10. A ball of mass 0.745 kg is dropped from rest from a height of 3.75 m. It rebounds from the ground and reaches a height of 1.78 m, what i

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10. A ball of mass 0.745 kg is dropped from rest from a height of 3.75 m. It rebounds from the ground and reaches a height of 1.78 m, what impulse (specify magnitude and direction) was delivered to the ball by the ground?

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Nguyệt Ánh 4 years 2021-07-29T03:27:57+00:00 1 Answers 18 views 0

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  1. haidang
    0
    2021-07-29T03:29:21+00:00
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    Answer:

    10.77kgm/s

    Explanation:

    Impulse is defined as change in momentum of a body.

    From Newton’s second law

    F = ma

    F = m(v-u)/t

    Ft = m(v-u) = Impulse

    where;

    M is the mass of the object

    v is the final velocity of the object

    u is the initial velocity of the object.

    Using the formula to time the impulse

    Impulse = m(v-u)

    m = 0.745kg

    Since the ball dropped from rest at a height of 3.75m, it’s initial velocity can be calculated according to the equation of motion

    g is positive (since it dropped from a height I.e downward motion)

    v²= u²+2gH

    v² = 0²+2(9.8)(3.75)

    v² = 73.5

    v = √73.5

    v = 8.57m/s(initial velocity of the ball)

    If the ball rebounds to a height if 1.78m, g will be negative (since it is an upward motion)

    In this case

    v² = u²-2gH

    0² = u²-2(9.8)(1.78)

    -u² = -34.89

    u = √34.89

    u = 5.91m/s (final velocity of the ball)

    Impulse = 0.745(5.91-(-8.57))

    Impulse = 0.745(5.91+8.57)

    Impulse = 0.745(14.48)

    Impulse = 10.77kgm/s

    Note that the initial velocity was negated while calculating impulse due to rebounce.

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