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

Options

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

Answer

To solve the problem, we need to determine the distance covered by the smaller body (mass M) just before collision with the larger body (mass 5M). Here’s a step-by-step solution:

### Step 1: Understand the system
We have two spherical bodies:
- Body 1: Mass = M, Radius = R
- Body 2: Mass = 5M, Radius = 2R

The initial separation between their centers is given as 12R.

### Step 2: Define the distances traveled
Let:
- \( x_1 \) = distance traveled by the smaller body (mass M) before collision
- \( x_2 \) = distance traveled by the larger body (mass 5M) before collision

### Step 3: Apply conservation of momentum
Since there are no external forces acting on the system, the total momentum is conserved. Initially, both bodies are at rest, so the initial momentum is zero.

The final momentum can be expressed as:
\[ 0 = M \cdot v_1 - 5M \cdot v_2 \]
Where \( v_1 \) and \( v_2 \) are the velocities of the smaller and larger body, respectively.

### Step 4: Relate velocities to distances
The velocities can be expressed in terms of the distances traveled:
- \( v_1 = \frac{x_1}{t} \)
- \( v_2 = \frac{x_2}{t} \)

Substituting these into the momentum equation gives:
\[ 0 = M \cdot \left(\frac{x_1}{t}\right) - 5M \cdot \left(\frac{x_2}{t}\right) \]
Cancelling \( M \) and \( t \) (since they are not zero) results in:
\[ x_1 = 5x_2 \]
(Equation 1)

### Step 5: Apply the distance condition
The total distance covered by both bodies before they collide is equal to the initial separation minus the sum of their radii:
\[ x_1 + x_2 = 12R - (R + 2R) = 12R - 3R = 9R \]
(Equation 2)

### Step 6: Substitute Equation 1 into Equation 2
Substituting \( x_1 = 5x_2 \) into Equation 2:
\[ 5x_2 + x_2 = 9R \]
This simplifies to:
\[ 6x_2 = 9R \]
Thus,
\[ x_2 = \frac{9R}{6} = 1.5R \]

### Step 7: Find \( x_1 \)
Now substituting \( x_2 \) back into Equation 1 to find \( x_1 \):
\[ x_1 = 5x_2 = 5 \cdot 1.5R = 7.5R \]

### Final Answer
The distance covered by the smaller body just before collision is:
\[ \boxed{7.5R} \]