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23 October, 02:25

A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 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. 23 October, 05:30
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    complete question:

    A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 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:

    F = 1776 N

    Explanation:

    mass of ball = 60 g = 0.06 kg

    velocity of downward direction = 22 m/s = v1

    velocity of upward direction = 15 m/s = v2

    Δt = 1/800 = 0.00125 s

    Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.

    p = mv

    When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse.

    I = pf - pi = ∆p

    F = ∆p/∆t = I/∆t

    let the upward velocity be the positive

    Δp = mv2 - m (-v1)

    Δp = mv2 - m (-v1)

    Δp = m (v2 + v1)

    Δp = 0.06 (15 + 22)

    Δp = 0.06 (37)

    Δp = 2.22 kg m/s

    ∆t = 0.00125

    F = ∆p/∆t

    F = 2.22/0.00125

    F = 1776 N
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