267_Physics ProblemsTechnical Physics

# 267_Physics ProblemsTechnical Physics - Chapter 9 P9.48 (a)...

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Chapter 9 269 P9.48 (a) Conservation of momentum for the two-ball system gives us: 0 200 1 50 0 300 0 400 0 200 0 300 12 .. . . . . kg m s kg m s kg kg bg b g +− = + vv ff Relative velocity equation: 21 190 −= . m s Then 0 300 0 120 0 200 0 300 1 90 11 ... . . + + di v f 1 0780 =− m s v f 2 112 = . m s vi 1 f . ± ms 2 f = . ± (b) Before, v ii CM kg m s kg m s kg = 0 200 1 50 0 300 0 400 0500 ± ± . b g CM = 0360 . ± Afterwards, the center of mass must move at the same velocity, as momentum of the system is conserved. Section 9.7 Rocket Propulsion P9.49 (a) Thrust = v dM dt e Thrust × = × 2 60 10 1 50 10 3 90 10 34 7 . kg s N ej e j (b) FM g M a y = Thrust : 3 90 10 3 00 10 9 80 3 00 10 76 6 . . ×− × = × a f a a = 320 m s 2 *P9.50 (a) The fuel burns at a rate dM dt == × 12 7 668 10 3 . . g 1.90 s kg s Thrust = v dM dt e : 5 26 6 68 10 3 N k g s v e v e = 787 m s (b) vvv M M fie i f F H G I K J ln : v f + F H G I K J 07 9 7 53 5 25 5 25 5 12 7 g g 53.5 g g g ln v f =
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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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