A spring is compressed between two toy carts to masses `m_1` and `m_2`. When the toy carts are released, the spring exerts on each toy cart equal and opposite forces for the same small time t. If the coefficients of friction `mu` between the ground and the toy carts are equal, then the magnitude of displacements of the toy carts are in the ratio

Options

A. (a) `S_1/S_2=m_2/m_1`
B. (b) `S_1/S_2=m_1/m_2`
C. (c) `S_1/S_2=(m_2/m_1)^2`
D. (d) `S_1/S_2=(m_1/m_2)^2`

Explanation

To solve the problem, we need to analyze the situation involving two toy carts with masses \( m_1 \) and \( m_2 \) that are separated by a compressed spring. When the spring is released, it exerts equal and opposite forces on both carts for a small time \( t \). We also know that the coefficients of friction \( \mu \) between the ground and the toy carts are equal.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Carts:**
When the spring is released, it exerts a force \( F \) on each cart. According to Newton's Third Law, the force exerted by the spring on cart 1 (mass \( m_1 \)) is equal in magnitude and opposite in direction to the force exerted on cart 2 (mass \( m_2 \)).

2. **Calculate the Acceleration of Each Cart:**
The acceleration \( a_1 \) of cart 1 can be expressed using Newton's second law:
\[ a_1 = \frac{F}{m_1} \]
Similarly, the acceleration \( a_2 \) of cart 2 is:
\[ a_2 = \frac{F}{m_2} \]

3. **Determine the Change in Momentum:**
The change in momentum \( \Delta p \) for each cart during the time \( t \) can be given by:
\[ \Delta p_1 = F \cdot t \quad \text{and} \quad \Delta p_2 = F \cdot t \]
Since the forces are equal and opposite, the momentum gained by each cart will be equal in magnitude but opposite in direction.

4. **Apply the Work-Energy Theorem:**
The work done against friction on each cart can be expressed as:
\[ W_{\text{friction}} = -\mu m g s \]
where \( s \) is the displacement of each cart. The work done by the spring will equal the change in kinetic energy minus the work done against friction.

5. **Set Up the Energy Equation:**
The work done by the spring on each cart can be equated to the change in kinetic energy:
\[ W_{\text{spring}} - W_{\text{friction}} = \Delta KE \]
For cart 1:
\[ \frac{1}{2} m_1 v_1^2 - \mu m_1 g s_1 = 0 \]
For cart 2:
\[ \frac{1}{2} m_2 v_2^2 - \mu m_2 g s_2 = 0 \]

6. **Relate Displacements to Masses:**
From the above equations, we can express the displacements \( s_1 \) and \( s_2 \):
\[ s_1 = \frac{v_1^2}{2 \mu g} \quad \text{and} \quad s_2 = \frac{v_2^2}{2 \mu g} \]

7. **Using Conservation of Momentum:**
Since the momentum is conserved and equal for both carts:
\[ m_1 v_1 = m_2 v_2 \]
Rearranging gives:
\[ v_1 = \frac{m_2}{m_1} v_2 \]

8. **Substituting for Displacements:**
Substitute \( v_1 \) into the expression for \( s_1 \):
\[ s_1 = \frac{\left(\frac{m_2}{m_1} v_2\right)^2}{2 \mu g} = \frac{m_2^2 v_2^2}{2 \mu g m_1^2} \]
Now, taking the ratio of displacements:
\[ \frac{s_1}{s_2} = \frac{\frac{m_2^2 v_2^2}{2 \mu g m_1^2}}{\frac{v_2^2}{2 \mu g}} = \frac{m_2^2}{m_1^2} \]

9. **Final Ratio of Displacements:**
Thus, the ratio of the displacements of the toy carts is:
\[ \frac{s_1}{s_2} = \frac{m_2^2}{m_1^2} \]

### Conclusion:
The final answer is that the magnitude of displacements of the toy carts is in the ratio \( \frac{s_1}{s_2} = \frac{m_2^2}{m_1^2} \).