Centre of Mass & Collisions: Advanced JEE Main Guide

Centre of Mass and Collisions is deceptively rich territory for JEE Main. The basics — COM formula and momentum conservation — are straightforward. But the questions that actually discriminate between 80-percentile and 99-percentile students involve systems where external forces are absent and the COM is therefore stationary, or require you to combine momentum conservation with kinetic energy analysis. This guide focuses on the advanced ideas that JEE returns to every year.

COM of Continuous Bodies and the Key Result

For a system of n particles: x_COM = Σ(mᵢxᵢ)/Σmᵢ. For uniform bodies, COM coincides with the geometric centroid: rod (midpoint), rectangle (intersection of diagonals), triangle (centroid, 1/3 height from each base), uniform solid/hollow sphere and cylinder (geometric centre), semicircle (4R/3π from centre), hemisphere (3R/8 from flat face). The last two are consistently tested — memorise both and know that the hemisphere result uses direct integration.

The crucial result: if no external force acts on a system, the COM does not move. This is the key to a whole class of JEE problems where a person walks on a boat, or two blocks on a frictionless surface interact via a spring. In all such cases, Δx_COM = 0, giving m₁Δx₁ + m₂Δx₂ = 0. The COM displacement equation — not momentum conservation directly — is the cleanest path to the answer. Practise COM problems with a free mock to build the reflex for spotting these problems.

Elastic, Inelastic, and Perfectly Inelastic Collisions

Elastic collision: both momentum and kinetic energy are conserved. For equal masses: velocities exchange (v₁_final = u₂, v₂_final = u₁). For unequal masses: v₁' = (m₁−m₂)u₁/(m₁+m₂) and v₂' = 2m₁u₁/(m₁+m₂) (if m₂ is initially at rest). These two results are derived from the two conservation equations; knowing them by heart saves two minutes per question. If m₁ << m₂ (light ball hitting heavy wall), v₁' ≈ −u₁ (ball bounces back). If m₁ >> m₂ (heavy ball hitting light ball), v₂' ≈ 2u₁.

Perfectly inelastic collision: objects stick together, momentum is conserved, kinetic energy is not. v_final = (m₁u₁ + m₂u₂)/(m₁+m₂). Loss in KE = ½μu_rel² where μ = m₁m₂/(m₁+m₂) is the reduced mass and u_rel is the relative approach speed. JEE frequently asks for the KE lost — use the reduced-mass formula directly instead of computing KE before and after separately. For related topics, see our work-energy-power guide and our mechanics master guide.

Coefficient of Restitution and Oblique Collisions

The coefficient of restitution e = relative speed of separation / relative speed of approach along the line of impact. e = 1: elastic; e = 0: perfectly inelastic; 0 < e < 1: partially inelastic. Given e and masses, find post-collision velocities by solving: (i) momentum conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂; (ii) restitution: v₂ − v₁ = e(u₁ − u₂). These two equations in two unknowns give v₁ and v₂ directly. JEE uses this setup in about half of all collision questions.

Oblique collisions (ball bouncing off a wall at an angle): the normal component of velocity changes by factor e; the tangential component is unchanged (frictionless wall). If ball hits at angle θ, normal component = u cosθ → eu cosθ after; tangential = u sinθ unchanged. The resulting angle with the wall after impact: tan φ = eu cosθ/(u sinθ) = e cotθ. This result appears once every two years in JEE Main.

Impulse and Variable Mass Problems

Impulse J = ΔP = F_avg × Δt. The impulse-momentum theorem is more useful than F = ma when force is impulsive (large force, short time). Variable mass systems (rocket, conveyor belt): thrust force on rocket = v_rel × (dm/dt), where v_rel is exhaust velocity relative to rocket. The rocket equation: m(dv/dt) = −v_rel(dm/dt). JEE asks for velocity after burning a fraction of fuel — integrate or use v_f = v_i + v_rel ln(m_i/m_f). For deeper context on Newton's laws applications, see our Newton's laws guide.