College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1011: 35


(a) $p = 6.7\times 10^9~kg~m/s$ (b) $t = 1.6\times 10^7~s$

Work Step by Step

(a) We can find the momentum: $p = \gamma~m~v$ $p = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}~m~v$ $p = \frac{1}{\sqrt{1-\frac{(0.87~c)^2}{c^2}}}~(12.6~kg)~(0.87~c)$ $p = \frac{1}{\sqrt{1-(0.87)^2}}~(12.6~kg)~(0.87)(3.0\times 10^8~m/s)$ $p = 6.7\times 10^9~kg~m/s$ (b) We can find the required time the force must be applied: $F~t = p$ $t = \frac{p}{F}$ $t = \frac{6.7\times 10^9~kg~m/s}{424.6~N}$ $t = 1.6\times 10^7~s$
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