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7 March, 18:44

A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnitude of the impulse exerted on the ball by the floor

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  1. 7 March, 21:03
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    9.6 Ns

    Explanation:

    Note: From newton's second law of motion,

    Impulse = change in momentum

    I = m (v-u) ... Equation 1

    Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

    Given: m = 2.4 kg, v = 2.5 m/s, u = - 1.5 m/s (rebounds)

    Substitute into equation 1

    I = 2.4[2.5 - (-1.5) ]

    I = 2.4 (2.5+1.5)

    I = 2.4 (4)

    I = 9.6 Ns
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