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1 Answer, Key { Homework 6 { Rubin H Landau 1 This print-out should have 24 questions. Check that it is complete before leaving the printer. Also, multiple-choice questions may continue on the next column or page: nd all choices before making your selection. Note, only a few (usually 4) of the problems will have their scores kept for a grade. You may make multiple tries to get a problem right, although it's worth less each time. Worked solutions to a number of these problems (even some of the scored ones) may be found on the Ph211 home page. AP M 1993 MC 16 09:01, calculus, multiple choice, < 1 min. 001 A balloon of mass M is oating motionless in the air. A person of mass less than M is on a rope ladder hanging from the balloon. The person begins to climb the ladder at a uniform speed v relative to the ground. How does the balloon move relative to the ground? 1. Up with speed v 2. Up with a speed less than v 3. Down with a speed less than v correct 4. Down with speed v 5. The balloon does not move. Let the mass of the person be m. Total momentum is conserved (because the exterior forces on the system are balanced), especially the component in the vertical direction. When the person begins to move, we have mv + Mv M 0; ) v M, m M v<v ) jv M j m M v<v; since m<m) m M < 1: Thus the balloon moves in the opposite direction. Return to the Shuttle 09:01, trigonometry, numeric, > 1 min. 002 A(n) 63:1 kg astronaut becomes separated from the shuttle, while on a spacewalk. She nds herself 70:2 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0:543 kg camera in her hand and decides to get back to the shuttle by throwing the camera at a speed of 12 ms in the direction away from the shuttle. How long will it take for her to reach the shuttle? Correct answer: 11:3301 min. Basic Concepts: Conservation of Linear Momentum Because of conservation of linear momentum, we have 0MV + m(,v) mv MV where V is the velocity of the astronaut and it has a direction toward the shuttle. So V mv M 0: ms And the time it takes for her to reach the shuttle t d V 70:2 m 0: ms 679:807 s 11:3301 min h min s i 0: mins (1) 50 M 63:1 kg (2) d 70:2 m (3) 75 0:5 m 0:543 kg (4) 0:95 v 12:0 ms (5) V mv (6) M h0:543ih12i h63:1i

2 Answer, Key { Homework 6 { Rubin H Landau 2 0: ms hmsi hkgihmsi hkgi t d (7) V h70:2i h0:103265i 679:807 s hsi hmi hmsi t u t h min s i (8) h679:807ih0: i 11:3301 min hmini hsihminsi Force on a Golf Ball 09:02, calculus, numeric, > 1 min. 003 A golf ball ( m 72:3 g ) is struck a blow that makes an angle of 45:6 with the horizontal. The drive lands 187 m away on a at fairway. If the golf club and ball are in contact for 3:59 ms, what is the average force of impact? (Neglect air resistance.) Correct answer: 862:234 N. Note that the range of the golf ball is given by: L v2 sin 2 g Then the initial velocity of the ball is: v 0 p Lg sin 2 42:8136 ms The average force exerted is the change in its momentum over the time of contact: F mv 0 t (72:3 g)(0:001 kgg)(42:8136 ms) (3:59 ms)(0:001 sms) 862:234 N h kg g i 0:001 kgg (1) hmsi s 0:001 sms (2) g 9:8 ms 2 (3) m 72:3 g 18:4 45:6 18 L 187 m 80 t 3:59 ms 2:8 v 0 hmsi s vu u t 73:6 (4) 72 (5) 320 (6) Lg sin, 2:0 180:0 sin 42:8136 ms vu u 11:2 (7) (8) h187ih9:8i i t hmihms2 sin F mhkg g iv 0 th s msi 2:0h45:6ih3: i hih ihi hi 180:0 h72:3ih0:001ih42:8136i h3:59ih0:001i 862:234 N hni hgihkggihmsi hmsihsmsi (9) Bullet Moves a Block 09:03, trigonometry, numeric, > 1 min. 004 A(n) 18:7 g bullet is shot into a(n) 6310 g wooden block standing on a frictionless surface. The block, with the bullet in it, acquires avelocity of0:869 ms. Calculate the velocity of the bullet before striking the block. Correct answer: 294:098 ms. Basic concepts: Momentum of any object is p mv The collision is inelastic, and by conservation of momentum, p before p after m b v b +0(m b + m w )v f v b (m b + m w )v f m b

3 m b 18:7 g 10 m w 6310 g 4500 v f 0:869 ms 0:75 Answer, Key { Homework 6 { Rubin H Landau 3 30 (1) 6500 (2) 2 (3) v (m b + m w ) v f (4) m b (h18:7i + h6310i) h0:869i h18:7i 294:098 ms (hgi + hgi) hmsi hmsi hgi AP B 1993 MC 7 09:04, trigonometry,multiple choice, > 1 min. 005 I Two pucks are attached by a stretched spring and are initially held at rest on a frictionless surface, as shown above. The pucks are then released simultaneously. If puck I has three times the mass of puck II, which of the following quantities is the same for both pucks as the spring pulls the two pucks toward each other? 1. Speed 2. Velocity 3. Acceleration 4. Magnitude of momentum correct 5. Kinetic energy The total momentum is conserved here (no friction). II 0P i P f P I + P II ) P I,P II jp I j jp II j The magnitude of momentum is the same for both pucks. Let's look at the other choices: (m I 3m II ) Speed: m I v I P I P II m II v II ) v II 3v I. Velocity: m I v I P I,P II,m II v II ) v II,3v I. Acceleration: a I F I,F II, 1 m I 3m II 3 Kinetic energy: K I F II m II 1 2 m Iv 2 I (3m II) 3 v II m IIv 2 II 1 3 K II:, 1 3 a II: 2 Collision With the Earth 09:04, calculus, numeric, > 1 min. 006 A 11:1 kg bowling ball initially at rest is dropped from a height of 4:32 m. What is the speed of the Earth coming up to meet the ball just before the ball hits the ground? Use 5: kg as the mass of the Earth. Correct answer: 1: ,23 ms. The speed of the ball just before impact is: v p 2gh 9:20174 ms From conservation of momentum p 0,or Therefore jv j vm b M e M e V + m b v 0 (9:20174 ms)(11:1 kg)5: kg 1: ,23 ms g9:8 ms 2 (1) m11:1 kg 2:8 11:2 (2) 1:2 h4:32 m 4:8 (3) M5: kg (4)

4 Answer, Key { Homework 6 { Rubin H Landau 4 v p 2:0 gh (5) p 2:0 h9:8ih4:32i 9:20174 ms q hmsi hi hms 2 ihmi V mv M h11:1ih9:20174i h5: i 1: ,23 ms hmsi hkgihmsi hkgi (6) AP B 1993 MC 10 09:04, calculus, numeric, > 1 min. 009 Which of the following is true when an object of mass m moving on a horizontal frictionless surface hits and sticks to an object of mass M > m, which is initially at rest on the surface? 1. The speed of the objects that are stuck together will be less than the initial speed of the less-massive object. correct 2. All of the initial kinetic energy of the less-massive object is lost. 3. The momentum of the objects that are stuck together has a smaller magnitude than the initial momentum of the less-massive object. 4. The collision is elastic. 5. The direction of motion of the objects that are stuck together depends on whether the hit is a head-on collision. The total momentum is conserved here (no friction). mv i P i P f (m + M)v f ) v f m m + M v i 0 < jv f j < jv i j The speed of objects that are stuck together is less than the initial speed of the less-massive object. Let's look at the other choices: It is obviously an inelastic collision since the two objects stick together after the collision. jv i j > 0 ) K f > 0. Only part of the initial kinetic energy is lost. The momentum is conserved. The nal momentum has the same magnitude as the initial momentum. v f m m + M v i So the direction of the motion of the stucktogether is the same as the direction of the initial motion of the less-massive object. It does not depend on whether the hit is a headon collision or not. 010 The mass of a star like our Sun is Earth masses, and the mean distance from the center of this star to the center of a planet like our Earth is 8: km. Treating this planet and star as particles, with each mass concentrated at its respective geometric center, how far from the center of the star is the center of mass of the planet-star system? Correct answer: 1719:29 km. From P mi x i x cm P mi in this case, taking the origin to be the center of the star M E L x cm M E + M E L (8:82 km) 108 ( ) 1719:29 km : The center-of-mass of the planet-star system is well inside the star.

5 Answer, Key { Homework 6 { Rubin H Landau 5 n (1) L 8: :19610 km 8 (2) L x c (3) 1:0+n h8:82 i 108 1:0+h513000i 1719:29 km hkmi hkmi hi + hi BlockonaWedge 09:05, trigonometry,multiple choice, > 1 min. 012 A large wedge (with a rough inclined surface and a smooth lower surface) rests on a horizontal frictionless surface. A block with a rough bottom starts from rest and slides down the inclined surface of the wedge. The coecient for friction between the block and inclined wedge surface is :1 < block incline <:5 and that between the lower wedge surface and horizontal surface is zero. 7. moves both horizontally and vertically with increasing speed 8. cannot be determined Basic Concepts: Momentum The motion of the center of mass of a system is only aected by the external forces acting on the system, and not by any internal forces such as the friction between the block and wedge. The net external force on the system acts only in the vertical direction, so the center of mass will not move in a horizontal direction. The fact that the block starts from rest tells us that it must be accelerating. Since one component of the system (the block) has a component of acceleration in the vertical direction and the other component (the wedge) does not, we know that the center of mass must be accelerating. The net external force on this system acts only in the vertical direction. Therefore, while the blockismoving, the center of mass of the block and wedge must be moving downward with increasing speed. While the block is sliding down the inclined surface, the combined center of mass of the block and wedge: 1. does not move 2. moves vertically with increasing speed correct 3. moves horizontally with constant speed 4. moves horizontally with increasing speed 5. moves vertically with constant speed 6. moves both horizontally and vertically with constant speed AP M 1993 MC 20 09:06, calculus, numeric, > 1 min. 013 A turntable that is initially at rest is set in motion with a constant angular acceleration. What is the angular velocity of the turntable after it has made one complete revolution? 1. p 2 2. p 2 3. p 4 correct Similar to uniform linear accelerated motion, we have! f 2! :

10 Answer, Key { Homework 6 { Rubin H Landau a M2 <a m1 3. a M2 a m1 correct 4. Not enough information is available. The rope does not stretch, so the acceleration of both masses must be equal. Atwood with Massive Pulley 10:05, trigonometry,multiple choice, < 1 min. 027 The center of mass of m 1 + M 2 1. accelerates up. 2. accelerates down. correct 3. does not accelerate. 4. Not enough information is available. The larger mass accelerates down and the smaller mass accelerates up, so the center of mass of the system accelerates down. The acceleration of the center of mass is a cm M 2 a M2 + m 1 a m1 M 2 + m 1 : Using \up" as the positive direction, then a M2,a and a m1 a. From Eq. (3), we have a cm,2 M 2 (M 2, m 1 ), m 1 (M 2, m 1 ) (M 2 + m 1 )[M p +2(M 2 + m 1 )] g,2 (M 2, m 1 ) 2 (M 2 + m 1 )[M p +2(M 2 + m 1 )] g: The negative sign indicates the center of mass acceleration is down. Atwood with Massive Pulley 10:06, trigonometry, numeric, > 1 min. 028 The relationship between the magnitudes of the tension T 2 and T 1 is 1. T 2 >T 1 correct 2. T 2 <T 1 3. T 2 T 1 4. Not enough information is available. Substituting a from Eq. (3) into Eq. (1) gives T 1 m 1 (g + a) m1 M p +4m 1 M 2 M p +2(M 2 + m 1 ) and substituting a into Eq. (2) gives T 2 M 2 (g, a) M2 M p +4m 1 M 2 M p +2(M 2 + m 1 ) Therefore T 2 >T 1 by the amount (M2, m 1 ) M p T T 2, T 1 M p +2(M 2 + m 1 ) g (4) g: (5) Alternative Solution: The torque equation is X : I(T2, T1) r: Since is positive, T 2 > T 1 ; however, is positive since T 2 > T 1, so this argument is circular. 029 The relationship between the magnitudes of the tension T 3, T 2, and T 1 is 1. T 1 + T 2 >T 3 2. Not enough information is available. 3. T 1 + T 2 T 3 4. T 1 + T 2 <T 3 correct Using a free body diagram for the pulley alone and one sees T 3 g: T 1 + T 2 + M p g >T 1 + T 2 : (6)

11 Answer, Key { Homework 6 { Rubin H Landau 11 Atwood with Massive Pulley 10:07, calculus, numeric, > 1 min. 030 The relations between the magnitudes of the tension T 3 and m 1, M 2, and M p is 1. T 3 > (m 1 + M 2 + M p ) g 2. T 3 (m 1 + M 2 + M p ) g 3. T 3 < (m 1 + M 2 + M p ) g correct 4. Not enough information is available. Using Eqs. (4), (5), & (6), we have T 3 T 1 + T 2 + M p g (M 2 + m 1 + M p ) g, (M 2, m 1 ) a < (M 2 + m 1 + M p ) g: Since M 2 > m 1 the center of mass of the blocks is accelerating down and the tension T 3 is reduced.

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