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 - Page 164: 49


The proof is below.

Work Step by Step

We know the following equation for elastic collisions: $v_{1f}=\frac{m_1-m_2}{m_1+m_2}v_{1i}+\frac{2m_2}{m_1+m_2}v_{2i}$ Thus, we find: $v_{1f}=\frac{3m-m}{4m}v+\frac{2m}{4m}(-v)=0$ Momentum is conserved, so it follows: $m(-v)+3mv=mv_f$ $v_f=2v$ Thus, the second mass rebounds at twice the speed.
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