Question

A 0.145 kg ball moving in the +x direction at 14 m/s is hit by a bat. The ball's final velocity is 20.0 m/s in the -x direction. The bat acts on the ball for a duration of 0.010 s.
Find the average force exerted on the ball by the bat. Include the sign of the force to indicate the direction of the force.

Answer 1

Answer:

Approximately $(-4.9\times 10^{2}\; {\rm N})$ (rounded to two significant figures.)

Explanation:

The impulse $\mathbf{J}$ on an object is equal to the change in the momentum $\mathbf{p}$ of this object. Dividing impulse by duration $\Delta t$ of the contact would give the average external force that was exerted on this object.

If an object of mass $m$ is moving at a velocity of $\mathbf{v}$, the momentum of that object would be $\mathbf{p} = m\, \mathbf{v}$.

Let $m$ denote the speed of this ball. Let $\mathbf{v}_{0}$ denote the velocity of this ball before the contact, and let $\mathbf{v}_{1}$ denote the velocity of the ball after the contact.

Momentum $\mathbf{p}_{0}$ of this ball before the contact: $\mathbf{p}_{0} = m\, \mathbf{v}_{0}$.

Momentum $\mathbf{p}_{1}$ of this ball after the contact: $\mathbf{p}_{1} = m\, \mathbf{v}_{1}$.

Impulse, which is equal to the change in momentum:

$\mathbf{J} = \mathbf{p}_{1} - \mathbf{p}_{0} = m\, \mathbf{v}_{1} - m\, \mathbf{v}_{0}$.

Since the velocity of the ball after the contact is in the $(-x)$ direction, $\mathbf{v}_{1} = -20.0\; {\rm m\cdot s^{-1}}$. The average external force on this ball would be:

$\begin{aligned} \mathbf{F} &= \frac{\mathbf{J}}{\Delta t} \\ &= \frac{m\, \mathbf{v}_{1} - m\, \mathbf{v}_{0}}{\Delta t} \\ &= \frac{m\, (\mathbf{v}_{1} - \mathbf{v}_{0})}{\Delta t} \\ &= \frac{0.145\; {\rm kg} \times ((-20.0\; {\rm m\cdot s^{-1}}) - 14\; {\rm m\cdot s^{-1}})}{0.010\; {\rm s}} \\ &\approx -4.9 \times 10^{2}\; {\rm N}\end{aligned}$.

(Rounded to two significant figures.)