Answer to Question #142132 in Physics for Ranxe

Question

A baseball of 145 grams is pitched at a speed of 30 m/s hit by a baseball bat causing the baseball to drive straight back horizontally at a speed of 45 m/s. The contact time between baseball and bat is 5.0 milliseconds, what is the magnitude of the force between the baseball and bat?

Answer 1

We can find the magnitude of the force between the baseball and bat from the impulse-momentum change equation:

$$F_{avg}\Delta t=m\Delta v,$$ $$F_{avg}\Delta t=m(v_f-v_i),$$

here, $F_{avg}$ is the average force between the baseball and bat, $\Delta t=5.0\cdot 10^{-3}\ s$ is the contact time between baseball and bat, $m=0.145\ kg$ is the mass of the baseball, $v_i=30\ \dfrac{m}{s}$ is the initial velocity of the baseball, $v_f=-45\ \dfrac{m}{s}$ is the final velocity of the baseball.

Then, from this equation we can find the average force between the baseball and bat:

$$F_{avg}=\dfrac{m(v_f-v_i)}{\Delta t},$$ $$F_{avg}=\dfrac{0.145\ kg\cdot((-45\ \dfrac{m}{s})-30\ \dfrac{m}{s})}{5.0\cdot 10^{-3}\ s}=-2175\ N.$$

The magnitude of the average force between the baseball and bat is 2175 N. The sign minus means that the average force between the baseball and bat directed toward the pitcher.

Answer:

$F_{avg}=2175\ N,$ toward the pitcher.