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 - General Problems - Page 769: 66

Answer

a) $p=3.51\times10^{-18}kg\frac{m}{s}$ b) $0$ c) $p_{rel}=4.98\times10^{-17}kg\frac{m}{s}$

Work Step by Step

a) $p=\gamma mv=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{(1.67\times10^{-27}kg)(0.990c)}{\sqrt{1-\frac{(0.990c)^2}{c^2}}}$ $=3.51\times10^{-18}kg\frac{m}{s}$ b) The two protons are moving in opposite direction, so the total momentum is $0$ c) $v_{rel}=\frac{0.990c+0.990c}{1+\frac{(0.990c)(0.990c)}{c^2}}=0.99995c$ $p_{rel}=\gamma mv=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{(1.67\times10^{-27}kg)(0.99995c)}{\sqrt{1-\frac{(0.99995c)^2}{c^2}}}$ $=4.98\times10^{-17}kg\frac{m}{s}$
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