2010 Semifinal Exam AAPT UNITEDSTATES PHYSICS TEAM AIP 2010 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 3-4),and Part B(pages 6-7).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 March 31.2010. 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 C2010 American Association of Physics Teachers
2010 Semifinal Exam 1 AAPT UNITED STATES PHYSICS TEAM AIP 2010 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 3-4), and Part B (pages 6-7). 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 March 31, 2010. • 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 2010 American Association of Physics Teachers
2010 Semifinal Exam Cover Sheet 2 AAPT UNITEDSTATES PHYSICS TEAM AIP 2010 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 school 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, School 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 March 31, 2010. 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=1/4π60=8.99×109N.m2/C2 km=40/4r=10-7T.m/A c=3.00×108m/s k3=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=6.63×10-34Js=4.14×10-15eV,s me=9.109×10-31kg=0.511MeV/c2(1+x)"≈1+nz for<1 sin0≈0-ae3 for<1 cos0≈1-202forl<1 Copyright C2010 American Association of Physics Teachers
2010 Semifinal Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP 2010 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 school 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, School 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 March 31, 2010. 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 2010 American Association of Physics Teachers
2010 Semifinal Exam Part A 2 Part A Question Al An object of mass m is sitting at the northernmost edge of a stationary merry-go-round of radius R.The merry-go-round begins rotating clockwise (as seen from above)with constant angular acceleration of a.The coefficient of static friction between the object and the merry-go-round is μs a.Derive an expression for the magnitude of the object's velocity at the instant when it slides off the merry-go-round in terms of us,R,a,and any necessary fundamental constants. b.For this problem assume that us =0.5,a =0.2 rad/s2,and R=4 m.At what angle,as measured clockwise from north,is the direction of the object's velocity at the instant when it slides off the merry-go-round?Report your answer to the nearest degree in the range 0 to 360°. Solution The object will begin to slide when the force required to keep it accelerating according to the motion of the merry-go-round exceeds the maximum static force of friction. The angular speed w of the merry-go-round is w=at. Since F=ma,we need consider the magnitude of the acceleration of the object.There are two components,the tangential and radial. at aR ar =W2R a Var2+at2 We also have for the static frictional force Ff<usN,where N is the normal force and N =mg, where g is the acceleration of gravity.Consequently, a,2+at2=42g2 is the condition for slipping.Combining the above expressions, a2R2+w4R2=u2g2, a2R2(1+a2t4=u2g2, μs2g2 t aVa2R2-1 Since the object moves off with a tangential velocity once it begins to slip,we have vt Rw Rat =VRVus2g2-a2R2 The angular position made by the merry-go-round is then given by 0=at2,so 1 0= 2aRVH2g2-a2R2 For the numbers given, 0=20.2④V0.5(9.8=o.2y④=3021ad, or 173 degrees.Add 90 for the angle as measured from north,or 263 degrees. Copyright C2010 American Association of Physics Teachers
2010 Semifinal Exam Part A 3 Part A Question A1 An object of mass m is sitting at the northernmost edge of a stationary merry-go-round of radius R. The merry-go-round begins rotating clockwise (as seen from above) with constant angular acceleration of α. The coefficient of static friction between the object and the merry-go-round is µs. a. Derive an expression for the magnitude of the object’s velocity at the instant when it slides off the merry-go-round in terms of µs, R, α, and any necessary fundamental constants. b. For this problem assume that µs = 0.5, α = 0.2 rad/s2 , and R = 4 m. At what angle, as measured clockwise from north, is the direction of the object’s velocity at the instant when it slides off the merry-go-round? Report your answer to the nearest degree in the range 0 to 360◦ . Solution The object will begin to slide when the force required to keep it accelerating according to the motion of the merry-go-round exceeds the maximum static force of friction. The angular speed ω of the merry-go-round is ω = αt. Since F = ma, we need consider the magnitude of the acceleration of the object. There are two components, the tangential and radial. at = αR ar = ω 2R a = p ar 2 + at 2 We also have for the static frictional force Ff ≤ µsN, where N is the normal force and N = mg, where g is the acceleration of gravity. Consequently, ar 2 + at 2 = µs 2 g 2 is the condition for slipping. Combining the above expressions, α 2R 2 + ω 4R 2 = µs 2 g 2 , α 2R 2 1 + α 2 t 4 = µs 2 g 2 , t = s 1 α r µs 2g 2 α2R2 − 1 Since the object moves off with a tangential velocity once it begins to slip, we have vt = Rω = Rαt = q R p µs 2g 2 − α2R2 The angular position made by the merry-go-round is then given by θ = 1 2 αt2 , so θ = 1 2αR p µs 2g 2 − α2R2 For the numbers given, θ = 1 2(0.2)(4) p (0.5)2(9.8)2 − (0.2)2(4)2 = 3.021rad, or 173 degrees. Add 90 for the angle as measured from north, or 263 degrees. Copyright c 2010 American Association of Physics Teachers
2010 Semifinal Exam Part A 4 Question A2 A spherical shell of inner radius a and outer radius b is made of a material of resistivity p and negligible dielectric activity.A single point charge go is located at the center of the shell.At time t =0 all of the material of the shell is electrically neutral,including both the inner and outer surfaces.What is the total charge on the outer surface of the shell as a function of time for t>0? Ignore any effects due to magnetism or radiation;do not assume that b-a is small. Solution The material of the shell will remain electrically neutral,although a charge-Q will build up on the inner surface while a charge of+Q will build up on the outer surface.By spherical symmetry and Gauss's law we can conclude that the electric field in the material of the shell will be given by 190-Q(t) E= 4TEO r2 This will cause a current density E 1 go-Q(t) J=上= P 4Teop r2 and therefore a current 1=JA= -(90-Q) EO where Q is still a function of time.But I=dQ/dt,so dQ dt go-Q coP which can easily be integrated to yield n 90 90-Q/ Q=1-e-t/cop 4o Copyright C2010 American Association of Physics Teachers
2010 Semifinal Exam Part A 4 Question A2 A spherical shell of inner radius a and outer radius b is made of a material of resistivity ρ and negligible dielectric activity. A single point charge q0 is located at the center of the shell. At time t = 0 all of the material of the shell is electrically neutral, including both the inner and outer surfaces. What is the total charge on the outer surface of the shell as a function of time for t > 0? Ignore any effects due to magnetism or radiation; do not assume that b − a is small. Solution The material of the shell will remain electrically neutral, although a charge −Q will build up on the inner surface while a charge of +Q will build up on the outer surface. By spherical symmetry and Gauss’s law we can conclude that the electric field in the material of the shell will be given by E = 1 4π0 q0 − Q(t) r 2 . This will cause a current density J = E ρ = 1 4π0ρ q0 − Q(t) r 2 . and therefore a current I = JA = 1 0ρ (q0 − Q) where Q is still a function of time. But I = dQ/dt, so dQ q0 − Q = dt 0ρ which can easily be integrated to yield ln q0 q0 − Q = t 0ρ or Q q0 = 1 − e −t/0ρ Copyright c 2010 American Association of Physics Teachers
2010 Semifinal Exam Part A 5 Question A3 A cylindrical pipe contains a movable piston that traps 2.00 mols of air.Originally,the air is at one atmosphere of pressure,a volume Vo,and at a temperature of To =298 K.First (process A) the air in the cylinder is compressed at constant temperature to a volume of Vo.Then(process B)the air is allowed to expand adiabatically to a volume of V =15.0 L.After this (process C) this piston is withdrawn allowing the gas to expand to the original volume Vo while maintaining a constant temperature.Finally (process D)while maintaining a fixed volume,the gas is allowed to return to the original temperature To.Assume air is a diatomic ideal gas,no air flows into,or out of,the pipe at any time,and that the temperature outside the remains constant always.Possibly useful information:Cp=R,C=3R,1 atm 1.01 x 105 Pa. a.Draw a P-V diagram of the whole process. b.How much work is done on the trapped air during process A? c.What is the temperature of the air at the end of process B? Solution a.Isotherm,adiabat,isotherm,isochoric b. 片6 For an ideal gas along an isotherm:PV =nRT PM=P乃6→ W=nRTIm Vo W=2 mols.8.31- J ol.K .298K.ln W=-6866 J (work done by the gas) W =6866 J (work done on the gas) c.Adiabatic process TV-1=2 T1=To =298 K-from isothermal process in part b γ-1=8- C2-1= Copyright C2010 American Association of Physics Teachers
2010 Semifinal Exam Part A 5 Question A3 A cylindrical pipe contains a movable piston that traps 2.00 mols of air. Originally, the air is at one atmosphere of pressure, a volume V0, and at a temperature of T0 = 298 K. First (process A) the air in the cylinder is compressed at constant temperature to a volume of 1 4 V0. Then (process B) the air is allowed to expand adiabatically to a volume of V = 15.0 L. After this (process C) this piston is withdrawn allowing the gas to expand to the original volume V0 while maintaining a constant temperature. Finally (process D) while maintaining a fixed volume, the gas is allowed to return to the original temperature T0. Assume air is a diatomic ideal gas, no air flows into, or out of, the pipe at any time, and that the temperature outside the remains constant always. Possibly useful information: Cp = 7 2R, Cv = 5 2R, 1 atm = 1.01 × 105 Pa. a. Draw a P-V diagram of the whole process. b. How much work is done on the trapped air during process A? c. What is the temperature of the air at the end of process B? Solution a. Isotherm, adiabat, isotherm, isochoric b. V1 = 1 4 V0 For an ideal gas along an isotherm: PV = nRT P1V1 = P0V0 =⇒ V1 V0 = P0 P1 W = nRT ln V1 V0 W = 2 mols · 8.31 J mol·K · 298 K · ln 1 4 W = −6866 J (work done by the gas) W = 6866 J (work done on the gas) c. Adiabatic process T1V γ−1 1 = T2V γ−1 2 T1 = T0 = 298 K – from isothermal process in part b T2 = T0 V1 V2 γ−1 γ − 1 = Cp Cv − 1 = 7 2 R 5 2 R − 1 = 7 5 − 1 = 2 5 Copyright c 2010 American Association of Physics Teachers
2010 Semifinal Exam Part A 6 乃=s(得) PoVo =nRTo Vo=nRTo Po %=1atm=1.01.105Pa 2 mols8.31 molK298K %= 1.01·105Pa %=0.0490m3 4-7%=}00490m3=0123m 1 2=15.0L=15.0L. 103 mL 1 cm3 1 m3 IL 1 mL 106cm3=0.0150m3 T2=298K 0.01231 0.0150 T2=275K d.Lowest Pressure From the P-V diagram,the lowest pressure occurs at the end of the second isothermal process. P3V3=nRT3 A-0 T3=T2=275K 3=%=0.0490m3 P3= 2.00 mol.1 m75K 0.0490m3 3=9.33.104Pa Copyright C2010 American Association of Physics Teachers
2010 Semifinal Exam Part A 6 T2 = 298 K V1 V2 2 5 P0V0 = nRT0 V0 = nRT0 P0 P0 = 1 atm = 1.01 · 105 Pa V0 = 2 mols · 8.31 J mol·K · 298 K 1.01 · 105 Pa V0 = 0.0490 m3 V1 = 1 4 V0 = 1 4 · 0.0490 m3 = 0.0123 m3 V2 = 15.0 L = 15.0 L · 103 mL 1 L · 1 cm3 1 mL · 1 m3 106 cm3 = 0.0150 m3 T2 = 298 K 0.0123 0.01502 5 T2 = 275 K d. Lowest Pressure From the P-V diagram, the lowest pressure occurs at the end of the second isothermal process. P3V3 = nRT3 P3 = nRT3 V3 T3 = T2 = 275 K V3 = V0 = 0.0490 m3 P3 = 2.00 mols · 8.31 J mol·K · 275 K 0.0490 m3 P3 = 9.33 · 104 Pa Copyright c 2010 American Association of Physics Teachers
2010 Semifinal Exam Part A Question A4 The energy radiated by the Sun is generated primarily by the fusion of hydrogen into helium-4.In stars the size of the Sun,the primary mechanism by which fusion takes place is the proton-proton chain.The chain begins with the following reactions: 2p-X1+e++X2(0.42MeV) (A4-1) p+X1→Xg+Y(5.49MeV) (A42) The amounts listed in parentheses are the total kinetic energy carried by the products,including gamma rays.p is a proton,et is a positron,y is a gamma ray,and X1,X2,and X3 are particles for you to identify. The density of electrons in the Sun's core is sufficient that the positron is annihilated almost immediately,releasing an energy x: e++e→2Y(x) (A4-3) Subsequently,two major processes occur simultaneously.The "pp I branch"is the single reaction 2X3→4He+2X4(y): (A44) which releases an energy y.The "pp II branch"consists of three reactions: X3+4He→X5+y (A4-5) X5+e-→X6+X7(z) (A4-6) X6+X4→24He (A4-7) where z is the energy released in step A4-6. a.Identify XI through X7.X2 and X7 are neutral particles of negligible mass.It is useful to know that the first few elements,in order of atomic number,are H,He,Li,Be,B,C,N,O. b.The mass of the electron is 0.51 MeV/c2,the mass of the proton is 938.27 MeV/c2,and the mass of the helium-4 nucleus is 3727.38 MeV/c2.Find the energy released during the production of one helium-4 nucleus,including the kinetic energy of all products and all energy carried by gamma rays. c.Find the unknown energies x and y above. d.Step (A4-6)does not proceed as follows because there is insufficient energy. X5→X6+e++X7 What constraint does this fact place on 2? e.In which of the reaction steps is the energy carried by any given product the same every time the step occurs?Assume that the kinetic energy carried in by the reactants in each step is negligible,and that the products are in the ground state. Copyright C2010 American Association of Physics Teachers
2010 Semifinal Exam Part A 7 Question A4 The energy radiated by the Sun is generated primarily by the fusion of hydrogen into helium-4. In stars the size of the Sun, the primary mechanism by which fusion takes place is the proton-proton chain. The chain begins with the following reactions: 2 p → X1 + e+ + X2 (0.42 MeV) (A4-1) p + X1 → X3 + γ (5.49 MeV) (A4-2) The amounts listed in parentheses are the total kinetic energy carried by the products, including gamma rays. p is a proton, e+ is a positron, γ is a gamma ray, and X1, X2, and X3 are particles for you to identify. The density of electrons in the Sun’s core is sufficient that the positron is annihilated almost immediately, releasing an energy x: e + + e− → 2 γ (x) (A4-3) Subsequently, two major processes occur simultaneously. The “pp I branch” is the single reaction 2 X3 → 4He + 2 X4 (y), (A4-4) which releases an energy y. The “pp II branch” consists of three reactions: X3 + 4He → X5 + γ (A4-5) X5 + e− → X6 + X7 (z) (A4-6) X6 + X4 → 2 4He (A4-7) where z is the energy released in step A4-6. a. Identify X1 through X7. X2 and X7 are neutral particles of negligible mass. It is useful to know that the first few elements, in order of atomic number, are H, He, Li, Be, B, C, N, O. b. The mass of the electron is 0.51 MeV/c 2 , the mass of the proton is 938.27 MeV/c 2 , and the mass of the helium-4 nucleus is 3 727.38 MeV/c 2 . Find the energy released during the production of one helium-4 nucleus, including the kinetic energy of all products and all energy carried by gamma rays. c. Find the unknown energies x and y above. d. Step (A4-6) does not proceed as follows because there is insufficient energy. X5 → X6 + e+ + X7 What constraint does this fact place on z? e. In which of the reaction steps is the energy carried by any given product the same every time the step occurs? Assume that the kinetic energy carried in by the reactants in each step is negligible, and that the products are in the ground state. Copyright c 2010 American Association of Physics Teachers
2010 Semifinal Exam Part A 8 Solution a.We know that in all nuclear processes,total charge is conserved,lepton number(electrons plus neutrinos minus positrons minus antineutrinos),and baryon number(neutrons plus protons) are conserved.As X2 is a neutral particle of negligible mass,XI must have charge +1 and contain two baryons.Thus it is2H.X2 is then a neutral particle with lepton number +1 and is an electron neutrino. Similar reasoning shows that X3 is 3He,X4 is a proton (or H,X5 is 7Be,X6 is 7Li,and X7 is an electron neutrino. b.The overall reaction (not including gamma rays)is 4p+2e-4 He+2ve as can be seen from conservation considerations or by combining the given reactions.The energy released is the difference in mass between the reactants and the products;using the given values,this is 26.72 MeV. c.x is simply twice the mass of the electron,1.02 MeV.To compute y,note that we can sum the other known energies to obtain the result from the previous problem: 2(0.42MeV+5.49MeV+1.02MeV)+y=26.72MeV where the factor of 2 arises because two 3He are produced in the course of the combined reaction.Solving,y=12.86 MeV. d.The forbidden reaction produces an energy of z-2(0.51 MeV),as it differs from the naturally occurring one by the consumption of one fewer electron and the production of an additional positron.Since it is forbidden,we know that z-2(0.51 MeV)<0;in other words,z< 1.02MeV. e.It is a well-known result that reactions with two products have a single set of product energies, whereas those with three or more products produce a spectrum of output energies.Students may quote this result;alternatively,observe that in a two-product reaction the conservation of momentum and of energy give two equations in the two unknowns,fixing their values, whereas there are insufficient equations in the case of three or more products. The reactions with only two products are (A4-2),(A4-3),(A4-5),and(A4-7). Copyright C2010 American Association of Physics Teachers
2010 Semifinal Exam Part A 8 Solution a. We know that in all nuclear processes, total charge is conserved, lepton number (electrons plus neutrinos minus positrons minus antineutrinos), and baryon number (neutrons plus protons) are conserved. As X2 is a neutral particle of negligible mass, X1 must have charge +1 and contain two baryons. Thus it is 2H. X2 is then a neutral particle with lepton number +1 and is an electron neutrino. Similar reasoning shows that X3 is 3He, X4 is a proton (or 1H, X5 is 7Be, X6 is 7Li, and X7 is an electron neutrino. b. The overall reaction (not including gamma rays) is 4 p + 2 e− →4 He + 2 νe as can be seen from conservation considerations or by combining the given reactions. The energy released is the difference in mass between the reactants and the products; using the given values, this is 26.72 MeV. c. x is simply twice the mass of the electron, 1.02 MeV. To compute y, note that we can sum the other known energies to obtain the result from the previous problem: 2(0.42 MeV + 5.49 MeV + 1.02 MeV) + y = 26.72 MeV where the factor of 2 arises because two 3He are produced in the course of the combined reaction. Solving, y = 12.86 MeV. d. The forbidden reaction produces an energy of z−2(0.51 MeV), as it differs from the naturally occurring one by the consumption of one fewer electron and the production of an additional positron. Since it is forbidden, we know that z − 2(0.51 MeV) < 0; in other words, z < 1.02 MeV. e. It is a well-known result that reactions with two products have a single set of product energies, whereas those with three or more products produce a spectrum of output energies. Students may quote this result; alternatively, observe that in a two-product reaction the conservation of momentum and of energy give two equations in the two unknowns, fixing their values, whereas there are insufficient equations in the case of three or more products. The reactions with only two products are (A4-2), (A4-3), (A4-5), and (A4-7). Copyright c 2010 American Association of Physics Teachers
2010 Semifinal Exam Part A 9 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 C2010 American Association of Physics Teachers
2010 Semifinal Exam Part A 9 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 2010 American Association of Physics Teachers
2010 Semifinal Exam Part B 10 Part B Question Bl A thin plank of mass M and length L rotates about a pivot at its center.A block of mass m<M slides on the top of the plank.The system moves without friction.Initially,the plank makes an angle 0o with the horizontal,the block is at the upper end of the plank,and the system is at rest. Throughout the problem you may assume that 6<1,and that the physical dimensions of the block are much,much smaller than the length of the plank. L/2 Let z be the displacement of the block along the plank,as measured from the pivot,and let 0 be the angle between the plank and the horizontal.You may assume that centripetal acceleration of the block is negligible compared with the linear acceleration of the block up and down the plank. a.For a certain value of 0o,x=ko throughout the motion,where k is a constant.What is this value of 0o?Express your answer in terms of M,m,and any fundamental constants that you require. b.Given that 0o takes this special value,what is the period of oscillation of the system?Express your answer in terms of M,m,and any fundamental constants that you require. c.Determine the maximum value of the ratio between the centripetal acceleration of the block and the linear acceleration of the block along the plank,writing your answer in terms of m and M,therefore justifying our approximation. Solution Eventually we will need to compute the rotational inertia of a long plank.The answer is 1=立M and it is acceptable to just write it down with no work. Assume the block is at point x away from the central pivot.The magnitude of the torque on the plank is then given by T=ymg cos0≈mgx so the angular acceleration is mg Copyright C2010 American Association of Physics Teachers
2010 Semifinal Exam Part B 10 Part B Question B1 A thin plank of mass M and length L rotates about a pivot at its center. A block of mass m M slides on the top of the plank. The system moves without friction. Initially, the plank makes an angle θ0 with the horizontal, the block is at the upper end of the plank, and the system is at rest. Throughout the problem you may assume that θ 1, and that the physical dimensions of the block are much, much smaller than the length of the plank. x L/2 Let x be the displacement of the block along the plank, as measured from the pivot, and let θ be the angle between the plank and the horizontal. You may assume that centripetal acceleration of the block is negligible compared with the linear acceleration of the block up and down the plank. a. For a certain value of θ0, x = kθ throughout the motion, where k is a constant. What is this value of θ0? Express your answer in terms of M, m, and any fundamental constants that you require. b. Given that θ0 takes this special value, what is the period of oscillation of the system? Express your answer in terms of M, m, and any fundamental constants that you require. c. Determine the maximum value of the ratio between the centripetal acceleration of the block and the linear acceleration of the block along the plank, writing your answer in terms of m and M, therefore justifying our approximation. Solution Eventually we will need to compute the rotational inertia of a long plank. The answer is I = 1 12 ML2 and it is acceptable to just write it down with no work. Assume the block is at point x away from the central pivot. The magnitude of the torque on the plank is then given by τ = ymg cos θ ≈ mgx so the angular acceleration is α = − mg I x Copyright c 2010 American Association of Physics Teachers
