Quantities and Units

Quantities and Units

:::info Board Coverage AQA Paper 1 & 3 | Edexcel CP1 | OCR (A) Paper 1 | CIE P1 :::

1. Physical Quantities and the SI System

We begin with the most fundamental question in physics: how do we measure things?

A physical quantity is a property of a phenomenon that can be quantified — assigned a numerical Value and compared with other instances of the same quantity. Physical quantities come in two Varieties:

• Base quantities are irreducible; they cannot be expressed in terms of other quantities. The SI system defines seven base quantities.
• Derived quantities are expressed as products and quotients of base quantities.

The Seven SI Base Units

Base Quantity	Symbol	SI Unit	Unit Symbol
Length	$l$	metre	m
Mass	$m$	kilogram	kg
Time	$t$	second	s
Electric current	$I$	ampere	A
Temperature	$T$	kelvin	K
Amount of substance	$n$	mole	mol
Luminous intensity	$I_v$	candela	cd

:::info Note The candela is rarely encountered in A Level Physics. Focus on the first six. :::

Derived Units

A derived unit is obtained by combining base units according to the physical relationship. We denote The dimensions of a quantity using square brackets.

Definition. The dimension of a physical quantity $Q$Written $[Q]$Is its expression in Terms of the base dimensions $\mathsf{M}$ (mass), $\mathsf{L}$ (length), $\mathsf{T}$ (time), $\mathsf{I}$ (current), $\mathsf{\Theta}$ (temperature), $\mathsf{N}$ (amount of substance), $\mathsf{J}$ (luminous intensity).

Examples.

• Velocity: $[v] = \frac{[s]}{[t]} = \mathsf{L}\mathsf{T}^{-1}$
• Acceleration: $[a] = \frac{[v]}{[t]} = \mathsf{L}\mathsf{T}^{-2}$
• Force (from $F = ma$): $[F] = \mathsf{M}\mathsf{L}\mathsf{T}^{-2}$
• Pressure (from $P = F/A$): $[P] = \mathsf{M}\mathsf{L}^{-1}\mathsf{T}^{-2}$
• Energy (from $W = Fd$): $[E] = \mathsf{M}\mathsf{L}^{2}\mathsf{T}^{-2}$

Some derived units have special names:

Derived Quantity	Name	Symbol	In Base Units
Force	newton	N	$\mathrm{kg m s}^{-2}$
Energy	joule	J	$\mathrm{kg m}^2\mathrm{s}^{-2}$
Power	watt	W	$\mathrm{kg m}^2\mathrm{s}^{-3}$
Pressure	pascal	Pa	$\mathrm{kg m}^{-1}\mathrm{s}^{-2}$
Charge	coulomb	C	$\mathrm{A s}$
Voltage	volt	V	$\mathrm{kg m}^2\mathrm{s}^{-3}\mathrm{A}^{-1}$
Resistance	ohm	$\Omega$	$\mathrm{kg m}^2\mathrm{s}^{-3}\mathrm{A}^{-2}$

2. Dimensional Analysis

Definition. An equation is dimensionally homogeneous (or dimensionally consistent) if Every term on each side has the same dimensions.

This is a necessary condition for any physically meaningful equation. If the dimensions do not Balance, the equation is certainly wrong. If they do balance, the equation may still be wrong (dimensional analysis cannot reveal dimensionless constants), but it is at least plausible.

Checking the SUVAT Equation

We prove that $v^2 = u^2 + 2as$ is dimensionally valid.

• $[v^2] = (\mathsf{L}\mathsf{T}^{-1})^2 = \mathsf{L}^2\mathsf{T}^{-2}$
• $[u^2] = (\mathsf{L}\mathsf{T}^{-1})^2 = \mathsf{L}^2\mathsf{T}^{-2}$
• $[2as] = [\mathsf{L}\mathsf{T}^{-2}][\mathsf{L}] = \mathsf{L}^2\mathsf{T}^{-2}$

Since $[v^2] = [u^2] = [2as] = \mathsf{L}^2\mathsf{T}^{-2}$The equation is dimensionally Homogeneous. $\square$

Determining the Form of an Equation

Suppose we wish to find how the period $T$ of a simple pendulum depends on its length $l$ and the Gravitational field strength $g$. We assume:

$T = k \cdot l^a \cdot g^b$

Where $k$ is a dimensionless constant. By dimensional homogeneity:

$\mathsf{T} = \mathsf{L}^a \cdot (\mathsf{L}\mathsf{T}^{-2})^b = \mathsf{L}^{a+b} \cdot \mathsf{T}^{-2b}$

Equating powers:

$$\begin{aligned} \mathsf{T}:&\quad 1 = -2b \implies b = -\frac{1}{2} \\[4pt] \mathsf{L}:&\quad 0 = a + b \implies a = \frac{1}{2} \end{aligned}$$

Therefore $T = k\sqrt{l/g}$. Full analysis reveals $k = 2\pi$.

:::tip Exam Technique Dimensional analysis is invaluable for checking your working. Get into the Habit of verifying dimensions for every formula you derive in an exam. :::

3. SI Prefixes

Prefix	Symbol	Factor
tera	T	$10^{12}$
giga	G	$10^9$
mega	M	$10^6$
kilo	k	$10^3$
centi	c	$10^{-2}$
milli	m	$10^{-3}$
micro	$\mu$	$10^{-6}$
nano	n	$10^{-9}$
pico	p	$10^{-12}$
femto	f	$10^{-15}$

4. Scalars and Vectors

Definition. A scalar is a physical quantity that has magnitude only. A vector is a Physical quantity that has both magnitude and direction.

Scalar Examples	Vector Examples
Mass, temperature, speed, energy, time, distance	Displacement, velocity, acceleration, force, weight, momentum

Vector Operations

Addition. Vectors are added using the triangle rule or the parallelogram rule. Given vectors $\mathbf{a}$ and $\mathbf{b}$The resultant $\mathbf{R} = \mathbf{a} + \mathbf{b}$ is found by Placing the tail of $\mathbf{b}$ at the head of $\mathbf{a}$.

Resolving. Any vector $\mathbf{F}$ can be resolved into perpendicular components. If $\mathbf{F}$ makes an angle $\theta$ with the horizontal:

$F_x = F\cos\theta, \qquad F_y = F\sin\theta$

Magnitude. Given components $F_x$ and $F_y$:

$|\mathbf{F}| = \sqrt{F_x^2 + F_y^2}$

Direction. The angle with the horizontal is $\theta = \arctan\left(\frac{F_y}{F_x}\right)$.

5. Uncertainty and Error Analysis

Types of Error

• Systematic error: A consistent deviation from the true value, caused by a flaw in the apparatus or method. It affects accuracy (closeness to true value) but not precision (repeatability). Example: a zero error on a micrometer.
• Random error: Unpredictable fluctuations in measured values, caused by limitations in resolution or environmental factors. It affects precision but not accuracy. Random errors are reduced by taking repeated measurements.

:::caution Common Pitfall Do not confuse precision with accuracy. A precise measurement is Repeatable; an accurate measurement is close to the true value. You can have one without the other. :::

Absolute, Fractional, and Percentage Uncertainty

Definition. If a quantity is measured as $x \pm \Delta x$Then:

• Absolute uncertainty: $\Delta x$
• Fractional uncertainty: $\frac◆LB◆\Delta x◆RB◆◆LB◆x◆RB◆$
• Percentage uncertainty: $\frac◆LB◆\Delta x◆RB◆◆LB◆x◆RB◆ \times 100\%$

Combining Uncertainties

We now derive the rules for propagating uncertainty through calculations. Consider a quantity $z = f(x, y)$ where $x$ and $y$ have uncertainties $\Delta x$ and $\Delta y$.

Rule 1: Addition and Subtraction

If $z = x + y$ or $z = x - y$Then:

$\Delta z = \Delta x + \Delta y$

Derivation. The worst-case scenario for $z = x + y$ is that both errors push $z$ in the same Direction. The maximum possible value is $z_{\max} = (x + \Delta x) + (y + \Delta y)$And the Minimum is $z_{\min} = (x - \Delta x) + (y - \Delta y)$. Hence:

$\Delta z = \frac◆LB◆z_{\max} - z_{\min}◆RB◆◆LB◆2◆RB◆ = \Delta x + \Delta y$

The same argument applies for subtraction. $\square$

Rule 2: Multiplication and Division

If $z = xy$ or $z = x/y$Then:

$\frac◆LB◆\Delta z◆RB◆◆LB◆z◆RB◆ = \frac◆LB◆\Delta x◆RB◆◆LB◆x◆RB◆ + \frac◆LB◆\Delta y◆RB◆◆LB◆y◆RB◆$

In words: when multiplying or dividing, add the fractional uncertainties.

Derivation for $z = xy$. We have:

$$\begin{aligned} Z_{\max} &= (x + \Delta x)(y + \Delta y) = xy + x\Delta y + y\Delta x + \Delta x \Delta y \\ Z_{\min} &= (x - \Delta x)(y - \Delta y) = xy - x\Delta y - y\Delta x + \Delta x \Delta y \end{aligned}$$

For small uncertainties, $\Delta x \Delta y$ is negligible:

$$\begin{aligned} \Delta z &= \frac◆LB◆z_{\max} - z_{\min}◆RB◆◆LB◆2◆RB◆ \approx x\Delta y + y\Delta x \\[4pt] \frac◆LB◆\Delta z◆RB◆◆LB◆z◆RB◆ &= \frac◆LB◆x\Delta y + y\Delta x◆RB◆◆LB◆xy◆RB◆ = \frac◆LB◆\Delta y◆RB◆◆LB◆y◆RB◆ + \frac◆LB◆\Delta x◆RB◆◆LB◆x◆RB◆ \end{aligned}$$

$\square$

Rule 3: Powers

If $z = x^n$Then:

$\frac◆LB◆\Delta z◆RB◆◆LB◆z◆RB◆ = |n| \cdot \frac◆LB◆\Delta x◆RB◆◆LB◆x◆RB◆$

Derivation. Write $z = \underbrace◆LB◆x \cdot x \cdots x◆RB◆_{n \mathrm{ times}}$. Applying the Multiplication rule repeatedly:

$\frac◆LB◆\Delta z◆RB◆◆LB◆z◆RB◆ = \underbrace◆LB◆\frac{\Delta x}{x} + \frac{\Delta x}{x} + \cdots + \frac{\Delta x}{x}◆RB◆_{n \mathrm{ terms}} = n \cdot \frac◆LB◆\Delta x◆RB◆◆LB◆x◆RB◆$

For negative or fractional powers, the result generalises via logarithmic differentiation (see 01-wave-properties for the general propagation of error formula). $\square$

The General Propagation of Error Formula

For any function $z = f(x_1, x_2, \ldots, x_n)$:

$\Delta z = \sqrt◆LB◆\sum_{i=1}^{n}\left(\frac{\partial f}{\partial x_i} \Delta x_i\right)^2◆RB◆$

This is the statistical (root-sum-square) combination, which gives the most probable uncertainty Rather than the worst case. For A Level exams, use the simpler worst-case rules above unless Instructed otherwise.

Significant Figures and Uncertainty

The number of significant figures in a quoted result should be consistent with the uncertainty.

Rule. A result should be quoted to the same number of significant figures as the uncertainty, And the uncertainty should be quoted to at most 2 significant figures.

Example. If a length is measured as $12.3 \pm 0.4$ cm, we quote two significant figures (matching the uncertainty’s one significant figure). We do not write $12.30 \pm 0.4$ cm — the Trailing zero implies precision we do not have.

:::tip Exam Technique When you compute $g = 9.78 \pm 0.15 \mathrm{ m s}^{-2}$Write $9.8 \pm 0.2$ m S$^{-2}$ (round the uncertainty to 1 s.f. And match the result). This is what examiners expect. :::

6. Determining Uncertainty from Repeated Measurements

When $n$ repeated measurements $x_1, x_2, \ldots, x_n$ are taken of the same quantity:

$\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i$

The absolute uncertainty is the half-range:

$\Delta x = \frac◆LB◆x_{\max} - x_{\min}◆RB◆◆LB◆2◆RB◆$

For large datasets, the standard deviation of the mean is more appropriate:

$\Delta x = \frac◆LB◆\sigma◆RB◆◆LB◆\sqrt{n}◆RB◆, \qquad \sigma = \sqrt◆LB◆\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2◆RB◆$

7. Graphical Analysis of Uncertainties

When determining a physical constant from the gradient of a straight-line graph, we use the line Of best fit and the worst acceptable line (the steepest and shallowest lines consistent with The error bars).

The uncertainty in the gradient is:

$\Delta m = \frac◆LB◆|m_{\mathrm{best}} - m_{\mathrm{worst}}|◆RB◆◆LB◆2◆RB◆$

A similar procedure applies to the $y$-intercept.

Problem Set

Problem 1

A student measures the diameter of a sphere five times and obtains: 2.04 cm, 2.06 cm, 2.05 cm, 2.03 cm, 2.07 cm. Calculate the mean diameter and its absolute uncertainty.

Answer. $\bar{d} = \frac{2.04 + 2.06 + 2.05 + 2.03 + 2.07}{5} = 2.05$ cm. The range is $2.07 - 2.03 = 0.04$ cm, so $\Delta d = 0.02$ cm. Result: $d = 2.05 \pm 0.02$ cm.

If you get this wrong, revise: Determining Uncertainty from Repeated Measurements

Problem 2

The kinetic energy of a particle is given by $E_k = \frac{1}{2}mv^2$. Show that this expression is dimensionally consistent with the definition of work $W = Fd$.

Answer. $[E_k] = [\mathrm{mass}][\mathrm{velocity}]^2 = \mathsf{M}(\mathsf{L}\mathsf{T}^{-1})^2 = \mathsf{M}\mathsf{L}^2\mathsf{T}^{-2}$. Meanwhile $[W] = [F][d] = (\mathsf{M}\mathsf{L}\mathsf{T}^{-2})(\mathsf{L}) = \mathsf{M}\mathsf{L}^2\mathsf{T}^{-2}$. The dimensions match. $\square$

If you get this wrong, revise: Derived Units

Problem 3

A force $F = 12.0 \pm 0.3$ N acts over a distance $d = 3.45 \pm 0.05$ m. Calculate the work done and its percentage uncertainty.

Answer. $W = Fd = 12.0 \times 3.45 = 41.4$ J. The fractional uncertainties are $\frac{0.3}{12.0} = 0.025$ and $\frac{0.05}{3.45} = 0.0145$. By the multiplication rule: $\frac◆LB◆\Delta W◆RB◆◆LB◆W◆RB◆ = 0.025 + 0.0145 = 0.0395$So $\Delta W = 41.4 \times 0.0395 \approx 1.6$ J. Result: $W = 41.4 \pm 1.6$ J (3.9% uncertainty).

If you get this wrong, revise: Rule 2: Multiplication and Division

Problem 4

Use dimensional analysis to show that the expression $v = \sqrt◆LB◆\frac{2\Delta E}{m}◆RB◆$ is dimensionally valid, where $\Delta E$ is energy and $m$ is mass.

Answer. $[v] = \mathsf{L}\mathsf{T}^{-1}$. $[\Delta E/m] = \frac◆LB◆\mathsf{M}\mathsf{L}^2\mathsf{T}^{-2}◆RB◆◆LB◆\mathsf{M}◆RB◆ = \mathsf{L}^2\mathsf{T}^{-2}$. $[\sqrt◆LB◆2\Delta E/m◆RB◆] = (\mathsf{L}^2\mathsf{T}^{-2})^{1/2} = \mathsf{L}\mathsf{T}^{-1} = [v]$. $\square$

If you get this wrong, revise: Dimensional Analysis

Problem 5

The density of a cylinder is $\rho = \frac◆LB◆m◆RB◆◆LB◆\pi r^2 h◆RB◆$. The mass $m = 150.0 \pm 0.5$ g, radius $r = 1.20 \pm 0.05$ cm, and height $h = 5.00 \pm 0.02$ cm. Calculate $\rho$ and its uncertainty.

Answer. $\rho = \frac◆LB◆150.0◆RB◆◆LB◆\pi(1.20)^2(5.00)◆RB◆ = \frac{150.0}{22.62} = 6.63$ g cm$^{-3}$.

Fractional uncertainties: $\frac{0.5}{150.0} = 0.0033$, $\frac{2(0.05)}{1.20} = 0.0833$ (power Rule), $\frac{0.02}{5.00} = 0.004$.

Total fractional uncertainty: $0.0033 + 0.0833 + 0.004 = 0.0907$. $\Delta\rho = 6.63 \times 0.0907 = 0.60$ g cm$^{-3}$. Result: $\rho = 6.6 \pm 0.6$ g cm$^{-3}$.

If you get this wrong, revise: Rule 3: Powers

Problem 6

Two vectors are given: $\mathbf{A} = 3\mathbf{i} + 4\mathbf{j}$ N and $\mathbf{B} = -2\mathbf{i} + 5\mathbf{j}$ N. Find the magnitude and direction of $\mathbf{A} + \mathbf{B}$.

Answer. $\mathbf{A} + \mathbf{B} = (3-2)\mathbf{i} + (4+5)\mathbf{j} = \mathbf{i} + 9\mathbf{j}$. Magnitude: $|\mathbf{A}+\mathbf{B}| = \sqrt{1^2 + 9^2} = \sqrt{82} = 9.06$ N. Direction: $\theta = \arctan(9/1) = 83.7^\circ$ above the positive $x$-axis.

If you get this wrong, revise: Vector Operations

Problem 7

A student proposes the formula for the period of a mass on a spring: $T = 2\pi\sqrt◆LB◆\frac{k}{m}◆RB◆$Where $k$ is the spring constant and $m$ is the mass. Use dimensional analysis to show this formula is incorrect, and find the correct form.

Answer. $[T] = \mathsf{T}$. $[k/m] = \frac◆LB◆[\mathrm{force}]/[\mathrm{displacement}]◆RB◆◆LB◆[\mathrm{mass}]◆RB◆ = \frac◆LB◆\mathsf{M}\mathsf{L}\mathsf{T}^{-2}/\mathsf{L}◆RB◆◆LB◆\mathsf{M}◆RB◆ = \mathsf{T}^{-2}$. So $[\sqrt{k/m}] = \mathsf{T}^{-1} \neq \mathsf{T}$. The formula is dimensionally wrong. The correct Form is $T = 2\pi\sqrt◆LB◆\frac{m}{k}◆RB◆$Which gives $[\sqrt{m/k}] = \sqrt◆LB◆\frac{\mathsf{M}}{\mathsf{T}^{-2}}◆RB◆ = \mathsf{T}$. $\square$

If you get this wrong, revise: Determining the Form of an Equation

Problem 8

Convert 0.0000450 nm to SI base units with appropriate significant figures.

Answer. $0.0000450$ nm $= 4.50 \times 10^{-5}$ nm $= 4.50 \times 10^{-5} \times 10^{-9}$ m $= 4.50 \times 10^{-14}$ m. The result has 3 significant figures, matching the original.

If you get this wrong, revise: SI Prefixes

Problem 9

A resistance is calculated from $R = \frac{V}{I}$ where $V = 6.00 \pm 0.05$ V and $I = 0.50 \pm 0.02$ A. Find $R$ and its absolute uncertainty.

Answer. $R = \frac{6.00}{0.50} = 12.0$ $\Omega$. Fractional uncertainties: $\frac{0.05}{6.00} = 0.0083$ and $\frac{0.02}{0.50} = 0.04$. Total: $0.0083 + 0.04 = 0.048$. $\Delta R = 12.0 \times 0.048 = 0.58 \approx 0.6$ $\Omega$. Result: $R = 12.0 \pm 0.6$ $\Omega$.

If you get this wrong, revise: Rule 2: Multiplication and Division

Problem 10

Explain the difference between a systematic error and a random error, giving one example of each from a measurement of the acceleration of free fall using a simple pendulum.

Answer. A systematic error is a consistent offset from the true value. Example: the bob is not Perfectly point-like, effectively increasing the pendulum length. A random error causes scatter in Repeated readings. Example: human reaction time when timing oscillations with a stopwatch — it Varies unpredictably from trial to trial.

If you get this wrong, revise: Types of Error

Problem 11

The gravitational potential energy is given by $E_p = -\frac{GMm}{r}$. Use dimensional analysis to determine the SI units of the gravitational constant $G$.

Answer. $[E_p] = \mathsf{M}\mathsf{L}^2\mathsf{T}^{-2}$. $[Mm/r] = \frac◆LB◆\mathsf{M} \cdot \mathsf{M}◆RB◆◆LB◆\mathsf{L}◆RB◆ = \mathsf{M}^2\mathsf{L}^{-1}$. Since $E_p = -\frac{GMm}{r}$: $[G] = \frac{[E_p][r]}{[Mm]} = \frac◆LB◆\mathsf{M}\mathsf{L}^2\mathsf{T}^{-2} \cdot \mathsf{L}◆RB◆◆LB◆\mathsf{M}^2◆RB◆ = \mathsf{M}^{-1}\mathsf{L}^3\mathsf{T}^{-2}$. In SI units: m$^3$ kg$^{-1}$ s$^{-2}$.

If you get this wrong, revise: Derived Units

Problem 12

A quantity $Q$ is measured as $Q = \frac◆LB◆a^2 b◆RB◆◆LB◆\sqrt{c}◆RB◆$ where $a = 4.0 \pm 0.2$$b = 3.0 \pm 0.1$$c = 9.0 \pm 0.3$. Calculate $Q$ and its percentage uncertainty.

Answer. $Q = \frac◆LB◆16.0 \times 3.0◆RB◆◆LB◆3.0◆RB◆ = 16.0$.

Fractional uncertainties: $\frac{2(0.2)}{4.0} = 0.10$ (power rule for $a^2$), $\frac{0.1}{3.0} = 0.033$ (for $b$), $\frac{1}{2} \cdot \frac{0.3}{9.0} = 0.0167$ (power rule for $c^{1/2}$).

Total: $0.10 + 0.033 + 0.0167 = 0.150$. Percentage uncertainty: 15.0%. $\Delta Q = 16.0 \times 0.150 = 2.4$. Result: $Q = 16.0 \pm 2.4$ (15%).

If you get this wrong, revise: Rule 3: Powers

8. SI Base Units in Detail

Understanding how each SI base unit is defined is essential for experimental physics and for Interpreting measurements correctly.

Base Quantity	SI Unit	Current Definition (SI 2019)
Length	metre (m)	Defined by fixing the speed of light $c = 299\,792\,458$ m s$^{-1}$
Mass	kilogram (kg)	Defined by fixing the Planck constant $h = 6.626\,070\,15 \times 10^{-34}$ J s
Time	second (s)	Defined by fixing the caesium frequency $\Delta\nu_{Cs} = 9\,192\,631\,770$ Hz
Electric current	ampere (A)	Defined by fixing the elementary charge $e = 1.602\,176\,634 \times 10^{-19}$ C
Temperature	kelvin (K)	Defined by fixing the Boltzmann constant $k_B = 1.380\,649 \times 10^{-23}$ J K$^{-1}$
Amount of substance	mole (mol)	Defined by fixing the Avogadro constant $N_A = 6.022\,140\,76 \times 10^{23}$ mol$^{-1}$
Luminous intensity	candela (cd)	Defined by fixing the luminous efficacy $K_{cd} = 683$ lm W$^{-1}$

:::info Note Since 2019, all SI base units are defined in terms of fundamental physical constants. The values of c$$h$$e$$k_B$$N_AAnd $\Delta\nu_{Cs}$ are now exact defined quantities, While the unit values are derived from them. :::

9. Dimensional Analysis: Extended Worked Examples

9.1 Checking Formula Validity

Example. A student proposes that the pressure at depth $h$ in a fluid is $P = \rho g h^2$. Check Whether this is dimensionally valid.

Answer. $[P] = \mathsf{M}\mathsf{L}^{-1}\mathsf{T}^{-2}$. $[\rho g h^2] = (\mathsf{M}\mathsf{L}^{-3})(\mathsf{L}\mathsf{T}^{-2})(\mathsf{L}^2) = \mathsf{M}\mathsf{L}^0\mathsf{T}^{-2} = \mathsf{M}\mathsf{T}^{-2}$. This does not match $[P]$. The correct formula is $P = \rho g h$Which gives $[\rho g h] = \mathsf{M}\mathsf{L}^{-3} \cdot \mathsf{L}\mathsf{T}^{-2} \cdot \mathsf{L} = \mathsf{M}\mathsf{L}^{-1}\mathsf{T}^{-2} = [P]$. $\square$

9.2 Verifying the Ideal Gas Equation

Example. Show that the ideal gas equation $PV = nRT$ is dimensionally consistent, where $R$ is The molar gas constant with units J mol$^{-1}$ K$^{-1}$.

Answer. $[PV] = (\mathsf{M}\mathsf{L}^{-1}\mathsf{T}^{-2})(\mathsf{L}^3) = \mathsf{M}\mathsf{L}^2\mathsf{T}^{-2}$.

$[nRT] = (\mathrm{mol})(\mathsf{M}\mathsf{L}^2\mathsf{T}^{-2}\mathrm{ mol}^{-1}\mathsf{\Theta}^{-1})(\mathsf{\Theta}) = \mathsf{M}\mathsf{L}^2\mathsf{T}^{-2}$.

Both sides have dimensions $\mathsf{M}\mathsf{L}^2\mathsf{T}^{-2}$So the equation is dimensionally Consistent. $\square$

9.3 Determining the Form of an Equation

Example. The centripetal force on an object moving in a circle of radius $r$ at speed $v$ is Assumed to depend on mass $m$Speed $v$And radius $r$. Find the form of the equation.

Answer. Assume $F = k \cdot m^a \cdot v^b \cdot r^c$.

$[F] = \mathsf{M}\mathsf{L}\mathsf{T}^{-2}$. $[m^a v^b r^c] = \mathsf{M}^a (\mathsf{L}\mathsf{T}^{-1})^b \mathsf{L}^c = \mathsf{M}^a \mathsf{L}^{b+c} \mathsf{T}^{-b}$.

Equating dimensions:

$\mathsf{M}: \quad a = 1, \qquad \mathsf{L}: \quad b + c = 1, \qquad \mathsf{T}: \quad -b = -2 \implies b = 2$

Therefore $c = 1 - 2 = -1$Giving $F = k \cdot m v^2 / r$. Full analysis gives $k = 1$. $\square$

9.4 Unit Conversions Using Dimensional Analysis

Example. Convert a density of $5.4$ g cm$^{-3}$ to kg m$^{-3}$.

Answer. $5.4$ g cm$^{-3} = 5.4 \times 10^{-3}$ kg $\times (10^{-2}$ M$)^{-3} = 5.4 \times 10^{-3} \times 10^6$ kg m$^{-3} = 5400$ kg m$^{-3}$.

Example. Convert $120$ km h$^{-1}$ to m s$^{-1}$.

Answer. $120$ km h$^{-1} = 120 \times 10^3$ m / $3600$ s $= 33.3$ m s$^{-1}$.

10. Systematic Errors vs Random Errors: Detailed Comparison

Feature	Systematic Error	Random Error
Cause	Faulty apparatus or method	Environmental fluctuations, resolution limits
Effect on accuracy	Reduces accuracy (shifts mean away from true value)	Does not affect mean accuracy
Effect on precision	Does not affect precision	Reduces precision (increases scatter)
Detection	Compare with accepted value or use different method	Take repeated measurements; large scatter indicates random error
Reduction	Calibrate instruments, improve method	Take more readings, use instruments with finer resolution
Effect on mean	Shifts the mean consistently	Mean approaches true value as $n$ increases
Effect on uncertainty	Not reduced by averaging	Reduced by $\Delta x = \sigma / \sqrt{n}$

Identifying Systematic Errors in Practice

Example: Zero Error. A micrometer reads $0.02$ mm when the jaws are fully closed. Every Measurement will be $0.02$ mm too large. Correction: subtract $0.02$ mm from all readings.

Example: Calibration Error. A stopwatch runs consistently fast by $0.1$ s per minute. Every Timed interval will be overestimated. Correction: apply a proportional correction factor.

Example: Methodological Error. In a pendulum experiment, measuring from the top of the bob Rather than its centre of mass introduces a consistent offset in the effective length, shifting all Calculated values of $g$ in the same direction.

11. Uncertainty Propagation: Extended Worked Examples

11.1 Uncertainty in a Power Relationship

Example. The area of a circle is $A = \pi r^2$. If $r = 4.50 \pm 0.05$ cm, find $A$ with its Uncertainty.

Answer. $A = \pi(4.50)^2 = 63.62$ cm$^2$.

Fractional uncertainty in $r$: $\Delta r / r = 0.05 / 4.50 = 0.0111$.

Since $A \propto r^2$, $\Delta A / A = 2 \times 0.0111 = 0.0222$.

$\Delta A = 63.62 \times 0.0222 = 1.41$ cm$^2$.

Result: $A = 63.6 \pm 1.4$ cm$^2$.

11.2 Combining Addition and Multiplication Rules

Example. The volume of a rectangular block is $V = l \times w \times h$. The measurements are $l = 5.00 \pm 0.02$ cm, $w = 3.00 \pm 0.02$ cm, $h = 2.00 \pm 0.01$ cm. Find $V$ with its percentage Uncertainty.

Answer. $V = 5.00 \times 3.00 \times 2.00 = 30.00$ cm$^3$.

Fractional uncertainties: $\Delta l/l = 0.02/5.00 = 0.004$, $\Delta w/w = 0.02/3.00 = 0.0067$ $\Delta h/h = 0.01/2.00 = 0.005$.

Total fractional uncertainty: $0.004 + 0.0067 + 0.005 = 0.0157$.

Percentage uncertainty: $1.57\%$. $\Delta V = 30.00 \times 0.0157 = 0.47$ cm$^3$.

Result: $V = 30.0 \pm 0.5$ cm$^3$.

11.3 Uncertainty in a Formula with Roots

Example. The speed of a wave on a string is $v = \sqrt◆LB◆T/\mu◆RB◆$Where $T$ is the tension and $\mu$ is the mass per unit length. Given $T = 10.0 \pm 0.2$ N and $\mu = 0.0250 \pm 0.0005$ kg M$^{-1}$Find $v$ and its uncertainty.

Answer. $v = \sqrt{10.0 / 0.0250} = \sqrt{400} = 20.0$ m s$^{-1}$.

Since $v = T^{1/2}\mu^{-1/2}$The fractional uncertainty is:

$\Delta v/v = \frac{1}{2}(\Delta T/T) + \frac{1}{2}(\Delta\mu/\mu) = \frac{1}{2}(0.2/10.0) + \frac{1}{2}(0.0005/0.0250) = 0.010 + 0.010 = 0.020$.

$\Delta v = 20.0 \times 0.020 = 0.40$ m s$^{-1}$.

Result: $v = 20.0 \pm 0.4$ m s$^{-1}$.

11.4 Mixed Operations: Adding Quantities Then Multiplying

Example. Two lengths are measured as $l_1 = 1.20 \pm 0.02$ m and $l_2 = 0.80 \pm 0.02$ m. Their Sum is multiplied by a width $w = 0.50 \pm 0.01$ m to find an area $A = (l_1 + l_2) \times w$. Find $A$ and its uncertainty.

Answer. First, $l_1 + l_2 = 2.00$ m. By the addition rule: $\Delta(l_1 + l_2) = 0.02 + 0.02 = 0.04$ m.

Now $A = 2.00 \times 0.50 = 1.00$ m$^2$. By the multiplication rule:

$\Delta A/A = 0.04/2.00 + 0.01/0.50 = 0.020 + 0.020 = 0.040$.

$\Delta A = 1.00 \times 0.040 = 0.040$ m$^2$.

Result: $A = 1.00 \pm 0.04$ m$^2$.

12. Common Pitfalls

1. Mixing absolute and percentage uncertainty when combining quantities. When adding or subtracting, use absolute uncertainties. When multiplying or dividing, use fractional (or percentage) uncertainties. Applying the wrong rule is a frequent source of error.

2. Forgetting the power rule. If $z = x^3$Then $\Delta z / z = 3(\Delta x / x)$Not $\Delta z / z = \Delta x / x$. A common mistake is treating all operations as simple multiplication.

3. Quoting too many significant figures. If the uncertainty is $0.3$The result should be quoted to one decimal place. Writing $9.814 \pm 0.3$ is wrong; write $9.8 \pm 0.3$. Match the result to the uncertainty.

4. Assuming dimensional consistency implies correctness. An equation can be dimensionally correct but still wrong (e.g., missing a factor of $\pi$ or a numerical constant). Dimensional analysis is a necessary but not sufficient check.

5. Confusing precision with accuracy. A precise set of readings (small scatter) can still be inaccurate if there is an undetected systematic error. Always consider both.

6. Ignoring the resolution uncertainty for single readings. If you take only one reading with a ruler (smallest division $1$ mm), the uncertainty is $\pm 0.5$ mm (half the smallest division for analogue instruments), not zero.

7. Using the wrong rule for digital instruments. For a digital instrument, the uncertainty equals the smallest division (the last digit), not half the smallest division.

13. Extension Problem Set

Problem 1

The escape velocity from a planet of mass $M$ and radius $R$ is given by $v_e = \sqrt{2GM/R}$. Use dimensional analysis to determine the SI units of the gravitational constant $G$.

Answer. $[v_e] = \mathsf{L}\mathsf{T}^{-1}$. $[2GM/R] = [G][M]/[R] = [G]\mathsf{M}\mathsf{L}^{-1}$.

Setting $[v_e]^2 = [2GM/R]$: $\mathsf{L}^2\mathsf{T}^{-2} = [G]\mathsf{M}\mathsf{L}^{-1}$.

$[G] = \mathsf{L}^3\mathsf{M}^{-1}\mathsf{T}^{-2}$. In SI units: m$^3$ kg$^{-1}$ s$^{-2}$.

If you get this wrong, revise: Derived Units

Problem 2

A student proposes the formula for the frequency of a mass-spring system: $f = \frac◆LB◆1◆RB◆◆LB◆2\pi◆RB◆\sqrt◆LB◆\frac{m}{k}◆RB◆$Where $k$ is the spring constant. Use dimensional analysis to determine whether this formula is correct.

Answer. $[f] = \mathsf{T}^{-1}$. $[m/k] = \frac◆LB◆\mathsf{M}◆RB◆◆LB◆[\mathrm{force}]/[\mathrm{displacement}]◆RB◆ = \frac◆LB◆\mathsf{M}◆RB◆◆LB◆\mathsf{M}\mathsf{L}\mathsf{T}^{-2}/\mathsf{L}◆RB◆ = \frac◆LB◆\mathsf{M}◆RB◆◆LB◆\mathsf{M}\mathsf{T}^{-2}◆RB◆ = \mathsf{T}^2$.

$[\sqrt{m/k}] = \mathsf{T} \neq \mathsf{T}^{-1}$. The formula is incorrect. The correct form is $f = \frac◆LB◆1◆RB◆◆LB◆2\pi◆RB◆\sqrt◆LB◆\frac{k}{m}◆RB◆$Giving $[\sqrt{k/m}] = \mathsf{T}^{-1} = [f]$. $\square$

If you get this wrong, revise: Determining the Form of an Equation

Problem 3

A force $F = 8.0 \pm 0.4$ N acts on an object of mass $m = 2.00 \pm 0.05$ kg. Calculate the acceleration $a = F/m$ with its percentage uncertainty.

Answer. $a = 8.0 / 2.00 = 4.0$ m s$^{-2}$.

Fractional uncertainties: $\Delta F/F = 0.4/8.0 = 0.050$, $\Delta m/m = 0.05/2.00 = 0.025$.

Total fractional uncertainty: $0.050 + 0.025 = 0.075 = 7.5\%$.

$\Delta a = 4.0 \times 0.075 = 0.30$ m s$^{-2}$.

Result: $a = 4.0 \pm 0.3$ m s$^{-2}$ (7.5%).

If you get this wrong, revise: Rule 2: Multiplication and Division

Problem 4

Show that the equation $v^2 = u^2 + 2as$ is dimensionally valid, and determine the SI base units of a quantity with dimensions $\mathsf{M}^{1/2}\mathsf{L}^{3/2}\mathsf{T}^{-1}$.

Answer. $[v^2] = (\mathsf{L}\mathsf{T}^{-1})^2 = \mathsf{L}^2\mathsf{T}^{-2}$. $[u^2] = \mathsf{L}^2\mathsf{T}^{-2}$. $[2as] = (\mathsf{L}\mathsf{T}^{-2})(\mathsf{L}) = \mathsf{L}^2\mathsf{T}^{-2}$. All terms match. $\square$

For $\mathsf{M}^{1/2}\mathsf{L}^{3/2}\mathsf{T}^{-1}$: $\sqrt◆LB◆\mathrm{kg}◆RB◆ \cdot \mathrm{m}^{3/2} \cdot \mathrm{s}^{-1}$. An example is $\sqrt{G} \cdot M / r$ where $G$ has units m$^3$ kg$^{-1}$ s$^{-2}$Giving $[\sqrt{G}] = \mathrm{m}^{3/2}\mathrm{ kg}^{-1/2}\mathrm{ s}^{-1}$ and $[\sqrt{G} \cdot M] = \mathrm{m}^{3/2}\mathrm{ kg}^{1/2}\mathrm{ s}^{-1}$.

If you get this wrong, revise: Derived Units

Problem 5

The kinetic energy of a particle is $E_k = \frac{1}{2}mv^2$. Given $m = 0.150 \pm 0.005$ kg and $v = 3.20 \pm 0.04$ m s$^{-1}$Calculate $E_k$ with its uncertainty.

Answer. $E_k = 0.5 \times 0.150 \times 3.20^2 = 0.5 \times 0.150 \times 10.24 = 0.768$ J.

Fractional uncertainties: $\Delta m/m = 0.005/0.150 = 0.0333$. For $v^2$: $2\Delta v/v = 2(0.04/3.20) = 0.025$.

Total fractional uncertainty: $0.0333 + 0.025 = 0.0583$.

$\Delta E_k = 0.768 \times 0.0583 = 0.045$ J.

Result: $E_k = 0.77 \pm 0.04$ J.

If you get this wrong, revise: Rule 3: Powers

Problem 6

A student measures the period of a pendulum as $T = 2.05 \pm 0.05$ s and the length as $L = 1.00 \pm 0.01$ m. Using $g = 4\pi^2 L / T^2$Calculate $g$ with its absolute uncertainty. Comment on whether the result is consistent with the accepted value of $9.81$ m s$^{-2}$.

Answer. $g = 4\pi^2 \times 1.00 / (2.05)^2 = 39.48 / 4.2025 = 9.395$ m s$^{-2}$.

Fractional uncertainties: $\Delta L/L = 0.01/1.00 = 0.010$. For $T^2$: $2\Delta T/T = 2(0.05/2.05) = 0.0488$.

Total fractional uncertainty: $0.010 + 0.0488 = 0.0588$.

$\Delta g = 9.395 \times 0.0588 = 0.55$ m s$^{-2}$.

Result: $g = 9.4 \pm 0.6$ m s$^{-2}$.

The accepted value $9.81$ m s$^{-2}$ falls within the range $9.4 \pm 0.6$ (i.e., $8.8$ to $10.0$ m S$^{-2}$), so the result is consistent with the accepted value. $\square$

If you get this wrong, revise: Rule 3: Powers

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Summary

This topic covers the fundamental principles of quantities and units, including the key equations, experimental methods, and applications relevant to the specification.

Key concepts include:

• fundamental principles and equations
• SI units and dimensional analysis
• mathematical modelling of physical phenomena
• experimental techniques and measurement
• applications to real-world problems

A strong understanding of these principles, combined with regular practice of quantitative problems and past paper questions, is essential for success in examinations.