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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.
First, let's find the change in velocity of the baseball. The ball was initially moving in one direction, and then it was hit back in the opposite direction. So when calculating the change in velocity, we get: Change in velocity = Final velocity - Initial velocity = -15.2 m/s - 28.2 m/s = -43.4 m/s Negative sign indicates the direction of the velocity has changed. The change in momentum of the ball (which is also called Impulse) is found by multiplying the mass of the ball by its change in velocity. Momentum is always conserved, meaning the total momentum before the collision (between the bat and the ball) is equal to the total momentum after the collision. So, the change in momentum (Impulse) is: Impulse = mass * change in velocity = 0.157 kg * -43.4 m/s = -6.81 kgm/s The average force exerted on the ball by the bat can be found using the formula: Force = change in momentum / time = Impulse / time = -6.81 kg*m/s / 0.00274 s = -2485.4 N The negative sign indicates the force is applied in the opposite direction to the initial motion of the baseball. So the average force exerted by the bat on the ball is approximately 2485.4 Newtons.
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