A 0.272-kg volleyball approaches a player horizontally with a speed of 12.8 m/s.

Question

A 0.272-kg volleyball approaches a player horizontally with a speed of 12.8 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 23.0 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. Need Help? Read It

Explanation / Answer

Part A

Impulse is given by:

J = Change in momentum = dP

J = m*dV

J = m*(Vf - Vi)

m = 0.272 kg = mass of volleyball

Vf = +23.0 m/sec

Vi = -12.8 m/sec

So,

J = 0.272*(23.0 - (-12.8))

J = 9.738 kg-m/sec

Part B

Average froce is given by:

F = J/dt

F = 9.738/0.0600

F = 162.3 N

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