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A ball of mass 0.5 kg drops from 10 m high and rebounds to a height of 4 m. The time ofcontact with the ground is 0.05 s. Find the change in velocity during the collision the averageforce acting on the ball by the ground.change in velcity / ms-1 : reaction / NA. 23.1 235B. 14.1 ... 顯示更多 A ball of mass 0.5 kg drops from 10 m high and rebounds to a height of 4 m. The time of contact with the ground is 0.05 s. Find the change in velocity during the collision the average force acting on the ball by the ground. change in velcity / ms-1 : reaction / N A. 23.1 235 B. 14.1 235 C. 8.9 230 D. 23.1 230 ans: a

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Assume no energy is lost throughout the process. Take the direction downward be positive By the law of conservation of energy, Loss of P.E. = Gain of K.E. mghi = 1/2 mu^2 ghi = 1/2 u^2 (10)(10) = 1/2 u^2 u = 14.14 ms^-1 So, the velocity of the ball just before the collision is 14.14 ms^-1 downward. By the law of conservation of energy, Loss of K.E. = Gain of P.E. 1/2 mv^2 = mghf 1/2 v^2 = ghf 1/2 v^2 = (10)(4) v = 8.94 ms^-1 So, the velocity of the ball just after the collision is 8.94 ms^-1 upward. Therefore, change in velocity during the collision = u - v = 14.14 - (-8.94) = 23.1 ms^-1 Let F be the average force acting on the ball by the ground, Weight of the ball, mg = (0.5)(10) = 5 N By impulse = change of momentum (F - mg)t = m(u - v) (F - 5)(0.05) = 0.5(23.1) F - 5 = 230 F = 235 N Therefore, the reaction is 235 N. So, the answer is A.

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