Two balls placed almost on top of one another are dropped from a height of 1 m.
ID: 1464915 • Letter: T
Question
Two balls placed almost on top of one another are dropped from a height of 1 m. The bottom ball hits the ground and bounces with a coefficient of restitution of 0.5. An instant later (i.e., after having lost none of its new upward velocity), it hits the second ball, which is still traveling down. The collision between the balls has a coefficient of restitution of 0.75. The mass of the bottom ball is nine times that of the top ball. What is the upward speed of the top ball after its collision? Assume in this question that the balls have zero radius (i.e., each one drops the full meter before its contact), and gravity is 10 m s^-2.Explanation / Answer
Height of dropping = 1 m
So velocity just before hitting the ground = (2gH)1/2 where g = 9.8 m/s2
= (19.6 )1/2 = 4.4 m/s
Coefficient of restitution = 0.5
So velocity just aafter bouncing = 2.2 m/s
The mass of bottom ball is 9 times that of top ball And coefficients of restitution = 0.75
So 1st Momentum conservation
9m(2.2) - m(4.4) = 9mV1 + m V2 where V1 and V2 are upward velocities of heavy and smaller ball respectevely.
2nd V2 -V1 = 0.75 (4.4 + 2.2)
Solving both the above equations we get
V1 = 1.05 m/s
V2 = 6.0 m/s