Impulse and force–time area
Key idea: H2 Physics lessons on force–time impulse, one-dimensional momentum conservation, elastic interactions and kinetic-energy change.
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The core idea
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Big question: Why does the area of a force–time graph measure a change in motion?
Impulse is the time integral of resultant force and equals change in momentum. For a constant force J = FΔt; for a changing force, signed area under the F–t graph gives J. A longer collision time can reduce peak force for the same momentum change, which explains crumple zones and follow-through.
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.
Key ideas to keep
- Use resultant force when linking impulse to one object's momentum change.
- Area below the time axis gives negative impulse.
- Equal impulse does not require equal peak force or collision time.
See 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.
Another worked model
Question
A force rises linearly from 0 to 400 N in 0.020 s, stays at 400 N until 0.060 s, then falls linearly to zero at 0.080 s. It acts on a 0.40 kg ball initially moving at −30 m s⁻¹. Find its final velocity.
Check the worked solution
Impulse is the two triangles plus rectangle: ½(0.020)(400) + (0.040)(400) + ½(0.020)(400) = 24 N s. Initial momentum is 0.40(−30) = −12 kg m s⁻¹, so final momentum is −12 + 24 = +12 kg m s⁻¹ and v = +30 m s⁻¹.
Use a hint if needed
Practise with support
Try this
A triangular force pulse below the time axis has base 0.050 s and magnitude 240 N. A 0.30 kg object initially moves at +12 m s⁻¹. Find its final velocity.
Hint: The graph area is negative for the declared positive direction.
Check your answer
J = −½(0.050)(240) = −6.0 N s. Initial momentum = 0.30(12) = 3.6 kg m s⁻¹, so final momentum = −2.4 kg m s⁻¹ and v = −8.0 m s⁻¹.
Now work without the hint
Practise independently
Your turn
A force–time graph has +12 N s of area above the axis followed by 5.0 N s below it. Find the net impulse and the momentum change.
Check your answer
Net signed area is +12 − 5.0 = +7.0 N s. Therefore the body’s momentum change is +7.0 kg m s⁻¹ in the declared positive direction.
Avoid these traps
Common mistakes
Common mistake
Impulse is the maximum force shown on a force–time graph.
What is wrong with this reasoning?
Show better thinking
Impulse is the signed area between the force curve and the time axis. Its unit is N s and it equals Δp; graph height alone is force.
Common mistake
Area below the force–time axis contributes a positive impulse.
What is wrong with this reasoning?
Show better thinking
Choose a positive force direction first. Area below the time axis is negative and must be combined algebraically with positive area.
Write for the examiner
Exam guidance
Read the graph scale carefully and state a direction before assigning the sign of impulse.
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.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
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.
Come back in three days
Check what stayed with you
Recall question 1
What graph quantity gives impulse?
Check the answer
The signed area under a force–time graph.
Recall question 2
Give two equivalent impulse units.
Check the answer
N s and kg m s⁻¹.
Recall question 3
For fixed Δp, how can average force be reduced?
Check the answer
Increase the interaction time.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. The coefficient of restitution is not required. Use momentum conservation, the perfectly elastic relative-speed condition and a direct comparison of total kinetic energy.
- GCE A-Level H2 PhysicsTopic 6(a) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027