2008 Semifinal Exam AAPT UNITEDSTATES PHYSICS TEAM AIP 2008 Semifinal Exam 6 QUESTIONS-Several MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Show all your work.Partial credit will be given. Start each question on a new sheet of paper.Put your name in the upper right-hand corner of each page,along with the question number and the page number/total pages for this problem.For example, Doe,Jamie Prob.1-P.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 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. Each of the four questions in part A are worth 25 points.Each of the two questions in part B are worth 50 points.The questions are not necessarily of the same difficulty.Good luck! In order to maintain exam security,do not communicate any information about the questions(or their answers or solutions)on this contest until after April 10,2008. DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Copyright C2008 American Association of Physics Teachers
Solutions 2008 Semifinal Exam 1 AAPT UNITED STATES PHYSICS TEAM AIP 2008 Semifinal Exam 6 QUESTIONS - Several MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Show all your work. Partial credit will be given. • Start each question on a new sheet of paper. Put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example, Doe, Jamie Prob. 1 - P. 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 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. • Each of the four questions in part A are worth 25 points. Each of the two questions in part B are worth 50 points. The questions are not necessarily of the same difficulty. Good luck! • In order to maintain exam security, do not communicate any information about the questions (or their answers or solutions) on this contest until after April 10, 2008. DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Copyright c 2008 American Association of Physics Teachers
2008 Semifinal Exam 2 Part A Question Al Four square metal plates of area A are arranged at an even spacing d as shown in the diagram.(Assume that A>>d2.) Plate ate 4 Plates 1 and 4 are first connected to a voltage source of magnitude Vo,with plate 1 positive.Plates 2 and 3 are then connected together with a wire,which is subsequently removed.Finally,the voltage source attached between plates 1 and 4 is replaced with a wire.The steps are summarized in the diagram below. (a) (b) (c) What is the resulting potential difference between a.Plates 1 and 2 (Call it Vi). b.Plates 2 and 3(Call it V2),and c.Plates 3 and 4(Call it V3). Assume,in each case,that a positive potential difference means that the top plates is at a high potential than the bottom plate. Solution There are two fairly easy ways to do this problem,one rather straightforward application of capacitors,the other a more elegant,and much,much short,application of boundary conditions in electric fields. The first method involves treating the problem as three series capacitors.Each has an identical capaci- tance C.The figure below then show the three steps. C2 C3 C3 Since C2 is shorted out originally,then effectively there are only two capacitors in series,so the voltage drop across each is Vo/2,where the a positive potential difference means that the top plate of any given capacitor is positive.The top plate of Ci will then have a positive charge of go=CVo/2.Note that this means that the bottom plate of the top capacitor will have a negative charge of-go.Removing the shorting wire across C2 will not change the charges or potential drops across the other two capacitors.Removing the source Vo will also make no difference. Copyright C2008 American Association of Physics Teachers
Solutions 2008 Semifinal Exam 2 Part A Question A1 Four square metal plates of area A are arranged at an even spacing d as shown in the diagram. (Assume that A >> d 2 .) Plate 1 Plate 2 Plate 3 Plate 4 ddd Plates 1 and 4 are first connected to a voltage source of magnitude V0, with plate 1 positive. Plates 2 and 3 are then connected together with a wire, which is subsequently removed. Finally, the voltage source attached between plates 1 and 4 is replaced with a wire. The steps are summarized in the diagram below. (a) (b) (c) What is the resulting potential difference between a. Plates 1 and 2 (Call it V1), b. Plates 2 and 3 (Call it V2), and c. Plates 3 and 4 (Call it V3). Assume, in each case, that a positive potential difference means that the top plates is at a high potential than the bottom plate. Solution There are two fairly easy ways to do this problem, one rather straightforward application of capacitors, the other a more elegant, and much, much short, application of boundary conditions in electric fields. The first method involves treating the problem as three series capacitors. Each has an identical capacitance C. The figure below then show the three steps. C1 C2 C3 C1 C2 C3 C1 C2 C3 Since C2 is shorted out originally, then effectively there are only two capacitors in series, so the voltage drop across each is V0 /2, where the a positive potential difference means that the top plate of any given capacitor is positive. The top plate of C 1 will then have a positive charge of q 0 = CV0 /2. Note that this means that the bottom plate of the top capacitor will have a negative charge of − q 0. Removing the shorting wire across C2 will not change the charges or potential drops across the other two capacitors. Removing the source V0 will also make no difference. Copyright c 2008 American Association of Physics Teachers
2008 Semifinal Exam 3 Shorting the top plate of Ci with the bottom plate of C3 will make a difference.Positive charge will flow out of top plate of C into the bottom plate of C3.Also,negative charge will flow out of the bottom plate of Ci into the top plate of C2.The result is that Ci will acquire a potential difference of Vi,C2 a potential difference of V2,and C3 a potential difference of V3.Let the final charge on the top plate of each capacitor also be labeled as q1,q2,and 93. The last figure implies that M+2+=0. By symmetry,we have =. So 2VM=-2. By charge conservation between the bottom plate of Ci and the top plate of C2 we have -q0=-91+2: But g=CV,so -6=-+6 Combining the above we get +2, Finally,solving for Vi,we get Vi=Vo/6. Alternatively,we could focus on the plate arrangement and the fact that across a boundary AEL= lo/col,a consequence of Gauss's Law.Also,we have,for parallel plate configurations,AV=Ed.Since co and d are the same for each of the three regions,it is sufficient to simply look at the behavior of the electric fields. EO E2 EO In the first picture we require that 2Eo =Vo/d.The charge density on the second plate requires that AE Eo.In the last picture we have 2E1+E2=0,since the potential between the top plate and the bottom plate is zero.But we also have,on the second plate,AE=E1-E2.Combining;Eo =-E2-E2 =-E2, and therefore V2 =-Vo,and Vi Vo/6. Copyright C2008 American Association of Physics Teachers
Solutions 2008 Semifinal Exam 3 Shorting the top plate of C 1 with the bottom plate of C3 will make a difference. Positive charge will flow out of top plate of C 1 into the bottom plate of C3. Also, negative charge will flow out of the bottom plate of C 1 into the top plate of C2. The result is that C 1 will acquire a potential difference of V1 , C2 a potential difference of V2, and C3 a potential difference of V3. Let the final charge on the top plate of each capacitor also be labeled as q 1 , q 2, and q 3 . The last figure implies that V1 + V2 + V3 = 0 . By symmetry, we have V1 = V3 . so 2 V1 = − V2 . By charge conservation between the bottom plate of C 1 and the top plate of C2 we have − q 0 = − q 1 + q 2 . But q = CV , so − 12 V0 = − V1 + V2 Combining the above we get − 12 V0 = 12 V2 + V2 , − 13 V0 = V2 . Finally, solving for V1, we get V1 = V0 /6. Alternatively, we could focus on the plate arrangement and the fact that across a boundary |∆ E ⊥| = |σ/ǫ0|, a consequence of Gauss’s Law. Also, we have, for parallel plate configurations, |∆V | = |Ed|. Since ǫ0 and d are the same for each of the three regions, it is sufficient to simply look at the behavior of the electric fields. E0 E0 E2 E1 E1 In the first picture we require that 2 E 0 = V0/d. The charge density on the second plate requires that ∆ E = E 0. In the last picture we have 2 E 1 + E 2 = 0, since the potential between the top plate and the bottom plate is zero. But we also have, on the second plate, ∆ E = E 1 − E 2. Combining, E 0 = − 12 E 2 − E 2 = − 32 E 2 , and therefore V2 = − 13 V0, and V1 = V0 /6. Copyright c 2008 American Association of Physics Teachers
2008 Semifinal Exam Question A2 A simple heat engine consists of a piston in a cylinder filled with an ideal monatomic gas.Initially the gas in the cylinder is at a pressure Po and volume Vo.The gas is slowly heated at constant volume.Once the pressure reaches 32Po the piston is released,allowing the gas to expand so that no heat either enters or escapes the gas as the piston moves.Once the pressure has returned to Po thethe outside of the cylinder is cooled back to the original temperature,keeping the pressure constant.For the monatomic ideal gas you should assume that the specific heat capacity at constant volume is given by Cy =inR,where n is the number of moles of the gas present and R is the ideal gas constant.You may express your answers in fractional form or as decimals.If you choose decimals,keep three significant figures in your calculations. The diagram below is not necessarily drawn to scale. 32P0 ainssald Po Vmax Volume a.Let Vmax be the maximum volume achieved by the gas during the cycle.What is Vimax in terms of Vo?If you are unable to solve this part of the problem,you may erpressyour answers to the remaining parts in terms of Vmax without further loss of points. b.In terms of Po and Vo determine the heat added to the gas during a complete cycle. c.In terms of Po and Vo determine the heat removed from the gas during a complete cycle. d.Defining efficiency e as the net work done by the gas divided by the heat added to the gas,what is the efficiency of this cycle? e.Determine the ratio between the maximum and minimum temperatures during this cycle. Solution It is convenient to construct two tables and solve this problem in a manner similar to a Sudoku puzzle. Defining point 1 to be the initial point,and measuring P,V,and T in terms of Po,Vo,and To,while measuring Q,W,and AU in terms of nRTo PoVo,we have initially Process Q W△U Point P V T 1→2 0 111 321 2→30 3→1 3 net 0 Copyright C2008 American Association of Physics Teachers
Solutions 2008 Semifinal Exam 4 Question A2 A simple heat engine consists of a piston in a cylinder filled with an ideal monatomic gas. Initially the gas in the cylinder is at a pressure P0 and volume V0. The gas is slowly heated at constant volume. Once the pressure reaches 32 P0 the piston is released, allowing the gas to expand so that no heat either enters or escapes the gas as the piston moves. Once the pressure has returned to P0 the the outside of the cylinder is cooled back to the original temperature, keeping the pressure constant. For the monatomic ideal gas you should assume that the specific heat capacity at constant volume is given by C V = 32 nR, where n is the number of moles of the gas present and R is the ideal gas constant. You may express your answers in fractional form or as decimals. If you choose decimals, keep three significant figures in your calculations. The diagram below is not necessarily drawn to scale. Pressure P00 V0 32P Vmax Volume a. Let Vmax be the maximum volume achieved by the gas during the cycle. What is Vmax in terms of V0 ? If you are unable to solve this part of the problem, you may expressyour answers to the remaining parts in terms of Vmax without further loss of points. b. In terms of P0 and V0 determine the heat added to the gas during a complete cycle. c. In terms of P0 and V0 determine the heat removed from the gas during a complete cycle. d. Defining efficiency e as the net work done by the gas divided by the heat added to the gas, what is the efficiency of this cycle? e. Determine the ratio between the maximum and minimum temperatures during this cycle. Solution It is convenient to construct two tables and solve this problem in a manner similar to a Sudoku puzzle. Defining point 1 to be the initial point, and measuring P , V , and T in terms of P0 , V0, and T0, while measuring Q , W, and ∆ U in terms of nRT0 = P0 V0, we have initially Point P V T 1 1 1 1 2 32 1 3 1 Process Q W ∆ U 1 → 2 0 2 → 3 0 3 → 1 net 0 Copyright c 2008 American Association of Physics Teachers
2008 Semifinal Exam 5 The "obvious"values have been filled in:the initial conditions,and zeroes corresponding to Q=0 along an adiabat,W=0 along a constant volume process,and finally AU =0 for a net process. The convention that will be used here is Q+W=AU. The ideal gas law,PV/T =nR,can be used to quickly determine T2,since P/T is a constant for that process.One can then use Q=Cv△T to find Q1-2.The table values will now read Point P V T Process QW△ 1→2 310 1 111 0 2 32 132 2→3 3→1 3 1 net 0 For the adiabatic process we have PVy is a constant,where y=Cp/Cy.Students who don't know this can derive it.although it will take some time. The derivation is straightforward enough.Along an adiabatic process,Q=0,so from Q+W=6U -PdV=3 之=工之P) Rearranging, 0=5P dv+3V dp or 0=+dp 3元+F Integrating, Constant nV+nP 3 which can be written in the more familiar form PVY=Constant. The factor of 32 was chosen so that the results are nice answers. One can then find V3 (and,for that matter,Viax)by using this,and get 1/y =(32)3/5=8 Putting this in the table,and then quickly applying the ideal gas law to find T3 then enables the finding of Q3-1,since along this process Q-CPAT =5 nR△T 2 The tables now look like Point P V T Process QW△U 1→2 111 310 2→3 0 2 32132 3→1 188 -号7 net 0 It is now possible to determine Qnet and,from Q+W =AU,Wnet. So the tables now look like Point P V T Process Q W△U 1→2 111 2310 2+3 0 2 32132 188 3→1 -57 net 29-290 Copyright C2008 American Association of Physics Teachers
Solutions 2008 Semifinal Exam 5 The “obvious” values have been filled in: the initial conditions, and zeroes corresponding to Q = 0 along an adiabat, W = 0 along a constant volume process, and finally ∆ U = 0 for a net process. The convention that will be used here is Q + W = ∆ U . The ideal gas law, P V /T = nR, can be used to quickly determine T2, since P/T is a constant for that process. One can then use Q = C V ∆ T to find Q 1 → 2. The table values will now read Point P V T 1 1 1 1 2 32 1 32 3 1 Process Q W ∆ U 1 → 2 32 31 0 2 → 3 0 3 → 1 net 0 For the adiabatic process we have P V γ is a constant, where γ = C P /C V . Students who don’t know this can derive it, although it will take some time. The derivation is straightforward enough. Along an adiabatic process, Q = 0, so from Q + W = δU −P dV = 32 nR dT = 32 (P dV + V dP ) Rearranging, 0 = 5P dV + 3V dP or 0 = 53 dVV + dPP Integrating, Constant = 53 ln V + ln P which can be written in the more familiar form P V γ = Constant . The factor of 32 was chosen so that the results are nice answers. One can then find V3 (and, for that matter, Vmax) by using this, and get V3 V2 = P2 P3 1/γ = (32) 3 / 5 = 8 Putting this in the table, and then quickly applying the ideal gas law to find T3 then enables the finding of Q 3 → 1, since along this process Q = C P ∆ T = 52 nR ∆T. The tables now look like Point P V T 1 1 1 1 2 32 1 32 3 1 8 8 Process Q W ∆ U 1 → 2 32 31 0 2 → 3 0 3 → 1 − 52 7 net 0 It is now possible to determine Qnet and, from Q + W = ∆ U , Wnet . So the tables now look like Point P V T 1 1 1 1 2 32 1 32 3 1 8 8 Process Q W ∆ U 1 → 2 32 31 0 2 → 3 0 3 → 1 − 52 7 net 29 −29 0 Copyright c 2008 American Association of Physics Teachers