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p0503

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Unformatted text preview: 5.3. CHAPTER 5, PROBLEM 3 229 5.3 Chapter 5, Problem 3 Problem: A series of n identical balls of mass m lie on a frictionless surface. Ball 1 has initial speed v1 = V and all of the other balls are initially at rest. Ball 1 collides with Ball 2, which in turn collides with Ball 3, etc. The coefficient of restitution for all of the balls is e. Determine the velocity of Ball n, vn , in terms of V , e and n. Assuming e = 0.9, compute vn /V for n = 5, 10 and 20. h xV .............. . . n . 1 .............. 2 3 4 .. ..................................................................................................................................................................................................................................................................................................................................................................................................................... . .. .. . . . . . . . . . . . . . . . . . . . .. . ..... ...... ..... ...... ..... .......................................................................................................................................................................................................................................................................................... .. . .. . . . . ... . ... . .. ..... ... . .. ... . . . . . ......... . . ............ . . .............. ............................. . . . . . . . . . . . . . . .. .................................................................................................................................................................................................................................................................................................................. .. ............................................................................................................................................................................................................................................................................................................................... . . . .. .. .. .. .. .. .. .. .. .. . .. ..... .... .. . .. ..... .. .. .. .. ............ . . ................. . ....... ........ .. ................ .. ................ .. ....... .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . ............................................................................................................................................................................................................................................................................................................................... . .. . ................................................................................................................................................................................................................................................ ..................................................................................................................................................................................... . .. . .. . . .. .. ............ .............. .... . . . . .. ....................... ....... . . . ...................... . . ........... . . .............. ......... . ........... . . . . h x h x h x ··· h x Solution: This is a direct central impact with the line of impact for the billiard balls parallel to the x axis. Balls 1 and 2. Using the fact that the initial speeds of Balls 1 and 2 are v1 = V and v2 = 0, momentum conservation along the line of impact tells us that mV + 0 = mv1 + mv2 Dividing through by m yields v1 + v2 = V For a coefficient of restitution equal to e, we have v2 − v1 = e (v1 − v2 ) = eV Now, we add the equations resulting from momentum conservation and the impact relation, which yields 2v2 = (1 + e)V =⇒ v2 = 1+e V 2 Balls 2 and 3. Dropping the prime on v2 for simplicity, the initial speeds of Balls 2 and 3 are v2 = 1 (1+e)V 2 and v3 = 0. Momentum conservation along the line of impact tells us that m Dividing through by m yields v2 + v3 = For a coefficient of restitution equal to e, we have v3 − v2 = e (v2 − v3 ) = e 1+e V 2 2 1+e V + 0 = mv2 + mv3 2 1+e V 2 Now, we add the equations resulting from momentum conservation and the impact relation, which yields 2v3 = (1 + e)2 V 2 =⇒ v3 = 1+e 2 V Balls 3 and 4. Dropping the prime on v3 for simplicity, the speeds of Balls 3 and 4 are v3 = 1 (1 + e)2 V 4 and v4 = 0. Momentum conservation along the line of impact tells us that m 1+e 2 2 V + 0 = mv3 + mv4 230 Dividing through by m yields v3 + v4 = For a coefficient of restitution equal to e, we have v4 − v3 = e (v3 − v4 ) = e 1+e 2 CHAPTER 5. IMPULSE AND MOMENTUM 2 V 1+e 2 2 V Now, we add the equations resulting from momentum conservation and the impact relation, which yields 2v4 = (1 + e) Generalization. We have shown that v2 = 1+e V, 2 v3 = 1+e 2 2 1+e 2 2 V =⇒ v4 = 1+e 2 3 V V, v4 = 1+e 2 3 V Consequently, for Ball n, we conclude that vn = 1+e 2 n−1 V Finally, for e = 0.9, a simple computation shows that v5 = 0.81, v10 = 0.63, v20 = 0.38 ...
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