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19 August, 19:30

A 0.45 kg ball is moving horizontally with a speed of 5.3 m/s when it strikes a vertical wall. The ball rebounds with a speed of 2.4 m/s. What is the magnitude of the change in linear momentum of the ball

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Answers (2)
  1. 19 August, 20:58
    0
    -3.465 kgm/s

    Explanation:

    Momentum; This can be defined as the product of the mass of a body and its change in velocity. The S. I unit of momentum is kgm/s

    From the question,

    ΔM = m (v-u) ... Equation 1

    Where ΔM = change in momentum, m = mass of the ball, v = final velocity, u = initial velocity

    Note: Let the direction of the initial velocity be the positive direction

    Given: m = 0.45 kg, v = - 2.4 m/s (Rebound), u = 5.3 m/s

    Substitute into equation 1

    ΔM = 0.45 (-2.4-5.3)

    ΔM = 0.45 (-7.7)

    ΔM = - 3.465 kgm/s

    The negative sign means that the change in momentum is along the direction of the final velocity
  2. 19 August, 21:33
    0
    1.31Kgms-1

    Explanation:

    ∆p = m∆v

    Where:

    ∆p = change in momentum

    m = mass of the ball

    ∆v = change in velocity of the ball = (5.3-2.4) = 2.9ms-1

    Therefore, substituting appropriately with the values above:

    ∆p = 0.45*2.9 = 1.31Kgms-1
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