Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 9 - Section 9.6 - Elastic Collisions - Example: 9.10

Answer

$\frac{K_2}{K_1} = \frac{4m_1m_2}{(m_2+m_1)^2}$

Work Step by Step

We know the following equations for each kinetic energy: $K_1 = \frac{1}{2}m_1v_1^2 $ $K_2 = \frac{1}{2}m_2v_2^2 = \frac{2m_1^2v_1^2m_2}{(m_2+m_1)^2}$ Thus, it follows: $\frac{K_2}{K_1} = \frac{4m_1m_2}{(m_2+m_1)^2}$
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