Consider a system of two particles having masses 2kg and 5 kg the particle of mass 2 kg is pushed towards the centre of mass of particles through a distance 5 m,by what distance would particle of mass 5kg move so as to keep the centre of mass of particles at the original position?

Options

A. 2 m
B. 12.5m
C. 8m
D. 5m

Answer

To solve the problem, we need to find out how far the 5 kg particle moves in the opposite direction when the 2 kg particle is pushed towards the center of mass.

### Step-by-step Solution:

1. **Identify the masses and distances**:
- Let \( m_1 = 2 \, \text{kg} \) (mass of the first particle).
- Let \( m_2 = 5 \, \text{kg} \) (mass of the second particle).
- The distance \( \Delta x_1 = 5 \, \text{m} \) (the distance the 2 kg particle is pushed towards the center of mass).

2. **Define the center of mass (CM)**:
The center of mass \( x_{cm} \) for two particles is given by the formula:
\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
where \( x_1 \) and \( x_2 \) are the positions of the masses.

3. **Set up the equation for the center of mass**:
Since we want the center of mass to remain in the same position, we can express the change in position of the center of mass as:
\[ m_1 \Delta x_1 + m_2 \Delta x_2 = 0 \]
Here, \( \Delta x_1 \) is the distance moved by the 2 kg mass (which is -5 m, since it moves towards the center), and \( \Delta x_2 \) is the distance moved by the 5 kg mass (which we need to find).

4. **Substitute the known values**:
Substituting the values into the equation:
\[ 2 \times (-5) + 5 \times \Delta x_2 = 0 \]
This simplifies to:
\[ -10 + 5 \Delta x_2 = 0 \]

5. **Solve for \( \Delta x_2 \)**:
Rearranging the equation gives:
\[ 5 \Delta x_2 = 10 \]
Dividing both sides by 5:
\[ \Delta x_2 = 2 \, \text{m} \]

6. **Determine the direction**:
Since the 2 kg mass was pushed towards the center of mass, the 5 kg mass must move in the opposite direction. Therefore, the 5 kg mass moves 2 m away from the center of mass.

### Final Answer:
The particle of mass 5 kg would move **2 meters** in the opposite direction to keep the center of mass at the original position.