Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 9 - Section 9.6 - Elastic Collisions - Example - Page 166: 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: $\frac{K_2}{K_1} = \frac{4m_1m_2}{(m_2+m_1)^2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.