Q.Two particles $P$ and $Q$ of mass 1 kg and 3 kg respectively start moving towards each other from rest under their mutual gravitational attraction. What is the velocity of their center of mass?

Step 1. Identify the nature of the force involved. The only force acting on this two-particle system is the mutual gravitational attraction between $P$ and $Q$ — this is an internal force of the $P$–$Q$ system (Newton's third-law pair: $P$ pulls $Q$, and $Q$ pulls $P$ back with an equal and opposite force). No external force acts on the system at all.

Step 2. Apply the center-of-mass motion result. For any system, however its individual particles interact internally, the center of mass obeys $\vec F_{ext}=M\vec a_{CM}$. Since $\vec F_{ext}=0$ here, $\vec a_{CM}=0$ — the center of mass can never accelerate, only move at constant velocity (or stay at rest, whichever it started at).

Step 3. Use the given initial condition. Both particles start from rest ($v_1=v_2=0$ initially), so the initial velocity of the center of mass is

$v_{CM}=\frac{m_1v_1+m_2v_2}{m_1+m_2}=\frac{(1)(0)+(3)(0)}{1+3}=0.$ …

• Trying to compute the individual accelerations/velocities of P and Q using Newton's law of gravitation, which is unnecessary (and much harder) — the question only asks about the center of mass, which is governed entirely by the (absent) external force.
• Assuming that because the particles are accelerating, the center of mass must also be accelerating — this confuses the motion of the individual particles with the motion of their combined center of mass. …