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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 calculate the average force, one must first find the change in momentum (final momentum minus initial momentum) and then divide it by the time during which the change occurred. The initial momentum, p_initial (before the ball is hit), is given by the product of the mass and the initial velocity: p_initial = (0.157 kg) * (28.2 m/s) = 4.4274 kg.m/s. The final momentum, p_final (after the ball is hit), is given by the product of the mass and the final velocity: p_final = (0.157 kg) * (-15.2 m/s), we used negative sign because the final direction of the ball is opposite to the initial direction; so we get p_final = -2.384 kg.m/s. The change in momentum, ∆p, is then found by subtracting the initial momentum from the final momentum: ∆p = p_final - p_initial = -2.384 kg.m/s - 4.4274 kg.m/s = -6.8114 kg.m/s. The average force, F_avg, exerted by the bat on the ball is then given by the change in momentum divided by the change in time: F_avg = ∆p / ∆t = -6.8114 kg.m/s / 0.00274 s ≈ -2485.4 N. Therefore, the average force exerted by the bat on the ball is about -2485.4 N. The minus sign indicates that the force is in the direction opposite to the initial motion of the baseball.

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