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Unformatted text preview: kuruvila (lk5992) HW 8 opyrchal (11113) 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) 70 . 8 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 31 . 8 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0 . 556 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: 5 . 62411 min. Explanation: Let : M = 70 . 8 kg , m = 0 . 556 kg , d = 31 . 8 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 . 556 kg) (12 m / s) 70 . 8 kg = 0 . 0942373 m / s . And the time it takes for her to reach the shuttle t = d V = 31 . 8 m . 0942373 m / s 1 min 60 s = 5 . 62411 min . 002 10.0 points A child bounces a 49 g superball on the side walk. The velocity change of the superball is from 28 m / s downward to 17 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: 1764 N. Explanation: Let : m = 49 g = 0 . 049 kg , v u = 17 m / s , v d = 28 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 = (49 g) (17 m / s + 28 m / s) . 00125 s = 1764 N . 003 (part 1 of 2) 10.0 points A(n) 20 g object moving to the right at 24 cm / s overtakes and collides elastically with a 33 g object moving in the same direction at 11 cm / s. Find the velocity of the slower object after the collision. Correct answer: 7 . 81132 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 ) . kuruvila (lk5992) HW 8 opyrchal (11113) 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 (20 g) + (33 g) bracketleftBig (20 g)(24 cm / s) + (33 g)(11 cm / s) (33 g) { (24 cm / s) (11 cm / s) } bracketrightBig = 7 . 81132 cm / s ....
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This note was uploaded on 10/28/2010 for the course PHYS 111 taught by Professor Moro during the Fall '08 term at NJIT.
 Fall '08
 moro

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