A baseball has mass 0.145 kg.

(a) If the velocity of a pitched ball has a magnitude of and the batted ball's velocity is in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

(b) If the ball remains in contact with the bat for 2.00 ms, find the magnitude of the average force applied by the bat.

a). p=15.67kg*m/s

b). F=7.83N

Explanation:

change in momentum is the subtraction from "after momentum" of the "before momentum" as momentum is a vector quantity this is a vector subtraction.

initial momentum of ball

m1v1 = (0.145) (44.0) = 6.38 kg-m/s

after momentum

m1v2 = (0.145) (64.0) = 9.28 kg-m/s

since these momentums are 180° opposite one must be called negative so their difference

6.38+9.28 = 15.67 kg-m/s

change in momentum = 15.67 kg-m/s ANS a1

Impulse = change in momenutm = 15.67 ANS a2

b)

Impulse = Favg*t

I = (2.00) Favg

Impulse from (a2) = 15.67

Favg = 15.67/2.00 =

F = 7.83 N