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Physics is a quantitative science. Every law, every derivation, every experimental result is ultimately a statement about a number and the unit it is attached to. Get the unit wrong and the number is meaningless — a value of "10" for the resistance of a wire is useful only if we know it is 10 ohms and not 10 kilo-ohms or 10 milli-ohms. The International System of Units (SI) is the universal language that physicists use to make their results unambiguous, comparable and reproducible.
This opening lesson of the AQA A-Level Physics course establishes the foundation on which every subsequent topic — mechanics, electricity, fields, quantum phenomena — will be built. We will look at the seven base quantities and their SI units, the derived units that flow from them, and the powerful technique of dimensional analysis that lets us check the validity of an equation or even derive a result for an unfamiliar physical quantity.
Spec mapping: This lesson addresses content from AQA 7408 §3.1 covering the use of SI base units and derived units, and the ability to convert between units and to derive units for unfamiliar quantities (refer to the official AQA specification document for exact wording).
Synoptic links:
- Mechanics (§3.4): every quantity from displacement to power is built from the SI base units introduced here; dimensional analysis is your first sanity check on any kinematics or dynamics derivation.
- Electricity (§3.5): the volt, ohm, watt and farad are all derived units — being able to express them in base units is examined explicitly when the spec asks you to "show that the unit of capacitance reduces to A² s⁴ kg⁻¹ m⁻²".
- Required practicals: every required practical demands that measurements be recorded with correct SI units and consistent prefixes; markschemes regularly penalise candidates who give a current in "amps" without the symbol A or who mix kilometres with metres in the same calculation.
The SI defines seven physical quantities as base quantities. These are taken as fundamental — the units of every other physical quantity are derived from them by multiplication, division and exponentiation. You are expected to know all seven, their symbols, and the rough physical idea each one captures.
| Quantity | SI Unit | Symbol | Defines... |
|---|---|---|---|
| Mass | kilogram | kg | inertia and gravitational coupling |
| Length | metre | m | spatial extent |
| Time | second | s | duration |
| Electric current | ampere | A | rate of charge flow |
| Temperature | kelvin | K | absolute thermal energy scale |
| Amount of substance | mole | mol | number of elementary entities |
| Luminous intensity | candela | cd | visible radiant power |
Two of these (the kelvin and the mole) are sometimes overlooked because they appear less frequently in everyday A-Level mechanics, but they are examined explicitly in thermal physics and in radioactive-decay / N_A questions.
The kilogram is the only SI base unit whose name and symbol already contain a prefix ("kilo-"). This is a historical accident: when the SI was formalised, the gram was already in use as the standard mass unit, and the practical reference standard was the International Prototype Kilogram. A practical consequence for exam work is that you must derive units from kilograms, not grams — e.g. the joule is kg m² s⁻², not g m² s⁻².
In May 2019 the kilogram, ampere, kelvin and mole were redefined in terms of fixed numerical values of the Planck constant, the elementary charge, the Boltzmann constant and the Avogadro constant respectively. The metre and second remain defined via the speed of light and the caesium hyperfine transition. You will not be examined on the formal definitions, but it is useful background: it is no longer the case that the kilogram is "whatever the platinum-iridium cylinder in Paris weighs".
All other physical quantities are expressed in terms of the seven base units. These combinations are called derived units. Many derived units have special names (newton, joule, watt, volt, ohm, ...) but every one of them can be rewritten in base units.
| Derived quantity | Special-name unit | In base units |
|---|---|---|
| Frequency | hertz, Hz | s⁻¹ |
| Force | newton, N | kg m s⁻² |
| Pressure / stress | pascal, Pa | kg m⁻¹ s⁻² |
| Energy / work | joule, J | kg m² s⁻² |
| Power | watt, W | kg m² s⁻³ |
| Electric charge | coulomb, C | A s |
| Electric potential | volt, V | kg m² s⁻³ A⁻¹ |
| Resistance | ohm, Ω | kg m² s⁻³ A⁻² |
| Capacitance | farad, F | A² s⁴ kg⁻¹ m⁻² |
| Magnetic flux density | tesla, T | kg s⁻² A⁻¹ |
A few derived quantities (e.g. strain, refractive index, the radian) are dimensionless because they are defined as ratios of two like quantities. Note that "dimensionless" does not mean "unitless on the page" — the radian and steradian are SI named units even though they are dimensionally pure numbers.
graph TD
BASE[7 SI base units] --> M[metre m]
BASE --> KG[kilogram kg]
BASE --> S[second s]
BASE --> A[ampere A]
M --> AREA[m² area]
M --> VOL[m³ volume]
M --> N["m s⁻² acceleration"]
KG --> N2["N = kg m s⁻² force"]
M --> N2
S --> N2
N2 --> J["J = N m energy"]
M --> J
J --> W["W = J s⁻¹ power"]
S --> W
A --> C["C = A s charge"]
S --> C
W --> V["V = W A⁻¹ potential"]
A --> V
V --> OHM["Ω = V A⁻¹ resistance"]
A --> OHM
style BASE fill:#1f4e79,color:#fff
style N2 fill:#27ae60,color:#fff
style J fill:#27ae60,color:#fff
style V fill:#27ae60,color:#fff
The diagram shows how a small set of base units fans out to all the named derived units you will meet in the course. Whenever you are unsure whether two quantities are dimensionally consistent, trace each one back to base units along edges like these.
Every physical equation must be dimensionally homogeneous — every additive term must have the same dimensions, and the dimensions on the two sides of an equals sign must match. This is a powerful sanity check that catches algebraic slips before they earn you a wrong final answer.
Worked example. A student writes the equation s = ut + ½at³ for displacement under constant acceleration. Is it dimensionally consistent?
The left-hand side s has units of m.
The first term ut has units (m s⁻¹)(s) = m. ✓
The second term ½at³ has units (m s⁻²)(s³) = m s. ✗
The second term has units of m s, not m, so the equation cannot be correct. (The correct expression is s = ut + ½at², with t² giving m s⁻² × s² = m.) Dimensional analysis has caught the error without us needing to know the kinematic derivation.
Worked example. A student claims that the energy stored in an inductor is E = ½LI², where L is in henries (V s A⁻¹) and I is in amperes. Show that this is dimensionally consistent with energy in joules.
Units of L = V s A⁻¹ = (kg m² s⁻³ A⁻¹) × s × A⁻¹ = kg m² s⁻² A⁻²
Units of LI² = (kg m² s⁻² A⁻²) × A² = kg m² s⁻²
These are the base units of the joule. ✓
A common exam question asks you to derive the SI base units for a quantity that does not have a special-name unit, or to express a special-name unit (like the volt or the ohm) entirely in base units. The approach is always the same: start from a defining equation, substitute base units for each symbol, and simplify.
Worked example — Young modulus. E = stress / strain = (F/A) / (ΔL/L).
Units of F = kg m s⁻². Units of A = m². Strain is dimensionless (m/m).
So units of E = (kg m s⁻²) / m² = kg m⁻¹ s⁻² = Pa.
Worked example — electrical resistance. R = V/I. Voltage V = W/Q = energy/charge. Energy has units kg m² s⁻²; charge Q = It has units A s.
So V has units (kg m² s⁻²) / (A s) = kg m² s⁻³ A⁻¹.
Therefore R = V/I has units (kg m² s⁻³ A⁻¹) / A = kg m² s⁻³ A⁻² (the ohm Ω).
Worked example — capacitance. C = Q/V. Q has units A s; V has units kg m² s⁻³ A⁻¹.
So C has units (A s) / (kg m² s⁻³ A⁻¹) = A² s⁴ kg⁻¹ m⁻² (the farad F).
Worked example — magnetic flux density (the tesla). Use F = BIL, so B = F/(IL). F is kg m s⁻²; IL is A m.
Therefore B has units (kg m s⁻²) / (A m) = kg s⁻² A⁻¹ (the tesla T).
When answers involve negative powers, present them with negative indices rather than fractions in formal physics writing (kg m⁻¹ s⁻², not kg / (m s²)). Both notations are mathematically correct, but the negative-index style is universal in published physics and is what AQA mark schemes use.
Beyond unit-checking, dimensional analysis can sometimes be used to derive the form of an unknown equation. The classic example is the period of a simple pendulum.
Worked example. A simple pendulum has a bob of mass m, a string of length L, in a gravitational field of strength g. What is the period T?
We guess T depends on some combination T ∝ mᵃ Lᵇ gᶜ. The units on each side must match.
| Quantity | Units |
|---|---|
| T | s |
| m | kg |
| L | m |
| g | m s⁻² |
So units of mᵃ Lᵇ gᶜ = kgᵃ × mᵇ × (m s⁻²)ᶜ = kgᵃ mᵇ⁺ᶜ s⁻²ᶜ.
Equating to s¹:
So T ∝ L^(1/2) g^(−1/2) = √(L/g). The result agrees with the full derivation T = 2π√(L/g) up to the dimensionless prefactor, which dimensional analysis alone cannot give. Mass m does not appear — a result that the dimensional argument delivers immediately.
This technique appears in mark schemes for "show that" questions, especially in fields and oscillations, where candidates can confirm the form of a result without re-deriving it from first principles.
A car engine produces a power of 75 kW. Express this in fundamental SI base units. The kilowatt is 10³ W, and the watt is the derived unit J s⁻¹ = (kg m² s⁻²) × s⁻¹ = kg m² s⁻³.
So 75 kW = 75 × 10³ kg m² s⁻³ = 7.5 × 10⁴ kg m² s⁻³.
Worked example. The pressure inside a car tyre is 220 kPa. Express this in base units.
Pa = kg m⁻¹ s⁻², so 220 kPa = 2.2 × 10⁵ kg m⁻¹ s⁻².
A satisfying illustration of how far dimensional analysis can take you without invoking the underlying dynamics is the derivation of Kepler's third law for a planet of mass m orbiting a star of mass M at semi-major-axis a. The empirical observation is that the orbital period T depends only on a, M and the gravitational constant G, in some combination T = k Mᵃ aᵇ Gᶜ where k is a dimensionless constant of order unity. Dimensional analysis fixes the exponents a, b, c without any reference to Newton's second law or to the conic-section nature of the orbit.
We need T to have units of seconds. The units of each input quantity are:
| Quantity | SI base units |
|---|---|
| T | s |
| M | kg |
| a | m |
| G | kg⁻¹ m³ s⁻² |
The units of Mᵃ aᵇ Gᶜ are kgᵃ × mᵇ × (kg⁻¹ m³ s⁻²)ᶜ = kg^(a−c) × m^(b+3c) × s^(−2c).
Equating to s¹ (which is kg⁰ m⁰ s¹):
So a = −½ and b = +3/2. Substituting:
T = k × M^(−1/2) × a^(3/2) × G^(−1/2) = k × √(a³ / GM)
Squaring both sides gives the famous form of Kepler's third law:
T² = (k² / GM) × a³ — that is, T² ∝ a³ for any given star.
Newtonian dynamics fixes the constant: k² = 4π², so T² = 4π² a³ / (GM). Dimensional analysis alone cannot recover the 4π² prefactor — dimensionless numbers are invisible to this technique — but it gives you the entire functional form for free. The same procedure works for the Bohr radius (where m_e, e, ε₀ and ℏ combine to give a length), for the Schwarzschild radius (where G, M and c combine to give a length 2GM/c²), and for the speed of waves on a stretched string (where T, μ combine to give v = √(T/μ)).
The wider lesson is that whenever a problem admits only one dimensionally-consistent combination of the input quantities, dimensional analysis pins down the entire form of the answer up to a numerical prefactor. Examiner-style questions on this topic appear in the AQA "fields" content and in the optional astrophysics module; preparing the technique now will repay you across the course.
Specimen question modelled on the AQA paper format. Total: 6 marks.
"The escape speed v from the surface of a planet of mass M and radius R is given by v = √(2GM/R), where G is the gravitational constant.
(a) State the SI base units of v. (1 mark)
(b) The SI base units of G are kg⁻¹ m³ s⁻². Using the formula for v, show by dimensional analysis that the equation is dimensionally consistent. (3 marks)
(c) A student claims a different formula v = √(GM/R²) for the same quantity. By dimensional analysis, show that this formula is dimensionally incorrect. (2 marks)"
| Part | AO | Marks | What is rewarded |
|---|---|---|---|
| (a) | AO1 (knowledge) | 1 | Stating m s⁻¹ |
| (b) | AO2 (application of physics) | 3 | Substituting base units, simplifying, showing both sides reduce to m s⁻¹ |
| (c) | AO2 (application) + AO3 (analysis) | 2 | Showing the candidate formula yields m s⁻² and recognising it cannot be a velocity |
Grade C response (typical 3-4 marks out of 6).
(a) The SI base units of v are m s⁻¹.
(b) GM has units kg⁻¹ m³ s⁻² × kg = m³ s⁻². Dividing by R gives m³ s⁻² / m = m² s⁻². Taking the square root gives m s⁻¹. This matches v, so the equation is dimensionally consistent.
(c) GM/R² has units m³ s⁻² / m² = m s⁻². The square root of m s⁻² is not m s⁻¹, so the formula is wrong.
Examiner commentary: The candidate scores M1 for (a) and the structural marks for (b) — substituting G in base units (M1), reducing the inside of the square root (M1), recognising the final m s⁻¹ (M1). In (c) they identify the unit of GM/R² correctly (M1) but write only "the square root is not m s⁻¹" without explicitly working it through, losing the second mark. Total: 5/6.
Grade A* response (full 6 marks).
(a) v has SI base units of m s⁻¹.
(b) Inside the square root we have 2GM/R. The factor 2 is dimensionless. GM has units (kg⁻¹ m³ s⁻²)(kg) = m³ s⁻². Dividing by R (units m) gives m² s⁻². The square root of m² s⁻² is m s⁻¹, which is identical to the SI base units of v stated in (a). The equation is therefore dimensionally homogeneous.
(c) GM has units m³ s⁻² as above. Dividing by R² (units m²) gives m³ s⁻² / m² = m s⁻². The square root of m s⁻² is m^(1/2) s⁻¹, which is not the unit of a velocity. The proposed formula therefore cannot be dimensionally correct for an escape speed. (Note: m s⁻² is the dimension of acceleration; the wrong index on R has produced an expression for the square root of an acceleration.)
Examiner commentary: The A* response scores every mark — clean unit working with the dimensionless 2 noted, full simplification, and the additional physical insight in (c) that the wrongly-indexed formula has the dimensions of √(acceleration). Examiners reward the interpretation step ("m s⁻² is the dimension of acceleration") as evidence of fluent reasoning rather than just unit chasing.
Specimen question modelled on the AQA paper format. Total: 6 marks.
"The capacitance C of a parallel-plate capacitor is defined by Q = CV, where Q is the charge stored on the plates and V is the potential difference across them.
(a) State the SI base units of charge Q and of potential difference V. (2 marks)
(b) Hence show that the unit of capacitance (the farad, F) reduces to A² s⁴ kg⁻¹ m⁻² when expressed in SI base units. (3 marks)
(c) The time constant of an RC circuit is the product τ = RC. By expressing the ohm and the farad in base units, show that τ has the units of seconds. (1 mark)"
| Part | AO | Marks | What is rewarded |
|---|---|---|---|
| (a) | AO1 | 2 | Q in A s; V in kg m² s⁻³ A⁻¹ |
| (b) | AO2 | 3 | Substituting, dividing Q by V, simplifying to A² s⁴ kg⁻¹ m⁻² |
| (c) | AO2 + AO3 | 1 | Showing the kg, m and A indices cancel, leaving s¹ |
Grade C response (typical 3-4 marks out of 6).
(a) Q is in A s and V is in kg m² s⁻³ A⁻¹.
(b) C = Q / V = (A s) / (kg m² s⁻³ A⁻¹) = A s × A / (kg m² s⁻³) = A² s × s³ / (kg m²) = A² s⁴ / (kg m²) = A² s⁴ kg⁻¹ m⁻². So the unit of the farad in base units is A² s⁴ kg⁻¹ m⁻².
(c) Ohm is V/A = (kg m² s⁻³ A⁻¹) / A = kg m² s⁻³ A⁻². Farad is A² s⁴ kg⁻¹ m⁻². Multiplying: kg m² s⁻³ A⁻² × A² s⁴ kg⁻¹ m⁻² = s. So τ = RC is in seconds.
Examiner commentary: The candidate scores both knowledge marks in (a) (M1 + M1). In (b) the algebra is shaky in the middle — moving the A⁻¹ from the denominator of V to the numerator of C is correct but expressed inelegantly — yet the final answer is right (M1 + M1, losing the third mark for the messy intermediate step). In (c) the cancellation is correct and the units of seconds emerge cleanly (M1). Total: 5/6.
Grade A* response (full 6 marks).
(a) From Q = It, the unit of charge is the ampere-second: Q has SI base units of A s. Potential difference is energy per unit charge: V = E/Q = J / C. Since J = kg m² s⁻² and C = A s, V has SI base units of kg m² s⁻² / (A s) = kg m² s⁻³ A⁻¹.
(b) Capacitance is defined by C = Q / V, so the SI base units of C are (A s) / (kg m² s⁻³ A⁻¹).
Inverting the denominator: (A s) × (A s³ kg⁻¹ m⁻²) = A² × s × s³ × kg⁻¹ × m⁻² = A² s⁴ kg⁻¹ m⁻².
Therefore F = A² s⁴ kg⁻¹ m⁻², as required.
(c) Ω = V / A = (kg m² s⁻³ A⁻¹) / A = kg m² s⁻³ A⁻². Then τ = RC has base units (kg m² s⁻³ A⁻²) × (A² s⁴ kg⁻¹ m⁻²) = kg^(1−1) m^(2−2) s^(−3+4) A^(−2+2) = kg⁰ m⁰ s¹ A⁰ = s, confirming τ is a time as expected.
Examiner commentary: The A* response is identifiable by the explicit two-line derivation of V from J/C (rather than the bare statement of the ohm in base units), the clean inversion step in (b), and the index-by-index cancellation table in (c). Examiners reward this layered presentation: each step is justified, every intermediate unit is named, and the final dimension is shown to emerge by cancellation rather than asserted. The "show that" command word is satisfied with no ambiguity.
Dimensional analysis is the foundation of a subject called dimensional homogeneity in fluid dynamics (the Buckingham π theorem, undergraduate engineering). It also underpins natural units in particle physics, where you set ℏ = c = 1 and read off the dimensions of every quantity in powers of energy. For an introduction, see chapter 1 of Tipler and Mosca's Physics for Scientists and Engineers or any first-year fluid-mechanics text.
The Buckingham π theorem generalises the technique used above for Kepler's third law to any problem with n physical variables and k fundamental dimensions. The theorem states that the dependence between those n variables can be re-expressed in terms of (n − k) independent dimensionless products, often called π-groups. In aeronautical and chemical engineering this is the standard route to identifying the dimensionless numbers that govern a problem — the Reynolds number Re = ρvL/μ for fluid flow, the Mach number M = v/c_s for compressible flow, the Froude number for free-surface waves. Once you know the relevant dimensionless groups, scaled-down models in wind tunnels and water flumes can be used to predict full-scale behaviour without ever solving the underlying differential equations. Cambridge and Imperial engineering departments examine this technique in their first-year fluid-mechanics papers, and it is a frequent topic in materials-science and biophysics interview discussions for top-tier university applicants. The connection to A-Level physics is direct: Kepler's law derivation above is a Buckingham-π exercise with two π-groups (T²GM/a³ being the dimensionless combination that comes out as a constant).
A subtle A-Level mistake is treating the radian as a "dimensionless unit that disappears". In angular kinematics ω = v/r, the units of v/r are m s⁻¹ / m = s⁻¹. The "rad" is conventional — strictly the SI rules treat rad s⁻¹ and s⁻¹ as equivalent. This becomes important when checking the units of equations involving ω and centripetal acceleration: a = ω²r has units (s⁻¹)²(m) = m s⁻², which is correct, even though many students worry that the radian "should appear somewhere".
A second subtle dimensional-checking error concerns strain, the fractional extension Δx / x in stress-strain problems. Although strain is a ratio of two lengths and is dimensionally m / m, students who write "strain has no dimensions" are technically slightly off-piste — the correct statement is that strain is a dimensionless quantity (or, equivalently, has dimensions of 1, written as M⁰ L⁰ T⁰). The distinction matters when checking the units of Young's modulus E = stress / strain: stress carries units of Pa = kg m⁻¹ s⁻², and dividing by the dimensionless strain leaves E in the same units as stress — Pa, not Pa per metre or Pa per anything-else. A small minority of candidates write E in "Pa / strain" or "Pa per fractional extension"; this is dimensionally identical to Pa but is not accepted in markschemes as a final answer.
A third related subtlety is the angle in radians appearing in formulae. The small-angle approximation sin θ ≈ θ assumes θ is in radians, and the formula for arc length s = rθ similarly requires radians. In dimensional terms the radian is dimensionless, so s = rθ has units m × 1 = m, consistent. The formula booklet writes the radian symbol explicitly in some equations (e.g. ω in rad s⁻¹) as a notational convenience, but a strict dimensional check treats rad as M⁰ L⁰ T⁰. The same convention applies to the steradian (solid angle) and to the radian's appearance in rotational quantities such as moment of inertia times angular acceleration.
| Concept | Key Point |
|---|---|
| SI base units | 7 fundamental: kg, m, s, A, K, mol, cd |
| Derived units | All other units are products/quotients of base units |
| Special-name units | N, J, W, Pa, V, Ω, F, T etc. can each be expanded into base units |
| Dimensional homogeneity | Both sides of an equation must have the same units; every additive term too |
| Deriving units | Start from a defining equation, substitute base units, simplify |
| Pendulum-style derivation | Dimensional analysis can pin down the exponents but not the dimensionless prefactor |
This content is aligned with the AQA A-Level Physics (7408) specification.