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1116_Physics ProblemsTechnical Physics

1116_Physics ProblemsTechnical Physics - Chapter 39*P39.71...

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Chapter 39 457 *P39.71 Take the two colliding protons as the system E K mc 1 2 = + E mc 2 2 = E p c m c 1 2 1 2 2 2 4 = + p 2 0 = . In the final state, E K Mc f f = + 2 : E p c M c f f 2 2 2 2 4 = + . By energy conservation, E E E f 1 2 + = , so E E E E E f 1 2 1 2 2 2 2 2 + + = p c m c K mc mc m c p c M c f 1 2 2 2 4 2 2 2 4 2 2 2 4 2 + + + + = + e j By conservation of momentum, p p f 1 = . Then M c Kmc m c Km c mc m c 2 4 2 2 4 2 4 2 2 4 2 4 4 2 4 = + = + Mc mc K mc 2 2 2 2 1 2 = + . By contrast, for colliding beams we have In the original state, E K mc 1 2 = + E K mc 2 2 = + . In the final state, E Mc f = 2 E E E f 1 2 + = : K mc K mc Mc + + + = 2 2 2 Mc mc K mc 2 2 2 2 1 2 = + F H G I K J . FIG. P39.71 *P39.72 Conservation of momentum γ mu : mu u c m u u c Mv v c mu u c f f 1 3 1 1 2 3 1 2 2 2 2 2 2 2 2 + = = a f . Conservation of energy γ mc 2 :
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