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 1060: 33

Answer

$v = 0.8~c$

Work Step by Step

We can use the expression for relativistic momentum to find the particle's speed: $p = \gamma ~m ~v$ $p = \frac{m ~v}{\sqrt{1-\frac{v^2}{c^2}}}$ $p~\sqrt{1-\frac{v^2}{c^2}} = m~v$ $p^2~(1-\frac{v^2}{c^2}) = m^2~v^2$ $p^2 = m^2~v^2+\frac{p^2~v^2}{c^2}$ $p^2 = (m^2+\frac{p^2}{c^2})~v^2$ $v^2 = \frac{p^2}{m^2+\frac{p^2}{c^2}}$ $v^2 = \frac{(4.0\times 10^5~kg~m/s)^2}{(1.0\times 10^{-3}~kg)^2+\frac{(4.0\times 10^5~kg~m/s)^2}{(3.0\times 10^8~m/s)^2}}$ $v^2 = \frac{(4.0\times 10^5~kg~m/s)^2}{2.78\times 10^{-6}~kg^2}$ $v = \frac{4.0\times 10^5~kg~m/s}{\sqrt{2.78\times 10^{-6}~kg^2}}$ $v = 2.4\times 10^8~m/s$ $v = 0.8~c$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.