Impulse
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
A force acting over time changes an object’s momentum. That change is the impulse: J = FavgΔt = Δp. On a force–time graph, impulse is the area under the graph.
1. Same impulse, different forces
2. Bouncing doubles the change
Airbags, helmets, and bending your knees on landing all work the same way: they make Δt longer, so the force needed for the same Δp is smaller.
A ball that bounces back has a bigger momentum change than one that just stops, because its velocity reverses sign. Stopping takes away all its momentum; bouncing takes it away and gives it momentum the other way.
3. Example, then try it
A 0.50 kg ball hits a wall at 6.0 m/s. Find Δp if it (a) stops, (b) bounces straight back at 6.0 m/s.
Toward the wall = +. (a) Δp = 0.50(0 − 6.0) = −3.0 kg·m/s.
(b) Δp = 0.50(−6.0 − 6.0) = −6.0 kg·m/s, twice as much.
A 70 kg passenger moving at 15 m/s is stopped in a crash. Find the average force on her if she is stopped (a) by an airbag in 0.15 s, (b) by the dashboard in 0.010 s.
Momentum change: Δp = 70(0 − 15) = −1050 kg·m/s. Same for both.
(a) Airbag: Favg = Δp ÷ Δt = 1050 ÷ 0.15 = 7000 N.
(b) Dashboard: Favg = 1050 ÷ 0.010 = 105,000 N.
Answer: 7000 N with the airbag versus 105,000 N against the dashboard: 15 times less force for the same impulse.
4. Common mistakes
The same egg is dropped from the same height onto concrete and onto a pillow. Both times it stops. How do the impulses on the egg compare?
The egg’s momentum changes by the same amount either way, so the impulse is the same. The pillow spreads it over a longer time, so the force is smaller, and the egg survives.
Ready to practice?
5 questions on Impulse, each with a hint and an explanation for every wrong answer.