Physical quantities and SI units

AS · 10 min

Every number in physics comes with a unit, and every unit can be broken down into a handful of base units. This topic gives you the language the rest of the course is written in: SI base units, derived units, prefixes, and the trick of checking an equation by its units (homogeneity). It is examined directly in Paper 1 multiple-choice questions and in the first part of many Paper 2 questions, and it quietly earns or loses marks in every calculation you ever do.

Physical quantities

A physical quantity is anything that can be measured: length, mass, current, temperature, pressure. Writing it down needs two things.

Definition

Every physical quantity consists of a numerical magnitude and a unit. For example, in $m = 3.5\ \text{kg}$ the magnitude is $3.5$ and the unit is the kilogram.

"The mass is $3.5$ " is meaningless: $3.5$ grams and $3.5$ kilograms differ by a factor of a thousand. In an exam a final answer without its unit usually loses the mark, even if the number is right.

The unit behaves like an algebraic factor. $3.5\ \text{kg}$ really means $3.5 \times (1\ \text{kg})$ , so units multiply, divide and cancel exactly like symbols. That is the whole idea behind derived units and homogeneity below.

SI base quantities and units

The Système International (SI) builds every unit from a small set of base units, each defined independently. You must recall these five.

Key result

Base quantity	Symbol	SI base unit	Unit symbol
mass	$m$	kilogram	$\text{kg}$
length	$l$	metre	$\text{m}$
time	$t$	second	$\text{s}$
electric current	$I$	ampere	$\text{A}$
thermodynamic temperature	$T$	kelvin	$\text{K}$

There are two more SI base units (the mole for amount of substance and the candela for luminous intensity). The mole appears at A Level with ideal gases; neither is on the AS list you have to recall.

Watch out

Charge is not a base quantity, and the coulomb is not a base unit. Current is the base quantity; charge is derived from it ( $Q = It$ , so $1\ \text{C} = 1\ \text{A s}$ ). Equally, force, energy, voltage and resistance are all derived. A favourite multiple-choice question asks "which of these is a base unit?" and lists the coulomb, the newton, the volt and the kelvin: the answer is the kelvin.

Derived units

A derived unit is a product or quotient of base units, found from the defining equation of the quantity. The method never changes.

Method

Finding a derived unit in SI base units

Write down an equation that defines the quantity (or any correct equation containing it).
Replace every quantity on the other side with its base units.
Simplify by collecting powers of $\text{kg}$ , $\text{m}$ , $\text{s}$ , $\text{A}$ and $\text{K}$ .

Here are the derived units you will meet at AS, worked out from their defining equations. You do not need to memorise the right-hand column; you need to be able to produce it.

Key result

Quantity	Defining equation	Named unit	In base units
velocity	$v = s/t$	—	$\text{m s}^{-1}$
acceleration	$a = \Delta v/t$	—	$\text{m s}^{-2}$
force	$F = ma$	newton, $\text{N}$	$\text{kg m s}^{-2}$
momentum	$p = mv$	— ( $\text{N s}$ )	$\text{kg m s}^{-1}$
work, energy	$W = Fs$	joule, $\text{J}$	$\text{kg m}^{2}\ \text{s}^{-2}$
power	$P = W/t$	watt, $\text{W}$	$\text{kg m}^{2}\ \text{s}^{-3}$
pressure, stress	$p = F/A$	pascal, $\text{Pa}$	$\text{kg m}^{-1}\ \text{s}^{-2}$
density	$\rho = m/V$	—	$\text{kg m}^{-3}$
frequency	$f = 1/T$	hertz, $\text{Hz}$	$\text{s}^{-1}$
charge	$Q = It$	coulomb, $\text{C}$	$\text{A s}$
potential difference	$V = W/Q$	volt, $\text{V}$	$\text{kg m}^{2}\ \text{s}^{-3}\ \text{A}^{-1}$
resistance	$R = V/I$	ohm, $\Omega$	$\text{kg m}^{2}\ \text{s}^{-3}\ \text{A}^{-2}$
resistivity	$\rho = RA/L$	— ( $\Omega\ \text{m}$ )	$\text{kg m}^{3}\ \text{s}^{-3}\ \text{A}^{-2}$
spring constant	$k = F/x$	— ( $\text{N m}^{-1}$ )	$\text{kg s}^{-2}$

Strain, efficiency, refractive index and other ratios of like quantities have no unit: the units cancel.

The volt in base units

Express the volt in SI base units.

Solution

Potential difference is energy transferred per unit charge, $V = W/Q$ .

Energy: $W = Fs$ , and $F = ma$ , so $[W] = \text{kg} \times \text{m s}^{-2} \times \text{m} = \text{kg m}^{2}\ \text{s}^{-2}$ .

Charge: $Q = It$ , so $[Q] = \text{A s}$ .

[V] = \frac{\text{kg m}^{2}\ \text{s}^{-2}}{\text{A s}} = \text{kg m}^{2}\ \text{s}^{-3}\ \text{A}^{-1}

Square brackets around a quantity, $[W]$ , mean "the units of $W$ ". It is a convenient shorthand in working.

Homogeneity of equations

An equation is homogeneous if the base units of every term are the same on both sides. A physically correct equation must be homogeneous: you cannot add metres to seconds, or equate a force to an energy.

This gives you a quick check on any equation you derive or half-remember. It also lets you find the unit of an unknown constant.

Method

Checking homogeneity

Write the base units of each term separately (terms are the parts joined by $+$ , $-$ or $=$ ).
Pure numbers such as $2$ , $\tfrac{1}{2}$ and $\pi$ have no units; drop them.
If every term reduces to the same base units, the equation is homogeneous.

Checking an equation of motion

Show that $v^2 = u^2 + 2as$ is homogeneous.

Solution

$[v^2] = (\text{m s}^{-1})^2 = \text{m}^2\ \text{s}^{-2}$

$[u^2] = \text{m}^2\ \text{s}^{-2}$

$[2as] = \text{m s}^{-2} \times \text{m} = \text{m}^2\ \text{s}^{-2}$ (the $2$ has no unit)

All three terms have units $\text{m}^2\ \text{s}^{-2}$ , so the equation is homogeneous.

Watch out

Homogeneity is necessary but not sufficient. An equation can be homogeneous and still wrong, because units cannot detect a missing or wrong pure number. $v^2 = u^2 + as$ and $s = ut + at^2$ are both homogeneous and both incorrect. If a question asks "explain why a homogeneous equation may not be correct", say: the equation may have a wrong numerical constant (or a missing dimensionless factor), which the units cannot reveal.

Units of a constant

The drag force on a sphere moving slowly through a fluid is $F = 6\pi \eta r v$ , where $r$ is the radius and $v$ the speed. Find the SI base units of $\eta$ (the viscosity).

Solution

Make $\eta$ the subject: $\eta = \dfrac{F}{6\pi r v}$ . The $6\pi$ has no units.

[\eta] = \frac{\text{kg m s}^{-2}}{\text{m} \times \text{m s}^{-1}} = \frac{\text{kg m s}^{-2}}{\text{m}^2\ \text{s}^{-1}} = \text{kg m}^{-1}\ \text{s}^{-1}

You do not need to know anything about viscosity to answer that: this is typical of how Cambridge tests the idea, using an unfamiliar equation and asking only about units.

Prefixes

Prefixes scale a unit by a power of ten, so very large and very small values stay readable.

Key result

Prefix	Symbol	Factor	Prefix	Symbol	Factor
pico	$\text{p}$	$10^{-12}$	deci	$\text{d}$	$10^{-1}$
nano	$\text{n}$	$10^{-9}$	kilo	$\text{k}$	$10^{3}$
micro	$\mu$	$10^{-6}$	mega	$\text{M}$	$10^{6}$
milli	$\text{m}$	$10^{-3}$	giga	$\text{G}$	$10^{9}$
centi	$\text{c}$	$10^{-2}$	tera	$\text{T}$	$10^{12}$

The danger comes with squared and cubed units. The prefix belongs to the unit, and the power applies to the prefix too:

1\ \text{cm}^2 = (10^{-2}\ \text{m})^2 = 10^{-4}\ \text{m}^2, \qquad 1\ \text{mm}^3 = (10^{-3}\ \text{m})^3 = 10^{-9}\ \text{m}^3

Watch out

$1\ \text{cm}^3$ is $10^{-6}\ \text{m}^3$ , not $10^{-2}\ \text{m}^3$ . Likewise a density of $1\ \text{g cm}^{-3}$ is $1000\ \text{kg m}^{-3}$ . The safest habit: replace the prefix by its power of ten inside a bracket, then apply the power to the whole bracket.

Prefixes with powers

A copper wire has diameter $0.28\ \text{mm}$ . Steel has density $7.9\ \text{g cm}^{-3}$ . (a) Find the cross-sectional area of the wire in $\text{m}^2$ . (b) Express the density of steel in $\text{kg m}^{-3}$ .

Solution

(a) Radius $r = 0.14\ \text{mm} = 0.14 \times 10^{-3}\ \text{m}$ .

A = \pi r^2 = \pi (0.14 \times 10^{-3})^2 = 6.2 \times 10^{-8}\ \text{m}^2

(b)

7.9\ \text{g cm}^{-3} = \frac{7.9 \times 10^{-3}\ \text{kg}}{(10^{-2}\ \text{m})^3} = \frac{7.9 \times 10^{-3}}{10^{-6}}\ \text{kg m}^{-3} = 7.9 \times 10^{3}\ \text{kg m}^{-3}

Making estimates

You are expected to make reasonable estimates of quantities in the syllabus. Paper 1 usually includes a question such as "Which estimate is reasonable?" with four options spread over powers of ten. You are not expected to know exact values, only the right order of magnitude. Learn a set of anchor values and reason from them.

Quantity	Reasonable value
mass of an adult	$70\ \text{kg}$
mass of an apple	$0.1\ \text{kg}$
mass of a car	$1000\ \text{kg}$
height of a room	$3\ \text{m}$
diameter of an atom	$10^{-10}\ \text{m}$
diameter of a nucleus	$10^{-15}\ \text{m}$ (a few fm)
wavelength of visible light	$400$ to $700\ \text{nm}$
walking speed	$1.5\ \text{m s}^{-1}$
speed of a car on a motorway	$30\ \text{m s}^{-1}$
speed of sound in air	$330\ \text{m s}^{-1}$
atmospheric pressure	$1.0 \times 10^{5}\ \text{Pa}$
density of water	$1000\ \text{kg m}^{-3}$
density of air	$1.2\ \text{kg m}^{-3}$
power of a kettle	$2\ \text{kW}$
current in a filament lamp	$0.1$ to $1\ \text{A}$
weight of an adult	$700\ \text{N}$
frequency of mains electricity	$50\ \text{Hz}$
time for one human heartbeat	$1\ \text{s}$

Estimating an energy

Which is the best estimate of the kinetic energy of a sprinter running at top speed?

A. $40\ \text{J}$ B. $400\ \text{J}$ C. $4000\ \text{J}$ D. $40\,000\ \text{J}$

Solution

A sprinter has mass about $70\ \text{kg}$ and top speed about $10\ \text{m s}^{-1}$ .

E_K = \tfrac{1}{2}mv^2 \approx \tfrac{1}{2} \times 70 \times 10^2 = 3500\ \text{J}

The answer is C. Only the order of magnitude matters: the other options are ten times too small or too large.

Finding unknown powers by homogeneity

A harder use of the same idea: if you are told a quantity depends on others as a product of powers, homogeneity fixes the powers.

Speed of a wave on a string

The speed $v$ of a transverse wave on a stretched string depends on the tension $T$ and the mass per unit length $\mu$ according to $v = k\,T^{a}\mu^{b}$ , where $k$ is a constant with no unit. Use base units to find $a$ and $b$ .

Solution

Base units: $[v] = \text{m s}^{-1}$ , $[T] = \text{kg m s}^{-2}$ (a force), $[\mu] = \text{kg m}^{-1}$ .

\text{m}^{1}\ \text{s}^{-1} = (\text{kg m s}^{-2})^{a}\,(\text{kg m}^{-1})^{b} = \text{kg}^{a+b}\ \text{m}^{a-b}\ \text{s}^{-2a}

Compare the powers of each base unit:

$\text{kg}$ : $0 = a + b$
$\text{m}$ : $1 = a - b$
$\text{s}$ : $-1 = -2a$

From the third, $a = \tfrac{1}{2}$ . Then $b = -\tfrac{1}{2}$ , and the second equation checks: $\tfrac{1}{2} - (-\tfrac{1}{2}) = 1$ .

So $v = k\sqrt{T/\mu}$ . (In fact $k = 1$ , but units alone can never tell you that.)

Exam tip

"Express in SI base units" means only $\text{kg}$ , $\text{m}$ , $\text{s}$ , $\text{A}$ , $\text{K}$ may appear in the answer. An answer containing $\text{N}$ , $\text{J}$ or $\text{C}$ scores nothing.
"Show that the equation is homogeneous" needs the base units of each term written separately, then a statement that they are the same. Simply cancelling everything to " $1 = 1$ " does not show it.
When a question asks you to "state the base units of $X$ ", write them as a product with negative powers ( $\text{kg m}^{-1}\ \text{s}^{-1}$ ), not as a fraction with a slash.
In estimate questions, eliminate the options that are absurd by a factor of ten or more. Usually only one survives.

Summary

A physical quantity is a numerical magnitude times a unit; a missing unit loses marks.
The five AS base quantities: mass (kg), length (m), time (s), current (A), temperature (K).
Every other unit is derived from its defining equation; work it out step by step.
An equation must be homogeneous (same base units in every term), but a homogeneous equation can still be wrong because pure numbers have no units.
Use homogeneity to find the unit of a constant or the powers in a relationship.
Prefixes p, n, μ, m, c, d, k, M, G, T: learn the powers of ten, and apply squares and cubes to the prefix too.
Keep a set of anchor values for estimates and reason to the right order of magnitude.

Practice

Question

Which of the following is an SI base unit: coulomb, kelvin, newton, volt? Explain why each of the others is not.
Express the joule and the pascal in SI base units.
Show that the ohm is equivalent to $\text{kg m}^{2}\ \text{s}^{-3}\ \text{A}^{-2}$ .
The power $P$ dissipated in a resistor is given by $P = I^2R$ . Show that this equation is homogeneous.
Convert: (a) $250\ \text{cm}^3$ to $\text{m}^3$ ; (b) $4.5\ \text{mm}^2$ to $\text{m}^2$ ; (c) $0.36\ \text{GW}$ to $\text{W}$ ; (d) $680\ \text{nm}$ to $\text{m}$ .
The force $F$ needed to stretch a material is related to its extension $x$ by $F = \dfrac{EAx}{L}$ , where $A$ is a cross-sectional area and $L$ a length. Find the SI base units of $E$ .
Estimate the weight of a 1 litre bottle of water, and the pressure it exerts on a table if its base has area $50\ \text{cm}^2$ .
The drag force on a car is $F = \tfrac{1}{2} C \rho A v^2$ , where $\rho$ is the density of air, $A$ the frontal area and $v$ the speed. Show that $C$ has no unit.
The period $T$ of a simple pendulum is thought to depend on its length $l$ , the mass $m$ of the bob and the acceleration of free fall $g$ , in the form $T = k\,l^{x} m^{y} g^{z}$ with $k$ a dimensionless constant. Find $x$ , $y$ and $z$ .
A student writes the equation $s = ut + at^2$ for uniformly accelerated motion. (a) Show that it is homogeneous. (b) Explain why it may still be incorrect, and state the correct equation.

Answers

The kelvin. The coulomb is an ampere second ( $Q = It$ ); the newton is $\text{kg m s}^{-2}$ ( $F = ma$ ); the volt is $\text{J C}^{-1}$ . All three are derived.
$W = Fs$ : $[\text{J}] = \text{kg m s}^{-2} \times \text{m} = \text{kg m}^{2}\ \text{s}^{-2}$ . $p = F/A$ : $[\text{Pa}] = \text{kg m s}^{-2} / \text{m}^2 = \text{kg m}^{-1}\ \text{s}^{-2}$ .
$R = V/I$ and $[V] = \text{kg m}^{2}\ \text{s}^{-3}\ \text{A}^{-1}$ (see the worked example), so $[R] = \text{kg m}^{2}\ \text{s}^{-3}\ \text{A}^{-1} / \text{A} = \text{kg m}^{2}\ \text{s}^{-3}\ \text{A}^{-2}$ .
$[P] = \text{J s}^{-1} = \text{kg m}^{2}\ \text{s}^{-3}$ . $[I^2R] = \text{A}^2 \times \text{kg m}^{2}\ \text{s}^{-3}\ \text{A}^{-2} = \text{kg m}^{2}\ \text{s}^{-3}$ . Both sides are the same, so the equation is homogeneous.
(a) $250 \times 10^{-6} = 2.5 \times 10^{-4}\ \text{m}^3$ ; (b) $4.5 \times 10^{-6}\ \text{m}^2$ ; (c) $3.6 \times 10^{8}\ \text{W}$ ; (d) $6.8 \times 10^{-7}\ \text{m}$ .
$E = \dfrac{FL}{Ax}$ , so $[E] = \dfrac{\text{kg m s}^{-2} \times \text{m}}{\text{m}^2 \times \text{m}} = \text{kg m}^{-1}\ \text{s}^{-2}$ (the same as the pascal: $E$ is the Young modulus).
Mass $\approx 1\ \text{kg}$ , so weight $\approx 10\ \text{N}$ (more precisely $9.81\ \text{N}$ ). Area $= 50 \times 10^{-4} = 5.0 \times 10^{-3}\ \text{m}^2$ , so $p = F/A \approx 10 / 5.0 \times 10^{-3} = 2000\ \text{Pa}$ (about $2 \times 10^{3}\ \text{Pa}$ ).
$C = \dfrac{2F}{\rho A v^2}$ . $[\rho A v^2] = \text{kg m}^{-3} \times \text{m}^2 \times \text{m}^2\ \text{s}^{-2} = \text{kg m s}^{-2}$ , the same as $[F]$ . So the units cancel and $C$ has no unit.
$\text{s} = \text{m}^{x}\ \text{kg}^{y}\ (\text{m s}^{-2})^{z} = \text{kg}^{y}\ \text{m}^{x+z}\ \text{s}^{-2z}$ . Comparing: $\text{kg}$ : $y = 0$ ; $\text{s}$ : $-2z = 1$ so $z = -\tfrac{1}{2}$ ; $\text{m}$ : $x + z = 0$ so $x = \tfrac{1}{2}$ . Hence $T = k\sqrt{l/g}$ and the period does not depend on the mass.
(a) $[s] = \text{m}$ ; $[ut] = \text{m s}^{-1} \times \text{s} = \text{m}$ ; $[at^2] = \text{m s}^{-2} \times \text{s}^2 = \text{m}$ . Every term is in metres, so it is homogeneous. (b) Homogeneity cannot detect a wrong dimensionless constant. The $at^2$ term should have a factor $\tfrac{1}{2}$ : $s = ut + \tfrac{1}{2}at^2$ .