2013 Semifinal Exam 7 AAPT UNITEDSTATES PHYSICS TEAM AIP CEE 2013 Semifinal 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 3-9),and Part B(pages 11-17).Examinees should be provided parts A and B individually,although they may keep the cover sheet. 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 1.2013. 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. Please provide the examinees with graph paper for Part A.A straight edge or ruler could also be useful. Copyright C2013 American Association of Physics Teachers
2013 Semifinal Exam 1 AAPT UNITED STATES PHYSICS TEAM AIP CEE 2013 Semifinal 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 3-9), and Part B (pages 11-17). Examinees should be provided parts A and B individually, although they may keep the cover sheet. • 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 1, 2013. • 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. • Please provide the examinees with graph paper for Part A. A straight edge or ruler could also be useful. Copyright c 2013 American Association of Physics Teachers
2013 Semifinal Exam Cover Sheet 2 AAPT UNITEDSTATES PHYSICS TEAM AIP CEE 2013 Semifinal 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 1,2013. Possibly Useful Information.You may use this sheet for both parts of the exam. g=9.8 N/kg G=6.67×10-11Nm2/kg2 k=1/4re0=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)m≈1+na for<1 sin0≈0-03forl9l≤1 cos0≈1-02forl0l<1 Copyright C2013 American Association of Physics Teachers
2013 Semifinal Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP CEE 2013 Semifinal 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 1, 2013. 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 2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 2 Part A Question Al The flow of heat through a material can be described via the thermal conductivity K.If the two faces of a slab of material with thermal conductivity K,area A,and thickness d are held at temperatures differing by AT,the thermal power P transferred through the slab is P=RAAT d A heat exchanger is a device which transfers heat between a hot fluid and a cold fluid;they are common in industrial applications such as power plants and heating systems.The heat exchanger shown below consists of two rectangular tubes of length l,width w,and height h.The tubes are separated by a metal wall of thickness d and thermal conductivity K.Originally hot fluid flows through the lower tube at a speed v from right to left,and originally cold fluid flows through the upper tube in the opposite direction(left to right)at the same speed.The heat capacity per unit volume of both fluids is c. The hot fluid enters the heat exchanger at a higher temperature than the cold fluid;the difference between the temperatures of the entering fluids is ATi.When the fluids exit the heat exchanger the difference has been reduced to ATf.(It is possible for the exiting originally cold fluid to have a higher temperature than the exiting originally hot fluid,in which case ATf <0.) h d h Assume that the temperature in each pipe depends only on the lengthwise position,and consider transfer of heat only due to conduction in the metal and due to the bulk movement of fluid.Under the assumptions in this problem,while the temperature of each fluid varies along the length of the exchanger,the temperature difference across the wall is the same everywhere.You need not prove this. Find ATf in terms of the other given parameters. Solution Suppose the temperature difference across the wall is AT.Since the total area of the wall is simply lw,the power transferred across the wall is Copyright C2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 3 Part A Question A1 The flow of heat through a material can be described via the thermal conductivity κ. If the two faces of a slab of material with thermal conductivity κ, area A, and thickness d are held at temperatures differing by ∆T, the thermal power P transferred through the slab is P = κA∆T d A heat exchanger is a device which transfers heat between a hot fluid and a cold fluid; they are common in industrial applications such as power plants and heating systems. The heat exchanger shown below consists of two rectangular tubes of length l, width w, and height h. The tubes are separated by a metal wall of thickness d and thermal conductivity κ. Originally hot fluid flows through the lower tube at a speed v from right to left, and originally cold fluid flows through the upper tube in the opposite direction (left to right) at the same speed. The heat capacity per unit volume of both fluids is c. The hot fluid enters the heat exchanger at a higher temperature than the cold fluid; the difference between the temperatures of the entering fluids is ∆Ti . When the fluids exit the heat exchanger the difference has been reduced to ∆Tf . (It is possible for the exiting originally cold fluid to have a higher temperature than the exiting originally hot fluid, in which case ∆Tf < 0.) l h d h w v v Assume that the temperature in each pipe depends only on the lengthwise position, and consider transfer of heat only due to conduction in the metal and due to the bulk movement of fluid. Under the assumptions in this problem, while the temperature of each fluid varies along the length of the exchanger, the temperature difference across the wall is the same everywhere. You need not prove this. Find ∆Tf in terms of the other given parameters. Solution Suppose the temperature difference across the wall is ∆Tw. Since the total area of the wall is simply lw, the power transferred across the wall is P = κlw d ∆Tw Copyright c 2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 4 In a time dt,the energy transferred is therefore dE =Pdt=kilw d △Tedt Meanwhile,suppose the red fluid enters at temperature T,and the blue fluid at temperature T.The red fluid then exits at temperature T+AT,so the overall temperature change of the red fluid is △T,=T,-(T6+△Tu)=△T-△Tw In a time dt,a volume of red fluid vwh dt flows through the pipe;the heat capacity of this amount of fluid is vwhe dt,so the energy transferred out of it is therefore dE=vwhc dt△T,=vwhc(△T:-△T)dt (Note that the same result is obtained for the heat transferred into the blue fluid;if the flow rates or heat capacities were not the same,this would not hold,exposing the fact that AT is not constant in that case.) Equating. △Tn=vuhd(△T,-△Tw) d △T: △Tu= 1+品 Because the red fluid exits at T+AT,and the blue fluid at Tr-AT △Tf=(T%+△Tw)-(Tr-△Tw)=-△T+2△Tw 2 1+ The performance of the heat exchanger is controlled entirely by the dimensionless parameteras might be intuitive,a long tube and high conductivity are beneficial,whereas a thick wall,high flow rate,and high heat capacity is not.The poorest performance,unsurprisingly,reflects essentially no heat transfer,with the red fluid and blue fluid exiting with the same temperatures they started with.Interestingly,the limit of performance is a complete reversal in the temperatures of the two uids,with△Tf→-△T. Copyright C2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 4 In a time dt, the energy transferred is therefore dE = P dt = κlw d ∆Tw dt Meanwhile, suppose the red fluid enters at temperature Tr and the blue fluid at temperature Tb. The red fluid then exits at temperature Tb + ∆Tw, so the overall temperature change of the red fluid is ∆Tr = Tr − (Tb + ∆Tw) = ∆Ti − ∆Tw In a time dt, a volume of red fluid vwh dt flows through the pipe; the heat capacity of this amount of fluid is vwhc dt, so the energy transferred out of it is therefore dE = vwhc dt ∆Tr = vwhc(∆Ti − ∆Tw) dt (Note that the same result is obtained for the heat transferred into the blue fluid; if the flow rates or heat capacities were not the same, this would not hold, exposing the fact that ∆Tw is not constant in that case.) Equating, κlw d ∆Tw = vwhc(∆Ti − ∆Tw) ∆Tw = ∆Ti 1 + κl dvhc Because the red fluid exits at Tb + ∆Tw, and the blue fluid at Tr − ∆Tw, ∆Tf = (Tb + ∆Tw) − (Tr − ∆Tw) = −∆Ti + 2∆Tw ∆Tf = ∆Ti 2 1 + κl dvhc − 1 ! The performance of the heat exchanger is controlled entirely by the dimensionless parameter κl dvhc ; as might be intuitive, a long tube and high conductivity are beneficial, whereas a thick wall, high flow rate, and high heat capacity is not. The poorest performance, unsurprisingly, reflects essentially no heat transfer, with the red fluid and blue fluid exiting with the same temperatures they started with. Interestingly, the limit of performance is a complete reversal in the temperatures of the two fluids, with ∆Tf → −∆Ti . Copyright c 2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 5 Question A2 A solid round object of radius R can roll down an incline that makes an angle 6 with the horizontal. Assume that the rotational inertia about an axis through the center of mass is given by I=BmR2. The coefficient of kinetic and static friction between the object and the incline is u.The object moves from rest through a vertical distance h. a.If the angle of the incline is sufficiently large,then the object will slip and roll;if the angle of the incline is sufficiently small,then the object with roll without slipping.Determine the angle 0c that separates the two types of motion. b.Derive expressions for the linear acceleration of the object down the ramp in the case of i.Rolling without slipping,and ii.Rolling and slipping. Solution As the object rolls down the incline,there is a torque about the center of mass given by T=Rf where f is the force of friction.The angular acceleration is then T f a=I=BmR or,as will be more useful, f=aBmR The object experience a linear acceleration down the incline given by ma mgsin0-f We have to cases to consider.Either the object rolls without slipping so that f umgcos and a =oR,or the object rolls while slipping so that f=umg cos6 and a >aR. Rolling without slipping Combining the equalities,we get ma mg sin0-Bma or sin a=91+B Rolling while slipping Combining the equalities,we get ma=ng sin6-μng cos6 or a=g(sin8-μcos0) Copyright C2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 5 Question A2 A solid round object of radius R can roll down an incline that makes an angle θ with the horizontal. Assume that the rotational inertia about an axis through the center of mass is given by I = βmR2 . The coefficient of kinetic and static friction between the object and the incline is µ. The object moves from rest through a vertical distance h. a. If the angle of the incline is sufficiently large, then the object will slip and roll; if the angle of the incline is sufficiently small, then the object with roll without slipping. Determine the angle θc that separates the two types of motion. b. Derive expressions for the linear acceleration of the object down the ramp in the case of i. Rolling without slipping, and ii. Rolling and slipping. Solution As the object rolls down the incline, there is a torque about the center of mass given by τ = Rf where f is the force of friction. The angular acceleration is then α = τ I = f βmR or, as will be more useful, f = αβmR The object experience a linear acceleration down the incline given by ma = mg sin θ − f We have to cases to consider. Either the object rolls without slipping so that f ≤ µmg cos θ and a = αR, or the object rolls while slipping so that f = µmg cos θ and a > αR. Rolling without slipping Combining the equalities, we get ma = mg sin θ − βma or a = g sin θ 1 + β Rolling while slipping Combining the equalities, we get ma = mg sin θ − µmg cos θ or a = g (sin θ − µ cos θ) Copyright c 2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 6 The motion is changes at an angle where the static friction is greatest,or when both conditions are equalities: f umg cos0 and a=aR In that case sin0c 1+B =sin0e-4cos0。 or tan Oc Question A3 A beam of muons is maintained in a circular orbit by a uniform magnetic field.Neglect energy loss due to electromagnetic radiation. The mass of the muon is 1.88x 10-28 kg,its charge is-1.602 x 10-19 C,and its half-life is 1.5234s. a.The speed of the muons is much less than the speed of light.It is found that half of the muons decay during each full orbit.What is the magnitude of the magnetic field? b.The experiment is repeated with the same magnetic field,but the speed of the muons is increased;it is no longer much less than the speed of light.Does the fraction of muons which decay during each full orbit increase,decrease,or stay the same? The following facts about special relativity may be useful: The Lorentz factor for a particle moving at speed v is Y=- 1-u2/c2 The Lorentz factor gives the magnitude of time dilation;that is,a clock moving at speed v in a given reference frame runs slow by a factor y in that frame. The momentum of a particle is given by p=Ymi where m does not depend on v. The Lorentz force law in the form 正=g(E+元× d continues to hold. Copyright C2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 6 The motion is changes at an angle where the static friction is greatest, or when both conditions are equalities: f = µmg cos θ and a = αR In that case sin θc 1 + β = sin θc − µ cos θc or tan θc = 1 β + 1 µ Question A3 A beam of muons is maintained in a circular orbit by a uniform magnetic field. Neglect energy loss due to electromagnetic radiation. The mass of the muon is 1.88 × 10−28 kg, its charge is −1.602 × 10−19 C, and its half-life is 1.523 µs. a. The speed of the muons is much less than the speed of light. It is found that half of the muons decay during each full orbit. What is the magnitude of the magnetic field? b. The experiment is repeated with the same magnetic field, but the speed of the muons is increased; it is no longer much less than the speed of light. Does the fraction of muons which decay during each full orbit increase, decrease, or stay the same? The following facts about special relativity may be useful: • The Lorentz factor for a particle moving at speed v is γ = 1 p 1 − v 2/c2 • The Lorentz factor gives the magnitude of time dilation; that is, a clock moving at speed v in a given reference frame runs slow by a factor γ in that frame. • The momentum of a particle is given by ~p = γm~v where m does not depend on v. • The Lorentz force law in the form d~p dt = q(E~ + ~v × B~ ) continues to hold. Copyright c 2013 American Association of Physics Teachers
2013 Semifinal Exam Part A Solution For brevity we simply present the full relativistic solution. The relationships for circular motion - ai 2nr =T are purely a matter of mathematics,and thus continue to hold under special relativity.Meanwhile, since l is constant for circular motion,is constant as well.Thus we can take magnitudes in the Lorentz force law (and set E=0)to find di m =qB Combining these relationships, 2n m=9B If half of the muons decay during each orbit,in the muons'frame of reference each orbit takes one half-life Ti/2.In the lab frame,then, T=YT and so 2T -m =gB T1/2 B= 2rm qT12 Numerically, B=4.85mT The speed of the muons is irrelevant to the fraction which decay per orbit,even in the relativistic case.(The orbits take longer,but the muons live longer,both by the same factor y.) Question A4 A graduated cylinder is partially filled with water;a rubber duck floats at the surface.Oil is poured into the graduated cylinder at a slow,constant rate,and the volume marks corresponding to the surface of the water and the surface of the oil are recorded as a function of time. Copyright C2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 7 Solution For brevity we simply present the full relativistic solution. The relationships for circular motion d~v dt = |~v| 2 r 2πr = |~v| T are purely a matter of mathematics, and thus continue to hold under special relativity. Meanwhile, since |~v| is constant for circular motion, γ is constant as well. Thus we can take magnitudes in the Lorentz force law (and set E~ = 0) to find γm d~v dt = q |~v| B Combining these relationships, 2π T γm = qB If half of the muons decay during each orbit, in the muons’ frame of reference each orbit takes one half-life T1/2 . In the lab frame, then, T = γT1/2 and so 2π γT1/2 γm = qB B = 2πm qT1/2 Numerically, B = 4.85 mT The speed of the muons is irrelevant to the fraction which decay per orbit, even in the relativistic case. (The orbits take longer, but the muons live longer, both by the same factor γ.) Question A4 A graduated cylinder is partially filled with water; a rubber duck floats at the surface. Oil is poured into the graduated cylinder at a slow, constant rate, and the volume marks corresponding to the surface of the water and the surface of the oil are recorded as a function of time. Copyright c 2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 8 500 x Volume mark at water surface 450 o Volume mark at oil surface 400 350 300 0 250 200 150 100 50 0 0 1 2 4 6 78910 Time (min) Water has a density of 1.00 g/mL;the density of air is negligible,as are surface effects.Find the density of the oil. Solution As the oil is poured in,more and more of the weight of the duck is supported by oil,and it rises out of the water,reducing the water level.Eventually this stops,either because the duck is fully submerged in oil or because it is floating entirely above the water.At all times,the weight of the water that is no longer displaced equals the weight of the newly displaced oil: pog△Vo=pug△Vn With this understanding many approaches are possible;we illustrate one.The change in the volume of displaced water is easily read off the graph as the distance between the dotted and dashed lines;it is 143 mL.Finding the volume of displaced oil requires us to take into account the increasing amount of oil in the cylinder.We know there is no oil at t=0,because the oil level and water level coincide,and we know that the rate of change of the oil level for t>6 min is the pour rate,because the water level is not changing.Extrapolating to t=0 we conclude that the volume of oil in the container at any time is given by the height of the shaded region.The volume of displaced oil can then be read as the distance between the solid and dashed lines;it is 186 mL. The density of the oil is △Ye=1.00g/mL186ml 143mL po=Pw△Va Copyright C2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 8 0 1 2 3 4 5 6 7 8 9 10 0 50 100 150 200 250 300 350 400 450 500 Time (min) Volume (mL) Volume mark at water surface Volume mark at oil surface Water has a density of 1.00 g/mL; the density of air is negligible, as are surface effects. Find the density of the oil. Solution As the oil is poured in, more and more of the weight of the duck is supported by oil, and it rises out of the water, reducing the water level. Eventually this stops, either because the duck is fully submerged in oil or because it is floating entirely above the water. At all times, the weight of the water that is no longer displaced equals the weight of the newly displaced oil: ρog∆Vo = ρwg∆Vw With this understanding many approaches are possible; we illustrate one. The change in the volume of displaced water is easily read off the graph as the distance between the dotted and dashed lines; it is 143 mL. Finding the volume of displaced oil requires us to take into account the increasing amount of oil in the cylinder. We know there is no oil at t = 0, because the oil level and water level coincide, and we know that the rate of change of the oil level for t > 6 min is the pour rate, because the water level is not changing. Extrapolating to t = 0 we conclude that the volume of oil in the container at any time is given by the height of the shaded region. The volume of displaced oil can then be read as the distance between the solid and dashed lines; it is 186 mL. The density of the oil is ρo = ρw ∆Vw ∆Vo = (1.00 g/mL)143 mL 186 mL Copyright c 2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 9 Po 0.77 g/mL 500 x Volume mark at water surface 450 o Volume mark at oil surface 400 350 0 目 300 250 200 150 100 50 0 0123 45678910 Time (min) Copyright C2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 9 ρo = 0.77 g/mL 0 1 2 3 4 5 6 7 8 9 10 0 50 100 150 200 250 300 350 400 450 500 Time (min) Volume (mL) Volume mark at water surface Volume mark at oil surface Copyright c 2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 10 STOP:Do Not Continue to Part B If there is still time remaining for Part A,you should review your work for Part A,but do not continue to Part B until instructed by your exam supervisor. Copyright C2013 American Association of Physics Teachers
2013 Semifinal Exam Part A 10 STOP: Do Not Continue to Part B If there is still time remaining for Part A, you should review your work for Part A, but do not continue to Part B until instructed by your exam supervisor. Copyright c 2013 American Association of Physics Teachers