1106_Physics ProblemsTechnical Physics

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

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Chapter 39 447 P39.37 Em c m c == γ 22 2 or = 2 . Thus, u c =− F H G I K J = 1 13 2 2 or u c = 3 2 . The momentum is then pm um cm c c F H G I K J = F H G I K J 2 3 2 3 2 p cc = F H G I K J 938 3 3 1 63 10 3 . . MeV MeV . P39.38 (a) Using the classical equation, Km v × = × 1 2 1 2 78 0 1 06 10 4 38 10 25 2 11 .. . kg m s J bg ej . (b) Using the relativistic equation, K vc mc = F H G G I K J J 1 1 1 2 2 . K = −× × L N M M M M O Q P P P P ×= × 1 1 1 06 10 2 998 10 1 780 2998 10 438 10 58 2 8 2 11 . kg m s J When v c << 1 , the binomial series expansion gives 11 1 2 2 12 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 2 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 . 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) c R × × = −− 7 8 2 10 1 67 10
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