Q: What does A-Level Physics: 6) Collisions & Impulse Guide cover? A: Understanding impulse, momentum conservation, and the energy landscape of elastic and inelastic collisions is essential for top grades in H2 Physics.
TL;DR Collisions questions funnel down to three imperatives: Step 1: Find the impulse - integrate the force-time graph or use J=F⋅Δt=Δp. Step 2: Box the momentum budget - total p before equals total p
the relative speed of approach equals that of separation, the collision is perfectly elastic; otherwise energy bleeds away as heat, sound or deformation.
Concrete example: how to use this page
If two trolleys collide, write total momentum before and after for the two-trolley system. Only bring in kinetic energy after deciding whether the question says the collision is elastic.
Keep reviewing adjacent mechanics chapters (projectiles, circular motion, oscillations) via the H2 Physics notes hub so each new impulse/momentum drill reinforces earlier checkpoints.
Route map: choose the collision method first
This map separates the first equation from the final classification. The common mistake is to test kinetic energy before building the momentum table.
Treating peak force as average force without evidence
"collide", "stick together", "move off together"
What is the closed system and sign convention?
Momentum before equals momentum after
Dropping the negative sign for the body moving left
"perfectly elastic"
Is kinetic energy also conserved?
Momentum equation plus relative-speed or energy equation
Assuming all bouncy-looking collisions are elastic
"inelastic" or "deformation"
What happened to kinetic energy?
Momentum conservation, then compare Ek before and after
Saying energy vanished instead of transferring to deformation, heat, or sound
"explosion" or "separate from rest"
What was the total momentum before separation?
Total momentum after equals the initial total momentum
Forgetting that opposite momenta can cancel even when speeds differ
1 Syllabus snapshot
Section II Mechanics, 6 Collisions lists two content bullets - Impulse and Conservation of momentum and energy - plus five learning outcomes (a-e).
Parents: these learning outcomes are commonly tested in Papers 1 and 2, so it pays to make the impulse-momentum link and momentum-table workflow fluent early.
2 Impulse - the momentum injector
2.1 Definition and units
Impulse J is the product of a force and the time for which it acts, and it equals the change in momentum: J=F⋅Δt=Δp.(1)
The SI unit is N⋅s (numerically identical to kg⋅m⋅s−1).
2.2 Area under the force-time graph
When the force varies, J is the area beneath the F−t curve.
Exam hack: sketch rectangles/triangles under the curve and sum the areas; avoid trapezium-rule mishaps.
Force-time graph area checkpoint
Before calculating impulse from a graph, split the shaded region into simple shapes and check whether the vertical axis is force, not momentum.
Graph section
Area to use
What it means physically
Common trap
Horizontal force plateau
Rectangle: force times time interval
Constant force gives steady impulse accumulation.
Using the peak force without multiplying by time.
Straight rise from zero
Triangle: half base times height
Force builds up during contact.
Treating it as a full rectangle.
Straight sloping section between two non-zero forces
Trapezium or average force times time interval
Force changes linearly during contact.
Forgetting to average the two force values.
Section below the time axis
Negative area if the chosen direction is positive
Impulse is opposite to the positive direction.
Adding the magnitude when the question asks for vector change in momentum.
Worked check: a force-time graph that rises linearly from 0N to 20N over 0.10s, stays at 20N for 0.20s, then falls to zero over 0.10s, has impulse 21(0.10)(20)+(0.20)(20)+21(0.10)(20)=6.0N⋅s.
2.3 Mini-drill
A hockey stick delivers an average 650N over 8.0⋅10−3s.
Calculate J.
If the puck's mass is 170g and it was initially at rest, find its exit speed.
3 Conservation of momentum
3.1 Principle
In an isolated system (no external net force) the vector sum of momenta remains constant: ∑pbefore=∑pafter.(2)
3.2 Worked example - “trolley and projectile”
Take the trolley moving right at 1.2m⋅s−1 and the clay moving left at 12m⋅s−1 (opposite directions). After sticking:
v=2.0+0.060(2.0)(1.2)+(0.060)(−12)≈0.82m⋅s−1.
Momentum is conserved. Kinetic energy is not: it drops from ≈5.76J to ≈0.69J (about an 88% decrease), with the “missing” energy converted into deformation/heat/sound - classic perfectly inelastic behaviour.
Momentum table sign checkpoint
Before solving a collision, build a signed momentum table. This prevents the most common algebra error: treating opposite directions as if they were both positive.
Table row
What to write
Why it matters
Common trap
Sign convention
Choose rightwards or the initial motion of one body as positive.
Every velocity must follow one shared direction rule.
Changing the positive direction halfway through the calculation.
Before collision
List each mass and signed initial velocity.
The total before momentum is ∑mu.
Dropping the minus sign for the body moving opposite to the chosen positive direction.
After collision
List each signed final velocity, using one symbol for the unknown.
The total after momentum is ∑mv.
Writing a shared final speed when the question does not say the bodies stick together.
Equation line
Set total before momentum equal to total after momentum.
Momentum conservation applies to the whole closed system.
Equating momentum for each object separately during a collision.
Worked check: if A of mass 0.50kg moves right at 3.0m⋅s−1 and B of mass 0.30kg moves left at 2.0m⋅s−1, choosing right as positive gives total momentum
(0.50)(3.0)+(0.30)(−2.0)=0.90kg⋅m⋅s−1.
If they stick together after collision, (0.50+0.30)v=0.90, so v=1.125m⋅s−1 to the right.
Misconception check: momentum is a vector. A smaller mass moving left can reduce, cancel, or reverse the total momentum depending on its speed.
4 Elastic versus inelastic collisions
Property
Perfectly elastic
Inelastic / perfectly inelastic
Momentum p
Conserved
Conserved
Kinetic energy Ek
Conserved
Decreases
Relative speed
vapproach=vseparation
vseparation<vapproach
The speed criterion stems from combining Eq. (2) with Ek conservation; its proof is examinable.
4.1 Quick test
Two identical steel balls collide head-on: one is stationary, the other approaches at 5.0m⋅s−1.
Elastic → the incident ball stops and the target departs at 5.0m⋅s−1.
Inelastic → both share speed 2.5m⋅s−1 (same p, less Ek).
5 Energy housekeeping
Momentum is always conserved for a closed system, but Ek usually leaks into deformation, sound or heat. Car-crash crumple zones lengthen Δt, reducing peak force via Eq. (1) while sacrificing kinetic energy irreversibly.
6 WA timing rules (Collisions flavour)
Label before/after clearly - one row per body, two columns (px, optionally py).
Check the extras - for elastic cases, write a second equation equating Ek or use the relative-speed shortcut.
7 Parent corner - why impulse matters in practicals
Paper 4 often supplies a force-time print-out from a data logger. Students must:
Count squares to find impulse.
Divide by mass to obtain Δv.
Compare predicted range against measured.
Missing the area-under-curve idea can cost marks. A short home drill on estimating areas (rectangles/triangles/trapezia) pays dividends.
Need structured practice on Collisions? Our H2 Physics tuition programme covers this topic with weekly problem sets and Paper 4 practical drills.
Comprehensive revision pack
9478 Section II, Topic 6 Syllabus outcomes
Candidates should be able to:
(a) recall that impulse is given by the area under the force-time graph for a body and use this to solve problems.
(b) state the principle of conservation of momentum.
(c) apply the principle of conservation of momentum to solve simple problems including inelastic and (perfectly) elastic interactions between two bodies in one dimension (knowledge of the concept of coefficient of restitution is not required).
(d) show an understanding that, for a (perfectly) elastic collision between two bodies, the relative speed of approach is equal to the relative speed of separation.
(e) show an understanding that, whilst the momentum of a closed system is always conserved in interactions between bodies, some change in kinetic energy usually takes place.
Concept map (in words)
Collisions live at the intersection of momentum (vector, conserved) and energy (scalar, sometimes conserved). Start by drawing system boundaries, then decide whether forces are impulsive (short duration, large magnitude) or continuous. Link impulse to the area under an F−t graph, use conservation of momentum to relate pre- and post-impact velocities, and finally classify the collision by testing kinetic energy or the coefficient of restitution. For 2D scenarios, resolve components and use geometry (e.g., right-angle scattering).
Key definitions & formulae
Quantity / relation
Expression / meaning
Units
Impulse
J=Favg⋅Δt=Δp
N⋅s
Linear momentum
p=mv (vector)
kg⋅m⋅s−1
Coefficient of restitution
e=vapproachvseparation
Perfectly elastic criterion
vseparation=vapproach; Ek
Kinetic energy comparison
Compare ∑21mv2 before vs after collision
J
Impulse from graph
Area under F-t graph (count rectangles or trapezia)
-
Centre-of-mass velocity of pair
vcm=∑m∑mv
Derivations & reasoning you must know
Impulse-momentum theorem: integrate Newton's 2nd law F=dtdp over the collision window.
Relative speed form of e: combine momentum and kinetic-energy conservation to show e=1 for perfectly elastic impacts, then derive e=vapproachvseparation.
Ballistic pendulum split-phase analysis: treat the collision stage with momentum conservation and the swing stage with mechanical energy conservation.
Oblique collision on a smooth wall: resolve normal and tangential components; only the normal component flips and scales by −e.
Worked example 1 - coefficient of restitution
Two trolleys A(0.80kg) and B(1.20kg) approach each other on a smooth track with speeds 1.5m⋅s−1 and 0.9m⋅s−1 respectively. After impact, A rebounds at 0.30m⋅s−1.
A smooth proton of mass m travelling at 3.0×106m⋅s−1 strikes an identical stationary proton. After collision, proton A deflects at 30∘ above the original line. Find proton B's velocity vector, assuming an elastic collision.
Approach: for an elastic collision of identical masses where one is initially at rest, the two outgoing velocity vectors are perpendicular. So if proton A is at 30∘, proton B is at 60∘ below the original line.
Hence vB has magnitude 1.5×106m⋅s−1 at −60∘ (components: vBx=0.75×106m⋅s−1, vBy=−1.30×106m⋅s−1).
Practical & data tasks to rehearse
Use motion sensors or high-frame-rate video to capture F−t profiles for cart collisions; integrate numerically to verify impulse values.
Investigate crumple zones by adding foam buffers and measuring peak force reduction for the same momentum change.
Carry out a ballistic pendulum experiment; separate the collision maths from the pendulum energy conversion and document uncertainties.
Common misconceptions and exam traps
Treating momentum as scalar and dropping direction signs.
Assuming energy is conserved for every collision - only momentum is guaranteed without external forces.
Forgetting that e is defined using speeds along the line of impact, not arbitrary velocity components.
Mixing up internal and external impulses (e.g., mistaking normal reaction for external force when dealing with colliding gliders).
Quick self-check quiz
During an inelastic collision, which quantity must remain constant for the system? - Total linear momentum.
What does the area under a force-time graph quantify? - Impulse (change in momentum).
If e=0, what type of collision has occurred? - Perfectly inelastic; bodies coalesce.
Why do airbags reduce injury in crashes? - They increase collision duration, lowering peak force for the same momentum change.
Two equal masses collide elastically head-on. What happens to their velocities? - They exchange velocities; one stops while the other takes the incident speed.
Revision workflow
Re-derive e=vapproachvseparation for elastic collisions starting from conservation laws.
Complete at least two SEAB Paper 2 questions featuring impulse graphs; practise estimating area quickly.
Summarise typical collision archetypes (perfectly elastic, partially inelastic, perfectly inelastic, explosions) in a single-page storyboard.
Pair up with a classmate to quiz each other on step-by-step momentum tables - speed comes from repetition.
Practice Quiz
Test yourself on the key concepts from this guide.
Parents: schedule a hands-on “collision cart” demo - cheap tracks are available for home practice.
Students: memorise Eq. (1), Eq. (2) and the speed criterion; they compress whole MCQs into three lines of working.
Last updated 14 Jul 2025. Next review on the 2027 syllabus draft release.
