IGCSE Physics Formula Practice
Worked Examples for Every Formula | Cambridge IGCSE Physics (0625)
Master calculations with step-by-step solutions
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Worked examples for all 37 physics formulas + 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 - marks are awarded for method, not just answers
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 = d / t
Worked Example:
A cyclist travels 240 metres in 20 seconds. Calculate the cyclist's speed.
v = d / 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
Acceleration measures how quickly velocity changes. A negative value means deceleration.
Acceleration = Change in Velocity ÷ Time
a = (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 = 10 N/kg).
Show Answer
W = 1.5 × 10 = 15 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)
The most important force equation. Use it to find force, mass, or acceleration.
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 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)
Springs extend proportionally to force applied (within the elastic limit). k is a constant 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 × Distance from Pivot
M = F × d
Worked Example:
A 50 N force is applied 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
Force = rate of change of momentum. Longer impact time = smaller force (why we use airbags and crumple zones).
Force = Change in Momentum ÷ Time
F = Δp / t
Worked Example:
A 0.4 kg ball moving at 10 m/s is caught in 0.1 s. Calculate the average force.
Δp = mv - mu = 0 - (0.4 × 10) = -4 kg m/s
F = Δp / t = 4 / 0.1
F = 40 N
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 force 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
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
Energy stored by an object's position in a gravitational field. Higher = more GPE.
GPE = Mass × g × Height
ΔEₚ = m × g × Δh
Worked Example:
A 5 kg box is lifted 3 m. Calculate the gain in gravitational PE. (g = 10 N/kg)
Eₚ = m × g × h
Eₚ = 5 × 10 × 3
Eₚ = 150 J
Practice Problem:
A ball is dropped from 20 m. Using energy conservation, calculate its speed just before hitting the ground. (g = 10 N/kg)
Show Answer
PE lost = KE gained → mgh = ½mv² → gh = ½v²
v² = 2gh = 2×10×20 = 400
v = 20 m/s
Energy Conservation: GPE ↔ Kinetic Energy
When objects fall, GPE converts to KE. The mass cancels out - all objects fall at the same rate!
GPE lost = KE gained (and vice versa)
mgh = ½mv²
Worked Example:
A 2 kg ball is dropped from a height of 45 m. Calculate its speed just before hitting the ground. (g = 10 N/kg)
Step 1: GPE at top = KE at bottom
mgh = ½mv²
Step 2: Cancel mass (m) from both sides
gh = ½v²
Step 3: Rearrange for v²
v² = 2gh = 2 × 10 × 45 = 900
Step 4: Take square root
v = √900 = 30 m/s
Practice Problem:
A roller coaster car (mass 500 kg) starts from rest at the top of a 25 m hill. Calculate: (a) its GPE at the top, (b) its speed at the bottom of the hill. (g = 10 N/kg)
Show Answer
(a) GPE = mgh = 500 × 10 × 25 = 125,000 J
(b) GPE = KE → 125,000 = ½ × 500 × v²
v² = 125,000 × 2 / 500 = 500
v = √500 = 22.4 m/s
Calculating Work Done
Work is energy transferred by a force. Work done = energy transferred (in joules).
Work = Force × Distance
W = F × d
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
No machine is 100% efficient - some energy is always wasted (usually as heat).
Efficiency = (Useful Output ÷ Total Input) × 100%
η = (useful out / total in) × 100%
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: Pa (pascals) = 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 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 Pa = 3 MPa
Calculating Pressure in Liquids
Liquid pressure increases with depth. It depends on depth, density, and gravity - not the shape of the container.
Pressure = Density × g × Depth
Δp = ρ × g × Δh
Worked Example:
Calculate the pressure due to water at a depth of 10 m. (ρ = 1000 kg/m³, g = 10 N/kg)
Δp = ρ × g × h
Δp = 1000 × 10 × 10
Δp = 100,000 Pa = 100 kPa
Practice Problem:
A diver experiences a pressure of 300,000 Pa due to the water. Calculate the depth. (ρ = 1000 kg/m³, g = 10 N/kg)
Show Answer
h = Δp / (ρ × g) = 300000 / (1000 × 10) = 30 m
7. Thermal Physics
Heat energy causes temperature changes OR state changes, but not both at the same time.
Calculating Energy for Temperature Change
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
Calculating Energy for State Change
Latent heat (L) is energy for melting (Lf) or boiling (Lv) - temperature stays constant during state change.
Energy = Mass × Specific Latent Heat
E = m × L
Worked Example:
Calculate the energy needed to melt 0.5 kg of ice at 0°C.
(Lf = specific latent heat of fusion for ice = 334,000 J/kg)
E = m × L
E = 0.5 × 334,000
E = 167,000 J = 167 kJ
Practice Problem:
2.3 MJ of energy is used to vaporise water at 100°C. Calculate the mass of water vaporised.
(Lv = specific latent heat of vaporisation for water = 2,300,000 J/kg)
Show Answer
m = E / L = 2,300,000 / 2,300,000 = 1 kg
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)
Light bends when entering a denser medium. The refractive index tells you how much it bends.
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)
The most important 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
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
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: components share current. Parallel: components share voltage. 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
In parallel, total resistance is always less than the smallest resistor. Current has more paths.
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
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