Physics · System Of Particles And Rotational Motion · NEET
If two particles of mass m1 and m2 sit at positions x1 and x2 on the x-axis, then x_cm = (m1x1 + m2x2)/(m1 + m2). It is a mass-weighted average of the positions, not a simple average. Only when m1 = m2 does it reduce to the midpoint (x1 + x2)/2.
Because each position is multiplied by its own mass in the numerator. The heavier mass 'pulls' the weighted average toward itself. Physically, the COM is the balance point: to balance a heavy mass and a light mass on a rod, the pivot must sit near the heavy end so the turning effects match.
Take the two masses a distance L apart. The COM distance from mass m1 is d1 = m2L/(m1 + m2), and from mass m2 it is d2 = m1L/(m1 + m2). Notice the cross-mass: distance from m1 uses m2 on top. This is the fastest form for NEET single-line problems.
Exactly at the midpoint of the line joining them. Set m1 = m2 = m in the formula: x_cm = (m·x1 + m·x2)/(2m) = (x1 + x2)/2. Mass cancels, so only the geometry decides.
No. You can pick any origin. The COM position you get will be measured from that same origin. If you place the origin on one of the particles, x1 = 0 and the formula collapses directly to the distance shortcut d = m2L/(m1 + m2).
Two particles of mass 5 kg and 10 kg are attached to the ends of a rigid massless rod of length 1 m. The distance of the centre of mass from the 5 kg particle is nearly:
Two objects of mass 10 kg and 20 kg are connected to the ends of a rigid massless rod of length 10 m. The distance of the centre of mass from the 10 kg mass is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. For two positive masses the COM always lies on the line segment joining them, somewhere between the two, never outside. It reaches an end only in the limit where one mass is infinitely larger than the other.
No. The rod is described as 'massless' or 'light', so it contributes nothing to the mass distribution. Only the two point masses matter. This is why NEET problems specify a massless rod — it keeps the calculation purely about the two particles.
Yes. Apply it separately to each coordinate: X = (m1x1 + m2x2)/(m1+m2), Y = (m1y1 + m2y2)/(m1+m2), and similarly for Z. Each axis is handled independently with the same mass weighting.
The two-particle formula is the foundation for the n-particle centre of mass, motion of the centre of mass, and momentum conservation. NEET almost every year asks a quick single-line COM-distance question worth easy marks, and getting the mass-weighting right avoids the midpoint trap.
Place your origin right on one of the masses. Then that particle's position becomes zero and the formula reduces to d = (other mass × length)/(total mass). One line, no sign errors, and you can read off which mass the answer is measured from.
