Centre of Mass & Collisions: JEE Main Guide

Centre of mass and collisions appears every year in JEE Main — typically two to three questions — and rewards students who treat momentum conservation as a unifying principle rather than a collection of disconnected formulas. The chapter connects to rotational mechanics, Newton's third law, and energy methods, so mastering it strengthens multiple chapters simultaneously.

Centre of Mass: Definition and Standard Results

The centre of mass is the weighted average position of all mass in the system. For a two-particle system: x_CM = (m₁x₁ + m₂x₂)/(m₁ + m₂). JEE Main expects you to know CM positions of standard bodies: uniform rod at midpoint, solid hemisphere at 3R/8 from base, hemispherical shell at R/2 from base, semicircular arc at 2R/π from centre, solid cone at H/4 from base. Derive each once to lock it in memory, then use the result directly under exam pressure.

Key theorem: if no external force acts on a system, the CM does not accelerate. A bomb exploding in mid-air has fragments that scatter, but the CM continues along the original parabola. JEE Main uses this principle frequently — when internal forces complicate the picture, track the CM and apply only external forces. For the mechanics foundation underlying this chapter, see our Mechanics Master Guide.

Linear Momentum and Impulse

Momentum p = mv is conserved when net external impulse is zero. Impulse J = F·Δt = Δp. The graphical version: impulse equals the area under the F-t curve. JEE Main uses F-t graphs to ask for change in momentum — recognise this as an area calculation and execute directly. The coefficient of restitution e = (relative separation speed)/(relative approach speed): e = 1 for elastic, e = 0 for perfectly inelastic, 0 < e < 1 for partially inelastic.

Elastic and Inelastic Collisions

For a head-on elastic collision between m₁ (velocity u₁) and m₂ (at rest): v₁ = (m₁−m₂)u₁/(m₁+m₂), v₂ = 2m₁u₁/(m₁+m₂). Three special cases JEE Main tests repeatedly: equal masses — first stops, second moves at u₁; heavy hits light — heavy barely slows, light flies at approximately 2u₁; light hits heavy — light bounces back at −u₁, heavy barely moves. Commit these to memory as they appear directly in MCQ options.

Perfectly inelastic collision: objects stick together, use only momentum conservation. Energy lost = ½·m₁m₂/(m₁+m₂)·(u₁−u₂)². This energy-loss formula is a direct JEE Main question type. Take a free mock test on collision problems to practise recognising which conservation law applies.

Two-Dimensional and Oblique Collisions

In oblique collisions, momentum conservation applies independently along x and y. For smooth sphere collisions: decompose velocities along the line of centres (apply e and momentum conservation here) and perpendicular (velocity component unchanged for smooth spheres). JEE Main almost exclusively uses 30-60-90 and 45-45-90 geometry — practise resolving at these standard angles until it becomes automatic.

Variable Mass: Rocket Propulsion

Thrust = v_exhaust × |dm/dt|. Net upward force = thrust − mg. JEE Main tests this mostly qualitatively: thrust acts opposite to exhaust direction, and this follows directly from momentum conservation on a variable-mass system. Numerical problems do appear in the integer-answer section occasionally.

Exam Strategy

Step 1: identify collision type from the problem text. Step 2: draw before/after velocity diagrams. Step 3: write momentum conservation. Step 4: for elastic collisions, use the velocity formulas directly rather than re-deriving. Step 5: maintain consistent sign conventions throughout. Most errors in collision problems are sign errors, not formula errors. For a complete revision schedule, see our 30-Day Physics Revision Plan. Upgrade for ₹149/month for 200+ collision and CM problems with step-by-step solutions.