Collisions — The Three Families (NCERT 5.11)
1. What Counts as a Collision?
A collision is any brief, violent interaction — billiard balls, a hammer and nail, a nucleus absorbing a neutron — where the contact forces spike enormously for milliseconds and then vanish. Two consequences define the whole topic. First, because the interaction is brief, any external impulse (gravity's, friction's) is negligible during it, so the total momentum of the colliding pair is conserved across the impact — the momentum ledger from Chapter 4 closes exactly. Second, because the bodies deform and heat, the pair's kinetic energy is negotiable: fully preserved only in idealised hard impacts. Collisions are therefore classified by what happens to K, and that classification is this page's entire syllabus.
2. Complete Theory: The Three Families and Their Formulas
Family one — elastic (idealised hard balls, atoms): both momentum and kinetic energy are conserved. Two equations, two unknowns, and solving them together (from p: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂; from K: ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂²) yields the 1-D workhorses:
v₁′ = [(m₁ − m₂)u₁ + 2m₂u₂] / (m₁ + m₂) · v₂′ = [(m₂ − m₁)u₂ + 2m₁u₁] / (m₁ + m₂)
Own these as derivations, because their special cases are the exam. Equal masses (m₁ = m₂): the formulas collapse to v₁′ = u₂ and v₂′ = u₁ — the bodies exchange velocities, head-on or not. Heavy on light at rest (m₁ ≫ m₂, u₂ = 0): the heavy one barely notices (v₁′ ≈ u₁) while the light one departs at ≈ 2u₁ — the bowling-pin launch. Light on heavy at rest (m₁ ≪ m₂): the light one rebounds at ≈ −u₁ while the heavy one barely stirs — a ball off a wall.
Family two — inelastic: momentum conserved, K partly lost (crumpled bumpers, thuds). Class 11 treats it qualitatively or through the restitution coefficient below.
Family three — perfectly inelastic: the bodies stick and move with one common velocity. Momentum alone fixes it:
v = (m₁u₁ + m₂u₂)/(m₁ + m₂), K lost = ½ μ vrel², μ = m₁m₂/(m₁ + m₂)
The loss formula says the kinetic energy of the relative motion dies; the centre-of-mass motion survives untouched. Sticking destroys the maximum K compatible with momentum conservation — that sentence is a standing Assertion–Reason item.
The coefficient of restitution grades family two between the extremes: e = (speed of separation)/(speed of approach). e = 1: elastic (relative speed fully restored). e = 0: putty. A ball dropped onto a fixed floor (infinite mass) simplifies to rebound speed = e × impact speed, and the rebound height comes out e² × drop height — two drops of a 0.5-restitution ball lose three-quarters of the height.
Two dimensions (glancing): conserve momentum as separate x and y equations. The one result NTA asks: when equal masses collide elastically with one initially at rest, the post-collision velocities emerge at a right angle — the billiards 90° rule, provable from px + py bookkeeping plus K conservation.
3. Visualising: The Elastic Exchange
Figure 5.8 — Equal masses, elastic: a clean swap
Top: before — 2 kg at 6 m/s approaches a resting 2 kg. Bottom: after — velocities exchanged. Audit both panels: p = 12 kg m/s, K = 36 J each time.
Exam read-out: for equal masses the elastic formulas degenerate into "give your velocity to your partner" — the fastest answer in collisions. When the masses are unequal, resist memory and re-derive with the two conservation lines; NTA's altered-mass options hunt for formula recall without understanding.
4. Solved Examples
Example 1 — Unequal masses, elastic: run the formulas
A 1 kg body moving at 4 m s⁻¹ collides head-on and elastically with a 2 kg body at rest. Find both velocities afterwards.
Solution (step by step): v₁′ = [(1 − 2)(4) + 2 × 2 × 0]/3 = −4/3 ≈ −1.33 m/s (the 1 kg body rebounds). v₂′ = [(2 − 1)(0) + 2 × 1 × 4]/3 = 8/3 ≈ +2.67 m/s. Audit both ledgers: p = 1(−4/3) + 2(8/3) = 12/3 = 4 kg m/s = before ✓; K = ½(1)(16/9) + ½(2)(64/9) = 8/9 + 64/9 = 8 J = ½(1)(16) ✓. Both books balance — the collision really was elastic.
Example 2 — Perfectly inelastic: common velocity and the loss
A 4 kg body moving at 5 m s⁻¹ overtakes and sticks to a 6 kg body at rest. Find the common velocity and the kinetic energy lost.
Solution: Momentum: 4 × 5 + 0 = (4 + 6)v → v = 2 m/s in the original direction. Kinetic ledger: before = ½ × 4 × 25 = 50 J; after = ½ × 10 × 4 = 20 J; loss = 30 J. Reduced-mass check: μ = 4 × 6/10 = 2.4 kg, vrel = 5 → ΔK = ½ × 2.4 × 25 = 30 J ✓. The 30 J went into permanent dent and heat — momentum never noticed, as always.
5. Practice Questions
Q1. Two identical 1 kg pucks collide head-on elastically on ice: puck A at 4 m s⁻¹ meets puck B moving toward it at 2 m s⁻¹ (opposite direction). Find both velocities after the collision.
ANSWER: Equal masses + elastic ⇒ velocity exchange: puck A leaves with B's incoming velocity and vice versa. With rightward positive: u₁ = +4, u₂ = −2 → v₁′ = −2 m/s (A rebounds leftward) and v₂′ = +4 m/s (B carries on rightward). Audit: p = 4 − 2 = 2 before; −2 + 4 = 2 after ✓; K = ½(16) + ½(4) = 10 J both sides ✓.
Q2 (MCQ). A ball hits the ground at 20 m s⁻¹ and rebounds with e = 0.5. The rebound speed and the fraction of the drop height it returns to are: (a) 10 m/s, 1/2 (b) 10 m/s, 1/4 (c) 5 m/s, 1/4 (d) 10 m/s, 3/4
ANSWER: (b) — rebound speed = e × impact = 10 m/s; height scales as v², so h′/h = (10/20)² = 1/4. Each bounce of an e = 0.5 ball forfeits three-quarters of its height — the e² rule behind every "how many bounces" sequence question.
Q3. A moving billiard ball strikes an identical stationary ball with a glancing (not head-on) elastic blow. State the angle between the two balls' velocities afterwards and name the conservation principles that force it.
ANSWER: 90°. Reasoning: equal masses + elastic ⇒ both momentum (as vectors: x and y components separately) and kinetic energy (as a scalar) survive. Writing p⃗ before = p⃗₁′ + p⃗₂′ and squaring while using 2mK = p² shows (p₁′ + p₂′)² = p₁′² + p₂′² only when p₁′·p₂′ = 0 — the two outgoing momenta (hence velocities) are perpendicular. The standard pool-hall geometry NTA sketches once a year.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ | Every isolated collision — the only universal law |
| v₁′ = [(m₁−m₂)u₁ + 2m₂u₂]/(m₁+m₂) | 1-D elastic; partner formula swaps the subscripts |
| Equal masses: v₁′ = u₂, v₂′ = u₁ | Velocity exchange, any u₂ — elastic 1-D only |
| m₁ ≫ m₂: v₂′ ≈ 2u₁ · m₁ ≪ m₂: v₁′ ≈ −u₁ | Heavy-on-light launch; light-on-heavy rebound |
| v = (m₁u₁ + m₂u₂)/(m₁ + m₂) | Perfectly inelastic: bodies stick, common velocity |
| ΔK = ½ μ vrel², μ = m₁m₂/(m₁+m₂) | Max possible KE loss on sticking; relative-motion KE dies |
| e = separation/approach speed | e = 1 elastic · e = 0 putty · rebound height = e² × drop |
| Equal-mass glancing elastic → 90° | Conserve px, py separately + K; the billiards rule |
Collisions close the chapter by marrying its two conservation laws: momentum from Chapter 4 (never breaks) and kinetic energy from this chapter (breaks on request). Read each question for which one holds — that reading is the answer key.
7. Frequently Asked Questions
Which quantities are conserved in a collision?
Linear momentum is conserved in every collision of an isolated pair — the contact forces are internal third-law pairs and the impact is too brief for external impulses to matter. Kinetic energy is conserved only in elastic collisions; in inelastic ones part of it converts to heat, sound and deformation. So the safe exam sentence is: momentum always, kinetic energy only if elastic. The perfectly inelastic (bodies stick) case loses the maximum kinetic energy consistent with momentum conservation.
State the one-dimensional elastic collision formulas.
For bodies of masses m1 and m2 with initial velocities u1 and u2 colliding elastically head-on, solving momentum conservation and kinetic-energy conservation together gives v1' = ((m1 − m2)u1 + 2 m2 u2)/(m1 + m2) and v2' = ((m2 − m1)u2 + 2 m1 u1)/(m1 + m2). Special cases NTA recycles: equal masses simply exchange velocities (v1' = u2, v2' = u1); a heavy body hitting a light one at rest keeps going barely slowed while the light one departs at about twice the heavy body's speed; a light body hitting a heavy one at rest bounces back at nearly its incoming speed.
What is the coefficient of restitution?
The coefficient of restitution e measures a collision's elasticity through relative speeds: e = (relative velocity of separation)/(relative velocity of approach), taken as a positive ratio. For e = 1 the collision is elastic (relative speed fully restored); for e = 0 the bodies stick together and move with a common velocity; real collisions sit between. For a ball bouncing off a fixed floor (infinite mass), e reduces to rebound speed over impact speed, and the rebound height after a drop is e² times the drop height.
How much kinetic energy is lost in a perfectly inelastic collision?
For two bodies colliding head-on and sticking, the loss is ΔK = ½ μ v_rel², where μ = m1 m2/(m1 + m2) is the reduced mass and v_rel = u1 − u2 is the initial relative speed. Example: a 4 kg body at 5 m/s hitting a 6 kg body at rest gives μ = 2.4 kg and v_rel = 5 m/s, so ΔK = ½ × 2.4 × 25 = 30 J. The formula shows the loss is largest when the relative speed is largest and vanishes when the bodies merely touch with no relative motion — it is the kinetic energy of the relative motion that dies.
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