Multiple choice questions [60 points]

Transcription

1 1 Multiple choice questions [60 points] Answe all o the ollowing questions. Read each question caeully. Fill the coect bubble on you scanton sheet. Each question has exactly one coect answe. All questions ae woth the same amount o points. 1. Susana ascends a mountain via a shot, steep tail. Sean ascends the same mountain via a long, gentle tail. Which o the ollowing statements is tue? A. Susana gains moe gavitational potential enegy than Sean. B. Susana gains less gavitational potential enegy than Sean. C. Susana gains the same gavitational potential enegy as Sean. U = mgh same h o both (note that we need to assume that Sean and Susana have the same mass). The best answe (not listed) would be: to compae gavitational potential enegies, we must know the masses o Sean and Susana. This question was counted coect o eveyone. D. To compae gavitational potential enegies, we must know the height o the mountain. E. To compae gavitational potential enegies, we must know the lengths o the two tails.

2 . When the potential enegy U() is given as in Figue A, then the oce is given in Figue B by cuve A. 1 B. du F = Since the gaph o U() is a staight line with a positive d slope, F is constant and negative. C. 3 D. 4 E. 5

3 3 3. The gaph shows a plot o the gavitational potential enegy U o a 1- kg body as a unction o its height h above the suace o a planet. The acceleation due to gavity at the suace o the planet is A. 0 m/s B. 9.8 m/s C. 6 m/s D. 3 m/s Fo instance, the change o gavitational potential enegy o the 1kg mass between 0 and 0 m is 60 U = U ( 0) U (0) = 60 J also U = mgh = 0 g g = 0 E. None o these is coect.

4 4 4. Two unequal masses hang om eithe end o a massless cod that passes ove a ictionless pulley. Which o the ollowing is tue about the gavitational potential enegy (U) and the kinetic enegy (K) o the system consisting o the two masses ate the masses ae eleased om est? A. U < 0 and K > 0 Only consevative oces do wok (namely the weight o the two masses). The wok by the tensions on both masses cancel out. The mechanical enegy o the system (made o the masses + Eath) is constant: K+U=constant Since the velocity o both masses inceases, K > 0. And since K =- U, U<0 B. U = 0 and K > 0 C. U < 0 and K = 0 D. U = 0 and K = 0 E. U > 0 and K < 0

5 5 5. x The suace shown in the igue is ictionless. I the block is eleased om est, it will compess the sping at the oot o the incline Hint: use the consevation o mechanical enegy. You will need to solve a quadatic equation. A. 4.0 m B. 3.4 m C m D m The mechanical enegy o the system sping + mass + Eath is constant o Initially: Emech = K + U = 0 + mgxi sinθ ( with xi = 4m and θ = 30 ) 1 Finally: E mech = K + U = 0 + mgx sinθ + kx 1 Solve mgx sinθ + kx = mgxi sinθ 9.8x + 50x = 39. x = 0.99m o m Only the negative oot makes sense since the sping must be compessed. E. 0.5 m

6 6 6. The mass o the ectangle in the igue is M, the mass o the ing is M, and the mass o the cicle is 3M. The cente o mass o the system (consisting o the ectangle, the cicle and the ing) with espect to the oigin O is located at point A. 1 The cente o mass must be inside the tiangle linking the 3 B. centes o masses o the thee objects. Since the cicle is moe massive (3M as compaed to M), the cente o mass should be close to the cicle than to the othe two objects. C. 3 D. 4 E. 5

7 7 7. A 1.0-kg mass is acted on by a net oce o 4.0 N and a 3.0-kg mass is acted on by a net oce o 3.0 N, in the diections shown. The acceleation o the cente o mass o this system is appoximately A. 1.5 m/s, 53º N o E = ma = 4a F net Also: F net = 4 yˆ + 3xˆ The angle with the x axis is B m/s, 45º N o E C m/s, due noth D m/s, N o E E. 7.5 m/s, 53º N o E tan 1 = 3, thus a = = 1.5 m / s o 53.1

8 8 8. A boy and gil initially at est on ice skates ace each othe. The gil has a mass o 0 kg and the boy has a mass o 30 kg. The boy pushes the gil backwad at a speed o 3.0 m/s. As a esult o the push, the speed o the boy is A. 0 m/s.0 m/s Use consevation o momentum: B. mgil p + pgil = 0 vboy = vgil = m mboy C. 3.0 m/s D. 4.5 m/s E. 9.0 m/s boy / s 9. I a body moves in such a way that its linea momentum is constant, then A. its kinetic enegy is zeo. B. the sum o all the oces acting on the body is constant and nonzeo. C. its acceleation is geate than zeo and is constant. D. its cente o mass emains at est. E. the sum o all the oces acting on it must be zeo. dp = = 0 dt F net 10. While in hoizontal light at a speed o 0 m/s, a baseball o mass 0.11 kg is stuck by a bat. Ate leaving the bat, the baseball has a speed o 9 m/s in a diection opposite to its oiginal diection. The magnitude o the impulse given the ball is A kg m/s B. 3. kg m/s C.. kg m/s D. 5.4 kg m/s J = p = m( v vi ) = 0.11 (9xˆ ( 0xˆ)) = 5.39xˆ kg. m / s unit vecto in the diection o v E kg m/s whee xˆ is a

9 9 11. The gaph shows the momentum o a body as a unction o time. The time at which the oce acting on the body is geatest is A. 0.5 s B..5 s C. 4.0 s D. 1.5 s dp F = The oce is maximum whee the slope o p(t) is maximum. dt E. 5.0 s

10 10 1. A bullet, m = kg, taveling with a velocity v stikes and embeds itsel in the bob o a ballistic pendulum, M = 9.50 kg. The combined masses ise to a height h = 1.8 m. The velocity v o the bullet is A m/s B. 50 m/s C. 100 m/s The velocity V o the bob + bullet just ate the collision is (using consevation o momentum) m mv = ( m + M ) V V = v m + M Use the wok enegy theoem between the instant just ate the collision and the instant when the bob+bullet system eaches the height h 1 0 ( m + M ) V = ( m + M ) gh V = gh It ollows: m + M 10 v = gh = = 100. m / s m 0.5 D. 50 m/s E. 75 m/s

11 11 PROBLEM [40 points] Two masses m 1 =.0 kg and m =5.0 kg ae on a hoizontal ictionless suace. Mass m 1 is moving to the ight with velocity v1 i = 10 m / s and m with velocity v = i 3.0 m / s. As shown in the igue a massless sping o oce constant k = 110 N/kg is attached to m. v 1i =10.0 m/s m 1 =.0kg k = 110 N/kg v i =3.0 m/s m =5.0 kg 1). [15 pts] Ate the masses collide and sepaate completely om one anothe, what ae the inal velocities o the two masses? (Hint: the collision is elastic). Use the consevation o momentum and the consevation o the kinetic enegy o the system made o the masses. m1v1i + mvi = m1v1 + mv m1v1 i + mvi = m1v1 + mv Using the appoach descibed in the text p 04, we ind m1v1 i + mvi = m1v1 + mv v1 + 5v = 35 v1 = 0 m / s v1 i + v1 = vi + v v1 v = 7 v = 7 m / s ). [10 pts] What ae the inal kinetic enegies o each o the two masses (K 1 and K )? K K 1 1 = m1v1 1 = mv = 0 J = 1.5 J 3). [15 pts] What is the maximum potential enegy stoed in the sping duing the collision? (Hint: at maximum compession, both masses have the same velocity equal to the cente o mass velocity. Use also that the mechanical enegy o m 1 +sping +m is constant). m1v1 i + mvi The velocity o the cente o mass is vcm = = = 5 m / s m1 + m 7 Just beoe the collision the mechanical enegy is Emech = K1 i + K i = 1. 5 J When the sping is at maximum compession, both masses have a velocity equal to the cente o mass velocity. The mechanical enegy is 1 1 Emech = m1 vcm + mvcm + U el = U el U el = 35 J