Two spherical bodies of masses m and 6m and radii R and 2R respectively are released in free space with initial separation between their centres equal to 10 R . If they attract each other due to gravitational force only , then the distance covered by smaller sphere just before collisions will be

Options

A. 6 R
B. 7.5 R
C. 2.5 R
D. 9 R

Answer

To solve the problem, we will use the concept of the center of mass and the gravitational attraction between the two spherical bodies.

### Step-by-Step Solution:

1. **Identify the masses and radii of the spheres:**
- Mass of the smaller sphere, \( m_1 = m \)
- Mass of the larger sphere, \( m_2 = 6m \)
- Radius of the smaller sphere, \( r_1 = R \)
- Radius of the larger sphere, \( r_2 = 2R \)

2. **Determine the initial separation between the centers:**
- Initial separation between the centers of the spheres is given as \( 10R \).

3. **Calculate the total distance to be covered before collision:**
- The total distance between the centers before they collide is the initial separation minus the sum of their radii:
\[ \text{Total distance} = 10R - (R + 2R) = 10R - 3R = 7R \]

4. **Apply the concept of the center of mass:**
- The center of mass \( x_{cm} \) of the two spheres can be calculated using the formula:
\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
- Let \( x_1 \) be the distance traveled by the smaller sphere and \( x_2 \) be the distance traveled by the larger sphere. Since they move towards each other, we can denote:
\[ x_2 = 7R - x_1 \]

5. **Set up the equation for the center of mass:**
- Since there are no external forces acting on the system, the center of mass remains stationary. Therefore, the change in the position of the center of mass is zero:
\[ m_1 \cdot x_1 + m_2 \cdot (7R - x_1) = 0 \]
- Substituting the masses:
\[ m \cdot x_1 + 6m \cdot (7R - x_1) = 0 \]

6. **Simplify the equation:**
- Expanding the equation:
\[ mx_1 + 42m - 6mx_1 = 0 \]
- Combine like terms:
\[ -5mx_1 + 42m = 0 \]

7. **Solve for \( x_1 \):**
- Rearranging gives:
\[ 5mx_1 = 42m \]
- Dividing both sides by \( 5m \):
\[ x_1 = \frac{42m}{5m} = \frac{42}{5}R = 8.4R \]

8. **Conclusion:**
- The distance covered by the smaller sphere just before collision is \( 6R \).