Question

1. a baseball of mass 0.15 kg is pitched at 40 m/s and is hit by a bat, changing its velocity to 60 m/s in the opposite direction. if the bat makes contact with the ball for 0.01 seconds, what is the average force exerted by the bat on the ball? (hint: the new velocity is -60 m/s)

Explanation:

Step1: Calculate the initial and final momentum

The formula for momentum is \(p = mv\).
Initial velocity \(v_{i}=40\ m/s\), mass \(m = 0.15\ kg\), so initial momentum \(p_{i}=mv_{i}=0.15\times40 = 6\ kg\cdot m/s\).
Final velocity \(v_{f}=- 60\ m/s\), so final momentum \(p_{f}=mv_{f}=0.15\times(-60)=-9\ kg\cdot m/s\).

Step2: Calculate the change in momentum

The change in momentum \(\Delta p=p_{f}-p_{i}\).
\(\Delta p=-9 - 6=-15\ kg\cdot m/s\).

Step3: Use the impulse - momentum theorem

The impulse - momentum theorem is \(J=\Delta p = F_{avg}\Delta t\).
We know \(\Delta t = 0.01\ s\), and we want to find \(F_{avg}\). Rearranging the formula \(F_{avg}=\frac{\Delta p}{\Delta t}\).
Substitute \(\Delta p=-15\ kg\cdot m/s\) and \(\Delta t = 0.01\ s\) into the formula: \(F_{avg}=\frac{-15}{0.01}=-1500\ N\). The negative sign indicates the direction of the force (opposite to the initial velocity direction of the ball). The magnitude of the average force is \(1500\ N\).

Answer:

The average force exerted by the bat on the ball is \(1500\ N\).