Derived unit
Physical Quantities & Units - OCR A-Level Physics
Key Definition
Derived unit
A unit built from a combination of SI base units. The combination comes from the defining equation of the quantity. Example: the newton is defined by $F = ma$, so $1 \text{ N} = 1 \text{ kg m s}^{-2}$.
A unit built from a combination of SI base units. The combination comes from the defining equation of the quantity. Example: the newton is defined by $F = ma$, so $1 \text{ N} = 1 \text{ kg m s}^{-2}$.
Procedure: deriving any derived unit
Step 1: write the defining equation. Step 2: replace each quantity on the right-hand side with its SI base units. Step 3: if any sub-unit is itself derived (such as $\text{N}$ inside $\text{Pa} = \text{N}/\text{m}^{2}$), substitute another equation until everything is in $\text{kg}$, $\text{m}$, $\text{s}$, $\text{A}$, $\text{K}$, $\text{mol}$. Step 4: simplify the indices.
$$1 \text{ Pa} = \frac{1 \text{ N}}{1 \text{ m}^{2}} = \frac{1 \text{ kg m s}^{-2}}{1 \text{ m}^{2}} = 1 \text{ kg m}^{-1} \text{ s}^{-2}$$
- VelocityRate of change of displacement. A vector quantity. Measured in $\text{m s}^{-1}$.: $\text{m s}^{-1}$ (from $v = s/t$).
- AccelerationRate of change of velocity. A vector quantity. Measured in $\text{m s}^{-2}$.: $\text{m s}^{-2}$ (from $a = \Delta v / t$).
- Force (newton, $\text{N}$): $\text{kg m s}^{-2}$ (from $F = ma$).
- EnergyThe capacity to do work. Measured in joules (J). (joule, $\text{J}$): $\text{kg m}^{2} \text{ s}^{-2}$ (from $W = Fd$).
- PowerThe rate of energy transfer. Measured in watts (W). (watt, $\text{W}$): $\text{kg m}^{2} \text{ s}^{-3}$ (from $P = W/t$).
- PressureForce per unit area. Measured in pascals (Pa), where $1 \text{ Pa} = 1 \text{ N m}^{-2}$. (pascal, $\text{Pa}$): $\text{kg m}^{-1} \text{ s}^{-2}$ (from $p = F/A$).
- Charge (coulomb, $\text{C}$): $\text{A s}$ (from $Q = It$).
- Potential difference (volt, $\text{V}$): $\text{kg m}^{2} \text{ s}^{-3} \text{ A}^{-1}$ (from $V = W/Q$).
- Resistance (ohm, $\Omega$): $\text{kg m}^{2} \text{ s}^{-3} \text{ A}^{-2}$ (from $R = V/I$).
Common Mistake
HIGH
Wrong: Stopping at $\text{N m}^{-2}$ when asked for the base units of the pascal. The newton is itself a derived unit, so the answer is incomplete.
Right: Keep substituting until only the six base symbols ($\text{kg}$, $\text{m}$, $\text{s}$, $\text{A}$, $\text{K}$, $\text{mol}$) remain. $1 \text{ Pa} = 1 \text{ N m}^{-2} = 1 \text{ kg m}^{-1} \text{ s}^{-2}$.
Right: Keep substituting until only the six base symbols ($\text{kg}$, $\text{m}$, $\text{s}$, $\text{A}$, $\text{K}$, $\text{mol}$) remain. $1 \text{ Pa} = 1 \text{ N m}^{-2} = 1 \text{ kg m}^{-1} \text{ s}^{-2}$.
Examiner Tips and Tricks
- To find the base units of any quantity, substitute its defining equation step by step.
- Worked: $V = W/Q = (\text{kg m}^{2}\text{ s}^{-2}) / (\text{A s}) = \text{kg m}^{2}\text{ s}^{-3}\text{ A}^{-1}$.
- Show every substitution in the working. The mark scheme typically awards 1 mark for the correct starting equation and 1 mark for the simplified base units.