A-Level Physics Formula Guide
Cambridge A-Level Physics (9702) | Understanding & Application
Some formulas are given in your exam, not all
Learn the ones the sheet leaves out, and understand when and how to use each formula
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Featured Equations
Across mechanics, fields, oscillations & electricityu = Initial velocity (m/s)
a = Acceleration (m/s²)
t = Time (s)
m = Mass (kg)
a = Acceleration (m/s²)
Δp = Change in momentum (kg m/s)
Δt = Time interval (s)
m = Mass (kg)
v = Speed (m/s)
m = Mass (kg)
v = Speed (m/s)
r = Radius (m)
G = 6.67×10⁻¹¹ N m²/kg²
M = Source mass (kg)
r = Distance from centre (m)
ω = Angular frequency (rad/s)
x = Displacement from equilibrium (m)
I = Current (A)
R = Resistance (Ω)
How to Use Formulas in Exams
- Identify what you're given (known values with units)
- Identify what you need to find (the unknown)
- Choose a formula that connects these quantities
- Rearrange to make the unknown the subject
- Substitute values (with consistent units)
- Calculate and state the answer with units
Physical Constants
Given on page 2 of Papers 1, 2 and 4. You don't need to memorize these.
Kinematics (Uniformly Accelerated Motion)
Use when acceleration is constant. Choose formula based on which variable you DON'T have.
Displacement from initial velocity, acceleration, time
Use when: You don't know final velocity (v)
Variables: s = displacement (m), u = initial velocity (m/s), a = acceleration (m/s²), t = time (s)
Tip: This is a quadratic in t - can have two solutions
Worked Example:
A ball is thrown vertically upward at 20 m/s. How high does it rise? (g = 9.81 m/s²)
1. Given: u = 20 m/s, v = 0 m/s (at max height), a = −9.81 m/s²
2. Find: max height s
3. Formula: v² = u² + 2as (no time given)
4. Rearrange: s = (v² − u²) / (2a)
5. Substitute: s = (0 − 20²) / (2 × −9.81) = −400 / −19.62
Answer: s = 20.4 m
Practice:
A car accelerates from rest at 3.0 m/s² for 5.0 s. Find the distance travelled.
Show Answer
1. Given: u = 0 m/s (from rest), a = 3.0 m/s², t = 5.0 s
2. Find: distance s
3. Formula: s = ut + ½at²
4. Substitute: s = 0 × 5.0 + ½ × 3.0 × (5.0)²
5. Calculate: s = 0 + ½ × 3.0 × 25 = 37.5 m
Answer: s = 37.5 m
Final velocity without knowing time
Use when: You don't know time (t)
Rearrangements: s = (v² - u²)/2a, a = (v² - u²)/2s
Tip: Very useful for projectile problems at max height (v = 0)
Worked Example:
An object is dropped from 45 m. Find its speed when it hits the ground.
1. Given: u = 0 m/s (dropped from rest), s = 45 m, a = 9.81 m/s²
2. Find: final speed v
3. Formula: v² = u² + 2as
4. Substitute: v² = 0 + 2 × 9.81 × 45 = 882.9 m²/s²
5. Calculate: v = √882.9
Answer: v = 29.7 m/s
Practice:
A car brakes from 25 m/s to rest over 50 m. Find the deceleration.
Show Answer
1. Given: u = 25 m/s, v = 0 m/s (rest), s = 50 m
2. Find: deceleration a
3. Formula: v² = u² + 2as
4. Rearrange: a = (v² − u²) / (2s)
5. Substitute: a = (0 − 25²) / (2 × 50)
6. Calculate: a = −625 / 100 = −6.25 m/s²
Answer: a = −6.25 m/s² (deceleration 6.25 m/s²)
Pressure & Fluids
For liquids and gases at rest or in equilibrium.
Hydrostatic pressure
Use when: Calculating pressure difference due to depth in a fluid
Variables: Δp = pressure change (Pa), ρ = density (kg/m³), g = 9.81 m/s², Δh = depth change (m)
Note: This gives pressure DIFFERENCE. Total pressure = atmospheric + ρgΔh
Worked Example:
Find the pressure at 10 m depth in seawater (ρ = 1025 kg/m³).
1. Given: ρ = 1025 kg/m³, g = 9.81 m/s², Δh = 10 m
2. Find: pressure difference Δp
3. Formula: Δp = ρgΔh
4. Substitute: Δp = 1025 × 9.81 × 10
5. Calculate: Δp = 100 553 Pa
Answer: Δp ≈ 1.01 × 10⁵ Pa (≈ 1 atm extra)
Practice:
A mercury barometer reads 760 mm. Find the atmospheric pressure. (ρ_Hg = 13600 kg/m³)
Show Answer
1. Given: h = 760 mm = 0.760 m, ρ = 13 600 kg/m³, g = 9.81 m/s²
2. Find: atmospheric pressure p
3. Formula: p = ρgh
4. Substitute: p = 13 600 × 9.81 × 0.760
5. Calculate: p = 101 396 Pa
Answer: p ≈ 1.01 × 10⁵ Pa (≈ 1 atm)
Upthrust (Archimedes' principle)
Use when: Finding buoyancy force on submerged/floating object
Key insight: V is volume of fluid DISPLACED, not volume of object (unless fully submerged)
For floating objects: Upthrust = Weight
Worked Example:
A 0.50 m³ block is fully submerged in water. Find the upthrust.
1. Given: V = 0.50 m³ (fully submerged), ρ_water = 1000 kg/m³, g = 9.81 m/s²
2. Find: upthrust F
3. Formula: F = ρgV (volume of fluid displaced = volume of object)
4. Substitute: F = 1000 × 9.81 × 0.50
Answer: F = 4905 N ≈ 4.9 kN
Practice:
A 2.0 kg object floats with 75% of its volume submerged. Find its density.
Show Answer
1. Given: m = 2.0 kg, fraction submerged = 0.75, ρ_water = 1000 kg/m³
2. Find: density of object ρ
3. Formula: Floating ⇒ Upthrust = Weight ⇒ ρ_water × (0.75V) × g = ρ × V × g
4. Rearrange: ρ = 0.75 × ρ_water
5. Substitute: ρ = 0.75 × 1000
6. Calculate: ρ = 750 kg/m³
Answer: ρ = 750 kg/m³
Circular Motion (Paper 4)
Objects moving in circles with constant speed still accelerate towards the center.
Key insight: Angular velocity is constant for uniform circular motion
Note: v is tangent to the circle, ω is same for all points on rigid body
Worked Example:
A wheel of radius 0.30 m rotates at 5.0 rad/s. Find the speed of a point on the rim.
1. Given: r = 0.30 m, ω = 5.0 rad/s
2. Find: tangential speed v
3. Formula: v = rω
4. Substitute: v = 0.30 × 5.0
5. Calculate: v = 1.5 m/s
Answer: v = 1.5 m/s
Practice:
A satellite orbits at 7500 m/s at radius 6.8 × 10⁶ m. Find its angular velocity.
Show Answer
1. Given: v = 7500 m/s, r = 6.8 × 10⁶ m
2. Find: angular velocity ω
3. Formula: v = rω
4. Rearrange: ω = v / r
5. Substitute: ω = 7500 / (6.8 × 10⁶)
6. Calculate: ω = 1.103 × 10⁻³ rad/s
Answer: ω ≈ 1.1 × 10⁻³ rad/s
Use when: Finding acceleration needed to maintain circular path
Worked Example:
A car travels at 20 m/s around a bend of radius 50 m. Find the centripetal acceleration.
1. Given: v = 20 m/s, r = 50 m
2. Find: centripetal acceleration a
3. Formula: a = v² / r
4. Substitute: a = (20)² / 50
5. Calculate: a = 400 / 50 = 8.0 m/s²
Answer: a = 8.0 m/s²
Practice:
An electron orbits at ω = 2.0 × 10⁶ rad/s at r = 5.0 × 10⁻¹¹ m. Find its acceleration.
Show Answer
1. Given: r = 5.0 × 10⁻¹¹ m, ω = 2.0 × 10⁶ rad/s
2. Find: centripetal acceleration a
3. Formula: a = rω²
4. Substitute: a = 5.0 × 10⁻¹¹ × (2.0 × 10⁶)²
5. Calculate: a = 5.0 × 10⁻¹¹ × 4.0 × 10¹² = 2.0 × 10² m/s²
Answer: a = 2.0 × 10² m/s² (200 m/s²)
Examples: Tension, friction, gravity, or normal force can provide this
Worked Example:
A 1200 kg car travels at 15 m/s around a bend of radius 45 m. Find the friction force required.
1. Given: m = 1200 kg, v = 15 m/s, r = 45 m
2. Find: centripetal (friction) force F
3. Formula: F = mv² / r
4. Substitute: F = 1200 × (15)² / 45
5. Calculate: F = 1200 × 225 / 45 = 270 000 / 45 = 6000 N
Answer: F = 6000 N (6.0 kN)
Practice:
A 0.50 kg ball on a smooth horizontal table is held by a 0.80 m string fixed to a pin at the centre, and moves in a circle at 4.0 m/s. Find the tension.
Show Answer
1. Given: m = 0.50 kg, r = 0.80 m, v = 4.0 m/s
2. Find: tension T (= centripetal force)
3. Formula: F = mv² / r
4. Substitute: F = 0.50 × (4.0)² / 0.80
5. Calculate: F = 0.50 × 16 / 0.80 = 8.0 / 0.80 = 10 N
Answer: T = 10 N
Waves & Doppler Effect
For sound waves from a moving source heard by a stationary observer (a moving observer is not required).
Doppler effect for sound (moving source)
Use when: Sound source moves toward/away from stationary observer
Variables: f_o = observed frequency (Hz), f_s = source frequency (Hz), v = wave speed (m/s), v_s = source speed (m/s)
Sign convention: Source approaching → use (v - v_s) → f_o increases | Source receding → use (v + v_s) → f_o decreases
Worked Example:
An ambulance siren emits 800 Hz and approaches at 30 m/s. Find the observed frequency. (v_sound = 340 m/s)
1. Given: f_s = 800 Hz, v = 340 m/s, v_s = 30 m/s (approaching → v - v_s)
2. Find: observed frequency f_o
3. Formula: f_o = f_s × v / (v - v_s)
4. Substitute: f_o = 800 × 340 / (340 - 30)
5. Calculate: f_o = 272 000 / 310 = 877.4 Hz
Answer: f_o ≈ 877 Hz (higher pitch)
Practice:
The same ambulance now moves away at 30 m/s. Find the observed frequency.
Show Answer
1. Given: f_s = 800 Hz, v = 340 m/s, v_s = 30 m/s (receding → v + v_s)
2. Find: observed frequency f_o
3. Formula: f_o = f_s × v / (v + v_s)
4. Substitute: f_o = 800 × 340 / (340 + 30)
5. Calculate: f_o = 272 000 / 370 = 735.1 Hz
Answer: f_o ≈ 735 Hz (lower pitch)
Electricity
Current flow and resistance in circuits.
Current from drift velocity
Use when: Relating microscopic electron movement to macroscopic current
Variables: I = current (A), A = cross-sectional area (m²), n = number density (charge carriers per m³), v = drift velocity (m/s), q = charge per carrier (C)
Tip: Drift velocities are surprisingly slow (often mm/s) despite fast electrical signals
Worked Example:
A copper wire (A = 1.0 × 10⁻⁶ m², n = 8.5 × 10²⁸ m⁻³) carries 2.0 A. Find the drift velocity.
1. Given: I = 2.0 A, A = 1.0 × 10⁻⁶ m², n = 8.5 × 10²⁸ m⁻³, q = 1.60 × 10⁻¹⁹ C
2. Find: drift velocity v
3. Formula: I = Anvq
4. Rearrange: v = I / (Anq)
5. Substitute: v = 2.0 / (1.0 × 10⁻⁶ × 8.5 × 10²⁸ × 1.60 × 10⁻¹⁹)
6. Calculate: v = 2.0 / (1.36 × 10⁴) = 1.47 × 10⁻⁴ m/s
Answer: v ≈ 1.47 × 10⁻⁴ m/s (0.15 mm/s)
Practice:
If the current doubles, what happens to drift velocity?
Show Answer
1. Given: Worked example v₁ = 1.47 × 10⁻⁴ m/s ≈ 0.15 mm/s at I₁ = 2.0 A; now I₂ = 2 × I₁
2. Find: new drift velocity v₂
3. Formula: I = Anvq ⇒ v = I / (Anq)
4. Reasoning: A, n, q are constant ⇒ v ∝ I
5. Substitute: v₂ = 2 × v₁ = 2 × 0.15 mm/s
6. Calculate: v₂ ≈ 0.30 mm/s
Answer: drift velocity doubles to ≈ 0.30 mm/s
Resistors in series
Key principle: Same current through each; voltages add up
Total resistance is always LARGER than any individual resistor
Worked Example:
Three resistors (10Ω, 20Ω, 30Ω) are connected in series. Find the total resistance.
1. Given: R₁ = 10 Ω, R₂ = 20 Ω, R₃ = 30 Ω (series)
2. Find: total resistance R
3. Formula: R = R₁ + R₂ + R₃
4. Substitute: R = 10 + 20 + 30
5. Calculate: R = 60 Ω
Answer: R = 60 Ω
Resistors in parallel
Key principle: Same voltage across each; currents add up
Total resistance is always SMALLER than the smallest individual resistor
Shortcut for 2 resistors: R = R₁R₂/(R₁+R₂)
Worked Example:
Two resistors (6Ω and 12Ω) are connected in parallel. Find the total resistance.
1. Given: R₁ = 6 Ω, R₂ = 12 Ω (parallel)
2. Find: total resistance R
3. Formula: R = R₁R₂ / (R₁ + R₂) (shortcut for 2 resistors)
4. Substitute: R = (6 × 12) / (6 + 12)
5. Calculate: R = 72 / 18 = 4 Ω
Answer: R = 4 Ω (less than smallest, as expected)
Practice:
Three 12Ω resistors are connected in parallel. Find the total resistance.
Show Answer
1. Given: R₁ = R₂ = R₃ = 12 Ω in parallel
2. Find: total resistance R
3. Formula: 1/R = 1/R₁ + 1/R₂ + 1/R₃
4. Substitute: 1/R = 1/12 + 1/12 + 1/12
5. Calculate: 1/R = 3/12 = 1/4 ⇒ R = 4 Ω
Answer: R = 4 Ω
Gravitational Fields (Paper 4)
For orbital motion and gravity beyond uniform field approximation.
Gravitational potential
Use when: Finding potential energy per unit mass at distance r from center of mass M
Variables: φ = potential (J kg⁻¹), G = 6.67 × 10⁻¹¹ N m² kg⁻², M = mass of body (kg), r = distance from center (m)
Potential is NEGATIVE - zero at infinity, decreases (more negative) closer to mass
Worked Example:
Find the gravitational potential at Earth's surface. (M = 6.0 × 10²⁴ kg, r = 6.4 × 10⁶ m)
1. Given: G = 6.67 × 10⁻¹¹ N m² kg⁻², M = 6.0 × 10²⁴ kg, r = 6.4 × 10⁶ m
2. Find: gravitational potential φ
3. Formula: φ = −GM / r
4. Substitute: φ = −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (6.4 × 10⁶)
5. Calculate: φ = −4.002 × 10¹⁴ / 6.4 × 10⁶ = −6.25 × 10⁷ J kg⁻¹
Answer: φ ≈ −6.25 × 10⁷ J kg⁻¹
Practice:
Find the potential at height 400 km above Earth's surface (ISS orbit).
Show Answer
1. Given: M = 6.0 × 10²⁴ kg, R_Earth = 6.4 × 10⁶ m, altitude = 400 km = 4.0 × 10⁵ m, G = 6.67 × 10⁻¹¹ N m² kg⁻²
2. Find: gravitational potential φ
3. Formula: φ = −GM / r, where r = R_Earth + altitude
4. Substitute: r = 6.4 × 10⁶ + 4.0 × 10⁵ = 6.8 × 10⁶ m; φ = −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (6.8 × 10⁶)
5. Calculate: φ = −4.002 × 10¹⁴ / 6.8 × 10⁶ = −5.885 × 10⁷ J kg⁻¹
Answer: φ ≈ −5.89 × 10⁷ J kg⁻¹
Gravitational potential energy
Use when: Finding energy in orbital/escape problems
Relationship: E_P = mφ (potential energy = mass × potential)
For escape velocity: KE + PE = 0 → ½mv² = GMm/r → v = √(2GM/r). Escape velocity is not named in the syllabus; it is an energy-conservation use of E_P = −GMm/r.
Worked Example:
Find the escape velocity from Earth's surface.
1. Given: G = 6.67 × 10⁻¹¹ N m² kg⁻², M = 6.0 × 10²⁴ kg, r = 6.4 × 10⁶ m
2. Find: escape velocity v
3. Formula: ½mv² = GMm / r ⇒ v = √(2GM / r)
4. Substitute: v = √(2 × 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ / 6.4 × 10⁶)
5. Calculate: v = √(1.25 × 10⁸) ≈ 1.12 × 10⁴ m/s
Answer: v ≈ 1.12 × 10⁴ m/s (11.2 km/s)
Practice:
How much energy is needed to move a 1000 kg satellite from Earth's surface to 400 km altitude?
Show Answer
1. Given: m = 1000 kg; from previous results, φ_surface = −6.253 × 10⁷ J/kg, φ_400km = −5.885 × 10⁷ J/kg (keep 4 figures for the subtraction)
2. Find: energy required ΔE
3. Formula: ΔE = m × Δφ = m × (φ_final − φ_initial)
4. Substitute: Δφ = −5.885 × 10⁷ − (−6.253 × 10⁷) = +3.68 × 10⁶ J/kg; ΔE = 1000 × 3.68 × 10⁶
5. Calculate: ΔE = 3.68 × 10⁹ J
Answer: ΔE ≈ 3.7 × 10⁹ J (3.7 GJ)
Ideal Gases (Paper 4)
Kinetic theory linking microscopic and macroscopic properties.
Worked Example:
2.0 mol of gas at 300 K occupies 0.050 m³. Find the pressure.
1. Given: n = 2.0 mol, R = 8.31 J mol⁻¹ K⁻¹, T = 300 K, V = 0.050 m³
2. Find: pressure p
3. Formula: pV = nRT
4. Rearrange: p = nRT / V
5. Substitute: p = (2.0 × 8.31 × 300) / 0.050
6. Calculate: p = 4986 / 0.050 = 9.97 × 10⁴ Pa
Answer: p ≈ 9.97 × 10⁴ Pa (≈ 100 kPa)
Practice:
0.50 mol of gas at 2.0 × 10⁵ Pa and 400 K. Find the volume.
Show Answer
1. Given: n = 0.50 mol, p = 2.0 × 10⁵ Pa, T = 400 K, R = 8.31 J mol⁻¹ K⁻¹
2. Find: volume V
3. Formula: pV = nRT
4. Rearrange: V = nRT / p
5. Substitute: V = (0.50 × 8.31 × 400) / (2.0 × 10⁵)
6. Calculate: V = 1662 / (2.0 × 10⁵) = 8.31 × 10⁻³ m³
Answer: V ≈ 8.3 × 10⁻³ m³
Relationship: N = nNₐ and R = kNₐ (where Nₐ = Avogadro constant)
Alternative form: p = ⅓ρ⟨c²⟩ where ρ = Nm/V = density
Note: ⟨c²⟩ ≠ ⟨c⟩² (mean square speed ≠ square of mean speed)
Worked Example:
Find the mean KE of a gas molecule at 300 K.
1. Given: k = 1.38 × 10⁻²³ J K⁻¹, T = 300 K
2. Find: mean translational KE per molecule
3. Formula: KE = (3/2) kT
4. Substitute: KE = 1.5 × 1.38 × 10⁻²³ × 300
5. Calculate: KE = 6.21 × 10⁻²¹ J
Answer: KE ≈ 6.21 × 10⁻²¹ J
Practice:
Find the root mean square speed of N₂ (m = 4.65 × 10⁻²⁶ kg) at 300 K.
Show Answer
1. Given: m = 4.65 × 10⁻²⁶ kg (N₂), T = 300 K, k = 1.38 × 10⁻²³ J K⁻¹
2. Find: root-mean-square speed c_rms
3. Formula: ½m⟨c²⟩ = (3/2)kT ⇒ ⟨c²⟩ = 3kT / m
4. Substitute: ⟨c²⟩ = 3 × 1.38 × 10⁻²³ × 300 / (4.65 × 10⁻²⁶)
5. Calculate: ⟨c²⟩ = 1.242 × 10⁻²⁰ / 4.65 × 10⁻²⁶ = 2.67 × 10⁵ m²/s²
6. c_rms: √(2.67 × 10⁵) ≈ 517 m/s
Answer: c_rms ≈ 517 m/s
Thermodynamics (Paper 4)
Internal energy, work done on or by a gas, and the first law.
First law of thermodynamics
Use when: Analyzing energy changes in a gas system
Variables: ΔU = change in internal energy (J), q = heat added TO system (J), W = work done ON system (J)
Sign convention: +ve when energy enters system | For ideal gas: ΔU ∝ ΔT (isothermal → ΔU = 0)
Worked Example:
A gas receives 500 J of heat and does 200 J of work on its surroundings. Find ΔU.
1. Given: heat added q = +500 J; work done BY gas = 200 J ⇒ work ON gas W = −200 J
2. Find: change in internal energy ΔU
3. Formula: ΔU = q + W
4. Substitute: ΔU = 500 + (−200)
5. Calculate: ΔU = +300 J
Answer: ΔU = +300 J (internal energy increases)
Practice:
In an isothermal expansion, 800 J of work is done by the gas. How much heat enters?
Show Answer
1. Given: isothermal expansion (T constant); work done BY gas = 800 J ⇒ W_on = −800 J
2. Find: heat q entering the gas
3. Formula: ΔU = q + W; isothermal ⇒ ΔU = 0
4. Rearrange: 0 = q + W ⇒ q = −W
5. Substitute: q = −(−800)
6. Calculate: q = +800 J
Answer: q = +800 J (heat enters to keep T constant)
Work done BY gas at constant pressure
Use when: Gas expands or compresses at constant pressure (isobaric process)
Variables: W = work done BY gas (J), p = pressure (Pa), ΔV = volume change (m³)
On p-V diagram, work = area under curve (for any process, not just isobaric)
Worked Example:
A gas at 2.0 × 10⁵ Pa expands from 0.010 m³ to 0.025 m³. Find the work done by the gas.
1. Given: p = 2.0 × 10⁵ Pa, V₁ = 0.010 m³, V₂ = 0.025 m³
2. Find: work W done BY the gas
3. Formula: W = pΔV
4. Substitute: ΔV = 0.025 − 0.010 = 0.015 m³; W = 2.0 × 10⁵ × 0.015
5. Calculate: W = 3000 J
Answer: W = 3000 J (done BY the gas)
Practice:
A gas is compressed from 0.030 m³ to 0.010 m³ at 1.5 × 10⁵ Pa. What work is done on the gas?
Show Answer
1. Given: p = 1.5 × 10⁵ Pa, V₁ = 0.030 m³, V₂ = 0.010 m³ (compression)
2. Find: work done ON the gas
3. Formula: W_BY = pΔV; W_ON = −W_BY
4. Substitute: ΔV = 0.010 − 0.030 = −0.020 m³; W_BY = 1.5 × 10⁵ × (−0.020)
5. Calculate: W_BY = −3000 J ⇒ W_ON = +3000 J
Answer: W_ON = +3000 J
Simple Harmonic Motion (Paper 4)
Oscillations where restoring force ∝ displacement from equilibrium.
Defining equation of SHM
Key insight: Acceleration proportional to displacement, directed towards equilibrium
Variables: a = acceleration (m/s²), ω = angular frequency (rad s⁻¹) = 2πf = 2π/T, x = displacement (m)
The negative sign indicates acceleration is opposite to displacement direction
Worked Example:
A pendulum oscillates at 0.50 Hz. Find the acceleration when displaced 0.10 m from equilibrium.
1. Given: f = 0.50 Hz, x = 0.10 m
2. Find: acceleration a
3. Formula: a = −ω²x, with ω = 2πf
4. Substitute: ω = 2π × 0.50 = π rad/s; a = −π² × 0.10
5. Calculate: a = −9.87 × 0.10 = −0.987 m/s²
Answer: a ≈ −0.99 m/s² (towards equilibrium)
Practice:
A mass-spring system has T = 0.40 s and amplitude 0.15 m. Find the maximum acceleration.
Show Answer
1. Given: T = 0.40 s, amplitude x₀ = 0.15 m
2. Find: maximum acceleration a_max
3. Formula: a_max = ω²x₀, with ω = 2π / T
4. Substitute: ω = 2π / 0.40 = 5π rad/s; a_max = (5π)² × 0.15
5. Calculate: ω² ≈ 246.7; a_max = 246.7 × 0.15 ≈ 37.0 m/s²
Answer: a_max ≈ 37 m/s² (at the extremes)
Velocity as function of displacement
Use when: You know position and amplitude, need velocity
Key points: v = 0 at x = ±x₀ (extremes), v = ±ωx₀ at x = 0 (equilibrium)
Maximum speed: v_max = ωx₀ = 2πfx₀ (occurs at equilibrium)
Worked Example:
A mass oscillates with amplitude 0.20 m and f = 2.0 Hz. Find the speed when x = 0.12 m.
1. Given: x₀ = 0.20 m, f = 2.0 Hz, x = 0.12 m
2. Find: speed v at displacement x
3. Formula: v = ω√(x₀² − x²), with ω = 2πf
4. Substitute: ω = 2π × 2.0 = 4π rad/s; v = 4π × √(0.20² − 0.12²)
5. Calculate: √(0.04 − 0.0144) = √0.0256 = 0.16; v = 4π × 0.16 ≈ 2.01 m/s
Answer: v ≈ 2.0 m/s
Practice:
The same oscillator: find the maximum speed.
Show Answer
1. Given: x₀ = 0.20 m, f = 2.0 Hz ⇒ ω = 4π rad/s
2. Find: maximum speed v_max
3. Formula: v_max = ωx₀ (occurs at x = 0)
4. Substitute: v_max = 4π × 0.20
5. Calculate: v_max ≈ 2.513 m/s
Answer: v_max ≈ 2.5 m/s (at equilibrium)
Displacement as function of time
Choice depends on starting point: sin if starting from equilibrium, cos if starting from extreme
Variables: x₀ = amplitude (m), ω = 2πf = 2π/T (rad/s), t = time (s)
Worked Example:
An oscillator starts at equilibrium, amplitude 0.08 m, period 0.50 s. Find displacement at t = 0.125 s.
1. Given: x₀ = 0.08 m, T = 0.50 s, t = 0.125 s; starts at equilibrium ⇒ use sin
2. Find: displacement x at time t
3. Formula: x = x₀ sin(ωt), with ω = 2π / T
4. Substitute: ω = 2π / 0.50 = 4π rad/s; ωt = 4π × 0.125 = π/2
5. Calculate: x = 0.08 × sin(π/2) = 0.08 × 1 = 0.08 m
Answer: x = 0.08 m (at maximum displacement)
Electric Fields (Paper 4)
Point charges and radial electric fields.
Electric potential around a point charge
Use when: Finding potential at distance r from a point charge
Variables: V = potential (V or J C⁻¹), Q = charge (C), ε₀ = 8.85 × 10⁻¹² F/m, r = distance (m)
Unlike gravity, V can be +ve (positive Q) or -ve (negative Q)
Worked Example:
Find the electric potential 0.30 m from a +5.0 μC charge.
1. Given: Q = +5.0 μC = 5.0 × 10⁻⁶ C, r = 0.30 m, k = 1/(4πε₀) = 8.99 × 10⁹ m F⁻¹ (on the data sheet)
2. Find: electric potential V
3. Formula: V = kQ / r
4. Substitute: V = 8.99 × 10⁹ × 5.0 × 10⁻⁶ / 0.30
5. Calculate: V = 4.50 × 10⁴ / 0.30 = 1.5 × 10⁵ V
Answer: V = 1.5 × 10⁵ V (150 kV)
Practice:
Find the potential at 0.20 m from a -3.0 μC charge.
Show Answer
1. Given: Q = −3.0 μC = −3.0 × 10⁻⁶ C, r = 0.20 m, k = 8.99 × 10⁹ m F⁻¹
2. Find: electric potential V
3. Formula: V = kQ / r
4. Substitute: V = 8.99 × 10⁹ × (−3.0 × 10⁻⁶) / 0.20
5. Calculate: V = −2.7 × 10⁴ / 0.20 = −1.35 × 10⁵ V
Answer: V = −1.35 × 10⁵ V (negative, as Q is negative)
Electric potential energy
Use when: Finding work done moving charge q near charge Q
Sign: +ve for like charges (repulsion - work needed to bring together), -ve for unlike charges (attraction)
E_P = qV (similar to gravitational: E_P = mφ)
Worked Example:
Two protons are 2.0 × 10⁻¹⁰ m apart. Find their electric potential energy.
1. Given: Q = q = +1.60 × 10⁻¹⁹ C, r = 2.0 × 10⁻¹⁰ m, k = 8.99 × 10⁹ m F⁻¹
2. Find: electric potential energy E_P
3. Formula: E_P = kQq / r
4. Substitute: E_P = 8.99 × 10⁹ × (1.60 × 10⁻¹⁹)² / (2.0 × 10⁻¹⁰)
5. Calculate: E_P = 8.99 × 10⁹ × 2.56 × 10⁻³⁸ / 2.0 × 10⁻¹⁰ = 1.15 × 10⁻¹⁸ J
Answer: E_P ≈ 1.15 × 10⁻¹⁸ J (≈ 7.2 eV)
Practice:
A proton and electron are 5.0 × 10⁻¹¹ m apart. Find their potential energy.
Show Answer
1. Given: Q = +e, q = −e, e = 1.60 × 10⁻¹⁹ C; r = 5.0 × 10⁻¹¹ m; k = 8.99 × 10⁹ m F⁻¹
2. Find: electric potential energy E_P
3. Formula: E_P = kQq / r
4. Substitute: E_P = 8.99 × 10⁹ × (+e)(−e) / (5.0 × 10⁻¹¹) = −8.99 × 10⁹ × (1.60 × 10⁻¹⁹)² / (5.0 × 10⁻¹¹)
5. Calculate: E_P = −2.30 × 10⁻²⁸ / 5.0 × 10⁻¹¹ ≈ −4.6 × 10⁻¹⁸ J
Answer: E_P ≈ −4.6 × 10⁻¹⁸ J (negative ⇒ bound system)
Capacitors (Paper 4)
Combinations and discharge through resistors.
Capacitors in series
Key principle: Same charge on each capacitor; voltages add
Note: Opposite of resistors! Series capacitance is SMALLER than smallest individual
For 2 capacitors: C = C₁C₂/(C₁+C₂)
Worked Example:
Two capacitors (4.0 μF and 6.0 μF) are connected in series. Find the total capacitance.
1. Given: C₁ = 4.0 μF, C₂ = 6.0 μF (series)
2. Find: total capacitance C
3. Formula: 1/C = 1/C₁ + 1/C₂ ⇒ C = C₁C₂ / (C₁ + C₂)
4. Substitute: C = (4.0 × 6.0) / (4.0 + 6.0)
5. Calculate: C = 24 / 10 = 2.4 μF
Answer: C = 2.4 μF (smaller than 4.0 μF)
Capacitors in parallel
Key principle: Same voltage across each; charges add
Note: Parallel capacitance is LARGER (opposite of resistors)
Worked Example:
The same capacitors (4.0 μF and 6.0 μF) are connected in parallel. Find the total capacitance.
1. Given: C₁ = 4.0 μF, C₂ = 6.0 μF (parallel)
2. Find: total capacitance C
3. Formula: C = C₁ + C₂
4. Substitute: C = 4.0 + 6.0
5. Calculate: C = 10 μF
Answer: C = 10 μF (larger than either)
Practice:
Three 12 μF capacitors in parallel, then in series with 6 μF. Find total C.
Show Answer
1. Given: three 12 μF capacitors in parallel; result then in series with 6 μF
2. Find: total capacitance C
3. Formula: parallel: C_p = C₁+C₂+C₃; series: C = C_pC₄ / (C_p + C₄)
4. Substitute: C_p = 12 + 12 + 12 = 36 μF; C = 36 × 6 / (36 + 6)
5. Calculate: C = 216 / 42 ≈ 5.14 μF
Answer: C ≈ 5.1 μF
Capacitor discharge (exponential decay)
Use when: Finding Q, V, or I during discharge
Variables: x = charge, voltage, or current at time t; x₀ = initial value; τ = RC = time constant (s)
At t = RC: value = x₀/e ≈ 0.37x₀ (37% remaining). At t = 5RC: essentially discharged (0.7%)
Worked Example:
A 100 μF capacitor charged to 12 V discharges through 50 kΩ. Find V after 3.0 s.
1. Given: V₀ = 12 V, R = 50 kΩ = 5.0 × 10⁴ Ω, C = 100 μF = 1.0 × 10⁻⁴ F, t = 3.0 s
2. Find: voltage V after t = 3.0 s
3. Formula: V = V₀ e^(−t/RC)
4. Substitute: RC = 5.0 × 10⁴ × 1.0 × 10⁻⁴ = 5.0 s; V = 12 × e^(−3.0/5.0)
5. Calculate: V = 12 × e⁻⁰·⁶ = 12 × 0.5488 ≈ 6.59 V
Answer: V ≈ 6.6 V
Practice:
How long until the voltage drops to 3.0 V?
Show Answer
1. Given: V₀ = 12 V, V = 3.0 V, RC = 5.0 s
2. Find: time t
3. Formula: V = V₀ e^(−t/RC)
4. Rearrange: t = −RC × ln(V / V₀)
5. Substitute: t = −5.0 × ln(3.0 / 12) = −5.0 × ln(0.25)
6. Calculate: ln(0.25) = −1.386 ⇒ t = −5.0 × (−1.386) ≈ 6.93 s
Answer: t ≈ 6.9 s
Magnetic Fields (Paper 4)
Forces on currents and moving charges in magnetic fields.
Force on current-carrying conductor
Use when: Wire carrying current in a magnetic field
Variables: B = magnetic flux density (T), I = current (A), L = length in field (m), θ = angle between I and B
Direction: Fleming's left-hand rule (First = B, seCond = I, thuMb = Motion). Max force when θ = 90°
Worked Example:
A 0.20 m wire carries 5.0 A perpendicular to a 0.30 T field. Find the force.
1. Given: B = 0.30 T, I = 5.0 A, L = 0.20 m, θ = 90°
2. Find: magnetic force F
3. Formula: F = BIL sinθ
4. Substitute: F = 0.30 × 5.0 × 0.20 × sin90°
5. Calculate: sin90° = 1; F = 0.30 × 5.0 × 0.20 = 0.30 N
Answer: F = 0.30 N
Practice:
The same wire makes a 30° angle with the field. What is the new force?
Show Answer
1. Given: B = 0.30 T, I = 5.0 A, L = 0.20 m, θ = 30°
2. Find: force F
3. Formula: F = BIL sinθ
4. Substitute: F = 0.30 × 5.0 × 0.20 × sin30°
5. Calculate: sin30° = 0.5; F = 0.30 × 0.5 = 0.15 N
Answer: F = 0.15 N
Force on moving charge
Use when: Charged particle moving through a magnetic field
Variables: q = charge (C), v = velocity (m/s), θ = angle between v and B
Force ⊥ velocity → no work done, only direction changes → circular motion. Radius: r = mv/(Bq)
Worked Example:
An electron (m = 9.11 × 10⁻³¹ kg) moves at 2.0 × 10⁷ m/s perpendicular to a 0.50 T field. Find (a) the force and (b) the radius of its circular path.
1. Given: m = 9.11 × 10⁻³¹ kg, q = 1.60 × 10⁻¹⁹ C, v = 2.0 × 10⁷ m/s, B = 0.50 T, θ = 90°
2. Find: (a) force F, (b) circular-path radius r
3. Formula: F = Bqv (sin90° = 1); r = mv / (Bq)
4. Substitute (a): F = 0.50 × 1.60 × 10⁻¹⁹ × 2.0 × 10⁷
5. Calculate (a): F = 1.6 × 10⁻¹² N
6. Substitute (b): r = (9.11 × 10⁻³¹ × 2.0 × 10⁷) / (0.50 × 1.60 × 10⁻¹⁹)
7. Calculate (b): r = 1.822 × 10⁻²³ / 8.0 × 10⁻²⁰ ≈ 2.28 × 10⁻⁴ m
Answer: F ≈ 1.6 × 10⁻¹² N; r ≈ 2.3 × 10⁻⁴ m (0.23 mm)
Practice:
A proton (m = 1.67 × 10⁻²⁷ kg) enters the same field at the same speed. Find its radius.
Show Answer
1. Given: m = 1.67 × 10⁻²⁷ kg, q = 1.60 × 10⁻¹⁹ C, v = 2.0 × 10⁷ m/s, B = 0.50 T
2. Find: radius r of circular path
3. Formula: r = mv / (Bq)
4. Substitute: r = (1.67 × 10⁻²⁷ × 2.0 × 10⁷) / (0.50 × 1.60 × 10⁻¹⁹)
5. Calculate: r = 3.34 × 10⁻²⁰ / 8.0 × 10⁻²⁰ ≈ 0.418 m
Answer: r ≈ 0.42 m (much larger than electron's path due to greater mass)
Electromagnetic Induction (Paper 4)
Faraday's and Lenz's laws - generating EMF from changing magnetic flux.
Units: 1 Wb = 1 T m² = 1 V s
Lenz: The negative sign - EMF opposes the change producing it
Worked Example:
A coil of 200 turns has flux through it changing from 5.0 mWb to 2.0 mWb in 0.10 s. Find the induced EMF.
1. Given: N = 200; Φ₁ = 5.0 mWb = 5.0 × 10⁻³ Wb, Φ₂ = 2.0 × 10⁻³ Wb; Δt = 0.10 s
2. Find: magnitude of induced EMF ε
3. Formula: ε = −N (ΔΦ / Δt)
4. Substitute: ΔΦ = 2.0 × 10⁻³ − 5.0 × 10⁻³ = −3.0 × 10⁻³ Wb; ε = −200 × (−3.0 × 10⁻³) / 0.10
5. Calculate: ε = 0.60 / 0.10 = 6.0 V
Answer: ε = 6.0 V
Practice:
A 500-turn coil in a 0.40 T field has area 0.020 m². It rotates 90° in 0.050 s. Find the average EMF.
Show Answer
1. Given: N = 500, B = 0.40 T, A = 0.020 m², rotates 0° → 90°, Δt = 0.050 s
2. Find: average induced EMF ε
3. Formula: Φ = BA cosθ; ε = −N (ΔΦ / Δt)
4. Substitute: ΔΦ = BA (cos90° − cos0°) = 0.40 × 0.020 × (0 − 1) = −8.0 × 10⁻³ Wb; ε = −500 × (−8.0 × 10⁻³) / 0.050
5. Calculate: ε = 4.0 / 0.050 = 80 V
Answer: ε = 80 V (average)
Derivation: From ε = -dΦ/dt where dΦ = B × L × (v dt). Not a separate syllabus equation; it follows from Faraday's law
Worked Example:
A 0.50 m rod moves at 4.0 m/s perpendicular to a 0.30 T field. Find the induced EMF.
1. Given: B = 0.30 T, L = 0.50 m, v = 4.0 m/s (B ⊥ v)
2. Find: induced EMF ε
3. Formula: ε = BLv
4. Substitute: ε = 0.30 × 0.50 × 4.0
5. Calculate: ε = 0.60 V
Answer: ε = 0.60 V
Practice:
An aircraft with 30 m wingspan flies at 250 m/s in Earth's 5.0 × 10⁻⁵ T field. Find the EMF between wingtips.
Show Answer
1. Given: B = 5.0 × 10⁻⁵ T, L = 30 m (wingspan), v = 250 m/s
2. Find: EMF ε between the wingtips
3. Formula: ε = BLv (assuming horizontal flight, B vertical component)
4. Substitute: ε = 5.0 × 10⁻⁵ × 30 × 250
5. Calculate: ε = 5.0 × 10⁻⁵ × 7500 = 0.375 V
Answer: ε ≈ 0.375 V
Hall Effect (Paper 4)
Voltage developed across a conductor in a magnetic field.
Hall voltage
Use when: Finding B-field using Hall probe, or determining charge carrier type
Variables: V_H = Hall voltage (V), B = flux density (T), I = current (A), n = number density (m⁻³), t = thickness (m), q = charge (C)
V_H is larger for semiconductors (small n) than metals (large n) - that's why semiconductor Hall probes are used
Worked Example:
A Hall probe (t = 0.50 mm, n = 2.0 × 10²⁵ m⁻³) carries 20 mA in a 0.40 T field. Find V_H.
1. Given: B = 0.40 T, I = 20 mA = 0.020 A, n = 2.0 × 10²⁵ m⁻³, t = 0.50 mm = 5.0 × 10⁻⁴ m, q = 1.60 × 10⁻¹⁹ C
2. Find: Hall voltage V_H
3. Formula: V_H = BI / (ntq)
4. Substitute: V_H = (0.40 × 0.020) / (2.0 × 10²⁵ × 5.0 × 10⁻⁴ × 1.60 × 10⁻¹⁹)
5. Calculate: V_H = 8.0 × 10⁻³ / (1.6 × 10³) = 5.0 × 10⁻⁶ V
Answer: V_H = 5.0 × 10⁻⁶ V (5.0 μV)
Practice:
The Hall voltage is 12 μV in a 0.30 T field. Find B when V_H = 20 μV (same current).
Show Answer
1. Given: V_H₁ = 12 μV at B₁ = 0.30 T; V_H₂ = 20 μV (same I, n, t, q)
2. Find: new flux density B₂
3. Formula: V_H = BI/(ntq) ⇒ V_H ∝ B
4. Rearrange: B₂ = B₁ × (V_H₂ / V_H₁)
5. Substitute: B₂ = 0.30 × (20 / 12)
6. Calculate: B₂ = 0.30 × 1.667 = 0.50 T
Answer: B₂ = 0.50 T
Alternating Current (Paper 4)
Sinusoidal variation of current and voltage.
AC voltage or current
Use when: Finding instantaneous value of AC signal at time t
Variables: x = instantaneous value (V or A), x₀ = peak value, ω = 2πf (rad/s), f = frequency (Hz)
UK mains: V₀ ≈ 325 V (peak), f = 50 Hz, so V = 325sin(100πt)
Worked Example:
An AC supply has peak voltage 340 V and frequency 50 Hz. Find the voltage at t = 2.5 ms.
1. Given: V₀ = 340 V, f = 50 Hz, t = 2.5 ms = 2.5 × 10⁻³ s
2. Find: instantaneous voltage V
3. Formula: V = V₀ sin(ωt), with ω = 2πf
4. Substitute: ω = 2π × 50 = 100π rad/s; ωt = 100π × 2.5 × 10⁻³ = π/4
5. Calculate: sin(π/4) = 0.7071; V = 340 × 0.7071 ≈ 240.4 V
Answer: V ≈ 240 V
RMS (root mean square) value
Use when: Calculating power, or comparing AC to equivalent DC (sinusoidal AC only)
Key insight: RMS value gives the same power in a resistor as a DC value of the same size
UK mains: V_rms = 230 V, so V₀ = 230 × √2 ≈ 325 V peak
Worked Example:
Find the average power dissipated in a 100 Ω resistor connected to 230 V rms supply.
1. Given: V_rms = 230 V, R = 100 Ω
2. Find: average power P
3. Formula: P = V_rms² / R
4. Substitute: P = (230)² / 100
5. Calculate: P = 52 900 / 100 = 529 W
Answer: P = 529 W
Practice:
An AC current has peak value 5.0 A. Find the RMS current.
Show Answer
1. Given: peak current I₀ = 5.0 A
2. Find: RMS current I_rms
3. Formula: I_rms = I₀ / √2
4. Substitute: I_rms = 5.0 / √2
5. Calculate: I_rms = 5.0 / 1.414 ≈ 3.54 A
Answer: I_rms ≈ 3.5 A
Quantum Physics (Paper 4)
Wave-particle duality and the photoelectric effect.
Worked Example:
Find the energy of a photon with wavelength 500 nm (green light).
1. Given: λ = 500 nm = 5.00 × 10⁻⁷ m, h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s
2. Find: photon energy E
3. Formula: E = hc / λ
4. Substitute: E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (5.00 × 10⁻⁷)
5. Calculate: E = 1.989 × 10⁻²⁵ / 5.00 × 10⁻⁷ = 3.98 × 10⁻¹⁹ J
Answer: E ≈ 3.98 × 10⁻¹⁹ J (≈ 2.49 eV)
Practice:
Find the frequency of a 4.0 eV photon. (1 eV = 1.60 × 10⁻¹⁹ J)
Show Answer
1. Given: E = 4.0 eV (convert: 4.0 × 1.60 × 10⁻¹⁹ = 6.4 × 10⁻¹⁹ J); h = 6.63 × 10⁻³⁴ J s
2. Find: frequency f
3. Formula: E = hf
4. Rearrange: f = E / h
5. Substitute: f = 6.4 × 10⁻¹⁹ / 6.63 × 10⁻³⁴
6. Calculate: f ≈ 9.65 × 10¹⁴ Hz
Answer: f ≈ 9.7 × 10¹⁴ Hz
Worked Example:
UV light (λ = 200 nm) hits a metal with work function 4.5 eV. Find the max KE of photoelectrons.
1. Given: λ = 200 nm = 2.00 × 10⁻⁷ m, Φ = 4.5 eV; h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s
2. Find: max KE of photoelectrons
3. Formula: hf = Φ + KE_max ⇒ KE_max = hc/λ − Φ
4. Substitute: hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (2.00 × 10⁻⁷) = 9.95 × 10⁻¹⁹ J ≈ 6.22 eV
5. Calculate: KE_max = 6.22 − 4.5 = 1.72 eV (= 1.72 × 1.60 × 10⁻¹⁹ ≈ 2.75 × 10⁻¹⁹ J)
Answer: KE_max ≈ 1.72 eV (≈ 2.75 × 10⁻¹⁹ J)
Practice:
A metal has threshold wavelength 450 nm. Find the work function in eV.
Show Answer
1. Given: threshold λ₀ = 450 nm = 4.50 × 10⁻⁷ m; h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s
2. Find: work function Φ (in eV)
3. Formula: Φ = hc / λ₀
4. Substitute: Φ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (4.50 × 10⁻⁷)
5. Calculate: Φ = 1.989 × 10⁻²⁵ / 4.50 × 10⁻⁷ = 4.42 × 10⁻¹⁹ J
6. Convert: Φ = 4.42 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ ≈ 2.76 eV
Answer: Φ ≈ 2.76 eV
Use when: Electron diffraction (evidence for the wave nature of particles)
Worked Example:
Find the de Broglie wavelength of an electron accelerated through 100 V.
1. Given: V = 100 V; e = 1.60 × 10⁻¹⁹ C, m_e = 9.11 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J s
2. Find: de Broglie wavelength λ
3. Formula: KE = eV; v = √(2KE / m); λ = h / (mv)
4. Substitute KE: KE = 1.60 × 10⁻¹⁹ × 100 = 1.60 × 10⁻¹⁷ J
5. Substitute v: v = √(2 × 1.60 × 10⁻¹⁷ / 9.11 × 10⁻³¹) ≈ 5.93 × 10⁶ m/s
6. Calculate λ: λ = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 5.93 × 10⁶) ≈ 1.23 × 10⁻¹⁰ m
Answer: λ ≈ 1.23 × 10⁻¹⁰ m (0.123 nm)
Practice:
A neutron (m = 1.67 × 10⁻²⁷ kg) has wavelength 0.10 nm. Find its speed.
Show Answer
1. Given: m = 1.67 × 10⁻²⁷ kg, λ = 0.10 nm = 1.0 × 10⁻¹⁰ m, h = 6.63 × 10⁻³⁴ J s
2. Find: speed v of the neutron
3. Formula: λ = h / (mv)
4. Rearrange: v = h / (mλ)
5. Substitute: v = 6.63 × 10⁻³⁴ / (1.67 × 10⁻²⁷ × 1.0 × 10⁻¹⁰)
6. Calculate: v = 6.63 × 10⁻³⁴ / 1.67 × 10⁻³⁷ ≈ 3970 m/s
Answer: v ≈ 3.97 × 10³ m/s (≈ 4000 m/s)
Emission: Electron drops to lower level, photon released
Nuclear Physics (Paper 4)
Mass-energy equivalence and radioactive decay.
Shortcut: 1 u of mass ≈ 931 MeV of energy (not on the data sheet and not required; the data-sheet values 1 u = 1.66 × 10⁻²⁷ kg and c = 3.00 × 10⁸ m s⁻¹ give about 934 MeV, so working from E = c²Δm in joules can differ in the third figure)
Worked Example:
A nuclear reaction releases 0.0035 u of mass. Find the energy released in MeV.
1. Given: Δm = 0.0035 u; mass-energy shortcut: 1 u ↔ 931 MeV
2. Find: energy released E (in MeV)
3. Formula: E = Δm c² (using 1 u = 931 MeV/c²)
4. Substitute: E = 0.0035 × 931
5. Calculate: E ≈ 3.26 MeV
Answer: E ≈ 3.26 MeV
Practice:
In U-235 fission, 200 MeV is released per nucleus. Find the mass converted to energy per fission.
Show Answer
1. Given: E = 200 MeV per fission; 1 u ↔ 931 MeV/c²
2. Find: mass converted to energy Δm
3. Formula: E = Δm c² ⇒ Δm = E / 931 (in u, when E in MeV)
4. Substitute: Δm = 200 / 931
5. Calculate: Δm ≈ 0.215 u; converting: 0.215 × 1.66 × 10⁻²⁷ ≈ 3.57 × 10⁻²⁸ kg
Answer: Δm ≈ 0.215 u (≈ 3.57 × 10⁻²⁸ kg)
Binding energy: Minimum energy needed to separate a nucleus into its individual nucleons
Worked Example:
A sample contains 2.0 × 10²⁰ atoms with half-life 5.0 days. Find the initial activity.
1. Given: N = 2.0 × 10²⁰ atoms, t½ = 5.0 days = 5.0 × 24 × 3600 s = 4.32 × 10⁵ s
2. Find: initial activity A
3. Formula: λ = ln2 / t½; A = λN
4. Substitute λ: λ = 0.693 / 4.32 × 10⁵ ≈ 1.60 × 10⁻⁶ s⁻¹
5. Substitute A: A = 1.60 × 10⁻⁶ × 2.0 × 10²⁰
6. Calculate: A = 3.21 × 10¹⁴ Bq
Answer: A ≈ 3.2 × 10¹⁴ Bq
Practice:
A source has activity 8.0 × 10⁸ Bq and decay constant 2.0 × 10⁻⁶ s⁻¹. How many nuclei remain?
Show Answer
1. Given: A = 8.0 × 10⁸ Bq, λ = 2.0 × 10⁻⁶ s⁻¹
2. Find: number of undecayed nuclei N
3. Formula: A = λN
4. Rearrange: N = A / λ
5. Substitute: N = 8.0 × 10⁸ / 2.0 × 10⁻⁶
6. Calculate: N = 4.0 × 10¹⁴
Answer: N = 4.0 × 10¹⁴ nuclei
Use when: Finding remaining amount/activity after time t
Note: 0.693 = ln(2)
Medical Physics - Ultrasound (Paper 4)
Reflection at tissue boundaries.
Intensity reflection coefficient
Use when: Calculating reflection at tissue boundaries in ultrasound imaging
Variables: I_R/I₀ = fraction of intensity reflected, Z = acoustic impedance = ρc (kg m⁻² s⁻¹)
If Z₁ = Z₂: no reflection (matched impedance). If Z₁ >> Z₂ or Z₁ << Z₂: nearly total reflection
Worked Example:
Find the fraction of ultrasound reflected at a muscle-bone boundary. (Z_muscle = 1.7 × 10⁶, Z_bone = 6.4 × 10⁶ kg m⁻² s⁻¹)
1. Given: Z₁ = Z_muscle = 1.7 × 10⁶, Z₂ = Z_bone = 6.4 × 10⁶ kg m⁻² s⁻¹
2. Find: reflected intensity fraction I_R / I₀
3. Formula: I_R / I₀ = (Z₁ − Z₂)² / (Z₁ + Z₂)²
4. Substitute: I_R / I₀ = (1.7 − 6.4)² × 10¹² / (1.7 + 6.4)² × 10¹² = (−4.7)² / (8.1)²
5. Calculate: I_R / I₀ = 22.09 / 65.61 ≈ 0.337
Answer: I_R / I₀ ≈ 0.34 (≈ 34% reflected)
Practice:
Why is gel used between the probe and skin? (Z_air ≈ 400, Z_skin = 1.7 × 10⁶)
Show Answer
1. Given: Z_air ≈ 400, Z_skin = 1.7 × 10⁶ kg m⁻² s⁻¹
2. Find: reflected intensity fraction (without gel)
3. Formula: I_R / I₀ = (Z₁ − Z₂)² / (Z₁ + Z₂)²
4. Substitute: I_R / I₀ = (400 − 1.7 × 10⁶)² / (400 + 1.7 × 10⁶)²
5. Calculate: ≈ (1.6996 × 10⁶)² / (1.7004 × 10⁶)² ≈ 0.9991
Answer: I_R / I₀ ≈ 0.999 (≈ 99.9% reflected by air gap; gel matches impedance to skin)
Astronomy & Cosmology (Paper 4)
Stars, radiation, and the expanding universe.
Use when: Finding star's power output from size and temperature
Use when: Finding peak wavelength from temperature, or vice versa
Worked Example:
A star has surface temperature 6000 K. Find the wavelength at which it emits most intensely.
1. Given: T = 6000 K; Wien's constant = 2.898 × 10⁻³ m K
2. Find: peak emission wavelength λ_max
3. Formula: λ_max × T = 2.898 × 10⁻³ m K
4. Rearrange: λ_max = 2.898 × 10⁻³ / T
5. Substitute: λ_max = 2.898 × 10⁻³ / 6000
6. Calculate: λ_max = 4.83 × 10⁻⁷ m (483 nm)
Answer: λ_max ≈ 4.83 × 10⁻⁷ m (483 nm, blue-green)
Practice:
A red giant has peak emission at 700 nm. Find its surface temperature.
Show Answer
1. Given: λ_max = 700 nm = 7.00 × 10⁻⁷ m; Wien's constant = 2.898 × 10⁻³ m K
2. Find: surface temperature T
3. Formula: λ_max × T = 2.898 × 10⁻³
4. Rearrange: T = 2.898 × 10⁻³ / λ_max
5. Substitute: T = 2.898 × 10⁻³ / 7.00 × 10⁻⁷
6. Calculate: T ≈ 4140 K
Answer: T ≈ 4140 K
Sign: Redshift (+ve Δλ) = moving away, Blueshift (-ve Δλ) = moving towards
Key insight: Universe is expanding - more distant galaxies recede faster
Worked Example:
A galaxy is 1.5 × 10²⁴ m away. Find its recession velocity. (H₀ = 2.3 × 10⁻¹⁸ s⁻¹)
1. Given: d = 1.5 × 10²⁴ m, H₀ = 2.3 × 10⁻¹⁸ s⁻¹
2. Find: recession velocity v
3. Formula: v ≈ H₀ d
4. Substitute: v = 2.3 × 10⁻¹⁸ × 1.5 × 10²⁴
5. Calculate: v = 3.45 × 10⁶ m/s
Answer: v ≈ 3.5 × 10⁶ m/s
Practice:
A galaxy recedes at 2.1 × 10⁷ m/s. How far away is it? (H₀ = 2.3 × 10⁻¹⁸ s⁻¹)
Show Answer
1. Given: v = 2.1 × 10⁷ m/s, H₀ = 2.3 × 10⁻¹⁸ s⁻¹
2. Find: distance d
3. Formula: v ≈ H₀ d
4. Rearrange: d = v / H₀
5. Substitute: d = 2.1 × 10⁷ / 2.3 × 10⁻¹⁸
6. Calculate: d ≈ 9.13 × 10²⁴ m
Answer: d ≈ 9.1 × 10²⁴ m
More A-Level Resources
LumiExams.com | Cambridge A-Level Physics (9702) | Syllabus 2025-2027
Frequently Asked Questions
Do I need to memorize formulas for A-Level Physics?
Yes, many of them. The data sheet in Papers 1, 2 and 4 gives some formulas, but the syllabus expects you to recall others, such as F = ma. You also need to understand what each formula means, when to use it, and how to apply it to solve problems. This guide focuses on that.
Which formulas are given in A-Level Physics exams?
Papers 1, 2 and 4 carry a data and formulae page (page 2), with constants and a short list of formulae: the uniformly accelerated motion equations, hydrostatic pressure, upthrust, the Doppler effect for sound, electric current (I = Anvq) and resistors in series and in parallel. Paper 4 adds page 3, with gravitational potential and gravitational potential energy, the pressure of an ideal gas, simple harmonic motion (a = −ω²x, v = v₀ cos ωt and v = ±ω√(x₀² − x²)), electric potential and electric potential energy, capacitors in series and in parallel, capacitor discharge, Hall voltage, alternating current, radioactive decay and the decay constant, the intensity reflection coefficient, the Stefan–Boltzmann law and Doppler redshift. Equations that are not on these pages, such as F = ma, F = mv²/r, g = GM/r², pV = nRT and E = hf, you need to recall.
How do I know which formula to use in A-Level Physics?
Identify what you're given (known values) and what you need to find (unknown). Choose a formula that connects these quantities. For example, if given initial velocity, acceleration, and time, and asked for displacement, use s = ut + ½at². This guide explains when to use each formula.
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