MixedJEE Physics · Original learning card10 original chapter questions

Centre of Mass of Discrete Particles

For discrete particles, the centre-of-mass position is the vector sum of each mass times its position divided by total mass; this point governs translation even when the particles move relative to one another.

Why this shows up in the exam

Locating the balance point of separated masses · Reducing a many-particle translation problem to one point · Finding centre-of-mass velocity from particle velocities

Learn the idea

A system balances translationally as though its total mass were concentrated at the mass-weighted average position. A system balances translationally as though its total mass were concentrated at the mass-weighted average position. Start from a clear axis, origin, body, and reference frame; the geometry and constraints then decide which rotational law is safe to use.

🧠 Memory hook: Heavy masses pull the average position closer.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • r_cm = (sum m_i r_i)/(sum m_i) — Mass-weighted position for particles measured in one inertial frame.
  • M v_cm = sum m_i v_i — Total linear momentum relation for a fixed-mass system.

How to approach it

  1. 1Choose one origin and axes
  2. 2Form each mass-position product componentwise
  3. 3Divide by total mass and check that the result lies in the expected region

Common slip-ups that cost marks

  • •Averaging coordinates without mass weights
  • •Mixing position vectors from different origins
  • •Using signed coordinates as unsigned distances

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 10

Masses 1 kg and 3 kg lie at x = 0 and x = 4 m. Find the x-coordinate of their centre of mass.

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