College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 26 - Problems - Page 1012: 62


$K = \frac{p^2}{2m}$

Work Step by Step

Note that if $K \lt \lt E_0$, then the expression $K^2+2KE_0 \approx 2KE_0$ We can find an expression for $K$: $(pc)^2 = K^2+2KE_0$ $(pc)^2 \approx 2KE_0$ $K = \frac{(pc)^2}{2E_0}$ $K = \frac{p^2c^2}{2mc^2}$ $K = \frac{p^2}{2m}$
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.