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: 26

Answer

2.7 GeV, 3.5 GeV/c.

Work Step by Step

Find $\gamma$ for this proton. Note that v/c = 0.9667. $$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}=\frac{1}{\sqrt{1-0.9667^2}}=3.9075$$ The kinetic energy is given by equation 26–5b. $$KE=(\gamma-1)mc^2=(3.9075-1)(938.3MeV)$$ $$=2728.2MeV\approx2.7GeV$$ The momentum is given by equation 26–4. $$p=\gamma mv=3.9075(938.3MeV/c^2)(0.9667c)$$ $$=3554MeV/c\approx 3.5GeV/c$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.