AS Physics Formula Sheet
Cambridge AS Physics (9702) | All Required Equations
Topics 1-11 | Syllabus 2025-2027
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1. 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
2. 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 = 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)
3. Forces, Density & Pressure
M = Moment (N m)
F = Force (N)
d = Perpendicular distance from pivot (m)
ρ (rho) = Density (kg/m³)
m = Mass (kg)
V = Volume (m³)
p = Pressure (Pa = N/m²)
F = Force (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
V is volume of FLUID displaced, not volume of 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
4. 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. Use for vehicles at constant speed
η (eta) = Efficiency (%)
Works with energy (J) or power (W)
5. 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)
Higher E = stiffer material. Gradient of 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
6. 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
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)
f₀ = fₛv/(v - vₛ) = 500 × 340/(340 - 30)
f₀ = 170000/310
f₀ = 548 Hz (higher pitch)
Practice:
The same ambulance moves away at 30 m/s. Find the observed frequency.
Show Answer
f₀ = 500 × 340/(340 + 30) = 170000/370 = 459 Hz (lower pitch)
7. 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 = 3.33×10⁻⁶/600×10⁻⁹ = 5.55 → n = 5 (max whole number)
8. 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)
9. 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
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