Expert: James J. Kovalcin - 11/13/2006

Question
A baseball (m = 149 g) approaches a bat horizontally at a speed of 43.0 m/s (96 mi/h) and is hit straight back at a speed of 49.8 m/s (111 mi/h). If the ball is in contact with the bat for a time of 1.10 ms, what is the average force exerted on the ball by the bat? Neglect the weight of the ball, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.

Answer
According to the impulse equation:
F*Dt=m*Dv where Dt is the time interval over which the change occurs, Dv is the change in velocity [always the final velocity minus the initial velocity] and m is the mass of the baseball.
In this case solving for F:
F=m*Dv/Dt=m*[Vf-Vo]/Dt
F=0.149kg*[-49.8m/s-43.0m/s]/[1.10x10^3s]