College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1013: 72


The sum of the kinetic energies is $0.77~MeV$

Work Step by Step

We can find the missing mass $M$: $1.00866~u = 1.00728~u+0.00055~u+M$ $M = 1.00866~u - 1.00728~u- 0.00055~u$ $M = 0.00083~u$ We can assume that the missing mass is converted into kinetic energy. We can find the energy $E$: $E = Mc^2$ $E = (0.00083~u)~c^2$ $E = (0.00083)~(931.5~MeV/c^2)~(c^2)$ $E = 0.77~MeV$ The sum of the kinetic energies is $0.77~MeV$.
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