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Unformatted text preview: h ic e8 e k# 8 i b iC a aha 8 Problem 7: The impulse provided is the change in momentum, or J = mv f mv = (0 . 0450 kg)(25 . 00 m / s) = 1 . 12 kg m / s . This is also equal to the average force times the time, so J = Ft F = J t = 560 N . The weight of the ball is 0.441 N, which is negligible in comparison. 8 Problem 10 ( a ): The impulse is J = F t = 104 000 kg m / s . ( b ): The shuttles change in momentum is p = J = 104 000 kg m / s. ( c ): We have p = m v v = p m = 1 . 1 m / s . ( d ): The change in the kinetic energy is 1 2 m ( v 2 f v 2 ) = 1 2 m ( v f v ) ( v f + v ) = m v parenleftbigg v f + v 2 parenrightbigg = J v ave . The change in the kinetic energy is the dot product of the impulse with the average velocity. As we dont know the average velocity, we dont know the change in the kinetic energy. 8 Problem 16 ( a ): Let m be the mass of the ball and M be your mass. The initial and final momenta are the same, so conservation of momentum tells us that mv = ( M + m ) v f v f = m M + m v = 0 . 0568 m / s = 5 . 68 cm / s . ( b ): The ball goes from v = 10 . 0 m/s to v f, ball = 8 . 0 m/s. Thus conservation of momentum tells us that mv = Mv f, you + mv f, ball v f, you = m ( v v f, ball ) M = 0 . 103 m / s = 10 . 3 cm / s ....
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This note was uploaded on 07/12/2008 for the course PHY 121 taught by Professor Chamberlin during the Summer '08 term at ASU.
 Summer '08
 Chamberlin
 Force, Momentum, Work

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