A particle of mass m is revolving around a planet in a circular orbit of radius R. At the instant the particle has velocity `vec V` another particle of mass `m/2` moving at velocity ` (vec v)/2` collides perfectly in-elastically with the first particle. The new path of the combined body will take is

Options

A. Circular
B. Elliptical
C. Straight line
D. Fall directly below on the ground

Explanation

To solve the problem, we need to analyze the situation step by step, focusing on the conservation of momentum and the implications of the collision on the motion of the combined mass.

### Step 1: Understand the Initial Conditions
- We have a particle of mass \( m \) revolving in a circular orbit with radius \( R \) and velocity \( \vec{V} \).
- Another particle of mass \( \frac{m}{2} \) is moving with velocity \( \frac{\vec{V}}{2} \).

### Step 2: Apply Conservation of Momentum
Since the collision is perfectly inelastic, we can apply the conservation of momentum before and after the collision. The total momentum before the collision is equal to the total momentum after the collision.

**Before the collision:**
- Momentum of the first particle: \( m \vec{V} \)
- Momentum of the second particle: \( \frac{m}{2} \cdot \frac{\vec{V}}{2} = \frac{m \vec{V}}{4} \)

**Total momentum before collision:**
\[ \text{Total momentum} = m \vec{V} + \frac{m \vec{V}}{4} = \frac{4m \vec{V}}{4} + \frac{m \vec{V}}{4} = \frac{5m \vec{V}}{4} \]

**After the collision:**
Let \( \vec{V}' \) be the velocity of the combined mass after the collision. The total mass after the collision is \( m + \frac{m}{2} = \frac{3m}{2} \).

Using conservation of momentum:
\[ \frac{5m \vec{V}}{4} = \frac{3m}{2} \vec{V}' \]

### Step 3: Solve for the New Velocity \( \vec{V}' \)
To find \( \vec{V}' \):
\[ \frac{5m \vec{V}}{4} = \frac{3m}{2} \vec{V}' \]

Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{5 \vec{V}}{4} = \frac{3}{2} \vec{V}' \]

Now, solving for \( \vec{V}' \):
\[ \vec{V}' = \frac{5 \vec{V}}{4} \cdot \frac{2}{3} = \frac{5 \vec{V}}{6} \]

### Step 4: Analyze the New Velocity
- The new velocity \( \vec{V}' = \frac{5}{6} \vec{V} \) is less than the original velocity \( \vec{V} \).
- Since the particle was initially in circular motion, a decrease in velocity indicates that the centripetal force is no longer sufficient to maintain circular motion.

### Step 5: Determine the Path After Collision
- When the velocity of the combined mass is less than the original orbital velocity, the particle will not continue in a circular path.
- Instead, it will start to fall towards the center of the planet, following an elliptical path.

### Conclusion
The new path of the combined body after the collision will be an **elliptical path**.