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Edexcel GCSE Physics Equations Sheet
Edexcel GCSE Physics Equations Sheet
Pearson Edexcel GCSE Physics (1PH0) | All 33 Required Equations
Based on Pearson Edexcel 1PH0 specification, Issue 4 (March 2024), Appendix 2. Current for May 2026 exams.
Featured Equations (8 of 33)
Across motion, energy, waves & electricity #1x = v × t
Memorise §2.6b
Distance travelled: distance travelled = average speed × time
x = Distance travelled (m)
v = Average speed (m/s)
t = Time (s)
📝 Worked example
A jogger maintains an average speed of 2.5 m/s for 180 s. How far has she travelled?
- Write the formula: x = v × t
- Substitute: x = 2.5 × 180
- Calculate: x = 450 m
Answer: 450 m
#3F = m × a
Memorise §2.15
Resultant force (Newton's 2nd law): force = mass × acceleration
F = Resultant force (N)
m = Mass (kg)
a = Acceleration (m/s²)
📝 Worked example
A 1200 kg car experiences a 3600 N resultant forward force. Calculate its acceleration.
- Rearrange for a: a = F / m
- Substitute: a = 3600 / 1200
- Calculate: a = 3.0 m/s²
Answer: 3.0 m/s²
#4W = m × g
Memorise §2.16
Weight: weight = mass × gravitational field strength
W = Weight (N)
m = Mass (kg)
g = Gravitational field strength (N/kg (use 10 N/kg on Earth — Edexcel rounds))
📝 Worked example
A backpack has a mass of 6.0 kg. Calculate its weight on Earth. (g = 10 N/kg)
- Write the formula: W = m × g
- Substitute: W = 6.0 × 10
- Calculate: W = 60 N
Answer: 60 N
#6ΔGPE = m × g × Δh
Memorise §3.1, 8.8
Change in gravitational PE: change in GPE = mass × gravitational field strength × change in vertical height
ΔGPE = Change in GPE (J)
m = Mass (kg)
g = Gravitational field strength (N/kg (10 on Earth))
Δh = Vertical height gained (m)
📝 Worked example
A 0.40 kg ball is lifted 2.5 m straight up. Calculate the change in GPE. (g = 10 N/kg)
- Write the formula: ΔGPE = m × g × Δh
- Substitute: ΔGPE = 0.40 × 10 × 2.5
- Calculate: ΔGPE = 10 J
Answer: 10 J
#7KE = ½ × m × v²
Memorise §3.2, 8.9
Kinetic energy: kinetic energy = ½ × mass × (speed)²
KE = Kinetic energy (J)
m = Mass (kg)
v = Speed (m/s)
📝 Worked example
A 60 kg cyclist rides at 7.0 m/s. Calculate her kinetic energy.
- Write the formula: KE = ½ × m × v²
- Square v first: v² = 7.0² = 49 m²/s²
- Substitute: KE = 0.5 × 60 × 49
- Calculate: KE = 1470 J
Answer: 1470 J (≈ 1.5 kJ)
#9v = f × λ
Memorise §4.6
Wave speed (frequency form): wave speed = frequency × wavelength
v = Wave speed (m/s)
f = Frequency (Hz)
λ = Wavelength (m)
📝 Worked example
A sound wave has frequency 320 Hz and wavelength 1.0 m. Calculate the wave speed.
- Write the formula: v = f × λ
- Substitute: v = 320 × 1.0
- Calculate: v = 320 m/s
Answer: 320 m/s
#15Q = I × t
Memorise §10.9
Charge: charge = current × time
Q = Charge (C)
I = Current (A)
t = Time (s)
📝 Worked example
A 2.0 A current flows for 5.0 minutes. Calculate the charge transferred.
- Convert time to seconds: t = 5.0 × 60 = 300 s
- Write the formula: Q = I × t
- Substitute: Q = 2.0 × 300
- Calculate: Q = 600 C
Answer: 600 C
#16V = I × R
Memorise §10.13
Potential difference (Ohm's law): potential difference = current × resistance
V = Potential difference (V)
I = Current (A)
R = Resistance (Ω)
📝 Worked example
A 0.30 A current flows through a 20 Ω resistor. Calculate the pd across it.
- Write the formula: V = I × R
- Substitute: V = 0.30 × 20
- Calculate: V = 6.0 V
Answer: 6.0 V
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Unlock the Remaining 25 Equations
All 33 Edexcel GCSE Physics equations organised by spec topic, with worked examples and HT markers. Free, printable PDF.
Edexcel splits its 33 equations into two lists. List A (22 equations) must be recalled and applied — you should memorise these. List B (11 equations) is printed in an Equation Booklet at the end of the exam paper — you only need to select the right one. Higher-tier-only equations are flagged. Equations are ordered by Edexcel's own topic structure (Topic 2 → Topic 15), so this matches the spec chapter-for-chapter.
Memorise (22)
Given in Equation Booklet (11)
HT only
Topic 2 — Motion and forces
#1x = v × t
Memorise §2.6b
Distance travelled: distance travelled = average speed × time
x = Distance travelled (m)
v = Average speed (m/s)
t = Time (s)
💡 Average speed assumes constant motion. For non-uniform motion, use the area under a speed-time graph.
📝 Worked example
A jogger maintains an average speed of 2.5 m/s for 180 s. How far has she travelled?
- Write the formula: x = v × t
- Substitute: x = 2.5 × 180
- Calculate: x = 450 m
Answer: 450 m
⚠ Speed must be in m/s and time in seconds. Convert km/h or minutes first.
#2a = (v − u) / t
Memorise §2.8
Acceleration: acceleration = change in velocity ÷ time taken
a = Acceleration (m/s²)
v = Final velocity (m/s)
u = Initial velocity (m/s)
t = Time taken (s)
💡 Negative value means deceleration. If u = 0 (starts from rest), this reduces to a = v/t.
📝 Worked example
A skater speeds up from 1.5 m/s to 5.0 m/s in 7.0 s. Calculate her acceleration.
- Find Δv: Δv = 5.0 − 1.5 = 3.5 m/s
- Write the formula: a = Δv / t
- Substitute: a = 3.5 / 7.0
- Calculate: a = 0.50 m/s²
Answer: 0.50 m/s²
⚠ Always subtract initial from final velocity, in that order. Reversing gives a wrong sign.
#3F = m × a
Memorise §2.15
Resultant force (Newton's 2nd law): force = mass × acceleration
F = Resultant force (N)
m = Mass (kg)
a = Acceleration (m/s²)
💡 F here is the RESULTANT force — the net of all forces acting. Balanced forces give a = 0.
📝 Worked example
A 1200 kg car experiences a 3600 N resultant forward force. Calculate its acceleration.
- Rearrange for a: a = F / m
- Substitute: a = 3600 / 1200
- Calculate: a = 3.0 m/s²
Answer: 3.0 m/s²
⚠ Use the resultant force. If 5000 N pushes forward and 1400 N of friction resists, resultant = 3600 N (not 5000).
#4W = m × g
Memorise §2.16
Weight: weight = mass × gravitational field strength
W = Weight (N)
m = Mass (kg)
g = Gravitational field strength (N/kg (use 10 N/kg on Earth — Edexcel rounds))
💡 Edexcel papers usually use g = 10 N/kg (not 9.8 like AQA). Check the question; it's stated when needed.
📝 Worked example
A backpack has a mass of 6.0 kg. Calculate its weight on Earth. (g = 10 N/kg)
- Write the formula: W = m × g
- Substitute: W = 6.0 × 10
- Calculate: W = 60 N
Answer: 60 N
⚠ Weight is a force (newtons). Don't write the answer in kg.
#5p = m × v
Memorise HT only §2.24
Momentum: momentum = mass × velocity
p = Momentum (kg m/s)
m = Mass (kg)
v = Velocity (m/s)
💡 Momentum has direction — opposite-moving objects have opposite-signed momenta.
📝 Worked example
A 1500 kg car drives at 18 m/s. What is its momentum?
- Write the formula: p = m × v
- Substitute: p = 1500 × 18
- Calculate: p = 27 000 kg m/s
Answer: 27 000 kg m/s
⚠ Don't confuse the symbol p for momentum with p for pressure.
#23v² − u² = 2 × a × x
Given §2.9
SUVAT (no-time form): (final velocity)² − (initial velocity)² = 2 × acceleration × distance
v = Final velocity (m/s)
u = Initial velocity (m/s)
a = Acceleration (m/s²)
x = Distance (m)
💡 Use this when time is NOT given. Both velocities are squared.
📝 Worked example
A car accelerates from rest at 3.0 m/s² over a distance of 24 m. Calculate its final speed.
- Write the formula: v² − u² = 2 × a × x
- Note u = 0: v² = 2 × 3.0 × 24
- Simplify: v² = 144
- Take square root: v = 12 m/s
Answer: 12 m/s
⚠ Both velocities are squared. A common error is squaring only v.
#24F = (mv − mu) / t
Given HT only §2.26
Force from change in momentum: force = change in momentum ÷ time
F = Force (N)
m = Mass (kg)
v = Final velocity (m/s)
u = Initial velocity (m/s)
t = Time of contact (s)
💡 Crumple zones and airbags work by lengthening t to reduce F for the same Δ(mv).
📝 Worked example
A 0.10 kg ball changes velocity from 5.0 m/s to 0 m/s in 0.040 s when caught. Find the average force.
- Find Δp = m(v − u): Δp = 0.10 × (0 − 5.0) = −0.50 kg m/s
- Write the formula: F = Δp / t
- Substitute: F = −0.50 / 0.040
- Calculate: F = −12.5 N (magnitude 12.5 N opposing motion)
Answer: −12.5 N (magnitude 12.5 N)
⚠ Use signed velocities. If the ball reverses direction, Δv is bigger than just the speed change.
Topic 3 — Conservation of energy
#6ΔGPE = m × g × Δh
Memorise §3.1, 8.8
Change in gravitational PE: change in GPE = mass × gravitational field strength × change in vertical height
ΔGPE = Change in GPE (J)
m = Mass (kg)
g = Gravitational field strength (N/kg (10 on Earth))
Δh = Vertical height gained (m)
💡 Δh is the VERTICAL height, not the path length up a ramp.
📝 Worked example
A 0.40 kg ball is lifted 2.5 m straight up. Calculate the change in GPE. (g = 10 N/kg)
- Write the formula: ΔGPE = m × g × Δh
- Substitute: ΔGPE = 0.40 × 10 × 2.5
- Calculate: ΔGPE = 10 J
Answer: 10 J
⚠ Reference level matters — GPE is measured relative to a chosen baseline (usually the ground).
#7KE = ½ × m × v²
Memorise §3.2, 8.9
Kinetic energy: kinetic energy = ½ × mass × (speed)²
KE = Kinetic energy (J)
m = Mass (kg)
v = Speed (m/s)
💡 Doubling speed quadruples KE (because of the v²). This is why crash damage scales sharply with speed.
📝 Worked example
A 60 kg cyclist rides at 7.0 m/s. Calculate her kinetic energy.
- Write the formula: KE = ½ × m × v²
- Square v first: v² = 7.0² = 49 m²/s²
- Substitute: KE = 0.5 × 60 × 49
- Calculate: KE = 1470 J
Answer: 1470 J (≈ 1.5 kJ)
⚠ Square v BEFORE multiplying. A common slip is to compute ½mv then square.
#8efficiency = useful energy out / total energy in
Memorise §3.11, 8.15
Efficiency: efficiency = useful energy transferred by the device / total energy supplied to the device
useful energy out = Useful output energy (J (or any matching energy unit))
total energy in = Total input energy (J)
💡 Efficiency is a ratio (no units). Multiply by 100 to express as a percentage. Always ≤ 1.
📝 Worked example
A motor takes in 500 J of electrical energy and produces 350 J of useful kinetic energy. Calculate the efficiency.
- Write the formula: efficiency = useful E out / total E in
- Substitute: = 350 / 500
- Calculate: = 0.70 (or 70%)
Answer: 0.70 or 70%
⚠ Efficiency cannot exceed 100%. If your answer is > 1, you've inverted the fraction.
Topic 4 — Waves
#9v = f × λ
Memorise §4.6
Wave speed (frequency form): wave speed = frequency × wavelength
v = Wave speed (m/s)
f = Frequency (Hz)
λ = Wavelength (m)
💡 Same wave speed → frequency up means wavelength down. Light in vacuum is always 3 × 10⁸ m/s.
📝 Worked example
A sound wave has frequency 320 Hz and wavelength 1.0 m. Calculate the wave speed.
- Write the formula: v = f × λ
- Substitute: v = 320 × 1.0
- Calculate: v = 320 m/s
Answer: 320 m/s
⚠ If frequency is given in kHz or MHz, convert to Hz first.
#10v = x / t
Memorise §4.6
Wave speed (distance form): wave speed = distance ÷ time
v = Wave speed (m/s)
x = Distance travelled by the wave (m)
t = Time taken (s)
💡 This treats the wave as a moving object. Useful for time-of-flight calculations (e.g. echo distances).
📝 Worked example
An ultrasound pulse takes 0.20 s to travel 300 m through water. Calculate its speed.
- Write the formula: v = x / t
- Substitute: v = 300 / 0.20
- Calculate: v = 1500 m/s
Answer: 1500 m/s
⚠ For echo problems (sonar, ultrasound), use the round-trip distance — out and back.
Topic 8 — Energy — forces doing work
#11E = F × d
Memorise §8.6
Work done: work done = force × distance moved in the direction of the force
E = Work done (= energy transferred) (J)
F = Force (N)
d = Distance moved in direction of force (m)
💡 1 J = 1 N moved through 1 m. Force perpendicular to motion does no work.
📝 Worked example
A worker pushes a 40 N force on a crate for 6.0 m in the same direction. Calculate the work done.
- Write the formula: E = F × d
- Substitute: E = 40 × 6.0
- Calculate: E = 240 J
Answer: 240 J
⚠ Use distance moved in the SAME direction as the force, not the diagonal path length.
#12P = E / t
Memorise §8.13
Power (from work): power = work done ÷ time taken
P = Power (W)
E = Work done (J)
t = Time (s)
💡 1 watt = 1 joule per second. Power is the rate of doing work.
📝 Worked example
A motor does 900 J of work in 30 s. Calculate its power output.
- Write the formula: P = E / t
- Substitute: P = 900 / 30
- Calculate: P = 30 W
Answer: 30 W
⚠ Two cars can do the same work; the more powerful one does it faster (smaller t).
Topic 9 — Forces and their effects
#13moment = F × d
Memorise §9.7P
Moment of a force: moment of a force = force × distance normal to the direction of the force
moment = Moment (Nm)
F = Force (N)
d = Perpendicular distance from pivot to line of action of the force (m)
💡 d must be PERPENDICULAR to the force. If the force is at 90° to the lever, just use the lever length.
📝 Worked example
A 25 N force is applied perpendicular to a spanner 0.18 m from the nut. Calculate the moment.
- Write the formula: moment = F × d
- Substitute: moment = 25 × 0.18
- Calculate: moment = 4.5 Nm
Answer: 4.5 Nm
⚠ If the force is applied at an angle, you must use the perpendicular component of distance, which is shorter.
Topic 10 — Electricity and circuits
#14E = Q × V
Memorise §10.6
Energy transferred (E = QV): energy transferred = charge moved × potential difference
E = Energy transferred (J)
Q = Charge moved (C)
V = Potential difference (V)
💡 Volts measure joules per coulomb. So 1 V × 1 C = 1 J.
📝 Worked example
A charge of 15 C passes through a 9.0 V battery. How much energy is transferred?
- Write the formula: E = Q × V
- Substitute: E = 15 × 9.0
- Calculate: E = 135 J
Answer: 135 J
⚠ Q is charge in coulombs, NOT current. Use Q = It first if only current and time are given.
#15Q = I × t
Memorise §10.9
Charge: charge = current × time
Q = Charge (C)
I = Current (A)
t = Time (s)
💡 1 ampere = 1 coulomb per second. Current is the rate at which charge flows.
📝 Worked example
A 2.0 A current flows for 5.0 minutes. Calculate the charge transferred.
- Convert time to seconds: t = 5.0 × 60 = 300 s
- Write the formula: Q = I × t
- Substitute: Q = 2.0 × 300
- Calculate: Q = 600 C
Answer: 600 C
⚠ Always convert time to seconds first. Minutes or hours will give wrong answers.
#16V = I × R
Memorise §10.13
Potential difference (Ohm's law): potential difference = current × resistance
V = Potential difference (V)
I = Current (A)
R = Resistance (Ω)
💡 Use the 'VIR triangle': cover what you want — V on top, I × R underneath.
📝 Worked example
A 0.30 A current flows through a 20 Ω resistor. Calculate the pd across it.
- Write the formula: V = I × R
- Substitute: V = 0.30 × 20
- Calculate: V = 6.0 V
Answer: 6.0 V
⚠ Watch small currents — slipping a decimal is the most common error here.
#17P = E / t
Memorise §10.29
Power (from energy): power = energy transferred ÷ time taken
P = Power (W)
E = Energy transferred (J)
t = Time (s)
💡 Same formula as P = work / time (8.13) — Edexcel lists it twice because power applies both to mechanics and electricity.
📝 Worked example
A heater transfers 120 000 J of energy in 60 s. Calculate its power.
- Write the formula: P = E / t
- Substitute: P = 120 000 / 60
- Calculate: P = 2000 W
Answer: 2000 W (2.0 kW)
⚠ Use time in seconds, not minutes — particularly easy to slip with electrical 'kWh' contexts.
#18P = I × V
Memorise §10.31
Electrical power (P = I × V): electrical power = current × potential difference
P = Power (W)
I = Current (A)
V = Potential difference (V)
💡 Use when voltage and current are known. UK mains is 230 V — multiply by current to get appliance power.
📝 Worked example
A 230 V iron draws 5.0 A. Calculate its power.
- Write the formula: P = I × V
- Substitute: P = 5.0 × 230
- Calculate: P = 1150 W
Answer: 1150 W (1.15 kW)
⚠ P = IV gives watts directly; if you need joules, multiply by time.
#19P = I² × R
Memorise §10.31
Electrical power (P = I² × R): electrical power = current squared × resistance
P = Power (W)
I = Current (A)
R = Resistance (Ω)
💡 Use when current and resistance are known (no voltage). Doubling current quadruples power dissipated.
📝 Worked example
A 0.40 A current flows through a 5.0 Ω resistor. Calculate the power dissipated.
- Write the formula: P = I² × R
- Square the current: I² = 0.40² = 0.16
- Substitute: P = 0.16 × 5.0
- Calculate: P = 0.80 W
Answer: 0.80 W
⚠ Square the current BEFORE multiplying by resistance — not after.
#25E = I × V × t
Given §10.27
Energy (E = IVt): energy transferred = current × potential difference × time
E = Energy transferred (J)
I = Current (A)
V = Potential difference (V)
t = Time (s)
💡 This is P × t expanded into electrical terms. Use when you have I, V, and t directly.
📝 Worked example
A 230 V appliance draws 3.0 A for 120 s. How much electrical energy does it transfer?
- Write the formula: E = I × V × t
- Substitute: E = 3.0 × 230 × 120
- Calculate: E = 82 800 J
Answer: 82 800 J (≈ 83 kJ)
⚠ Watch the time unit. Seconds give joules; hours would give Wh (which isn't asked for here).
Topic 12 — Magnetism and the motor effect
#26F = B × I × l
Given HT only §12.13
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 90° to the field. Use Fleming's left-hand rule for direction.
📝 Worked example
A 0.25 m wire carries 2.0 A through a 0.30 T field, at right angles. Calculate the force.
- Write the formula: F = B × I × l
- Substitute: F = 0.30 × 2.0 × 0.25
- Calculate: F = 0.15 N
Answer: 0.15 N
⚠ If the wire is parallel to the field, force is zero. The equation gives maximum force, at 90°.
Topic 13 — Electromagnetic induction
#27Vₚ / Vₛ = Nₚ / Nₛ
Given HT only §13.7P
Transformer turns ratio: primary pd / secondary pd = primary turns / secondary turns
Vₚ = Primary (input) pd (V)
Vₛ = Secondary (output) pd (V)
Nₚ = Number of turns on primary coil ((count))
Nₛ = Number of turns on secondary coil ((count))
💡 More turns on the output side = step-up. Fewer turns = step-down.
📝 Worked example
A transformer with 800 primary turns and 200 secondary turns is connected to a 240 V supply. Find the secondary pd.
- Rearrange for Vₛ: Vₛ = Vₚ × (Nₛ / Nₚ)
- Substitute: Vₛ = 240 × (200 / 800)
- Calculate: Vₛ = 60 V
Answer: 60 V (step-down)
⚠ Match the ratios: primary on top with primary on top — don't cross them.
#28Vₚ × Iₚ = Vₛ × Iₛ
Given §13.10
Transformer power (ideal): primary pd × primary current = secondary pd × secondary current (100% efficient)
Vₚ = Primary pd (V)
Iₚ = Primary current (A)
Vₛ = Secondary pd (V)
Iₛ = Secondary current (A)
💡 Steps voltage up → steps current down (and vice versa). Total power is conserved if 100% efficient.
📝 Worked example
An ideal transformer with primary pd 240 V, secondary pd 12 V, and secondary current 5.0 A. Find the primary current.
- Write the law: Vₚ × Iₚ = Vₛ × Iₛ
- Substitute: 240 × Iₚ = 12 × 5.0
- Simplify: 240 × Iₚ = 60
- Solve for Iₚ: Iₚ = 60 / 240 = 0.25 A
Answer: 0.25 A
⚠ If voltage steps down, current must step UP — they never both change in the same direction.
Topic 14 — Particle model
#20ρ = m / V
Memorise §14.2
Density: density = mass ÷ volume
ρ = Density (rho) (kg/m³ (or g/cm³))
m = Mass (kg)
V = Volume (m³)
💡 Water is 1000 kg/m³ (= 1 g/cm³). Anything denser sinks; less dense floats.
📝 Worked example
A metal sample has a mass of 540 g and a volume of 200 cm³. Calculate its density in g/cm³.
- Write the formula: ρ = m / V
- Substitute: ρ = 540 / 200
- Calculate: ρ = 2.7 g/cm³
Answer: 2.7 g/cm³
⚠ Units must match: kg with m³, or g with cm³. Mixing them is a frequent slip.
#29ΔQ = m × c × Δθ
Given §14.8
Specific heat capacity: change in thermal energy = mass × specific heat capacity × change in temperature
ΔQ = 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). High SHC = takes more energy to heat — why water is good for radiators.
📝 Worked example
How much energy is needed to raise 2.0 kg of water by 25 °C? (c = 4200 J/(kg °C))
- Write the formula: ΔQ = m × c × Δθ
- Substitute: ΔQ = 2.0 × 4200 × 25
- Calculate: ΔQ = 210 000 J
Answer: 210 000 J (210 kJ)
⚠ Δθ is the CHANGE, not the final temperature. Going 20 °C → 45 °C means Δθ = 25, not 45.
#30Q = m × L
Given §14.9
Specific latent heat: thermal energy for a change of state = mass × specific latent heat
Q = Thermal energy (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 0.20 kg of ice at 0 °C? (L_f of ice = 334 000 J/kg)
- Write the formula: Q = m × L
- Substitute: Q = 0.20 × 334 000
- Calculate: Q = 66 800 J
Answer: 66 800 J (≈ 67 kJ)
⚠ During a change of state, temperature is constant — don't also apply ΔQ = mcΔθ in parallel.
#31P₁ × V₁ = P₂ × V₂
Given §14.19P
Gas law (Boyle's law): for a fixed mass of gas at constant temperature, pressure × volume = constant
P₁ = Initial pressure (Pa (any pressure unit, must match P₂))
V₁ = Initial volume (m³ (any volume unit, must match V₂))
P₂ = Final pressure (Pa)
V₂ = Final volume (m³)
💡 Halve the volume at constant T → pressure doubles. Only valid for a fixed mass at constant T.
📝 Worked example
A gas at 200 kPa occupies 0.040 m³. It is compressed at constant temperature to 0.010 m³. Find the new pressure.
- Write the law: P₁ × V₁ = P₂ × V₂
- Substitute: 200 × 0.040 = P₂ × 0.010
- Simplify: 8.0 = P₂ × 0.010
- Solve for P₂: P₂ = 8.0 / 0.010 = 800 kPa
Answer: 800 kPa
⚠ Units must be consistent on both sides (Pa with Pa, or kPa with kPa). They don't have to be SI.
Topic 15 — Forces and matter
#21F = k × x
Memorise §15.3
Force on a spring: force exerted on a spring = spring constant × extension
F = Force (N)
k = Spring constant (N/m)
x = Extension (or compression) (m)
💡 Valid only up to the limit of proportionality. Convert extension to metres first.
📝 Worked example
A spring (k = 80 N/m) extends 0.15 m. What's the force on it?
- Write the formula: F = k × x
- Substitute: F = 80 × 0.15
- Calculate: F = 12 N
Answer: 12 N
⚠ Extension must be in metres. 15 cm = 0.15 m (not 15).
#22P = F / A
Memorise §15.11P
Pressure: pressure = force normal to surface ÷ area of surface
P = Pressure (Pa)
F = Force normal to the surface (N)
A = Area (m²)
💡 Same force, smaller area = bigger pressure. Sharp knives cut by concentrating force into tiny area.
📝 Worked example
A 400 N force is spread over a 0.20 m² surface. Calculate the pressure.
- Write the formula: P = F / A
- Substitute: P = 400 / 0.20
- Calculate: P = 2000 Pa
Answer: 2000 Pa (2 kPa)
⚠ Area must be in m². Convert from cm² by dividing by 10 000.
#32E = ½ × k × x²
Given §15.4
Elastic potential energy: energy transferred in stretching = 0.5 × spring constant × (extension)²
E = Elastic potential energy (J)
k = Spring constant (N/m)
x = Extension (or compression) (m)
💡 Doubling extension quadruples energy stored (because of x²). Valid up to limit of proportionality only.
📝 Worked example
A spring (k = 150 N/m) is stretched 0.080 m. How much elastic PE is stored?
- Write the formula: E = ½ × k × x²
- Square x first: x² = 0.080² = 0.0064 m²
- Substitute: E = 0.5 × 150 × 0.0064
- Calculate: E = 0.48 J
Answer: 0.48 J
⚠ Square the extension BEFORE multiplying — and remember to convert to metres first.
#33P = h × ρ × g
Given HT only §15.14P
Pressure in a fluid column: pressure due to a column of liquid = height of column × density × gravitational field strength
P = Pressure due to the column (Pa)
h = Vertical height of the column (m)
ρ = Density of the liquid (kg/m³)
g = Gravitational field strength (N/kg (10 N/kg on Earth))
💡 Depth, not total volume, determines pressure. 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 3.0 m. (ρ_water = 1000 kg/m³, g = 10 N/kg)
- Write the formula: P = h × ρ × g
- Substitute: P = 3.0 × 1000 × 10
- Calculate: P = 30 000 Pa
Answer: 30 000 Pa (30 kPa)
⚠ h is the vertical depth from the surface, not horizontal distance or volume of liquid.
Equation list and topic groupings are based on the Pearson Edexcel GCSE Physics (1PH0) specification, Issue 4 (March 2024), Appendix 2. Verified against the official Equation Booklet insert provided in 1PH0/1H past papers. This is the active specification for May 2026 exams. Worked examples and explanations are original LumiExams content.
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LumiExams.com | Edexcel GCSE Physics (1PH0) | Spec Issue 4 (2024-03)
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