Question

During a tennis volley, a ball arrives at a player at 40 m/s is struck by the racket and returned at 40 m/s. The other player, realizing the ball is out of bounds, catches it in her hand. Compare the force exerted by the player with racket with the force exerted by the second player assuming that the time interval of contact is the same in both cases.

Answer 1

Answer:

The force exerted by the racket player is greater than the force exerted by the catcher.

Explanation:

So the average force exerted by both players is the impulse divided by the time of impact

$F = \frac{\Delta P}{\Delta t}$

with the time of impact being the same for both cases, the impulse is what dictates the average force.

Impulse is the change in momentum, which is the product of mass and change in velocity.

For the racket-struck case, the change in velocity is from 40m/s to -40m/s in opposite direction, a change of 80m/s

Fotr the hand-catching case, the change in velocity is from 40m/s to 0m/s (rest), a change of 40m/s

Both cases have the same mass. So the change of momentum, and impulse, of the racket-struck's case is larger than the hand-catching's case, same for the average force.