Question

A 0.140-kg baseball is thrown with a velocity of 27.1 m/s. It is struck by the bat with an average force of 5000 N, which results in a velocity of 37.0 m/s in the opposite direction from the original velocity. How long were the bat and ball in contact?

Answer 1

Answer:

About $1.795 \times 10^{-3}$ seconds

Explanation:

$\Delta p=F \Delta t$, where delta p represents the change in momentum, F represents the average force, and t represents the change in time.

The change of velocity is:

$37-(-27.1)=64.1m/s$

Meanwhile, the mass stays the same, meaning that the change in momentum is:

$64.1\cdot 0.14kg=8.974$

Plugging this into the equation for impulse, you get:

$8.974=5000\cdot \Delta t \\\\\\\Delta t= \dfrac{8.974}{5000}\approx 1.795 \times 10^{-3}s$

Hope this helps!