Two paricles of equal mass `m` go round a circle of radius R under the action of their mutual gravitional attraction. The speed of each particle is

Options

A. `(1)/(sqrt3)v_(0)`
B. `(sqrt3)/(2)v_(0)`
C. `(2)/(sqrt3)v_(0)`
D. `(sqrt7)/(3)v_(0)`

Answer

D

Explanation

`COLM rArr 2mv_(0) = 3mv_(1) rArr v_(1) = (2)/(3)v_(0)`
`COME rArr 2 xx (1)/(2)mv_(0)^(2) = 2 xx (1)/(2)m(v_(1)^(2) + v_(2)^(2)) + (1)/(2)mv_(1)^(2)`
`rArr 2mv_(0)^(2) = 2m((2)/(3)v_(0))^(2) + 2mv_(2)^(2) + m((2)/(3)v_(0))^(2)`
`rArr 2v_(2)^(2) = 2v_(0)^(2) - 3 ((2)/(3)v_(0))^(2) rArr v_(2) = (v_(0))/(sqrt3)`
Velocity of particle `= sqrt(v_(1)^(2) + v_(2)^(2)) = sqrt((4v_(0)^(2))/(9) + (v_(0)^(2))/(3)) = (sqrt7)/(3)v_(0)`