A 0.15 kg baseball collides with a 1.0 kg bat. The ball has a velocity of 40 m/s immediately before the collision. The center of mass of the bat also has a velocity of 40 m/s, but in the opposite direction, just before the collision. The coefficient of restitution between the bat and the ball is 0.50. Estimate how fast the baseball is moving as it leaves the bat following the collision.

The final velocity of the baseball as it leaves the bat is 40 m/s

Explanation:

The given parameters of the baseball and bat are;

The mass of the baseball = 0.15 kg

The mass of the bat = 1.0 kg

The velocity of the ball before collision, v₁ = 40 m/s

The velocity of the bat before collision, v₂ = -40 m/s

The coefficient of restitution, e = 0.50

Let, 'v₃', and 'v₄' represent the final velocity of the ball and the bat respectively after collision, we have;

Taking the final velocity of the bat, v₄ = 0 m/s

According to Newton's Law of restitution

e = (v₃ - v₄)/(v₁ - v₂)

∴ 0.5 = (v₃ - v₄)/(40 - (-40))

80 × 0.5 = 40 = (v₃ - v₄)

v₃ - v₄ = 40

v₃ =  40 + v₄ = 40 + 0 = 40

The final velocity of the baseball as it leaves the bat, v₃ = 40 m/s.

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