Mq6, Mq7 - Building models from dimensions | OCR A-Level Further Mathematics B (MEI)

Use dimensions to build and use a proposed model. Mq6 determines unknown powers; Mq7 calibrates a dimensionless constant and predicts changes, while keeping clear what dimensions alone cannot establish.

Turn dimensions into equations for powers

Can you predict the effect of doubling a pendulum's length without knowing its full period formula? If a proposed power-law model depends only on mass, length and gravitational acceleration, dimensions determine the length and gravity powers. They even force the mass power to be zero.

You already know

Dimensions multiply using laws of indices. Length has dimension $L$ , mass $M$ , gravitational acceleration $LT^{-2}$ , period $T$ and frequency $T^{-1}$ . A coefficient has dimension $1$ only when it is stated or established to be dimensionless.

Start from the model actually proposed, rather than writing a familiar formula from memory. Replace quantities by dimensions, collect the total powers of $M$ , $L$ and $T$ , then equate those powers on the two sides.

Matching dimensional powers

If a proposed relationship is dimensionally consistent and its numerical coefficient $k$ is dimensionless, its two sides must have equal exponents of each independent base dimension. For example,

M^aL^bT^c=M^pL^qT^r \quad\Longrightarrow\quad a=p,\quad b=q,\quad c=r

This determines only what follows from the stated variables and form of model.

Not in the formula booklet: learn it.

Use separate lines for the three dimensions. Zero powers are equations too: a quantity with no mass factor has total mass exponent $0$ .

Worked example 1: a proposed pendulum-period model

Question. A model assumes that the period $P$ of a pendulum satisfies $P=km^a\ell^b g^c$ , where $m$ is its mass, $\ell$ its length, $g$ gravitational acceleration and $k$ dimensionless. Determine $a$ , $b$ and $c$ .

Plan. Substitute dimensions and match the mass, length and time exponents with those of a period.

\begin{aligned} T&=M^aL^b(LT^{-2})^c &&\text{substitute dimensions} \\ &=M^aL^{b+c}T^{-2c} &&\text{collect powers} \\ a&=0 &&\text{mass} \\ b+c&=0 &&\text{length} \\ -2c&=1 &&\text{time} \end{aligned}

Hence $c=-1/2$ and $b=1/2$ .

Why this step? $[P]=T$ , not $1$ : matching the time power to $0$ would erase the quantity you are trying to model.

Answer. $a=0$ , $b=1/2$ , $c=-1/2$ , giving $P=k\sqrt{\ell/g}$ within the proposed model.

Check. $[\ell/g]=L/(LT^{-2})=T^2$ , so its square root has dimension $T$ .

Change only the quantity being modelled from period to frequency. The input quantities stay the same, but the target time dimension reverses.

Worked example 2: a proposed frequency model

Question. Suppose instead that $f=cm^a\ell^b g^d$ , where $f$ is frequency and $c$ is dimensionless. Determine the powers.

Plan. Match the same input dimensions to $T^{-1}$ , the dimensions of frequency.

\begin{aligned} T^{-1}&=M^aL^{b+d}T^{-2d} &&\text{substitute and collect} \\ a&=0,\qquad b+d=0,\qquad -2d=-1 &&\text{match all three dimensions} \\ d&=\frac12,\qquad b=-\frac12 &&\text{solve the equations} \end{aligned}

Why this step? Frequency counts cycles per time; the mass and length equations are unchanged, but the time equation has right-hand side $-1$ .

Answer. $a=0$ , $b=-1/2$ , $d=1/2$ , so $f=c\sqrt{g/\ell}$ .

Check. Since frequency is the reciprocal of period, this agrees with the previous model when $c=1/k$ .

The quantity on the left decides the target time exponent.

Watch out

Wrong: $-2d=1$ for the frequency model $f=cm^a\ell^b g^d$ . That is the equation for a period; frequency has dimension $T^{-1}$ . Correct: $-2d=-1$ , giving $d=1/2$ .

The mass-independent result is conditional on the variables and form supplied. It does not prove that air resistance or other omitted effects are irrelevant to a real pendulum.

Exam tip

Read each symbol's stated meaning before assigning dimensions. A letter $a$ could be a dimensionless exponent here, an amplitude with dimension $L$ elsewhere, or an acceleration with dimension $LT^{-2}$ in another question.

The next frequency model replaces gravitational acceleration by a force magnitude. This changes the dimensional equations, including the mass equation.

Keep all three dimensional equations

With density, area and speed in a model, more powers contribute to the length dimension. Write the collected powers carefully before solving; most errors happen between a correct dimensional statement and an incorrect set of equations.

Worked example 3: a force model

Question. A resistance-force model has the form $F=k\rho^a A^b v^c$ , where $\rho$ is density, $A$ area, $v$ speed and $k$ dimensionless. Determine the powers.

Plan. Substitute the dimensions of density, area and speed, then match all three exponents with force.

\begin{aligned} MLT^{-2}&=(ML^{-3})^a(L^2)^b(LT^{-1})^c &&\text{substitute} \\ &=M^aL^{-3a+2b+c}T^{-c} &&\text{collect powers} \\ a&=1,\qquad -3a+2b+c=1,\qquad -c=-2 &&\text{match }M,L,T \\ c&=2,\qquad -3+2b+2=1 &&\text{substitute the known powers} \\ b&=1 &&\text{solve the length equation} \end{aligned}

Why this step? Density contributes $-3a$ to the length power, not $-3$ , until $a$ has been found.

Answer. $a=1$ , $b=1$ , $c=2$ , giving $F=k\rho Av^2$ .

Check. $(ML^{-3})L^2(L^2T^{-2})=MLT^{-2}$ .

Change only the output from force to power. The model's inputs stay the same, so the new target dimensions determine the new speed exponent.

Worked example 4: a power model

Question. A proposed power model is $W=k\rho^a A^b v^c$ , with the same meanings of $\rho$ , $A$ , $v$ and dimensionless $k$ . Determine the powers.

Plan. Match the collected input powers to $[W]=ML^2T^{-3}$ .

\begin{aligned} ML^2T^{-3}&=M^aL^{-3a+2b+c}T^{-c} &&\text{same input dimensions} \\ a&=1,\qquad -3a+2b+c=2,\qquad -c=-3 &&\text{new target powers} \\ c&=3,\qquad -3+2b+3=2 &&\text{substitute} \\ b&=1 &&\text{solve} \end{aligned}

Why this step? Multiplying a force by speed produces power, so an extra factor of $v$ should appear; dimensional matching makes that expectation precise.

Answer. $a=1$ , $b=1$ , $c=3$ , giving $W=k\rho Av^3$ . The dimensionless constant need not have the same value as in a different model.

Check. $(ML^{-3})L^2(L^3T^{-3})=ML^2T^{-3}$ .

These proposed force and power models are not in the formula booklet: obtain their powers from the information in the question. The method is the reusable result.

Watch out

Wrong: solve $-3a+2b+c=0$ for every model. Right: the right-hand side is the output's length exponent: $1$ for force and $2$ for power. Write the output dimensions first.

Change the output once more. A volume-flow rate is a volume per unit time, so its target dimensions are given explicitly.

Use measurements to find the constant

Dimensions can fix powers but cannot distinguish $k=0.3$ from $k=3$ . Measurements can calibrate the model: substitute a measured case, solve for $k$ , then use that value for another case under the same modelling assumptions.

Put all measurements in a consistent unit system first. A dimensionless constant should have the same value in any coherent unit system; mixing centimetres with metres inside one substitution can create a spurious value.

Worked example 5: calibrating a resistance model

Question. The model $F=k\rho Av^2$ gives a force of $18\,\mathrm{N}$ when $\rho=1.2\,\mathrm{kg\,m^{-3}}$ , $A=0.5\,\mathrm{m^2}$ and $v=10\,\mathrm{m\,s^{-1}}$ . Find $k$ .

Plan. Substitute the SI values into the model and divide by $\rho Av^2$ .

\begin{aligned} 18&=k(1.2)(0.5)(10^2) &&\text{consistent SI units} \\ k&=\frac{18}{60}=0.3 &&\text{isolate the constant} \end{aligned}

Why this step? The squared speed must be evaluated before the product in the denominator; replacing $v^2$ by $v$ would change the calibration tenfold here.

Answer. $k=0.3$ , dimensionless.

Check. $0.3\times1.2\times0.5\times100=18$ . The dimensions of the denominator are force, so its ratio with $F$ is dimensionless.

Now change the unknown from $k$ to speed. Keep the calibrated constant, density and area fixed.

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Mq6, Mq7 - Building models from dimensions

Turn dimensions into equations for powers

Keep all three dimensional equations

Use measurements to find the constant

Continue this lesson.

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