A 0.265-kg volleyball approaches a player horizontally with a speed of 17.4 m/s.

Question

A 0.265-kg volleyball approaches a player horizontally with a speed of 17.4 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 21.7 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

From the defintion of Impulse:

Impulse = chnage in momentum

(a)

Initial momentum = mu = ( 0.265-kg ) (17.4 m/s) = 4.611 kg m/s

Final momentum = mv= ( 0.265-kg )( -21.7 ) = - 5.7505 kgm/s

Therefore, the impulse:

Impulse = 4.611 kg m/s - (- 5.7505 kgm/s)= 10.36 kg m/s

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(b) Avergae force = impulse /time = (10.36 kg m/s) / ( 0.060 s) =172.7 N

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