A baseball with a mass of 151 g is thrown horizontally with a speed of 40.3 m/s (90 mi/h) at a bat. The ball is in contact with the bat for 1.15 ms and then travels straight back at a speed of 46.0 m/s (103 mi/h). Determine the average force (in N) exerted on the ball by the bat. Neglect the weight of the ball (it is much smaller than the force of the bat) and choose the direction of the incoming ball to be positive. (Indicate the direction with the sign of your answer.)

A baseball (m= 149g) approaches a bat horizontally at a speed of 40.2 m/s (90 mi/h) and is hit straight back at a speed of 45.6m/s (102mi/h). If the ball is in contact with the bat for a time of 1.10ms, 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.

Explanation:

Use the impulse equation (a form of Newton's 2nd Law): FΔt = Δ(mv) where Δ means "change in"

The change in momentum is mBB(vf - vi) = (.150 kg)(-46.9 m/s - 40.5 m/s)

Divide this by the time interval and you get F exerted by the bat in Newtons.

Take care.

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.