2007 Semifinal Exam Solutions 1 AAPT UNITEDSTATES PHYSICS TEAM AIP 2007 Solutions to Problems Part A Question 1 a.There is a high degree of symmetry present.Points b,c,and e are at the same potential; similarly,points d,f,and g are at the same potential.The circuit then reduces to a series connection of three parallel resistor clusters. The three parallel clusters have effective resistances of 1 8/5 and 1/3 The effective resistance of the circuit is then 44/15 b.After a long time no current will flow through the branches of the circuits containing capaci- tors.The circuit then reduces to a parallel connection of three series resistor clusters. The effective resistance of the circuit is 4 the current through the circuit is then (12V)/(42)=3A. (A1-1) The three branches are identical;each then carries 1 A. The potential drop across each capacitor is the same as the potential drop across the 8 resistors,so Vc=(1A)(82)=8V. (A1-2) Finally,the charge on each capacitor is Q=(15F)(8V)=120μC. (A1-3) Question 2 a.Two parts,solved individually. i.Isochoric compression: P1-T P-T (A2-1) so T Po (A2-2) Copyright C2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 1 AAPT UNITED STATES PHYSICS TEAM AIP 2007 Solutions to Problems Part A Question 1 a. There is a high degree of symmetry present. Points b, c, and e are at the same potential; similarly, points d, f, and g are at the same potential. The circuit then reduces to a series connection of three parallel resistor clusters. The three parallel clusters have effective resistances of 1 Ω, 8/5 Ω, and 1/3 Ω. The effective resistance of the circuit is then 44/15 Ω. b. After a long time no current will flow through the branches of the circuits containing capacitors. The circuit then reduces to a parallel connection of three series resistor clusters. The effective resistance of the circuit is 4 Ω, the current through the circuit is then (12 V)/(4 Ω) = 3 A. (A1-1) The three branches are identical; each then carries 1 A. The potential drop across each capacitor is the same as the potential drop across the 8 Ω resistors, so VC = (1 A)(8 Ω) = 8 V. (A1-2) Finally, the charge on each capacitor is Q = (15 µF)(8 V) = 120 µC. (A1-3) Question 2 a. Two parts, solved individually. i. Isochoric compression: Pf Pi = Tf Ti , (A2-1) so T = Pcr P0 T0. (A2-2) Copyright c 2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 2 ii.Adiabatic compression: PVY const, (A2-3) where y=Cp/Cv =(Cv+1)/Cv =5/3 for a monatomic gas.Consequently PoLo Per LY, (A2-4) and then Po 5/3 L=Lo (A2-5) b.The normal pressure on the bullet comes from or E. P=- (A2-6) Therefore,the normal force on the bullet is N=rE×(2mrh, (A2-7) and finally the force of friction is uFN.The force due to the pressure difference between the inside of the barrel and the outside must equal the normal force,so (xr2)(Per -Po)=2whu 6r E, (A2-8) and then P-=乃+2E (A2-9) Question 3 a.If the sphere has radius r,it has charge 4 9= 3Tpr3 (A3-1) and thus its surface is at electrostatic potential V=9 02 (A3-2) 4πe0r3e0 To increase the radius by dr,an additional charge dg =4xr2dr must be brought in from infinity,requiring work dU =V dq dr 3e0 (A3-3) Thus to grow the sphere from r=0 to r =R requires U= hrip dr 0 3e0 150 (A3-4) Copyright C2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 2 ii. Adiabatic compression: P V γ = const, (A2-3) where γ = CP /CV = (CV + 1)/CV = 5/3 for a monatomic gas. Consequently P0L γ 0 = PcrL γ , (A2-4) and then L = L0 P0 Pcr 5/3 . (A2-5) b. The normal pressure on the bullet comes from P = δr rc E. (A2-6) Therefore, the normal force on the bullet is FN = δr rc E × (2πrch), (A2-7) and finally the force of friction is µFN . The force due to the pressure difference between the inside of the barrel and the outside must equal the normal force, so (πr2 c )(Pcr − P0) = 2πhµ δr E, (A2-8) and then Pcr = P0 + 2µEh r 2 c δr. (A2-9) Question 3 a. If the sphere has radius r, it has charge q = 4 3 πρr3 (A3-1) and thus its surface is at electrostatic potential V = q 4π0r = ρr2 30 (A3-2) To increase the radius by dr, an additional charge dq = 4πr2dr must be brought in from infinity, requiring work dU = V dq = 4πr4ρ 2 30 dr (A3-3) Thus to grow the sphere from r = 0 to r = R requires U = Z R 0 4πr4ρ 2 30 dr = 4πR5ρ 2 150 (A3-4) Copyright c 2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 3 b.Each drop has volume Va=R3,so the number of drops is n二Va (A3-5) Since we are ignoring inter-drop forces,the total energy of the drops is simply the sum of the energies of each individual drop: Ue.tot =nU 、V_4标RE_Ry 专πR3150 (A3-6) 5e0 c.Each drop has surface area 4n R2 and thus surface tension energy 4 R2y.As before,the total energy due to surface tension is just the sum of the energies of the individual drops: U,=4Rm=4nR元=7 (A3-7) d.The total potential energy from both sources is Utot R2p2 (A3-8) 5e0 Equilibrium is reached when the total energy is a minimum;since U-oo at both R0 and R-oo,it must have an interior minimum. d Utot 2Rp2 R2 (A3-9) dR 5e0 Setting this equal to zero, 2Rp2 1 2 (A3-10) 5e0 R3= 15y0 2p2 (A3-11) R 15Ye0 2p2 (A3-12) Question 4 a.The electric field between the plates is given by E=V/d.The force on the charged ball is then F=Eg=Va/d.The acceleration of the ball is a =Vq/md. Kinematics gives us d=at2/2 for the time of flight.So t=V2d/a V2md2/qV. (A41) b.The kinetic energy collected by a ball will be K=qV as it moves between the plates.That's what will be dissipated. Copyright C2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 3 b. Each drop has volume Vd = 4 3 πR3 , so the number of drops is n = Vf Vd = Vf 4 3 πR3 (A3-5) Since we are ignoring inter-drop forces, the total energy of the drops is simply the sum of the energies of each individual drop: Ue,tot = nU = Vf 4 3 πR3 4πR5ρ 2 150 = R2ρ 2 50 Vf (A3-6) c. Each drop has surface area 4πR2 and thus surface tension energy 4πR2γ. As before, the total energy due to surface tension is just the sum of the energies of the individual drops: Us,tot = 4πR2 γn = 4πR2 γ Vf 4 3 πR3 = 3γ R Vf (A3-7) d. The total potential energy from both sources is Utot = R2ρ 2 50 + 3γ R ! Vf (A3-8) Equilibrium is reached when the total energy is a minimum; since U → ∞ at both R → 0 and R → ∞, it must have an interior minimum. d dRUtot = 2Rρ2 50 − 3γ R2 ! Vf (A3-9) Setting this equal to zero, 2Rρ2 50 = 3γ R2 (A3-10) R 3 = 15γ0 2ρ 2 (A3-11) R = 15γ0 2ρ 2 1 3 (A3-12) Question 4 a. The electric field between the plates is given by E = V /d. The force on the charged ball is then F = Eq = V q/d. The acceleration of the ball is a = V q/md. Kinematics gives us d = at2/2 for the time of flight. So t = q 2d/a = q 2md2/qV . (A4-1) b. The kinetic energy collected by a ball will be K = qV as it moves between the plates. That’s what will be dissipated. Copyright c 2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 4 c.The current is given by I=AQ/At.The total number of balls is N =noA,where A is the surface area of a plate.The charge AQ is then AQ noqA,so the current is 1-兴- n09.A (A42) We can't stop here,since this is not in terms of the allowed variables.The problem is A and d,but since C=eoA/d,we have n09.A V2mdqV (A43) A qV (A44) 2m1 C qv = 2m (A4-5) EO d.R=V/I,so R=7= EOV 2m (A4-6) Cnoq We can simplify,slightly,with R= EO 2mV Cnoq (A4-7) 9 e.P=VI,so P=vEnoQV qV Eo2n02C2q3V3 (A4-8) EO 2m 2m Part B Question 1 a.To not slip,from a free-body diagram,we must have umg cos6≥mg sin0 (B1-1) so μ≥tana. (B1-2) Therefore ue=tan 6 and hence tan = 2 (B1-3) b.In one cycle the energy input into the system is MgLsin0, (B1-4) the energy of the block dropping. Copyright C2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 4 c. The current is given by I = ∆Q/∆t. The total number of balls is N = n0A, where A is the surface area of a plate. The charge ∆Q is then ∆Q = n0qA, so the current is I = ∆Q ∆t = n0qA p 2md2/qV . (A4-2) We can’t stop here, since this is not in terms of the allowed variables. The problem is A and d, but since C = 0A/d, we have I = n0qA p 2md2/qV , (A4-3) = A d n0q s qV 2m , (A4-4) = C 0 n0q s qV 2m . (A4-5) d. R = V /I, so R = V I = 0V Cn0q s 2m qV . (A4-6) We can simplify, slightly, with R = 0 Cn0q s 2mV q . (A4-7) e. P = V I, so P = V C 0 n0q s qV 2m = s 0 2n0 2C2q 3V 3 2m . (A4-8) Part B Question 1 a. To not slip, from a free-body diagram, we must have µmg cos θ ≥ mg sin θ (B1-1) so µ ≥ tan θ. (B1-2) Therefore µc = tan θ and hence µ = tan θ 2 . (B1-3) b. In one cycle the energy input into the system is MgLsin θ, (B1-4) the energy of the block dropping. Copyright c 2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 5 The energy loss on the way up is Lμng cos8 (B1-5) and the energy loss on the way down is Lμ(m+M)gcos8 (B1-6) Thus MgLsin0 Lumg cos0+Lu(m+M)g cos0 (B1-7） and since 2u cos sin6, M m+M 2 2 (B1-8) =2m, (B1-9) R=M/m=2. (B1-10) c.The period of a mass m oscillating on a spring of spring constant k is T=2mV下1 m (B1-11) In this case,the friction force is constant on both the up and down trips,and so each trip is simple harmonic (with different equilibrium points).Hence T6=1 m 3m (B1-12) T= m 2mVk (B1-13) T0/T'= 1+V3 2 (B1-14) d.As mentioned in part (c),both the up and down trips are simple harmonic,this time with a mass of m both ways.The equilibrium points for the two trips are different,however.On the up trip,the equilibrium point is clearly at a distance L/2 from B,since the plate stops at both B and A and hence those are the endpoints of the oscillation and the equilibrium is halfway between.For the trip down,the equilibrium point will shift by a distance y such that ky=2μng cos0=ng sin0 (B1-15) because 2umg cos6 is the difference between the friction forces on the trip up and the trip down. The place where the plate finally comes to a stop is the first place that is at the end of an oscillation (either up or down)and where the total force being exerted by gravity and the spring is less than the maximal force of friction.For that to happen,the plate needs to not have gone past the other equilibrium point during that oscillation. So we start by determining where the endpoints of the oscillations are.For the first trip up these are B and A.For the following trip down,the plate stops at a distance of(2mgsin)/k from B (because the equilibrium shifts up by (mgsin)/k.For the following trip up,the Copyright C2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 5 The energy loss on the way up is Lµmg cos θ (B1-5) and the energy loss on the way down is Lµ(m + M)g cos θ (B1-6) Thus MgLsin θ = Lµmg cos θ + Lµ(m + M)g cos θ (B1-7) and since 2µ cos θ = sin θ, M = m 2 + m + M 2 , (B1-8) = 2m, (B1-9) R = M/m = 2. (B1-10) c. The period of a mass m oscillating on a spring of spring constant k is T = 2π rm k . (B1-11) In this case, the friction force is constant on both the up and down trips, and so each trip is simple harmonic (with different equilibrium points). Hence T0 = π rm k + π r 3m k , (B1-12) T 0 = 2π rm k , (B1-13) T0/T0 = 1 + √ 3 2 . (B1-14) d. As mentioned in part (c), both the up and down trips are simple harmonic, this time with a mass of m both ways. The equilibrium points for the two trips are different, however. On the up trip, the equilibrium point is clearly at a distance L/2 from B, since the plate stops at both B and A and hence those are the endpoints of the oscillation and the equilibrium is halfway between. For the trip down, the equilibrium point will shift by a distance y such that ky = 2µmg cos θ = mg sin θ (B1-15) because 2µmg cos θ is the difference between the friction forces on the trip up and the trip down. The place where the plate finally comes to a stop is the first place that is at the end of an oscillation (either up or down) and where the total force being exerted by gravity and the spring is less than the maximal force of friction. For that to happen, the plate needs to not have gone past the other equilibrium point during that oscillation. So we start by determining where the endpoints of the oscillations are. For the first trip up these are B and A. For the following trip down, the plate stops at a distance of (2mg sin θ)/k from B (because the equilibrium shifts up by (mg sin θ)/k. For the following trip up, the Copyright c 2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 6 plate stops a distance L-(2mgsin)/k from B,since the equilibrium point is again in the middle of the incline.And so forth. Thus the stopping points are located at n(2mg sin0)/k and L-n(2mgsin0)/k (B1-16) for integer n.The plate will stop permanently once either n(2mgsin)/k L/2 (B1-17） or L-n(2mg sin0)/k<L/2+(mgsin0)/k, (B1-18) whichever happens first.(The first condition corresponds to going down and ending up above the midpoint at the end of the down trip,the second condition corresponds to going up and stopping below the upper equilibrium.)The second condition can be rewritten as n+2 (2mg sin0)/k L/2. (B1-19) Question 2 a.Magnetic Moments i.From Coulomb's Law, F= e 4TEo R2 (B2-1) ii.For circular motion, F=mev2 =me Rwo; (B2-2) R The force is provided by the Coulomb force,so meRwg e2 = 4mcoR2' (B2-3) 2 w0 = 4ncomeR3 (B2-4) iii.From the law of Biot and Savart, Be toi f ds× 3, (B2-5) Be 012rR R 4π z2+r2)3/21 (B2-6) HoiR2 223 (B2-7) For the current,i,we can write (B2-8) 2π Then Be= HoewoR2 4r3· (B2-9) Copyright C2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 6 plate stops a distance L − (2mg sin θ)/k from B, since the equilibrium point is again in the middle of the incline. And so forth. Thus the stopping points are located at n(2mg sin θ)/k and L − n(2mg sin θ)/k (B1-16) for integer n. The plate will stop permanently once either n(2mgsin)/k > L/2 (B1-17) or L − n(2mg sin θ)/k L/2. (B1-19) Question 2 a. Magnetic Moments i. From Coulomb’s Law, F = e 2 4π0R2 (B2-1) ii. For circular motion, F = mev 2 R = meRω2 0 , (B2-2) The force is provided by the Coulomb force, so meRω2 0 = e 2 4π0R2 , (B2-3) ω0 = s e 2 4π0meR3 (B2-4) iii. From the law of Biot and Savart, B~ e = µ0i 4π I d~s × ~r r 3 , (B2-5) Be = µ0i 4π 2πR R (z 2 + r 2) 3/2 , (B2-6) ≈ µ0iR2 2z 3 . (B2-7) For the current, i, we can write i = q t = eω0 2π . (B2-8) Then Be = µ0eω0R2 4πz3 . (B2-9) Copyright c 2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 7 iv.By substitution, ewoR m=2 (B2-10) b.Diamagnetism i.If half go one way and half go the other,M=0. ii.Additional force from magnetism, FB quBo eRwBo (B2-11) modifies previous central force problem to give meRw2= o 4noR±eRwoBo, (B2-12) where the positive sign corresponds to anticlockwise motion,the negative to clockwise motion. A little math, meR(uw2-w)=±eRwBo, (B2-13) me(w-wo)(w+wo)=tewBo; (B2-14) me(△w)(2wo）=±ewo Bo, (B2-15) where in the last line we have used the approximation wwo.Then Aw=±eBo -2me (B2-16) iii.The emf is given by △Φ△n 8-nAt-Ai (B2-17) but An/At is a measure of the number of turns made by the electron in a time interval △t,so △n_w0R_w0 △t=2xR=2元 (B2-18) Then ￡=BAmR2= oboR2 (B2-19) 2T iv.The change in kinetic energy is given by △K= △（m2r) (B2-20) meR2w△w, (B2-21) meR2wo△w, (B2-22) mewo R2 eBo (B2-23) 2me =eE. (B2-24) Copyright C2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 7 iv. By substitution, m = eω0R 2 . (B2-10) b. Diamagnetism i. If half go one way and half go the other, M = 0. ii. Additional force from magnetism, FB = qvB0 = eRωB0 (B2-11) modifies previous central force problem to give meRω2 = e 2 4π0R2 ± eRω0B0, (B2-12) where the positive sign corresponds to anticlockwise motion, the negative to clockwise motion. A little math, meR(ω 2 − ω 2 0 ) = ±eRωB0, (B2-13) me(ω − ω0)(ω + ω0) = ±eωB0, (B2-14) me(∆ω)(2ω0) = ±eω0B0, (B2-15) where in the last line we have used the approximation ω ≈ ω0. Then ∆ω = ± eB0 2me . (B2-16) iii. The emf is given by E = n ∆Φ ∆t = ∆n ∆t Φ, (B2-17) but ∆n/∆t is a measure of the number of turns made by the electron in a time interval ∆t, so ∆n ∆t = ω0R 2πR = ω0 2π . (B2-18) Then E = ω0 2π B0πR2 = 1 2 ω0b0R 2 . (B2-19) iv. The change in kinetic energy is given by ∆K = ∆ 1 2 meω 2R 2 , (B2-20) = meR 2ω ∆ω, (B2-21) ≈ meR 2ω0 ∆ω, (B2-22) = meω0R 2 ± eB0 2me , (B2-23) = eE. (B2-24) Copyright c 2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 8 v.AM Nom,where N is the number of atoms,and Am the change in magnetic moment in each.The change is △m=△( ewoR 2 (B2-25) eR △w, (B2-26) 2 e2R2Bo Ame (B2-27) 80 AM N2R2Bo (B2-28) Ame vi.Repelled,by Lenz's law. Copyright C2007 American Association of Physics Teachers
2007 Semifinal Exam Solutions 8 v. ∆M = Nδm, where N is the number of atoms, and ∆m the change in magnetic moment in each. The change is ∆m = ∆ eω0R 2 . (B2-25) = eR 2 ∆ω, (B2-26) = e 2R2B0 4me , (B2-27) so ∆M = N e 2R2B0 4me . (B2-28) vi. Repelled, by Lenz’s law. Copyright c 2007 American Association of Physics Teachers