Three particles of masses 1 kg, 2 kg and 3 kg are situated at the corners of an equilateral triangle of side b in the x-y plane with mass 1 kg at the origin and 2 kg on the x-axis. The coordinates of the centre of mass are

Options

A. `((7b)/(12) , (3 sqrt3 b)/(12))`
B. `((3 sqrt3 b)/(12) , (7b)/(12))`
C. `((b)/(12) , (3 sqrt3 b)/(12))`
D. `((7b)/(12) , (sqrt3b)/(12))`

Answer

A

Explanation

The coordinates of points A , B and C are (0 , 0) (b , 0) and `((b)/(2) , (bsqrt3)/(2))` respectively
`X_(CM) = (1 xx 0 + 2 xx b + 3 (b //2))/(1 + 2 + 3) = (7b)/(12)`
`Y_(CM) = (1 xx 0 + 2 xx 0 + 3sqrt3 (b //2))/(1 + 2 + 3) = (3sqrt3 b)/(12)`
So , the coordinates of centre of mass are `((7b)/(12) , (3 sqrt3b)/(12))`