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1107_Physics ProblemsTechnical Physics

# 1107_Physics ProblemsTechnical Physics - 448 P39.41...

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448 Relativity P39.41 We must conserve both energy and relativistic momentum of the system of fragments. With subscript 1 referring to the 0 868 . c particle and subscript 2 to the 0 987 . c particle, γ 1 2 1 1 0 868 2 01 = = . . a f and γ 2 2 1 1 0 987 6 22 = = . . a f . Conservation of energy gives E E E 1 2 + = total which is γ γ 1 1 2 2 2 2 2 m c m c m c + = total or 2 01 6 22 3 34 10 1 2 27 . . . m m + = × kg. This reduces to: m m 1 2 27 3 09 1 66 10 + = × . . kg . (1) Since the final momentum of the system must equal zero, p p 1 2 = gives γ γ 1 1 1 2 2 2 m u m u = or 2 01 0 868 6 22 0 987 1 2 . . . . a fa f a fa f c m c m = which becomes m m 1 2 3 52 = . . (2) FIG. P39.41 Solving (1) and (2) simultaneously, m 1 28 8 84 10 = × . kg and m 2 28 2 51 10 = × . kg . *P39.42 Energy conservation: 1 1 0 1 400 900 1 0 85 1 2 2 2 2 2 2 2 + = kg kg c c Mc v c . 3 108 1 2 2 kg = v c M . Momentum conservation: 0 900 0 85 1 0 85 1 2 2 2 + = kg . . c Mv v c a f 1 452 1 2 2 kg = v c Mv c . (a) Dividing gives v c = = 1 452 3 108 0 467
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