homework 19-solutions

homework 19-solutions - Bodet (ngb299) – homework 19 –...

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Unformatted text preview: Bodet (ngb299) – homework 19 – turner – (57340) 1 This print-out should have 17 questions. Multiple-choice 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 1070 kg have in order to have the same mo- mentum as a 2390 kg pickup truck traveling at 24 m / s to the east? Correct answer: 53 . 6075 m / s. Explanation: Let : m 1 = 1070 kg , m 2 = 2390 kg , and v 2 = 24 m / s to the east . !p = m 1 !v 1 = m 2 !v 2 v 1 = m 2 v 2 m 1 = (2390 kg) (24 m / s) 1070 kg = 53 . 6075 m / s to the east. 002 10.0 points Note: Take East as the positive direction. A(n) 82 kg fisherman jumps from a dock into a 139 kg rowboat at rest on the West side of the dock. If the velocity of the fisherman is 3 . 5 m / s to the West as he leaves the dock, what is the final velocity of the fisherman and the boat? Correct answer:- 1 . 29864 m / s. Explanation: Let West be negative: Let : m 1 = 82 kg kg , m 2 = 139 kg kg , and v i, 1 =- 3 . 5 m / s m / s . The boat and fisherman have the same final speed, and v i, 2 = 0 m/s, so m 1 !v i, 1 + m 2 !v i, 2 = ( m 1 + m 2 ) !v f m 1 !v i, 1 = ( m 1 + m 2 ) !v f v f = m 1 v i m 1 + m 2 = (82 kg) (- 3 . 5 m / s) 82 kg + 139 kg =- 1 . 29864 m / s , which is 1 . 29864 m / s to the West. 003 10.0 points A(n) 598 N man stands in the middle of a frozen pond of radius 14 . 8 m. He is unable to get to the other side because of a lack of fric- tion between his shoes and the ice. To over- come this difficulty, he throws his 1 kg physics textbook horizontally toward the north shore, at a speed of 5 . 3 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: 170 . 223 s. Explanation: Let : W m = 598 N , r = 14 . 8 m , m b = 1 kg , and v b = 5 . 3 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 kg) 598 N (5 . 3 m / s) =- . 0869448 m / s . Bodet (ngb299) – homework 19 – turner – (57340) 2 The time to travel the 14 . 8 m to shore is t = Δ x | v m | = 14 . 8 m . 0869448 m / s = 170 . 223 s . 004 10.0 points An 73 . 7 kg object moving to the right at 36 . 4 cm / s overtakes and collides elastically with a second 55 . 4 kg object moving in the same direction at 20 . 7 cm / s. Find the velocity of the second object after the collision. Correct answer: 38 . 6255 cm / s. Explanation: Let : m 1 = 73 . 7 kg , m 2 = 55 . 4 kg , v 1 = 36 . 4 cm / s , and v 2 = 20 . 7 cm / s . Momentum conservation gives m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f (1) For head-on elastic collisions the center-of- mass motion remains constant- ( v 1 f- v 2 f ) = v 1 i- v 2 i , that is v 2 f = v 1 i- v 2 i + v 1 f , or v 1 f = v 2 i- v 1 i + v 2 f . (2) Substituting Eq. 2 into Eq. 1, we have m 1 v 1 i + m 2 v 2 i = m 1 v 2 i- m 1 v 1 i +( m 1 + m 2 ) v 2 f 2 m 1 v...
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This note was uploaded on 10/06/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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homework 19-solutions - Bodet (ngb299) – homework 19 –...

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