A 0.278-kg volleyball approaches a player horizontally with a speed of 16.2 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 22.4 m/s.

Question

(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

Answer 1

Force = change in momentum/time = m a when m is constant
so
Force * time = change of momentum, called impulse

impulse = force * time = change of momentum

final momentum = .278 (22.4)
initial momentum = .278 (-16.2)
change = .278 (22.4 - {- 16.2} )
= .278 (38.6)
= 10.7 kg m/s (part a)

force * time = 10.7
force = 10.7/.06 = 179 Newtons (part b)