p0503 - 5.3. CHAPTER 5, PROBLEM 3 229 5.3 Chapter 5,...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

This note was uploaded on 12/16/2010 for the course AME 301 at USC.

Ask a homework question - tutors are online