A 0.278-kg volleyball approaches a player horizontally with a speed of 12.5 m/s.

Question

A 0.278-kg volleyball approaches a player horizontally with a speed of 12.5 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 21.6 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

part a )

Impulse = change in momentum

I = m(v2-v1)

given :

v2 = 21.6 m/s

v1 = -12.5 m/s

m = 0.278 kg

I = 0.278 * (21.6-(-12.5))

I = 9.4798 kg.m/s

part b )

I = F.dt

F = I/dt

dt = 0.0600 s

F = 157.997 N = 158 N

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