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13 August, 07:15

A child bounces a 56 g superball on the sidewalk. the velocity change of the superball is from 20 m/s downward to 13 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. 13 August, 08:41
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    Using the impulse-momentum theorem and taking this down as positive, we will get:

    FΔt = Δp

    F = Δp / Δt

    = m (v - v₀) / t

    = 0.056kg [13m/s - ( - 20m/s) / 0.00125s

    = 1478.4 N when properly rounded off is the answer we are looking for in this problem.
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