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP 2017 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 proctor's AAPT ID,your AAPT ID, 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, Doe,Jamie 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 8,2017. 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=1/4re0=8.99×109N.m2/C2 km=40/4π=10-7T·m/A c=3.00×10ǒm/s kB=1.38×10-23J/K NA=6.02×1023(mol)-1 R=NAB=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+n.r for<1 mp 1.673 x 10-27 kg 938 MeV/c2 In(1+)for 1 sin0≈0-a03forl9l<1 cos0≈1-号02forl0<1 Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Cover Sheet 2 AAPT AIP 2017 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 proctor’s AAPT ID, your AAPT ID, 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, Doe, Jamie 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 8, 2017. 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 mp = 1.673 × 10−27 kg = 938 MeV/c 2 ln(1 + x) ≈ x 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 2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 3 Part A Question Al A pair of wedges are located on a horizontal surface.The coefficient of friction (both sliding and static)between the wedges is u,the coefficient of friction between the bottom wedge B and the horizontal surface is u,and the angle of the wedge is 0.The mass of the top wedge A is m,and the mass of the bottom wedge B is M=2m.A horizontal force F directed to the left is applied to the bottom wedge as shown in the figure. A B Determine the range of values for F so that the the top wedge does not slip on the bottom wedge.Express your answer(s)in terms of any or all of m,g,0,and u. Solution Solution 1.Assume the block does not slip.Considering the horizontal forces on the entire system gives F-3mg=3ma→a=3 F -ug. When F is small,the top wedge wants to slide downward,so static friction points up the ramp. Considering the horizontal and vertical forces on the block gives N cos0+f sin0 mg,Nsin0-f cos0 ma. When the minimal force is applied,the friction is maximal,f=uN.Eliminating N gives sin9-μcos0 ma mg- cos0+usin0 and plugging in our first equation gives Fmin =3mg sin0-μcos0\ u+ cos0+μsin0 =3mg (1+μ2)tan0 1+μtan9 When F is large,the top wedge wants to slide upward,so static friction points down the ramp,and N cos0-f sin0 mg, N sin0+f cos0 ma. Now setting f=uN gives Fmax =3mg sin 6+u cos 2μ+(1-r2)tan0 cos0-μsinθ =3mg 1-μtan0 Therefore,naively the range of forces so that the block will not slip is F∈[Fmin,Fmax: Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 3 Part A Question A1 A pair of wedges are located on a horizontal surface. The coefficient of friction (both sliding and static) between the wedges is µ, the coefficient of friction between the bottom wedge B and the horizontal surface is µ, and the angle of the wedge is θ. The mass of the top wedge A is m, and the mass of the bottom wedge B is M = 2m. A horizontal force F directed to the left is applied to the bottom wedge as shown in the figure. B A F θ Determine the range of values for F so that the the top wedge does not slip on the bottom wedge. Express your answer(s) in terms of any or all of m, g, θ, and µ. Solution Solution 1. Assume the block does not slip. Considering the horizontal forces on the entire system gives F − 3µmg = 3ma ⇒ a = F 3m − µg. When F is small, the top wedge wants to slide downward, so static friction points up the ramp. Considering the horizontal and vertical forces on the block gives N cos θ + f sin θ = mg, N sin θ − f cos θ = ma. When the minimal force is applied, the friction is maximal, f = µN. Eliminating N gives ma = mg sin θ − µ cos θ cos θ + µ sin θ and plugging in our first equation gives Fmin = 3mg µ + sin θ − µ cos θ cos θ + µ sin θ = 3mg (1 + µ 2 ) tan θ 1 + µ tan θ . When F is large, the top wedge wants to slide upward, so static friction points down the ramp, and N cos θ − f sin θ = mg, N sin θ + f cos θ = ma. Now setting f = µN gives Fmax = 3mg µ + sin θ + µ cos θ cos θ − µ sin θ = 3mg 2µ + (1 − µ 2 ) tan θ 1 − µ tan θ . Therefore, naively the range of forces so that the block will not slip is F ∈ [Fmin, Fmax]. Copyright c 2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A However,to get full credit,students must account for two edge cases.First,when u>tan,no force is required at all to keep the block in place,so the minimum force is zero.Second,when u>cot0,the block will not slip up under any circumstances,so there is no maximal force. Solution 2.The problem can also be solved geometrically.In general,the no slip condition is u>tano where o is the angle between the vertical and the normal to the plane.Working in the noninertial reference frame of the plane,the fictitious force due to the acceleration is equivalent to a tilting of the gravity vector by an angle tanB= g where,as in solution 1, F a≠3m -μg Then the top block will not slip as long as 0-B<o.At the minimum acceleration amin,B=6-0, and taking the tangent of both sides gives amin tan0-μ 9 1+utan0 This is only meaningful for tan0>u,otherwise the answer is simply amin =0.At the maximum acceleration amax,B=0+o,which gives dmax tanf+μ 9 1-utan This is only meaningful for cot 6>u,otherwise the answer is simply amax=oo. Question A2 Consider two objects with equal heat capacities C and initial temperatures T and T2.A Carnot engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures. Assume that the temperature changes of both the hot and cold reservoirs is very small compared to the temperature during any one cycle of the Carnot engine. a.Find the final temperature Tf of the two objects,and the total work W done by the engine. Solution Since a Carnot engine is reversible,it produces no entropy, dS1 +ds2 d+d2=0. T T2 Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 4 However, to get full credit, students must account for two edge cases. First, when µ > tan θ, no force is required at all to keep the block in place, so the minimum force is zero. Second, when µ > cot θ, the block will not slip up under any circumstances, so there is no maximal force. Solution 2. The problem can also be solved geometrically. In general, the no slip condition is µ > tan φ where φ is the angle between the vertical and the normal to the plane. Working in the noninertial reference frame of the plane, the fictitious force due to the acceleration is equivalent to a tilting of the gravity vector by an angle tan β = a g where, as in solution 1, a = F 3m − µg. Then the top block will not slip as long as |θ−β| ≤ φ. At the minimum acceleration amin, β = θ−φ, and taking the tangent of both sides gives amin g = tan θ − µ 1 + µ tan θ . This is only meaningful for tan θ > µ, otherwise the answer is simply amin = 0. At the maximum acceleration amax, β = θ + φ, which gives amax g = tan θ + µ 1 − µ tan θ . This is only meaningful for cot θ > µ, otherwise the answer is simply amax = ∞. Question A2 Consider two objects with equal heat capacities C and initial temperatures T1 and T2. A Carnot engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures. Assume that the temperature changes of both the hot and cold reservoirs is very small compared to the temperature during any one cycle of the Carnot engine. a. Find the final temperature Tf of the two objects, and the total work W done by the engine. Solution Since a Carnot engine is reversible, it produces no entropy, dS1 + dS2 = dQ1 T1 + dQ2 T2 = 0. Copyright c 2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 5 By the definition of heat capacity,dQi=CdTi,so dT T Integrating this equation shows that TT is constant,so the final temperature is Ti=VTT2 The change in thermal energy of the objects is C(G-I)+C(4-)=C[2V西-T- By the First Law of Thermodynamics,the missing energy has been used to do work,so W=C[口+五-2V西 Now consider three objects with equal and constant heat capacity at initial temperatures T1 100 K,T2=300 K,and T3 300 K.Suppose we wish to raise the temperature of the third object. To do this,we could run a Carnot engine between the first and second objects,extracting work W.This work can then be dissipated as heat to raise the temperature of the third object.Even better,it can be stored and used to run a Carnot engine between the first and third object in reverse,which pumps heat into the third object. Assume that all work produced by running engines can be stored and used without dissipation b.Find the minimum temperature TL to which the first object can be lowered. Solution By the Second Law of Thermodynamics,we must have TL=100K.Otherwise,we would have a process whose sole effect was a net transfer of heat from a cold body to a warm one. c.Find the maximum temperature TH to which the third object can be raised. Solution The entropy of an object with constant heat capacity is dl =CInT. Copyright C2017 American Association of Physics Teachers
2017 USA Physics Olympiad Exam Part A 5 By the definition of heat capacity, dQi = CdTi , so dT1 T1 = − dT2 T2 . Integrating this equation shows that T1T2 is constant, so the final temperature is Tf = p T1T2. The change in thermal energy of the objects is C(Tf − T1) + C(Tf − T2) = C h 2 p T1T2 − T1 − T2 i . By the First Law of Thermodynamics, the missing energy has been used to do work, so W = C h T1 + T2 − 2 p T1T2 i . Now consider three objects with equal and constant heat capacity at initial temperatures T1 = 100 K, T2 = 300 K, and T3 = 300 K. Suppose we wish to raise the temperature of the third object. To do this, we could run a Carnot engine between the first and second objects, extracting work W. This work can then be dissipated as heat to raise the temperature of the third object. Even better, it can be stored and used to run a Carnot engine between the first and third object in reverse, which pumps heat into the third object. Assume that all work produced by running engines can be stored and used without dissipation. b. Find the minimum temperature TL to which the first object can be lowered. Solution By the Second Law of Thermodynamics, we must have TL = 100K. Otherwise, we would have a process whose sole effect was a net transfer of heat from a cold body to a warm one. c. Find the maximum temperature TH to which the third object can be raised. Solution The entropy of an object with constant heat capacity is S = Z dQ T = C Z dT T = C ln T. Copyright c 2017 American Association of Physics Teachers