Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 29 - Relativity - Problems and Conceptual Exercises - Page 1044: 83

Answer

a) $v=0.6427c$ b) $P=2.29\times 10^{-22}Kg.m/s$ c) $P=2.29\times 10^{-22}Kg.m/s$ d) $v=9.85Km/s$

Work Step by Step

(a) We know that $\frac{v}{c}=\sqrt{1-\frac{1}{(1+\frac{K}{m_{\circ}c^2})^2}}$ We plug in the known values to obtain: $\frac{v}{c}=\sqrt{1-\frac{1}{(1+\frac{156KeV}{511KeV})^2}}$ This simplifies to: $v=0.6427c$ (b) We can find the required momentum as follows: $P=\frac{m_{\circ}v}{\sqrt{1-\frac{v^2}{c^2}}}$ We plug in the known values to obtain: $P=\frac{(9.11\times 10^{-31}Kg)(0.6427c)}{\sqrt{1-(\frac{0.6427c}{c})^2}}$ $\implies P=2.29\times 10^{-22}Kg.m/s$ (c) We know that $P_e+P_{\beta}=0$ $\implies P_e=-P_{\beta}$ $\implies P_e=-(2.29\times 10^{-22}Kg.m/s)$ (d) We know that $v=c[1+\frac{m_{\circ}^2c^2}{p^2}]^{-\frac{1}{2}}$ We plug in the known values to obtain: $v=(3\times 10^8m/s)[1+\frac{(14.003242)^2(1.66\times 10^{-27}Kg)^2(3\times 10^8m/s)^2}{(-2.29\times 10^{-22}Kg.m/s)^2}]^{-\frac{1}{2}}$ $v=9.85Km/s$
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