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Unformatted text preview: hiarker (srh959) – homework 19 – Turner – (58220) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What velocity must a car with a mass of 1320 kg have in order to have the same mo mentum as a 2300 kg pickup truck traveling at 27 m / s to the east? Correct answer: 47 . 0455 m / s. Explanation: Let : m 1 = 1320 kg , m 2 = 2300 kg , and v 2 = 27 m / s to the east . vectorp = m 1 vectorv 1 = m 2 vectorv 2 v 1 = m 2 v 2 m 1 = (2300 kg) (27 m / s) 1320 kg = 47 . 0455 m / s to the east. 002 10.0 points Note: Take East as the positive direction. A(n) 87 kg fisherman jumps from a dock into a 121 kg rowboat at rest on the West side of the dock. If the velocity of the fisherman is 4 . 1 m / s to the West as he leaves the dock, what is the final velocity of the fisherman and the boat? Correct answer: − 1 . 7149 m / s. Explanation: Let West be negative: Let : m 1 = 87 kg kg , m 2 = 121 kg kg , and v i, 1 = − 4 . 1 m / s m / s . The boat and fisherman have the same final speed, and v i, 2 = 0 m/s, so m 1 vectorv i, 1 + m 2 vectorv i, 2 = ( m 1 + m 2 ) vectorv f m 1 vectorv i, 1 = ( m 1 + m 2 ) vectorv f v f = m 1 v i m 1 + m 2 = (87 kg) ( − 4 . 1 m / s) 87 kg + 121 kg = − 1 . 7149 m / s , which is 1 . 7149 m / s to the West. 003 10.0 points A(n) 783 N man stands in the middle of a frozen pond of radius 6 . 5 m. He is un able to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1 . 2 kg physics textbook horizontally toward the north shore, at a speed of 4 . 8 m / s. The acceleration of gravity is 9 . 81 m / s 2 . How long does it take him to reach the south shore? Correct answer: 90 . 0707 s. Explanation: Let : W m = 783 N , r = 6 . 5 m , m b = 1 . 2 kg , and v ′ b = 4 . 8 m / s . The mass of the man is m m = W m g . From conservation of momentum, m m v m + m b v b = m m v ′ m + m b v ′ b 0 = m m v ′ m + m b v ′ b v ′ m = − m b m m v ′ b = − g m b W m v ′ b = − ( 9 . 81 m / s 2 ) (1 . 2 kg) 783 N (4 . 8 m / s) = − . 0721655 m / s . hiarker (srh959) – homework 19 – Turner – (58220) 2 The time to travel the 6 . 5 m to shore is t = Δ x  v ′ m  = 6 . 5 m . 0721655 m / s = 90 . 0707 s . 004 10.0 points An 56 . 1 kg object moving to the right at 35 . 6 cm / s overtakes and collides elastically with a second 37 . 3 kg object moving in the same direction at 17 . 4 cm / s....
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This note was uploaded on 04/23/2010 for the course PHY 303K taught by Professor Nui during the Spring '09 term at University of TexasTyler.
 Spring '09
 NUI
 Physics, Mass, Work

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