Question

A baseball of mass 0.145 kg sits on a tee. A kid's bat hits the ball and maintains contact for

0.0500 s. The result is that the ball leaves the tee with velocity 15.0 m/s, traveling exactly
horizontally.
(a) Find the momentum of the ball as it leaves the tee.
(b) Find the net impulse.
(c) Find the net force that acted on the ball.

Answer 1

Step-by-step explanation:

(a) To find the momentum of the ball as it leaves the tee, we can use the equation:

Momentum = mass * velocity

Given:

Mass of the ball (m) = 0.145 kg

Velocity of the ball (v) = 15.0 m/s (since it is traveling horizontally)

Momentum = 0.145 kg * 15.0 m/s

Momentum = 2.175 kg·m/s

Therefore, the momentum of the ball as it leaves the tee is 2.175 kg·m/s.

(b) To find the net impulse, we can use the impulse-momentum theorem, which states that:

Impulse = Change in momentum

Since the ball starts from rest (initial momentum is zero) and ends up with a momentum of 2.175 kg·m/s, the change in momentum is:

Change in momentum = Final momentum - Initial momentum

Change in momentum = 2.175 kg·m/s - 0 kg·m/s

Change in momentum = 2.175 kg·m/s

Therefore, the net impulse acting on the ball is 2.175 kg·m/s.

(c) The net force acting on the ball can be found using Newton's second law of motion:

Force = Change in momentum / Time

Given:

Change in momentum = 2.175 kg·m/s (from part (b))

Time (Δt) = 0.0500 s

Force = 2.175 kg·m/s / 0.0500 s

Force = 43.5 N

Therefore, the net force that acted on the ball is 43.5 Newtons.