Two blocks of masses `m_1` and `m_2 (m_1 gt m_2) ` in contact with each other on frictionless, horizontal surface. If a horizontal force F is given on` m_1` set into motion with acceleration a, then reaction force on mass `m_1` by `m_2` is

Explanation

Let us consider two blocks of masses `m_(1)` and `m_(2) (m_(1) gt m_(2))` . Kept in contact with each other on a smooth horizontal frictional surface as shown in the figure

By the application of a horizontal force F both the blocks are set into motion with acceleration a simultaneously in the direction of the force F .
To find the acceleration `vec(a)` Newton's second law has to be applied to the system mass `m=m_(1)+ m_(2)`
If we select the motion of the two masses along the positive x direction `F hat(j) = mahat(i)`
The acceleration of the system is given by
`:. , a= (F )/( m_(1) + m_(2))`
The force exerted by teh block `m_(1)` on `m_(2)` due to its motion is called force of contact `(f_(n))` According to Newton's third law the block `m_(2)` will exert an equivalent opposite reaction
Figure shows the free body diagram of block `m_(1)`

`Fhat(i) - f_(12) hat(i) = m_(1) a hat(i)`
By comparing the components of the above equation we get
`F-f_(12) = m_(1) a`
`f_(12) = F - m_(1) a`
Substituting the value of acceleration from equation (1) in (2) we get
` f_(12) = f- m_(1) (F)/(m_(1)+m_(2))`
`F_(12) = (Fm_(2))/(m_(1) +m_(2))`
Equation (3) shows that the magnitude of contact force depends on mass `m_(2)` which provides the reaction force . Note that this force is acting along the negative x direction `m_(1) ` is given by `f_(12) = - (Fm_(2))/( m_(1) + m_(2))`
For `m_(2)` there is only one force force acting on it the x direction and it is denoted by `f_(21)` . this force is exerted by mass `m_(1)` .the free body diagram for mass `m_(2)` is shown in the figure .

Applying Newton' second law for mass `m_(2)`
`f_(21) hat(i) = m_(2) hat(i)`
By comparing the components on both sides of the above equation
`f_(21) = m_(2) a`
Substituting for acceleration from equation
(1) in equation (4) we get
`f_(21) = (Fm_(2))/( m_(1) +m_(2))`
In this case the magnitude of the contact force is `f_(21) = (Fm)/(m_(1) + m_(2))`