College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1012: 54


$p = 1.45~MeV/c$

Work Step by Step

$mc^2 = 0.511~MeV$ $mc = 0.511~MeV/c$ Since the total energy is 3.00 times the rest energy, the total energy $E = 3.00~mc^2$ We can find the momentum $p$: $E^2 = p^2c^2+m^2c^4$ $(3.00~mc^2)^2 = p^2c^2+m^2c^4$ $9.00~m^2c^4 = p^2c^2+m^2c^4$ $p^2c^2 = 8.00~m^2c^4$ $p^2 = 8.00~m^2c^2$ $p = \sqrt{8.00}~mc$ $p = (\sqrt{8.00})~(0.511~MeV/c)$ $p = 1.45~MeV/c$
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