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Unformatted text preview: hernandez (ejh742) homework 19 Turner (58120) 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 1030 kg have in order to have the same mo mentum as a 2160 kg pickup truck traveling at 27 m / s to the east? Correct answer: 56 . 6214 m / s. Explanation: Let : m 1 = 1030 kg , m 2 = 2160 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 = (2160 kg) (27 m / s) 1030 kg = 56 . 6214 m / s to the east. 002 10.0 points Note: Take East as the positive direction. A(n) 86 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 3 . 4 m / s to the West as he leaves the dock, what is the final velocity of the fisherman and the boat? Correct answer: 1 . 41256 m / s. Explanation: Let West be negative: Let : m 1 = 86 kg kg , m 2 = 121 kg kg , and v i, 1 = 3 . 4 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 = (86 kg) ( 3 . 4 m / s) 86 kg + 121 kg = 1 . 41256 m / s , which is 1 . 41256 m / s to the West. 003 10.0 points A(n) 866 N man stands in the middle of a frozen pond of radius 11 . 8 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 . 5 kg physics textbook horizontally toward the north shore, at a speed of 4 . 1 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: 169 . 378 s. Explanation: Let : W m = 866 N , r = 11 . 8 m , m b = 1 . 5 kg , and v b = 4 . 1 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 . 5 kg) 866 N (4 . 1 m / s) = . 0696669 m / s . hernandez (ejh742) homework 19 Turner (58120) 2 The time to travel the 11 . 8 m to shore is t = x  v m  = 11 . 8 m . 0696669 m / s = 169 . 378 s . 004 10.0 points An 58 . 3 kg object moving to the right at 54 . 9 cm / s overtakes and collides elastically with a second 39 . 1 kg object moving in the same direction at 36 . 9 cm / s....
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This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Mass, Work

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