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6 July, 14:16

A cue stick hits a cue ball with an average force of 22 N for a duration of 0.029 s. If the mass of the ball is 0.15 kg, how fast is it moving after being struck?

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Answers (2)
  1. 6 July, 14:44
    0
    4.25 m/s

    Explanation:

    An impulse is defined as a change in momentum, and can be found my multiplying an applied force by its duration. In this case,

    that's 22*0.029 = 0.638 kg-m/s. Momentum is also mass times velocity, and your velocity is unknown.

    Momentum P = mv

    0.15 v = 0.638

    v = 4.25 m/s
  2. 6 July, 14:44
    0
    4.25 m/s

    Explanation:

    Force, F = 22 N

    Time, t = 0.029 s

    mass, m = 0.15 kg

    initial velocity of the cue ball, u = 0

    Let v be the final velocity of the cue ball.

    Use newton's second law

    Force = rate of change on momentum

    F = m (v - u) / t

    22 = 0.15 (v - 0) / 0.029

    v = 4.25 m/s

    Thus, the velocity of cue ball after being struck is 4.25 m/s.
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