AS Physics Formula Sheet
Cambridge AS Physics (9702) | Key Equations by Topic
AS Level | Syllabus 2025-2027
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Key AS Equations
One from each of Topics 2-9Papers 1 and 2 print a short data and formulae page, but most AS equations are not on it. Each card says whether the equation is on the data sheet or one you need to learn. Subscribing unlocks the rest of the sheet, topic by topic, with worked examples and the printable PDF.
u = Initial velocity (m/s)
a = Acceleration (m/s²)
s = Displacement (m)
m = Mass (kg)
a = Acceleration (m/s²)
ρ = Density of the fluid (kg/m³)
g = Acceleration of free fall (m/s²)
Δh = Change in depth (m)
m = Mass (kg)
v = Speed (m/s)
F = Force (N)
A = Cross-sectional area (m²)
x = Extension (m)
L = Original length (m)
f = Frequency (Hz)
λ = Wavelength (m)
θ = Angle of the nth-order maximum
n = Order (0, 1, 2, ...)
λ = Wavelength (m)
V = Potential difference (V)
I = Current (A)
R = Resistance (Ω)
2. Kinematics
v = Speed (m/s)
s = Distance (m)
t = Time (s)
a = Acceleration (m/s²)
v = Final velocity (m/s)
u = Initial velocity (m/s)
t = Time (s)
Use when: you don't know final velocity
Worked Example:
A ball is thrown vertically upward at 20 m/s. Find its displacement after 3.0 s. (g = 9.81 m/s²)
u = 20 m/s, a = -9.81 m/s², t = 3.0 s
s = ut + ½at²
s = (20 × 3.0) + ½ × (-9.81) × 3.0²
s = 60 - 44.1
s = 15.9 m (above starting point)
Practice:
A car starts from rest and accelerates at 2.5 m/s² for 8.0 s. Find the distance travelled.
Show Answer
s = 0 + ½ × 2.5 × 8.0² = 80 m
Use when: time not given; great for "dropped" problems (u=0)
Worked Example:
A stone is dropped from a cliff 45 m high. Find its speed just before hitting the ground.
u = 0 (dropped), s = 45 m, a = 9.81 m/s²
v² = u² + 2as
v² = 0 + 2 × 9.81 × 45 = 882.9
v = 29.7 m/s
Practice:
A car brakes from 25 m/s to rest over 50 m. Find the deceleration.
Show Answer
a = (0 - 25²)/(2 × 50) = -625/100 = -6.25 m/s²
s = Displacement (m)
u = Initial velocity (m/s)
v = Final velocity (m/s)
t = Time (s)
Use when: you know both velocities and time
3. Dynamics
Vector quantity - direction matters in collisions
Worked Example:
A 1200 kg car travels at 15 m/s. Calculate its momentum.
m = 1200 kg, v = 15 m/s
p = mv = 1200 × 15
p = 18000 kg m/s
Practice:
A 0.45 kg football is kicked at 20 m/s. Find its momentum.
Show Answer
p = 0.45 × 20 = 9.0 kg m/s
F and a are always in the same direction
Worked Example:
A 800 kg car accelerates at 2.5 m/s². Find the resultant force.
m = 800 kg, a = 2.5 m/s²
F = ma = 800 × 2.5
F = 2000 N
Practice:
A 60 kg sprinter accelerates from rest to 10 m/s in 2.0 s. Find the average force.
Show Answer
a = 10/2 = 5 m/s², F = 60 × 5 = 300 N
F = Resultant force (N)
Δp = Change in momentum (kg m/s)
Δt = Time interval (s)
More general form of F=ma; use for collisions, rockets
W = Weight (N)
m = Mass (kg)
g = 9.81 m/s² (on Earth)
4. Forces, Density & Pressure
M = Moment (N m)
F = Force (N)
d = Perpendicular distance from the pivot to the line of action of the force (m)
ρ (rho) = Density (kg/m³)
m = Mass (kg)
V = Volume (m³)
p = Pressure (Pa = N/m²)
F = Force normal (perpendicular) to the surface (N)
A = Area (m²)
This gives pressure DIFFERENCE. Total = atmospheric + ρgh
Worked Example:
A diver is 25 m below the surface. Find the water pressure at this depth. (ρ = 1000 kg/m³)
ρ = 1000 kg/m³, g = 9.81 m/s², Δh = 25 m
Δp = ρgΔh = 1000 × 9.81 × 25
Δp = 245000 Pa = 245 kPa
Total pressure = 101 + 245 = 346 kPa
Practice:
Mercury has density 13600 kg/m³. Find the pressure at 0.76 m depth in a mercury barometer.
Show Answer
Δp = 13600 × 9.81 × 0.76 = 101400 Pa ≈ 1 atm
ρ is the density of the FLUID; V is the volume of fluid displaced (the immersed volume), not the volume of the object
Worked Example:
A 0.50 m³ block is fully submerged in water. Find the upthrust. (ρ = 1000 kg/m³)
ρ = 1000 kg/m³, g = 9.81 m/s², V = 0.50 m³
F = ρgV = 1000 × 9.81 × 0.50
F = 4905 N ≈ 4900 N
Practice:
A wooden block floats with 60% submerged. If V = 0.02 m³, find its weight.
Show Answer
Floating: W = Upthrust = 1000 × 9.81 × (0.6 × 0.02) = 118 N
5. Work, Energy & Power
W = Work (J)
F = Force (N)
s = Displacement (m)
Double velocity = 4× kinetic energy (v² relationship)
Worked Example:
A 1500 kg car travels at 20 m/s. Calculate its kinetic energy.
m = 1500 kg, v = 20 m/s
Eₖ = ½mv² = ½ × 1500 × 20²
Eₖ = ½ × 1500 × 400
Eₖ = 300000 J = 300 kJ
Practice:
A 70 kg runner has 2450 J of KE. Find their speed.
Show Answer
v² = 2Eₖ/m = 2×2450/70 = 70, v = 8.4 m/s
Worked Example:
A 2.0 kg ball is dropped from 5.0 m. Find its speed just before hitting the ground (energy method).
PE lost = KE gained
mgΔh = ½mv²
v² = 2gΔh = 2 × 9.81 × 5.0 = 98.1
v = 9.9 m/s
Practice:
A 500 kg lift rises 12 m. Find the work done against gravity.
Show Answer
W = ΔEₚ = 500 × 9.81 × 12 = 58860 J ≈ 59 kJ
P = Power (W)
W = Work done (J)
t = Time (s)
P = Power (W)
F = Force (N)
v = Velocity (m/s)
Derived from P = W/t = Fs/t = Fv. Gives the power at any instant; at constant speed, F equals the total resistive force
η (eta) = Efficiency (%)
Works with energy (J) or power (W)
6. Deformation of Solids
Only valid up to the limit of proportionality
Worked Example:
A spring extends by 8.0 cm when a 4.0 N force is applied. Find the spring constant.
F = 4.0 N, x = 8.0 cm = 0.080 m
k = F/x = 4.0/0.080
k = 50 N/m
Practice:
A spring with k = 200 N/m has a 5.0 kg mass hung on it. Find the extension.
Show Answer
F = mg = 5 × 9.81 = 49.05 N, x = F/k = 49.05/200 = 0.245 m = 24.5 cm
σ (sigma) = Stress (Pa or N/m²)
ε (epsilon) = Strain (no unit - it's a ratio)
ΔL = Extension (m)
L = Original length (m)
Within the limit of proportionality. Higher E = stiffer material. Gradient of the straight part of the stress-strain graph
Worked Example:
A 2.0 m copper wire (A = 1.0×10⁻⁶ m²) extends 1.6 mm under 80 N. Find Young modulus.
σ = F/A = 80/(1.0×10⁻⁶) = 8.0×10⁷ Pa
ε = ΔL/L = 0.0016/2.0 = 8.0×10⁻⁴
E = σ/ε = (8.0×10⁷)/(8.0×10⁻⁴)
E = 1.0×10¹¹ Pa = 100 GPa
Practice:
Steel has E = 2.0×10¹¹ Pa. A 3.0 m wire (A = 2.0×10⁻⁶ m²) is under 600 N. Find extension.
Show Answer
σ = 3×10⁸ Pa, ε = σ/E = 1.5×10⁻³, ΔL = εL = 1.5×10⁻³ × 3 = 4.5 mm
Eₚ = Elastic PE (J)
F = Force (N)
k = Spring constant (N/m)
x = Extension (m)
Area under force-extension graph
7. Waves
v = Wave speed (m/s)
f = Frequency (Hz)
λ (lambda) = Wavelength (m)
f = Frequency (Hz)
T = Period (s)
I = Intensity (W/m²)
P = Power (W)
A = Area (m²)
Double amplitude = 4× intensity
I = Transmitted intensity (W/m²)
I₀ = Incident intensity (W/m²)
θ = Angle between polarisation direction and filter axis
fₒ = observed frequency, fₛ = source frequency, v = speed of sound, vₛ = speed of source. Use (v - vₛ) approaching; (v + vₛ) receding
Worked Example:
An ambulance siren (fₛ = 500 Hz) approaches at 30 m/s. Find observed frequency. (v = 340 m/s)
Source approaching → use (v - vₛ)
fₒ = fₛv/(v - vₛ) = 500 × 340/(340 - 30)
fₒ = 170000/310
fₒ = 548 Hz (higher pitch)
Practice:
The same ambulance moves away at 30 m/s. Find the observed frequency.
Show Answer
fₒ = 500 × 340/(340 + 30) = 170000/370 = 459 Hz (lower pitch)
8. Superposition
λ = Wavelength (m)
a = Slit separation (m)
x = Fringe spacing (m)
D = Distance to screen (m)
d = 1/(lines per metre). Max order when sin θ = 1
Worked Example:
Light (λ = 600 nm) hits a grating with 300 lines/mm. Find the angle for the 2nd order maximum.
d = 1/(300000) = 3.33×10⁻⁶ m
λ = 600×10⁻⁹ m, n = 2
sin θ = nλ/d = 2 × 600×10⁻⁹ / 3.33×10⁻⁶
sin θ = 0.360
θ = 21.1°
Practice:
Find the maximum order visible for this grating and wavelength.
Show Answer
n_max = d/λ = 3.333×10⁻⁶/600×10⁻⁹ = 5.56 → n = 5 (max whole number)
9. Electricity
Q = Charge (C)
I = Current (A)
t = Time (s)
I = Current (A)
A = Cross-sectional area (m²)
n = Number density (m⁻³)
v = Drift velocity (m/s)
q = Charge per carrier (C)
V = Voltage (V)
W = Energy (J)
Q = Charge (C)
1 volt = 1 joule per coulomb
Worked Example:
A 12 V supply drives 0.50 A through a resistor. Find its resistance.
V = 12 V, I = 0.50 A
R = V/I = 12/0.50
R = 24 Ω
Practice:
A 470 Ω resistor has 9.0 V across it. Find the current in mA.
Show Answer
I = V/R = 9.0/470 = 0.0191 A = 19.1 mA
Use I²R for power loss in cables
Worked Example:
A kettle draws 10 A from 230 V. Find (a) power and (b) energy used in 3 minutes.
(a) P = VI = 230 × 10 = 2300 W
(b) E = Pt = 2300 × 180 = 414000 J = 414 kJ
Practice:
A 60 W bulb runs on 12 V. Find (a) current and (b) resistance.
Show Answer
(a) I = P/V = 60/12 = 5.0 A, (b) R = V/I = 12/5 = 2.4 Ω
R = Resistance (Ω)
ρ (rho) = Resistivity (Ω m)
L = Length (m)
A = Cross-sectional area (m²)
10. DC Circuits
Same current through each
Voltages add up
Same voltage across each
Currents add up
For 2 resistors: R = R₁R₂/(R₁+R₂) is often faster
ε = EMF (V)
I = Current (A)
R = External resistance (Ω)
r = Internal resistance (Ω)
V = Terminal voltage (V)
ε = EMF (V)
I = Current (A)
r = Internal resistance (Ω)
When I = 0, terminal p.d. = EMF
Use LDR/thermistor as R₁ or R₂ to make sensors
Worked Example:
A 12 V supply connects to R₁ = 4.0 kΩ and R₂ = 2.0 kΩ in series. Find V₂ (across R₂).
V = 12 V, R₁ = 4000 Ω, R₂ = 2000 Ω
V₂ = V × R₂/(R₁+R₂)
V₂ = 12 × 2000/(4000+2000)
V₂ = 12 × 2000/6000
V₂ = 4.0 V
Practice:
An LDR (R₁) has 10 kΩ in dark, 1 kΩ in light. With R₂ = 5 kΩ and 9 V supply, find V₂ in each condition.
Show Answer
Dark: V₂ = 9×5/(10+5) = 3.0 V. Light: V₂ = 9×5/(1+5) = 7.5 V
More AS Physics Resources
LumiExams.com | Cambridge AS Physics (9702) | Syllabus 2025-2027
Frequently Asked Questions
What formulas are given in AS Physics exams?
Papers 1 and 2 include a data and formulae page (page 2). It lists constants such as g, c, e, h, the unified atomic mass unit and the proton and electron masses, plus a short list of formulae: s=ut+½at², v²=u²+2as, hydrostatic pressure (Δp=ρgΔh), upthrust (F=ρgV), the Doppler effect for sound (fₒ=fₛv/(v±vₛ)), electric current (I=Anvq) and resistors in series and in parallel. Everything else on this sheet, such as F=ma, V=IR, P=VI and v=fλ, is not given, so you need to know it. This guide explains when and how to use each formula.
How do I know which physics formula to use?
Identify what quantities you're given and what you need to find. Choose a formula that connects these. For example, if you know the initial velocity, final velocity and acceleration and need the displacement, with no time given, use v²=u²+2as. Our formula sheet shows when to use each equation.
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