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1104_Physics ProblemsTechnical Physics

# 1104_Physics ProblemsTechnical Physics - Chapter 39 mu = mu...

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Chapter 39 445 P39.30 Using the relativistic form, p mu u c mu = = 1 2 b g γ we find the difference p from the classical momentum, mu : p mu mu mu = = γ γ 1 b g . (a) The difference is 1.00% when γ γ = 1 0 010 0 b g mu mu . : γ = = 1 0 990 1 1 2 . u c b g thus 1 0 990 2 2 F H G I K J = u c . a f , and u c = 0 141 . . (b) The difference is 10.0% when γ γ = 1 0 100 b g mu mu . : γ = = 1 0 900 1 1 2 . u c b g thus 1 0 900 2 2 F H G I K J = u c . a f and u c = 0 436 . . P39.31 p mu mu mu mu mu = = γ γ 1 : γ = + F H G I K J = F H G I K J 1 1 1 1 1 1 2 1 1 2 2 2 2 u c u c u c b g p mu mu = × F H G I K J = × 1 2 90 0 3 00 10 4 50 10 8 2 14 . . . m s m s P39.32 p mu u c = 1 2 b g becomes 1 2 2 2 2 2 = u c m u p which gives: 1 1 2 2 2 2 = + F H G I K J u m p c or c u m c p 2 2 2 2 2 1 = + F H G I K J and u c m c p = + 2 2 2 1 e j . P39.33 Relativistic momentum of the system of fragments must be conserved. For total momentum to be zero after as it was before, we must have, with subscript 2 referring to the heavier fragment, and
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