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?

Respuesta :

Answer:

About [tex]1.795 \times 10^{-3}[/tex] seconds

Explanation:

[tex]\Delta p=F \Delta t[/tex], 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:

[tex]37-(-27.1)=64.1m/s[/tex]

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

[tex]64.1\cdot 0.14kg=8.974[/tex]

Plugging this into the equation for impulse, you get:

[tex]8.974=5000\cdot \Delta t \\\\\\\Delta t= \dfrac{8.974}{5000}\approx 1.795 \times 10^{-3}s[/tex]

Hope this helps!