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hw19 - alexander(jra2623 homework 19 Turner(92510 This...

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alexander (jra2623) – homework 19 – Turner – (92510) 1 This print-out should have 13 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 1290 kg have in order to have the same mo- mentum as a 2370 kg pickup truck traveling at 24 m / s to the east? Correct answer: 44 . 093 m / s. Explanation: Let : m 1 = 1290 kg , m 2 = 2370 kg , and v 2 = 24 m / s to the east . vectorp = m 1 vectorv 1 = m 2 vectorv 2 v 1 = m 2 v 2 m 1 = (2370 kg) (24 m / s) 1290 kg = 44 . 093 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 137 kg rowboat at rest on the West side of the dock. If the velocity of the fisherman is 3 . 6 m / s to the West as he leaves the dock, what is the final velocity of the fisherman and the boat? Correct answer: 1 . 32719 m / s. Explanation: Let West be negative: Let : m 1 = 80 kg kg , m 2 = 137 kg kg , and v i, 1 = 3 . 6 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) ( 3 . 6 m / s) 80 kg + 137 kg = 1 . 32719 m / s , which is 1 . 32719 m / s to the West. 003 10.0 points A(n) 871 N man stands in the middle of a frozen pond of radius 8 . 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 . 2 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: 124 . 02 s. Explanation: Let : W m = 871 N , r = 8 . 8 m , m b = 1 . 5 kg , and v b = 4 . 2 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) 871 N (4 . 2 m / s) = 0 . 0709564 m / s .
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alexander (jra2623) – homework 19 – Turner – (92510) 2 The time to travel the 8 . 8 m to shore is t = Δ x | v m | = 8 . 8 m 0 . 0709564 m / s = 124 . 02 s . 004 10.0 points An 67 . 1 kg object moving to the right at 47 . 2 cm / s overtakes and collides elastically with a second 51 . 1 kg object moving in the same direction at 28 . 1 cm / s. Find the velocity of the second object after the collision. Correct answer: 49 . 7854 cm / s. Explanation: Let : m 1 = 67 . 1 kg , m 2 = 51 . 1 kg , v 1 = 47 . 2 cm / s , and v 2 = 28 . 1 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 1 i + ( m 2 m 1 ) v 2 i = ( m 1 + m 2 ) v 2 f v 2 f = bracketleftbigg 2 m 1 m 1 + m 2 bracketrightbigg v 1 + bracketleftbigg m 2 m 1 m 1 + m 2 bracketrightbigg v 2 = bracketleftbigg 2 (67 . 1 kg) (67 . 1 kg) + (51 .
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