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 - Exercises and Problems - Page 172: 66



Work Step by Step

We first use the fact that half of the kinetic energy is lost to find: $\frac{1}{2}m_1v^2 + \frac{1}{2}m_2v^2=(m_1+m_2)v_f^2$ $v_f = \sqrt{\frac{\frac{1}{2}(m_1+m_2)v^2}{m_1+m_2}}$ We use conservation of momentum to find: $m_1v-m_2v=(m_1+m_2)v_f$ $m_1v-m_2v=(m_1+m_2)\sqrt{\frac{\frac{1}{2}(m_1+m_2)v^2}{m_1+m_2}}$ $(m_1-m_2)v=(m_1+m_2)\sqrt{\frac{1}{2}}v$ $(m_1-m_2)=\sqrt{\frac{1}{2}}m_1+m_2\sqrt{\frac{1}{2}}$ $\frac{m_1}{m_2}=\frac{1-\sqrt{\frac{1}{2}}}{\sqrt{\frac{1}{2}}+1}$
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