Questions: A 0.164-kg baseball is moving horizontally to the left at 2 m / s when it is hit by a bat. The ball flies off in the exact opposite direction. If the bat hits the ball with an average force of 508.1 N and contact time of 23.3 ms, what is the final speed of the ball?

Solution Steps

We need to find the final speed of a baseball after it is hit by a bat. We know the initial speed, the mass of the baseball, the average force applied by the bat, and the contact time.

The impulse-momentum theorem states that the change in momentum of an object is equal to the impulse applied to it. Mathematically, this is expressed as: \[ F \cdot \Delta t = m \cdot \Delta v \] where \( F \) is the force, \( \Delta t \) is the contact time, \( m \) is the mass, and \( \Delta v \) is the change in velocity.

First, calculate the impulse: \[ F \cdot \Delta t = 508.1 \, \text{N} \times 0.0233 \, \text{s} = 11.8373 \, \text{Ns} \]

Next, use the impulse to find the change in velocity: \[ \Delta v = \frac{F \cdot \Delta t}{m} = \frac{11.8373 \, \text{Ns}}{0.164 \, \text{kg}} = 72.186 \, \text{m/s} \]

The initial velocity of the baseball is \( -2 \, \text{m/s} \) (since it is moving to the left). The change in velocity is positive, indicating a reversal in direction. Therefore, the final velocity \( v_f \) is: \[ v_f = v_i + \Delta v = -2 \, \text{m/s} + 72.186 \, \text{m/s} = 70.186 \, \text{m/s} \]

Final Answer