Physics I: Mechanics › Momentum › full formula sheet

v1f = (m₁−m₂)/(m₁+m₂) v1i,  v2f = 2m₁/(m₁+m₂) v1i

Say it: “in an elastic collision both momentum and kinetic energy are conserved — the objects bounce without losing energy”

Elastic collisions in one dimension

The perfect bounce: momentum and kinetic energy both conserved — the formulas, and the three landmark cases.

Formulas shown for target at rest (v2i = 0) — the most common setup. m₁ is the incoming mass, m₂ the target. Both momentum and KE are conserved here.

Before this lesson: Conservation

Where it comes from

Two billiard balls click: the cue ball stops dead and the target ball takes off at the cue ball's old speed. No heat, no dent, no sound-waste to speak of — the collision is elastic, conserving kinetic energy as well as momentum.

Before reading on: equal masses, target at rest — you saw the cue ball stop dead. Now the incoming ball is heavier than the target. Does the heavy ball stop, plow through, or bounce back?

It plows through — slowed, but still moving forward, while the light target rockets ahead. And a light ball hitting a heavy wall? It bounces straight back at nearly its incoming speed. Three landmark cases fall out of one derivation:

m₁ = m₂
⇒
v1f = 0, v2f = v1i
Equal masses: the swap — incoming stops dead, target takes all the velocity.
m₁ << m₂
⇒
v1f ≈ −v1i
Light hits heavy: bounces back at nearly full speed (the wall barely notices).
m₁ >> m₂
⇒
v1f ≈ v1i, v2f ≈ 2v1i
Heavy hits light: plows on nearly unaffected; the light target flies off at ~twice the incoming speed.

Derivation

Two conservation laws, two unknowns (v1f, v2f). Target at rest (v2i = 0):

mom.
=
m₁v1i = m₁v1f + m₂v2f
Step 1 — conserve momentum. Before-total equals after-total.
KE
=
½m₁v1i² = ½m₁v1f² + ½m₂v2f²
Step 2 — conserve kinetic energy. Elastic means no KE lost. The ½s cancel.
⇒
v1i − 0 = −(v1f − v2f)
Step 3 — the elegant shortcut. Rearranging both equations (factor differences of squares) gives: relative velocity reverses. The approach speed equals the separation speed.
solve
=
v1f = (m₁−m₂)/(m₁+m₂)·v1i, v2f = 2m₁/(m₁+m₂)·v1i
Step 4 — solve the linear system. Two equations (momentum + relative-velocity reversal), two unknowns. ∎

The relative-velocity trick (Step 3) is worth memorizing: in any 1D elastic collision, v1i − v2i = −(v1f − v2f). It replaces the quadratic KE equation with a linear one.

v1i − v2i = −(v1f − v2f)approach speed equals separation speed — the elastic collision's fingerprintSay it: “the relative velocity reverses in an elastic collision”

How to use it

The procedure, every time:

  1. Confirm elastic. “Elastic collision,” billiard balls, atomic collisions — KE conserved. If they stick or deform, it's inelastic instead.
  2. Label 1 = incoming, 2 = target. The formulas assume v2i = 0. If the target moves, shift to its rest frame or use the general form.
  3. Check a landmark first. Equal masses? Expect the swap. Wall? Expect the bounce-back. Often no formula needed.
  4. Otherwise plug in. v1f = (m₁−m₂)/(m₁+m₂)·v1i; v2f = 2m₁/(m₁+m₂)·v1i.
  5. Verify both conservations. Check momentum and KE before/after — two independent checks catch sign slips.
Common mistake: using the elastic formulas on a sticky collision. If the problem says “stick together” or mentions denting/heat, KE is NOT conserved — momentum alone governs.

Worked examples

Four problems, easiest first. Try the landmark cases before the formula.

Example 1 — the swap: m₁ = m₂ = 2 kg, v1i = 6 m/s, target at rest

  1. Landmark: equal masses. v1f = 0, v2f = v1i — no formula needed.
  2. Answer. v1f = 0 m/s, v2f = 6 m/s.
  3. Verify. Momentum: 12 = 0 + 12 ✓. KE: 36 = 0 + 36 ✓.
Common mistake: “they bounce apart equally” — splitting the velocity 3/3. Equal masses don't share; the incoming stops dead. (That's the billiard click you hear.)
Your turn — equal 1 kg masses, v1i = 10 m/s, target at rest. Finals?

Answer: v1f = 0, v2f = 10 m/s. The swap, every time.

Example 2 — formula: m₁ = 3 kg at 8 m/s hits m₂ = 1 kg at rest

  1. v1f. (3−1)/(3+1) × 8 = (2/4) × 8 = 4 m/s.
  2. v2f. (2×3)/(3+1) × 8 = (6/4) × 8 = 12 m/s.
  3. Verify. Momentum: 24 = 12 + 12 ✓. KE: 96 = 24 + 72 ✓.
Common mistake: swapping the numerators — v2f = (m₁−m₂)/(…). v2f's numerator is 2m₁ (the incoming mass doubled), not the difference.
Your turn — m₁ = 4 kg at 9 m/s, m₂ = 2 kg at rest. Finals?

Answer: v1f = 3 m/s, v2f = 12 m/s. (4−2)/6×9 = 3; (8/6)×9 = 12.

Example 3 — the wall: 0.5 kg ball at 4 m/s hits a massive wall

  1. Landmark: light vs heavy. v1f ≈ −v1i.
  2. Answer. v1f ≈ −4 m/s — bounces straight back at full speed. (The wall's recoil is m₁/m₂ ≈ 0.)
Common mistake: v1f = 0 (“the wall stops it”). Stopping would destroy KE — the wall can't absorb it elastically. Bounce-back preserves both conservations.
Your turn — same ball at 6 m/s vs the wall. v1f ≈ ?

Answer: ≈ −6 m/s. Bounce-back at full speed.

Example 4 — heavy hitter: m₁ = 10 kg at 5 m/s vs m₂ = 1 kg at rest

  1. v1f. (10−1)/11 × 5 = (9/11) × 5 ≈ 4.09 m/s — barely slowed.
  2. v2f. 20/11 × 5 ≈ 9.09 m/s — nearly twice the incoming speed.
  3. Verify. Momentum: 50 ≈ 40.9 + 9.09 ✓. KE: 125 ≈ 83.7 + 41.3 ✓.
Common mistake: expecting the light target to leave at ~5 m/s (“it can't go faster than what hit it”). It can — nearly 2×. The heavy ball keeps most of its KE; the light one takes a big slice of speed.
Your turn — m₁ = 6 kg at 8 m/s, m₂ = 2 kg at rest. Finals?

Answer: v1f = 4 m/s, v2f = 12 m/s. (6−2)/8×8 = 4; (12/8)×8 = 12.

Memorization tips

  • Say it aloud: “elastic means momentum AND kinetic energy both conserved.”
  • Three landmarks: equal → swap; light-vs-wall → bounce back; heavy-vs-light → plow through, target at ~2×. Most problems are landmarks in disguise.
  • Relative velocity reverses: approach speed = separation speed. The linear shortcut that dodges the quadratic.
  • Label 1 = incoming. The formulas assume v2i = 0. Wrong labels, wrong answers.
  • Double-check with both laws: verify momentum AND KE after. Two checks, zero sign slips.
  • Elastic ≠ sticky: 'stick together', 'dent', 'heat' → inelastic. Read the adjectives.

Final challenge

Five mixed questions — landmarks, formulas, and the elastic/inelastic line. Score 5/5 and elastic collisions are yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

Are real collisions ever perfectly elastic?

Almost: billiard balls, steel bearings, and atomic/subatomic collisions are very close. Most macroscopic crashes lose some KE to heat, sound, and deformation — treat 'elastic' as the ideal limit.

Why does the light target leave at ~2x the incoming speed?

Momentum: the heavy ball barely slows, so it keeps ~its momentum; energy: the leftover must go somewhere. Solving both gives v2f ≈ 2v1i — the light mass converts a little momentum into a lot of speed.

What if the target is moving too?

Shift to the target's rest frame (subtract v2i from all velocities), apply the formulas, then shift back. Or use the general form — the relative-velocity reversal still holds.

How do I know a problem is elastic?

Magic words: 'elastic collision', 'billiard', 'bounces', 'steel balls'. Red flags: 'stick', 'couple', 'dent', 'heat', 'completely inelastic' — those are the inelastic page.

Does the relative-velocity rule work in 2D?

Along the line of impact, yes — decompose into normal and tangential components. The tangential velocities don't change (no friction); the normal components obey the 1D reversal rule.

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