Two particles of mass `m_(1)` and `m_(2)` are approaching towards each other under their mutual gravitational field. If the speed of the particle is v, the speed of the center of mass of the system is equal to :

Options

A. `(m_(1))/(m_(2))v`
B. Zero
C. `((m_(1)-m_(2))v)/(m_(1)+m_(2))`
D. `(m_(2))/(m_(1))v`

Answer

To find the speed of the center of mass of the system consisting of two particles with masses \( m_1 \) and \( m_2 \) that are approaching each other with speed \( v \), we can follow these steps:

### Step 1: Define the velocities of the particles
Assume that particle \( m_1 \) is moving towards the right with speed \( v \) and particle \( m_2 \) is moving towards the left with speed \( v \). We can assign their velocities as:
- Velocity of \( m_1 \) (towards the right): \( v_1 = v \)
- Velocity of \( m_2 \) (towards the left): \( v_2 = -v \)

### Step 2: Write the formula for the velocity of the center of mass
The velocity of the center of mass \( V_{cm} \) of a system of particles is given by the formula:
\[ V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \]

### Step 3: Substitute the velocities into the formula
Substituting the velocities of the particles into the center of mass formula:
\[ V_{cm} = \frac{m_1 (v) + m_2 (-v)}{m_1 + m_2} \]

### Step 4: Simplify the expression
This simplifies to:
\[ V_{cm} = \frac{m_1 v - m_2 v}{m_1 + m_2} \]
\[ V_{cm} = \frac{(m_1 - m_2) v}{m_1 + m_2} \]

### Final Answer
Thus, the speed of the center of mass of the system is:
\[ V_{cm} = \frac{(m_1 - m_2) v}{m_1 + m_2} \]

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