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Physics 221 2005S Final Exam Solutions PHYSICS 221 Spring 2005 Final Exam Solutions May 3 2005 2:15pm—4:15pm Part A (18 Questions/Lectures 1-42) [56] In the figure below there are two vectors, 1 and 2. There exists a cross-product (“vector product”), C, of the vectors (i.e., C = 1 × 2). Calculate the direction of C. Note: • represents a vector pointing out of the page. × represents a vector pointing into the page. 1 × 2 Which of the following boxes best describes the direction of C ? Box B Box A Box C Both × • and correctly describe the direction of C Box D Both Box A and Box B correctly describe the direction of C Box E The cross-product has a magnitude equal to zero. Answer [B]: Using the right hand rule, if you put your right thumb in direction 1 and your index finger in direction 2, then your middle finger naturally points in the direction shown in box B. Page 1 of 19 Physics 221 2005S Final Exam Solutions [57] A particle of mass m = 10 g moves along the x-axis with acceleration ax = (3m / s 3 ) t The velocity of the particle at t = 0 is 3.0 m/s in the +x direction. What is the velocity of the particle at t = 4 s? (A) 12 m/s (B) 27 m/s (C) 37 m/s (D) 41 m/s (E) 51 m/s Answer [B]: The velocity at t=4 is given by 4 v(t ) = v(0) + ∫ a dt 0 [ ] = 3m / s + (3m / s 3 ) 12 t 2 t =4s t =0 = 27m / s The following information applies to questions [58] and [59]: A 50-g dry sock rotates stuck to the inner wall of a dryer drum. As a first approximation, you can think of the drum as made of a 2-kg hollow cylinder of radius R = 30 cm with a 1-kg disk of radius R = 30 cm attached to it on one side (see figure). Both the disk and the hollow cylinder have uniform density. The drum rotates at a constant angular of 100 rpm. Disk (cap) (1 kg, radius 30 cm) Hollow cylinder (2 kg, radius 30 cm) Open front (no cap) Page 2 of 19 Physics 221 2005S Final Exam Solutions [58] Determine the force on the sock by the drum wall when the sock is at the top of the drum. (A) 0.5 N pointing down (B) 0.5 N pointing up (C) 1.2 N pointing down (D) 1.2 N pointing up (E) 1.6 N pointing down Answer [C]: The angular momentum of the drum is ω = (100)(2π radians ) /(60s ) = 10.47 radians / s The acceleration of the sock is centripetal a=v²/r so the net force is Fnet=(mv²/r)(-j). The two forces on the sock are gravity and the normal force so Fnet = N − j mg Thus N = ˆjm g − rω 2 ( ) ( ) Putting in the numbers N = ˆj (.05kg ) (9.8m / s 2 ) − (0.3m)(10.46s −1 ) 2 = −1.15 ˆj It is therefore downwards and indeed must be downwards because the normal force can only push not pull. [59] What is the moment of inertia of the drum? (A) 0.135 kg m² (B) 0.180 kg m² (C) 0.270 kg m² (D) 0.225 kg m² (E)0.0675 kg m² Answer [D]: The back plate is a solid cylinder so the moment of inertia is I back = 12 mr 2 = 12 (1kg )(0.3m) 2 = 0.045 kg m 2 . The curved cylinder is a hollow cylinder so the moment of inertia is I curvedk = mr 2 = (2kg )(0.3m) 2 = 0.180 kg m 2 adding these two together I = I curved + I back = (.045 + .180)kg m 2 = .225kg m 2 Page 3 of 19 Physics 221 2005S Final Exam Solutions [60] The horizontal beam shown has uniform density and weighs 250N. It is connected to the wall with an ideal massless string as shown and supports a 500N weight from the end. What is the magnitude of the net force that the wall exerts on the beam if the system is in equilibrium? (A)1200N (D)1505N (B)1375N (E)1625N 13m 5m (C)1400N 250N θ 12m 500N Answer [D]: First we need to find the tension T in the 13m string. Let q be the angle between the string and the beam. We use the condition that the net torque about the hinge must be 0: τ = 0 = T (12m) sin θ − (12m)(500 N ) − (250 N )(6m) since sinθ=5/13 and cosθ=12/13 we can solve for T: T=1625N Now we can figure out the net force on the hinge. This can be calculated using the fact that the net force on the beam must be 0. x-component: 0 = Fnet x = Fhinge x − T cosθ y-component: 0 = Fnet y = Fhinge y + T sin θ − (500 N ) − (250 N ) Solving these two equations for the force of the hinge: Fhinge x = (1625 N )(12 / 13) = 1500 N Fhinge y = (750 N ) − (1625 N )(5 / 13) = 125 N The magnitude of the vector is thus Fhinge = (1500 N ) 2 + (125 N ) 2 = 1505 N Page 4 of 19 Physics 221 2005S Final Exam Solutions [61] A flywheel has moment of inertia I=8kg m2. At t=0 it rotates with a period of T=1.5s. A motor gradually speeds up the flywheel such that the period, T, shortens at a rate of dT/dt= −2.0x10−3. What is the rate of change of the kinetic energy of the flywheel at t=0? (A)94mW (B)140mW (C)187mW (D)376mW (E)423mW Answer [C]: Following the method in Problem 9.98 on Written Assignment 5, the total rotational energy is: I T2 The rate of change in kinetic energy is obtained using the chain rule: K = 12 Iω 2 = 2π 2 4π 2 dT P = K = 2π I d (1 / T ) / dt = − 3 T dt Plugging in the numbers from the question d dt P=− 2 ( 2 ) 4π 2 dT 4π 2 (8kg m 2 ) = − (−2.0 × 10− 3 ) = 187mW 3 3 T dt (1.5s) [62] A girl stands in an elevator that has a constant upwards acceleration over a distance of 20m. During the acceleration the normal force exerted by the floor of the elevator does 10.0kJ of work on the girl while gravity does –8.0kJ of work on her. What is the acceleration of the elevator? (A)2.45m/s² (B)4.90 m/s² (C)7.28 m/s² (E)Cannot be determined from the information given. (D)12.27 m/s² Answer [A]: Following question 6.60 from Written Assignment 4, Gravity does –8kJ of work over 20m so the girl’s weight is W=(8kJ/20m)=400N. The girl’s mass is m=W/g=40.8kg. The work done by the normal force is 10kJ so that the normal force is (10kJ)/(20m)=500N. The net force on the girl is thus Fnet=500N-400N=100N. The acceleration is thus given by Newton’s second law a=Fnet/m=2.45m/s². Page 5 of 19 Physics 221 2005S Final Exam Solutions [63] A cannon, located 60m from the base of a vertical 25m tall cliff, shoots a 15kg shell at an angle of 45° above the horizontal toward the cliff at a speed of 20m/s. What is the speed of the shell when it lands on the top of the cliff? Neglect air resistance. (A) 0m/s (B) 12.1m/s (C) 13.7m/s (E) The shell does not land on the top of the cliff (D) 20.0m/s Answer [E]: The kinetic energy of the shell is K=(1/2)(15kg)(20m/s)²=3kJ. The potential energy (taking U=0 at the bottom of the cliff) of the shell at the top of the cliff, 25m above the bottom is U=mgh=(15kg)(9.8m/s²)(25m)=3.675kJ>3kJ. The shell can therefore never make it to the top of the cliff. G G G G [64] If A = iˆ + 2 ˆj + 4 kˆ and B = 4iˆ − 4 ˆj + kˆ , what is the angle between A and B in radians? (A) 0 (B) 3π/4 (C) π/4 (D) π (E) π/2 G G Answer[E]: Taking the dot product: A ⋅ B = (iˆ + 2 ˆj + 4kˆ) ⋅ (4iˆ − 4 ˆj + kˆ) = 0 . The two vectors are therefore perpendicular and the angle between them is thus π/2. [65] Consider a physical pendulum consisting of a thin rod of uniform density, length L=2m and mass m=4kg. The rod is suspended from one end. What is the period of this pendulum? (A) 1.64s (B) 2.32s (C) 2.84s (D) 3.28s Answer [B]: The period of a physical pendulum is given by T = 2π d=L/2 and I=(1/3)mL² so T = 2π I (1 / 3)mL2 2L = 2π = 2π = 2.32s mgd (1 / 2)mgL 3g Page 6 of 19 (E) 4.02s I . In this case mgd Physics 221 2005S Final Exam Solutions [66] A spaceship travels through empty space at 300 km/s in the positive x direction and splits into two pieces. One of the pieces has twice the mass of the other one. They both come out at 20° with the initial direction (see figure). Find the velocity of the center of mass of the system after the separation. (A)+150 km/s (D)+460 km/s iˆ iˆ (B)+282 km/s (E)+564 km/s iˆ iˆ (C)+300 km/s iˆ y Part 1 spaceship 20° x 20° Part 2 Answer [C]: there is no external force on the system so the center of mass velocity does not change. Initially the center of mass velocity is +300 km/s iˆ so after the separation the center of mass velocity is also +300 km/s iˆ . [67] A 1kg car and a 2 kg car are constrained to move along a one dimensional air track. Initially the 1kg car has a velocity of + 3m / s while the 2kg car has a velocity of − 3m / s . The two cars collide and the collision is fully elastic. What is the velocity of the 1kg car after the collision? (A) −5m/s (B) −4m/s (C) −3m/s (D) −2m/s Answer [A]: The velocity of the center of mass is (1kg )(3m / s ) + (2kg )(−3m / s ) vcm = = −1m / s 1kg + 2kg The final velocity after a 1D elastic collision is given by vf=2vcm−vi=2(−1m/s)−3m/s=−5m/s. Page 7 of 19 (E) −1m/s Physics 221 2005S Final Exam Solutions [68]The point charges in the figures below are placed at distance s from points A and B. Compare the magnitude of the net electric field and the electric potential at points A and B. Assume that each system is very far from any other charges. Take the potential to be zero at infinity. Q + Q + s s 2Q + •A s •B (A) EA = EB ; VA = VB (B) EA > EB ; VA = VB (C) EA = EB ; VA > VB (D) EA < EB ; VA = VB (E) EA = EB ; VA < VB Answer[D]: Think of the 2Q object as two Q charges on top of each other. At point B both of the field vectors from the two Q charges are aligned so the net field is just the sum. At point A they are not aligned so the electric field is less. The contribution to the potential depends only on the distance so the potentials in both cases are the same. [69] A parallel plate capacitor is connected to a battery with EMF=V. If the space between the plates is filled with air the energy stored in the capacitor is Uair. If the space is filled with a dielectric having dielectric constant κ, the energy stored in the capacitor is Udielectric. What is the ratio Udielectric/Uair? (A) κ (B) κ2 (C) 1/κ (D) 1/κ2 (E) 1 (same energy in both cases) Answer[A]: If C is the capacitance with air then the capacitance with dielectric is κC. Here the voltage in the capacitor is the same in both cases so the energy stored in the air capacitor is Uair=(1/2)CV² while the energy stored in the dielectric capacitor is Udielectric=(1/2)(κC)V². The ratio Udielectric/Uair=κ. Page 8 of 19 Physics 221 2005S Final Exam Solutions [70] Determine the escape speed from the surface of a spherical asteroid of constant density with a radius of 700 m and a mass of 4×1012 kg. (A) 0.44m/s (B) 0.62m/s (C) 0.87m/s (D) 1.1m/s (E)1.2m/s Answer[C]: The escape speed is given by ve = 2GM / R = 2(6.7 × 10−11 N m 2 / kg 2 )(4 × 1012 kg ) /(700m) = 0.873m / s [71] Consider the following three systems: : Spherical Gaussian surface inside the shells Q Q Q A Empty conducting spherical shell with charge Q > 0 B A conducting spherical shell, uncharged, with a point-charge Q > 0 in its center. C Empty insulating spherical shell with charge Q > 0 uniformly distributed throughout the volume of the insulator Let ΦA, ΦB and ΦC be the net electric flux through a Gaussian spherical surface inside the shells (dotted lines in the figure). Which of the following is true? (A) ΦA = ΦB = ΦC = 0 (B) ΦA = ΦB = 0; ΦC > 0 (C) ΦA = 0; ΦB > 0; ΦC > 0 (D) ΦA > 0; ΦB = 0; ΦC > 0 (E) ΦA > 0; ΦB >0; ΦC > 0 Answer[B]: Within a conductor the electric field is 0 so the flux in case A and B are 0. In case C the surface encloses a positive finite amount of charge and therefore ΦA = ΦB = 0; ΦC > 0. Page 9 of 19 Physics 221 2005S Final Exam Solutions [72] In the circuit shown, what is the power dissipated by the 2Ω resistor? (A) 16W (B)32W (C)64W (D)128W (E)256W 2Ω 20V 1Ω 0.5Ω 1Ω Answer[D]: The two 1Ω resistors are in parallel and are equivalent to 0.5Ω. The equivalent resistance to the whole circuit is 2.5Ω. By ohms law, the current through the circuit is 8A=(20V)/(2.5Ω). All that current goes through the 2Ω resistor so that the power dissipated is P=I²R=(8A)²(2Ω)=128W. [73][Extra Credit] Two identical springs have spring constant k=100N/m. These springs are connected end to end as shown below, with the left end connected to a fixed wall. How much force does it take to extend the system of two springs 1cm beyond their unstretched length? (A)5N (B)2N (C)1N (D)0.5N k=100N/m (E)0.25N k=100N/m Force=? Unstreched Length 1cm Answer[D]: Following the logic of 13.96 from written assignment 6, the force which each spring exerts is the same. By symmetry, each of the springs stretches 0.5cm. Since the spring constant is 100N/m the force exerted by each spring is (100N/m)(.005cm)=0.5N. Page 10 of 19 Physics 221 2005S Final Exam Solutions Part B (6 Questions/Lectures 29-42) [74] In Figure 1, two infinite parallel conducting plates carrying charge densities σ and −σ are separated by a distance d. Let E1 be the magnitude of the electric field half way between the two plates. In Figure 2 the two identical plates are separated by a distance 2d and the electric field half way between the plates is E2. What is the ratio between E1 and E2? (A)E1:E2=4:1 −σ (B)E1:E2=2:1 E1 (C)E1:E2=1:1 (D)E1:E2=1:2 +σ −σ d (E)E1:E2=1:4 E2 2d Figure 1 Figure 2 Answer[C]: Just as in problem 23.38a on written assignment 8, note that the electric field of due to an infinite sheet of charge is independent of distance so E1=E2. [75] A particle with charge 1mC and mass 1kg is in a uniform electric field directed to the left. It is released from rest and moves to the left; after it has moved 50cm, the velocity of the particle is 2m/s. What is the magnitude of the electric field? (A) 250 V/m (B)500 V/m (C)1000 V/m (D)2000V/m (E)4000V/m Answer[E]: Just as in Problem 23.18 we see that the particle has gained a kinetic energy of K=(1/2)mv²=2J which is the work W that the electric field. The electric force is therefore F=W/d=(2J)/(0.5m)=4N. In terms of the electric field, E=F/q=(4N)/(1mC)=4000 V/m Page 11 of 19 +σ Physics 221 2005S Final Exam Solutions [76] A satellite of mass m orbits in a circular, near earth orbit (radius of orbit≈RE=radius of Earth). What is the minimum amount of work that must be done on the satellite in order to boost it to a circular orbit of radius 2 RE ? (ME=mass of Earth) 1 1 1 (A) 0 (B) GmME/RE (C) GmME/RE (D) GmME/RE (E) GmME/RE 2 4 8 Answer[D]: From the result of problem 12.66 on Assignment 7, the mechanical energy of a object in circular orbit is exactly half the potential energy. The mechanical energy of 1 the satellite in low earth orbit is thus − GmME/RE while the mechanical energy of the 2 1 satellite in the 2R orbit is − GmME/RE. The amount of work which must be done to 4 1 raise the satellite is thus GmME/RE. 4 Page 12 of 19 Physics 221 2005S Final Exam Solutions [77] In the circuit below, R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω and V0 = 100 V. Determine the magnitude of the current through R1. (A) 1.4 A (D) 2.4 A (B) 1.8 A (E) 2.9 A (C) 2.0 A Answer[B]: R1 and R2 are in parallel with equivalent resistance R12=1/(1/10Ω+1/20Ω)=6.67Ω. R3 is in series with that system so the equivalent resistance of the three resistors is R12+R3=36.67Ω. The current is therefore (100V)/(36.67Ω)=2.73A giving a voltage drop across R3 of (2.73A)(30Ω)=81.82V. The voltage across R1 is thus 18.18V=100V-81.82V hence the current through it is (18.18V)/(10Ω)=1.82A. Page 13 of 19 Physics 221 2005S Final Exam Solutions [78] Consider the two circuits depicted below. In circuit A the three capacitors are all 1pF and are arranged in a delta network between terminals P, Q and R. In circuit B three capacitors, each of which has capacitance CY , are arranged in a Y-network between terminals P, Q and R. For what value of CY are the two capacitor networks equivalent circuits? (A) CY=4pF (B) CY=3pF (C) CY=2pF (D) CY=1pF (E) CY=0.33pF Q P 1pF P Q CY CY 1pF 1pF CY R R Circuit B Circuit A Answer[B]: Following the method of problem 24.75 on assignment 8, if we connect P and Q in circuit A to an external EMF and disconnect R the equivalent capacitance is Ceq=1pF+1/(1/1pF+1/1pF)=1.5pF. If we hook up B the same way we must get the same equivalent capacitance. In circuit B we the capacitor attached to R is irrelevant so it is just two CY capacitors in series. Thus Ceq=CY/2. CY/2=1.5pF so CY=3pF. [79] [Extra Credit] In this problem, consider the electric field produced by two separate systems depicted in the figures below System A is a very long, solid insulating cylinder with radius R that has a cylindrical hole with radius R/2 bored along its entire length. The axis of the hole is parallel to the axis of the insulating cylinder and is at distance R/2 from the axis of the cylinder. The solid material of the cylinder has a uniform charge density ρ. Let X be the point at the center of the hole and EX is the magnitude of the electric field at point X. System B is a very long, solid insulating cylinder with radius R with no hole. The substance of system B also has uniform charge density ρ. In system B the point Y is located at distance R/2 from the center of the axis of the cylinder and EY is the magnitude of the electric field at point Y. Page 14 of 19 Physics 221 2005S Final Exam Solutions What is the correct relation between EX and EY? (A) EX :EY=1 : 1 (B) EX:EY=0 : 1 (C) EX:EY=1 : 2 (D) EX:EY=1 : 4 (E) EX:EY=2 : 1 R/2 R/2 R R R/2 X Y System A: End view of cylinder with hole System B: End view of cylinder without hole Answer[A]: Using the same method as problem 22.62, we can think of the charge distribution in system A as being a superposition of system B with a cylinder of charge density −ρ centered at X with radius R/2. By symmetry, the electric field produced by the negative cylinder at its center is 0 so the electric field at X is the same as Y. Page 15 of 19 Physics 221 2005S Final Exam Solutions Part C (4 Questions/Lab Final) [80]An electrophorus is equipped with an insulating plate that becomes positively charged when rubbed with tissue. Which metal of the following best describes the change in the charge on plate the insulating plate, when the metal plate gains 1 µC of charge as the electrophorus is operated normally. insulating handle (A) The insulating plate gains approximately 1 µC of positive charge. (B) The insulating plate gains approximately 1 µC of negative charge. (C) The insulating plate loses approximately 1 µC of positive charge. (D) The insulating plate loses approximately 1 µC of negative charge. (E) The charge on the insulating plate will be relatively unchanged, compared to 1 µC. insulating plate ELECTROPHORUS Answer[E]: The bulk of the charge on the metal plate is placed there by electrostatic induction. When the metal plate is placed on the charged insulating plate and grounded, the 1µC flows into the metal from the ground. The insulating plate is only responsible for the separation of charges between the metal plate and the ground but no significant charge is transferred directly from the insulating plate to the metal plate. Page 16 of 19 Physics 221 2005S Final Exam Solutions [81] As done in lab, a student, seat, chain, and two load cells are at rest and supported by cables attached to the upper (sensitive) fittings of the two load cells. The cell readings and those of the large protractors are given in the protractors table below. load cells LEFT RIGHT Cell reading 531 N 600 N Protractor reading 60° 50° 0 Y seat Use this information to estimate the X component of the force exerted by the left cell upon the end of the left chain. (A) 0 (The force must be zero, since nothing is accelerating.) (B) +266 N (C) −266 N (D) +460 N (E) −460 N Answer [E]: From the figure it should be clear that the force which the cell exerts on the chain is generally up and to the left so the answer should be negative. The magnitude of the total force is F=531N and paying attention to which way the protractor reads, the x component of that force is Fx=−Fsin(60º)=−460N. Page 17 of 19 0 chain X Physics 221 2005S Final Exam Solutions [82] Consider the motion of a cart like the one you used in lab, and which is adjusted to have significant frictional drag. Assume that the friction is well described by the "laws" of (dry) friction that you studied this semester. Also assume that the cart is given an initial velocity toward the left on the horizontal track and then released. (Assume the sensor gives positions relative to the X axis illustrated in the figure). Which of the following graphs best illustrates the velocity, VX , of the cart versus time after it leaves your hand? Vx Vx t 0 A Vx Vx t B Vx t C t D t E Answer[E]: Since it is initially moving to the left, the velocity must start at a negative value. The drag decreases the magnitude of the velocity so the curve has a positive slope and is linear since the acceleration due to friction should be independent of velocity. Only E looks like that. Page 18 of 19 Physics 221 2005S Final Exam Solutions [83] In the rotational motion experiment, the following graph was obtained when the wheel of the apparatus which you used was rotated by hand. Provide the most likely explanation for this particular graph. A) The wheel was rotated slowly, and the angular encoder on the wheel axis has a resolution of 1/200 of a revolution. B) The wheel was rotated at different velocities, increasing in steps as time went on. C) The wheel was rotated slowly, and the angular encoder on the wheel axis has a resolution of 1/20 of a revolution. D) The wheel was rotated rapidly then stopped for a second or two, then rotated rapidly, then stopped again, and so on. E) The apparatus likely was malfunctioning. Answer[A]:The jumps in the graph are artifacts of the way the angular encoder works, the output of the encoder is advanced only at discrete angles. Looking at the graph we see that 10 jumps in angular position correspond to about 0.31 radians so each jump corresponds to about .031radians. A full circle would thus give about (2π)/(0.031)=203 jumps. Choice A is thus the most plausible possibility. Page 19 of 19