IGCSE Physics Formula Practice
Worked Examples for Key Topics | Cambridge IGCSE Physics (0625)
Master calculations with step-by-step solutions
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Worked examples by topic + printable PDF
Calculating Speed
Speed = Distance ÷ Time
How to Use This Guide
- Read the formula and understand what each variable means
- Study the example - note how values are substituted
- Try the practice problem before looking at the answer
- Check your working - write the equation, the substitution, then the answer with its unit
- Items marked (Extended) are Supplement content. Core candidates (Papers 1 and 3) are not tested on them
1. Motion
Motion describes how objects move through space and time. These formulas help you calculate speed, velocity, and acceleration.
Calculating Speed
Speed tells you how fast something is moving. It's the most basic motion calculation.
Speed = Distance ÷ Time
v = s / t
Worked Example:
A cyclist travels 240 metres in 20 seconds. Calculate the cyclist's speed.
v = s / t
v = 240 / 20
v = 12 m/s
Practice Problem:
A train travels 1500 m in 50 s. Calculate the speed of the train.
Show Answer
v = 1500 / 50 = 30 m/s
Calculating Acceleration (Extended)
Acceleration measures how quickly velocity changes. A negative value means deceleration.
Acceleration = Change in Velocity ÷ Time Taken
a = Δv / Δt = (v - u) / t
Worked Example:
A car accelerates from rest to 25 m/s in 5 seconds. Calculate its acceleration.
u = 0 m/s (rest), v = 25 m/s, t = 5 s
a = (v - u) / t
a = (25 - 0) / 5
a = 5 m/s²
Practice Problem:
A motorcycle slows down from 30 m/s to 10 m/s in 4 seconds. Calculate the deceleration.
Show Answer
a = (10 - 30) / 4 = -20 / 4 = -5 m/s² (deceleration = 5 m/s²)
2. Mass, Weight & Density
Mass stays constant, but weight depends on gravity. Density relates mass to volume - crucial for floating/sinking problems.
Calculating Weight
Weight is the gravitational force on an object. It changes with location (Earth vs Moon).
Weight = Mass × Gravitational Field Strength
W = m × g
Worked Example:
Calculate the weight of a 70 kg person on Earth (g = 9.8 N/kg).
W = m × g
W = 70 × 9.8
W = 686 N
Practice Problem:
A bag of flour has a mass of 1.5 kg. Calculate its weight on Earth (g = 9.8 N/kg).
Show Answer
W = 1.5 × 9.8 = 14.7 N
Calculating Density
Density tells you how compact a material is. Objects less dense than water float.
Density = Mass ÷ Volume
ρ = m / V
Worked Example:
A metal block has a mass of 540 g and a volume of 200 cm³. Calculate its density.
ρ = m / V
ρ = 540 / 200
ρ = 2.7 g/cm³
Practice Problem:
A liquid has a mass of 800 g and occupies a volume of 1000 cm³. Calculate its density. Is this liquid water?
Show Answer
ρ = 800 / 1000 = 0.8 g/cm³ (Not water - less dense than water which is 1 g/cm³)
3. Forces
Forces cause objects to accelerate, stretch, or turn. These formulas link force to motion and deformation.
Calculating Force (Newton's 2nd Law) (Extended)
A key force equation. Use it to find resultant force, mass, or acceleration. The force and the acceleration are in the same direction.
Resultant Force = Mass × Acceleration
F = m × a
Worked Example:
A 1200 kg car accelerates at 2.5 m/s². Calculate the resultant force.
F = m × a
F = 1200 × 2.5
F = 3000 N
Practice Problem:
A resultant force of 500 N acts on a trolley of mass 25 kg. Calculate the acceleration.
Show Answer
a = F / m = 500 / 25 = 20 m/s²
Calculating Spring Extension (Hooke's Law) (Extended)
Up to the limit of proportionality, extension is proportional to the force applied. The spring constant k is the force per unit extension (k = F / x), and is fixed for each spring.
Force = Spring Constant × Extension
F = k × x
Worked Example:
A spring has a spring constant of 40 N/m. Calculate the force needed to extend it by 0.15 m.
F = k × x
F = 40 × 0.15
F = 6 N
Practice Problem:
A spring extends by 8 cm when a force of 20 N is applied. Calculate the spring constant.
Show Answer
k = F / x = 20 / 0.08 = 250 N/m
Calculating Moments
Moments cause rotation. For balance problems, clockwise moments = anticlockwise moments.
Moment = Force × Perpendicular Distance from Pivot
M = F × d
Worked Example:
A 50 N force is applied at right angles to a spanner, 0.3 m from the pivot. Calculate the moment.
M = F × d
M = 50 × 0.3
M = 15 Nm
Practice Problem:
A seesaw is balanced. A child weighing 400 N sits 2 m from the pivot. How far from the pivot should a 500 N adult sit?
Show Answer
Moment₁ = Moment₂ → 400 × 2 = 500 × d → d = 800/500 = 1.6 m
4. Momentum (Extended)
Momentum is conserved in collisions. Total momentum before = total momentum after. Extended content only.
Calculating Momentum
Momentum combines mass and velocity into one quantity. It has direction (a vector).
Momentum = Mass × Velocity
p = m × v
Worked Example:
A 1500 kg car travels at 20 m/s. Calculate its momentum.
p = m × v
p = 1500 × 20
p = 30,000 kg m/s
Practice Problem:
A 0.5 kg ball has a momentum of 15 kg m/s. Calculate its velocity.
Show Answer
v = p / m = 15 / 0.5 = 30 m/s
Calculating Force from Momentum Change
Resultant force = rate of change of momentum. Longer impact time = smaller force (why we use airbags and crumple zones).
Resultant Force = Change in Momentum ÷ Time Taken
F = Δp / Δt
Worked Example:
A 0.4 kg ball moving at 10 m/s is caught in 0.1 s. Calculate the average force on the ball.
Δp = mv - mu = 0 - (0.4 × 10) = -4 kg m/s
F = Δp / Δt = -4 / 0.1 = -40 N
Average force = 40 N
The minus sign shows the force acts opposite to the ball's motion.
Practice Problem:
A 60 kg person jumps and lands on a hard floor in 0.02 s vs. a soft mat in 0.5 s. If their velocity just before landing is 4 m/s, calculate the average resultant force on them in each case.
Show Answer
Δp = 60 × 4 = 240 kg m/s
Hard floor: F = 240/0.02 = 12,000 N
Soft mat: F = 240/0.5 = 480 N
5. Energy, Work & Power
Energy cannot be created or destroyed, only transferred. These formulas track energy changes in systems.
Calculating Kinetic Energy (Extended)
Energy due to motion. Doubles when velocity doubles? No - it quadruples!
Kinetic Energy = ½ × Mass × Velocity²
Eₖ = ½mv²
Worked Example:
Calculate the kinetic energy of a 2 kg ball moving at 5 m/s.
Eₖ = ½ × m × v²
Eₖ = ½ × 2 × 5²
Eₖ = ½ × 2 × 25
Eₖ = 25 J
Practice Problem:
A 1000 kg car has kinetic energy of 125,000 J. Calculate its speed.
Show Answer
v² = 2Eₖ/m = 2×125000/1000 = 250
v = √250 = 15.8 m/s
Calculating Gravitational Potential Energy (Extended)
Energy stored by an object's position in a gravitational field. Higher = more GPE.
Change in GPE = Mass × g × Change in Height
ΔEₚ = m × g × Δh
Worked Example:
A 5 kg box is lifted 3 m. Calculate the gain in gravitational PE. (g = 9.8 N/kg)
ΔEₚ = m × g × Δh
ΔEₚ = 5 × 9.8 × 3
ΔEₚ = 147 J
Practice Problem:
A ball is dropped from 20 m. Using energy conservation, calculate its speed just before hitting the ground. Ignore air resistance. (g = 9.8 N/kg)
Show Answer
PE lost = KE gained → mgΔh = ½mv² → gΔh = ½v²
v² = 2gΔh = 2×9.8×20 = 392
v = √392 = 19.8 m/s
Energy Conservation: GPE ↔ Kinetic Energy (Extended)
When objects fall, GPE converts to KE. The mass cancels out - if air resistance is ignored, all objects fall at the same rate!
GPE lost = KE gained (and vice versa)
mgΔh = ½mv²
Worked Example:
A 2 kg ball is dropped from a height of 45 m. Calculate its speed just before hitting the ground. Ignore air resistance. (g = 9.8 N/kg)
Step 1: GPE lost = KE gained
mgΔh = ½mv²
Step 2: Cancel mass (m) from both sides
gΔh = ½v²
Step 3: Rearrange for v²
v² = 2gΔh = 2 × 9.8 × 45 = 882
Step 4: Take square root
v = √882 = 29.7 m/s
Practice Problem:
A roller coaster car (mass 500 kg) starts from rest at the top of a 25 m hill. Calculate: (a) the GPE it loses going down to the bottom of the hill, (b) its speed at the bottom of the hill. Ignore friction and air resistance. (g = 9.8 N/kg)
Show Answer
(a) ΔEₚ = mgΔh = 500 × 9.8 × 25 = 122,500 J
(b) GPE lost = KE gained → 122,500 = ½ × 500 × v²
v² = 122,500 × 2 / 500 = 490
v = √490 = 22.1 m/s
Calculating Work Done
Work is energy transferred by a force. Work done = energy transferred (in joules).
Work Done = Force × Distance Moved (in the direction of the force)
W = F × d = ΔE
Worked Example:
A person pushes a box with a force of 80 N over a distance of 5 m. Calculate the work done.
W = F × d
W = 80 × 5
W = 400 J
Practice Problem:
A crane does 50,000 J of work lifting a container 25 m. Calculate the lifting force.
Show Answer
F = W / d = 50000 / 25 = 2000 N
Calculating Power
Power is the rate of energy transfer. Measured in watts (W) = joules per second.
Power = Work Done ÷ Time
P = W / t
Worked Example:
A motor does 6000 J of work in 30 seconds. Calculate its power.
P = W / t
P = 6000 / 30
P = 200 W
Practice Problem:
A 60 W light bulb is on for 2 minutes. Calculate the energy used.
Show Answer
E = P × t = 60 × 120 = 7200 J
Calculating Efficiency (Extended)
No machine is 100% efficient - some energy is always wasted (usually as heat). Core candidates only need to understand efficiency in words.
Efficiency = (Useful Output ÷ Total Input) × 100%
η = (useful out / total in) × 100%
Use energy for both, or power for both
Worked Example:
A motor uses 500 J of electrical energy and produces 350 J of useful work. Calculate its efficiency.
η = (useful output / total input) × 100%
η = (350 / 500) × 100%
η = 70%
Practice Problem:
A power station has an efficiency of 40%. How much fuel energy is needed to produce 200 MJ of electrical energy?
Show Answer
Input = Output / Efficiency = 200 / 0.4 = 500 MJ
6. Pressure
Pressure is force spread over an area. Smaller area = higher pressure (why knives are sharp).
Calculating Pressure
The basic pressure formula. Units: N/m² or N/cm². Extended candidates also use the pascal: 1 Pa = 1 N/m².
Pressure = Force ÷ Area
p = F / A
Worked Example:
A box weighing 200 N has a base area of 0.5 m². Calculate the pressure on the floor.
p = F / A
p = 200 / 0.5
p = 400 N/m² (400 Pa)
Practice Problem:
A woman weighing 600 N wears stiletto heels with a combined heel area of 2 cm². Calculate the pressure under her heels.
Show Answer
A = 2 cm² = 0.0002 m²
p = 600 / 0.0002 = 3,000,000 N/m² (3 MPa)
Calculating Pressure in Liquids (Extended)
Liquid pressure increases with depth. It depends on depth, density, and gravity - not the shape of the container. Core candidates describe this in words; Extended candidates calculate it.
Change in Pressure = Density × g × Change in Depth
Δp = ρ × g × Δh
Worked Example:
Calculate the pressure due to water at a depth of 10 m. (ρ = 1000 kg/m³, g = 9.8 N/kg)
Δp = ρ × g × Δh
Δp = 1000 × 9.8 × 10
Δp = 98,000 Pa = 98 kPa
Practice Problem:
A diver experiences a pressure of 300,000 Pa due to the water. Calculate the depth. (ρ = 1000 kg/m³, g = 9.8 N/kg)
Show Answer
Δh = Δp / (ρ × g) = 300000 / (1000 × 9.8) = 30.6 m
7. Thermal Physics
Thermal energy supplied to a substance either raises its temperature, or melts or boils it with no change in temperature.
Calculating Energy for Temperature Change (Extended)
Specific heat capacity (c) tells you how much energy is needed to heat 1 kg by 1°C.
Energy = Mass × Specific Heat Capacity × Temp Change
ΔE = m × c × Δθ
Worked Example:
Calculate the energy needed to heat 2 kg of water from 20°C to 100°C. (c = 4200 J/(kg °C))
ΔE = m × c × Δθ
ΔE = 2 × 4200 × (100 - 20)
ΔE = 2 × 4200 × 80
ΔE = 672,000 J = 672 kJ
Practice Problem:
A 0.5 kg metal block is heated with 5000 J, and its temperature rises from 25°C to 50°C. Calculate the specific heat capacity.
Show Answer
c = ΔE / (m × Δθ) = 5000 / (0.5 × 25) = 400 J/(kg °C)
Melting and Boiling
Specific latent heat calculations (E = mL) are not part of the 2026-2028 syllabus. You do need to describe melting and boiling as energy going in with no change in temperature, and know that water melts at 0°C and boils at 100°C at standard atmospheric pressure (Core).
8. Waves & Light
Waves transfer energy without transferring matter. Light is an electromagnetic wave that can be reflected and refracted.
Calculating Wave Speed
The wave equation works for all waves - sound, light, water, and radio waves.
Wave Speed = Frequency × Wavelength
v = f × λ
Worked Example:
A wave has a frequency of 500 Hz and wavelength of 0.68 m. Calculate its speed.
v = f × λ
v = 500 × 0.68
v = 340 m/s
Practice Problem:
A radio wave travels at 3 × 10⁸ m/s with a frequency of 100 MHz. Calculate its wavelength.
Show Answer
λ = v / f = (3 × 10⁸) / (100 × 10⁶) = 3 m
Calculating Refractive Index (Snell's Law) (Extended)
Light changes direction when it passes at an angle from air into a material such as glass or water, because its speed changes. The refractive index tells you how much it bends (i is the angle in air, r the angle in the material).
Refractive Index = sin(incidence) ÷ sin(refraction)
n = sin i / sin r
Worked Example:
Light enters glass at 45° and refracts to 28°. Calculate the refractive index.
n = sin i / sin r
n = sin 45° / sin 28°
n = 0.707 / 0.469
n = 1.51
Practice Problem:
Light enters water (n = 1.33) at an angle of 50°. Calculate the angle of refraction.
Show Answer
sin r = sin i / n = sin 50° / 1.33 = 0.766 / 1.33 = 0.576
r = sin⁻¹(0.576) = 35.2°
9. Electricity
Current is the flow of charge. Voltage pushes it, resistance opposes it. These formulas connect them all.
Calculating Voltage (Ohm's Law)
A key electricity equation. Memorise the triangle: V at top, I and R at bottom.
Voltage = Current × Resistance
V = I × R
Worked Example:
A current of 2 A flows through a 6 Ω resistor. Calculate the voltage across it.
V = I × R
V = 2 × 6
V = 12 V
Practice Problem:
A 240 V supply is connected to a 60 Ω heater. Calculate the current.
Show Answer
I = V / R = 240 / 60 = 4 A
Calculating Electrical Power
Electrical power = rate of energy transfer. A 2000 W kettle uses 2000 J every second.
Power = Current × Voltage
P = I × V
Worked Example:
A kettle draws 10 A from a 230 V supply. Calculate its power.
P = I × V
P = 10 × 230
P = 2300 W = 2.3 kW
Practice Problem:
A 2000 W microwave operates on a 250 V supply. Calculate the current drawn.
Show Answer
I = P / V = 2000 / 250 = 8 A
Calculating Electrical Energy
Energy bills use kWh (kilowatt-hours): 1 kWh = 1 kW running for 1 hour = 3.6 MJ.
Energy = Power × Time
E = P × t (or E = I × V × t)
Worked Example:
A 100 W light bulb is on for 3 hours. Calculate the energy used in kWh.
E = P × t
E = 0.1 kW × 3 h
E = 0.3 kWh
Practice Problem:
Electricity costs $0.15 per kWh. Calculate the cost of running a 2 kW heater for 5 hours.
Show Answer
E = 2 × 5 = 10 kWh
Cost = 10 × $0.15 = $1.50
Calculating Power Loss in Cables (Extended)
Cables have resistance, so current causes heating. This is why we use high voltages for transmission.
Power Loss = Current² × Resistance
P = I² × R
Worked Example:
A current of 3 A flows through a 10 Ω resistor. Calculate the power dissipated.
P = I² × R
P = 3² × 10
P = 9 × 10
P = 90 W
Practice Problem:
Explain why high-voltage transmission lines reduce power loss. (Hint: If power transmitted is the same, how does doubling voltage affect current?)
Show Answer
Doubling V halves I (since P = IV is constant).
Power loss = I²R, so halving I gives ¼ the power loss.
Higher voltage → lower current → much less power loss in cables
10. Resistance & Circuits
Series: the same current flows through each component and the p.d. is shared. Parallel: each branch has the same p.d. and the current splits. Know how to combine resistances.
Calculating Series Resistance
In series, resistances simply add up. Total resistance is always bigger than any single resistor.
Total Resistance = R₁ + R₂ + R₃ + ...
Rₜ = R₁ + R₂ + ...
Worked Example:
Three resistors of 4 Ω, 6 Ω, and 10 Ω are connected in series. Calculate the total resistance.
Rₜ = R₁ + R₂ + R₃
Rₜ = 4 + 6 + 10
Rₜ = 20 Ω
Practice Problem:
A 12 V battery is connected to three 4 Ω resistors in series. Calculate the current.
Show Answer
Rₜ = 4 + 4 + 4 = 12 Ω
I = V / R = 12 / 12 = 1 A
Calculating Parallel Resistance (Extended)
In parallel, total resistance is always less than the smallest resistor. Current has more paths. Core candidates need to know that the combined resistance of two resistors in parallel is less than either one; Extended candidates calculate it.
1/Total = 1/R₁ + 1/R₂
1/Rₜ = 1/R₁ + 1/R₂
Worked Example:
Two resistors of 6 Ω and 3 Ω are connected in parallel. Calculate the total resistance.
1/Rₜ = 1/R₁ + 1/R₂
1/Rₜ = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2
Rₜ = 2 Ω
Practice Problem:
Two 10 Ω resistors are connected in parallel. Calculate the total resistance.
Show Answer
1/Rₜ = 1/10 + 1/10 = 2/10 = 1/5
Rₜ = 5 Ω
(Shortcut: two equal resistors in parallel = half of one)
11. Transformers
Transformers change AC voltage using electromagnetic induction. More secondary turns = step-up, fewer = step-down.
Calculating Transformer Voltage
The turns ratio equals the voltage ratio. Cross-multiply to solve for unknowns.
Primary Voltage / Secondary Voltage = Primary Turns / Secondary Turns
Vₚ / Vₛ = Nₚ / Nₛ
Worked Example:
A transformer has 500 turns on the primary and 50 turns on the secondary. If the input voltage is 230 V, calculate the output voltage.
Vₚ/Vₛ = Nₚ/Nₛ
230/Vₛ = 500/50
230/Vₛ = 10
Vₛ = 230/10
Vₛ = 23 V (step-down transformer)
Practice Problem:
A step-up transformer increases 110 V to 11,000 V. If the primary coil has 100 turns, how many turns does the secondary have?
Show Answer
Nₛ/Nₚ = Vₛ/Vₚ
Nₛ/100 = 11000/110 = 100
Nₛ = 10,000 turns
More Formula Sheets
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Frequently Asked Questions
How do I use physics formulas in IGCSE exams?
Follow these steps: 1) Identify the correct formula, 2) Write out the formula, 3) Substitute the given values with units, 4) Calculate the answer, 5) Include the correct unit. Always show your working, so your method is clear even if you slip on the final number.
Which physics calculations should I practise first?
Start with the Core ones: v = s/t and R = V/I. Extended candidates add F = ma, Ek = ½mv² and ΔEp = mgΔh. Practise them until choosing and rearranging them is automatic.
How do I convert energy between GPE and KE?
Use energy conservation (the Ek and ΔEp equations are Extended): if no energy is lost to air resistance, the GPE lost equals the KE gained. So mgΔh = ½mv². The mass cancels, giving gΔh = ½v², so v = √(2gΔh), where Δh is the height fallen. This is useful for falling objects and roller coaster problems.
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