Elastic vs Inelastic Collisions — Momentum Conservation
The single most misunderstood fact about collisions is which quantity is conserved in which case. Momentum is conserved in every collision of an isolated system — elastic or not. What changes is kinetic energy, and only elastic collisions keep that constant too. This guide works through a perfectly inelastic collision, two elastic collisions (equal and unequal masses), and a real partially-elastic collision using the coefficient of restitution.
Conservation of momentum — always true
This holds for any collision between two bodies with no external force acting — elastic, inelastic, or anything in between. u₁, u₂ are the velocities before collision; v₁, v₂ are the velocities after.
Elastic vs inelastic — what actually differs
| Momentum | Kinetic energy | |
|---|---|---|
| Elastic collision | Conserved | Conserved |
| Inelastic collision | Conserved | Not conserved — some becomes heat, sound or deformation |
| Perfectly inelastic collision | Conserved | Maximum possible KE loss — the bodies stick together and move with one common velocity |
Elastic collision — final velocities (1D)
v₂ = [2m₁/(m₁+m₂)] u₁ + [(m₂−m₁)/(m₁+m₂)] u₂
These come from solving momentum conservation and kinetic-energy conservation together as simultaneous equations — they only apply when the collision is genuinely elastic.
Coefficient of restitution — real collisions
e is the ratio of the relative speed of separation to the relative speed of approach. e = 1 for a perfectly elastic collision, e = 0 for a perfectly inelastic one, and most real collisions fall somewhere between, 0 < e < 1.
Worked example 1 — perfectly inelastic collision
A 2 kg cart moving at u₁ = 5 m/s collides with a stationary 3 kg cart (u₂ = 0) and they stick together. Find the common velocity and the kinetic energy lost.
v = (m₁u₁ + m₂u₂) / (m₁+m₂) = (2×5 + 3×0) ÷ (2+3) = 10 ÷ 5 = 2 m/s
KEinitial = (1/2)(2)(5²) = 25 J. KEfinal = (1/2)(5)(2²) = 10 J. KE lost = 25 − 10 = 15 J (60% of the initial KE)
Momentum check: pi = 2×5 = 10 kg·m/s, pf = 5×2 = 10 kg·m/s ✓ — momentum is conserved even though 60% of the kinetic energy is gone.
Worked example 2 — elastic collision, equal masses
m₁ = 1 kg moving at u₁ = 6 m/s strikes a stationary m₂ = 1 kg (u₂ = 0), elastically.
v₁ = [(1−1)/(1+1)](6) + [2(1)/(1+1)](0) = 0 + 0 = 0 m/s
v₂ = [2(1)/(1+1)](6) + [(1−1)/(1+1)](0) = 1×6 + 0 = 6 m/s
The striking ball stops dead and the target ball moves off with all its speed — the classic equal-mass elastic-collision result seen in Newton's cradle.
Checks: momentum pi = 1×6 = 6, pf = 1×0+1×6 = 6 ✓. KE KEi = (1/2)(1)(36) = 18 J, KEf = 0 + (1/2)(1)(36) = 18 J ✓ — both conserved, confirming the collision really is elastic.
Worked example 3 — elastic collision, unequal masses
m₁ = 4 kg at u₁ = 3 m/s strikes a stationary m₂ = 1 kg elastically.
v₁ = [(4−1)/(4+1)](3) + 0 = (3/5)(3) = 1.8 m/s
v₂ = [2(4)/(4+1)](3) + 0 = (8/5)(3) = 4.8 m/s
Momentum check: pi = 4×3 = 12, pf = 4×1.8 + 1×4.8 = 7.2+4.8 = 12 ✓
KE check: KEi = (1/2)(4)(9) = 18 J. KEf = (1/2)(4)(1.8²) + (1/2)(1)(4.8²) = 6.48 + 11.52 = 18 J ✓ — both quantities match exactly, confirming the elastic-collision formulas were applied correctly.
Worked example 4 — a real collision, using the coefficient of restitution
Two 2 kg balls: ball 1 approaches at u₁ = 8 m/s, ball 2 is stationary (u₂ = 0). The coefficient of restitution for this collision is e = 0.6 (neither perfectly elastic nor perfectly inelastic). Find both final velocities.
Momentum: 2(8) + 2(0) = 2v₁ + 2v₂ → v₁ + v₂ = 8 …(i)
Restitution: e = (v₂−v₁)/(u₁−u₂) → 0.6 = (v₂−v₁)/8 → v₂ − v₁ = 4.8 …(ii)
Adding (i) and (ii): 2v₂ = 12.8 → v₂ = 6.4 m/s. From (i): v₁ = 8 − 6.4 = 1.6 m/s
Momentum check: 2(1.6) + 2(6.4) = 3.2 + 12.8 = 16 = 2(8) ✓
KEi = (1/2)(2)(64) = 64 J. KEf = (1/2)(2)(1.6²) + (1/2)(2)(6.4²) = 2.56 + 40.96 = 43.52 J. KE lost = 64 − 43.52 = 20.48 J — a partial loss, consistent with 0 < e < 1, as expected for a real (not perfectly elastic or perfectly inelastic) collision.
Common mistakes that cost marks
- Thinking momentum is conserved only in elastic collisions. It is conserved in every collision of an isolated system — it is kinetic energy that is exclusive to the elastic case.
- Using the elastic-collision velocity formulas for an inelastic collision. Those formulas are derived assuming KE is also conserved; they give the wrong answer the moment the collision is not elastic.
- Sign errors when objects move in opposite directions. Fix one positive direction before writing any equation, and keep every velocity's sign consistent with it throughout.
- Confusing "perfectly inelastic" with "inelastic" in general. Perfectly inelastic (e = 0) means the bodies stick together — it is one specific case, not a synonym for every non-elastic collision.
- Summing signed velocities for kinetic energy. KE uses v² (always positive), so it can never be added the way momentum (which keeps its sign) can.
Where collisions appear in exams
| Exam | Typical use |
|---|---|
| CBSE Class 11 | Work, Energy and Power — 1D elastic and inelastic collisions, momentum conservation |
| JEE Main & Advanced | 1D and 2D collision numericals, often combined with the coefficient of restitution |
| NEET | Direct perfectly-elastic and perfectly-inelastic substitution questions |
| GATE (Engineering) | Impact and restitution problems in engineering mechanics |
Verify momentum and KE conservation quickly. The Scientific Calculator makes checking a step like (1/2)(2)(6.4²) fast while you practise setting up the simultaneous equations above.
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