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Unformatted text preview: gutierrez (ig3472) HW8 ditmire (58216) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points Two balls have masses of 46 kg and 75 kg. The 46 kg ball has an initial velocity of 62 m / s (lefttoright is the positive direction, along a line joining the two balls), as shown in the figure below. The 75 kg ball has in initial velocity of 31 m / s. The two balls make a headon elastic collision with each other. 46 kg 62 m / s 75 kg 31 m / s What is the final velocity of the 46 kg ball? Correct answer: 53 . 2893 m / s. Explanation: Basic Concepts: Using Conservation of Energy, we have 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 = 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 m 1 bracketleftBig v 1 2 v 1 2 bracketrightBig = m 2 bracketleftBig v 2 2 v 2 2 bracketrightBig m 1 bracketleftbig v 1 v 1 bracketrightbig bracketleftbig v 1 + v 1 bracketrightbig = m 2 bracketleftbig v 2 v 2 bracketrightbig bracketleftbig v 2 + v 2 bracketrightbig . (1) Using Conservation of Momentum, we have m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 m 1 bracketleftbig v 1 v 1 bracketrightbig = m 2 bracketleftbig v 2 v 2 bracketrightbig . (2) Dividing Eq. 1 by Eq. 2, we have v 1 + v 1 = v 2 + v 2 , so (3) v 2 = v 1 + v 1 v 2 , or (4) v 1 = v 2 + v 2 v 1 . (5) Substituting v 2 from Eq. 4 into Eq. 2, we have m 1 [ v 1 v 1 ] = m 2 [ v 1 + v 1 2 v 2 ] v 1 [ m 1 + m 2 ] = [ m 1 m 2 ] v 1 + 2 m 2 v 2 v 1 = bracketleftbigg m 1 m 2 m 1 + m 2 bracketrightbigg v 1 + bracketleftbigg 2 m 2 m 1 + m 2 bracketrightbigg v 2 . (6) and substituting v 1 from Eq. 5 into Eq. 2, we have m 1 [2 v 1 v 2 v 2 ] = m 2 [ v 2 v 2 ] v 2 [ m 1 + m 2 ] = 2 m 1 v 1 + [ m 2 m 1 ] v 2 v 2 = bracketleftbigg 2 m 1 m 1 + m 2 bracketrightbigg v 1 + bracketleftbigg m 2 m 1 m 1 + m 2 bracketrightbigg v 2 . (7) All solutions can be determined using the above Eqs. 6 and 7. Let : m 1 = 46 kg , m 2 = 75 kg , M = m 1 + m 2 = 121 kg , v 1 = 62 m / s , and v 2 = 31 m / s . Conservation of momentum gives us m 1 v 1 + m 2 v 2 = ( m 1 + m 2 ) V cm . Therefore, the centerofmass velocity is V cm = m 1 v 1 + m 2 v 2 m 1 + m 2 (1) = m 1 v 1 + m 2 v 2 M = (46 kg) (62 m / s) (121 kg) + (75 kg) ( 31 m / s) (121 kg) = 4 . 35537 m / s . Solution: After the collision is shown in the figure below. m 1 53 . 2893 m / s m 2 39 . 7107 m / s Part 1: Using Eq. 6, we have v 1 = bracketleftbigg m 1 m 2 m 1 + m 2 bracketrightbigg v 1 + bracketleftbigg 2 m 2 m 1 + m 2 bracketrightbigg v 2 (6) = bracketleftbigg (46 kg) (75 kg) (46 kg) + (75 kg) bracketrightbigg (62 m / s) + bracketleftbigg 2 (75 kg) (46 kg) + (75 kg) bracketrightbigg ( 31 m / s) = 53 . 2893 m / s . gutierrez (ig3472) HW8 ditmire (58216) 2 Alternative Solution: Conservation of momentum gives us m 1 v 1 + m 2 v 2 = ( m 1 + m 2 ) V cm...
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This note was uploaded on 04/13/2010 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

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