Question

A baseball player hits a fastball that is moving at 37.1 m/s . Immediately after the bat-ball collision, the baseball has a speed of 45.9 m/s

​ in the opposite direction. The bat and ball are in contact for about 1.1 ms, and the mass of a baseball is 145 g. Assume that the ball and bat move horizontally during the collision. Calculate the magnitude of the average force exertted by the bat on the ball during the hit.

Answer 1

The magnitude of the average force exerted by the bat on the baseball during the hit is approximately 11027.27 Newtons, calculated by applying the principles of impulse and the change in momentum over the short collision time.

Step-by-step explanation:

To calculate the magnitude of the average force exerted by the bat on a baseball during the collision, we need to apply the principles of impulse and momentum. The impulse exerted on an object equals the change in momentum of the object. Given that the baseball's initial velocity is 37.1 m/s (in one direction) and its final velocity is 45.9 m/s (in the opposite direction), we can compute the change in velocity (Δv) as the sum of these two speeds since they are in opposite directions.

The mass of the baseball (m) is 145 g, which is 0.145 kg in SI units. The time of collision (t) is given as 1.1 ms, or 0.0011 seconds. We can now use the following equation to find the magnitude of the average force (Favg):

Favg = (m * Δv) / t

The change in velocity Δv is 37.1 m/s + 45.9 m/s = 83.0 m/s (since velocity is a vector quantity and the final velocity is in the opposite direction of the initial velocity).

So, the equation becomes:

Favg = (0.145 kg * 83.0 m/s) / 0.0011 s

Finally, we calculate:

Favg = (0.145 kg * 83.0 m/s) / 0.0011 s = 11027.27 N (rounded to two decimal places)

The magnitude of the average force exerted by the bat on the ball is approximately 11027.27 Newtons.