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: 69

Answer

$E_K=\frac{-2mc^2\pm\sqrt{4m^2c^4+4p^2c^2}}{2}$

Work Step by Step

$E^2=p^2c^2+m^2c^4$ $E=E_K+mc^2$ $E^2=E_K^2+2E_Kmc^2+m^2c^4$ $p^2c^2+m^2c^4=E_K^2+2E_Kmc^2+m^2c^4$ $E_K^2+2E_Kmc^2-p^2c^2=0$ $E_K=\frac{-2mc^2\pm\sqrt{4m^2c^4+4p^2c^2}}{2}$ When mass is $0$ $E_K=pc$
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