Three particles, each of mass m, are placed at the corners of a right angled triangle as shown in figure. If OA = a and OB = 5, the position vector of the centre of mass is (Here `hati` and `hatj` are unit vectors along x and m y axes respectively).

Options

A. `(1)/(3) ( ahati + bhatj)`
B. `(1)/(3) (a hati - b hatj)`
C. `(2)/(3) ( ahati + bhatj)`
D. `(2)/(3) (a hati - bhatj)`

Answer

A

Explanation

The (x,y) co-ordinates of the masses at O , A and B respectively are `(x_(1) = 0 , y_(1) = 0) , (x_(2) = a , y_(2) = 0)` and `(x_(3) = 0 , y_(3) = b)`
The (x , y) co-ordinates of the centre of the mass are
`X_(CM)= (m_(1) x_(1) + m_(2) x_(2) + m_(3) x_(3))/(m_(1) + m_(2) + m_(3)) = (m xx 0 + m xx a + m xx 0 )/( m + m + m) = (a)/(3)`
`Y_(CM) = (m_(1) y_(1) + m_(2) y_(2) + m_(3) y_(3))/(m_(1) + m_(2) + m_(3)) = (m xx 0 + m xx 0 + m xx b) /(m + m + m) = (b)/(3)`
The position vector of the centre of the mass is
`vecR_(CM) = X_(CM) hati + Y_(CM) hatj = (a)/(3) hati + (b)/(3) hatj = (1)/(3) (a hati + b hatj)`