Derived Units and Homogeneity of Equations

In Lesson 1 you met the seven SI base units. Almost every other physical quantity in the A-Level course — force, energy, pressure, voltage, resistance, magnetic flux — is expressed as a derived unit, built from base units through physical equations.

This lesson shows you how to derive units, how to express familiar quantities like the newton, joule and pascal in terms of base units, and — crucially — how to use the principle of homogeneity to check whether an equation you are trying to remember (or derive) is physically possible.

Homogeneity analysis is one of the most powerful techniques in A-Level physics. It costs almost no time, prevents a huge class of exam errors, and is explicitly rewarded in OCR mark schemes.

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What Is a Derived Unit?

A derived unit is any unit that is not one of the seven base units but can be expressed in terms of them via multiplication, division and powers. Every derived unit carries a physical meaning — it tells you what quantity the unit is measuring and how that quantity is defined.

Consider speed:

speed = distance / time ⇒ m / s = m s⁻¹

The unit m s⁻¹ is derived. There is no single symbol for it; we simply write the combination. For quantities that appear very frequently, scientists have given special names — newton, joule, pascal, etc. — but these special names are always equivalent to combinations of base units.

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Key Derived Units You Must Know

The OCR specification lists several derived units that you are expected to know and to express in base units. The table below is the most important single piece of information in this lesson.

Quantity	Defining Equation	Derived Unit	Base-Unit Form
Area	A = l²	m²	m²
Volume	V = l³	m³	m³
Density	ρ = m/V	kg m⁻³	kg m⁻³
Speed, velocity	v = s/t	m s⁻¹	m s⁻¹
Acceleration	a = Δv/Δt	m s⁻²	m s⁻²
Force	F = ma	newton (N)	kg m s⁻²
Pressure, stress	p = F/A	pascal (Pa)	kg m⁻¹ s⁻²
Energy, work	W = Fd	joule (J)	kg m² s⁻²
Power	P = W/t	watt (W)	kg m² s⁻³
Charge	Q = It	coulomb (C)	A s
Potential difference	V = W/Q	volt (V)	kg m² s⁻³ A⁻¹
Resistance	R = V/I	ohm (Ω)	kg m² s⁻³ A⁻²
Frequency	f = 1/T	hertz (Hz)	s⁻¹
Magnetic flux density	B = F/(IL)	tesla (T)	kg s⁻² A⁻¹

You should be able to derive any row of this table on demand. You should not try to memorise the base-unit column; instead, memorise the defining equation and derive the base units as needed. That way, if you forget the base units under exam pressure, you can recreate them in ten seconds.

Worked Derivation 1 — The Newton

Force is defined by Newton's second law:

F = ma

Substituting base units:

[F] = [m] × [a] = kg × m s⁻² = kg m s⁻²

The square brackets notation "[X]" means "the units of X". So 1 N ≡ 1 kg m s⁻².

Worked Derivation 2 — The Joule

Work is force times distance:

W = Fd

Substituting:

[W] = [F] × [d] = kg m s⁻² × m = kg m² s⁻²

So 1 J ≡ 1 kg m² s⁻². The same is true of all energy quantities — kinetic energy, gravitational potential energy, elastic strain energy.

Quick check with kinetic energy: E_k = ½ m v²

[E_k] = kg × (m s⁻¹)² = kg × m² s⁻² = kg m² s⁻² ✓

Both formulas give joules, exactly as they must.

Worked Derivation 3 — The Pascal

Pressure is force per area:

p = F/A

Substituting:

[p] = (kg m s⁻²) / m² = kg m⁻¹ s⁻²

So 1 Pa ≡ 1 kg m⁻¹ s⁻². Note the negative exponent on metres — this is where students often slip.

Worked Derivation 4 — The Watt

Power is rate of energy transfer:

P = W/t

Substituting:

[P] = (kg m² s⁻²) / s = kg m² s⁻³

So 1 W ≡ 1 kg m² s⁻³.

Worked Derivation 5 — The Volt

Potential difference is energy per unit charge:

V = W/Q, where Q = It

Substituting:

[V] = (kg m² s⁻²) / (A s) = kg m² s⁻³ A⁻¹

Five different base units, all required — the ampere appears because voltage is fundamentally an electrical quantity.

Worked Derivation 6 — The Ohm

From V = IR, resistance is voltage per unit current:

R = V/I

[R] = (kg m² s⁻³ A⁻¹) / A = kg m² s⁻³ A⁻²

The negative exponent on A² is the tricky part. Note that A divides twice: once in the definition of the volt, once here in Ohm's law.

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Homogeneity of Equations

Every physically valid equation must be homogeneous: the units on the left-hand side must exactly match the units on the right-hand side. Symbolically:

[LHS] = [RHS]

If they disagree, the equation is wrong — full stop. There is no exception.