The big idea: Catch a fast cricket ball and it stings — unless you pull your hands back as you catch, spreading the stop over more time so the force is gentler.
That's momentum (how much motion something has, mass × velocity) and impulse (the push that changes it, force × time). Same change in momentum, more time → less force.
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.
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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)
Momentum p = m × v (m = mass, v = velocity). Cover the one you want — two side by side → multiply; one above the other → divide.
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- 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)
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.
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How this is tested — impulse questions almost always ask for the average force or the resulting motion. Two flavours:
Paper 1A
- The area under a force–time graph is the impulse.
- Quick FΔt = Δp calculation.
Paper 2
- The average force in a collision (a ball bouncing off a wall).
- The kinetic energy gained after an impulse.
The classic trap: Momentum is a vector. A ball that bounces back reverses direction, so Δp = m(v − (−u)) = m(v + u) — you add the speeds, not subtract.
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 |
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.
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