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1 June, 11:47

A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wall and rebounds with 70.0% of its initial kinetic energy. what is the magnitude of the change in momentum of the stone?

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  1. 1 June, 12:02
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    Given:

    m = 0.5 kg, the mass of the ball

    v₁ = 20 m/s, the impact velocity

    The KE of the ball at impact is

    KE = (1/2) * (0.5 kg) * (20 m/s) ² = 100 J

    Because the ball rebounds with 70% of its KE, the rebound velocity, v₂, is

    (1/2) * (0.5 kg) * (v₂ m/s) ² = 0.7 * (100 J)

    0.25v₂² = 70

    v₂ = 16.733 m/s (in the opposite direction)

    The change in momentum of the ball is

    ΔP = (0.25 kg) * ) 20 - (-16.733) m/s)

    = 9.1833 (kg-m) / s

    Because momentum is conserved, this change in momentum applies to the stone as well.

    Answer: 9.2 (kg-m) / s
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