Thank you for reporting, we will resolve it shortly
Q.
Two blocks of $2\,kg$ and $1\,kg$ are in contact on a frictionless table. If a force of $3\,N$ is applied on $2\,kg$ block, then the force of contact between the two blocks will be :
Acceleration of both the blocks taken together
$=\frac{3}{[2+1]}=1\, m / s ^{2}$
Hence, the contact force between the two blocks is equal to
$1 \,kg \times 1\, m / s ^{2}=1 \,N$
