Two blocks A and B of masses 3 kg and 6 kg are connected by a massless spring of force constant `"1800 N m"^(-1)` and then they are placed on a smooth horizontal surface. The blocks are pulled apart to stretch the spring by 5 cm and then released. What is the relative velocity (in `ms^(-1)`) of the blocks when the spring comes to its natural length?

Answer

To solve the problem, we will use the principles of conservation of energy and conservation of momentum. Here’s the step-by-step solution:

### Step 1: Understand the Initial and Final Conditions
Initially, the spring is stretched by \( x = 5 \, \text{cm} = 0.05 \, \text{m} \). The potential energy stored in the spring when stretched is given by:
\[ PE = \frac{1}{2} k x^2 \]
where \( k = 1800 \, \text{N/m} \).

### Step 2: Calculate the Initial Potential Energy
Substituting the values into the potential energy formula:
\[ PE = \frac{1}{2} \times 1800 \times (0.05)^2 \]
Calculating this gives:
\[ PE = \frac{1}{2} \times 1800 \times 0.0025 = 2.25 \, \text{J} \]

### Step 3: Set Up the Conservation of Energy Equation
When the spring returns to its natural length, all the potential energy converts into kinetic energy. The kinetic energy of the two blocks can be expressed as:
\[ KE = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
where \( m_1 = 3 \, \text{kg} \) and \( m_2 = 6 \, \text{kg} \).

Setting the initial potential energy equal to the total kinetic energy:
\[ 2.25 = \frac{1}{2} \times 3 v_1^2 + \frac{1}{2} \times 6 v_2^2 \]
Simplifying this gives:
\[ 2.25 = \frac{3}{2} v_1^2 + 3 v_2^2 \]
Multiplying through by 2 to eliminate the fractions:
\[ 4.5 = 3 v_1^2 + 6 v_2^2 \quad \text{(Equation 1)} \]

### Step 4: Use Conservation of Momentum
Since there are no external forces acting on the system, we can apply the conservation of momentum:
\[ m_1 v_1 = m_2 v_2 \]
Substituting the masses gives:
\[ 3 v_1 = 6 v_2 \]
From this, we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = 2 v_2 \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 2 into Equation 1
Now substitute \( v_1 = 2 v_2 \) into Equation 1:
\[ 4.5 = 3 (2 v_2)^2 + 6 v_2^2 \]
This simplifies to:
\[ 4.5 = 3 \times 4 v_2^2 + 6 v_2^2 \]
\[ 4.5 = 12 v_2^2 + 6 v_2^2 \]
\[ 4.5 = 18 v_2^2 \]
Solving for \( v_2^2 \):
\[ v_2^2 = \frac{4.5}{18} = 0.25 \]
Taking the square root:
\[ v_2 = 0.5 \, \text{m/s} \]

### Step 6: Calculate \( v_1 \)
Using Equation 2:
\[ v_1 = 2 v_2 = 2 \times 0.5 = 1 \, \text{m/s} \]

### Step 7: Calculate the Relative Velocity
The relative velocity \( v_{rel} \) of block A with respect to block B is given by:
\[ v_{rel} = v_1 + v_2 = 1 + 0.5 = 1.5 \, \text{m/s} \]

### Final Answer
The relative velocity of the blocks when the spring comes to its natural length is:
\[ \boxed{1.5 \, \text{m/s}} \]