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 EKm c 1 2 =+ Em c 2 2 = Ep cm c 1 2 1 22 24 p 2 0 = . In the final state, EKM c ff 2 : cM c 2 2 4 . By energy conservation, EEE f 12 += , so EE E E E f 1 2 2 2 ++ = pc mc K m c m c Mc f 1 2 2 2 + + ej By conservation of momentum, pp f 1 = . Then Km c Km c mc 2 2 4 2 4 =+= + Mc mc K mc 2 21 2 . By contrast, for colliding beams we have In the original state, c 1 2 c 2 2 . In the final state, EM c f = 2 f : c Km c M c 2 Mc mc K mc 2 2 F H G I K J . FIG. P39.71 *P39.72 Conservation of momentum γ mu : mu uc mu Mv vc mu f f 13 1 1 2 31 2 2 + = = a f . Conservation of energy mc 2 : mc mc Mc mc f 2 2 2 2 1 1 4 + = = . To start solving we can divide:
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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