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A pitcher throws a 0.141-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just before it makes contact with the bat. The batter then hits the ball straight back at the pitcher with a speed of 50 m/s. Assume the ball travels along the same line leaving the bat as it followed before contacting the bat.

(a) What is the magnitude of the impulse delivered by the bat to the baseball?


(b) If the ball is in contact with the bat for 0.0046 s, what is the magnitude of the average force exerted by the bat on the ball?


(c) How does your answer to part (b) compare to the weight of the ball?
Fav/mg=

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1 answer

  1. It crossed home plate horizontally ? - clever vertical plane curve ball :)
    before:
    Vx = 42
    momentumx = Px = .141*42 = 5.922 kg m/s
    after:
    Vx = -50
    momentum = Px = -.141*50 = -7.05 kg m/s

    impulse = change of momentum = -7.05-5.922 = - 12.972 kg m/s
    so 13 kg m/s (part A)

    force = change of momentum/time
    = 13/.0046 = 2820 Newtons (part B)

    weight = m g = .141*9.81 = 1.38 Newtons
    2820/1.38 = 2038 times the weight

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