College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 26 - Problems - Page 1011: 43


The mass of the system increases by $1.0\times 10^{-14}~kg$

Work Step by Step

We can assume that the kinetic energy that is lost is converted into additional mass. We can find the initial kinetic energy: $K = \frac{1}{2}m_1~v^2 + \frac{1}{2}m_2~v^2$ $K = \frac{1}{2}(1.00~kg)(30.0~m/s)^2 + \frac{1}{2}(1.00~kg)~(30.0~m/s)^2$ $K = 900~J$ We can find the increase in mass: $E = 900~J$ $mc^2 = 900~J$ $m = \frac{900~J}{c^2}$ $m = \frac{900~J}{(3.0\times 10^8~m/s)^2}$ $m = 1.0\times 10^{-14}~kg$ The mass of the system increases by $1.0\times 10^{-14}~kg$.
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