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A batter strikes a baseball of mass 0.85 kg, and the time of impact between the bat and the ball is 0.01 s. If the ball acquires a velocity of 82 m/s after the hit, calculate the average force exerted on the ball.

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5 votes

Answer:

6970 N

Step-by-step explanation:

To calculate the average force exerted on the ball, we can use the impulse-momentum theorem, which states that the impulse on an object is equal to its change in momentum. In equation form:

Impulse = Δp

where Impulse is the force applied over a given time, and Δp is the change in momentum of the object.

We can calculate the momentum of the ball before the hit as:

p1 = m * v1

where m is the mass of the ball and v1 is its initial velocity (which we assume to be zero). Substituting the given values, we get:

p1 = (0.85 kg) * 0 m/s = 0 kg m/s

The momentum of the ball after the hit is:

p2 = m * v2

where v2 is the final velocity of the ball (82 m/s). Substituting the given values, we get:

p2 = (0.85 kg) * 82 m/s = 69.7 kg m/s

The change in momentum (Δp) is therefore:

Δp = p2 - p1 = 69.7 kg m/s - 0 kg m/s = 69.7 kg m/s

The impulse on the ball is equal to the change in momentum, so we have:

Impulse = Δp = 69.7 kg m/s

Finally, we can calculate the average force exerted on the ball using the formula:

Impulse = Force * time

Substituting the given values, we get:

69.7 kg m/s = Force * 0.01 s

Solving for Force, we get:

Force = 6970 N

Therefore, the average force exerted on the ball by the bat is 6970 Newtons.

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