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How do you use the dimensions of physical quantities to check equations and predict relationships?

The dimensions of physical quantities in terms of mass, length and time, checking equations for dimensional consistency, and using dimensional analysis to find the form of a relationship.

A focused answer to the AQA A-Level Further Mathematics dimensional analysis content, covering the dimensions of physical quantities in terms of mass, length and time, checking equations for dimensional consistency, and using dimensional analysis to predict the form of a physical relationship.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Dimensions of common quantities
Checking consistency
Finding the form of a relationship

What this dot point is asking

AQA wants you to express physical quantities in terms of the base dimensions mass $M$ , length $L$ and time $T$ , check whether an equation is dimensionally consistent, and use dimensional analysis to predict the form of an unknown relationship up to a dimensionless constant.

Dimensions of common quantities

Build the dimensions of a compound quantity from its defining equation rather than memorising every one. Velocity is displacement over time, so $LT^{-1}$ . Acceleration is velocity over time, so $LT^{-2}$ . Force is mass times acceleration, $MLT^{-2}$ . Work and energy are force times distance, $ML^2T^{-2}$ . Power is energy over time, $ML^2T^{-3}$ . Pressure is force over area, $ML^{-1}T^{-2}$ . Momentum is mass times velocity, $MLT^{-1}$ . Building each from its definition means you never need to recall a long table under exam pressure.

Checking consistency

An equation is dimensionally consistent if every additive term has the same dimensions. This is a fast first check on any derived formula: if two terms that are added or equated turn out to have different dimensions, the equation must be wrong. The method has a clear limitation, though, which examiners like to probe: it cannot detect a missing or wrong dimensionless factor, so a formula that should read $\frac{1}{2}mv^2$ would still pass the check if the $\frac{1}{2}$ were dropped, and it cannot tell apart two quantities that happen to share dimensions, such as work and torque.

Finding the form of a relationship

The more powerful use is to predict the form of an unknown law. Assume the target quantity is a product of the others raised to unknown powers, multiplied by a dimensionless constant. Write the dimension equation, then equate the powers of $M$ , $L$ and $T$ separately to get one linear equation per base dimension. Solving those simultaneous equations fixes the exponents, leaving only the dimensionless constant undetermined.

Predict a relationship by dimensional analysis

The period $T_p$ of a simple pendulum is assumed to depend on its length $l$ , the mass $m$ of the bob, and gravity $g$ . Find the form of the relationship.

Step 1: Write the assumed power-law form

The strategy is to assume the unknown quantity equals a product of the given quantities each raised to an unknown power, multiplied by a dimensionless constant $k$ . This is the most general form consistent with dimensional homogeneity.

T_p = k\, l^{a}\, m^{c}\, g^{b}

Step 2: Substitute dimensions and simplify

Write down the dimensions of each quantity: $[T_p] = T$ , $[l] = L$ , $[m] = M$ , $[g] = LT^{-2}$ . Then expand the right-hand side, collecting powers of each base dimension.

T = L^{a}\, M^{c}\, (LT^{-2})^{b} = M^{c}\, L^{a+b}\, T^{-2b}

Step 3: Equate powers of each base dimension

For the equation to be dimensionally consistent, the power of each base dimension must match on both sides. We read off three equations, one per dimension.

For $M$ : $c = 0$ , so the mass of the bob does not affect the period. For $T$ : $-2b = 1$ , giving $b = -\tfrac{1}{2}$ . For $L$ : $a + b = 0$ , giving $a = \tfrac{1}{2}$ .

Step 4: State the predicted relationship

Substituting the exponents back:

T_p = k\, l^{1/2}\, g^{-1/2} = k\sqrt{\dfrac{l}{g}}.

Final answer: $T_p = k\sqrt{\dfrac{l}{g}}$ . Dimensional analysis correctly predicts that the period is independent of mass and proportional to $\sqrt{l}$ , although it cannot determine the constant $k = 2\pi$ .

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20186 marksThe speed

v

of a wave on a stretched string is thought to depend on the tension

T

in the string (a force) and the mass per unit length

\mu

of the string. Assuming

v = k\, T^{a} \mu^{b}

for a dimensionless constant

k

, use dimensional analysis to find

a

and

b

and hence the form of the relationship.

Show worked answer →

Write the dimensions of each quantity. Speed $[v] = LT^{-1}$ . Tension is a force, so $[T] = MLT^{-2}$ . Mass per unit length is mass divided by length, so $[\mu] = ML^{-1}$ .

Substitute into $v = k\, T^{a}\mu^{b}$ (the constant $k$ is dimensionless):
$LT^{-1} = (MLT^{-2})^{a}(ML^{-1})^{b} = M^{a+b}L^{a-b}T^{-2a}$ .

Equate powers of each base dimension:
For $M$ : $a + b = 0$ . For $T$ : $-2a = -1$ , so $a = \frac{1}{2}$ . For $L$ : $a - b = 1$ .

From $a = \frac{1}{2}$ and $a + b = 0$ , $b = -\frac{1}{2}$ . Check $L$ : $\frac{1}{2} - (-\frac{1}{2}) = 1$ , correct.

So $v = k\, T^{1/2}\mu^{-1/2} = k\sqrt{\frac{T}{\mu}}$ .

Markers reward correct dimensions for all three quantities, the three power equations, solving for $a$ and $b$ , and the final form.

AQA 20214 marksDetermine whether the equation

T = 2\pi\sqrt{\dfrac{m}{k}}

for the period

T

of an oscillating mass on a spring is dimensionally consistent, where

m

is a mass and

k

is a spring constant (force per unit length).

Show worked answer →

Find the dimensions of each side. The left side is a time, $[T] = T$ (the time dimension).

The spring constant is force per unit length, so $[k] = \frac{MLT^{-2}}{L} = MT^{-2}$ .

Then $\frac{m}{k}$ has dimensions $\frac{M}{MT^{-2}} = T^{2}$ , and $\sqrt{\frac{m}{k}}$ has dimensions $\sqrt{T^2} = T$ .

The factor $2\pi$ is dimensionless, so the right side has dimensions $T$ , matching the left side.

The equation is dimensionally consistent. Markers reward the dimensions of $k$ , simplifying $\frac{m}{k}$ to $T^2$ , taking the square root, and the conclusion that both sides are times.

Related dot points

Sources & how we know this

AQA A-level Further Mathematics (7367) specification — AQA (2017)