Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 36 - Relativity - Exercises and Problems - Page 1062: 72


(a) $K = 8.2\times 10^{-14}~J$ (b) $v = 0.866~c$

Work Step by Step

(a) Initially there are two electrons. After the collision, there are four particles. Each electron must initially have at least the kinetic energy equal to the rest mass of an electron (or a positron). We can find the threshold kinetic energy: $K = mc^2$ $K = (9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2$ $K = 8.2\times 10^{-14}~J$ (b) We can find $\gamma$: $K = (\gamma-1)mc^2 = mc^2$ $\gamma = 2$ We can find the speed when $\gamma = 2$: $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ $\sqrt{1-\frac{v^2}{c^2}} = \frac{1}{\gamma}$ $1-\frac{v^2}{c^2} = \frac{1}{\gamma^2}$ $\frac{v^2}{c^2} = 1-\frac{1}{\gamma^2}$ $v^2 = (1-\frac{1}{\gamma^2})~c^2$ $v = \sqrt{1-\frac{1}{\gamma^2}}~c$ $v = \sqrt{1-\frac{1}{2^2}}~c$ $v = \sqrt{0.75}~c$ $v = 0.866~c$
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