Centre of Mass, Momentum & Collisions

Whenever an external force is absent or impulsive forces dominate, momentum conservation cracks the problem. Advanced deepens this with variable-mass systems (rockets), centre-of-mass (COM) frame analysis of collisions, and combined linear-plus-rotational impulse.

Key results

• COM: x_cm = Σm_i x_i / Σm_i; velocity of COM stays constant if net external force is zero.
• Momentum: conserved along any direction with no external impulse; impulse J = ∫F dt = Δp.
• Elastic 1-D collision: relative velocity reverses (e = 1). For equal masses, velocities exchange. General: v1' = ((m1−m2)u1 + 2m2u2)/(m1+m2).
• Coefficient of restitution: e = (v2'−v1')/(u1−u2); KE loss = ½·(m1m2/(m1+m2))·(1−e²)(u1−u2)².
• Rocket: thrust = v_rel·(dm/dt); v = v0 + v_rel·ln(m0/m) (Tsiolkovsky).

Collision	e	KE
Perfectly elastic	1	conserved
Perfectly inelastic	0	maximum loss
Real	0<e<1	partial loss

Exam Tricks & Tips

• 🎯 Work in the COM frame: total momentum is zero there, so in an elastic collision each speed simply reverses — collisions become trivial.
• 🎯 Equal-mass elastic collision = velocity swap; a moving ball hitting an identical stationary one stops dead.
• 🎯 COM does not accelerate under internal forces — an exploding shell's COM continues on the original parabola.
• 🎯 Use impulse–momentum for short, large forces (bat-ball, collisions) where you cannot resolve the force-time detail.
• 🎯 Fraction of KE transferred is maximised when masses are equal (perfectly elastic, target at rest).
• ❌ Common mistake: assuming KE is conserved in every collision — only elastic (e=1) conserves it; always check e.

Expected exam pattern

1–2 questions: a two-body collision (often with restitution) feeding into subsequent motion, or a COM/variable-mass conceptual. Frequently a numerical answer requiring e or KE-loss.

Quick recap

Conserve momentum whenever external impulse is absent; use e to relate approach and separation speeds. The COM moves at constant velocity under internal forces. Equal masses swap velocities in elastic collisions; remember Tsiolkovsky for rockets.

Cover the answer, recall, then check. 11 cards on COM, momentum and collisions for JEE Advanced.

Q1. Coefficient of restitution definition.
A1. e = (speed of separation)/(speed of approach) = (v2'−v1')/(u1−u2), along the line of impact.

Q2. Result of a 1-D elastic collision between equal masses (one at rest).
A2. They exchange velocities; the incoming mass stops, the target moves off with the original speed.

Q3. Does the COM accelerate when a shell explodes in flight?
A3. No — only internal forces act, so the COM continues on the original trajectory.

Q4. Impulse–momentum theorem.
A4. J = ∫F dt = Δp (change in linear momentum).

Q5. KE lost in a collision in terms of e and reduced mass.
A5. ½·(m1m2/(m1+m2))·(1−e²)(u1−u2)².

Q6. Rocket thrust formula.
A6. Thrust = v_rel·|dm/dt|, where v_rel is exhaust speed relative to the rocket.

Q7. Why is the COM frame convenient for collisions?
A7. Total momentum is zero there; for elastic collisions each particle's velocity simply reverses direction.

Q8. Perfectly inelastic collision — what is e and what happens to KE?
A8. e = 0; the bodies move together and KE loss is maximum (momentum still conserved).

Q9. Tsiolkovsky rocket equation.
A9. v = v0 + v_rel·ln(m0/m).

Q10. Position of COM of two masses m1, m2 separated by d (from m1).
A10. x = m2·d/(m1+m2).

Q11. In an elastic collision, what happens to relative velocity?
A11. Its magnitude is unchanged and its direction reverses (e = 1).

When external forces are absent or brief, momentum is the master variable. This topic teaches you to replace a swarm of interacting particles with a single point — the centre of mass — whose motion obeys one clean equation, and to handle collisions where energy may vanish but momentum never does.

Core concept: the centre of mass moves as if all mass were concentrated there and all external force acted on it: F_ext = M a_cm. Internal forces never move the CM.

Core ideas

Beginner — locating the CM

For discrete masses, r_cm = Σm_i r_i / Σm_i. For symmetric bodies the CM sits at the geometric centre. A powerful shortcut for bodies with a hole: treat the removed piece as negative mass, so a disc with a hole is (full disc) − (small disc), and r_cm = (M·0 − m·d)/(M − m).

Intermediate — momentum conservation

If F_ext = 0, total momentum p = Mv_cm is constant. This is why, when a shell explodes in flight, the CM continues on the original parabola even as fragments scatter. Internal explosions, recoils, and man-on-a-boat problems all yield to "CM doesn't move (or keeps its velocity)". For a man walking on a frictionless boat, the CM stays fixed, so boat and man shift in inverse mass ratio.

Advanced — collisions and the coefficient of restitution

In any collision, momentum is conserved. Define the coefficient of restitution e = (relative speed of separation)/(relative speed of approach), measured along the line of impact. e = 1 is perfectly elastic (KE conserved), e = 0 is perfectly inelastic (bodies stick). For a 1-D elastic collision of equal masses, the velocities exchange — a result worth memorising. General 1-D elastic: v₁' = ((m₁−m₂)u₁ + 2m₂u₂)/(m₁+m₂). For 2-D collisions, apply restitution along the line of impact and conserve momentum perpendicular to it (where no impulse acts, so that component is unchanged).

Energy lost in a collision = ½ μ (1−e²)(u₁−u₂)², where μ = m₁m₂/(m₁+m₂) is the reduced mass — a compact formula that saves time.

Worked example

A 2 kg ball moving at 6 m/s strikes a stationary 4 kg ball head-on with e = 0.5. Momentum: 2(6) = 2v₁ + 4v₂. Restitution: v₂ − v₁ = 0.5(6) = 3. From these: 12 = 2v₁ + 4v₂ and v₂ = v₁ + 3, so 12 = 2v₁ + 4v₁ + 12 ⇒ v₁ = 0, v₂ = 3 m/s. Energy lost = ½·(8/6)·(1−0.25)·36 = ½·(4/3)·0.75·36 = 18 J. Check with ½·2·36 − ½·4·9 = 36 − 18 = 18 J. ✓

How JEE Advanced tests this

Explosion/recoil problems using CM invariance; man-on-boat and variable-mass (rocket, falling-chain) setups; multi-collision sequences on a line; oblique collisions with restitution along the line of impact; and "find the hole's effect on CM" geometry.

Exam tricks & shortcuts

• Equal-mass elastic head-on collision ⇒ velocities swap.
• Removed mass = negative mass for CM of bodies with holes.
• Energy loss = ½μ(1−e²)(v_rel)² — one line instead of two KE computations.
• Mnemonic: "Momentum always, energy only if e = 1."

Key formulas

• Centre of mass: r⃗_cm = Σmᵢr⃗ᵢ/Σmᵢ; F_ext = M a_cm.
• Momentum conservation: Σp⃗ = constant (no external force).
• Elastic 1-D: exchange velocities if equal masses; e = 1.
• Coefficient of restitution: e = (v₂−v₁)/(u₁−u₂); perfectly inelastic e = 0.
• Inelastic: common velocity v = (m₁u₁+m₂u₂)/(m₁+m₂).

Momentum is always conserved in collisions; kinetic energy is conserved only in elastic ones.

Worked Example

Problem: A ball of mass 2 kg moving at 3 m/s undergoes a head-on elastic collision with a stationary ball of mass 4 kg. Find the velocities of both balls after the collision, and verify conservation of momentum and kinetic energy. Also state the velocity of the centre of mass.

Solution:
For a one-dimensional elastic collision (m₁ = 2, u₁ = 3, m₂ = 4, u₂ = 0), the standard results are:
v₁ = [(m₁ − m₂)/(m₁ + m₂)] u₁ = [(2 − 4)/6] × 3 = (−1/3)(3) = −1 m/s
v₂ = [2m₁/(m₁ + m₂)] u₁ = (4/6)(3) = 2 m/s.

So the lighter ball rebounds at 1 m/s and the heavier moves forward at 2 m/s.

Check momentum: before = 2×3 = 6 kg·m/s; after = 2(−1) + 4(2) = −2 + 8 = 6 kg·m/s. ✓
Check kinetic energy: before = ½(2)(9) = 9 J; after = ½(2)(1) + ½(4)(4) = 1 + 8 = 9 J. ✓

Velocity of the centre of mass (unchanged, no external force):
v_cm = (m₁u₁ + m₂u₂)/(m₁ + m₂) = 6/6 = 1 m/s.

Answer: v₁ = −1 m/s, v₂ = +2 m/s; v_cm = 1 m/s (constant throughout).