Physics I: Mechanics › Momentum › full formula sheet

J = FΔt = Δp

Say it: “impulse equals force times time, which equals the change in momentum”

Impulse

Why airbags save lives and boxers roll with punches — the same momentum change, stretched over more time, means less force.

J is impulse in N·s (= kg·m/s), F the (average) force, Δt the contact time, Δp = pf − pi the momentum change. Signs follow your direction convention.

Before this lesson: Linear momentum

Where it comes from

An airbag and a dashboard stop the same driver from the same speed. One kills, the other saves — with identical momentum change. The difference is time: the airbag stretches the stop over 0.4 s instead of 0.02 s, so the force drops twenty-fold.

Before reading on: stopping a 60 kg person from 10 m/s takes the same momentum change either way. If the airbag doubles the stopping time, what happens to the average force — halves, quarters, or stays the same?

It halves. Same Δp, twice the Δt — and since F = Δp/Δt, the force must halve. This trade — more time, less force, same momentum change — is the whole subject:

J = FΔt = Δpforce times time is the currency that buys momentum changeSay it: “impulse equals force times time equals delta p”

Derivation

From F = dp/dt, multiply through by dt:

F
=
dp/dt
Step 1 — Newton's law in momentum form (see linear momentum).
F dt
=
dp
Step 2 — multiply by dt. Force through an instant of time equals the momentum change in that instant.
∫ F dt
=
Δp
Step 3 — add it up. Integrate over the collision: total impulse = total momentum change. For constant (or average) force: J = FΔt = Δp. ∎

The graph picture: on an F-vs-t graph, impulse is the area under the curve — just as work is the area under F-vs-x. Same area, same impulse, whether the force is a tall spike or a low hump.

How to use it

The procedure, every time:

  1. Fix the positive direction. Δp = pf − pi with signs — a bounce from −8 to +10 m/s is a bigger change than stopping from +8.
  2. Compute Δp first. It's usually the easy side: m(vf − vi).
  3. Set J = Δp. Then J = FΔt (or FavgΔt) gives whichever of F, Δt you need.
  4. Read the trade. Fixed Δp: longer Δt → smaller F. Safety equipment is all about stretching time.
  5. Graphs: take the area. F-vs-t area = impulse. Triangles: ½ × base × height.
Common mistake: computing Δp as pi − pf (backwards). Delta always means final minus initial — a bounce from −8 to +10 is Δv = +18, not −18.

Worked examples

Four problems, easiest first. Compute Δp first, always.

Example 1 — basic: F = 10 N for Δt = 0.5 s

  1. Multiply. J = 10 × 0.5 = 5 N·s.
  2. Meaning. This force delivers 5 kg·m/s of momentum change — e.g. a 1 kg mass gains 5 m/s.
Common mistake: J = 10/0.5 = 20 — dividing instead of multiplying. Impulse is force times time: longer push, more impulse.
Your turn — F = 20 N for 0.3 s. J = ?

Answer: 6 N·s. 20 × 0.3 = 6 N·s.

Example 2 — a bounce: 0.15 kg ball, vi = −8 m/s, vf = +10 m/s

  1. Momentum change. Δp = m(vf − vi) = 0.15 × (10 − (−8)) = 0.15 × 18 = 2.7 N·s.
  2. Note. Bouncing reverses direction, so Δv = 18 m/s — bigger than just stopping (8 m/s). Bounces hit harder than sticks.
Common mistake: Δv = 10 − 8 = 2 (dropping the sign on vi). Final minus initial with signs: 10 − (−8) = 18.
Your turn — 0.2 kg ball, −5 m/s → +3 m/s. Δp = ?

Answer: 1.6 N·s. 0.2 × (3 − (−5)) = 0.2 × 8 = 1.6 N·s.

Example 3 — the airbag: 60 kg person at 10 m/s, stopped in Δt = 0.4 s

  1. Momentum to kill. Δp = 0 − 60×10 = −600 N·s (magnitude 600).
  2. Average force. Favg = Δp/Δt = −600/0.4 = −1500 N.
  3. Compare. Against a dashboard (Δt ≈ 0.02 s): F ≈ −30,000 N — twenty times worse. Time is the lifesaver.
Common mistake: F = 600 × 0.4 = 240 N — multiplying instead of dividing. Longer time reduces force: F = Δp/Δt.
Your turn — 80 kg at 8 m/s stopped in 0.2 s. Favg = ?

Answer: −3200 N (3200 N opposing motion). Δp = −640; F = −640/0.2 = −3200 N.

Example 4 — graph area: force spikes 0 → 40 N over 0.2 s (triangle)

  1. Read the graph. F-vs-t is a triangle: base 0.2 s, height 40 N.
  2. Area = impulse. J = ½ × 0.2 × 40 = 4 N·s.
  3. Equivalent. Same as a constant 20 N for 0.2 s — the average force is half the peak.
Common mistake: using the peak force (40 N) as if it acted the whole time: J = 8 N·s. The triangle's area has the ½ — only the average counts.
Your turn — triangular spike 0 → 60 N over 0.1 s. J = ?

Answer: 3 N·s. ½ × 0.1 × 60 = 3 N·s.

Memorization tips

  • Say it aloud: “impulse is force times time equals delta p.”
  • Δp first: m(vf − vi) with signs is usually the easy side — start there.
  • Final minus initial: the delta direction. Backwards deltas are the #1 impulse error.
  • The safety trade: same Δp, more time, less force. Airbags, crumple zones, boxing gloves, bending knees — all one idea.
  • Area under F-vs-t: the graph picture. Triangle → ½bh; rectangle → bh.
  • Bounces hit harder: reversing direction needs ~2× the impulse of just stopping. That's why balls bounce and eggs don't.

Final challenge

Five mixed questions — deltas, graphs, and the safety trade. Score 5/5 and impulse 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

What's the difference between impulse and momentum?

Momentum (p = mv) is the state — how much motion an object has. Impulse (J = FΔt) is the transfer — how much momentum a force delivers. J = Δp links them: impulse is the change-maker.

Why do airbags work?

Same Δp (you must stop), longer Δt (0.4 s vs 0.02 s) → F = Δp/Δt drops ~20×. The airbag doesn't change the momentum change — it stretches it.

Is impulse a vector?

Yes — J = FΔt inherits force's direction, and Δp is a vector difference. In 1D the sign carries the direction; keep it.

How is impulse related to work?

Both involve force, but differently: impulse is F×time (changes momentum), work is F×distance (changes kinetic energy). A long slow push and a short hard shove can give the same impulse with different work.

What if the force varies during the collision?

Use the average: J = FavgΔt, or the F-vs-t graph's area. The theorem J = Δp holds regardless — only the F computation changes.

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