An arrangement of two blocks placed one above the other is shown. There is no fricition any where. A horizontal force `F` acts on one of the blocks as shown. It is required to calculate the acceleration of centre of mass.

Explanation

Method-1 : The FBDs of two blocks in ground's frame are shown.

For `m_(2)`, `a_(2)=0`, `N_(2)=m_(2)g`
For `m_(1)`, `a_(1)=(F)/(m_(1))`, `N_(2)+m_(1)g=N_(1)`
`impliesN_(1)=(m_(1)+m_(2))g`
Now, `a_(CM)=(m_(1)a_(1)+m_(2)a_(2))/(m_(1)+m_(2))=(m_(1)xx(F)/(m_(1))+0)/(m_(1)+m_(2))`
`=(F)/(m_(1)+m_(2))`
Method-2 :
By definition of centre of mass, the total mass of the system `m_(1)+m_(2)` is situated at a point and the external forces are acting on it. The FBD of centre of mass is shown. Net force in vertical direction is zero.

`impliesN_(1)=(m_(1)+m_(2))g`
and `a_(CM)=(F)/(m_(1)+m_(2))`