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Unformatted text preview: 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 1110 kg have in order to have the same mo mentum as a 2140 kg pickup truck traveling at 28 m / s to the east? Correct answer: 53 . 982 m / s. Explanation: Let : m 1 = 1110 kg , m 2 = 2140 kg , and v 2 = 28 m / s to the east . vectorp = m 1 vectorv 1 = m 2 vectorv 2 v 1 = m 2 v 2 m 1 = (2140 kg) (28 m / s) 1110 kg = 53 . 982 m / s to the east. 002 10.0 points Note: Take East as the positive direction. A(n) 80 kg fisherman jumps from a dock into a 129 kg rowboat at rest on the West side of the dock. If the velocity of the fisherman is 4 . 7 m / s to the West as he leaves the dock, what is the final velocity of the fisherman and the boat? Correct answer: − 1 . 79904 m / s. Explanation: Let West be negative: Let : m 1 = 80 kg kg , m 2 = 129 kg kg , and v i, 1 = − 4 . 7 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 = (80 kg) ( − 4 . 7 m / s) 80 kg + 129 kg = − 1 . 79904 m / s , which is 1 . 79904 m / s to the West. 003 10.0 points A(n) 584 N man stands in the middle of a frozen pond of radius 11 . 2 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 . 4 kg physics textbook horizontally toward the north shore, at a speed of 5 . 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: 82 . 1118 s. Explanation: Let : W m = 584 N , r = 11 . 2 m , m b = 1 . 4 kg , and v ′ b = 5 . 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 . 4 kg) 584 N (5 . 8 m / s) = − . 136399 m / s . homework 22 – Turner – (58185) 1 hinojosa (jlh3938) – homework 22 – Turner – (58185) 2 The time to travel the 11 . 2 m to shore is t = Δ x  v ′ m  = 11 . 2 m . 136399 m / s = 82 . 1118 s . 004 10.0 points An 67 kg object moving to the right at 45 cm / s overtakes and collides elastically with a sec ond 49 . 7 kg object moving in the same direc tion at 25 . 2 cm / s....
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This note was uploaded on 05/02/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Mass, Work

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