Two point masses `m_(1)` and `m_(2)` at rest in gravity free space are released from distance d. Find the velocity of the CM of the system at the time of collision of particles

Options

A. Zero
B. 2
C. 3
D. None

Answer

To solve the problem, we need to find the velocity of the center of mass (CM) of a system consisting of two point masses \( m_1 \) and \( m_2 \) when they collide after being released from a distance \( d \) in gravity-free space.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- The two point masses \( m_1 \) and \( m_2 \) are at rest initially, which means their initial velocities are both zero.
- They are separated by a distance \( d \) in a gravity-free space.

2. **Identify the Forces Acting on the Masses**:
- Since the problem states that it is gravity-free space, there are no external forces acting on the masses.
- Therefore, the net force acting on the system is zero.

3. **Apply Newton's Second Law**:
- According to Newton's second law, if the net force acting on a system is zero, the acceleration of the system is also zero.
- This implies that the velocities of the masses will not change over time.

4. **Calculate the Velocity of the Center of Mass**:
- The velocity of the center of mass (CM) of a two-particle system is given by the formula:
\[ v_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \]
- Since both masses are initially at rest, we have \( v_1 = 0 \) and \( v_2 = 0 \).
- Substituting these values into the formula gives:
\[ v_{CM} = \frac{m_1 \cdot 0 + m_2 \cdot 0}{m_1 + m_2} = \frac{0}{m_1 + m_2} = 0 \]

5. **Conclusion**:
- The velocity of the center of mass at the time of collision is \( 0 \).

### Final Answer:
The velocity of the center of mass of the system at the time of collision of the particles is \( 0 \).

---