Impulse and force–time area

Key idea: Impulse measures the accumulated effect of force over time. On a force–time graph it is the signed area, and it equals the change in momentum.

GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 1 of 3

Check your understanding

By the end of this lesson, you should be able to

Find impulse from rectangular, triangular or composite force–time areas.
Use sign to connect impulse with momentum direction.
Explain how increasing collision time can reduce force for the same momentum change.

Learn the idea

Big question: How does the area under a force–time graph tell the complete story of a changing collision force?

Treat impulse as signed area

Impulse is the accumulated effect of force over time. For a changing force, it is the signed area under the force–time graph—not the peak force multiplied by the full duration.

Newton's second law F = dp/dt leads to impulse = Δp. Keep one positive direction throughout, because a negative area produces a negative momentum change and may reverse the motion.

Check your understanding: Can a large peak force give a small impulse?

Yes. If it acts for a very short time, the area under the graph can still be small.

Use time to manage force

For the same momentum change, average force equals Δp/Δt. Increasing collision time reduces average force, which is why airbags, crumple zones and bending the knees can reduce injury.

The forces on the two interacting bodies form a third-law pair, so their impulses are equal and opposite. Their momentum changes are therefore equal and opposite too.

Check your understanding: Does an airbag reduce the passenger's momentum change from moving to rest?

No. It increases the stopping time, reducing average force for the same momentum change.

Scroll diagram horizontally to read all labels.

For the same momentum change, increasing the contact time reduces the average and peak force; the force–time area remains equal.

Key ideas

Peak force is not impulse.
N s and kg m s⁻¹ are equivalent units.
Use final momentum minus initial momentum, with signs.

Relationships to know

impulse J = area under F–t graph
J = Δp = pf − pi
for constant force, J = FΔt

Follow the reasoning

Worked example

Read a force pulse and find rebound velocity

Question: A 0.20 kg ball initially moves at +12 m s⁻¹. A force opposite its motion rises linearly to 300 N in 0.010 s and falls linearly to zero over the next 0.020 s. Find its final velocity.

Step 1: Find signed impulse
Why: The two triangular parts make one triangle of total base 0.030 s.
Working: J = −½(300)(0.030) = −4.50 N s.
Step 2: Use impulse–momentum
Why: Impulse changes signed momentum, not speed directly.
Working: pi = 0.20(12) = +2.40 kg m s⁻¹; pf = pi + J = −2.10 kg m s⁻¹.
Step 3: Recover velocity
Why: The negative momentum shows a rebound.
Working: v = pf/m = −2.10/0.20 = −10.5 m s⁻¹.

Answer: The ball rebounds at 10.5 m s⁻¹ in the negative direction.

Check: The impulse magnitude exceeds the initial momentum, so reversal is expected.

Now try it with support

Practise with support

A constant force of −180 N acts for 0.050 s on a 0.30 kg ball initially moving at +20 m s⁻¹. Find its final velocity.

Hints

First find signed impulse.
Use pf = pi + J.

View the guided answer

J = −180(0.050) = −9.0 N s. pi = 0.30(20) = +6.0 kg m s⁻¹, so pf = −3.0 kg m s⁻¹ and vf = −10 m s⁻¹.

Your turn

Practise independently

A force rises uniformly from 0 to 400 N in 0.010 s, remains constant for 0.015 s, then falls to zero in 0.005 s. Find the impulse.

Check your answer

Impulse = rising triangle + rectangle + falling triangle = ½(400)(0.010) + 400(0.015) + ½(400)(0.005) = 2.0 + 6.0 + 1.0 = 9.0 N s.

Common mistakes and exam guidance

Watch out for

Multiplying peak force by the entire time for a non-rectangular graph.
Using momentum magnitudes and losing the direction of the change.

In an exam

Shade or split the graph area before calculating.
Carry the sign from force through impulse to momentum change.

Put the ideas together

Exam-style practice [6 marks]

A 70 kg passenger moving at 18 m s⁻¹ is brought to rest. With a rigid restraint this takes 0.060 s; with an airbag it takes 0.18 s. Find the impulse magnitude and average force in each case, then explain the safety benefit.

Plan before you answer

The momentum change is the same in both cases.
Divide by each stopping time.
Link lower force to longer time, not smaller impulse.

View the marking points and model answer

Marking points

Finds impulse magnitude 1260 N s.
Finds rigid-restraint average force 2.10 × 10⁴ N.
Finds airbag average force 7.00 × 10³ N.
States same momentum change/impulse.
States airbag triples stopping time.
Explains that lower average force reduces injury risk.

Model answer

The impulse magnitude is Δp = 70(18) = 1260 N s in both cases. Rigid-restraint force = 1260/0.060 = 2.10 × 10⁴ N. Airbag force = 1260/0.18 = 7.00 × 10³ N. The airbag gives the same stopping impulse over three times as long, so the average force is one third as large.

Finish from memory

Three-question recap

What graph quantity gives impulse?
Check
The signed area under a force–time graph.
Give two equivalent impulse units.
Check
N s and kg m s⁻¹.
For fixed Δp, how can average force be reduced?
Check
Increase the interaction time.

Try this next

Use impulse within a two-body collision, then apply momentum conservation to the complete system.

Continue to Momentum conservation and elastic interactions Return to Collisions

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H1 Physics
Edition: GCE A-Level H1 Physics 2027