AAPT UNITED STATES PHYSICS TEAM AIP 2012 DO NOT DISTRIBUTE THIS PAGE

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1 2012 Semifina Exam 1 AAPT UNITED STATES PHYSICS TEAM AIP 2012 Semifina 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 aowed 90 minutes. Part B has two questions and is aowed 90 minutes. The first page that foows 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 ony aowed to do one part at a time, and may not work on other parts, even if they have time remaining. Aow 90 minutes to compete Part A. Do not et students ook at Part B. Coect the answers to Part A before aowing the examinee to begin Part B. Examinees are aowed a 10 to 15 minute break between parts A and B. Aow 90 minutes to compete Part B. Do not et students go back to Part A. Ideay the test supervisor wi divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-4), and Part B (pages 6-7). Examinees shoud be provided parts A and B individuay, athough they may keep the cover sheet. The supervisor must coect a examination questions, incuding the cover sheet, at the end of the exam, as we 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 Apri 1, Examinees are aowed cacuators, but they may not use symboic math, programming, or graphic features of these cacuators. Cacuators may not be shared and their memory must be ceared of data and programs. Ce phones, PDA s or cameras may not be used during the exam or whie the exam papers are present. Examinees may not use any tabes, books, or coections of formuas. Pease provide the examinees with graph paper for Part A.

2 2012 Semifina Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP 2012 Semifina Exam INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Work Part A first. You have 90 minutes to compete a four probems. Each question is worth 25 points. Do not ook at Part B during this time. After you have competed Part A you may take a break. Then work Part B. You have 90 minutes to compete both probems. Each question is worth 50 points. Do not ook at Part A during this time. Show a your work. Partia credit wi 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/tota pages for this probem, in the upper right hand corner of each page. For exampe, AAPT ID # Doe, Jamie A1-1/3 A hand-hed cacuator may be used. Its memory must be ceared of data and programs. You may use ony the basic functions found on a simpe scientific cacuator. Cacuators may not be shared. Ce phones, PDA s or cameras may not be used during the exam or whie the exam papers are present. You may not use any tabes, books, or coections of formuas. Questions with the same point vaue are not necessariy of the same difficuty. In order to maintain exam security, do not communicate any information about the questions (or their answers/soutions) on this contest unti after Apri 1, Possiby Usefu Information. You may use this sheet for both parts of the exam. g = 9.8 N/kg G = N m 2 /kg 2 k = 1/4πɛ 0 = N m 2 /C 2 k m = µ 0 /4π = 10 7 T m/a c = m/s k B = J/K N A = (mo) 1 R = N A k B = 8.31 J/(mo K) σ = J/(s m 2 K 4 ) e = C 1eV = J h = J s = ev s m e = kg = MeV/c 2 (1 + x) n 1 + nx for x 1 sin θ θ 1 6 θ3 for θ 1 cos θ θ2 for θ 1

3 2012 Semifina Exam Part A 3 Part A Question A1 A newy discovered subatomic partice, the S meson, has a mass M. When at rest, it ives for exacty τ = seconds before decaying into two identica partices caed P mesons (peons?) that each have a mass of αm. a. In a reference frame where the S meson is at rest, determine i. the kinetic energy, ii. the momentum, and iii. the veocity of each P meson partice in terms of M, α, the speed of ight c, and any numerica constants. b. In a reference frame where the S meson traves 9 meters between creation and decay, determine i. the veocity and ii. kinetic energy of the S meson. Write the answers in terms of M, the speed of ight c, and any numerica constants. Soution a. Appy conservation of four momentum. For the S meson, we have p S c = (E s, 0) and for the two P mesons we have p P c = (E p, ±p), where p is the magnitude of the (reativistic) three momentum of the P mesons. This yieds E S = 2E P We must aso satisfy the reation E 2 = p 2 c 2 + m 2 c 4 for each partice, so and E S 2 = M 2 c 4 E P 2 = p 2 c 2 + α 2 M 2 c 4. Therefore, the kinetic energy of each P meson is K P = E P αmc 2 = 1 2 Mc2 αmc 2 = ( ) 1 2 α Mc 2.

4 2012 Semifina Exam Part A 4 Square the energy conservation expression, and combine with the momentum/energy/mass reations: 1 4 M 2 c 4 = p 2 c 2 + α 2 M 2 c 4, ( ) 1 4 α2 M 2 c 4 = p 2 c α2 Mc 2 = pc. The veocity of each P meson wi be found from the reativistic three momentum, p = mγv and the reativistic energy, so E = γmc 2 pc E = mcγv γmc 2 = β. Putting in the vaues for the P meson, v = c 1 4 α2 Mc 2 1 = c 1 4α 2 2 Mc2 b. From reativistic kinematics, so Ca this k for now. Then Combine, d = vt = vγτ, v c γ = d cτ k = βγ, k 2 = β 2 γ 2, k 2 = β 2 1 β 2, k 2 (1 β 2 ) = β 2, k 2 = (1 + k 2 )β 2, k k 2 = β 2. v = c d 2 c 2 τ 2 + d 2

5 2012 Semifina Exam Part A 5 so Then 9 2 v = c = c 2 γ = 1/ 1 1/2 = 2 It isn t much work to find the kinetic energy, c. This is a veocity addition probem, so K = (γ 1)Mc 2 = ( 2 1)Mc 2. v = v S + v S 1 + v S 2 /c 2 or, using the numbers from the first part of the probem, 1 4α 2 v = c 1 2α 2

6 2012 Semifina Exam Part A 6 Question A2 An idea (but not necessariy perfect monatomic) gas undergoes the foowing cyce. The gas starts at pressure P 0, voume V 0 and temperature T 0. The gas is heated at constant voume to a pressure αp 0, where α > 1. The gas is then aowed to expand adiabaticay (no heat is transferred to or from the gas) to pressure P 0 The gas is cooed at constant pressure back to the origina state. The adiabatic constant γ is defined in terms of the specific heat at constant pressure C p and the specific heat at constant voume C v by the ratio γ = C p /C v. a. Determine the efficiency of this cyce in terms of α and the adiabatic constant γ. As a reminder, efficiency is defined as the ratio of work out divided by heat in. b. A ab worker makes measurements of the temperature and pressure of the gas during the adiabatic process. The resuts, in terms of T 0 and P 0 are Pressure units of P Temperature units of T Pot an appropriate graph from this data that can be used to determine the adiabatic constant. c. What is γ for this gas? Soution a. Labe the end points as 0, 1, and 2. A quick appication of P V = nrt requires that T 1 = αt 0. It takes more work to do the process 1 2; it is acceptabe to simpy state the adiabatic aw of P V γ = constant; if you don t know this, you wi need to derive it. In the case that you know the adiabatic process aw, P 1 V γ 1 = P 2V γ 2 = αp 1V γ 2, so that V 2 = V 1 (α) 1 γ. Another quick appication of P V = nrt requires that T 2 = (α) 1 γ T 0. Heat enters the gas during isochoric process 0 1, so Q in = nc v T = nc v (α 1)T 0 Heat exits the system during process 2 0, so Q out = nc p T = nc p (α 1/γ 1)T 0

7 2012 Semifina Exam Part A 7 We ony consider absoute vaues, and insert negative signs ater as needed. The work done is the difference, so and the efficiency is then This can be greaty simpified to W = Q in Q out = nc v (α 1)T 0 nc p (α 1/γ 1)T 0 e = C v(α 1) C p (α 1/γ 1) C v (α 1) e = 1 γ α1/γ 1 α 1 b. Aong an adiabatic path, the reationship between pressure and temperature is given by so As such, Note that, for an idea gas, P V γ = constant P P T γ 1 γ γ γ 1 = = constant P T γ γ 1 ( T P ) γ C p/c v C p /C v 1 = C p R This means that we want to pot a og-og pot with og T horizonta and og P vertica. The sope of the graph wi be C p /R. For the data given, C p = (7/2)R, so γ = 7/5.

8 2012 Semifina Exam Part A 8 Question A3 This probem inspired by the 2008 Guangdong Province Physics Oympiad Two infinitey ong concentric hoow cyinders have radii a and 4a. Both cyinders are insuators; the inner cyinder has a uniformy distributed charge per ength of +λ; the outer cyinder has a uniformy distributed charge per ength of λ. An infinitey ong dieectric cyinder with permittivity ɛ = κɛ 0, where κ is the dieectric constant, has a inner radius 2a and outer radius 3a is aso concentric with the insuating cyinders. The dieectric cyinder is rotating about its axis with an anguar veocity ω c/a, where c is the speed of ight. Assume that the permeabiity of the dieectric cyinder and the space between the cyinders is that of free space, µ 0. a. Determine the eectric fied for a regions. b. Determine the magnetic fied for a regions. Soution a. Consider a Gaussian cyinder of radius r and ength centered on the cyinder axis. Gauss s Law states that E da = q enc ɛ 0 2πrE = λ enc ɛ 0 E = λ enc 2πrɛ 0 where λ enc is the encosed inear charge density. The fied due to the hoow cyinders aone is therefore 0 r < a E appied = λ 2πrɛ 0 a < r < 4a 0 r > 4a The fied within the dieectric is reduced by a factor κ, so that in tota 0 r < a E = λ 2πrɛ 0 λ 2πrκɛ 0 λ a < r < 2a 2a < r < 3a 2πrɛ 0 3 < r < 4a 0 r > 4a

9 2012 Semifina Exam Part A 9 b. We can appy the resuts of the previous section to obtain the encosed charge density λ enc as a function of radius: 0 r < a Defining λ λ enc = λ i = λ κ a < r < 2a 2a < r < 3a λ 3 < r < 4a 0 r > 4a ( 1 1 ) λ κ we concude that a charge density λ i exists on the inner surface of the dieectric, a charge density λ i on the outer surface, and no charge on the interior. As with the case of a very ong soenoid, we expect the magnetic fied to be entirey parae to the cyinder axis, and to go to zero for arge r. Consider an Amperian oop of ength extending aong a radius, the inner side of which is at radius r and the outer side of which is at a very arge radius. We have on this oop B d = µ 0 I enc Letting B now be the magnetic fied at radius r, B = µ 0 I enc B = µ 0I enc For r > 3a, I enc = 0, since the charge on the hoow cyinders is not moving. For 2a < r < 3a, the oop now encoses the outer surface of the dieectric. In time 2π ω a charge λ i passes through the oop, so the current due to the outer surface is I out = λ iω 2π and thus this is I enc for 2a < r < 3a. For r < 2a, the oop now encoses both surfaces of the dieectric; the inner surface contributes a current that exacty cances the outer one, so again I enc = 0. Putting this together, 0 r < 2a µ B = 0 ω 2π λ i 2a < r < 3a 0 r > 3a or, using our expression for λ i, 0 r < 2a ( ) B = 1 1 µ0 ωλ κ 2π 2a < r < 3a 0 r > 3a

10 2012 Semifina Exam Part A 10 Question A4 Two masses m separated by a distance are given initia veocities v 0 as shown in the diagram. The masses interact ony through universa gravitation. v 0 v 0 a. Under what conditions wi the masses eventuay coide? b. Under what conditions wi the masses foow circuar orbits of diameter? c. Under what conditions wi the masses foow cosed orbits? d. What is the minimum distance achieved between the masses aong their path? Soution a. In order for the masses to coide, the tota anguar momentum of the system must be zero, which ony occurs if v 0 = 0. b. In this case, the masses undergo uniform circuar motion with radius 2 and speed v 0, so that Gm 2 2 = mv 0 2 Gm v 0 2 = 2 c. The masses foow cosed orbits if they do not have enough energy to escape, i.e. if the tota energy of the system is negative. The tota energy of the system is 2 so that the condition required is mv 0 2 Gm2 mv 2 0 Gm2 < 0 Gm v 2 0 > 1

11 2012 Semifina Exam Part A 11 d. Note that the masses wi aways move symmetricay about the center of mass. Thus, in order to be at minimum separation, their veocities must be perpendicuar to the ine joining them (and wi be oppositey directed). Let the minimum separation be d, and et the speed of each mass at minimum separation be v. L = 2mv d 2 = mvd The initia anguar momentum is ikewise mv 0, and so by conservation of anguar momentum mvd = mv 0 By conservation of energy v = v 0 d mv 0 2 Gm2 = mv2 Gm2 d v 0 2 Gm = v 2 Gm d Combining these, so that v 2 0 Gm 2 = v 2 0 d 2 Gm d ( 1 Gm ) ( ) d 2 v 2 + Gm ( ) d 0 v 2 1 = 0 0 ( ) (( d 1 1 Gm ) ) d v = 0 d = or d = Gm v The second root is ony sensibe if Gm v 2 0 > 1, and is ony smaer than the first if Gm v 2 0 > 2. (Note that both of these resuts make sense in ight of the previous ones.) Thus the minimum separation is if Gm v and otherwise. Gm v 2 1 0

12 2012 Semifina Exam Part A 12 STOP: Do Not Continue to Part B If there is sti time remaining for Part A, you shoud review your work for Part A, but do not continue to Part B unti instructed by your exam supervisor.

13 2012 Semifina Exam Part B 13 Part B Question B1 A partice of mass m moves under a force simiar to that of an idea spring, except that the force repes the partice from the origin: F = +mα 2 x In simpe harmonic motion, the position of the partice as a function of time can be written Likewise, in the present case we have for some appropriate functions f 1 and f 2. x(t) = A cos ωt + B sin ωt x(t) = A f 1 (t) + B f 2 (t) a. f 1 (t) and f 2 (t) can be chosen to have the form e rt. What are the two appropriate vaues of r? b. Suppose that the partice begins at position x(0) = x 0 and with veocity v(0) = 0. What is x(t)? c. A second, identica partice begins at position x(0) = 0 with veocity v(0) = v 0. The second partice becomes coser and coser to the first partice as time goes on. What is v 0? a. We have Soution ma = mα 2 x d 2 x dt 2 α2 x = 0 As with the case of simpe harmonic motion, we insert a tria function, in this case x(t) = Ae rt : d 2 dt 2 Aert α 2 Ae rt = 0 r 2 Ae rt α 2 Ae rt = 0 r 2 α 2 = 0 r = ±α b. We have and therefore x(t) = Ae αt + Be αt v(t) = αae αt αbe αt Inserting our initia vaues, x(0) = A + B = x 0

14 2012 Semifina Exam Part B 14 v(0) = αa αb = 0 These equations have soution and therefore c. This time our initia vaues are A = B = x 0 2 x(t) = x 0 2 eαt + x 0 2 e αt x(0) = A + B = 0 v(0) = αa αb = v 0 with soution Therefore A = v 0 2α B = v 0 2α x(t) = v 0 2α eαt x 0 2α e αt After a ong time, the second (e αt ) term wi become negigibe. Thus, the second partice wi approach the first partice if the first term matches: x 0 2 eαt = v 0 2α eαt v 0 = αx 0

15 2012 Semifina Exam Part B 15 Question B2 For this probem, assume the existence of a hypothetica partice known as a magnetic monopoe. Such a partice woud have a magnetic charge q m, and in anaogy to an eectricay charged partice woud produce a radiay directed magnetic fied of magnitude B = µ 0 q m 4π r 2 and be subject to a force (in the absence of eectric fieds) F = q m B A magnetic monopoe of mass m and magnetic charge q m is constrained to move on a vertica, nonmagnetic, insuating, frictioness U-shaped track. At the bottom of the track is a wire oop whose radius b is much smaer than the width of the U of the track. The section of track near the oop can thus be approximated as a ong straight ine. The wire that makes up the oop has radius a b and resistivity ρ. The monopoe is reeased from rest a height H above the bottom of the track. Ignore the sef-inductance of the oop, and assume that the monopoe passes through the oop many times before coming to a rest. a. Suppose the monopoe is a distance x from the center of the oop. What is the magnetic fux φ B through the oop? b. Suppose in addition that the monopoe is traveing at a veocity v. What is the emf E in the oop? c. Find the change in speed v of the monopoe on one trip through the oop. d. How many times does the monopoe pass through the oop before coming to a rest? e. Aternate Approach: You may, instead, opt to find the above answers to within a dimensioness mutipicative constant (ike 2 3 or π2 ). If you ony do this approach, you wi be abe to earn up to 60% of the possibe score for each part of this question. You might want to make use of the integra 1 (1 + u 2 ) 3 du = 3π 8 or the integra π 0 sin 4 θ dθ = 3π 8 Soution Version 1 The magnetic fied around a monopoe is given by B = µ 0 q m 4π r 2

16 2012 Semifina Exam Part B 16 The fux through the oop wi then be Φ B = B d A = 1 2 µ 0q m sin θ dθ where θ is the ange between a ine aong the axis of the oop and a ine drawn between the monopoe and any point on the rim of the oop. It is easy to see that dφ B = 1 2 µ 0q m sin θ From Faraday s aw, we have that a changing fux wi induce a current I in the oop. dφ B dt = IR, where R is the resistance of the oop. We figure R out ater. The induced current wi create a magnetic fied that wi oppose the monopoe motion. We need to use the aw of Biot & Savart to find that fied. Aong the axis of the oop, we have d B = µ 0I 4π d r r 3, where r is a vector connecting the monopoe with some point on the rim of the oop. Ony components of B parae to the axis of the oop wi survive, so we can concern ourseves with db = µ 0I 4π d r 2 sin θ The integra is trivia; d is around the circumference; nothing ese changes, so B = µ 0Ib 2r 2 It is better to think in terms of b, the radius of the oop, than it is to dea with r, the distance from the rim of the oop to the monopoe. In that case, sin θ = b r, sin θ so The force on the monopoe is then B = µ 0I 2b (sin θ)3 µ 0 I F = q m B = q m 2b (sin µ 0 θ)3 = q m 2b (sin 1 dφ B θ)3 R dt Note than mutipying through by dt gives an expression that is reated to the change in momentum, Using the provided integra, µ 0 dp = q m 2b (sin 1 θ)3 R dφ 2 B = q µ 2 0 m 4bR (sin θ)4 dθ p = q m 2 µ 0 2 4bR 3π 8

17 2012 Semifina Exam Part B 17 as the monopoe moves from one side to the other. If the monopoe started from rest a distance H above the oop, then the initia energy of the system is mgh, and the initia momentum when passing through the oop (assuming there is no oop) is then p 0 = m 2gH The monopoe wi ose p from the momentum on each pass through the oop, so the number of times it passes through the oop N is or N = p ( 0 p = m qm 2 µ 2 0 2gH 4bR N = 32bRm 2gH 3πµ 0 2 q m 2 ) 1 3π 8 Oh, we sti need to do R. Since a b, we can treat it as a ong thin cyindrica wire, and so we finay get R = ρ2πb πa 2 N = 64b2 ρm 2gH 3πµ 0 2 q m 2 a 2 Version 2 Instead of force, focus on the power dissipated in the oop, which is The energy is ost from the partice: Combining these, and since v = dx dt, P = P = µ 0 2 q m 2 a 2 b 3 8ρ P = E2 R ( 1 2 µ 0q m b 2 (b 2 +x 2 ) 3 2 2πbρ πa 2 1 (b 2 + x 2 ) 3 P = d 1 dt 2 mv2 P = mv dv dt ) 2 dx dt ( ) dx 2 dt dv dt = µ 0 2 q 2 m a 2 b 3 1 dx 8ρm (b 2 + x 2 ) 3 dt Use the provided integra (or do a trig subsitution that gives the other provided integra), and continue.