Two particles A and B intiallly at rest, move towards each other under a mutual force of attraction. AT the instant when the speed of A is v and the speed of B is 2 v, the speed of the centre of mass of the system is

Options

A. zero
B. v
C. 1.5 v
D. 3 v

Explanation

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

### Step 1: Understand the System
We have two particles, A and B, initially at rest, moving towards each other under a mutual force of attraction. At a certain instant, the speed of A is \( v \) and the speed of B is \( 2v \).

### Step 2: Define the Masses
Let the mass of particle A be \( m_A \) and the mass of particle B be \( m_B \). According to the problem, we will find the relationship between their masses based on the forces acting on them.

### Step 3: Apply Newton's Second Law
The forces acting on the particles can be expressed as:
- For particle A: \( F_A = m_A \cdot a_A \)
- For particle B: \( F_B = m_B \cdot a_B \)

Since both particles are attracting each other, the magnitudes of the forces are equal:
\[ F_A = F_B \]

### Step 4: Express Accelerations
The acceleration of each particle can be expressed in terms of their velocities and the time taken to reach those velocities:
- For particle A: \( a_A = \frac{v}{t} \)
- For particle B: \( a_B = \frac{2v}{t} \)

### Step 5: Set Up the Equation
From the equality of forces:
\[ m_A \cdot \frac{v}{t} = m_B \cdot \frac{2v}{t} \]

### Step 6: Simplify the Equation
Cancelling \( v \) and \( t \) from both sides (assuming \( v \neq 0 \) and \( t \neq 0 \)):
\[ m_A = 2m_B \]

### Step 7: Calculate the Velocity of the Center of Mass
The velocity of the center of mass \( v_{cm} \) for two particles is given by:
\[ v_{cm} = \frac{m_A v_A + m_B v_B}{m_A + m_B} \]

Substituting \( v_A = v \) and \( v_B = -2v \) (negative because they are moving towards each other):
\[ v_{cm} = \frac{m_A v + m_B (-2v)}{m_A + m_B} \]

### Step 8: Substitute Masses
Substituting \( m_A = 2m_B \):
\[ v_{cm} = \frac{(2m_B)v + m_B(-2v)}{2m_B + m_B} \]
\[ = \frac{2m_B v - 2m_B v}{3m_B} \]
\[ = \frac{0}{3m_B} \]
\[ = 0 \]

### Conclusion
The speed of the center of mass of the system is \( 0 \).