A 0.266-kg volleyball approaches a player horizontally with a speed of 15.8 m/s.

Question

A 0.266-kg volleyball approaches a player horizontally with a speed of 15.8 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 21.9 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.) Answer in 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. Answer in N.

Explanation / Answer

m =0.266 kg , u = -15.8 m/s, v =21.9 m/s

(a) impulse = change in momentum = m(v- u)

= 0.266(21.9 +15.8)

impulse = 10.03 kg.m/s

(b) t =0.06 s

Force *time= impulse

F*0.06 = 10.03

F = 167.14 N

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