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
2008 Semifinal Exam 6 The negative net work reflects that the gas does work on the outside world.The efficiency of the process is then 2958 e= 932=5 It isn't much more work to fill in all of the values for both tables,a task that each student ought be able to do.One key point will be process 3-1,where W=-PAV.thethe rest are filled in by applications of Q+W=△U. Process W△U Point P V T 1→2 号31031 1 111 2→3 32132 0-36-36 2 3 188 3→1-号77-27 net 29-290 Solutions Copyright C2008 American Association of Physics Teachers
Solutions 2008 Semifinal Exam 6 The negative net work reflects that the gas does work on the outside world. The efficiency of the process is then e = 29 93 / 2 = 58 93 It isn’t much more work to fill in all of the values for both tables, a task that each student ought be able to do. One key point will be process 3 → 1, where W = − P ∆ V . the the rest are filled in by applications of Q + W = ∆ U . 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 32 31 2 → 3 0 -36 -36 3 → 1 − 52 7 7 − 32 7 net 29 −29 0 Copyright c 2008 American Association of Physics Teachers
2008 Semifinal Exam 7 Question A3 A certain planet of radius R is composed of a uniform material that,through radioactive decay,generates a net power P.This results in a temperature differential between the inside and outside of the planet as heat is transfered from the interior to the surface. The rate of heat transfer is governed by the thermal conductivity.The thermal conductivity of a ma- terial is a measure of how quickly heat flows through that material in response to a temperature gradient. Specifically,consider a thin slab of material of area A and thickness Ax where one surface is hotter than the other by an amount AT.Suppose that an amount of heat AQ flows through the slab in a time At.The thermal conductivity k of the material is then 214 △tA△T It is found that k is approximately constant for many materials;assume that it is constant for the planet. For the following assume that the planet is in a steady state;temperature might depend on position,but does not depend on time. a.Find an expression for the temperature of the surface of the planet assuming blackbody radiation,an emissivity of 1,and no radiation incident on the planet surface.You may express your answer in terms of any of the above variables and the Stephan-Boltzmann constant o. b.Find an expression for the temperature difference between the surface of the planet and the center of the planet.You may express your answer in terms of any of the above variables;you do not need to answer part (a)to be able to answer this part. Solution For the first question,apply the Boltzmann equation,and P=0AT where A is the surface area of the planet,and T the temperature at the center.Then 1/4 T81 \4ToR产 For the second question,it is reasonable to assume that the temperature depends on the distance form the center only.Then the definition of k gives for a spherical shell of thickness dr △Q1dr △t4πr2dT The heat through the shell depends on the power radiated from within the shell.Since the planet is uniform,this depends on the volume according to △Q 3 R3=P R3 so that rearrangement yields P dT=. AxkRr dr Integrating between the center and the surface, △T= P 8元kR which could be used to find the temperature of the interior. Copyright C2008 American Association of Physics Teachers
Solutions 2008 Semifinal Exam 7 Question A3 A certain planet of radius R is composed of a uniform material that, through radioactive decay, generates a net power P. This results in a temperature differential between the inside and outside of the planet as heat is transfered from the interior to the surface. The rate of heat transfer is governed by the thermal conductivity. The thermal conductivity of a material is a measure of how quickly heat flows through that material in response to a temperature gradient. Specifically, consider a thin slab of material of area A and thickness ∆ x where one surface is hotter than the other by an amount ∆ T . Suppose that an amount of heat ∆ Q flows through the slab in a time ∆ t. The thermal conductivity k of the material is then k = ∆ Q∆t 1A ∆ x ∆ T . It is found that k is approximately constant for many materials; assume that it is constant for the planet. For the following assume that the planet is in a steady state; temperature might depend on position, but does not depend on time. a. Find an expression for the temperature of the surface of the planet assuming blackbody radiation, an emissivity of 1, and no radiation incident on the planet surface. You may express your answer in terms of any of the above variables and the Stephan-Boltzmann constant σ . b. Find an expression for the temperature difference between the surface of the planet and the center of the planet. You may express your answer in terms of any of the above variables; you do not need to answer part (a) to be able to answer this part. Solution For the first question, apply the Boltzmann equation, and P = σAT 4s where A is the surface area of the planet, and Ts the temperature at the center. Then Ts = P 4πσR 2 1 / 4 For the second question, it is reasonable to assume that the temperature depends on the distance form the center only. Then the definition of k gives for a spherical shell of thickness dr k = ∆Q∆t 1 4πr2 dr dT . The heat through the shell depends on the power radiated from within the shell. Since the planet is uniform, this depends on the volume according to ∆Q∆t = P 43 πr 3 43 πR 3 = P r 3 R3 so that rearrangement yields dT = P 4πkR 3 r dr Integrating between the center and the surface,∆ T = P 8πkR , which could be used to find the temperature of the interior. Copyright c 2008 American Association of Physics Teachers
2008 Semifinal Exam 8 Question A4 A tape recorder playing a single tone of frequency fo is dropped from rest at a height h.You stand directly underneath the tape recorder and measure the frequency observed as a function of time.Here t=0s is the time at which the tape recorder was dropped. t (s)f(Hz) 2.0 581 4.0 619 6.0 665 8.0 723 10.0 801 The acceleration due to gravity is g=9.80 m/s2 and the speed of sound in air is va =340 m/s.Ignore air resistance.You might need to use the Doppler shift formula for co-linear motion of sources and observers in still air, f=f%土 Va士vs where fo is the emitted frequency as determined by the source,f is the frequency as detected by the observer, and va,vs,and vo are the speed of sounds in air,the speed of the source,and the speed of the observer.The positive and negative signs are dependent upon the relative directions of the source and the observer. a.Determine the frequency measured on the ground at time t,in terms of fo,g,h,and va. b.Verify graphically that your result is consistent with the provided data. c.What (numerically)is the frequency played by the tape recorder? d.From what height h was the tape recorder dropped? Solution The position of the tape recorder above the ground at a time t is given by and the speed of the tape recorder is given by Us =-gt The observer "hears"the sound emitted from the tape recorder a time 6t earlier,since it takes time for the sound to travel to the listener.In this case. y=Vadt So at time t the listener is hearing the tape recorder when it had emitted at time t'=t-ot,or t=t-上+9e2 Va 2va Solve this for t',first by rearranging, y-w+(t-)=0 and the by applying the quadratic formula t=a±Va2+2gh-2gat 9 Copyright C2008 American Association of Physics Teachers
Solutions 2008 Semifinal Exam 8 Question A4 A tape recorder playing a single tone of frequency f0 is dropped from rest at a height h. You stand directly underneath the tape recorder and measure the frequency observed as a function of time. Here t = 0s is the time at which the tape recorder was dropped. t (s) f (Hz) 2.0 581 4.0 619 6.0 665 8.0 723 10.0 801 The acceleration due to gravity is g = 9 .80 m / s 2 and the speed of sound in air is v a = 340 m /s. Ignore air resistance. You might need to use the Doppler shift formula for co-linear motion of sources and observers in still air, f = f0 v a ± v o v a ± v s where f0 is the emitted frequency as determined by the source, f is the frequency as detected by the observer, and v a , v s, and v o are the speed of sounds in air, the speed of the source, and the speed of the observer. The positive and negative signs are dependent upon the relative directions of the source and the observer. a. Determine the frequency measured on the ground at time t, in terms of f0 , g , h, and v a . b. Verify graphically that your result is consistent with the provided data. c. What (numerically) is the frequency played by the tape recorder? d. From what height h was the tape recorder dropped? Solution The position of the tape recorder above the ground at a time t is given by y = h − 12 gt 2 and the speed of the tape recorder is given by v s = −gt The observer “hears” the sound emitted from the tape recorder a time δt earlier, since it takes time for the sound to travel to the listener. In this case, y = v aδt So at time t the listener is hearing the tape recorder when it had emitted at time t ′ = t − δt, or t ′ = t − hva + g 2 v a ( t ′ ) 2 Solve this for t ′ , first by rearranging, g2 ( t ′ ) 2 − v a t ′ + ( v a t − h) = 0 and the by applying the quadratic formula t′ = v a ± p v a 2 + 2gh − 2gv a t g . Copyright c 2008 American Association of Physics Teachers
2008 Semifinal Exam 9 This might not look right,but in the limit of small h and large va,it does reduce to the expected t'=t if one keeps the negative result. Consequently, Us Vva2 +2gh-2gvat-va gives the velocity of that source had when it emitted the sound heard at time t.This result is negative, indicating motion down,and toward the observer,so one must use the positive sign in the denominator of the Doppler shift formula. Applying the Doppler shift formula, f=fo- Va Vva2 +2gh-2gvat which,in the limit of large va and small h,reduces to 1=(1+是） Keeping to the correct expression,we can rearrange it as 六-(+等- 29 which would graph as a straight line by plotting t horizontally and 1/f?vertically.The slope of the line would yield 29 Vafo2 while the vertical intercept would yield 2gh fo? Using this hint,a quick table of data to consider graphing would be t(s)f(Hz)、1/f2(×10-6s2) 2.0 581 2.96 4.0 619 2.61 6.0 665 2.26 8.0 723 1.91 10.0 801 1.56 The slope is -1.75 x 10-7 s Then 2(9.8) f0= -Hz=574Hz (1.75×10-7)(340 The intercept is 3.31 x 10-6 s2.This yields a height in meters given by h=(340 (3.31)(340)2 =533 (0.1752(9.8) Clearly,an impressive building;and a more impressive tape player,that it could be heard from such a distance! Copyright C2008 American Association of Physics Teachers
Solutions 2008 Semifinal Exam 9 This might not look right, but in the limit of small h and large v a, it does reduce to the expected t ′ = t if one keeps the negative result. Consequently, v s = p v a 2 + 2gh − 2gv a t − v a gives the velocity of that source had when it emitted the sound heard at time t. This result is negative, indicating motion down, and toward the observer, so one must use the positive sign in the denominator of the Doppler shift formula. Applying the Doppler shift formula, f = f0 p v a v a 2 + 2gh − 2gv a t which, in the limit of large v a and small h, reduces to f = f0 1 + gva t Keeping to the correct expression, we can rearrange it as 1f 2 = 1f02 1 + 2gh va2 − 2 g v a t which would graph as a straight line by plotting t horizontally and 1/f 2 vertically. The slope of the line would yield − 2 g v a f0 2 while the vertical intercept would yield 1f0 2 1 + 2gh va2 Using this hint, a quick table of data to consider graphing would be t (s) f (Hz) 1/f 2 (×10−6 s 2 ) 2.0 581 2.96 4.0 619 2.61 6.0 665 2.26 8.0 723 1.91 10.0 801 1.56 The slope is − 1 .75 × 10 − 7 s. Then f0 = s 2(9 .8) (1 .75 × 10 − 7)(340)Hz = 574Hz The intercept is 3 .31 × 10 − 6 s 2 . This yields a height in meters given by h = (340) (3.31) (0 .175) − (340) 2 2(9 .8) = 533 Clearly, an impressive building; and a more impressive tape player, that it could be heard from such a distance! Copyright c 2008 American Association of Physics Teachers
2008 Semifinal Exam 10 Part B Question B1 A platform is attached to the ground by an ideal spring of constant k;both the spring and the platform have negligible mass.Sitting on the platform is a rather large lump of clay of mass me.You then gently step onto the platform,and the platform settles down to a new equilibrium position,a vertical distance D below the original position.Assume that your mass is mp. h a.You then pick up the lump of clay and hold it a height h above the platform.Upon releasing the clay the you and the platform will oscillate up and down;you notice that the clay strikes the platform after the platform has completed exactly one oscillation.Determine h in terms of any or all of k,D,the masses mp and mc,the acceleration of free fall g,and any necessary numerical constants.You must express your answer in the simplest possible form. b.Assume the resulting collision between the clay and the platform is completely inelastic.Find the ratio of the amplitude of the oscillation of the platform before the collision (A:)and the amplitude of the oscillations of the platform after the collision (Af).Determine Af/Ai in terms of any or all of k,D, the masses mp and me,the acceleration of free fall g,and any necessary numerical constants. c.Sketch a graph of the position of the platform as a function of time,with t=0 corresponding to the moment when the clay is dropped.Show one complete oscillation after the clay has collided with the platform. d.The above experiment is only possible if the ratio me/mp is smaller than some critical value re, otherwise the clay will hit the platform before one complete oscillation.An estimate for the value of the critical ratio re can be obtained by assuming the clay hits the platform after exactly one-half of an oscillation.Assume that h is the same as is determined by part (a),and use this technique to determine re in terms of any or all of k,D,the mass mp,the acceleration of free fall g,and any necessary numerical constants. e.Is this estimate for re too large or too small?You must defend your answer with an appropriate diagram. Solution Stepping on the platform will lower it a distance D.This means that the spring constant of the platform spring is given by kD =mpg. Copyright C2008 American Association of Physics Teachers
Solutions 2008 Semifinal Exam 10 Part B Question B1 A platform is attached to the ground by an ideal spring of constant k; both the spring and the platform have negligible mass. Sitting on the platform is a rather large lump of clay of mass m c. You then gently step onto the platform, and the platform settles down to a new equilibrium position, a vertical distance D below the original position. Assume that your mass is m p . D h a. You then pick up the lump of clay and hold it a height h above the platform. Upon releasing the clay the you and the platform will oscillate up and down; you notice that the clay strikes the platform after the platform has completed exactly one oscillation. Determine h in terms of any or all of k , D, the masses m p and m c, the acceleration of free fall g, and any necessary numerical constants. You must express your answer in the simplest possible form. b. Assume the resulting collision between the clay and the platform is completely inelastic. Find the ratio of the amplitude of the oscillation of the platform before the collision ( A i) and the amplitude of the oscillations of the platform after the collision ( A f ). Determine A f /A i in terms of any or all of k , D , the masses m p and m c, the acceleration of free fall g, and any necessary numerical constants. c. Sketch a graph of the position of the platform as a function of time, with t = 0 corresponding to the moment when the clay is dropped. Show one complete oscillation after the clay has collided with the platform. d. The above experiment is only possible if the ratio m c/m p is smaller than some critical value r c , otherwise the clay will hit the platform before one complete oscillation. An estimate for the value of the critical ratio r c can be obtained by assuming the clay hits the platform after exactly one-half of an oscillation. Assume that h is the same as is determined by part (a), and use this techniqu e to determine r c in terms of any or all of k , D, the mass m p, the acceleration of free fall g, and any necessary numerical constants. e. Is this estimate for r c too large or too small? You must defend your answer with an appropriate diagram. Solution Stepping on the platform will lower it a distance D. This means that the spring constant of the platform spring is given by kD = m pg. Copyright c 2008 American Association of Physics Teachers
