Physics I: Mechanics › Momentum › full formula sheet

Σ pbefore = Σ pafter

Say it: “the total momentum before equals the total momentum after”

Conservation of momentum

The deepest law in mechanics: in a closed system, momentum is never created or destroyed — only exchanged.

Add every object's momentum with signs, before and after. Isolated system = no net external impulse (collisions are quick; external forces barely act).

Before this lesson: Linear momentum, Impulse

Where it comes from

Fire a gun and it kicks back. Two skaters push apart and glide off in opposite directions. A firework explodes and the fragments' momenta still add up to the rocket's. In every case, gains exactly balance losses — momentum moves around but never appears or vanishes.

Before reading on: a 4 kg gun fires a 0.01 kg bullet at 300 m/s. The gun was at rest, so total momentum was zero. After firing, the bullet carries +3 kg·m/s. What must the gun's momentum be — and which way does it move?

−3 kg·m/s — backward. Zero total before means zero total after: the gun must carry exactly the opposite momentum. Recoil isn't a side effect; it's the law balancing its books:

Σ pbefore = Σ pafterevery gain is someone else's loss — the total never changesSay it: “the total momentum before equals the total momentum after”

Derivation

Newton's third law, integrated over the collision time:

F₁₂
=
−F₂₁
Step 1 — third law. During the collision, object 1's force on 2 is equal and opposite to 2's force on 1 — at every instant.
J₁₂
=
−J₂₁
Step 2 — same Δt. Both forces act over the same contact time, so their impulses are also equal and opposite.
Δp₁
=
−Δp₂
Step 3 — impulse is Δp. So 1's momentum change is minus 2's: whatever 2 gains, 1 loses.
p₁ + p₂
=
constant
Step 4 — conserve. Δp₁ + Δp₂ = 0: the total never changes. External forces (friction, gravity) add their own impulse — the system must be isolated for the equality to hold. ∎

How to use it

The procedure, every time:

  1. Define the system. Which objects? (Gun + bullet; both skaters; the colliding cars.) Everything else is “external.”
  2. Check isolation. Collisions are brief — external impulses (friction, gravity over milliseconds) are usually negligible. If they're not, conservation fails.
  3. Fix the positive direction. Write it down. Every velocity gets a sign.
  4. Write the before-total. m₁v₁i + m₂v₂i + … (rest objects contribute 0).
  5. Write the after-total with the unknown. Set equal, solve. One equation — one unknown (for two objects).
Common mistake: forgetting an object's momentum — usually the one “at rest” after, or the recoil. Every object in the system appears on both sides of the equation.

Worked examples

Four problems, easiest first. Before-total equals after-total, signs and all.

Example 1 — recoil: 4 kg gun, 0.01 kg bullet at 300 m/s

  1. Before. Both at rest: total = 0.
  2. After. Bullet: 0.01 × 300 = +3. Gun: 4 × vg.
  3. Conserve. 0 = 3 + 4vg → vg = −0.75 m/s.
  4. Read it. The gun kicks back at 0.75 m/s — small speed, but momentum −3 exactly cancels +3. ✓
Common mistake: vg = +0.75 m/s (same direction as the bullet). The totals must cancel to zero — recoil is always opposite.
Your turn — 5 kg gun fires 0.02 kg at 400 m/s. Recoil speed?

Answer: −1.6 m/s. Bullet p = 8; vg = −8/5 = −1.6 m/s.

Example 2 — skaters push: 50 kg at +3 m/s, 70 kg = ?

  1. Before. Both at rest: total = 0.
  2. After. 50 × 3 + 70 × v = 0.
  3. Solve. v = −150/70 ≈ −2.14 m/s.
  4. Check. Heavier skater, slower recoil — momentum equal and opposite. ✓
Common mistake: “equal and opposite” misread as equal speeds. It's equal momenta: the heavier skater moves slower.
Your turn — 60 kg skater moves at −2.5 m/s; 40 kg skater's velocity?

Answer: +3.75 m/s. 60×(−2.5) + 40v = 0 → v = 150/40 = 3.75 m/s.

Example 3 — stick together: 1500 kg at 20 m/s hits stationary 1000 kg

  1. Before. 1500 × 20 + 1000 × 0 = 30,000.
  2. After. (1500 + 1000) × vf = 2500vf.
  3. Solve. vf = 30,000/2500 = 12 m/s.
  4. Note. Momentum conserved; kinetic energy is NOT (see inelastic collisions).
Common mistake: dividing by 1500 (forgetting the stuck mass adds): v = 20 m/s — as if the second car weren't there.
Your turn — 2000 kg at 15 m/s sticks to stationary 1000 kg. vf = ?

Answer: 10 m/s. 30,000/3000 = 10 m/s.

Example 4 — explosion: 3 kg firework splits; 1 kg flies at +12 m/s

  1. Before. At rest: total = 0.
  2. After. 1 × 12 + 2 × v = 0.
  3. Solve. v = −12/2 = −6 m/s.
  4. Check. The heavier piece recoils slower — momenta ±12 cancel. ✓
Common mistake: “explosions create momentum” — they don't. Chemical energy becomes kinetic, but the vector total stays zero.
Your turn — 4 kg splits: 1 kg at +20 m/s, 3 kg = ?

Answer: −6.67 m/s. v = −20/3 ≈ −6.67 m/s.

Memorization tips

  • Say it aloud: “the total momentum before equals the total momentum after.”
  • System first: name every object in the system before writing anything. Missing objects are missing momentum.
  • Zero is a number: 'at rest' contributes 0 — write the 0. It keeps the bookkeeping honest.
  • Recoil is opposite: zero before → after-totals cancel. If your recoil points the same way, flip it.
  • Heavier = slower: equal momenta means the bigger mass moves slower. Never 'equal and opposite speeds'.
  • Isolation check: quick collisions → external impulses negligible → conserve. Long pushes with friction → don't.

Final challenge

Five mixed questions — recoil, explosions, and sign traps. Score 5/5 and conservation is 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

Why is momentum conserved but kinetic energy often isn't?

Momentum conservation comes from Newton's third law — internal forces cancel pairwise, always. Kinetic energy can become heat, sound, or deformation, which the KE ledger doesn't track. Momentum has nowhere to hide; energy does.

Does conservation work in 2D?

Yes — conserve each component separately: Σpx before = Σpx after, and same for y. The vector equation splits into two scalar ones.

What counts as an 'isolated system'?

One with no net external impulse during the event. Collisions are over in milliseconds, so external forces (friction, gravity) barely contribute — that's why conservation works for crashes but not for long slides.

Can momentum be created in an explosion?

No — the vector total stays what it was (often zero). The explosion converts chemical potential into kinetic energy, but every fragment's momentum is balanced by the others'.

Is momentum conservation more fundamental than Newton's laws?

It's deeper: it follows from the translational symmetry of space (Noether's theorem) and survives in quantum mechanics and relativity where F = ma doesn't. Newton's third law is the classical shadow of it.

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