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29 March, 11:42

Darna rolls the 7.05-kg ball down the lane and it hits the 1.52-kg pin head on. The ball was moving at 8.24 m/s before the collision. The pin went flying forward at 13.2 m/s. Determine the post-collision speed of the ball.

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  1. 29 March, 11:47
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    The velocity of the ball after the collision is 5.39 m/s

    Explanation:

    Hi there!

    To solve this problem, we will use the conservation of momentum: the momentum of the system ball-pin remains the same before and after the collision. The momentum of the system is calculated as follows:

    momentum before the collision (initial momentum) = mb · vb1 + mp · vp1

    momentum after the collision (final momentum) = mb · vb2 + mp · vp2

    Where:

    mb = mass of the ball = 7.05 kg

    vb1 = velocity of the ball before the collision = 8.24 m/s

    mp = mass of the pin = 1.52 kg

    vp1 = velocity of the pin before the collision = 0 m/s.

    vb2 = velocity of the ball after the collsion = unknown.

    vp2 = velocity of the pin after the collision = 13.2 m/s

    Since momentum is conserved, then:

    initial momentum = final momentum

    mb · vb1 + mp · vp1 = mb · vb2 + mp · vp2

    Solving for vb2:

    mb · vb1 + mp · vp1 - mp · vp2 = mb · vb2

    (mb · vb1 + mp · vp1 - mp · vp2) / mb = vb2

    Since the pin is initially at rest, vp1 = 0:

    (mb · vb1 - mp · vp2) / mb = vb2

    (7.05 kg · 8.24 m/s - 1.52 kg · 13.2 m/s) / 7.05 kg = vb2

    vb2 = 5.39 m/s

    The velocity of the ball after the collision is 5.39 m/s
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