Question

A baseball approaches home plate at a speed of 48.0 m/s, moving horizontally just before being hit by a bat. The batter hits a pop-up such that after hitting the bat, the baseball is moving at 59.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 2.10 ms. What is the average vector force the ball exerts on the bat during their interaction

Answer 1

The average vector force exerted by the ball on the bat during their interaction is 758.1 N in the upward direction.

Step-by-step explanation:

To find the average vector force exerted by the bat on the baseball, we can use the impulse-momentum principle. The impulse experienced by the baseball is equal to the change in momentum, which can be calculated by subtracting the initial momentum from the final momentum.

First, let's convert the mass of the baseball from grams to kilograms: 145 g = 0.145 kg.

Using the formula for momentum (p = mv), we can calculate the initial momentum of the baseball:

pi = (mass)(initial velocity)

pi = (0.145 kg)(48.0 m/s) = 6.96 kg·m/s

Next, we can calculate the final momentum of the baseball:

pf = (mass)(final velocity)

pf = (0.145 kg)(59.0 m/s) = 8.555 kg·m/s

The change in momentum is Δp = pf - pi:

Δp = 8.555 kg·m/s - 6.96 kg·m/s = 1.595 kg·m/s

The average vector force can be calculated by dividing the change in momentum by the time of interaction:

average force = Δp / t

average force = 1.595 kg·m/s / 2.10 ms = 758.1 N

Therefore, the average vector force exerted by the ball on the bat during their interaction is 758.1 N in the upward direction.