1. Initial orbit of body A

Body $A$ of mass $m$ is moving in a circular orbit of radius $R$ around the planet.

For a circular orbit,

so

where $v$ is the orbital speed of $A$.

2. Body B before collision

Body $B$ has mass $\dfrac{m}{2}$ and collides with $A$ with velocity equal to half of $A$'s instantaneous velocity and in the same direction as $\vec v$.

So,

The collision is completely inelastic, so after collision the two bodies stick together.

3. Use conservation of linear momentum during collision

Since collision time is very small, external gravitational impulse can be neglected during the collision.

Initial momentum:

Total mass after sticking:

Let final velocity be $\vec u$. Then

Hence,

So the combined body moves tangentially with speed

4. Compare with circular-orbit speed and escape speed

At radius $R$:

• circular speed is $v$
• escape speed is $\sqrt{2}v$

Here,

So it is too slow to remain in the same circular orbit.

Also,

so it definitely does not escape.

Thus options C and D are false.

5. Will it fall vertically downward?

Immediately after collision, the velocity is still tangential, not zero.

So it does not fall vertically downward toward the planet.

Hence option B is false.

6. Nature of the new orbit

Since the body is at radius $R$ with tangential speed less than circular speed, it enters a bound orbit.

A bound orbit under gravity is an ellipse (with the planet at one focus).

Also, since the velocity is purely tangential at the instant and smaller than circular speed, the point of collision becomes the apogee of the ellipse.

Therefore, the combined body starts moving in an elliptical orbit around the planet.

7. Option check

• A: True
• B: False
• C: False
• D: False

So the correct answer is: