Two block of masses `m_(1) and m_(2)` are placed side by side on a smooth horizontal surface as shown in fig. A horizontal force F is applied on the block .

(a) Find the acceleration of each block.
(b) Find the normal reaction between the two blocks.

Explanation

Method 1: Since the two blocks always remain in contact with each other, they msut move with the same acceleration.

Using Newton's second law, we get
For block `m_(1):F-N=m_(1)a`
For block `m_(2): N = m_(2)a`
On adding the two equations, we get
`f=(m_(1)+m_(2)) a implies a=(F)/(m_(1)+m_(2))`
Substituting the value of a in (ii), we get
`N=m_(2)a = (Fm_(2))/(m_(1)+m_(2))`
Method 2 : The situation may be considered as follows : Instead of drawing the free-body diagrams of each block, we can draw the free-body diagram of both blocks together as shown in fig

The net force acting on the system is F, and the total mass of the system is `m_(1)+m_(2)`. Thus, `a=(F)/(m_(1)+m_(2))`
To find out the normal reaction N between the two blocks, we can imagine the following: Block `m_(2)` is moving with an acceleration a, therefore, the net force acting on it most be `m_(2)a` which is nothing but the normal reaction applied by the block `m_(1)`. Thus
`N=m_(2)a = (m_(2)F)/(m_(1)+m_(2))` .