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8 December, 02:22

A 0.50 kg ball that is traveling at 9.0 m/s collides head-on with a 1.00 kg ball moving in the opposite direction at a speed of 11.0 m/s. The 0.50 kg ball bounces backward at 15.0 m/s after the collision. Find the speed of the second ball after the collision.

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  1. 8 December, 03:55
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    1 m/s

    Explanation:

    From the law of conservation of momentum,

    Total momentum before collision = Total momentum after collision

    mu + m'u' = mv + m'v' ... Equation 1

    Where m = mass of the first ball m' = mass of the second ball, u = initial velocity of the first ball, u' = initial velocity of the second ball, v = final velocity of the first ball, v' = final velocity of the second ball.

    Note: Let the direction of the first ball be positive.

    Given: m = 0.5 kg, u = 9.0 m/s, m' = 1 kg, u' = - 11 m/s (opposite direction to the first ball), v = - 15 m/s (bounces backward).

    Substitute into equation 1

    0.5 (9) + 1 (-11) = 0.5 (-15) + 1 (v')

    Solving for v'

    4.5-11 = - 7.5 + v'

    -6.5 = - 7.5 + v'

    v' = 7.5-6.5

    v' = 1 m/s.

    Hence the speed of the second ball after collision = 1 m/s
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