Two trolleys of masses m and 3 m are connected by a spring. The spring is compressed and released the trolleys move off in opposite directions and come to rest after travelling distances s1, and s2, respectively. Assuming coefficient of friction is same for both the ratio of sto `s_(1)` to `s_(2)` is

Options

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

Answer

To solve the problem, we will use the principles of conservation of momentum and the work-energy theorem.

### Step-by-Step Solution:

1. **Identify the Masses**:
- Let the mass of the first trolley be \( m_1 = m \).
- Let the mass of the second trolley be \( m_2 = 3m \).

2. **Apply Conservation of Momentum**:
- When the spring is released, the trolleys move in opposite directions. According to the conservation of momentum:
\[ m_1 v_1 = m_2 v_2 \]
- This can be rearranged to give:
\[ \frac{m_1}{m_2} = \frac{v_1}{v_2} \]
- Substituting the values of \( m_1 \) and \( m_2 \):
\[ \frac{m}{3m} = \frac{v_1}{v_2} \implies \frac{1}{3} = \frac{v_1}{v_2} \]
- Thus, we have:
\[ v_1 = \frac{1}{3} v_2 \quad \text{(Equation 1)} \]

3. **Apply Work-Energy Principle for Trolley 1**:
- The work done by friction on trolley 1 can be expressed as:
\[ \frac{1}{2} m_1 v_1^2 = \mu m_1 g S_1 \]
- Rearranging gives:
\[ S_1 = \frac{v_1^2}{2\mu g} \]

4. **Apply Work-Energy Principle for Trolley 2**:
- Similarly, for trolley 2:
\[ \frac{1}{2} m_2 v_2^2 = \mu m_2 g S_2 \]
- Rearranging gives:
\[ S_2 = \frac{v_2^2}{2\mu g} \]

5. **Find the Ratio \( \frac{S_1}{S_2} \)**:
- Now, we can find the ratio of the distances:
\[ \frac{S_1}{S_2} = \frac{\frac{v_1^2}{2\mu g}}{\frac{v_2^2}{2\mu g}} = \frac{v_1^2}{v_2^2} \]
- Substituting \( v_1 = \frac{1}{3} v_2 \) into the equation:
\[ \frac{S_1}{S_2} = \left(\frac{v_1}{v_2}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]

6. **Final Ratio**:
- Therefore, the ratio \( S_1 : S_2 \) is:
\[ S_1 : S_2 = 1 : 9 \]

### Conclusion:
The ratio of distances \( S_1 \) to \( S_2 \) is \( 1 : 9 \).