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6 April, 23:08

In tae-kwon-do, a hand is slammed down onto a target at a speed of 10 m/s and comes to a stop during the 7.0 ms collision. Assume that during the impact the hand is independent of the arm and has a mass of 0.90 kg. What are the magnitudes of the (a) impulse and (b) average force on the hand from the target

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  1. 7 April, 02:01
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    (A) Impulse = 9Ns

    (B) F = 1286N

    Explanation:

    Impulse = change in momentum = m (v-u)

    v = 0 (the hand comes to a stop)

    u = - 10m/s

    Mass = 0.9kg

    Impulse = 0.9 * (0 - (-10))

    = 9Ns

    (B) F*t = Impulse

    F = Impulse / t

    t = 7ms = 7*10-³

    F = 9 / (7*10-³)

    F = 1286N.
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