Revision Notes
AQA GCSE Physics 8463: Forces – Revision Notes
Condensed AQA GCSE Physics 8463 Forces notes: every equation with units, method steps, graph rules, must-know contrasts and a 12-question self-test.
- Subject
- Physics
- Level
- GCSE
- Topic
- Forces
- Author
- Marlbridge Academic Team
- Updated
- Reviewed by
- Iftikhar Azeemi (what this means)
Aligned to AQA GCSE Physics (8463), For first teaching 2016. Official specification .
Syllabus page (what it covers and how it is assessed): AQA GCSE Physics.
Syllabus points this page covers
8463
- 4.5.1 Forces and their interactions
- 4.5.2 Work done and energy transfer
- 4.5.3 Forces and elasticity
- 4.5.4 Moments, levers and gears
- 4.5.5 Pressure and pressure differences in fluids
- 4.5.6 Forces and motion
- 4.5.7 Momentum
- 5 Forces (whole topic)
Found an error? Report a correction.
Need help with this topic? Request a free trial class for GCSE Physics (8463).
These notes condense section 4.5 Forces of the AQA GCSE Physics (8463) specification, Version 1.1 (30 September 2019), for teaching from September 2016 and exams from 2018 onwards. They cover 4.5.1 to 4.5.7. Forces is examined on Paper 2, at Foundation and Higher Tier; (HT only) content is marked Higher tier only. For full explanations and worked examples, read the Forces study guide first.
Then test yourself with the Forces practice questions. The course hub is AQA GCSE Physics, the printable checklist lists every point, and the free diagnostics show where your gaps are.
Equations
| Equation | Symbols and units | Recall or given? |
|---|---|---|
| W = mg | weight N, mass kg, g N/kg | Recall |
| W = Fs | work J, force N, distance m | Recall |
| F = ke | force N, spring constant N/m, extension m | Recall |
| Eₑ = ½ke² | elastic potential energy J | Equation sheet |
| M = Fd | moment N m, perpendicular distance m | Recall |
| p = F/A | pressure Pa, area m² | Recall |
| p = hρg | height m, density kg/m³ (Higher tier only) | Equation sheet |
| s = vt | distance m, speed m/s, time s | Recall |
| a = Δv/t | acceleration m/s² | Recall |
| v² − u² = 2as | final and initial velocity m/s | Equation sheet |
| F = ma | resultant force N | Recall |
| p = mv | momentum kg m/s (Higher tier only) | Recall |
| F = mΔv/Δt | force = rate of change of momentum (Higher tier only) | Equation sheet |
The value of g is always given in a calculation. Free-fall acceleration near the Earth is about 9.8 m/s².
4.5.1 Forces and their interactions
- Scalar: magnitude only. Vector: magnitude and direction; shown by an arrow (length = size).
- Contact forces: friction, air resistance, tension, normal contact force.
- Non-contact forces: gravitational, electrostatic, magnetic.
- Weight acts at the centre of mass; weight ∝ mass; measured with a newtonmeter.
- Resultant of forces in a line: add one way, subtract the other.
- Higher tier only: free body diagrams; resolving a force into two perpendicular components; scale drawings for resultants and equilibrium (balanced forces make a closed triangle).
4.5.2 Work done
- Work is done when a force causes a displacement. 1 J = 1 N m.
- Work done transfers energy. Work against friction raises the temperature.
4.5.3 Forces and elasticity
- Changing shape of a stationary object needs more than one force: with only one force it would just accelerate. A hanging spring is stretched by the load pulling down and the clamp pulling up; a ruler is bent by a push in the middle and supports at each end.
- Elastic deformation: returns to original shape. Inelastic: does not.
- Linear region: e ∝ F, up to the limit of proportionality. Then non-linear.
- Work done stretching = elastic potential energy stored (if not inelastically deformed).
Method: spring constant from data (required practical 6)
Step 1 extension = stretched length − original length
Step 2 convert cm or mm to m
Step 3 in the linear region, k = F / e (or the gradient of F against e)
Reminder: 2.0 N gives 0.050 m → k = 2.0 / 0.050 = 40 N/m.
4.5.4 Moments, levers and gears
- Moment = force × perpendicular distance from the pivot.
- Balanced: total clockwise moment = total anticlockwise moment.
- Levers and gears transmit the rotational effect of forces. Small gear driving a large one: bigger moment, slower turning.
Method: balanced beam
Step 1 list every force with its distance from the pivot
Step 2 clockwise moments on one side, anticlockwise on the other
Step 3 solve for the unknown force or distance
4.5.5 Pressure in fluids
- Fluid = liquid or gas. Pressure acts normal to surfaces.
- Higher tier only: pressure rises with depth and density (p = hρg); Δp = Δh × ρ × g. Upthrust = more pressure on the bottom surface than the top. Floating and sinking depend on weight, upthrust and density.
- Atmosphere: thin layer, less dense with height. Less air above a surface at greater height, so lower atmospheric pressure.
4.5.6 Forces and motion
Motion
- Distance and speed: scalars. Displacement and velocity: vectors.
- Typical speeds: walking ~1.5 m/s, running ~3 m/s, cycling ~6 m/s, sound in air ~330 m/s. Transport: car ~13 m/s in town and ~30 m/s on a motorway, train ~50 m/s, passenger jet ~250 m/s.
- Everyday accelerations: a car pulling away ~2 m/s²; free fall ~10 m/s².
- Higher tier only: circular motion at constant speed has changing velocity.
Graphs
| Graph | Gradient | Area under line |
|---|---|---|
| Distance–time | Speed (Higher tier only: tangent for a curve) | Not used |
| Velocity–time | Acceleration | Distance travelled (Higher tier only) |
- Horizontal d–t line: stationary. Horizontal v–t line: constant velocity.
- Terminal velocity on a v–t graph: curve that levels off. Weight = drag; resultant force zero.
Method: v² − u² = 2as
Step 1 list u, v, a, s; mark the unknown
Step 2 square both velocities before subtracting
Step 3 rearrange; a negative a means deceleration
Newton’s laws
- First: zero resultant force → no change in velocity. Higher tier only: inertia is the tendency to stay at rest or in uniform motion.
- Second: a ∝ F and a ∝ 1/m; F = ma. Higher tier only: inertial mass = F/a.
- Third: interacting objects exert equal and opposite forces on each other (different objects, same type of force).
- Required practical 7: trolley, pulley, hanging masses, light gates. Constant mass: move masses from trolley to hanger.
Stopping distances
- Stopping distance = thinking distance + braking distance.
- Reaction time typically 0.2–0.9 s; longer with tiredness, drugs, alcohol, distraction.
- Braking distance longer with higher speed, wet or icy roads, worn brakes or tyres.
- For the same reaction time and deceleration, doubling the speed doubles the thinking distance (s = vt) and quadruples the braking distance (v² = 2as).
- Brakes: friction does work, kinetic energy falls, brakes heat up. Large decelerations: overheating, loss of control.
- Higher tier only: estimate braking forces with F = ma.
4.5.7 Momentum (Higher tier only)
- Closed system: total momentum before = total momentum after.
- Safety features (air bags, seat belts, crash mats, helmets, soft surfaces) increase the time of the collision, so the rate of change of momentum, and so the force, is smaller.
Must-know distinctions
- Mass vs weight: kg, same everywhere vs N, depends on g.
- Distance vs displacement; speed vs velocity: direction makes the second a vector.
- Elastic vs inelastic deformation.
- Thinking vs braking distance: depends on the driver vs depends on the vehicle and road.
- Newton’s third law pair vs balanced forces: pair acts on two objects; balanced forces act on one.
Quick self-test
Use g = 9.8 N/kg and ρ(water) = 1000 kg/m³ where needed.
- Calculate the weight of a 2.5 kg bag.
- Find the resultant of 45 N to the right and 60 N to the left.
- A 25 N force moves an object 4.0 m along its line of action. Calculate the work done.
- A spring with k = 25 N/m is stretched 0.20 m. Find the force and the energy stored.
- Calculate the moment of a 30 N force acting 0.40 m from a pivot.
- A 150 N force acts on 0.050 m². Calculate the pressure.
- (Higher tier only) Calculate the pressure due to water 5.0 m deep.
- A walker covers 3.0 km in 20 minutes. Calculate the average speed in m/s.
- A cyclist speeds up from 4.0 m/s to 16 m/s in 6.0 s. Calculate the acceleration.
- A car starts from rest and accelerates at 2.5 m/s² for 80 m. Calculate its final velocity.
- Calculate the resultant force needed to accelerate a 0.80 kg trolley at 5.0 m/s².
- (Higher tier only) Calculate the momentum of a 0.060 kg ball at 25 m/s.
Answers
- 2.5 × 9.8 = 24.5 N
- 60 − 45 = 15 N to the left
- 25 × 4.0 = 100 J (100 N m)
- F = 25 × 0.20 = 5.0 N; Eₑ = 0.5 × 25 × 0.20² = 0.50 J
- 30 × 0.40 = 12 N m
- 150 / 0.050 = 3000 Pa
- 5.0 × 1000 × 9.8 = 49 000 Pa
- 3000 m / 1200 s = 2.5 m/s
- (16 − 4.0) / 6.0 = 2.0 m/s²
- v² = 0 + 2 × 2.5 × 80 = 400, so v = 20 m/s
- 0.80 × 5.0 = 4.0 N
- 0.060 × 25 = 1.5 kg m/s
Where marks are usually lost
- Giving weight in kg, or mass in N.
- Leaving the extension in cm when using F = ke or Eₑ = ½ke².
- Using the length along a slanted beam instead of the perpendicular distance for a moment.
- Converting minutes or kilometres wrongly before using s = vt.
- Writing v − u instead of v² − u² in the equation-sheet equation.
- Taking the gradient of a velocity–time graph as speed.
- (Higher tier only) Counting squares on a v–t graph but forgetting the scale of each square.
- Saying a skydiver at terminal velocity has “no forces” rather than balanced forces.
- Naming weight and normal contact force on one object as a third-law pair.
- (Higher tier only) Explaining a crash mat by “it absorbs the force” instead of “increases the time, so reduces the rate of change of momentum”.
Official syllabus
AQA GCSE Physics (8463) specification, Version 1.1, 30 September 2019, for teaching from September 2016 and exams from 2018 onwards (AQA), section 4.5 Forces.
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