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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. Steve
    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

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