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28 September, 16:51

A child bounces a 48 g superball on the sidewalk. The velocity change of the superball is from 28 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

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  1. 28 September, 19:27
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    F = 1.2*10⁻³ N

    Explanation:

    From the question,

    Applying newton's second law of motion,

    F = m (v-u) / t ... Equation 1

    Given: F = magnitude of the average force exerted on the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time of contact.

    Note: let downward be negative and upward be positive.

    Given: m = 48 g = 48/1000 = 0.048 kg, v = 17 m/s, u = - 28 m/s (downward),

    t = 1800 s

    Substitute into equation 1

    F = 0.048 (17-[28]) / 1800

    F = 1.2*10⁻³ N
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