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: 65

Answer

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Work Step by Step

The momentum is given by equation 26–4. $$p=\gamma mv = \frac{\gamma mc^2 v}{c^2}$$ The energy is given by equation 26–6b, $E=\gamma mc^2$. Make a substitution. $$p= \frac{E v}{c^2}$$ This can be solved for v. $$v=\frac{pc^2}{E}$$ This was the first part of the relation to be proven. Now combine this with equation 26–9. $$v=\frac{pc^2}{E}=\frac{pc^2}{\sqrt{p^2c^2+m^2c^4}}=\frac{pc}{\sqrt{p^2+m^2c^2}}$$ This is the relationship shown in the problem, which is now proven.
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