In a lecture demonstration - which you should watch on the course website if you

Question

In a lecture demonstration - which you should watch on the course website if you have not seen it in class - a small ball of mass m is held directly above a large ball of mass M with a small space between them, and the two balls arc dropped simultaneously from height H. (The height is much larger than the radius of each ball, so you may neglect the radius.) The large ball bounces elastically off the floor and the small ball bounces elastically off the large ball. For which value of the mass m, in terms of M, docs the large ball stop when it collides with the small ball? What final height, in terms of H, does the small ball reach?

Explanation / Answer

here,
mass of larger ball = M
mass of smaller ball = m
Vf = speed after collision

Part A:
Using Conservation of Energy during fall we have
Mgh = Mv^2/r
The speed of larger ball just before collision
Vi = sqrt(2gh) -----------(1)

After the collision with the floor, the large ball moves upward with the speed vi and collides with the small ball.Therefore
Vf = ( 2m*(-vi) + (M-m)*vi ) / M+m ----------(2)
as after collision Vf = 0

2m*vi/m+m = (M-m)*vi/m+m

M = 3m

Part B:
Using 2 speed of smaller ball after collision is:

vf = (m-M/m+M)(-vi) + (2M/m+M)vi

vf = 2vi
from eqn 1

hf = vf^2 / 2g
hf = 4h

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