Two blocks `A(3 kg)` and `B(2 kg)` resting on a smooth horizontal surface is connected by a spring of stiffness `480N//m`. Initially the spring is underformed and a velocity of `2 m//s` is imparted to A along the line of the spring away from B. The maximum extension in meters of the spring during subsequent motion is :-

Options

A. `(1)/(10)`
B. `(1)/(2sqrt10)`
C. `(1)/(2sqrt15)`
D. `0.15`

Explanation

To solve the problem step by step, we will use the principles of conservation of momentum and energy.

### Step 1: Calculate Initial Momentum
The initial momentum \( P_i \) of the system can be calculated using the formula:
\[ P_i = m_A \cdot v_A \]
where \( m_A = 3 \, \text{kg} \) (mass of block A) and \( v_A = 2 \, \text{m/s} \) (initial velocity of block A).

\[ P_i = 3 \, \text{kg} \cdot 2 \, \text{m/s} = 6 \, \text{kg m/s} \]

**Hint:** Remember that momentum is the product of mass and velocity.

### Step 2: Calculate Final Momentum
The final momentum \( P_f \) of the system after the blocks start moving can be expressed as:
\[ P_f = (m_A + m_B) \cdot v \]
where \( m_B = 2 \, \text{kg} \) (mass of block B) and \( v \) is the final velocity of both blocks after the spring is extended.

\[ P_f = (3 \, \text{kg} + 2 \, \text{kg}) \cdot v = 5v \, \text{kg m/s} \]

**Hint:** The final momentum is the total mass of the system multiplied by the common velocity after the spring has extended.

### Step 3: Apply Conservation of Momentum
According to the principle of conservation of momentum:
\[ P_i = P_f \]
Substituting the values we calculated:
\[ 6 \, \text{kg m/s} = 5v \]
Solving for \( v \):
\[ v = \frac{6}{5} = 1.2 \, \text{m/s} \]

**Hint:** Conservation of momentum states that the total momentum before an event must equal the total momentum after the event if no external forces act on the system.

### Step 4: Calculate Initial Kinetic Energy
The initial kinetic energy \( KE_i \) of the system can be calculated using:
\[ KE_i = \frac{1}{2} m_A v_A^2 \]
\[ KE_i = \frac{1}{2} \cdot 3 \cdot (2)^2 = \frac{1}{2} \cdot 3 \cdot 4 = 6 \, \text{J} \]

**Hint:** Kinetic energy is calculated using the formula \( \frac{1}{2} mv^2 \).

### Step 5: Calculate Final Kinetic Energy
The final kinetic energy \( KE_f \) of the system is:
\[ KE_f = \frac{1}{2} (m_A + m_B) v^2 \]
\[ KE_f = \frac{1}{2} \cdot 5 \cdot (1.2)^2 = \frac{1}{2} \cdot 5 \cdot 1.44 = 3.6 \, \text{J} \]

**Hint:** Ensure to use the total mass and the final velocity when calculating the final kinetic energy.

### Step 6: Calculate Change in Kinetic Energy
The change in kinetic energy, which is converted into potential energy in the spring, is:
\[ \Delta KE = KE_i - KE_f = 6 \, \text{J} - 3.6 \, \text{J} = 2.4 \, \text{J} \]

**Hint:** The energy lost in kinetic energy is stored as potential energy in the spring.

### Step 7: Relate Potential Energy to Spring Extension
The potential energy stored in the spring at maximum extension \( x \) is given by:
\[ PE = \frac{1}{2} k x^2 \]
Setting this equal to the change in kinetic energy:
\[ \frac{1}{2} \cdot 480 \cdot x^2 = 2.4 \]
Solving for \( x^2 \):
\[ 240 x^2 = 2.4 \implies x^2 = \frac{2.4}{240} = 0.01 \]
Taking the square root:
\[ x = \sqrt{0.01} = 0.1 \, \text{m} \]

**Hint:** The potential energy in the spring is proportional to the square of the extension.

### Final Answer
The maximum extension of the spring is:
\[ \boxed{0.1 \, \text{m}} \]