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28 March, 07:02

A 0.42 kg football is thrown with a velocity of 17 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.017 S. What is the force exerted on the receiver? Answer in units of N

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  1. 28 March, 08:47
    0
    420 N

    Explanation:

    m = 0.42 Kg, u = 17 m/s, v = 0 m/s, t = 0.017 s

    By first law of Newtons' laws of motion, the rate of change in momentum is force, F = m (v - u) / t

    F = 0.42 x (0 - 17) / 0.017

    F = - 420 N

    Negative sign shows hat the force is resistive that means the ball finally comes to rest.
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