A 0.060-kg handball is thrown straight toward a wall with a speed of 10 m/s. It

Question

A 0.060-kg handball is thrown straight toward a wall with a speed of 10 m/s. It rebounds straight backward at a speed of 9.2 m/s.
(a) What impulse is exerted on the wall?
magnitude kg·m/s
direction

(b) If the ball is in contact with the wall for 3.0 ms, what average force is exerted on the wall by the ball?
N

(c) The rebounding ball is caught by a player who brings it to rest. During the process, her hand moves back 0.40 m. What is the impulse received by the player?
magnitude N·s
direction

(d) What average force was exerted on the player by the ball?
N

Explanation / Answer

Data: Mass of the ball, m = 0.060 kg Initial speed, vi = 10 m/s Speed of rebound, vf = - 9.2 m/s [ since it is opposite to vi ] Solution: (a) Impulse, I = m ( vi - vf )                   = 0.060 * ( 10 + 9.2 )                   = 1.152 kg.m/s Ans: Magnitude of impulse, I = 1.152 kg.m/s Direction: towards the wall (b) Time, t = 3 ms              = 3 x 10^-3 s Impulse = Force * time Force, F = I / t                 = 1.152 / 3 x 10^-3                 = 384 N Ans: Force, F = 384 N (c) Initial speed, vi = 9.2 m/s Final speed, vf = 0 m/s Distance, s = 0.4 m Impulse, I = m ( vi - vf )                  = 0.06 * ( 9.2 - 0 )                  = 0.552 kg.m/s Ans: Magnitude of impulse, I = 0.552 N.s Direction: towards the player (d) s = [ ( vi + vf ) / 2 ] * t 0.4 = ( 9.2 / 2 ) * t t = 0.087 s Force, F = I / t                 = 0.552 / 0.087                 = 6.35 N Ans: Force, F = 6.35 N

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