Two blocks of masses 2 kg and 1 kg are in contact with each other on a frictionless table. When a horizontal force of `3.0` N is applied to the block of mass 2 kg the value of the force of contact between the two blocks is:

Options

A. 4 N
B. 3 N
C. 5 N
D. 1 N

Explanation

To solve the problem, we need to find the force of contact between the two blocks when a horizontal force of 3 N is applied to the block of mass 2 kg. Let's break it down step by step.

### Step 1: Understand the System
We have two blocks:
- Block A (mass = 2 kg)
- Block B (mass = 1 kg)

They are in contact with each other on a frictionless table. A horizontal force of 3 N is applied to Block A.

### Step 2: Calculate the Total Mass
The total mass of the system (both blocks) is:
\[ m_{total} = m_A + m_B = 2 \, \text{kg} + 1 \, \text{kg} = 3 \, \text{kg} \]

### Step 3: Calculate the Acceleration of the System
Using Newton's second law, the acceleration \( a \) of the entire system can be calculated as:
\[ F = m_{total} \cdot a \]
Where \( F \) is the applied force (3 N). Thus:
\[ 3 \, \text{N} = 3 \, \text{kg} \cdot a \]
Solving for \( a \):
\[ a = \frac{3 \, \text{N}}{3 \, \text{kg}} = 1 \, \text{m/s}^2 \]

### Step 4: Analyze the Forces on Block B
Now, let's analyze the forces acting on Block B (1 kg). The only horizontal force acting on Block B is the contact force \( N \) exerted by Block A. According to Newton's second law for Block B:
\[ N = m_B \cdot a \]
Substituting the values:
\[ N = 1 \, \text{kg} \cdot 1 \, \text{m/s}^2 = 1 \, \text{N} \]

### Step 5: Conclusion
The force of contact between the two blocks is:
\[ N = 1 \, \text{N} \]

Thus, the value of the force of contact between the two blocks is **1 N**.