Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 26 - The Special Theory of Relativity - Problems - Page 768: 19

Answer

$v=0.95c$

Work Step by Step

$p_1=p_2$ $p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2} }}$ $\frac{(6.68\times10^{-27}kg)(0.6c)}{\sqrt{1-\frac{0.36c^2}{c^2} }}=\frac{(1.67\times10^{-27}kg)v}{\sqrt{1-\frac{v^2}{c^2} }}$ $3c=\frac{v}{\sqrt{1-\frac{v^2}{c^2} }}$ $9c^2-9v^2=v^2$ $9c^2=10v^2$ $v=\sqrt{0.9}c=0.95c$
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