Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 9 - Exercises and Problems: 93

Answer

$.26v_0;.31v_0$

Work Step by Step

a) We find that the blocks, after the collision, have a velocity of: $v_{1f}=\frac{M-nM}{m+nM}v_{1i}+\frac{2nM}{M+nM}v_{2i}$ $v_{1f}=\frac{M-nM}{m+nM}v_{0}$ $v_{2f}=\frac{2M}{m_1+m_2}v_{1i}+\frac{M-nM}{M+nM}v_{2i}$ $v_{2f}=\frac{2M}{m+nM}v_{0}$ If n is less than three, we see than 2M will be greater than $|M-nM|$. This would make the second block faster, meaning the blocks will never again collide. b) If n is equal to 4, we see than 2M will be less than $|M-nM|$. This would make the first block faster, meaning the blocks will collide a second time. c) We find: $v_{1f}=\frac{-9M}{11M}v_{0}$ $v_{2f}=\frac{2M}{11M}v_{0}$ They will collide again. Continuing to use the equations $v_{1f}=\frac{M-nM}{m+nM}v_{1i}+\frac{2nM}{M+nM}v_{2i}$ and $v_{2f}=\frac{2M}{m_1+m_2}v_{1i}+\frac{M-nM}{M+nM}v_{2i}$ until $|v_{1f}|$
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