Two blocks of mass `m=1` kg and `M=2` kg are in contact on a frictionless table. A horizontal force `F(=3N)` is applied to m. Find the force of contact between the blocks. Will the force of contact remain same if F is applied to M?

Explanation

All the forces acting on the two blocks are shown in As the blocks are rigid under the action of a force F, both will move together with same acceleration,
`a=F/(m+M)=3/(1+2)=1m//s^(2)`
Now as the mass of larger block is M and its acceleration a, so force of contact i.e, action on it,
`A_(M)=Ma=(MF)/(M+m)=(2xx3)/(2+1)=2N`
if the force is applied to M, its action on m will be
`A_(m)=ma=(mF)/(M+m)`
`=(1xx3)/(2+1)=1N`
Note: From this problem it is clear that acceleration does not depend on the fact that whether the force is applied to m or M but force of contact does.