College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1011: 32


$p = 1.0\times 10^{-18}~kg~m/s$

Work Step by Step

We can find the momentum of the proton: $p = \gamma~m~v$ $p = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}~m~v$ $p = \frac{1}{\sqrt{1-\frac{(0.90~c)^2}{c^2}}}~(1.67\times 10^{-27}~kg)~(0.90~c)$ $p = \frac{1}{\sqrt{1-(0.90)^2}}~(1.67\times 10^{-27}~kg)~(0.90)(3.0\times 10^8~m/s)$ $p = 1.0\times 10^{-18}~kg~m/s$
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