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Unformatted text preview: hÝ × ic×ÙÑÑ eÖ8ÓÑ eÛ Ó Ök# 8 Ó ÐÙ Ø iÓÒ ×bÝ iÑC a ÐÐahaÒ § 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 ˆ kg m / s . ( b ): The shuttle’s change in momentum is Δ p = J = 104 000 ˆ kg m / s. ( c ): We have Δ p = m Δ v ⇒ Δ v = Δ p m = 1 . 1 ˆ 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 Δ 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 don’t know the average velocity, we don’t 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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 Summer '08
 Chamberlin
 Force, Mass, Momentum, Work, m/s, MD

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