The big idea: Momentum measures how hard something is to stop: mass × velocity.
A heavy or fast object has lots of momentum.
To change an object's momentum you push on it. The impulse is that push — the force × the time it acts for. A bigger impulse makes a bigger change in momentum.
Momentum (the state)
- p = mv — how much motion an object has
- unit: kg m s⁻¹
- a vector — it has a direction
Impulse (the change)
- J = FΔt — the push that changes the motion
- unit: N s (the same as kg m s⁻¹)
- equals the change in momentum, Δp
Spot it: Momentum = how much motion you have (mv). Impulse = the push that changes it (FΔt).
Key link: impulse = change in momentum, so FΔt = Δp.
The data booklet gives all three equations you need. Start with momentum:
- momentum (kg m s⁻¹)
- mass of the object (kg)
- velocity (m s⁻¹)
[Diagram: phys-formula-triangle] - Available in full study mode
- impulse — equals the change in momentum (N s, same as kg m s⁻¹)
- average force acting (N)
- time the force acts for (s)
The two are linked: Newton's second law (also given) connects them:
F = ma = Δp/Δt — the average force equals the change in momentum ÷ the time.
Multiply both sides by Δt and you get FΔt = Δp: impulse = change in momentum.
- average force (N)
- change in momentum (kg m s⁻¹)
- time the force acts for (s)
- mass (kg)
- acceleration (m s⁻²)
Worked example — impulse changes the momentum
A 0.50 kg ball is at rest. A bat pushes it with an average force of 40 N for 0.15 s. Find the ball's speed just after the hit.
Solution
- Start with the given impulse equation:
- Put in the numbers (F = 40, Δt = 0.15):
- Impulse equals the change in momentum, and it starts from rest (Δp = mv − 0):
- Solve for the speed — keep the unit:
Final answer
v = 12 m s⁻¹ — the impulse of 6.0 N s gave the ball 6.0 kg m s⁻¹ of momentum.
Stop wasting time on topics you know
Our AI identifies your weak areas and focuses your study time where it matters. No more overstudying easy topics.
How this is tested: Impulse questions almost always ask for the average force or the resulting motion.
- Paper 1A: a quick calculation — the area under a force–time graph is the impulse, or FΔt = Δp. - Paper 2: the average force in a collision (e.g. a ball bouncing off a wall), or the kinetic energy an object gains after an impulse.
Classic trap: momentum is a vector. A ball that bounces back reverses direction, so its change in momentum is m(v − (−u)) = m(v + u) — bigger than you'd expect, not v − u.
Bounce-back: mind the signs: Call the outgoing direction positive. A ball arriving at speed u has momentum −mu; leaving at speed v has momentum +mv.
The change is Δp = mv − (−mu) = m(v + u) — you add the speeds because the direction flips.
| Situation | Change in momentum Δp | Why |
|---|---|---|
| Stops dead (v = 0) | Δp = mu | loses all its momentum |
| Speeds up u → v (same way) | Δp = m(v − u) | subtract — same direction |
| Bounces back u → v (reversed) | Δp = m(v + u) | add — direction flips sign |
IB-style question — average force on a bouncing ball
A 0.20 kg ball hits a wall horizontally at 8.0 m s⁻¹ and bounces straight back at 6.0 m s⁻¹. The contact lasts 0.040 s. Find the average force the wall exerts on the ball.
Solution
- Take the rebound direction as positive. The ball comes in at −8.0 and leaves at +6.0, so use:
- Put in the numbers (m = 0.20, v = 6.0, u = 8.0):
- Average force comes from the given Newton's-second-law form:
- Put in Δp = 2.8 and Δt = 0.040:
Final answer
average force = 70 N. Note the speeds were ADDED (the ball reversed direction) — using 8.0 − 6.0 would give the wrong answer.