Question

A 0.140 kg baseball is thrown with a velocity of 33.6 m/s. It is struck with an average force of 5000.0 N, which results in a velocity of 37.0 m/s in the opposite direction. How long were the bat and ball in contact?

Answer 1

Answer:

0.00198 secs

Explanation:

Parameters given:

Mass of baseball, m = 0.14 kg

Initial velocity of baseball, u = 33.6 m/s

Force applied to baseball, F = -5000 N

(The force is applied in an opposite direction to the initial velocity)

Final velocity, v = -37 m/s

Using the impulse-momentum theory, we have that the impulse applied to the baseball is equal to the change in momentum of the baseball:

FΔt = m(v - u)

Time interval, Δt, will be given as:

Δt = $\frac{m(v - u)}{F}$

Δt = $\frac{0.14(-37.0 - 33.6)}{-5000}$

Δt = $\frac{0.14 * -70.6}{-5000}$

Δt = $\frac{-9.884}{5000}$

Δt = $0.00198 secs$

The bat and the baseball were in contact for 0.00198 secs.