## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 26 - Problems - Page 1012: 62

#### Answer

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

#### 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 $K$: $(pc)^2 = K^2+2KE_0$ $(pc)^2 \approx 2KE_0$ $K = \frac{(pc)^2}{2E_0}$ $K = \frac{p^2c^2}{2mc^2}$ $K = \frac{p^2}{2m}$

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