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28 May, 02:51

A ball of mass 2.0 kg falls vertically and hits the ground with speed 7.0 ms-1 as shown below.

The ball leaves the ground with a vertical speed 3.0 m/s. The magnitude of the change in momentum of the ball is

A. zero.

B. 8.0 Ns.

C. 10 Ns.

D. 20 Ns.

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Answers (2)
  1. 28 May, 03:18
    0
    Magnitude of the change in momentum of the ball is 8.0Ns

    Explanation:

    Momentum of a body is defined as the product of mass of the body and its change in velocity.

    Momentum = mass * change in velocity

    If change in velocity = final velocity - initial velocity

    Momentum = velocity (final velocity-initial velocity)

    Momentum = m (v-u)

    Given;

    Mass of the ball m = 2.0kg

    If the ball leaves the ground with a vertical speed 3.0 m/s,

    Initial velocity of the ball u = 3.0m/s

    If the ball falls vertically and hits the ground with speed 7.0 m/s,

    Final velocity of the ball v = 7.0m/s

    Substituting the given datas into the momentum formula will be equivalent to;

    Momentum = m (v-u)

    Momentum = 2.0 (7-3)

    Momentum = 2*4

    Momentum = 8.0kgm/s or 8.0Ns

    Magnitude of the change in momentum of the ball is 8.0Ns
  2. 28 May, 03:18
    0
    8.0 Ns

    Explanation:

    Change in momentum is given as:

    Final momentum - Initial momentum

    = m*v - m*u

    Where m = mass of ball

    v = final velocity

    u = initial velocity

    Change in momentum = (2.0 * 3.0) - (2.0 * 7.0)

    = 6.0 - 14.0 = - 8.0 Ns

    The magnitude will be |-8.0| = 8.0 Ns
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