A body A of mass m is moving in a circular orbit of radius R about a planet. Another body B of mass m/2 collides with A with a velocity which is half ( v⃗/2 ) the instantaneous velocityv⃗ of A. The collision is completely inelastic. Then, the combined body :

Solution

Physics sir · 3 steps

Find the new speed with momentum conservation, then compare it with the circular orbit speed at R.

1. Momentum conserved. The collision is very brief, so gravity does not change the momentum during it. B moves along A's velocity at v/2, and the two stick together, so the total mass becomes 3m/2.

   m v + m/2·v/2 = 3m/2 V ⇒ V = 5v/6

2. Circular speed. At radius R a circular orbit needs exactly v = √(GM/R), whatever the mass of the body. The combined body is still moving tangentially, but with 5v/6, which is less than v.

   Tangential speed at R	What happens
   0	falls straight down
   between 0 and v	ellipse, this point is the farthest point
   exactly v	circle
   between v and √2 v	ellipse, this point is the nearest point
   √2 v or more	escapes

3. So. With V = 5v/6, which is not zero and is below v, the combined body moves in an elliptical orbit. That is option (a).

The idea to keep. Orbit shape depends only on speed and direction at that point, never on mass. If the tangential speed is below √(GM/R) but not zero, the orbit becomes an ellipse.

Answer from the NTA final answer key. Solution by PrepWiser.