Answer:
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.