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26 September, 05:58

A baseball bat contacts a 0.145-kg baseball for 1.3*10-3 s The average force exerted by the bat on the ball is 8900 N. If the ball has an initial velocity of 29 m/s toward the bat and the force of the bat causes the ball's motion to reverse direction, what is the ball's speed as it leaves the bat?

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  1. 26 September, 07:34
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    Answer: v2 = - 50.8m/s

    Therefore, the speed of the baseball is 50.8m/s away from the bat. (Reverse direction)

    Explanation:

    Using the law of conservation of momentum:

    The impulse of the bat = change in momentum of the baseball

    Ft = ∆M

    Ft = m1v1 - m2v2

    Since m1 = m2 = m (the mass remains unchanged)

    Ft = m (v1 - v2) ... 1

    Given:

    Force F = 8900N

    Mass m = 0.145kg

    time t = 1.3 * 10^-3 s

    Initial velocity v1 = 29m/s

    Substituting into eqn 1

    8900 (1.3*10^-3) = 0.145 (29-v2)

    29-v2 = 79.79

    v2 = 29 - 79.79 = - 50.8m/s

    v2 = - 50.8m/s

    Therefore, the speed of the baseball is 50.8m/s away from the bat.
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