A baseball batter hits an incoming 45-m/s fastball.The ball leaves the bat at 50 m/s after a ball-on-bat contact time of 0.050 s.What is the force exerted on the 0.10-kg baseball?

Question

Quizplus · Accepted Answer

The force exerted on the baseball can be calculated using the impulse-momentum theorem: F = Δp/Δt. The change in momentum (Δp) is m(v_f - v_i) = 0.10 kg * (50 m/s - (-45 m/s)) = 0.10 kg * 95 m/s = 9.5 kg*m/s. The force is then F = 9.5 kg*m/s / 0.050 s = 190 N.

A Baseball Batter Hits an Incoming 45-M/s Fastball