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## GR0877 Problem 22

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (C) $E = \gamma m_e c^2 \equiv 4m_e c^2$. From this, $\gamma = 4$ and $v = c\sqrt{1 - 1/\gamma^2} = \sqrt{15}c/4$. The momentum is $p = \gamma m_e v = \sqrt{15} m_e c.$

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Categories: 0877 Tags: , , , ,
1. 09.11.2011 at 01:49

Nice! Another way is to use the super-useful expression: $E^2=(mc^2)^2 + (pc)^2$.

$(4mc^2)^2=(mc^2)^2 + (pc)^2$

$15(mc^2)^2=p^2 c^2$

$p = \sqrt{15}mc$

(Sorry for the ugly formatting).

Physicsworks:edited