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AQA GCSE Physics Equations Sheet
AQA GCSE Physics Equations Sheet
AQA GCSE Physics (8463) | Key Equations by Topic
Based on AQA 8463 specification v1.1 (Sept 2019), still the current version (checked November 2026).
For exams in 2025, 2026 and 2027 you get every equation on a sheet in the exam, so you do not need to memorise them (Department for Education and Ofqual, November 2024). In May 2026 Ofqual decided that equation sheets continue for exams from 2028, for the rest of the life of the current specification. The specification still sorts the equations into a recall list and a given list, shown below: learn to choose and apply each one. Ofqual's 2024 decision, Ofqual's decision for 2028 onwards and AQA's notice. Your teacher can confirm what you will get.
Featured Equations
Across forces, energy, electricity & waves #10Eₖ = ½ × m × v²
Recall list
Kinetic energy: kinetic energy = 0.5 × mass × speed squared
Eₖ = Kinetic energy (J)
m = Mass (kg)
v = Speed (m/s)
📝 Worked example
A 70 kg runner reaches 8.0 m/s. Calculate the runner's kinetic energy.
- Write the formula: Eₖ = ½ × m × v²
- Square v first: v² = 8.0² = 64 m²/s²
- Substitute: Eₖ = 0.5 × 70 × 64
- Calculate: Eₖ = 2240 J
Answer: 2240 J (or 2.24 kJ)
#11Eₚ = m × g × h
Recall list
Gravitational potential energy: gravitational potential energy = mass × gravitational field strength × height
Eₚ = GPE gained (J)
m = Mass (kg)
g = Gravitational field strength (N/kg (9.8 on Earth))
h = Vertical height raised (m)
📝 Worked example
A 0.50 kg book is lifted 1.2 m from the floor onto a shelf. Calculate the GPE gained. (g = 9.8 N/kg)
- Write the formula: Eₚ = m × g × h
- Substitute: Eₚ = 0.50 × 9.8 × 1.2
- Calculate: Eₚ = 5.88 J
Answer: 5.88 J (≈ 5.9 J to 2 s.f.)
#17Q = I × t
Recall list
Charge flow: charge flow = current × time
Q = Charge (C (coulombs))
I = Current (A (amperes / amps))
t = Time (s)
📝 Worked example
A current of 0.40 A flows through a wire for 30 s. Calculate the charge that flowed.
- Write the formula: Q = I × t
- Substitute: Q = 0.40 × 30
- Calculate: Q = 12 C
Answer: 12 C
#18V = I × R
Recall list
Potential difference (Ohm's law): potential difference = current × resistance
V = Potential difference (V (volts))
I = Current (A)
R = Resistance (Ω (ohms))
📝 Worked example
A current of 0.25 A flows through a 12 Ω resistor. Calculate the potential difference across it.
- Write the formula: V = I × R
- Substitute: V = 0.25 × 12
- Calculate: V = 3.0 V
Answer: 3.0 V
#23ρ = m / V
Recall list
Density: density = mass / volume
ρ = Density (Greek letter rho) (kg/m³ (or g/cm³))
m = Mass (kg)
V = Volume (m³)
📝 Worked example
A metal block has a mass of 270 g and a volume of 100 cm³. Calculate the density in g/cm³.
- Write the formula: ρ = m / V
- Substitute: ρ = 270 / 100
- Calculate: ρ = 2.7 g/cm³
Answer: 2.7 g/cm³ (this is aluminium)
#1W = m × g
Recall list
Weight: weight = mass × gravitational field strength
W = Weight (N (newtons))
m = Mass (kg)
g = Gravitational field strength (N/kg (the value is given in the exam; 9.8 N/kg on Earth))
📝 Worked example
A laptop has a mass of 1.4 kg. Calculate its weight on Earth.
- Write the formula: W = m × g
- Substitute (g = 9.8 N/kg on Earth): W = 1.4 × 9.8
- Calculate: W = 13.72 N
Answer: 13.72 N (≈ 13.7 N to 3 s.f.)
#8F = m × a
Recall list
Resultant force (Newton's 2nd law): resultant force = mass × acceleration
F = Resultant force (N)
m = Mass (kg)
a = Acceleration (m/s²)
📝 Worked example
A trolley of mass 2.5 kg experiences a resultant force of 5.0 N. Calculate its acceleration.
- Rearrange for a: a = F / m
- Substitute: a = 5.0 / 2.5
- Calculate: a = 2.0 m/s²
Answer: 2.0 m/s²
#16v = f × λ
Recall list
Wave speed: wave speed = frequency × wavelength
v = Wave speed (m/s)
f = Frequency (Hz (1/s))
λ = Wavelength (lambda) (m)
📝 Worked example
A water wave has a frequency of 4.0 Hz and a wavelength of 0.50 m. Calculate the wave speed.
- Write the formula: v = f × λ
- Substitute: v = 4.0 × 0.50
- Calculate: v = 2.0 m/s
Answer: 2.0 m/s
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Key AQA GCSE Physics equations organised by specification topic, with worked examples and HT markers. Free, printable PDF.
The specification sorts AQA's equations into two lists: equations to recall and apply, and equations to select and apply from the Physics equation sheet. For exams in 2025, 2026 and 2027 AQA prints every equation on the equation sheet you get in the exam, so focus on choosing the right one and applying it. Higher-tier-only equations are flagged. Equations follow the order of the sections in AQA's specification.
Recall list
Given list
HT only
4.1 Energy
4.1.1 — Energy changes in a system, and the ways energy is stored
#10Eₖ = ½ × m × v²
Recall list
Kinetic energy: kinetic energy = 0.5 × mass × speed squared
Eₖ = Kinetic energy (J)
m = Mass (kg)
v = Speed (m/s)
💡 Double the speed → four times the kinetic energy (because of the v²). This is why fast cars are so dangerous.
📝 Worked example
A 70 kg runner reaches 8.0 m/s. Calculate the runner's kinetic energy.
- Write the formula: Eₖ = ½ × m × v²
- Square v first: v² = 8.0² = 64 m²/s²
- Substitute: Eₖ = 0.5 × 70 × 64
- Calculate: Eₖ = 2240 J
Answer: 2240 J (or 2.24 kJ)
⚠ Square v before multiplying. A common slip is computing ½ × m × v then squaring at the end — that gives the wrong answer.
#11Eₚ = m × g × h
Recall list
Gravitational potential energy: gravitational potential energy = mass × gravitational field strength × height
Eₚ = GPE gained (J)
m = Mass (kg)
g = Gravitational field strength (N/kg (9.8 on Earth))
h = Vertical height raised (m)
💡 GPE is measured relative to a reference level (usually the ground). Use the vertical height, not the path length up a slope.
📝 Worked example
A 0.50 kg book is lifted 1.2 m from the floor onto a shelf. Calculate the GPE gained. (g = 9.8 N/kg)
- Write the formula: Eₚ = m × g × h
- Substitute: Eₚ = 0.50 × 9.8 × 1.2
- Calculate: Eₚ = 5.88 J
Answer: 5.88 J (≈ 5.9 J to 2 s.f.)
⚠ h is the vertical height gained, not the distance walked up a ramp.
#12P = E / t
Recall list
Power (from energy): power = energy transferred / time
P = Power (W (watts) = J/s)
E = Energy transferred (J)
t = Time (s)
💡 1 watt = 1 joule per second. A 60 W bulb transfers 60 J every second.
📝 Worked example
A kettle transfers 90 000 J of energy in 60 s. Calculate its power.
- Write the formula: P = E / t
- Substitute: P = 90 000 / 60
- Calculate: P = 1500 W
Answer: 1500 W (1.5 kW)
⚠ Power and energy are different quantities. Energy is in joules, power is the rate — joules per second.
#13P = W / t
Recall list
Power (from work): power = work done / time taken
P = Power (W)
W = Work done (J)
t = Time (s)
💡 Identical in structure to P = E/t. AQA lists both because 'energy transferred' and 'work done' are different names for the same thing (in J).
📝 Worked example
A motor does 4500 J of work lifting a load in 15 s. Calculate the motor's power output.
- Write the formula: P = W / t
- Substitute: P = 4500 / 15
- Calculate: P = 300 W
Answer: 300 W
⚠ Here W stands for work done (in joules). Don't confuse it with W for watts, the unit of power.
4.1.2 — Conservation and dissipation of energy
#14efficiency = useful energy out / total energy in
Recall list
Efficiency (from energy): efficiency = useful output energy transfer / total input energy transfer
useful energy out = Useful output energy (J (any energy unit, must match total energy in))
total energy in = Total input energy (J)
💡 Efficiency is a ratio (no units). Multiply by 100 to express as a percentage. Always less than 1 (or 100%).
📝 Worked example
A motor takes in 800 J of electrical energy and produces 600 J of useful kinetic energy. Calculate the efficiency.
- Write the formula: efficiency = useful E out / total E in
- Substitute: efficiency = 600 / 800
- Calculate ratio: = 0.75
- Convert to %: = 0.75 (or 75%)
Answer: 0.75 or 75%
⚠ Efficiency cannot exceed 100% — if your answer is more than 1, you've swapped numerator and denominator.
#15efficiency = useful power out / total power in
Recall list
Efficiency (from power): efficiency = useful power output / total power input
useful power out = Useful output power (W)
total power in = Total input power (W)
💡 Same as the energy version, just using power. Both forms are in the specification — use whichever matches the information the question gives.
📝 Worked example
A lamp draws 60 W of electrical power and emits 12 W of useful light. Calculate the efficiency.
- Write the formula: efficiency = useful P out / total P in
- Substitute: efficiency = 12 / 60
- Calculate: = 0.20 or 20%
Answer: 0.20 or 20%
⚠ Watch the wording: 'wasted as heat' is NOT the useful output. Only count the useful (intended) form.
4.2 Electricity
4.2.1 — Current, potential difference and resistance
#17Q = I × t
Recall list
Charge flow: charge flow = current × time
Q = Charge (C (coulombs))
I = Current (A (amperes / amps))
t = Time (s)
💡 1 ampere = 1 coulomb per second. Current is the rate at which charge flows.
📝 Worked example
A current of 0.40 A flows through a wire for 30 s. Calculate the charge that flowed.
- Write the formula: Q = I × t
- Substitute: Q = 0.40 × 30
- Calculate: Q = 12 C
Answer: 12 C
⚠ Always use time in seconds, not minutes.
#18V = I × R
Recall list
Potential difference (Ohm's law): potential difference = current × resistance
V = Potential difference (V (volts))
I = Current (A)
R = Resistance (Ω (ohms))
💡 The 'VIR triangle': cover what you want to find. V on top, I × R underneath.
📝 Worked example
A current of 0.25 A flows through a 12 Ω resistor. Calculate the potential difference across it.
- Write the formula: V = I × R
- Substitute: V = 0.25 × 12
- Calculate: V = 3.0 V
Answer: 3.0 V
⚠ Currents are often small (e.g. 0.05 A). Check your decimal placement before computing — slips here cause big errors.
4.2.4 — Energy transfers
#19P = V × I
Recall list
Electrical power (P = V × I): power = potential difference × current
P = Power (W)
V = Potential difference (V)
I = Current (A)
💡 Use this when you know voltage and current. For UK mains (230 V), a 1.0 A appliance dissipates 230 W.
📝 Worked example
A heater plugged into a 230 V supply draws 5.0 A of current. Calculate the heater's power.
- Write the formula: P = V × I
- Substitute: P = 230 × 5.0
- Calculate: P = 1150 W
Answer: 1150 W (1.15 kW)
⚠ Watts × seconds = joules, but volts × amps = watts directly (not joules). Stay clear on units.
#20P = I² × R
Recall list
Electrical power (P = I² × R): power = current squared × resistance
P = Power (W)
I = Current (A)
R = Resistance (Ω)
💡 Use this form when you know current and resistance (no voltage given). The I² means doubling current quadruples power dissipated.
📝 Worked example
A 4.0 Ω resistor carries a current of 0.50 A. Calculate the power dissipated.
- Write the formula: P = I² × R
- Square I first: I² = 0.50² = 0.25
- Substitute: P = 0.25 × 4.0
- Calculate: P = 1.0 W
Answer: 1.0 W
⚠ Square the current before multiplying by resistance — not after.
#21E = P × t
Recall list
Energy transferred (E = P × t): energy transferred = power × time
E = Energy transferred (J)
P = Power (W)
t = Time (s)
💡 Just the rearranged P = E/t. A 1000 W heater for 10 s transfers 10 000 J.
📝 Worked example
A 100 W lamp is left on for 60 s. How much energy does it transfer?
- Write the formula: E = P × t
- Substitute: E = 100 × 60
- Calculate: E = 6000 J
Answer: 6000 J (6 kJ)
⚠ If the question gives time in minutes or hours, convert to seconds first.
#22E = Q × V
Recall list
Energy transferred (E = Q × V): energy transferred = charge × potential difference
E = Energy transferred (J)
Q = Charge (C)
V = Potential difference (V)
💡 Volts measure energy per coulomb. So 1 V × 1 C = 1 J.
📝 Worked example
A charge of 8.0 C flows through a 12 V battery. How much energy is transferred?
- Write the formula: E = Q × V
- Substitute: E = 8.0 × 12
- Calculate: E = 96 J
Answer: 96 J
⚠ Don't confuse with E = P × t. This form uses charge (C), not time (s).
4.3 Particle model of matter
4.3.1 — Changes of state and the particle model
#23ρ = m / V
Recall list
Density: density = mass / volume
ρ = Density (Greek letter rho) (kg/m³ (or g/cm³))
m = Mass (kg)
V = Volume (m³)
💡 Water has a density of 1000 kg/m³ (or 1 g/cm³). Anything denser than water sinks; less dense floats.
📝 Worked example
A metal block has a mass of 270 g and a volume of 100 cm³. Calculate the density in g/cm³.
- Write the formula: ρ = m / V
- Substitute: ρ = 270 / 100
- Calculate: ρ = 2.7 g/cm³
Answer: 2.7 g/cm³ (this is aluminium)
⚠ Match units: kg with m³ gives kg/m³; g with cm³ gives g/cm³. Mixing them is a frequent slip.
4.3.2 — Internal energy and energy transfers
#28ΔE = m × c × Δθ
Given list
Specific heat capacity: change in thermal energy = mass × specific heat capacity × change in temperature
ΔE = Change in thermal energy (J)
m = Mass (kg)
c = Specific heat capacity (J/(kg °C))
Δθ = Change in temperature (°C)
💡 Water has c = 4200 J/(kg °C) — one of the highest, which is why it takes so long to heat up.
📝 Worked example
How much energy is needed to raise the temperature of 0.50 kg of water by 30 °C? (c_water = 4200 J/(kg °C))
- Write the formula: ΔE = m × c × Δθ
- Substitute: ΔE = 0.50 × 4200 × 30
- Calculate: ΔE = 63 000 J
Answer: 63 000 J (63 kJ)
⚠ Δθ is the change in temperature, not the final temperature. Going from 20 °C to 50 °C is Δθ = 30 °C, not 50.
#32E = m × L
Given list
Specific latent heat: energy for a change of state = mass × specific latent heat
E = Energy transferred (no temperature change) (J)
m = Mass changing state (kg)
L = Specific latent heat (J/kg)
💡 Use L_f for fusion (solid↔liquid), L_v for vaporisation (liquid↔vapour). L_v is always larger.
📝 Worked example
How much energy is needed to melt 0.30 kg of ice at 0 °C? (L_f of ice = 334 000 J/kg)
- Write the formula: E = m × L
- Substitute: E = 0.30 × 334 000
- Calculate: E = 100 200 J
Answer: 100 200 J (≈ 100 kJ)
⚠ During a change of state, temperature stays constant — don't try to also apply ΔE = mcΔθ during melting or boiling.
4.3.3 — Particle model and pressure
#35p × V = constant
Given list
Gas law (constant temperature, fixed mass): pressure × volume = constant (for a fixed mass of gas at constant temperature)
p = Pressure (Pa (or any pressure unit, both sides must match))
V = Volume (m³ (or any volume unit, both sides must match))
💡 Squeeze a gas to half the volume → pressure doubles. This is Boyle's law. Only valid at constant T and fixed mass.
📝 Worked example
A gas occupies 0.020 m³ at 100 kPa. It is compressed at constant temperature to 0.005 m³. What's the new pressure?
- Write the law: p₁ × V₁ = p₂ × V₂
- Substitute: 100 × 0.020 = p₂ × 0.005
- Simplify: 2.0 = p₂ × 0.005
- Solve for p₂: p₂ = 2.0 / 0.005 = 400 kPa
Answer: 400 kPa
⚠ Units must be consistent on both sides — but they don't have to be SI. Pa with Pa, or kPa with kPa, both work.
4.5 Forces
4.5.1 — Forces and their interactions
#1W = m × g
Recall list
Weight: weight = mass × gravitational field strength
W = Weight (N (newtons))
m = Mass (kg)
g = Gravitational field strength (N/kg (the value is given in the exam; 9.8 N/kg on Earth))
💡 Weight is a force, so the answer must be in newtons, never kilograms.
📝 Worked example
A laptop has a mass of 1.4 kg. Calculate its weight on Earth.
- Write the formula: W = m × g
- Substitute (g = 9.8 N/kg on Earth): W = 1.4 × 9.8
- Calculate: W = 13.72 N
Answer: 13.72 N (≈ 13.7 N to 3 s.f.)
⚠ Don't write the answer in kg. Weight is measured in newtons because it is a force.
4.5.2 — Work done and energy transfer
#2W = F × s
Recall list
Work done: work done = force × distance moved along the line of action of the force
W = Work done (J (joules))
F = Force (N)
s = Distance moved along the line of action of the force (m)
💡 1 joule = 1 newton-metre. If the force and the motion are at right angles, no work is done.
📝 Worked example
A student pushes a box with a steady force of 25 N across a smooth floor for 4.0 m. How much work is done?
- Write the formula: W = F × s
- Substitute: W = 25 × 4.0
- Calculate: W = 100 J
Answer: 100 J
⚠ Use the distance along the direction the force pushes, not the total path length the object travels.
4.5.3 — Forces and elasticity
#3F = k × e
Recall list
Force on a spring (Hooke's law): force applied to a spring = spring constant × extension
F = Force (N)
k = Spring constant (N/m)
e = Extension (or compression) from natural length (m)
💡 Only valid up to the limit of proportionality. Use 'e' in metres — convert from cm or mm first.
📝 Worked example
A spring with a spring constant of 50 N/m is stretched 0.08 m. What force is needed?
- Write the formula: F = k × e
- Substitute: F = 50 × 0.08
- Calculate: F = 4.0 N
Answer: 4.0 N
⚠ Extension must be in metres. A common error is leaving it in cm — converting 8 cm to 0.08 m is essential.
#27Eₑ = ½ × k × e²
Given list
Elastic potential energy: elastic potential energy = 0.5 × spring constant × extension squared
Eₑ = Elastic potential energy (J)
k = Spring constant (N/m)
e = Extension (or compression) (m)
💡 Only valid up to the limit of proportionality. Double the extension → four times the energy stored.
📝 Worked example
A spring with spring constant 200 N/m is stretched 0.10 m. How much elastic PE is stored?
- Write the formula: Eₑ = ½ × k × e²
- Square e first: e² = 0.10² = 0.01 m²
- Substitute: Eₑ = 0.5 × 200 × 0.01
- Calculate: Eₑ = 1.0 J
Answer: 1.0 J
⚠ Square the extension before multiplying — and convert to metres first if given in cm.
4.5.4 — Moments, levers and gears (physics only)
#4M = F × d
Recall list
Moment of a force: moment of a force = force × perpendicular distance from the pivot to the line of action of the force
M = Moment (Nm (newton-metres))
F = Force (N)
d = Perpendicular distance from pivot to line of action of force (m)
💡 'd' must be the perpendicular distance — drop a 90° line from the pivot to the force's line of action.
📝 Worked example
A spanner has a 20 cm handle. A student pushes the end with a force of 30 N at 90° to the handle. Calculate the moment about the nut.
- Convert distance: d = 20 cm = 0.20 m
- Write the formula: M = F × d
- Substitute: M = 30 × 0.20
- Calculate: M = 6.0 Nm
Answer: 6.0 Nm
⚠ If the force is not at 90° to the lever, you can't just use the lever length — you need the perpendicular distance, which is shorter.
4.5.5 — Pressure and pressure differences in fluids (physics only)
#5p = F / A
Recall list
Pressure: pressure = force normal to a surface / area of that surface
p = Pressure (Pa (pascals) = N/m²)
F = Force acting normal to the surface (N)
A = Area (m²)
💡 Pressure is force spread over an area. Sharp knives cut better because tiny area gives huge pressure for the same force.
📝 Worked example
A 600 N force is applied through a block of cross-sectional area 0.30 m². What pressure does it exert?
- Write the formula: p = F / A
- Substitute: p = 600 / 0.30
- Calculate: p = 2000 Pa
Answer: 2000 Pa (2 kPa)
⚠ Area must be in m². 30 cm² is not 30 m² — it's 0.0030 m² (divide by 10,000).
#24p = h × ρ × g
Given list HT only
Pressure in a column of liquid: pressure = height of the column × density of the liquid × gravitational field strength
p = Pressure due to the column (Pa)
h = Height of the column (m)
ρ = Density of the liquid (kg/m³)
g = Gravitational field strength (N/kg (9.8 on Earth))
💡 Depth, not total volume, determines pressure in a liquid. A tall thin glass and a wide pool have the same pressure at the same depth.
📝 Worked example
Calculate the water pressure at a depth of 5.0 m. (Density of water = 1000 kg/m³, g = 9.8 N/kg)
- Write the formula: p = h × ρ × g
- Substitute: p = 5.0 × 1000 × 9.8
- Calculate: p = 49 000 Pa
Answer: 49 000 Pa (49 kPa)
⚠ h is the vertical depth below the surface, not the horizontal distance or volume of liquid.
4.5.6 — Forces and motion
#6s = v × t
Recall list
Distance travelled at constant speed: distance travelled = speed × time
s = Distance (m)
v = Speed (m/s)
t = Time (s)
💡 If the speed changes, s = vt still works when v is the average speed. Higher tier: the distance is also the area under a velocity-time graph.
📝 Worked example
A cyclist travels at a constant 6.0 m/s for 45 s. How far does she travel?
- Write the formula: s = v × t
- Substitute: s = 6.0 × 45
- Calculate: s = 270 m
Answer: 270 m
⚠ If the question gives time in minutes, convert to seconds first. Speed is m/s, not m/min.
#7a = Δv / t
Recall list
Acceleration: acceleration = change in velocity / time taken for that change
a = Acceleration (m/s²)
Δv = Change in velocity (final − initial) (m/s)
t = Time taken (s)
💡 Δv means 'change in v', so Δv = v − u. A negative value means deceleration.
📝 Worked example
A car accelerates from 12 m/s to 20 m/s in 4.0 s. Find its acceleration.
- Find Δv: Δv = 20 − 12 = 8.0 m/s
- Write the formula: a = Δv / t
- Substitute: a = 8.0 / 4.0
- Calculate: a = 2.0 m/s²
Answer: 2.0 m/s²
⚠ Don't use the final velocity directly — you must subtract the starting velocity first.
#8F = m × a
Recall list
Resultant force (Newton's 2nd law): resultant force = mass × acceleration
F = Resultant force (N)
m = Mass (kg)
a = Acceleration (m/s²)
💡 It's the resultant force — the net force after adding all forces. If forces balance, a = 0.
📝 Worked example
A trolley of mass 2.5 kg experiences a resultant force of 5.0 N. Calculate its acceleration.
- Rearrange for a: a = F / m
- Substitute: a = 5.0 / 2.5
- Calculate: a = 2.0 m/s²
Answer: 2.0 m/s²
⚠ Use the resultant force, not just the driving force. If friction is 2 N and the push is 7 N, the resultant is 5 N.
#25v² − u² = 2 × a × s
Given list
Uniform acceleration (SUVAT): (final velocity)² − (initial velocity)² = 2 × acceleration × distance
v = Final velocity (m/s)
u = Initial velocity (m/s)
a = Acceleration (m/s²)
s = Distance (m)
💡 Use this when time is NOT given and not asked for. If the time is given, a = Δv / t is usually quicker.
📝 Worked example
A car accelerates from rest at 2.0 m/s² over a distance of 100 m. What's its final speed?
- Write the formula: v² − u² = 2 × a × s
- Note u = 0 (starts from rest): v² − 0 = 2 × 2.0 × 100
- Simplify: v² = 400
- Take square root: v = 20 m/s
Answer: 20 m/s
⚠ Both velocities are squared. A common error is squaring only v, not u.
4.5.7 — Momentum (HT only)
#9p = m × v
Recall list HT only
Momentum: momentum = mass × velocity
p = Momentum (kg m/s)
m = Mass (kg)
v = Velocity (m/s)
💡 Momentum is a vector — direction matters. Two objects moving towards each other have momenta with opposite signs.
📝 Worked example
A 0.060 kg tennis ball moves at 25 m/s. Calculate its momentum.
- Write the formula: p = m × v
- Substitute: p = 0.060 × 25
- Calculate: p = 1.5 kg m/s
Answer: 1.5 kg m/s
⚠ Don't confuse the symbol p for momentum with p for pressure — context tells you which one.
#26F = m × Δv / Δt
Given list HT only
Force from change in momentum: force = mass × change in velocity / change in time
F = Force (N)
m = Mass (kg)
Δv = Change in velocity (m/s)
Δt = Time over which velocity changes (s)
💡 Force is the rate of change of momentum. Crumple zones and airbags work by increasing Δt to reduce F.
📝 Worked example
A 0.16 kg cricket ball moving towards a bat at 30 m/s is hit straight back at 10 m/s. Taking the direction towards the bat as positive, its velocity changes from 30 m/s to −10 m/s. The contact lasts 0.020 s. Calculate the force on the ball.
- Find Δv: Δv = (−10) − 30 = −40 m/s
- Write the formula: F = m × Δv / Δt
- Substitute: F = 0.16 × (−40) / 0.020
- Calculate: F = −320 N (the minus sign shows direction)
Answer: −320 N (magnitude 320 N, directed away from the bat)
⚠ If the ball bounces back, Δv is bigger than just the speed change — direction reverses, so use signed values.
4.6 Waves
4.6.1 — Waves in air, fluids and solids
#16v = f × λ
Recall list
Wave speed: wave speed = frequency × wavelength
v = Wave speed (m/s)
f = Frequency (Hz (1/s))
λ = Wavelength (lambda) (m)
💡 Same wave speed: if frequency goes up, wavelength must go down. Light in a vacuum is always 3 × 10⁸ m/s.
📝 Worked example
A water wave has a frequency of 4.0 Hz and a wavelength of 0.50 m. Calculate the wave speed.
- Write the formula: v = f × λ
- Substitute: v = 4.0 × 0.50
- Calculate: v = 2.0 m/s
Answer: 2.0 m/s
⚠ If frequency is given in kHz or MHz, convert to Hz first (multiply by 1000 or 1 000 000).
#29T = 1 / f
Given list
Wave period: period = 1 / frequency
T = Period (time for one full wave) (s)
f = Frequency (Hz)
💡 T and f are reciprocals. A 50 Hz wave has a period of 1/50 = 0.02 s.
📝 Worked example
A sound wave has a frequency of 250 Hz. Calculate its period.
- Write the formula: T = 1 / f
- Substitute: T = 1 / 250
- Calculate: T = 1 / 250 = 0.004 s (or 4 ms)
Answer: 0.004 s (4 ms)
⚠ Don't mix up T and f: T = 1 / f and f = 1 / T. A frequency in Hz gives a period in seconds.
4.6.2 — Electromagnetic waves
#30magnification = image height / object height
Given list
Magnification: magnification = image height / object height
magnification = Magnification (no units, it's a ratio) (no units)
image height = Height of the image (mm (any length unit, must match object))
object height = Height of the object (mm)
💡 Magnification > 1 = image bigger than object. Magnification < 1 = image smaller. No units, because it's a ratio.
📝 Worked example
A lens produces an image 6.0 cm tall of an object that is 2.0 cm tall. Calculate the magnification.
- Write the formula: magnification = image height / object height
- Substitute: = 6.0 / 2.0
- Calculate: = 3.0
Answer: 3.0 (no units — the image is 3 times larger)
⚠ Image and object heights must be in the same unit. Don't write a unit on the final answer — magnification is a pure ratio.
4.7 Magnetism and electromagnetism
4.7.2 — The motor effect
#31F = B × I × l
Given list HT only
Force on a current-carrying conductor: force = magnetic flux density × current × length
F = Force (N)
B = Magnetic flux density (T (tesla))
I = Current (A)
l = Length of conductor inside the field (m)
💡 Valid only when the conductor is at right angles to the field. Use Fleming's left-hand rule for direction.
📝 Worked example
A 0.10 m wire carries a current of 3.0 A at right angles to a magnetic field of flux density 0.40 T. Calculate the force on the wire.
- Write the formula: F = B × I × l
- Substitute: F = 0.40 × 3.0 × 0.10
- Calculate: F = 0.12 N
Answer: 0.12 N
⚠ If the wire is parallel to the field, force is zero. The equation only gives the maximum, at 90°.
4.7.3 — Induced potential, transformers and the National Grid (physics only) (HT only)
#33Vₚ / Vₛ = nₚ / nₛ
Given list HT only
Transformer voltage ratio: primary potential difference / secondary potential difference = primary turns / secondary turns
Vₚ = Primary (input) potential difference (V)
Vₛ = Secondary (output) potential difference (V)
nₚ = Number of turns on primary coil (no units (count))
nₛ = Number of turns on secondary coil (no units (count))
💡 More turns on the output side = step-up (voltage increases). Fewer turns = step-down.
📝 Worked example
A transformer has 1000 primary turns and 50 secondary turns. The primary pd is 230 V. What's the secondary pd?
- Rearrange for Vₛ: Vₛ = Vₚ × (nₛ / nₚ)
- Substitute: Vₛ = 230 × (50 / 1000)
- Calculate: Vₛ = 11.5 V
Answer: 11.5 V (this is a step-down transformer)
⚠ Make sure the ratios are matched: primary on top with primary on top, not crossed.
#34Vₚ × Iₚ = Vₛ × Iₛ
Given list HT only
Transformer power (ideal): primary potential difference × primary current = secondary potential difference × secondary current
Vₚ = Primary pd (V)
Iₚ = Primary current (A)
Vₛ = Secondary pd (V)
Iₛ = Secondary current (A)
💡 Assumes 100% efficient transformer (no energy loss). Stepping voltage UP means current is stepped DOWN, and vice versa.
📝 Worked example
A transformer steps 230 V down to 12 V. If the secondary current is 4.0 A, what current flows in the primary coil?
- Write the formula: Vₚ × Iₚ = Vₛ × Iₛ
- Substitute: 230 × Iₚ = 12 × 4.0
- Simplify: 230 × Iₚ = 48
- Solve for Iₚ: Iₚ = 48 / 230 = 0.21 A
Answer: 0.21 A
⚠ Step-down transformers reduce voltage but raise current — never both in the same direction.
Equation list and topic groupings are based on the AQA GCSE Physics (8463) specification, version 1.1 (September 2019), Appendix A, still the current version (checked November 2026). Worked examples and explanations are original LumiExams content.
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LumiExams.com | AQA GCSE Physics (8463) | Spec v1.1 (2019-09-30)
Frequently Asked Questions
What equations do I need to memorise for AQA GCSE Physics?
For exams in 2025, 2026 and 2027, none: AQA gives you an equation sheet with every equation, so learn to choose the right one and apply it. Ofqual has decided that equation sheets continue for exams from 2028 for the rest of the life of the current specification. The specification still splits the equations into a recall list and a given list, and this page shows which is which.
Are AQA Physics equations given in the exam?
Yes. For exams in 2025, 2026 and 2027 AQA provides an equation sheet with every equation, including the ones the specification normally asks you to recall, such as Eₖ = ½mv² and V = IR. This follows a Department for Education and Ofqual decision of November 2024. In May 2026 Ofqual decided that equation sheets continue for exams from 2028.
Which AQA Physics equations are higher tier only?
These equations are needed for higher tier papers only: momentum (p = mv), pressure due to a column of liquid (p = hρg), force from change in momentum (F = mΔv/Δt), the force on a current-carrying conductor (F = BIl), and the two transformer equations. Foundation tier students do not need to apply these.
How is this AQA Physics equation sheet organised?
This page follows the order of the sections in AQA's specification: 4.1 Energy, 4.2 Electricity, 4.3 Particle model of matter, 4.5 Forces, 4.6 Waves and 4.7 Magnetism and electromagnetism. Atomic structure (4.4) and Space physics (4.8) do not appear because none of the specification's listed equations come from them.
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