Three identicle particle each of mass `1 kg` are placed with their centres on a straight line. Their centres are marked `A, B and C` respectively. The distance of centre of mass of the system from `A` is.

Options

A. `(AB+AC)/(2)`
B. `(AB+BC)/(2)`
C. `(AC-AB)/(3)`
D. `(AB+AC)/(3)`

Answer

D

Explanation

The distance is to be measured from A

`therefore` Origin will be at A
Now, for centre of mass `=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))`
For the above figure
Centre of mass `=(1xx0+1xxAB+1xxAC)/(1+1+1)=(AB+AC)/(3)`