An isolated smooth ring of mass M = 2m with two small beads each of mass m is as shown in the figure. Initially both the beads are at diametrically opposite points and have velocity `v_(0)` (for each) in same direction. The speed of the beads just before they collide for the first time is (complete system is placed on a smooth horizontal surface and assume each point of ring is touching the surface)

Options

A. `V_(0)`
B. `(2)/(sqrt(3))V_(0)`
C. `(V_(0))/(2)`
D. `(sqrt(3))/(2) V_(0)`

Answer

D

Explanation

Just before the collision, situation is as shwon
Let the beads have velocity `V_(1)` .w.r.t. the ring and the ring has velocity `V_(2)`.
Then, by momentum conservation `2mv_(0)= (M + 2m) V_(2)`
`rArr V_(2) = (V_(0))/(2)` …(i)
By mechanical energy conservation `(1)/(2) (2m) V_(0)^(2) = (1)/(2) 2mv^(2) + (1)/(2) 2m ((V_(0))/(2))^(2)`
where `v =` velocity of beads .w.r.t. ground
So, `V = (sqrt3)/(2)V_(0)`