College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1011: 39


$v = 1.55\times 10^7~m/s$

Work Step by Step

We can find the Lorentz factor $\gamma$: $K = (\gamma-1)~m~c^2$ $\gamma = 1+\frac{K}{m~c^2}$ $\gamma = 1+\frac{ 2.50~MeV}{1875.6~MeV}$ $\gamma = 1.001333$ We can find the speed: $\gamma = 1.001333$ $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = 1.001333$ $\sqrt{1-\frac{v^2}{c^2}} = \frac{1}{1.001333}$ $1-\frac{v^2}{c^2} = (\frac{1}{1.001333})^2$ $v = \sqrt{1-(\frac{1}{1.001333})^2}~c$ $v = \sqrt{1-(\frac{1}{1.001333})^2}~(3.0\times 10^8~m/s)$ $v = 1.55\times 10^7~m/s$
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