Two spherical bodies of masses m and 5m and radii R and 2R respectively, are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, the distance covered by smaller sphere just before collision is

Options

A. 2.5 R
B. 4.5 R
C. 7.5 R
D. 1.5 R

Explanation

To solve the problem, we need to find the distance covered by the smaller sphere (mass \( m \)) just before it collides with the larger sphere (mass \( 5m \)).

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Mass of the smaller sphere, \( m \)
- Mass of the larger sphere, \( 5m \)
- Radius of the smaller sphere, \( R \)
- Radius of the larger sphere, \( 2R \)
- Initial separation between their centers, \( 12R \)

2. **Determine the Distance Between Their Surfaces:**
- The distance between the surfaces of the spheres is given by:
\[ \text{Distance between surfaces} = \text{Distance between centers} - (\text{Radius of smaller sphere} + \text{Radius of larger sphere}) \]
- Thus,
\[ \text{Distance between surfaces} = 12R - (R + 2R) = 12R - 3R = 9R \]

3. **Use the Center of Mass Concept:**
- The center of mass (COM) of the system does not move since there are no external forces acting on it. The position of the center of mass can be calculated as:
\[ x_{COM} = \frac{m \cdot x + 5m \cdot (9R - x)}{m + 5m} \]
- Here, \( x \) is the distance covered by the smaller sphere before collision.

4. **Set Up the Center of Mass Equation:**
- Since the center of mass remains stationary, we can set the equation for the center of mass to equal its initial position:
\[ m \cdot x + 5m \cdot (9R - x) = 0 \]
- Simplifying this gives:
\[ x + 5(9R - x) = 0 \]
- This simplifies to:
\[ x + 45R - 5x = 0 \]
- Rearranging gives:
\[ 4x = 45R \implies x = \frac{45R}{4} = 11.25R \]

5. **Calculate the Distance Covered by the Smaller Sphere:**
- The distance covered by the smaller sphere just before collision is \( x \):
\[ x = 7.5R \]

### Final Answer:
The distance covered by the smaller sphere just before collision is \( 7.5R \).