Two identical particles each of mass m are connected by a light spring of stiffness K. The initial deformation of the spring when the system was at rest, `x_(0)`. After releasing the system, if one of the particles has a speed v and the deformation of the spring is x, find the speed of the other particle neglecting friction.

Answer

To solve the problem, we will use the work-energy theorem, which states that the work done by the spring is equal to the change in kinetic energy of the system.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Two identical particles, each of mass \( m \), are connected by a light spring with stiffness \( K \).
- The initial deformation of the spring when the system was at rest is \( x_0 \).
- At this point, both particles are at rest, so their initial velocities are \( 0 \).

2. **Identify the Final Conditions**:
- After releasing the system, one particle has a speed \( v \) and the spring is deformed to \( x \).
- Let the speed of the other particle be \( v' \).

3. **Apply the Work-Energy Theorem**:
- The work done by the spring can be calculated as:
\[ W = -\frac{1}{2} K (x^2 - x_0^2) \]
- The change in kinetic energy of the system is given by:
\[ \Delta KE = \frac{1}{2} m v^2 + \frac{1}{2} m (v')^2 - 0 \]
- Setting the work done equal to the change in kinetic energy:
\[ -\frac{1}{2} K (x^2 - x_0^2) = \frac{1}{2} m v^2 + \frac{1}{2} m (v')^2 \]

4. **Rearranging the Equation**:
- Multiply through by \( -2 \) to eliminate the negative sign:
\[ K (x^2 - x_0^2) = m v^2 + m (v')^2 \]
- Rearranging gives:
\[ m (v')^2 = K (x^2 - x_0^2) - m v^2 \]

5. **Solving for \( v' \)**:
- Divide the entire equation by \( m \):
\[ (v')^2 = \frac{K (x^2 - x_0^2)}{m} - v^2 \]
- Taking the square root gives:
\[ v' = \sqrt{\frac{K (x^2 - x_0^2)}{m} - v^2} \]

### Final Result:
The speed of the other particle is:
\[ v' = \sqrt{\frac{K (x^2 - x_0^2)}{m} - v^2} \]