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A 0.060 kg tennis ball moving at 15 m/s strikes a tennis racket and rebounds at 10 m/s in the opposite direction. The ball is in contact with the racket for 0.030 s. What is the magnitude of the average force exerted by the racket on the ball

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Answer:We can use the impulse-momentum theorem to find the average force exerted by the racket on the ball:

impulse = change in momentum

The impulse can be calculated as the product of the force and the time during which the force is applied:

impulse = force × time

We can rearrange the first equation to solve for the force:

force = impulse / time

We can find the impulse by calculating the change in momentum of the ball:

change in momentum = final momentum - initial momentum

The initial momentum of the ball is:

p1 = m × v1 = 0.060 kg × 15 m/s = 0.9 kg m/s

The final momentum of the ball is:

p2 = m × v2 = 0.060 kg × (-10 m/s) = -0.6 kg m/s

The negative sign indicates that the direction of the momentum is opposite to the initial direction, as the ball rebounds in the opposite direction.

The change in momentum is:

Δp = p2 - p1 = -0.6 kg m/s - 0.9 kg m/s = -1.5 kg m/s

The impulse is equal to the change in momentum:

impulse = Δp = -1.5 kg m/s

The time during which the force is applied is:

t = 0.030 s

Therefore, the average force exerted by the racket on the ball is:

force = impulse / time = (-1.5 kg m/s) / (0.030 s) = -50 N

The negative sign indicates that the force is in the opposite direction to the initial direction of the ball. The magnitude of the force is 50 N.

Step-by-step explanation:

User Rishabh Kumar
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