## Essential University Physics: Volume 1 (4th Edition)

Published by Pearson

# Chapter 9 - Exercises and Problems - Page 173: 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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