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Unformatted text preview: Muraj, Hamza – Homework 21 – Due: Mar 22 2006, 4:00 am – Inst: Florin 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 4) 10 points Two balls have masses m 1 and m 2 . Ball m 1 has an initial velocity v 1 > 0 (lefttoright is the positive direction, along a line joining the two balls), as shown in the figure below. Ball m 2 is at rest. Balls m 1 and m 2 make a headon elastic collision with each other. m 1 v 1 m 2 If m 1 = m 2 , what is the final velocity v 1 of the ball m 1 ? 1. v 1 < v 1 < 2. v 1 = +2 v 1 3. +2 v 1 < v 1 ≤ + ∞ 4. v 1 = v 1 5. v 1 = 0 correct 6. < v 1 < + v 1 7. + v 1 < v 1 < +2 v 1 8. v 1 = + v 1 9.∞≤ v 1 < v 1 Explanation: Basic Concepts: Conservation of Energy, where v 2 = 0 1 2 m 1 v 1 2 = 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 m 1 h v 1 2 v 1 2 i = m 2 v 2 2 m 1 £ v 1 v 1 / £ v 1 + v 1 / = m 2 v 2 2 . (1) Conservation of Momentum, where v 2 = 0 m 1 v 1 = m 1 v 1 + m 2 v 2 m 1 £ v 1 v 1 / = m 2 v 2 . (2) Dividing Eq. 1 by Eq. 2, we have v 2 = v 1 + v 1 , or (3) v 1 = v 2 v 1 . (4) Substituting v 2 from Eq. 3 into Eq. 2, we have m 1 [ v 1 v 1 ] = m 2 [ v 1 + v 1 ] v 1 [ m 1 + m 2 ] = v 1 [ m 1 m 2 ] v 1 = • m 1 m 2 m 1 + m 2 ‚ v 1 , (5) and substituting v 1 from Eq. 4 into Eq. 2, we have m 1 [ v 1 v 2 + v 1 ] = m 2 v 2 v 2 [ m 1 + m 2 ] = v 1 [2 m 1 ] v 2 = • 2 m 1 m 1 + m 2 ‚ v 1 . (6) All solutions can be determined using the above Eqs. 5 and 6. Solution: m 1 m 2 v 2 Think of playing pool. You hit the cue ball with velocity v 1 dead center on the 8 ball v 2 = 0. The cue ball stops and the 8 ball proceeds with the same velocity as the cue ball originally had. Part 1: Using Eq. 5, where m 2 = m 1 = m , v 1 = • m 1 m 2 m 1 + m 2 ‚ v 1 (5) = • m m m + m ‚ v 1 = . 002 (part 2 of 4) 10 points Muraj, Hamza – Homework 21 – Due: Mar 22 2006, 4:00 am – Inst: Florin 2 If m 1 = m 2 , what is the final velocity v 2 of the ball m 2 ? 1. < v 1 < + v 1 2. +2 v 1 < v 1 < ≤∞ 3. v 2 = + v 1 correct 4.∞≤ v 1 < v 1 5. v 2 = +2 v 1 6. v 2 = 0 7. v 1 < v 1 < 8. + v 1 < v 1 < +2 v 1 9. v 2 = v 1 Explanation: Part 2: Using Eq. 6, where m 2 = m 1 = m , v 2 = • 2 m 1 m 1 + m 2 ‚ v 1 (6) = • 2 m m + m ‚ v 1 = + v 1 003 (part 3 of 4) 10 points In the limit, when...
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This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas.
 Spring '09
 Turner

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