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Question: A(n) 0.157 kg baseball is thrown with a speed of 28.2 m/s. It is hit straight back at the pitcher with a final speed of 15.2 m/s.Find the average force exerted by the bat on the ball if the two are in contact for 0.00274 s.
To find the average force exerted by the bat on the ball, we first need to find the change in momentum which is calculated by the formula: Δp = mvf - mvi = m(vf - vi). Where m is the mass of the baseball, vf is the final velocity after the ball is hit, vi is the initial velocity before the ball is hit, Δp is the change in momentum. Then, to find the average force, we use the formula: F = Δp/Δt. Where F is the average force, Δt is the time of contact between the bat and the ball. Let's calculate: The change in momentum is: Δp = m(vf - vi) = 0.157 kg * (15.2 m/s - (-28.2 m/s)) = 0.157 kg * 43.4 m/s = 6.81 kg*m/s Next, the average force is: F = Δp/Δt = 6.81 kg*m/s / 0.00274 s F ≈ 2487.59 N. So the average force exerted by the bat on the ball is approximately 2488 N.
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