A 0.29kg billiard ball traveling at a speed of 20m/s strikes the side rail of a
ID: 1281786 • Letter: A
Question
A 0.29kg billiard ball traveling at a speed of 20m/s strikes the side rail of a pool table at an angle of 60 ? and rebounds at the same speed and angle. If the billiard ball is in contact with the rail for 0.016s , what is the magnitude of the average force exerted on the ball?
A 0.29kg billiard ball traveling at a speed of 20m/s strikes the side rail of a pool table at an angle of 60 ? and rebounds at the same speed and angle. If the billiard ball is in contact with the rail for 0.016s , what is the magnitude of the average force exerted on the ball?Explanation / Answer
impluse = change in momentum = m( vf - vi)
= 0.29 ( 20cos60 - (-20cos60))
= 5.8 kg.m/s
Impluse = F x time
5.8 = F x 0.016
F =362.5 N
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