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28 May, 04:19

slader A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s and it bounces back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball?

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  1. 28 May, 06:46
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    Answer: 800N

    Explanation:

    Given:

    Mass of ball = 0.8kg

    Contact time = 0.05 sec

    Final velocity = initial velocity = 25m/s

    Magnitude of the average force exerted on the wall by the ball is can be calculated using the relation;

    Force (F) = mass (m) * average acceleration (a)

    a = (initial velocity (u) + final velocity (v)) / t

    m = 0.8kg

    u = v = 25m/s

    t = contact time of the ball = 0.05s

    Therefore,

    a = (25 + 25) : 0.05 = 1000m/s^2

    Therefore,

    Magnitude of average force (F)

    F=ma

    m = mass of ball = 0.8

    a = 1000m/s^2

    F = 0.8 * 1000

    F = 800N
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