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Unformatted text preview: sindt (as53996) HW 8 opyrchal (11101) 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) 59 . 3 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 49 . 8 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0 . 848 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: 4 . 83678 min. Explanation: Let : M = 59 . 3 kg , m = 0 . 848 kg , d = 49 . 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 . 848 kg) (12 m / s) 59 . 3 kg = 0 . 171602 m / s . And the time it takes for her to reach the shuttle t = d V = 49 . 8 m . 171602 m / s 1 min 60 s = 4 . 83678 min . 002 10.0 points A child bounces a 51 g superball on the side walk. The velocity change of the superball is from 26 m / s downward to 15 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: 1672 . 8 N. Explanation: Let : m = 51 g = 0 . 051 kg , v u = 15 m / s , v d = 26 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) (15 m / s + 26 m / s) . 00125 s = 1672 . 8 N . 003 (part 1 of 2) 10.0 points A(n) 15 g object moving to the right at 24 cm / s overtakes and collides elastically with a 26 g object moving in the same direction at 13 cm / s. Find the velocity of the slower object after the collision. Correct answer: 10 . 0488 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 ) . sindt (as53996) HW 8 opyrchal (11101) 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 (15 g) + (26 g) bracketleftBig (15 g)(24 cm / s) + (26 g)(13 cm / s) (26 g) { (24 cm / s) (13 cm / s) } bracketrightBig = 10 . 0488 cm / s ....
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This note was uploaded on 11/09/2010 for the course PHYSICS 111 taught by Professor Wang during the Spring '09 term at NJIT.
 Spring '09
 WANG
 Physics

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