Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 7 - Linear Momentum - Problems: 10


The ratio of the masses is $\frac{2}{1}$

Work Step by Step

Let the masses be $m_1$ and $m_2$. By the law of conservation of momentum: $m_1v_1 = m_2v_2$ This can be written as: $v_1 = \frac{m_2v_2}{m_1}$ Let's assume that $m_1$ has twice the kinetic energy. Therefore, $\frac{1}{2}m_1v_1^2 = 2\times \frac{1}{2}m_2v_2^2$ $m_1(\frac{m_2v_2}{m_1})^2 = 2\times m_2v_2^2$ $\frac{m_2^2}{m_1} = 2m_2$ $\frac{m_2}{m_1} = 2$ The ratio of the masses is $\frac{2}{1}$, where the larger fragment with half as much kinetic energy has double the mass of the smaller fragment.
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