Two blocks of mass 4 kg and 6 kg are placed in contact with each other on a frictionless horizontal surface. If we apply a push of 5 N on the heavier mass, the force on the lighter mass will be

The correct option is 2 N.

Let's break down the problem step-by-step to understand how the force is transmitted between the two blocks.

Step 1: Identify the given values
Mass of the lighter block, $m_1=4\mathrm{kg}$
Mass of the heavier block, $m_2=6\mathrm{kg}$
Applied push (force), $F=5N$ (applied to the heavier mass, $m_2$)
The surface is frictionless, meaning there is no resistive force opposing the motion of the blocks.

Step 2: Calculate the total mass of the system
Since the two blocks are in contact and move together, they can be treated as a single system. The total mass ($M$) of this combined system is:
$M=m_1+m_2$
$M=4\mathrm{kg}+6\mathrm{kg}=10\mathrm{kg}$

Step 3: Calculate the common acceleration of the blocks
According to Newton's second law of motion, the acceleration ($a$) of the combined system is given by:
$a=\frac{F}{M}$
Substituting the values, we get:
$a=\frac{5N}{10\mathrm{kg}}=0.5m/s^2$

Step 4: Find the force acting on the lighter mass
The lighter block of mass $m_1=4\mathrm{kg}$ moves with the same acceleration $a=0.5m/s^2$. The contact force ($F_c$) exerted by the heavier block on the lighter block is the only horizontal force causing it to accelerate. Using Newton's second law for the lighter block:
$F_c=m_1\times a$
Substituting the values, we find:
$F_c=4\mathrm{kg}\times 0.5m/s^2=2N$

Thus, the force acting on the lighter mass is 2 N.