p0502 - 5.2. CHAPTER 5, PROBLEM 2 299 5.2 Chapter 5,...

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Unformatted text preview: 5.2. CHAPTER 5, PROBLEM 2 299 5.2 Chapter 5, Problem 2 Problem: Two identical balls of mass m and coefficient of restitution e approach each other with velocities v1 = v i and v2 = −u i. Ignore effects of friction. (a) Determine their velocities after the impact, v1 and v2 , in terms of u, v and e. (b) If v = 5 u and v1 = − 11 v2 , what is e? 4 16 . .. . . .. .... .. .. ............. .... ........... ..................... ................................ ..................................... . ..................... 1 ....................... .......................................... . ........................................... . ............................................ .................. .. . .................................................................. . ............................................ ........................................... . . . .......................................... ....................... ....................................... ........................................ ..................... ........................ ........................... .. .... ..... . . . ... . .................. ............ .. ....... . .. . . m v . ....... . ........... .................. ...... . . . ....... ......................................... ................................ .. .. .. .. 2 ............................................................................... .. .......................................... . .. .......................... .. . . ................................................................................... ............................................................. ............................................ ... . . .......................................... .......................................... . .......................................... .................................... .................................. . . ................................... ................. . ............................ ............. . . ......... .... . .. v m Solution: (a) This is a direct central impact with the line of impact for the balls being the x axis. Momentum Conservation. Using the fact that the initial velocity components for Balls 1 and 2 are v1 = v and v2 = −u, momentum conservation along the line of impact tells us that mv − mu = mv1 + mv2 Dividing through by m yields v1 + v2 = v − u Impact Relation. For a coefficient of restitution equal to e, we have v2 − v1 = e (v1 − v2 ) = e(v + u) To complete the solution, we add the equations resulting from momentum conservation and the impact relation, which yields 2v2 = (1 + e)v − (1 − e)u =⇒ v2 = 1 1 (1 + e)v − (1 − e)u 2 2 Similarly, subtracting the impact-relation equation from the momentum-conservation equation tells us that 2v1 = (1 − e)v − (1 + e)u =⇒ v1 = 1 1 (1 − e)v − (1 + e)u 2 2 Therefore, the puck velocity vectors after the impact are v1 v2 (b) We are given v = 5 u, wherefore 4 v1 = v2 = 1 5 1 1 (1 − e) u − (1 + e)u = (1 − 9e)u 2 4 2 8 1 5 1 1 (1 + e) u − (1 − e)u = (1 + 9e)u 2 4 2 8 = = 1 [(1 − e)v − (1 + e)u] i 2 1 [(1 + e)v − (1 − e)u] i 2 300 We are also given v2 = − 11 v1 . Hence, 16 1 11 1 (1 − 9e)u = − (1 + 9e)u 8 16 8 =⇒ CHAPTER 5. IMPULSE AND MOMENTUM 16(1 − 9e) = −11(1 + 9e) Solving for e, we conclude that coefficient of restitution is e= 3 5 ...
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This note was uploaded on 05/11/2011 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

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