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 CM ` = (m_(1) x_(1) + m_(2) x_(2) + m_(3) x_(3))/(m_(1) + m_(2) + m_(3))`
For the above figure
`CM = (1 xx 0 + 1 xx AB + 1 xx AC)/(1 + 1 + 1)`
`= (AB + AC)/(3)`