Physical quantities — fundamental and derived: Worked examples

Before any symbol appears, one reminder in plain words. When we write $[X]$ with square brackets, we mean "the units of $X$ " — nothing else. So $[speed] = m s^{- 1}$ is read "the units of speed are metres per second". Whenever units multiply or divide, their exponents add or subtract, exactly like ordinary algebra with letters. That single rule — treat units like letters — is the whole game.

One more tool we will lean on repeatedly. When you switch a quantity to a smaller unit, you need more copies of it, so the number grows. Written out:

The scenario matrix

Every dimensional-analysis question falls into one of these cells. The examples below are tagged with the cell they cover, and together they fill the whole table.

Cell	Case class	What makes it tricky	Example
A	Pure multiply (build a unit)	keep track of all base units	Ex 1 (Newton)
B	Divide → negative exponent	$m / m^{2} = m^{- 1}$	Ex 2 (Pascal)
C	Power of a quantity	exponent multiplies through	Ex 3 (area, volume)
D	Unit conversion (chain of ×1)	which fraction goes on top?	Ex 4 (density)
E	Dimensionless result	everything must cancel	Ex 5 (angle, strain)
F	Zero / degenerate input	does the unit survive when the number is 0?	Ex 6
G	Limiting / very-large value	prefixes and orders of magnitude	Ex 7 (prefix chain)
H	Real-world word problem	translate words → symbols first	Ex 8 (fuel economy)
I	Exam twist — find an unknown exponent	solve for a power by matching units	Ex 9
J	Exam trap — hidden inconsistency	spot the wrong equation by units	Ex 10

Cell A — build a unit by multiplying

The bar chart below shows how the exponents of the base units stack up for the newton — a visual "fingerprint" you will reuse for every derived unit on this page. The horizontal axis names the three base units; the vertical axis is the power each is raised to.

deepdives/dd-physics-1.1.01-d3-s02.png

Cell B — division makes a negative exponent

Worked example Ex 2 · Base units of the

pascal Statement: Pressure $P = \frac{F}{A}$ , force divided by area. Find its base units.

Forecast: guess the sign of the exponent on the metre — will it be positive or negative?

Units on top: $[F] = kg m s^{- 2}$ (from Ex 1). Why this step? The numerator of $P$ is a force, so we must first replace "force" by its already-known base units before we can cancel anything.
Units on the bottom: area is length × length, so $[A] = m^{2}$ . Why this step? Area is a derived quantity too; we can't leave it as "area", we need its base units.
Divide → subtract exponents of like units. $[P] = \frac{kg m s ^{- 2}}{m ^{2}} = kg m^{1 - 2} s^{- 2} = kg m^{- 1} s^{- 2}$ Why this step? $m^{1} \div m^{2} = m^{1 - 2} = m^{- 1}$ — the metre ends up on the bottom.

Verify: $1 Pa = 1 kg m^{- 1} s^{- 2}$ . Negative exponent means "metre in the denominator", which matches force spread over area. ✓

The figure contrasts the newton's fingerprint with the pascal's: two bars per base unit (a legend tells them apart), and you can watch the metre bar flip from $+ 1$ (up) to $- 1$ (down) when we divide by area. Axes are labelled just as in the previous figure.

deepdives/dd-physics-1.1.01-d3-s03.png

Cell C — raising a quantity to a power

The picture makes the "power multiplies through" idea literal: one length gives a line, two give a square, three give a cube — the exponent counts how many perpendicular copies of the metre you stacked.

deepdives/dd-physics-1.1.01-d3-s04.png

Cell D — unit conversion (a chain of multiply-by-one)

Worked example Ex 4 · Convert a

density from $g cm^{- 3}$ to $kg m^{- 3}$ Statement: Water has density $1 g cm^{- 3}$ . Express it in the SI derived unit $kg m^{- 3}$ .

Forecast: water in SI is famously $1000 kg m^{- 3}$ . Predict why the number jumps by exactly $1000$ .

Write the conversion factors, each equal to 1. $\frac{1 kg}{1000 g} = 1$ and $\frac{100 cm}{1 m} = 1$ . Why this step? By the identity $n_{1} u_{1} = n_{2} u_{2}$ defined above, $1 kg = 1000 g$ is the same mass, so their ratio is exactly the number $1$ . Multiplying by $1$ never changes the quantity — it only relabels units.
Handle the cubed length. Because the unit is $cm^{- 3}$ , the length factor must be cubed: $(\frac{100 cm}{1 m})^{3}$ . Why this step? The metre appears to the power $- 3$ , so its conversion applies three times.
Multiply it all out. $1 \frac{g}{cm ^{3}} \times \frac{1 kg}{1000 g} \times (\frac{100 cm}{1 m})^{3} = \frac{1 \times 10 0 ^{3}}{1000} \frac{kg}{m ^{3}} = \frac{1 0 ^{6}}{1 0 ^{3}} = 1000 kg m^{- 3}$ Why this step? We chain the factors so that grams cancel grams and centimetres cancel centimetres, leaving only the SI units; the leftover numbers ( $10 0^{3}$ on top, $1000$ underneath) then multiply out to the final figure.

Verify: $10 0^{3} = 1 0^{6}$ , divided by $1 0^{3}$ gives $1 0^{3} = 1000$ . Units left: $kg m^{- 3}$ . ✓ Matches the known value.

Cell E — a dimensionless quantity (everything cancels)

deepdives/dd-physics-1.1.01-d3-s01.png

Cell F — zero and degenerate inputs

Worked example Ex 6 · Does a

zero value still carry a unit? Statement: A car starts from rest: initial velocity $v_{0} = 0$ . In the equation $v = v_{0} + a t$ , does the $0$ have units?

Forecast: is " $0 m s^{- 1}$ " different from a bare " $0$ "?

Check the equation is unit-consistent. Every term of a valid physics equation must share the same units. Here every term must be $m s^{- 1}$ . Why this step? You cannot add quantities of different units — like adding metres to seconds.
Give the zero its unit. So $v_{0} = 0 m s^{- 1}$ , not a dimensionless $0$ . Why this step? The number is zero but the slot still measures a velocity; the unit tags what kind of nothing it is.
Degenerate check — zero area in Ex 2. If area $A \to 0$ in $P = F / A$ , pressure $\to \infty$ : units stay $Pa$ but the value blows up. Why this step? Zero in a denominator is a genuine limit, not a value — the unit survives, the number does not.

Verify: $v = 0 m s^{- 1} + (2 m s^{- 2}) (3 s) = 6 m s^{- 1}$ . All terms in $m s^{- 1}$ , and the zero pulled its weight unit-wise. ✓

Cell G — limiting / very-large values (prefix chains)

Worked example Ex 7 · A

very large length: light-year to metres via prefixes Statement: Light travels at $c = 3.0 \times 1 0^{8} m s^{- 1}$ . How many metres in one light-year (the distance light covers in $1$ year $\approx 3.15 \times 1 0^{7} s$ )? Express with a prefix.

Forecast: order-of-magnitude guess — is it closer to $1 0^{15}$ or $1 0^{20}$ metres?

Distance = speed × time. $[d] = (m s^{- 1}) (s) = m$ — the seconds cancel. Why this step? $s^{- 1} \times s = s^{0} = 1$ ; only the metre survives, as a distance should.
Multiply the numbers. $d = (3.0 \times 1 0^{8}) (3.15 \times 1 0^{7}) = 9.45 \times 1 0^{15} m$ Why this step? Numbers and powers of ten multiply separately: $3.0 \times 3.15 = 9.45$ handles the mantissas, while $1 0^{8} \times 1 0^{7} = 1 0^{15}$ adds the exponents — the same "add exponents when multiplying" rule we use for units.
Attach a prefix. $1 0^{15} = peta (P)$ , so $d \approx 9.45 Pm$ . Why this step? SI Units and Prefixes lets huge numbers ride on a prefix instead of scientific notation.

Verify: $3.0 \times 3.15 = 9.45$ ; $1 0^{8} \times 1 0^{7} = 1 0^{15}$ . So $9.45 \times 1 0^{15} m$ . ✓ Between $1 0^{15}$ and $1 0^{16}$ — the peta scale.

Cell H — real-world word problem

Worked example Ex 8 ·

Fuel economy — translate words, then units Statement: A car uses $6.0 litres$ of fuel per $100 km$ . Convert this to the "distance per volume" figure in $km L^{- 1}$ (kilometres per litre), then to base-SI $m m^{- 3} = m^{- 2}$ .

Forecast: "distance per volume" gives units $m^{- 2}$ — an inverse area. Guess why an inverse area appears.

Translate the words. "6.0 L per 100 km" is $\frac{6.0 L}{100 km}$ . We want distance per fuel, so invert: $\frac{100 km}{6.0 L} = 16.7 km L^{- 1}$ . Why this step? "distance per volume" puts distance on top; the question flips the given ratio.
To base SI. $1 km = 1 0^{3} m$ , and $1 L (litre) = 1 0^{- 3} m^{3}$ . Why this step? Each of these is again the identity $n_{1} u_{1} = n_{2} u_{2}$ in disguise: $1 km$ and $1 0^{3} m$ are the same distance, $1 L$ and $1 0^{- 3} m^{3}$ the same volume, so substituting them changes nothing physical — it only re-expresses each unit in base SI so the exponents can be combined in the next line.
Combine and cancel exponents. $16.7 \frac{km}{L} = 16.7 \times \frac{1 0 ^{3} m}{1 0 ^{- 3} m ^{3}} = 16.7 \times 1 0^{6} m m^{- 3} = 1.67 \times 1 0^{7} m^{- 2}$ Why this step? $\frac{m}{m ^{3}} = m^{1 - 3} = m^{- 2}$ . The reciprocal figure — volume per distance, $m^{3} / m = m^{2}$ — is literally the cross-sectional area of fuel the engine "sweeps up" per metre travelled; distance-per-volume is just its inverse, hence $m^{- 2}$ .

Verify: $100/6.0 = 16.67$ ; $1 0^{3} /1 0^{- 3} = 1 0^{6}$ ; $16.67 \times 1 0^{6} = 1.667 \times 1 0^{7} m^{- 2}$ . ✓

Cell I — exam twist: find an unknown exponent by matching units

Worked example Ex 9 · The

period of a pendulum — solve for the power Statement: A pendulum's period $T$ (a time, unit $s$ ) depends on length $ℓ$ (m) and gravity $g$ ( $m s^{- 2}$ ) as $T = k ℓ^{a} g^{b}$ , where $k$ is a dimensionless constant. Find $a$ and $b$ using Dimensional Analysis.

Forecast: guess the sign of $b$ — should stronger gravity make the swing faster or slower?

Write units of both sides. Left: $[T] = s = m^{0} s^{1}$ . Right: $[ℓ^{a} g^{b}] = (m)^{a} (m s^{- 2})^{b} = m^{a + b} s^{- 2 b}$ . Why this step? A valid equation must have identical units on both sides, letter by letter.
Match the exponent of each base unit.
- Metre: $a + b = 0$ .
- Second: $- 2 b = 1 \Rightarrow b = - \frac{1}{2}$ . Why this step? Two unknowns need two equations; each base unit contributes exactly one equation because its exponent must agree on the left and right.
Solve the two linked equations. The "second" equation already gives $b = - \frac{1}{2}$ on its own. Substitute that into the "metre" equation $a + b = 0$ : $a = - b = + \frac{1}{2}$ . Why this step? We solve them in sequence — the second-unit equation has only $b$ , so it pins $b$ first; then the metre equation, now containing only the unknown $a$ , is trivial to finish. This is standard back-substitution for two linked linear equations.

Result: $a = + \frac{1}{2}, b = - \frac{1}{2}$ , so $T = k ℓ^{1/2} g^{- 1/2} = k \frac{ℓ}{g} .$

Verify: This matches the true pendulum formula $T = 2 π ℓ / g$ (with $k = 2 π$ , which dimensional analysis cannot supply). $b < 0$ : bigger $g$ ⇒ smaller $T$ ⇒ faster swing, matching intuition. ✓

Cell J — exam trap: catch the inconsistent equation

Worked example Ex 10 · Which equation is

dimensionally impossible? Statement: Two students propose formulas for distance $s$ (unit m): (i) $s = v t + \frac{1}{2} a t^{2}$ , and (ii) $s = v t^{2} + \frac{1}{2} a t$ . Only one can be right — decide by units alone.

Forecast: guess which one fails before checking, just from the powers of $t$ .

Units of each term in (i). $v t = (m s^{- 1}) (s) = m$ . $a t^{2} = (m s^{- 2}) (s^{2}) = m$ . Both metres. ✓ Why this step? Adding requires matching units; both terms give metres, consistent with $s$ .
Units of each term in (ii). $v t^{2} = (m s^{- 1}) (s^{2}) = m s$ . $a t = (m s^{- 2}) (s) = m s^{- 1}$ . Why this step? Neither term is a metre, and they don't even match each other — you cannot add $m s$ to $m s^{- 1}$ .
Conclude. Equation (ii) is dimensionally impossible; (i) is the correct kinematic equation. Why this step? A formula is only valid if every added term shares the units of the left-hand side ( $m$ ); since (ii)'s terms are neither metres nor even equal to each other, it cannot describe a distance, so we discard it.

Verify: In (i) every term is $m$ ; in (ii) terms are $m s$ and $m s^{- 1}$ , neither equal to $m$ . Dimensional analysis rejects (ii). ✓ (It can rule formulas out, never confirm dimensionless constants — the $\frac{1}{2}$ is invisible to it.)

Recall

Recall One-line self-tests

Units of pressure in base SI ::: kg m⁻¹ s⁻² Why does dividing lengths give a dimensionless angle? ::: m ÷ m = m⁰ = 1, no unit left In v = v₀ + at with v₀ = 0, what unit does the 0 carry? ::: m s⁻¹ (the slot still measures velocity) Converting g cm⁻³ to kg m⁻³ multiplies the number by ::: 1000 For T = k ℓᵃ gᵇ, the exponents are ::: a = +½, b = −½ Which fails units: s = vt + ½at² or s = vt² + ½at? ::: the second (vt² + ½at)

Connections

Dimensional Analysis — Ex 9 and Ex 10 are pure dimensional-analysis technique.
SI Units and Prefixes — the multiply-by-one conversions in Ex 4, 7, 8.
Newton's Second Law — source of the newton in Ex 1.
Errors and Significant Figures — every numeric answer here is rounded consistently.
Vectors — velocity and force here are vectors; their units are scalars, handled identically.