GCSE Physics resources
AQA GCSE Forces
Use a direction convention for momentum. Conservation concerns a system, while a stopping-time calculation gives an average interaction force.
Subsection scope: Momentum and qualitative conservation: both courses Higher. Collision calculations and force from momentum change: Physics Higher only.
Start here
Use a direction convention for momentum. Conservation concerns a system, while a stopping-time calculation gives an average interaction force.
Before you revise
Choose an answer from memory. Your result tells you what to watch for in the guide.
Core revision guide
Use a direction convention for momentum. Conservation concerns a system, while a stopping-time calculation gives an average interaction force.
Scope: 8463 and 8464 · Higher only
Momentum p = mv is a vector in kg m/s. For a closed system with negligible external resultant, total momentum before equals total momentum after. Opposite velocities have opposite signs. Momentum may be conserved in a collision even when kinetic energy becomes thermal or deformation energy.
Scope: 8463 Physics only · Higher only
For a 2 kg trolley at 3 m/s joining a stationary 1 kg trolley, initial momentum is 6 kg m/s. Final moving mass is 3 kg, so v = 6/3 = 2 m/s in the original direction. Include both bodies when identifying the closed system.
Scope: 8463 Physics only · Higher only
For a fixed change in momentum, Faverage = Δp/Δt. Airbags and crumple zones extend stopping time and lower the average force on the occupant. This equation does not guarantee a lower peak for every force–time shape; peak comparisons require comparable profiles. State the same initial/final velocity and mass when comparing safety designs.
Misconception clinic
Read each belief, say what is wrong with it, then compare your correction.
Tempting idea: +4 kg m/s.
Use instead: Direction makes velocity negative. 0.50 × (−8) = −4.
Tempting idea: Every possible peak force must halve.
Use instead: Peak needs the force–time profile. Average force is Δp/Δt.
Retrieval practice
Explore the relationship
Change momentum or collision time, then calculate average force using F = Δp ÷ Δt.
Model scope: 8463 Physics only · Higher only
What to notice: use the readout to describe how one variable changes when the other is controlled.
Worked examples
Apply momentum and conservation in the worked reasoning, then try the changed-context questions below.
Scope: 8463 and 8464 · Higher only
p = 18 000 kg m/s.
Scope: 8463 Physics only · Higher only
Final velocity = 2.0 m/s in the original direction.
Scope: 8463 Physics only · Higher only
The airbag increases stopping time, so the same change in momentum happens with a smaller average force.
Exam precision
Do this: Define the system and positive direction before balancing momentum. In safety explanations keep momentum change fixed when comparing average stopping forces.
Independent practice
Write an answer before opening the marking guidance. Then edit the exact phrase or step that would gain the next mark.
Scope: 8463 and 8464 · Higher only
Original PhysicsUK mark scheme: Point 1 (1 mark): p = mv. Point 2 (1 mark): p = 0.40 × 6.0. Point 3 (1 mark): p = 2.4 kg m/s to the right. Award each distinct point once. Accept equivalent scientifically correct wording. No half marks.
Model answer: p = mv. p = 0.40 × 6.0. p = 2.4 kg m/s to the right.
Acceptable answers: Expected result: p = 2.4 kg m/s to the right. Accept the unrounded value or a correctly rounded equivalent to two or more significant figures; accept equivalent units if converted correctly. No additional measurement interval is implied.
Marking notes: Method marks remain available after an arithmetic error. Carry an earlier numerical error into a later valid step; award that method point if the unit conversion and physics are correct. A fully correct answer with unit earns all marks unless working is expressly requested.
Scope: 8463 Physics only · Higher only
Original PhysicsUK mark scheme: Point 1 (1 mark): |Faverage| = |Δp|/Δt. Point 2 (1 mark): |Faverage| = 600/0.30. Point 3 (1 mark): |Faverage| = 2000 N. Award each distinct point once. Accept equivalent scientifically correct wording. No half marks.
Model answer: |Faverage| = |Δp|/Δt. |Faverage| = 600/0.30. |Faverage| = 2000 N.
Acceptable answers: Expected result: |Faverage| = 2000 N. Accept the unrounded value or a correctly rounded equivalent to two or more significant figures; accept equivalent units if converted correctly. No additional measurement interval is implied.
Marking notes: Method marks remain available after an arithmetic error. Carry an earlier numerical error into a later valid step; award that method point if the unit conversion and physics are correct. A fully correct answer with unit earns all marks unless working is expressly requested.
Questions pupils ask
It reduces average force for the same momentum change. A peak comparison also requires information about the force-time profile.
Useful next steps