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 uc mu = = 1 2 bg γ we find the difference p from the classical momentum, mu : pm u m u m u =− = γγ 1 . (a) The difference is 1.00% when −= 10 0 1 0 0 mu mu . : == 1 0990 1 1 2 . thus 9 9 0 2 2 F H G I K J = u c . af , and = 0141 . . (b) The difference is 10.0% when 1 0 0 mu mu . : 1 0900 1 1 2 . thus 9 0 0 2 2 F H G I K J = u c . and = 0436 . . P39.31 u mu mu mu mu = 1: −≈+ F H G I K J F H G I K J 1 1 1 11 1 2 1 1 2 2 22 u c u c u mu = × F H G I K J 1 2 90 0 300 10 450 10 8 2 14 . . . ms P39.32 p mu = 1 2 becomes 1 2 2 2 u c mu p which gives: 1 1 2 2 =+ F H G I K J u m pc or cu mc p 2 1 F H G I K J and u c mc p = + 22 2 1 ej . 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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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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