IB Physics SL Tool 3 — Mathematics Paper 1 & 2 SI units & prefixes ~8 min read

Base & Derived Units

Every measurement in physics, from the mass of an electron to the distance across a galaxy, is built out of just seven base units. Everything else — the newton, the joule, the volt — is a derived unit, assembled from those seven like words from an alphabet. Learn how the building blocks fit together and you can check any equation, decode any unit, and never be thrown by an unfamiliar one again.

📘 What you need to know

The seven SI base units

These are the foundation. Each measures one physical quantity and can’t be broken down into anything simpler within the SI system.

QuantityUnit nameSymbol
lengthmetrem
masskilogramkg
timeseconds
electric currentampereA
temperaturekelvinK
amount of substancemolemol
luminous intensitycandelacd
A tiny exam-saver: the candela is the odd one out — it’s a genuine SI base unit, but IB Physics doesn’t use it. So when a question asks how many base units you’ll actually meet in the course, the answer is six, not seven.

Prefixes: scaling units up and down

When numbers get very large or very small, prefixes let you keep them tidy. Each prefix stands for a power of ten — a kilowatt is 10³ watts, a milligram is 10⁻³ grams. Here are the ones that come up most in physics.

PrefixSymbolValue
petaP1015
teraT1012
gigaG109
megaM106
kilok103
hectoh102
decada101
decid10−1
centic10−2
millim10−3
microμ10−6
nanon10−9
picop10−12
femtof10−15

Building derived units

A derived unit is just base units multiplied and divided together. The trick to finding one is to start from the quantity’s defining equation and swap each symbol for its base units. Take the newton, the unit of force. Force is mass × acceleration, and acceleration is metres per second squared, so:

force F = mass m × acceleration akg × m s⁻²combine ↓ 1 N = 1 kg m s⁻²
Swap each quantity for its base units, then multiply through: the newton is kilogram-metre-per-second-squared.

The same recipe unpacks every derived unit. Energy is ½mv², so the joule is kg × (m s⁻¹)² = kg m² s⁻². Pressure is force ÷ area, so the pascal is kg m s⁻² ÷ m² = kg m⁻¹ s⁻².

🧭 Finding a derived unit

  1. Write the defining equation for the quantity (from the data booklet if needed)
  2. Replace each symbol with its base units
  3. Multiply and divide the units, collecting powers of each base unit
  4. Simplify to the tidiest form — that’s your answer

Common derived units

These come up again and again. It’s worth being able to recognise the base-unit form, since a “state the fundamental units of…” question is a quick, easy mark.

Derived unitQuantityIn base SI units
newton (N)forcekg m s−2
pascal (Pa)pressurekg m−1 s−2
joule (J)energykg m2 s−2
watt (W)powerkg m2 s−3
hertz (Hz)frequencys−1
coulomb (C)chargeA s
volt (V)potential differencekg m2 s−3 A−1
ohm (Ω)resistancekg m2 s−3 A−2
Quick recap: seven base units (six used in IB); prefixes scale by powers of ten; derived units come from a defining equation — newton = kg m s⁻², joule = kg m² s⁻², pascal = kg m⁻¹ s⁻².

Handy non-SI units

Sometimes an SI unit is awkwardly big or small, so physicists use a convenient alternative. You need to know what each one means and how to convert it.

UnitUsed forConversion
electronvolt (eV)tiny energies1 eV = 1.60 × 10−19 J
kilowatt-hour (kWh)household energy1 kWh = 3.60 × 106 J
light year (ly)stellar distance1 ly = 9.46 × 1015 m
parsec (pc)stellar distance1 pc = 3.26 ly
astronomical unit (AU)Solar-System distance1 AU = 1.50 × 1011 m
WE 1

(a) A household uses 3200 kWh of electricity in a year. Express this energy in joules. (b) A star lies 40 pc from Earth. Express this distance in metres. Use 1 kWh = 3.60 × 10⁶ J, 1 pc = 3.26 ly and 1 ly = 9.46 × 10¹⁵ m.

Part (a) — energy in J 1 kWh = 3.60 × 10⁶ J 3200 × (3.60 × 10⁶) = 1.152 × 10¹⁰ ≈ 1.2 × 10¹⁰ J Part (b) — distance in m first pc → ly, then ly → m 40 × 3.26 × (9.46 × 10¹⁵) ≈ 1.2 × 10¹⁸ m Chain the conversions one step at a time — parsecs to light years, light years to metres — so no factor gets dropped.
WE 2

Show that the pascal, the SI unit of pressure, is equivalent to kg m⁻¹ s⁻² in base units.

Start from the definition of pressure and substitute the base units of force and area.

Definition pressure = force ÷ area, so Pa = N ÷ m² Substitute base units N = kg m s⁻² (force = mass × acceleration) Pa = (kg m s⁻²) ÷ m² Pa = kg m⁻¹ s⁻² Dividing by m² drops the length power from +1 to −1 — that’s where the m⁻¹ comes from.

💡 Top tips

⚠ Common mistakes

Up next: Using Dimensional Analysis — now that you can break any unit into base units, we’ll use that skill to check whether an equation is even possible, by testing that the units balance on both sides.

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