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Unformatted text preview: billing (cab4763) HW 8 opyrchal (11109) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A(n) 51 . 6 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 44 . 4 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0 . 933 kg camera in her hand and decides to get back to the shuttle by throwing the camera at a speed of 12 m / s in the direction away from the shuttle. How long will it take for her to reach the shuttle? Answer in minutes. Correct answer: 3 . 41051 min. Explanation: Let : M = 51 . 6 kg , m = 0 . 933 kg , d = 44 . 4 m , and v = 12 m / s . Because of conservation of linear momentum, we have 0 = M V + m ( v ) mv = M V where V is the velocity of the astronaut and it has a direction toward the shuttle. V = mv M = (0 . 933 kg) (12 m / s) 51 . 6 kg = 0 . 216977 m / s . And the time it takes for her to reach the shuttle t = d V = 44 . 4 m . 216977 m / s 1 min 60 s = 3 . 41051 min . 002 10.0 points A child bounces a 51 g superball on the side walk. The velocity change of the superball is from 22 m / s downward to 12 m / s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? Correct answer: 1387 . 2 N. Explanation: Let : m = 51 g = 0 . 051 kg , v u = 12 m / s , v d = 22 m / s , and t = 0 . 00125 s Choose the upward direction as positive. The impulse is I = F t = P = mv u m ( v d ) = m ( v u + v d ) F = m ( v u + v d ) t = (51 g) (12 m / s + 22 m / s) . 00125 s = 1387 . 2 N . 003 (part 1 of 2) 10.0 points A(n) 17 g object moving to the right at 30 cm / s overtakes and collides elastically with a 31 g object moving in the same direction at 17 cm / s. Find the velocity of the slower object after the collision. Correct answer: 13 . 2083 cm / s. Explanation: Basic Concepts: Momentum conserva tion gives m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f . Solution: For headon elastic collisions, we know that v 1 i v 2 i = ( v 1 f v 2 f ) . billing (cab4763) HW 8 opyrchal (11109) 2 For the relative velocities, we have v 1 v 2 = v 2 f v 1 f v 2 f = v 1 v 2 + v 1 f . Momentum is conserved, so m 1 v 1 + m 2 v 2 = m 1 v 1 f + m 2 v 2 f m 1 v 1 + m 2 v 2 = m 1 v 1 f + m 2 ( v 1 v 2 + v 1 f ) m 1 v 1 + m 2 v 2 m 2 ( v 1 v 2 ) = v 1 f ( m 1 + m 2 ) v 1 f = m 1 v 1 + m 2 v 2 m 2 { v 1 v 2 } m 1 + m 2 = 1 (17 g) + (31 g) bracketleftBig (17 g)(30 cm / s) + (31 g)(17 cm / s) (31 g) { (30 cm / s) (17 cm / s) } bracketrightBig = 13 . 2083 cm / s ....
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This note was uploaded on 11/23/2010 for the course PHYS PHYS111 taught by Professor Gayen during the Spring '10 term at NJIT.
 Spring '10
 gayen
 Physics

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