Derived Units and Homogeneity of Equations

Spec mapping: OCR H556 Module 2.1 — Physical quantities and units, specifically the derivation of units in terms of base units and the principle of homogeneity. (Refer to the official OCR H556 specification document for exact wording.)

In the previous lesson you met the seven SI base units (kg, m, s, A, K, mol, cd). Almost every other physical quantity that appears across the H556 specification — force, energy, pressure, voltage, resistance, frequency, magnetic flux density, capacitance, electric field strength — is expressed as a derived unit, built from those seven via multiplication, division and integer powers.

This lesson does two things. First, it works through the most important rows of the derived-units table that OCR expects you to be able to recall on demand. Second, and crucially, it develops the principle of homogeneity — the requirement that the units on the left-hand side of any physically valid equation match the units on the right-hand side. Homogeneity is one of the most powerful weapons in your A-Level toolkit: it costs almost no time, catches a huge class of equation-recall errors, and is explicitly rewarded in OCR mark schemes (typically as a 2–3 mark "show that" question on every paper).

Key definition: A derived unit is any unit that is not one of the seven SI base units but can be written as a product, quotient or power of them. A physical equation is homogeneous when the units on both sides — and on every additive term — are identical.

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What is a derived unit?

A derived unit is built from base units through the defining equation of the quantity it measures. Speed, for example, is defined as $v = s/t$v=s/t — a distance over a time — so its unit is metre per second, written $\text{m s}^{-1}$m s−1. There is no special symbol for the unit of speed; we simply read the combination.

For quantities that appear constantly throughout physics, the international community has given the derived unit a named symbol — newton for force, joule for energy, pascal for pressure, watt for power, hertz for frequency, ohm for resistance. These named units are still defined in terms of base units; the name is just a convenient shorthand. When OCR asks for "the unit of force in base units" the answer is $\text{kg m s}^{-2}$kg m s−2, not "the newton".

Why this matters: Many candidates lose marks at A-Level by answering "newton" when asked for the base-unit form of force. The newton is the named derived unit; its base-unit decomposition is kg m s⁻². The exam question wording usually says "in terms of base SI units" — read for that phrase before answering.

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The derived-units reference table

The table below is the single most important page of reference material in Module 2. You should aim to derive any row of it on demand from the defining equation, rather than memorising the right-hand column directly. That way, if exam pressure makes you forget that the volt is $\text{kg m}^{2}\text{ s}^{-3}\text{ A}^{-1}$kg m2 s−3 A−1, you can rebuild it in ten seconds from $V = W/Q = (\text{energy})/(\text{charge})$V=W/Q=(energy)/(charge).

Quantity	Defining equation	Named unit	Base-unit form
Area	$A = l^2$A=l2	m²	m²
Volume	$V = l^3$V=l3	m³	m³
Density	$\rho = m/V$ρ=m/V	kg m⁻³	kg m⁻³
Speed / velocity	$v = \Delta s / \Delta t$v=Δs/Δt	m s⁻¹	m s⁻¹
Acceleration	$a = \Delta v / \Delta t$a=Δv/Δt	m s⁻²	m s⁻²
Force	$F = ma$F=ma	newton (N)	kg m s⁻²
Pressure / stress	$p = F/A$p=F/A	pascal (Pa)	kg m⁻¹ s⁻²
Energy / work	$W = Fd$W=Fd	joule (J)	kg m² s⁻²
Power	$P = W/t$P=W/t	watt (W)	kg m² s⁻³
Charge	$Q = It$Q=It	coulomb (C)	A s
Potential difference	$V = W/Q$V=W/Q	volt (V)	kg m² s⁻³ A⁻¹
Resistance	$R = V/I$R=V/I	ohm ($\Omega$Ω)	kg m² s⁻³ A⁻²
Frequency	$f = 1/T$f=1/T	hertz (Hz)	s⁻¹
Magnetic flux density	$B = F/(IL)$B=F/(IL)	tesla (T)	kg s⁻² A⁻¹
Capacitance	$C = Q/V$C=Q/V	farad (F)	kg⁻¹ m⁻² s⁴ A²
Electric field strength	$E = V/d$E=V/d	V m⁻¹	kg m s⁻³ A⁻¹

The exotic ones at the bottom — capacitance, electric field strength — are exam favourites precisely because students rarely practise deriving them. You will not be asked to memorise the right-hand columns; you will be asked to derive them.

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Worked derivations from the table

The following six derivations are the most-tested rows of the table. Do them on paper twice through and they will stick.

Worked Example 1 — The newton

Force is defined by Newton's second law: $F = ma$F=ma. Substituting base units:

$[F] = [m] \times [a] = \text{kg} \times \text{m s}^{-2} = \text{kg m s}^{-2}$[F]=[m]×[a]=kg×m s−2=kg m s−2

So $1\,\text{N} \equiv 1\,\text{kg m s}^{-2}$1N≡1kg m s−2. The square-bracket notation $[X]$[X] is read "the units of $X$X" and is OCR-standard shorthand on every mark scheme.

Worked Example 2 — The joule

Work is force times distance moved in the direction of the force: $W = Fd$W=Fd. Substituting:

$[W] = [F] \times [d] = \text{kg m s}^{-2} \times \text{m} = \text{kg m}^{2}\text{ s}^{-2}$[W]=[F]×[d]=kg m s−2×m=kg m2 s−2

So $1\,\text{J} \equiv 1\,\text{kg m}^{2}\text{ s}^{-2}$1J≡1kg m2 s−2. The same answer should fall out of any energy formula. Quick check with kinetic energy $E_k = \tfrac{1}{2}mv^{2}$Ek​=21​mv2:

$[E_k] = \text{kg} \times (\text{m s}^{-1})^{2} = \text{kg m}^{2}\text{ s}^{-2}\,\checkmark$[Ek​]=kg×(m s−1)2=kg m2 s−2✓

Or gravitational potential energy $E_p = mgh$Ep​=mgh:

$[E_p] = \text{kg} \times \text{m s}^{-2} \times \text{m} = \text{kg m}^{2}\text{ s}^{-2}\,\checkmark$[Ep​]=kg×m s−2×m=kg m2 s−2✓

All energy units must agree. If they don't, you have used the wrong formula.

Worked Example 3 — The pascal

Pressure is force per unit area: $p = F/A$p=F/A. Substituting:

$[p] = \frac{\text{kg m s}^{-2}}{\text{m}^{2}} = \text{kg m}^{-1}\text{ s}^{-2}$[p]=m2kg m s−2​=kg m−1 s−2

So $1\,\text{Pa} \equiv 1\,\text{kg m}^{-1}\text{ s}^{-2}$1Pa≡1kg m−1 s−2. The negative exponent on metres is where students slip — they write "kg m s⁻²" (which is force, not pressure) by forgetting the division. The sign of the metre exponent is load-bearing.

Worked Example 4 — The watt

Power is the rate at which energy is transferred: $P = W/t$P=W/t. Substituting:

$[P] = \frac{\text{kg m}^{2}\text{ s}^{-2}}{\text{s}} = \text{kg m}^{2}\text{ s}^{-3}$[P]=skg m2 s−2​=kg m2 s−3

So $1\,\text{W} \equiv 1\,\text{kg m}^{2}\text{ s}^{-3}$1W≡1kg m2 s−3.

Worked Example 5 — The volt

Potential difference is energy per unit charge: $V = W/Q$V=W/Q. Since $Q = It$Q=It:

$[V] = \frac{\text{kg m}^{2}\text{ s}^{-2}}{\text{A s}} = \text{kg m}^{2}\text{ s}^{-3}\text{ A}^{-1}$[V]=A skg m2 s−2​=kg m2 s−3 A−1

Five different base units, all required, with the ampere appearing because voltage is fundamentally an electrical quantity. The volt is the first derived unit you meet that involves the ampere in a non-trivial way.

Worked Example 6 — The ohm

From Ohm's law $V = IR$V=IR, resistance is voltage per unit current:

$[R] = \frac{[V]}{[I]} = \frac{\text{kg m}^{2}\text{ s}^{-3}\text{ A}^{-1}}{\text{A}} = \text{kg m}^{2}\text{ s}^{-3}\text{ A}^{-2}$[R]=[I][V]​=Akg m2 s−3 A−1​=kg m2 s−3 A−2

The $\text{A}^{-2}$A−2 is the tricky part: A divides twice, once via the volt definition and once via Ohm's law itself.

Worked Example 7 — The farad (bonus)

Capacitance is charge per unit voltage: $C = Q/V$C=Q/V:

$[C] = \frac{\text{A s}}{\text{kg m}^{2}\text{ s}^{-3}\text{ A}^{-1}} = \text{kg}^{-1}\text{ m}^{-2}\text{ s}^{4}\text{ A}^{2}$[C]=kg m2 s−3 A−1A s​=kg−1 m−2 s4 A2

Four powers, four base units involved, and the only place in A-Level where $\text{s}^{4}$s4 appears. Many A* candidates have practised this once and can reproduce it in twenty seconds; many A candidates have not, and lose two marks reconstructing it under time pressure.

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The mermaid base-to-derived tree

flowchart TD
    A[SI base units kg, m, s, A] --> B[velocity m s⁻¹]
    A --> C[acceleration m s⁻²]
    C --> D[force F=ma → kg m s⁻²]
    D --> E[pressure p=F/A → kg m⁻¹ s⁻²]
    D --> F[work W=Fd → kg m² s⁻²]
    F --> G[power P=W/t → kg m² s⁻³]
    A --> H[charge Q=It → A s]
    G --> I[voltage V=P/I → kg m² s⁻³ A⁻¹]
    I --> J[resistance R=V/I → kg m² s⁻³ A⁻²]
    H --> K[capacitance C=Q/V → kg⁻¹ m⁻² s⁴ A²]
    I --> K

Every electrical and mechanical unit in the A-Level course can be traced back to base units along one of these arrows.

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Homogeneity of equations

A physically valid equation must be homogeneous — every additive term, on both sides of the equation, must have identical units. Formally:

$[\text{LHS}] = [\text{RHS}]$[LHS]=[RHS]

and, for any sum of terms $A + B + C$A+B+C, $[A] = [B] = [C]$[A]=[B]=[C]. If the units disagree, the equation is wrong — full stop. There is no exception. This is not a heuristic; it is a structural consequence of the (number × unit) framework. You cannot add joules to newtons any more than you can add a length to an area.

What homogeneity will not tell you is whether a dimensionless constant — like $\tfrac{1}{2}$21​, $\pi$π, or $2$2 — is correct. The two equations $E_k = mv^{2}$Ek​=mv2 and $E_k = \tfrac{1}{2}mv^{2}$Ek​=21​mv2 are both homogeneous; only the second is right. Homogeneity catches missing factors of length, time, mass etc., and catches wrong powers; it does not police numerical pre-factors.

Worked Example 8 — Checking a suvat equation

Is $v^{2} = u^{2} + 2as$v2=u2+2as homogeneous?

$[v^{2}] = (\text{m s}^{-1})^{2} = \text{m}^{2}\text{ s}^{-2}$[v2]=(m s−1)2=m2 s−2 $[u^{2}] = (\text{m s}^{-1})^{2} = \text{m}^{2}\text{ s}^{-2}$[u2]=(m s−1)2=m2 s−2 $[2as] = [a][s] = (\text{m s}^{-2})(\text{m}) = \text{m}^{2}\text{ s}^{-2}$[2as]=[a][s]=(m s−2)(m)=m2 s−2

All three terms carry units of $\text{m}^{2}\text{ s}^{-2}$m2 s−2. The equation passes the homogeneity test. (Homogeneity cannot confirm the factor of 2 — that comes from integrating Newton's laws — but we know there is no missing length or time factor anywhere.)

Worked Example 9 — Checking the pendulum period

The period of a simple pendulum is $T = 2\pi\sqrt{l/g}$T=2πl/g​. Verify homogeneity.

$[l/g] = \frac{\text{m}}{\text{m s}^{-2}} = \text{s}^{2}$[l/g]=m s−2m​=s2 $[\sqrt{l/g}] = \sqrt{\text{s}^{2}} = \text{s}$[l/g​]=s2​=s $[2\pi] = 1\,(\text{dimensionless})$[2π]=1(dimensionless)

So RHS = $1 \times \text{s} = \text{s}$1×s=s, matching LHS = $\text{s}$s. Homogeneous. ✓

Worked Example 10 — Spotting a wrong equation

A student writes $E_k = \tfrac{1}{2}mv$Ek​=21​mv. Use homogeneity to show this is wrong.

$[\tfrac{1}{2}mv] = \text{kg} \times \text{m s}^{-1} = \text{kg m s}^{-1}$[21​mv]=kg×m s−1=kg m s−1

But $[E_k] = \text{kg m}^{2}\text{ s}^{-2}$[Ek​]=kg m2 s−2 (the joule). The proposed RHS is short by a factor of $\text{m s}^{-1}$m s−1 — another velocity. The correct formula is $E_k = \tfrac{1}{2}mv^{2}$Ek​=21​mv2. Homogeneity caught the slip immediately.

Worked Example 11 — Spring potential energy

A student is unsure whether the elastic potential energy stored in a spring is $\tfrac{1}{2}kx$21​kx or $\tfrac{1}{2}kx^{2}$21​kx2. Resolve by homogeneity.

The spring constant $k$k has SI units N m⁻¹ — force per unit extension — equivalent to $\text{kg s}^{-2}$kg s−2.

• $[\tfrac{1}{2}kx] = \text{kg s}^{-2} \times \text{m} = \text{kg m s}^{-2}$[21​kx]=kg s−2×m=kg m s−2 — these are units of force, not energy.
• $[\tfrac{1}{2}kx^{2}] = \text{kg s}^{-2} \times \text{m}^{2} = \text{kg m}^{2}\text{ s}^{-2}$[21​kx2]=kg s−2×m2=kg m2 s−2 — these are units of energy. ✓

The correct formula is $E = \tfrac{1}{2}kx^{2}$E=21​kx2. A homogeneity check has rescued the candidate from a potentially expensive slip.

Worked Example 12 — Showing that the unit of momentum is kg m s⁻¹

From $p = mv$p=mv:

$[p] = \text{kg} \times \text{m s}^{-1} = \text{kg m s}^{-1}$[p]=kg×m s−1=kg m s−1

Cross-check via impulse $= F \Delta t$=FΔt:

$[F \Delta t] = \text{kg m s}^{-2} \times \text{s} = \text{kg m s}^{-1}\,\checkmark$[FΔt]=kg m s−2×s=kg m s−1✓

Both routes give the same unit, exactly as homogeneity demands. This kind of cross-check is the structural test that any physical equation must pass.

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Homogeneity can disprove but not prove

A subtle but important philosophical point. Homogeneity is a necessary condition for an equation to be correct, but it is not sufficient. An equation that fails the homogeneity test is definitely wrong; an equation that passes it might still be wrong (it could have the wrong numerical pre-factor, or be missing a dimensionless ratio).

For example, $E = mc^{2}$E=mc2 is correct, but so (homogeneously) is $E = 2mc^{2}$E=2mc2, $E = \pi mc^{2}$E=πmc2, or $E = \tfrac{1}{2}mc^{2}$E=21​mc2. Only experiment or a deeper theoretical derivation can fix the numerical pre-factor. This asymmetry — that dimensional analysis can falsify but not verify — is one of the central truths of physics and worth internalising at A-Level.

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Further derivations — the trickier units

The named units in the table above split cleanly into "easy" mechanical units (newton, joule, pascal, watt) and "tricky" electromagnetic units (tesla, farad, henry). The mechanical ones flow from $F = ma$F=ma via at most two further definitions; the electromagnetic ones require a careful chain through the definition of the volt and the ampere. The next three worked derivations cover the trickiest exam-favoured cases.

Worked Example 13 — The hertz

Frequency is defined as the reciprocal of period: $f = 1/T$f=1/T. Substituting:

$[f] = \frac{1}{[T]} = \frac{1}{\text{s}} = \text{s}^{-1}$[f]=[T]1​=s1​=s−1