In Dimensional Analysis you used units to convert from one unit to another, letting the units you didn't want cancel. Physics uses the same habit for a second job: checking equations. Before you trust a formula, or an answer, look at its units.
Key idea
Units multiply, divide and cancel just like algebra. Every term that is added, subtracted or set equal in an equation must have the same units. If they don't match, the equation is wrong.
Why it works
An equation says two amounts are the same quantity. A length can equal a length, but it can never equal a time, just as 3 meters can't be "the same as" 3 seconds. Adding works the same way: $3\ \text{m} + 2\ \text{s}$ has no meaning, so any formula that asks you to do it must be misremembered.
| Quantity | Symbol | SI unit |
|---|---|---|
| length, position, height | $x$, $h$, $d$ | m |
| time | $t$ | s |
| mass | $m$ | kg |
| velocity | $v$ | m/s |
| acceleration | $a$, $g$ | m/s² |
| force | $F$ | N = kg·m/s² |
| energy | $E$, $K$, $U$ | J = kg·m²/s² |
Worked examples
Worked example: Checking a motion equation
Check the units of $x = v_0 t + \tfrac12 a t^2$.
- Left side: $x$ is a position, in m.
- First term: $v_0 t$ has units $\dfrac{\text{m}}{\text{s}} \cdot \text{s} = \text{m}$.
- Second term: $\tfrac12$ has no units, so $a t^2$ has units $\dfrac{\text{m}}{\text{s}^2} \cdot \text{s}^2 = \text{m}$.
- Every term is in meters, so the equation passes the test.
Answer: The units match: m = m + m.
Worked example: Catching a slip
A student remembers the speed of a falling object as $v = 2gh$. Is that possible?
- Units of $2gh$: $\dfrac{\text{m}}{\text{s}^2} \cdot \text{m} = \dfrac{\text{m}^2}{\text{s}^2}$.
- A speed is in $\dfrac{\text{m}}{\text{s}}$, so the two sides don't match.
- Taking a square root fixes the units: $\sqrt{\text{m}^2/\text{s}^2} = \text{m/s}$. The real formula is $v = \sqrt{2gh}$.
Answer: No. It must be $v = \sqrt{2gh}$.
Worked example: Units of an answer
You compute $\dfrac{v^2}{a}$ with $v$ in m/s and $a$ in m/s². What kind of quantity did you find?
- $\dfrac{(\text{m/s})^2}{\text{m/s}^2} = \dfrac{\text{m}^2/\text{s}^2}{\text{m}/\text{s}^2}$.
- The $\text{s}^2$ cancel, and $\text{m}^2/\text{m} = \text{m}$.
Answer: A length, in meters (it shows up later as a stopping distance).
Common mistake
Adding quantities with different units, or keeping an answer whose units don't fit the question. If you are asked for a distance and your answer comes out in m/s, you used the wrong formula or dropped a factor of time. Quick check: write the units next to every number as you work, and look at the units of the final answer before you look at its size.
Matching units can't catch a missing $\tfrac12$ or a wrong factor of 2, because plain numbers have no units. Units prove an equation wrong; they can't prove it right.