HookThe airliner that ran out of fuel over a unit error
On 23 July 1983, Air Canada Flight 143 — a brand-new Boeing 767 — ran completely out of fuel at 41,000 feet, halfway between Montreal and Edmonton. Nothing had broken. The fuel gauges were faulty, so the ground crew worked out the fuel by dipping the tanks, reading a volume in litres, and multiplying by a density to get the mass the flight plan demanded. The density they used, \(1.77\), was in pounds per litre; the figure the aircraft actually needed, and the number the crew treated it as, was in kilograms. Because a kilogram is about \(2.2\) pounds, the aeroplane left the ground with under half the fuel it required. It became a glider over Manitoba and was landed, engines dead, on a disused runway at Gimli. Everyone survived — but the accident has a single cause, and it is the subject of this entire section: a quantity is meaningless without its unit.
Mechanics is built from a tiny set of measured quantities and the units that carry them. Here you meet the three fundamental (base) quantities of the S.I. system — length, mass and time — and the derived quantities assembled from them: velocity, acceleration, force, weight and moment. You will learn what a newton actually is in terms of kilograms, metres and seconds, why mass and weight are different things measured in different units, how to tell a scalar from a vector, and how checking the units of an equation — the discipline that would have saved Flight 143 — catches errors before they cost you a mark. Get the vocabulary and the units automatic and every later mechanics chapter inherits a solid base; blur them and, like the crew at Montreal, every calculation downstream carries the mistake.
ModelThe base quantities — the three atoms of the S.I. system
Every measurement in mechanics is ultimately built from three base quantities, each with one agreed S.I. unit. Length is measured in metres (\(\text{m}\)), mass in kilograms (\(\text{kg}\)) and time in seconds (\(\text{s}\)). These are the atoms of the system: they are defined independently, by international agreement, and everything else is manufactured from them. The full S.I. system has seven base units, but A-level mechanics lives almost entirely inside these three.
The reason a single agreed system exists is precisely the Gimli problem. If one engineer works in feet and pounds and another in metres and kilograms, a bare number handed between them is a trap. Fixing one coherent set of base units — and insisting every quantity is quoted with its unit attached — is what makes a calculation portable and checkable. When you write \(g=9.8\) you have written almost nothing; \(g=9.8\ \text{m s}^{-2}\) is a physical statement.
Note the notation, because it earns marks and enables the checking later in this section. Units multiply and divide just like algebra, and negative indices replace the division line: metres per second is written \(\text{m s}^{-1}\) and metres per second per second is \(\text{m s}^{-2}\). Writing units this way is not fussiness — it is what lets you treat them algebraically when you test an equation, which is the skill this whole topic is really training.
MechanismDerived quantities and the newton — units you can rebuild from a formula
A derived quantity is any quantity built by multiplying or dividing base quantities, and its unit is built the same way. Velocity is displacement per unit time, so its unit is \(\text{m s}^{-1}\). Acceleration is change of velocity per unit time, so its unit is \(\text{m s}^{-1}\) per second, i.e. \(\text{m s}^{-2}\). Force comes from Newton's second law \(F=ma\): mass times acceleration, so its unit is \(\text{kg}\times\text{m s}^{-2}=\text{kg m s}^{-2}\). That combination is given its own name, the newton (\(\text{N}\)): one newton is the resultant force that gives a mass of one kilogram an acceleration of one metre per second squared. Memorising \(1\ \text{N}=1\ \text{kg m s}^{-2}\) is the single most useful unit fact in mechanics.
From force, two more follow at once. Weight is the particular force gravity exerts on a mass, \(W=mg\), so weight is measured in newtons, never in kilograms. Moment — the turning effect of a force — is force times perpendicular distance, so its unit is \(\text{N}\times\text{m}=\text{N m}\). (Momentum, which you meet later, is mass times velocity, \(\text{kg m s}^{-1}\); do not confuse momentum with moment.) Each of these units is simply its defining formula written in symbols — which means if you ever forget a unit, you can rebuild it from the equation rather than guessing.
A \(1500\ \text{kg}\) car accelerates at \(3\ \text{m s}^{-2}\). Find the resultant force, and confirm its unit from base units.
By \(F=ma\), \(F=1500\times 3=4500\ \text{N}\), or \(4.5\ \text{kN}\). Now the unit check: \(m\) contributes \(\text{kg}\) and \(a\) contributes \(\text{m s}^{-2}\), so \(F\) has units \(\text{kg}\times\text{m s}^{-2}=\text{kg m s}^{-2}\), which is exactly the definition of the newton. The number \(4500\) and the unit \(\text{N}\) are two halves of one answer: examiners credit the unit separately, so a bare \(4500\) is an incomplete response even when the arithmetic is perfect.
ModelScalars and vectors — size only, or size and direction
Quantities split into two families, and knowing which is which decides how you are allowed to combine them. A scalar has size only: mass, time, distance, speed and energy are scalars, so \(5\ \text{kg}\) is already a complete statement. A vector has size and direction: displacement, velocity, acceleration, force and weight are vectors, and quoting only the magnitude throws away half the information. A velocity of \(20\ \text{m s}^{-1}\) is incomplete; \(20\ \text{m s}^{-1}\) due north is a vector.
The distinction has teeth. Scalars add arithmetically, but vectors add tip-to-tail, so two \(3\ \text{N}\) forces can produce a resultant of anything from \(0\) to \(6\ \text{N}\) depending on their directions. In one dimension — motion along a straight line — direction collapses to a sign: choose one way as positive and the other is negative, so a velocity of \(-4\ \text{m s}^{-1}\) means \(4\ \text{m s}^{-1}\) in the negative direction, not a velocity that is somehow less than nothing. In two dimensions the standard shorthand is \(\mathbf{i}\) and \(\mathbf{j}\), unit vectors along the horizontal and vertical axes, so a force written \((3\mathbf{i}+4\mathbf{j})\ \text{N}\) has components \(3\ \text{N}\) across and \(4\ \text{N}\) up.
Each scalar usually has a vector partner it is easy to confuse with: distance is the scalar partner of displacement, and speed is the scalar partner of velocity. Speed is the magnitude of velocity, and distance is the total path length regardless of direction — a runner who completes a \(400\ \text{m}\) lap has covered a distance of \(400\ \text{m}\) but has a displacement of zero, because they finish where they started.
MechanismMass versus weight — the most examined confusion in mechanics
Mass is the amount of matter in a body, measured in kilograms; it is a scalar and it does not change if you move the body to the Moon, to orbit, or nowhere at all. Weight is the gravitational force acting on that mass, \(W=mg\); it is a vector pointing down, measured in newtons, and it does change with location because the gravitational acceleration \(g\) changes.
At the Earth's surface \(g\approx 9.8\ \text{m s}^{-2}\) (many questions accept \(9.81\); a few use \(10\) for simplicity — always use the value the paper specifies). The symbol \(g\) is an acceleration, not a force: it is the acceleration a freely falling body has, equivalently the gravitational field strength of \(9.8\ \text{N kg}^{-1}\). Weight is the force you get by multiplying that field strength by the mass. This is why bathroom scales, which really measure the force you press on them with, can be marked in kilograms only by assuming Earth's \(g\): take the same scales to the Moon and the reading would be wrong by a factor of six, even though your mass had not changed at all.
A rover has a mass of \(60\ \text{kg}\). Find its weight on Earth (\(g=9.8\ \text{m s}^{-2}\)) and on the Moon (\(g_{\text{Moon}}=1.62\ \text{m s}^{-2}\)).
On Earth: \(W=mg=60\times 9.8=588\ \text{N}\). On the Moon: \(W=60\times 1.62=97.2\ \text{N}\). The weight has fallen to about a sixth, but the mass is \(60\ \text{kg}\) in both places — the quantity of matter is unchanged. A candidate who writes 'the mass on the Moon is \(97.2\ \text{kg}\)' has made exactly the base error the newton is designed to prevent: kilograms measure mass, newtons measure the force of gravity on it.
CaseChecking an equation with its units — the free error-catcher
A correct equation in mechanics must be dimensionally consistent: every term that is added or subtracted, and both sides of the equals sign, must carry the same units. You cannot add a length to a time any more than you can add apples to hours. This gives you a fast, free check on any formula you write or half-remember — if the units do not match, the equation is wrong, with no arithmetic required. It is the exact safeguard the Gimli crew skipped: had anyone asked 'is this number in kilograms or in pounds?', the flight would have been routine.
The method is to replace each symbol by its unit and simplify using ordinary index laws, ignoring pure numbers such as \(\tfrac12\), which carry no units. If every term reduces to the same combination of \(\text{m}\), \(\text{kg}\) and \(\text{s}\), the equation passes; if one term comes out in \(\text{m s}^{-1}\) while the rest are in \(\text{m}\), you have found a mistake. The test will not catch a missing numerical factor, but it reliably catches the far more damaging error of a mis-built formula.
Check that the constant-acceleration equation \(s=ut+\tfrac12 at^2\) is dimensionally consistent, where \(s\) is displacement, \(u\) a velocity, \(a\) an acceleration and \(t\) a time.
Take the terms one at a time. The left side \(s\) is a displacement, unit \(\text{m}\). First term on the right, \(ut\): velocity times time is \(\text{m s}^{-1}\times\text{s}=\text{m}\). Second term, \(\tfrac12 at^2\): the \(\tfrac12\) is a pure number and carries no unit, while \(at^2\) is \(\text{m s}^{-2}\times\text{s}^2=\text{m}\). Every term is in metres, so the equation is consistent — it passes. Notice the check would still pass if someone dropped the \(\tfrac12\), because pure numbers are invisible to it; what it guarantees is that the structure of the formula is sound, which is precisely the error class that grounds aeroplanes.
ModelModelling assumptions — the words that make the maths possible
Mechanics never studies the real world directly; it studies a model of it, stripped of complications. The vocabulary of those simplifications is examined explicitly — 'state one modelling assumption you have made' is a standard one-mark question — so learn what each word switches off. Treating a body as a particle means ignoring its size and shape and imagining all its mass at a single point, which lets you ignore rotation and forces spread over an area. A string described as light has negligible mass, so the tension is the same all along it; an inextensible string does not stretch, so two bodies it connects share one acceleration.
A surface called smooth exerts no friction, while a rough one does; a rigid body does not bend; a rod called uniform has its mass evenly spread, so its weight acts at the centre. Each assumption is a deliberate simplification that makes the mathematics tractable, and each has a cost in accuracy — a 'smooth' ramp overestimates how fast a real crate slides, and a 'particle' model ignores the spin that curves a real football. The higher-mark modelling questions ask you to name an assumption and say how relaxing it would change the answer, which is the same honesty about a model's limits that this whole subject depends on.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
This is the section where sloppy notation is punished hardest, so make three habits automatic. First, every numerical answer carries its unit, written in index form (\(\text{m s}^{-1}\), \(\text{m s}^{-2}\), \(\text{N}\)); examiners award the unit as a separate mark, and a bare number is incomplete. Second, when a question gives \(g\), quote the value you are using and keep it consistent throughout — mixing \(9.8\) and \(10\) in one solution loses accuracy marks.
Keep mass and weight rigidly apart: if a body's mass is \(5\ \text{kg}\), its weight is \(5g\approx 49\ \text{N}\), and it is the weight, in newtons, that goes on a force diagram — never the mass. When a 'show that' or modelling question appears, use the unit check as a silent sanity test on any formula before you trust it, and be ready to name a modelling assumption and its effect (for example, 'the sledge is modelled as a particle, so air resistance and its length are ignored; in reality it would decelerate slightly faster'). These are cheap marks that candidates routinely leave on the table by treating the opening mechanics section as too basic to revise.