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: 29

Answer

a. 4.50 GeV. b. 5.36 GeV/c.

Work Step by Step

a. By the work-energy theorem, the required work is the change in kinetic energy. The initial kinetic energy is 0, so find the final KE. Find $\gamma$ for this proton. $$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}=\frac{1}{\sqrt{1-0.985^2}}=5.795$$ The kinetic energy is given by equation 26–5b. $$KE=(\gamma-1)mc^2=(5.795-1)(938.3MeV)$$ $$=4499.4MeV\approx 4.50GeV$$ That is the work required. b. The momentum is given by equation 26–4. $$p=\gamma mv=5.795 (938.3MeV/c^2)(0.985c)$$ $$=5356MeV/c\approx 5.36GeV/c$$
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