Abaseball (m = 145 g) approaches a bat horizontally at a speed of 38.9 m/s (87 mi/h) and is hit straight back at a speed of 45.3 m/s (101 mi/h). estimating that the ball is in contact with the bat for a time of 1.00 ms, what is the average force exerted on the ball by the bat? choose the direction of the incoming ball as the positive direction.

Fav = -12209 N

Explanation:

let p1 be the initial momentum of the ball and p2 be the final momentum of the ball. let v1 be the initial velocity of the ball and v2  be the final velocity of the ball.

From Newton's second law of motion, we get that the avarage force acting on the ball is given by:

Fav = Δp/t

      = (p2 - p1)/t

      = m(v2 - v1)/t

      = [(145×10^-3)(-45.3 - 38.9)/(1×10^-3)

      = - 12209 N

Therefore, the average force applies on the ball by tha bat is 12209 in the direction opposite to the balls initial direction.