In a game of pool you strike a ball with a cue. The typical impact time of the c

Question

In a game of pool you strike a ball with a cue. The typical impact time of the cue and the ball is 10 ms. The speed of the ball immediately after the impact with the cue, measured by the speed gun was 3 m/s. How large was the average force you delivered to the ball. The cue-ball mas mass 0f 170 grams. Before you took a shot you determined that you need to hit the 8-ball with the cue head on to make the corner pocket. The 8-ball has mass of 156 grams. Assume that the cue ball lost about 20% of the speed before it hit the 8-ball. What will be the speed of the 8-ball immediately after it was hit by the cue-ball? If the 8-ball looses about 30% of energy as it rolls towards the pocket, what will be it's speed when it gets to the pocket? (Moment of inertia of the ball is I=2mr2/3 I 2 m r 2 3 . The diameter of the pool-ball is 5.71 cm 5.71 c m .) Assume that the collision between the cue-ball and the 8-ball is elastic.

Explanation / Answer

t = time of impact of cue and ball = 10 ms = 0.01 sec

Vi = initial speed before ball was hit = 0 m/s

Vf = final speed after the ball was hit = 3 m/s

F = force applied = m (Vf - Vi) /t = (0.170) (3 - 0) /(0.01) = 51 N

for cue ball and 8 ball collision

V1i = initial velocity of cue ball before collision = 0.80 x 3 = 2.4 m/s

V2i = initial velocity of 8 ball before collision = 0 m/s

V1f = Final velocity of cue ball after collision

V2f = Final velocity of 8 ball after collision

using conservation of momentum

m1 V1i + m2 V2i = m1 V1f + m2 V2f

(0.17) (2.4) + (0.156) (0) = (0.17) V1f + (0.156) V2f

(0.17) V1f + (0.156) V2f = 0.41    

V1f = (0.41 - (0.156) V2f )/(0.17)                           Eq-1

using conservation of kinetic energy

m1 V21i + m2 V22i = m1 V21f + m2 V22f

(0.17)(2.4)2 + (0.156) (0)2 = (0.17) ( (0.41 - (0.156) V2f )/(0.17) )2 + (0.156) V22f

V2f = 2.5 m/s

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