A) what is the impulse delivered to ball by player B) if t=.0600 what is magnitude of average force on players hand?

Question

A 0.258-kg volleyball approaches a player horizontally with a speed of 17.8 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 22.0 m/s.



Answer 1

Impulse J = Δp = m(v_f - v_i) = (0.258 kg) [22.0 -(-17.8)] = 10.2₇ kg m/s. Note the large change in momentum because the velocities are in opposite directions.

Also J = Δt, so average force = (10.2₇ kg m/s) / (0.0600 s) = 171 N.