College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1012: 49


$v = 0.60~c$

Work Step by Step

We can find the Lorentz factor $\gamma$: $E = \gamma~m~c^2$ $\gamma = \frac{E}{m~c^2}$ $\gamma = \frac{1.02\times 10^{-13}~J}{(9.1\times 10^{-31}~kg)~(3.0\times 10^8~m/s)^2}$ $\gamma = 1.245$ We can find the speed: $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ $\sqrt{1-\frac{v^2}{c^2}} = \frac{1}{\gamma}$ $1-\frac{v^2}{c^2} = (\frac{1}{\gamma})^2$ $\frac{v^2}{c^2} = 1-(\frac{1}{\gamma})^2$ $v = \sqrt{1-(\frac{1}{\gamma})^2}~c$ $v = \sqrt{1-(\frac{1}{1.245})^2}~c$ $v = 0.60~c$
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