A 0.277-kg volleyball approaches a player horizontally with a speed of 14.9 m/s.

Question

A 0.277-kg volleyball approaches a player horizontally with a speed of 14.9 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 21.8 m/s.

(a) What impulse is delivered to the ball by the player? (Take the direction of final velocity to be the positive direction. Indicate the direction with the sign of your answer.)
kg · m/s

(b) If the player's fist is in contact with the ball for 0.0600 s, find the magnitude of the average force exerted on the player's fist.
N

Explanation / Answer

(a)

If the direction of final velocity is the positive direction, then, initial momentum of the ball is,

P1 =  -(0.277 kg)(14.9 m/s)

and the final momentum is;

P2 = (0.277 kg)(21.8 m/s)

So impulse provided = P2 - P1 = (0.277 kg)(21.8 m/s) - -(0.277 kg)(14.9 m/s) =  (0.277 kg)[(21.8 m/s) + (14.9 m/s)]

or, impulse = +10.2 kg-m/s

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(b)

Force exerted by the player's fist is F = (+10.2 kg-m/s)/0.06s

or F = +169.43 N

So the force exerte on the player's fist is -F or -169.43 N.

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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