Question

When Reggie stepped up to the plate and hit a .15 kg fast ball traveling at 36m/s the impact caused the ball to leave his bat with a velocity of 45m/s in the oppsite direction. If the impact lasted 0.0020s, what force did Reggie exert on the ball

Answer 1

Assuming the initial direction of motion before impact as negative and final direction of motion as positive.

m = mass of the ball being hit = 0.15 kg

v₀ = initial velocity of the ball before the impact = - 36 m/s

v = final velocity of the ball after impact = 45 m/s

t = time of impact = 0.0020 s

F = force Reggie exerted on the ball

Using Impulse-change in momentum equation

F t = m (v - v₀)

F (0.0020) = (0.15) (45 - (- 36))

F = 6075 N