IB Physics SL Topic A.2 — Forces & Momentum Paper 1 & 2 Core equation ~7 min read

Impulse & Momentum

When a force acts on an object for a short time — a kick, a catch, a collision — it produces an impulse, and that impulse equals the object’s change in momentum. The same impulse can come from a big force acting briefly or a small force acting for longer, which is the secret behind seatbelts, crumple zones and a good cricket catch.

📘 What you need to know

What impulse is

An impulse is what you get when a resultant force acts for a short interval of time:

Impulse J = FΔt

For very short impacts it’s hard to measure the force and the contact time directly, so we measure impulse indirectly through the change in momentum it causes. Starting from Newton’s second law, F = Δp / Δt, rearranging gives Δp = FΔt — so the impulse is exactly the change in momentum:

Impulse = change in momentum J = Δp = mvmu

where u is the initial velocity and v the final velocity. (These apply when the force F is constant.) Following the units confirms it: since 1 N = 1 kg m s⁻², an impulse in N s is the same as a momentum in kg m s⁻¹.

Impulse as the area under a force–time graph

Because impulse is force × time, it is the area under a force–time graph. For a constant force that’s just a rectangle, F × Δt.

F t F impulse = F Δt Δt
The impulse is the area under the force–time graph — here simply F × Δt for a constant force.

Why a longer impact means a smaller force

Rearranging F = Δp / Δt shows that for a fixed change in momentum, a longer contact time means a smaller force. That’s exactly what a cricket fielder does when catching a fast ball — they draw their hands back to stretch out the impact time and reduce the force on their hands. The same idea protects you with airbags and crumple zones.

F t hard catch large F, short t soft catch (hands move back) small F, long t
Both impacts have the same area — the same impulse and change in momentum — but the longer, softer one needs a far smaller peak force.
Quick reference: J = FΔt = Δp = mvmu • impulse = area under an F–t graph • 1 N s = 1 kg m s⁻¹ • longer time → smaller force.

Worked examples

WE 1

A tennis ball struck back

A 58 g tennis ball moving left at 30 m s⁻¹ is hit by a racket and returns to the right at 20 m s⁻¹. (a) Find the impulse on the ball. (b) State its direction.

Take the initial motion (left) as positive u = +30, v = −20, m = 0.058 kg (a) J = m(v − u) = 0.058 × (−20 − 30) = 0.058 × (−50) J = −2.9 N s (b) Negative → opposite to the initial motion impulse is to the right
WE 2

From force and time to speed

A constant 150 N force acts on a 5.0 kg trolley, initially at rest, for 0.40 s. Find the impulse on the trolley and its final speed.

Impulse: J = F Δt = 150 × 0.40 = 60 N s This equals the change in momentum: Δp = mv − mu 60 = 5.0 × v − 0 v = 12 m s⁻¹

💡 Top tips

⚠ Common mistakes

Impulse and the force–momentum link are two views of the same equation. This page worked with J = Δp; the next one rearranges it to F = Δp / Δt to find the average force in an impact. Up next: Force & Momentum.

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