Question

Two particle masses are free to move inside a Motionless bowl of radius R, and both masses are initially at rest. The mass at the top side is three times the mass at the bottom. Assume the mass at the top side is released and collides with the bottom mass and they stick together. Determine how high from the bottom of the bowl the combined masses will go (in terms of radius R)

Answer 1

The top ball is at a height R from the ground.The velocity of the ball at the bottom can be found using the conservation of energy
Mg R = (1/2) M v²
Where M is the mass of the ball at the top. The ball at the top is 3 times massive than the ball at the top
v = sqrt (2 g R)
The top ball hits the bottom ball with this much of speed. We need to find the resultant speed of the combined body after the collision .We can use the conservation of linear momentum
3m v = (3m + m) V
Where v is the velocity of the top ball, 3m is its mass , the second mass was initially at rest ,so its momentum is zero. (3m + m) is the combined mass of the system after collision.V is the velocity with which the combined mass system will move .
V = 3 v / 4
Again using the conservation of energy we can find the height the system will reach.
(1/2) (4m) V²  = (4m) g h
h = V² / 2 g
h = 9 v² / 32 g
h = 9x 2 gR / 32 g
h = 9 R / 16