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 i

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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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Dulcie 3 years 2021-08-25T05:25:34+00:00 1 Answers 21 views 0

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  1. King
    0
    2021-08-25T05:27:06+00:00
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    Answer:

    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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