A baseball bat contacts a 0.145-kg baseball for 1.3*10-3 s The average force exerted by the bat on the ball is 8900 N. If the ball has an initial velocity of 29 m/s toward the bat and the force of the bat causes the ball's motion to reverse direction, what is the ball's speed as it leaves the bat?

Answer: v2 = - 50.8m/s

Therefore, the speed of the baseball is 50.8m/s away from the bat. (Reverse direction)

Explanation:

Using the law of conservation of momentum:

The impulse of the bat = change in momentum of the baseball

Ft = ∆M

Ft = m1v1 - m2v2

Since m1 = m2 = m (the mass remains unchanged)

Ft = m (v1 - v2) ... 1

Given:

Force F = 8900N

Mass m = 0.145kg

time t = 1.3 * 10^-3 s

Initial velocity v1 = 29m/s

Substituting into eqn 1

8900 (1.3*10^-3) = 0.145 (29-v2)

29-v2 = 79.79

v2 = 29 - 79.79 = - 50.8m/s

v2 = - 50.8m/s

Therefore, the speed of the baseball is 50.8m/s away from the bat.