A 0.2 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds at 20 m/s. The magnitude

Question

A 0.2 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds at 20 m/s. The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is _______.

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Thiên Hương 3 years 2021-08-15T04:35:50+00:00 1 Answers 429 views 0

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  1. diemkieu
    0
    2021-08-15T04:36:58+00:00
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    Answer:

    10 kgm/s

    Explanation:

    Change in momentum: This can be defined as the product of mass and change in velocity. The S.I unit of change in momentum is Kgm/s.

    From the question,

    ΔM = m(v-u)…………………. Equation 1

    Where ΔM = change in momentum, u = initial velocity, v = final velocity.

    Note: Let upward direction be negative, and downward direction be positive.

    Given: m = 0.2 kg, v = -20 m/s, u = 30 m/s

    Substitute into equation 1

    ΔM = 0.2(-20-30)

    ΔM = 0.2(-50)

    ΔM = -10 kgm/s.

    The negative sign shows that the change in momentum is Upward

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