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1106_Physics ProblemsTechnical Physics

1106_Physics ProblemsTechnical Physics - Chapter 39 P39.37...

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Chapter 39 447 P39.37 E mc mc = = γ 2 2 2 or γ = 2 . Thus, u c = F H G I K J = 1 1 3 2 2 γ or u c = 3 2 . The momentum is then p mu m c mc c = = F H G I K J = F H G I K J γ 2 3 2 3 2 p c c = F H G I K J = × 938 3 3 1 63 10 3 . . MeV MeV . P39.38 (a) Using the classical equation, K mv = = × = × 1 2 1 2 78 0 1 06 10 4 38 10 2 5 2 11 . . . kg m s J b g e j . (b) Using the relativistic equation, K v c mc = F H G G I K J J 1 1 1 2 2 b g . K = × × L N M M M M O Q P P P P × = × 1 1 1 06 10 2 998 10 1 78 0 2 998 10 4 38 10 5 8 2 8 2 11 . . . . . e j b g e j kg m s J When v c << 1 , the binomial series expansion gives 1 1 1 2 2 1 2 2 F H G I K J L N M M O Q P P + F H G I K J v c v c . Thus, 1 1 1 2 2 1 2 2 F H G I K J L N M M O Q P P F H G I K J v c v c . and the relativistic expression for kinetic energy becomes K v c mc mv F H G I K J = 1 2 1 2 2 2 2 . That is, in the limit of speeds much smaller than the speed of light, the relativistic and classical expressions yield the same results. P39.39 (a) E mc R = = × × = × = 2 27 8 2
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