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21 November, 20:39

A child bounces a 48 g superball on the sidewalk. the velocity change of the superball is from 26 m/s downward to 17 m/s upward. if the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? answer in units of n.

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  1. 21 November, 22:57
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    By definition we have the momentum is:

    P = m * v

    Where,

    m = mass

    v = speed

    Before the impact:

    P1 = (0.048) * (26) = 1.248 kg * m / s

    After the impact:

    P2 = (0.048) * ( - 17) = - 0.816 Kg * m / s.

    Then we have that deltaP is:

    deltaP = P2-P1

    deltaP = ( - 0.816) - (1,248)

    deltaP = - 2,064 kg * m / s.

    Then, by definition:

    deltaP = F * delta t

    Clearing F:

    F = (deltaP) / (delta t)

    Substituting the values

    F = ( - 2.064) / (1/800) = - 1651.2N

    answer:

    the magnitude of the average force exerted on the superball by the sidewalk is 1651.2N
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