HookThe physicist who worked out the bomb from a photograph
In 1947 the US government declassified a sequence of photographs of the first nuclear test at Trinity, New Mexico — each frame stamped with a time and printed with a distance scale. The yield of the bomb remained top secret. The British physicist G. I. Taylor looked at the pictures and realised the secret was printed on them. The radius of the expanding fireball, he reasoned, could depend only on three things: the energy \(E\) released, the density \(\rho\) of the surrounding air, and the time \(t\) since detonation. There is exactly one way to combine those three quantities into something with the dimensions of length. Reading radii and times off the photographs, Taylor published an energy estimate of roughly \(10^{14}\) joules — the equivalent of some 17 kilotons of TNT, embarrassingly close to the classified figure of about 20. He never saw a blueprint. He matched dimensions.
That is the whole method of this section, and it is the shortest, highest-leverage topic in Further Mechanics. Every mechanical quantity is built from three base dimensions — mass \(\mathrm{M}\), length \(\mathrm{L}\), time \(\mathrm{T}\) — and any equation of physics must balance in all three simultaneously. That single constraint lets you do two things AQA examines directly: check a proposed formula for consistency, and predict the form of an unknown formula by solving for the powers. It will not give you dimensionless constants, and it will not add terms together for you — but as Taylor showed, what it does give you, it gives you almost for free.
ModelThe three-letter alphabet: finding dimensions
Write the dimensions of a quantity \(X\) as \([X]\), built from powers of \(\mathrm{M}\), \(\mathrm{L}\) and \(\mathrm{T}\). You derive them, not memorise them, by unwinding definitions. Velocity is distance over time: \([v]=\mathrm{L}\mathrm{T}^{-1}\). Acceleration is velocity over time: \([a]=\mathrm{L}\mathrm{T}^{-2}\). Force comes from \(F=ma\): \([F]=\mathrm{M}\mathrm{L}\mathrm{T}^{-2}\). Energy from work \(=\) force \(\times\) distance: \([E]=\mathrm{M}\mathrm{L}^2\mathrm{T}^{-2}\). Power is energy per time: \(\mathrm{M}\mathrm{L}^2\mathrm{T}^{-3}\); density is mass per volume: \(\mathrm{M}\mathrm{L}^{-3}\); pressure is force per area: \(\mathrm{M}\mathrm{L}^{-1}\mathrm{T}^{-2}\). Each one is a thirty-second derivation from a formula you already know — and deriving them is exactly what the first mark of a question buys.
Some quantities have no dimensions. An angle in radians is arc length over radius, \(\mathrm{L}/\mathrm{L}=1\); the coefficient of restitution is a ratio of speeds; pure numbers like \(2\pi\) and \(\frac{1}{2}\) are invisible to the method. This cuts both ways: dimensionless quantities can never be detected by dimensional analysis, and arguments of functions like \(\sin\), \(e^x\) and \(\ln\) must always be dimensionless — you cannot take the sine of three metres.
An equation is dimensionally consistent when every added term carries identical dimensions and the two sides match. Consistency is a necessary condition for correctness, never sufficient: \(E=\frac{1}{2}mv^2\) and the (false) \(E=7mv^2\) pass the same check. But inconsistency is a death sentence — a single mismatched term proves an equation wrong without any experiment at all, which makes the check the cheapest error-detector in your toolkit.
Check whether \(v^2=u^2+2as\) is dimensionally consistent.
Left side: \([v^2]=(\mathrm{L}\mathrm{T}^{-1})^2=\mathrm{L}^2\mathrm{T}^{-2}\).
Right side, term by term: \([u^2]=\mathrm{L}^2\mathrm{T}^{-2}\); the \(2\) is dimensionless, and \([as]=(\mathrm{L}\mathrm{T}^{-2})(\mathrm{L})=\mathrm{L}^2\mathrm{T}^{-2}\).
All three terms are \(\mathrm{L}^2\mathrm{T}^{-2}\): consistent. Contrast \(v^2=u^2+2at\), where \([at]=\mathrm{L}\mathrm{T}^{-1}\ne\mathrm{L}^2\mathrm{T}^{-2}\) — one line of dimensions proves that equation cannot be right, no data required.
MechanismPredicting formulae: solve for the powers
The second skill runs the logic in reverse. Suppose a quantity \(Q\) plausibly depends on quantities \(a\), \(b\), \(c\). Write \(Q=k\,a^{\alpha}b^{\beta}c^{\gamma}\) with \(k\) a dimensionless constant, take dimensions of both sides, and equate the powers of \(\mathrm{M}\), \(\mathrm{L}\) and \(\mathrm{T}\) separately. That yields three simultaneous equations in the unknown powers — usually trivially solvable — and out falls the only combination the physics permits.
The classic is the simple pendulum. Guess that the period depends on the bob's mass \(m\), the string length \(l\) and gravity \(g\): \(T=k\,m^{\alpha}l^{\beta}g^{\gamma}\). The \(\mathrm{M}\) equation immediately forces \(\alpha=0\) — the mass cannot appear, because nothing else in the list carries mass to cancel it against. Dimensional analysis just rediscovered Galileo's observation that heavy and light pendulums keep the same time, from pure bookkeeping. That is the method at its best: it tells you what a formula is allowed to look like before any experiment runs.
Know the limits, because AQA asks about them. The constant \(k\) is beyond reach — only experiment (or a full derivation) reveals the \(2\pi\) in the pendulum formula. The method cannot produce sums of terms with different powers, so anything like \(s=ut+\frac{1}{2}at^2\) is invisible to it. And with three base dimensions you can pin down at most three unknown powers; choose the governing quantities badly and the equations are inconsistent or underdetermined — which is itself useful information that your list of variables is wrong.
The period \(T\) of a simple pendulum is assumed to depend on mass \(m\), length \(l\) and gravitational acceleration \(g\): \(T=k\,m^{\alpha}l^{\beta}g^{\gamma}\).
Dimensions: \(\mathrm{T}=\mathrm{M}^{\alpha}\,\mathrm{L}^{\beta}\,(\mathrm{L}\mathrm{T}^{-2})^{\gamma}=\mathrm{M}^{\alpha}\mathrm{L}^{\beta+\gamma}\mathrm{T}^{-2\gamma}\).
Equate powers. \(\mathrm{M}\): \(\alpha=0\). \(\mathrm{T}\): \(-2\gamma=1\), so \(\gamma=-\frac{1}{2}\). \(\mathrm{L}\): \(\beta+\gamma=0\), so \(\beta=\frac{1}{2}\).
Conclusion: \(T=k\sqrt{\dfrac{l}{g}}\). The mass drops out by necessity, and the true formula \(T=2\pi\sqrt{l/g}\) confirms the shape — the \(2\pi\) being exactly the dimensionless constant the method can never see.
CaseTaylor's blast wave: the method at full power
Now run Taylor's calculation. The fireball radius \(R\) depends on the energy released \(E\), the air density \(\rho\) and the elapsed time \(t\): \(R=k\,E^{a}\rho^{b}t^{c}\). Taking dimensions: \(\mathrm{L}=(\mathrm{M}\mathrm{L}^2\mathrm{T}^{-2})^{a}(\mathrm{M}\mathrm{L}^{-3})^{b}\mathrm{T}^{c}\). Equating powers — \(\mathrm{M}\): \(a+b=0\); \(\mathrm{L}\): \(2a-3b=1\); \(\mathrm{T}\): \(-2a+c=0\) — gives \(a=\frac{1}{5}\), \(b=-\frac{1}{5}\), \(c=\frac{2}{5}\). So \[R=k\left(\frac{Et^2}{\rho}\right)^{1/5},\] and rearranged for the secret: \(E\approx\dfrac{\rho R^5}{t^2}\) (taking \(k\approx 1\), which fuller theory justifies).
One declassified frame shows the fireball at roughly \(R=140\) metres a mere \(t=0.025\) seconds after detonation. With \(\rho\approx 1.2\) kg m\(^{-3}\): \(E\approx\dfrac{1.2\times 140^5}{0.025^2}\approx 1.0\times 10^{14}\) joules — about 20 kilotons of TNT equivalent, from a photograph, a ruler and three simultaneous equations. Notice the leverage and the fragility in the same formula: \(R\) enters at the fifth power, so a 10% error in reading the radius becomes a 60% error in energy. Fifth-root dependencies the other way explain why the fireball's growth looks so undramatic frame-to-frame — to double \(R\), the bomb would need \(32\) times the energy.
For the exam, this case is the full question template: justify the list of governing quantities in a sentence, set up the power equation, solve the three simultaneous equations cleanly, and interpret the result — including, if asked, what happens to the predicted quantity when one input is doubled, which is now just a matter of reading the fractional powers.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Set every answer out as bookkeeping the examiner can follow line by line. For a consistency check, write the dimensions of each term separately, then state the comparison in words — 'all terms are \(\mathrm{L}^2\mathrm{T}^{-2}\), so the equation is dimensionally consistent'. The stated conclusion is a mark; a table of symbols with no sentence is not a conclusion. Derive dimensions from defining equations on the page (\([F]=\mathrm{M}\mathrm{L}\mathrm{T}^{-2}\) from \(F=ma\)) rather than quoting them cold — if you misremember, the visible derivation still earns method.
For prediction questions, the rubric is fixed: state the assumed form \(Q=k\,a^{\alpha}b^{\beta}c^{\gamma}\) with \(k\) dimensionless, write the dimensional equation, then equate powers of \(\mathrm{M}\), \(\mathrm{L}\) and \(\mathrm{T}\) as three clearly labelled equations before solving. Two habits protect the accuracy marks: substitute your solved powers back into the \(\mathrm{L}\) equation as a check, and keep fractional indices as fractions — \(\frac{1}{5}\), not \(0.2\) — because the follow-up part is often 'what happens to \(R\) if \(E\) is doubled?', which wants \(2^{1/5}\), an exact answer. Finally, if the question invites comment, name the two blind spots explicitly: the dimensionless constant \(k\) cannot be determined, and any dimensionless dependence (an angle, a ratio) is undetectable. Those two sentences are the standard final marks of the topic.