College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1012: 50


(a) $E = 532.65~MeV$ (b) The total energy in the muon's frame is $106~MeV$

Work Step by Step

(a) We can find the total energy in the Earth observer's frame: $E = \gamma~m~c^2$ $E = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}~m~c^2$ $E = \frac{1}{\sqrt{1-\frac{(0.980~c)^2}{c^2}}}~(106~MeV)$ $E = (5.025)(106~MeV)$ $E = 532.65~MeV$ (b) In the muon's frame, the muon is at rest. Therefore, the total energy in the muon's frame is $106~MeV$
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