What is the magnitude of the average force exerted on a 0.150 kg baseball hit back at a pitcher at 180 km/h if the collision lasts for 5.0 milliseconds?

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To find the magnitude of the average force exerted by the bat on the baseball, we can use the impulse-momentum theorem. This theorem states that the impulse (force multiplied by time) is equal to the change in momentum of the object.

Step-by-Step Calculation

1. Convert Units:

  • First, we need to convert the speed of the baseball from kilometers per hour to meters per second.
  • 180 km/h = (180 * 1000 m) / (3600 s) = 50 m/s

2. Calculate Initial and Final Momentum:

  • The mass of the baseball, m = 0.150 kg.
  • Assuming the baseball was initially moving toward the pitcher at a certain speed (let’s assume it’s 0 m/s for simplicity as we are only observing the impact), the initial momentum is:
    p_initial = m * v_initial = 0.150 kg * 0 m/s = 0 kg·m/s
  • After being hit, the final momentum can be calculated:
    p_final = m * v_final = 0.150 kg * 50 m/s = 7.5 kg·m/s (in the opposite direction)

3. Calculate Change in Momentum:

The change in momentum is:

  • Δp = p_final – p_initial = 7.5 kg·m/s – 0 kg·m/s = 7.5 kg·m/s.

4. Calculate Average Force:

Now, we use the impulse-momentum theorem:

  • Impulse = Force * time = Δp
  • Where time = 5.0 ms = 5.0 * 10-3 s.
  • Force * (5.0 * 10-3) = 7.5 kg·m/s.

Rearranging the equation gives us:

  • Average Force = Δp / time = (7.5 kg·m/s) / (5.0 * 10-3 s) = 1500 N.

Conclusion:

The magnitude of the average force exerted by the bat on the baseball is 1500 Newtons.

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