A 450 g baseball is pitched, reaching the batter with a speed of 37 m/s, and is hit such that it returns on the same line at 48 m/s. If the ball was in contact with the bat for 0.22 seconds, what was the magnitude of the average net force experienced by the ball during the time of contact?
A) 170 N
B) 98 N
C) 77 N
D) 18 N

Question

Quizplus · Accepted Answer

The magnitude of the average net force experienced by the ball can be calculated using the formula for force, $ F = \frac{\Delta p}{\Delta t} $, where $ \Delta p $ is the change in momentum and $ \Delta t $ is the change in time. The change in momentum, $ \Delta p = m(v_f - v_i) $, where $ m $ is the mass of the baseball, $ v_f $ is the final velocity, and $ v_i $ is the initial velocity. Given that $ m = 0.45 $ kg (since 450 g = 0.45 kg), $ v_f = -48 $ m/s (the negative sign indicates the direction is opposite to the initial direction), and $ v_i = 37 $ m/s, we find $ \Delta p = 0.45 \times (-48 - 37) = -38.25 $ kg·m/s. The time of contact, $ \Delta t = 0.22 $ s, so the average net force is $ F = \frac{-38.25}{0.22} \approx -173.86 $ N. The magnitude of this force is approximately 174 N, which rounds to 170 N, making option A correct.

A 450 G Baseball Is Pitched, Reaching the Batter with a Speed