Two blocks of masses m and 3m on a horizontal surface are in contact with the ends of a horizontal massless spring. The coefficient of friction between m and surface and between 3m and surface is `mu` and `mu//3` respectively. The two blocks are moved towards each other to compress the spring and then released. The two blocks move off in opposite directions covering distances `S_1` and `S_2` before comming to rest. `S_1 : S_2` is

Options

A. `1 : 9`
B. `1 : 3`
C. `3 : 1`
D. `9 : 1`

Answer

To solve the problem, we need to analyze the motion of two blocks, one with mass \( m \) and the other with mass \( 3m \), after they are released from a compressed spring. We will derive the ratio of the distances \( S_1 \) and \( S_2 \) that the blocks travel before coming to rest.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have two blocks: Block A with mass \( m \) and Block B with mass \( 3m \).
- The coefficients of friction are \( \mu \) for Block A and \( \frac{\mu}{3} \) for Block B.

2. **Conservation of Momentum**:
- Initially, both blocks are at rest, so the initial momentum \( P_i = 0 \).
- When released, let \( V_1 \) be the velocity of Block A and \( V_2 \) be the velocity of Block B.
- By conservation of momentum:
\[ mV_1 = 3m(-V_2) \implies V_1 = 3V_2 \]

3. **Frictional Forces**:
- The frictional force \( F_A \) on Block A is:
\[ F_A = \mu mg \]
- The frictional force \( F_B \) on Block B is:
\[ F_B = \frac{\mu}{3} \cdot 3mg = \mu mg \]

4. **Acceleration due to Friction**:
- For Block A:
\[ m a_1 = \mu mg \implies a_1 = \mu g \]
- For Block B:
\[ 3m a_2 = \mu mg \implies a_2 = \frac{\mu g}{3} \]

5. **Using Kinematic Equations**:
- For Block A (initial velocity \( V_1 \), final velocity \( 0 \)):
\[ 0 = V_1^2 - 2a_1S_1 \implies S_1 = \frac{V_1^2}{2a_1} = \frac{V_1^2}{2\mu g} \]
- For Block B:
\[ 0 = V_2^2 - 2a_2S_2 \implies S_2 = \frac{V_2^2}{2a_2} = \frac{V_2^2}{2 \cdot \frac{\mu g}{3}} = \frac{3V_2^2}{2\mu g} \]

6. **Finding the Ratio \( S_1 : S_2 \)**:
- Substitute \( V_1 = 3V_2 \) into the equation for \( S_1 \):
\[ S_1 = \frac{(3V_2)^2}{2\mu g} = \frac{9V_2^2}{2\mu g} \]
- Now, we can find the ratio:
\[ \frac{S_1}{S_2} = \frac{\frac{9V_2^2}{2\mu g}}{\frac{3V_2^2}{2\mu g}} = \frac{9}{3} = 3 \]
- Thus, the ratio \( S_1 : S_2 = 3 : 1 \).

### Final Answer:
The ratio \( S_1 : S_2 \) is \( 3 : 1 \).