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Unformatted text preview: alexander (jra2623) – homework 18 – Turner – (92510) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points Tony (of mass 50 kg) coasts on his bicycle (of mass 5 kg) at a constant speed of 7 m / s, carrying a 9 kg pack. Tony throws his pack forward, in the direction of his motion, at 6 m / s relative to the speed of bicycle just before the throw. What is the initial momentum of the system (Tony, the bicycle, and the pack)? Correct answer: 448 kg · m / s. Explanation: Let : m T = 50 kg , m B = 5 kg , m P = 9 kg , and v i = 7 m / s . Since p = mv , p i = ( m T + m B + m P ) v i = (50 kg + 5 kg + 9 kg) (7 m / s) = 448 kg · m / s . 002 (part 2 of 3) 10.0 points What is the momentum of the system im mediately after the pack is thrown? Correct answer: 448 kg · m / s. Explanation: Since momentum is conserved, p f = p i = 448 kg · m / s . 003 (part 3 of 3) 10.0 points What is the bicycle speed immediately after the throw? Correct answer: 6 . 01818 m / s. Explanation: Let : v PB = 6 m / s . After the throw, the velocity of the pack with respect to the ground is v P = v i + v PB = 7 m / s + 6 m / s = 13 m / s so the momentum immediately after the throw is p f = ( m T + m B ) v B + m P v P p f m P v P = ( m T + m B ) v B v B = p f m P v P m T + m B v B = 448 kg · m / s (9 kg)(13 m / s) 50 kg + 5 kg = 6 . 01818 m / s ....
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 Summer '08
 Kaplunovsky
 Energy, Kinetic Energy, Mass, Momentum, Work, Correct Answer, m/s, Joule

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