SI base units and derived units

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Every measurement in physics is a number multiplied by a unit.

"The cable is 25" means nothing; "the cable is 25 m long" is a measurement.

This section sets out the small set of units everything else is built from, and shows how every other unit is made from them.

You will use this in every unit of the course, and Paper 1 regularly sets multiple-choice items on it.

You also need

No earlier unit: this is the first unit of the course. It assumes GCSE standard form, rearranging equations and the equations of motion and energy you met at GCSE.

Later units lean on it all the time: the newton, joule, watt, volt and ohm in units 3-4b and 3-5, the tesla and weber in unit 3-7c, and the mole and the kelvin in unit 3-6-2.

Fundamental (base) units

Scientists across the world agree on one system of units, the Système International (SI).

It picks out a few quantities as base quantities and gives each one a fundamental (base) unit.

Every other unit is defined from these.

You need to know the use of mass, length, time, amount of substance, temperature, electric current and their associated SI units:

Base quantity

Usual symbol

SI base unit

Unit symbol

mass

$m$

kilogram

length

$l$ , $x$ , $s$

metre

time

$t$

second

amount of substance

$n$

mole

mol

temperature

$T$

kelvin

electric current

$I$

ampere

The fundamental unit of light intensity, the candela, is excluded.

There are seven SI base units in all, but the candela is not part of this course, so you only ever work with the six in the table.

Three things in the table are easy to get wrong.

The kilogram is the base unit of mass, not the gram. It is the only base unit that already carries a prefix. A mass given in grams must be converted to kilograms before it goes into $F = ma$ , $E_k = \frac{1}{2}mv^2$ or any other equation.

The kelvin is the base unit of temperature. A temperature in degrees Celsius is converted to kelvin by adding 273 when the absolute temperature matters (the gas laws in unit 3-6-2). A temperature change is the same in both: a rise of 15 °C is a rise of 15 K, so you never add 273 to a difference. The symbol is a capital K; a lower-case k means kilo and is not accepted for the kelvin.

The ampere, not the coulomb, is the base unit for electricity. The coulomb is derived from it: one coulomb is one ampere for one second, $1\ \text{C} = 1\ \text{A s}$ .

Students are not expected to recall definitions of the fundamental quantities.

You will not be asked how the metre or the second is defined.

What you must know is which units are base units, and how to build every other unit from them.

Why it works (not examined): what fixes the base units

Since 20 May 2019 every SI base unit has been fixed by choosing an exact value for a constant of nature.

The second is set by the frequency of a particular transition in the caesium-133 atom (exactly 9 192 631 770 Hz).

The metre then follows from fixing the speed of light at exactly 299 792 458 m s⁻¹, and the kilogram from fixing the Planck constant.

Before 2019 the kilogram was the mass of one metal cylinder kept near Paris.

Fixing units to constants means any well-equipped laboratory can realise them, and they cannot drift.

None of this is examined, but it explains why the table above is short: a few agreed units, with everything else built from them.

The picture below shows what the old standard looked like: a cylinder of platinum-iridium alloy about 39 mm tall and 39 mm across, kept in air under three glass bell jars.

Any scratch, any trace of dirt and any cleaning changed its mass, and with it the kilogram everywhere in the world, which is why the definition moved to a constant.

Photograph-style view of a small polished metal cylinder, as tall as it is wide, under three nested glass bell jars: the kind of platinum-iridium cylinder that defined the kilogram before 2019.

SI units derived

Every other unit is a derived unit: a combination of base units, made by following the equation that defines the quantity.

Speed is distance divided by time, so its unit is m s⁻¹.

Acceleration is change of velocity divided by time, so m s⁻¹ divided by s gives m s⁻².

Many derived units get a name of their own.

The newton is the force that gives a mass of 1 kg an acceleration of 1 m s⁻², because $F = ma$ :

F = ma

$F$ is the resultant force in newtons (N);

$m$ is the mass in kilograms (kg);

$a$ is the acceleration in m s⁻².

AQA has also printed this equation in the Data and Formulae Booklet in recent exam series.

So $1\ \text{N} = 1\ \text{kg m s}^{-2}$ .

The joule follows from work done, force multiplied by distance moved in the direction of the force:

W = Fs\cos\theta

AQA has also printed this equation in the Data and Formulae Booklet in recent exam series.

So $1\ \text{J} = 1\ \text{N m} = 1\ \text{kg m}^2\,\text{s}^{-2}$ .

The cosine has no unit, so it does not change the unit.

Here are the derived units you will meet most, each built step by step from its defining equation:

Quantity

Defining equation

Named unit

In base units

force

$F = ma$

newton, N

kg m s⁻²

work, energy

$W = Fs$

joule, J = N m

kg m² s⁻²

power

$P = \Delta W/\Delta t$

watt, W = J s⁻¹

kg m² s⁻³

pressure, stress

$p = F/A$

pascal, Pa = N m⁻²

kg m⁻¹ s⁻²

momentum

$p = mv$

(N s)

kg m s⁻¹

frequency

$f = 1/T$

hertz, Hz

s⁻¹

density

$\rho = m/V$

(kg m⁻³)

kg m⁻³

charge

$Q = It$

coulomb, C

A s

potential difference

$V = W/Q$

volt, V = J C⁻¹

kg m² s⁻³ A⁻¹

resistance

$R = V/I$

ohm, Ω = V A⁻¹

kg m² s⁻³ A⁻²

Momentum, $p = mv$ (mass in kg multiplied by velocity in m s⁻¹), is not printed on a line of its own.

Not in the booklet: learn it.

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

Not in the booklet: learn it.

(The booklet prints tensile stress $= F/A$ under materials, which has the same form and the same unit.)

Read the table as a set of steps you can always reproduce, not as a list to memorise.

If you know the defining equation, you can always rebuild the unit.

Same base units, different quantities

Two different quantities can share base units while a third does not.

Put four quantities side by side:

Quantity

Built from

Named unit

In base units

work done, energy

force × distance moved along the force

J = N m

kg m² s⁻²

moment of a force

force × perpendicular distance from the pivot

N m

kg m² s⁻²

pressure

force ÷ area

Pa = N m⁻²

kg m⁻¹ s⁻²

Young modulus

tensile stress ÷ tensile strain

kg m⁻¹ s⁻²

Energy and the moment of a force have the same base units, kg m² s⁻², because both are a force multiplied by a distance.

They are still different quantities: in a moment the distance is perpendicular to the force, and nothing moves.

That is why a moment is always written in N m and never in joules.

Pressure and the Young modulus also share base units.

Strain is an extension divided by a length, m ÷ m, so it has no unit, and the Young modulus has the unit of stress, the pascal.

But energy and pressure do not share base units: kg m² s⁻² is not kg m⁻¹ s⁻².

Matching base units tells you an equation could be right; it never tells you two quantities are the same thing.

"An appropriate SI unit" and "fundamental (base) units"

Exam questions ask for units in two different ways, and they want different answers.

"State an appropriate SI unit for your answer" (or "State an appropriate unit for your answer"). Give the usual derived unit: N kg⁻¹ for a gravitational field strength, Ω m for resistivity, T for magnetic flux density, Wb for magnetic flux, Pa for pressure. A long string of base units in place of a named unit that is standard is not what is wanted. On a field-strength part the unit mark is often the easiest mark to earn, but only with the named unit: building it from base units when nobody asked for them is slow and rarely comes out right.

"State, in fundamental (base) units, the unit for your answer" (or "Deduce the fundamental base units for ..."). Now the base units are the point. The named unit alone scores nothing: "T" is not an answer in base units. Use the ampere for electricity, never the coulomb, and keep flux (Wb) and flux density (T) apart.

For example, the tesla follows from $F = BIl$ (unit 3-7c): $B = F/(Il)$ , so $\text{T} = \text{N A}^{-1}\,\text{m}^{-1} = \text{kg m s}^{-2}\,\text{A}^{-1}\,\text{m}^{-1} = \text{kg s}^{-2}\,\text{A}^{-1}$ .

Units of a constant: making an equation homogeneous

Dimensional analysis is not required.

You will not be asked to use the dimensions M, L and T to predict the form of an equation.

But every correct physics equation is homogeneous: every term in it has the same unit, and the two sides have the same unit.

You can use that to check an equation or to find the unit of a constant in it.

Recent papers have asked you to find the fundamental (base) units of a constant in an equation printed in the question, by making the equation homogeneous.

The method that scores:

1. Rearrange the equation to make the constant the subject.

2. Replace every quantity by its SI unit, one at a time.

3. Turn each named unit into base units (N = kg m s⁻², J = kg m² s⁻², and so on).

4. Collect the powers of each base unit. Join every step with an equals sign.

Keep quantity symbols and unit symbols apart.

In $F = ma$ the $m$ is a mass; in "kg m s⁻²" the m is a metre.

If you try to "cancel" letters as if the unit symbols were algebra from the equation, you lose the marks.

Replace the quantities by units first, then simplify the units only.

Two shortcuts help.

Pure numbers have no unit. The 2 in $2\pi$ , the $\frac{1}{2}$ in $\frac{1}{2}mv^2$ , $\pi$ , a cosine, a ratio such as strain or refractive index: none of them carry a unit.

The argument of an exponential has no unit. In $I = I_0 e^{-\mu x}$ , the power $\mu x$ must be a pure number, so if $x$ is in m then $\mu$ is in m⁻¹. If an equation contains $e^{B/T}$ with $T$ a temperature in K, then $B/T$ is a pure number and $B$ has the unit of temperature, K. The same is true inside a logarithm.

Key terms

Fundamental (base) units: the SI units of mass (kg), length (m), time (s), amount of substance (mol), temperature (K) and electric current (A), from which every other unit is derived.

Derived unit: a unit formed from base units by following the equation that defines the quantity, for example $1\ \text{N} = 1\ \text{kg m s}^{-2}$ .

Homogeneous equation: an equation in which every term, and both sides, have the same unit.

These units and the method for building one are not in the booklet.

Not in the booklet: learn it.

Say it this way

"The base units of energy are kg m² s⁻²."

"The unit of a moment is N m" (never J).

"The ampere is the base unit; the coulomb is a derived unit, A s."

Write the kelvin as capital K. Write units with negative indices: m s⁻², N kg⁻¹, kg m⁻³, Ω m, J kg⁻¹ K⁻¹.

Worked example 1: the pascal in base units

Express the pascal in SI base units.

1. MP1 Pressure is force per unit area, $p = F/A$ , so $1\ \text{Pa} = 1\ \text{N m}^{-2}$ , and from $F = ma$ , $1\ \text{N} = 1\ \text{kg m s}^{-2}$ ✓

2. MP2 $\text{Pa} = \text{kg m s}^{-2} \times \text{m}^{-2} = \text{kg m}^{-1}\,\text{s}^{-2}$ ✓

Answer: $\text{kg m}^{-1}\,\text{s}^{-2}$ .

Notice that the metre appears once in the newton and twice (as m⁻²) in the area, leaving m⁻¹.

Worked example 2: three quantities side by side

Show that work done and the moment of a force have the same SI base units, and that the Young modulus does not.

1. MP1 Work done $= Fs$ : $\text{N} \times \text{m} = \text{kg m s}^{-2} \times \text{m} = \text{kg m}^2\,\text{s}^{-2}$ ✓

2. MP2 Moment $= F \times d$ , with $d$ the perpendicular distance in m: the same product, $\text{kg m}^2\,\text{s}^{-2}$ ✓

3. MP3 Young modulus = tensile stress ÷ tensile strain; strain is m ÷ m, which has no unit, so the unit is that of stress, $\text{N m}^{-2} = \text{kg m}^{-1}\,\text{s}^{-2}$ , which differs from $\text{kg m}^2\,\text{s}^{-2}$ ✓

Answer: work done and moment are both $\text{kg m}^2\,\text{s}^{-2}$ ; the Young modulus is $\text{kg m}^{-1}\,\text{s}^{-2}$ .

So, in this case, sharing base units does not make work and moment the same quantity: a moment is still written in N m.

Worked example 3: the unit of a spring constant from an energy equation

The energy stored in a stretched spring is $E = \frac{1}{2}kx^2$ , where $x$ is the extension.

Deduce the SI base units of the spring constant $k$ .

1. MP1 Rearrange: $k = \dfrac{2E}{x^2}$ ; the 2 is a pure number ✓

2. MP2 Units: $k = \dfrac{\text{J}}{\text{m}^2} = \dfrac{\text{kg m}^2\,\text{s}^{-2}}{\text{m}^2}$ ✓

3. MP3 $= \text{kg s}^{-2}$ ✓

Answer: $\text{kg s}^{-2}$ .

As a check, the usual unit of a spring constant is N m⁻¹, and $\text{N m}^{-1} = \text{kg m s}^{-2} \times \text{m}^{-1} = \text{kg s}^{-2}$ , the same.

Worked example 4: a constant in an exponent

The intensity of light passing through seawater falls with depth $x$ as $I = I_0 e^{-\mu x}$ , where $I_0$ is the intensity at the surface.

State the SI unit of $\mu$ , and explain your answer.

1. MP1 The power of $e$ must be a pure number (it has no unit), so $\mu x$ has no unit ✓

2. MP2 $x$ is in m, so $\mu$ is in $\text{m}^{-1}$ ✓

Answer: m⁻¹.

The ratio $I/I_0$ has no unit either, so the equation is homogeneous whatever unit the intensity is in.

Worked example 5: base units of a constant in a printed equation

It can be shown that the volume of liquid flowing each second through a narrow tube is

Q = \frac{\pi p r^4}{8 \eta L}

where $Q$ is the volume flow rate in m³ s⁻¹, $p$ is the pressure difference between the ends of the tube, $r$ is the internal radius of the tube, $L$ is its length and $\eta$ is a constant for the liquid.

Deduce the fundamental base units of $\eta$ .

1. MP1 Rearrange: $\eta = \dfrac{\pi p r^4}{8 Q L}$ ; $\pi$ and 8 are pure numbers ✓

2. MP2 Substitute units: $\eta = \dfrac{\text{Pa} \times \text{m}^4}{\text{m}^3\,\text{s}^{-1} \times \text{m}} = \dfrac{\text{kg m}^{-1}\,\text{s}^{-2} \times \text{m}^4}{\text{m}^4\,\text{s}^{-1}}$ ✓

3. MP3 $= \text{kg m}^{-1}\,\text{s}^{-1}$ ✓

Answer: $\text{kg m}^{-1}\,\text{s}^{-1}$ .

Every step is a line of units joined by an equals sign, and the pascal is turned into base units before anything cancels.

Writing "Pa s" scores nothing here, because the stem asks for fundamental (base) units.

Exam tip

Read which kind of unit the stem wants before you write anything.

"An appropriate SI unit" wants the usual derived unit (Ω m, N kg⁻¹, T); "fundamental (base) units" wants a string of kg, m, s, A, K and mol, built line by line with an equals sign at every step, and the named unit on its own then earns nothing.

Common mistakes

Treating the gram as the base unit. The base unit of mass is the kilogram, so 250 g is 0.250 kg before it goes into any equation.

Giving the coulomb as a base unit. The ampere is the base unit; the coulomb is A s.

Writing a moment in joules. A moment has the same base units as energy, but it is a different quantity and its unit is N m.

Answering "the tesla" when the question asks for fundamental (base) units. A named unit is not a base unit: convert it, for example $\text{T} = \text{kg s}^{-2}\,\text{A}^{-1}$ .

Writing lower-case k for the kelvin, or "°K". The kelvin is K, with no degree sign.

Cancelling letters across a quantity and a unit (the mass $m$ against the metre m). Replace every quantity by its unit first, then simplify units only.

Giving a unit to the power of an exponential. Anything inside $e^{(\ )}$ or a logarithm is a pure number, so a constant inside it takes whatever unit cancels the other quantity.

Check your understanding

1. Which is an SI base unit? [1]

A coulomb

B kelvin

C newton

D volt

2. What are the SI base units of power? [1]

A kg m² s⁻¹

B kg m² s⁻²

C kg m² s⁻³

D kg m s⁻³

3. Which pair of quantities have the same SI base units? [1]

A force and momentum

B impulse and momentum

C energy and power

D pressure and force

4. At high speed, the drag force $F$ on a car moving through air of density $\rho$ is given by

F = \frac{1}{2} C \rho A v^2

where $A$ is the frontal area of the car, $v$ is its speed and $C$ is a constant for the car's shape.

Deduce whether $C$ has a unit. [3]

5. In the equation $E = \frac{1}{2}kx^2$ , a student writes the unit of $k$ as "J m⁻¹".

5.1 Explain why the student's unit is wrong. [2]

5.2 State an appropriate SI unit for $k$ . [1]

6. The rate $P$ at which energy is transferred by conduction through a window of area $A$ and thickness $d$ , when the temperature difference across it is $\Delta T$ , is

P = \frac{\lambda A \Delta T}{d}

where $\lambda$ is a constant for the glass.

6.1 Deduce the fundamental base units of $\lambda$ . [3]

6.2 State an appropriate SI unit for $\lambda$ that uses the watt. [1]

6.3 After the heating is switched off, the temperature difference falls as $\Delta T = \Delta T_0\, e^{-t/\tau}$ , where $t$ is the time.

State the SI unit of $\tau$ . [1]

7. A student is asked to give the resistivity of a metal in "an appropriate SI unit". She writes "kg m³ s⁻³ A⁻²". Explain why this answer may not earn the unit mark, and state the unit expected. [2]

8. Explain why the moment of a force is given in N m and not in J, even though both have the base units kg m² s⁻². [2]

Answers

1. B kelvin [1]

The coulomb is A s, the newton is kg m s⁻² and the volt is J C⁻¹: all three are derived units.

2. C kg m² s⁻³ [1]

$P = W/t$ : $\text{kg m}^2\,\text{s}^{-2} \div \text{s} = \text{kg m}^2\,\text{s}^{-3}$ .
A divides by s⁻¹ instead of s (multiplies by time). B is the joule, the unit of energy, not power. D drops a metre from the joule.

3. B impulse and momentum [1]

Impulse $F\Delta t$ : $\text{kg m s}^{-2} \times \text{s} = \text{kg m s}^{-1}$ ; momentum $mv$ : $\text{kg m s}^{-1}$ . They are equal.
A: force is kg m s⁻², momentum kg m s⁻¹ (a power of s differs). C: power is energy per second. D: pressure is force per area.

4. [3]

MP1 rearrange: $C = \dfrac{2F}{\rho A v^2}$ ✓
MP2 units: $\dfrac{\text{kg m s}^{-2}}{\text{kg m}^{-3} \times \text{m}^2 \times \text{m}^2\,\text{s}^{-2}} = \dfrac{\text{kg m s}^{-2}}{\text{kg m s}^{-2}}$ ✓
MP3 the units cancel completely, so $C$ has no unit (it is a pure number) ✓ (dependent on a correct MP2; a bare "no" scores 0)

5. [3]

5.1 MP1 rearranging gives $k = 2E/x^2$ , so the unit is J m⁻² (energy divided by a length squared) ✓; MP2 J m⁻¹ is the unit of energy per unit length, which is the newton, so it cannot be right ✓ (allow: J m⁻¹ = N, a force, not a stiffness)
5.2 N m⁻¹ ✓ (accept J m⁻², kg s⁻²)

6. [5]

6.1 MP1 $\lambda = \dfrac{Pd}{A\Delta T}$ ✓; MP2 $= \dfrac{\text{kg m}^2\,\text{s}^{-3} \times \text{m}}{\text{m}^2 \times \text{K}}$ ✓; MP3 $= \text{kg m s}^{-3}\,\text{K}^{-1}$ ✓ (condone the order of the base units; the watt must be converted)
6.2 W m⁻¹ K⁻¹ ✓
6.3 s ✓ ( $t/\tau$ is the power of $e$ , so it has no unit)

7. [2]

MP1 the base units are correct ( $\Omega\,\text{m} = \text{kg m}^2\,\text{s}^{-3}\,\text{A}^{-2} \times \text{m}$ ), but the question asks for an appropriate SI unit, which is the usual derived unit, and a string of base units is not accepted where a named unit is standard ✓
MP2 Ω m ✓

8. [2]

MP1 work done (energy) is force × distance moved in the direction of the force; a moment is force × perpendicular distance from the pivot, so the two are different physical quantities ✓
MP2 no energy is transferred by a moment on its own (nothing need move), so writing J would wrongly suggest an energy; the unit is kept as N m to show this ✓

SI base units and derived units

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