Two particles `A` and `B` initially at rest, move towards each other by mutual force of attraction. At the instant when the speed of `A` is `n` and the speed of `B` is `3n`, the speed of the centre of mass of the system is

Options

A. `3n`
B. `2Ns`
C. `1.5n`
D. `0`

Explanation

To find the speed of the center of mass of the system consisting of two particles A and B, we can follow these steps:

### Step 1: Understand the System
We have two particles A and B that are initially at rest and move towards each other due to a mutual force of attraction. At a certain instant, particle A has a speed of `n` and particle B has a speed of `3n`.

### Step 2: Identify the Masses
Let the mass of particle A be \( m_A \) and the mass of particle B be \( m_B \). The speeds of the particles are given as:
- Speed of A, \( v_A = n \)
- Speed of B, \( v_B = -3n \) (negative because it is moving in the opposite direction)

### Step 3: Use the Formula for Center of Mass Velocity
The velocity of the center of mass \( v_{CM} \) of a two-particle system is given by the formula:
\[ v_{CM} = \frac{m_A v_A + m_B v_B}{m_A + m_B} \]

### Step 4: Substitute the Values
Substituting the values of the speeds into the formula:
\[ v_{CM} = \frac{m_A (n) + m_B (-3n)}{m_A + m_B} \]
\[ v_{CM} = \frac{m_A n - 3m_B n}{m_A + m_B} \]

### Step 5: Factor Out Common Terms
We can factor out \( n \) from the numerator:
\[ v_{CM} = \frac{n (m_A - 3m_B)}{m_A + m_B} \]

### Step 6: Analyze the Result
Since there are no external forces acting on the system (only internal forces), the center of mass will not accelerate. The system is isolated, and thus the center of mass will remain stationary if it was initially at rest. Therefore, we can conclude that:
\[ v_{CM} = 0 \]

### Final Answer
Thus, the speed of the center of mass of the system is:
\[ \text{Speed of center of mass} = 0 \]

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