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 10 8 6 8 201 = = . . a f and 2 2 1 9 8 7 622 = = . . a f . Conservation of energy gives EEE 12 += total which is γγ 11 2 22 mc m c total or 334 10 27 .. . mm × kg. This reduces to: 27 3 09 1 66 10 × . . kg . (1) Since the final momentum of the system must equal zero, pp = gives 111 222 mu = or 2 01 0 868 6 22 0 987 a fa f a fa f cm = which becomes 352 = . . (2) FIG. P39.41 Solving (1) and (2) simultaneously, m 1 28 884 10 . k g and m 2 28 251 10 k g . *P39.42 Energy conservation: 1 1400 900 8 5 1 2 2 2 2 2 + = kg kg c c Mc vc . 3108 1 2 2 kg −= v c M . Momentum conservation: 0 900 0 85 8 5 1 2 + = kg . . c Mv af 1452 1 2 2 kg v c Mv c . (a) Dividing gives v c == 0467 . = (b) Now by substitution 3 108 1 0 467 2 75 10
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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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