4. a big league hitter attacks a fastball! the ball has a mass of 0.16 kg. it is pitched at 38 m/s. after the player hits the ball, it is now traveling 44 m/s in the opposite direction. the impact lasted 0.002 seconds. how big of a force did the ballplayer put on that ball?

Use conservation of linear momentum (p) and impulse (I)

Formulas:

p = m*v

I = Δp = f * t

1) Linear momentum of the ball before being hitted by the player:

p = m * v = 0.16 kg * ( - 38 m/s) = - 6.08 N*s

2) linear momentum of the ball after being hitted:

p = m * v = 0.16 kg * (44 m/s) = + 7.04 N*s

3) Change of linear momentum

Δp = + 7.04 n*s - ( -6.08 N*s) = 13.12 N*s

4) Impulse

I = Δp = f * t => f = Δp * t = 13.12 N*s / 0.002 s = 6,560 N

6,560 N

Ithink the answer would be a