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

Explanation

All the forces acting on the two blocks are shown in figure. As the blocks are rigid under the action 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`