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: 84

Answer

$\frac{K_t}{K_i}=\frac{m_1(\frac{2m_2}{m_1+m_2})^2}{m_2}$

Work Step by Step

$v_{1f}=\frac{m_1-m_2}{m_1+m_2}v_{1i}+\frac{2m_2}{m_1+m_2}v_{2i}$ This simplifies to: $v_{1f}=\frac{2m_2}{m_1+m_2}v_{2i}$ We know that the initial kinetic energy is: $K_i = \frac{1}{2}m_2v_{2i}^2 $ We know that the kinetic energy transferred is: $K_t=\frac{1}{2}m_1(\frac{2m_2}{m_1+m_2}v_{2i})^2$ The ratio is: $\frac{K_t}{K_i}=\frac{m_1(\frac{2m_2}{m_1+m_2})^2}{m_2}$
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