HookAn 80 kg mass, a 4.8 m fall, and a rope that must never pull harder than 12 kN
Every rope sold for lead climbing has passed the same brutal audition. In the standard UIAA drop test, an 80 kg steel mass falls about 4.8 m onto a short length of rope — a nastier fall than almost any climber will ever take — and the rope fails if the peak force it transmits to that mass ever exceeds 12 kN. Good ropes register 8 to 9 kN. Now swap the nylon for a steel cable rated to hold far more. Same mass, same fall, same momentum to get rid of — and the peak force explodes to well over 100 kN. The anchor rips out of the rock, the karabiner snaps, or the climber's body takes a load no spine survives. The cable is stronger, and it kills.
The difference is time. The momentum change in the catch is fixed; how long the stop takes is negotiable, and nylon negotiates by stretching. That one trade — a fixed change of momentum, spread over more time, means less force — sits at the centre of section 3.4, and everything else in the section builds towards it or explains it: vectors and moments, the suvat equations, projectiles and terminal speed, Newton's laws, momentum and impulse, work and energy, and then the materials half — Hooke's law, stress, strain and the Young modulus — which tells you precisely why nylon stretches metres while steel stretches millimetres. Two required practicals, measuring \(g\) and measuring the Young modulus, come with it.
ModelVectors — resolving, adding, and the equilibrium triangle
A scalar has magnitude only (mass, energy, speed); a vector has magnitude and direction (force, velocity, acceleration, momentum). Two perpendicular vectors combine by Pythagoras, with the direction from \(\tan\theta\); any single vector can be run in reverse — resolved — into perpendicular components \(F\cos\theta\) along one axis and \(F\sin\theta\) along the other. The skill is choosing the axes: for an object on a slope, resolve along and perpendicular to the slope, so weight splits into \(mg\sin\theta\) trying to slide it down and \(mg\cos\theta\) pressing it into the surface.
An object is in equilibrium when the resultant force is zero. For three coplanar forces that means the vectors drawn tip-to-tail close into a triangle — or, equivalently, the components in each of two perpendicular directions sum to zero separately. Two directions, two equations: that is the standard machinery for every 'find the tension in each cable' problem.
ModelMoments, couples and the centre of mass
The moment of a force about a point is the force multiplied by the perpendicular distance from the point to the force's line of action — and 'perpendicular' is where the marks hide, because on a leaning ladder or an angled crane arm that distance is not the length of the object. A couple is a pair of equal, antiparallel forces whose lines of action do not coincide: it produces rotation with no resultant force, and its moment is one of the forces times the separation of the pair — the reason two hands on a steering wheel turn it without pushing the car sideways.
For a body in equilibrium the principle of moments holds: total clockwise moment about any point equals total anticlockwise moment. Choose the pivot cunningly — take moments about a support and that support's unknown reaction contributes nothing, leaving one equation with one unknown. The centre of mass is the point through which the whole weight can be taken to act; for a uniform beam it is the midpoint, and a bridge, crane or wheelbarrow problem is just weights acting at centres of mass balanced against reactions.
ModelMotion in a straight line — suvat and the two graphs
Displacement, velocity and acceleration are the vector versions of distance and speed: \(v=\Delta s/\Delta t\), \(a=\Delta v/\Delta t\), with instantaneous values given by gradients of tangents rather than averages. When acceleration is uniform — check this before you touch them — the suvat equations connect the five quantities: \(v=u+at\), \(s=ut+\tfrac{1}{2}at^2\), \(v^2=u^2+2as\), \(s=\tfrac{(u+v)}{2}t\). List what you know, pick the equation missing the quantity you do not care about, and keep one sign convention from the first line to the last.
The graphs carry their own marks. On a displacement-time graph the gradient is velocity; on a velocity-time graph the gradient is acceleration and the area under the line is displacement — including negative areas when the object doubles back. Examiners routinely ask which feature you are using; name it.
The Highway Code quotes a braking distance of 75 m from 70 mph (\(u=31.3\ \text{m s}^{-1}\)). What deceleration does that assume, and how long does the stop take?
Use \(v^2=u^2+2as\) with \(v=0\): \[a=\frac{v^2-u^2}{2s}=\frac{0-31.3^2}{2\times 75}=-6.5\ \text{m s}^{-2}\] about two-thirds of \(g\) — roughly the best an ordinary car and dry tarmac can do. Then \(t=(v-u)/a=31.3/6.5\approx 4.8\ \text{s}\). Add the Code's 21 m of thinking distance (a 0.67 s reaction at 31.3 m s\(^{-1}\)) and the full stop is 96 m — the length of a football pitch, from one misjudged glance at a phone.
DataRequired practical 3 — measuring g by free fall
The classic arrangement: an electromagnet holds a steel ball; cutting the current releases the ball and starts an electronic timer; the ball breaks a light gate or knocks open a trapdoor at the bottom, stopping the clock. With \(u=0\), the fall obeys \(h=\tfrac{1}{2}gt^2\). The independent variable is the drop height \(h\) (measured centre of ball to gate with a metre rule), the dependent variable is \(t\), and the controls are the same ball and the same release mechanism throughout. Plot \(h\) against \(t^2\): the gradient is \(g/2\).
The error analysis is what AQA actually examines. Electronic timing exists because a stopwatch cannot help you — human reaction time of about 0.2 s swamps a fall that lasts 0.4 s. The residual magnetism of the electromagnet releases the ball slightly late, so every \(t\) is systematically long and a single-drop calculation of \(g=2h/t^2\) comes out low; the graph is the fix, because a constant timing offset bends the data away from the origin while leaving the gradient — and therefore \(g\) — clean. Percentage uncertainty in \(t\) is worst for the smallest drops, so use the largest range of heights the bench allows and repeat each one.
MechanismProjectiles, drag and terminal speed
A projectile is two independent problems wearing one trajectory. Horizontally there is no force (ignoring air resistance), so velocity is constant; vertically there is only gravity, so \(a=g\) downwards. The vertical motion alone fixes the time of flight; the horizontal motion then converts that time into range. Fire one ball horizontally and drop another at the same instant and they land together — the demonstration examiners love because it forces you to say the vertical motions are identical.
Real air spoils the symmetry, and AQA wants the qualitative story: drag acts against motion and grows rapidly with speed, lift acts perpendicular to the airflow, and a falling body accelerates until drag has grown to equal weight — zero resultant, no further acceleration, terminal speed. A belly-down skydiver reaches about 55 m s\(^{-1}\); opening the canopy multiplies the drag at that speed enormously, giving a large upward resultant, fierce deceleration, and a new, far lower terminal speed of around 5 m s\(^{-1}\) — safe to meet the ground at.
A stone is thrown horizontally at \(12\ \text{m s}^{-1}\) from a 45 m cliff. Vertical motion first: \(45=\tfrac{1}{2}\times 9.81\times t^2\) gives \(t=\sqrt{2\times 45/9.81}=3.0\ \text{s}\). Range: \(12\times 3.0=36\ \text{m}\). At impact the vertical velocity is \(v=gt=9.81\times 3.0=29.7\ \text{m s}^{-1}\), so the resultant speed is \(\sqrt{12^2+29.7^2}=32\ \text{m s}^{-1}\) at \(\tan^{-1}(29.7/12)=68^\circ\) below the horizontal. Notice what was never used: the stone's mass. It cancels, exactly as in free fall.
ModelNewton's three laws, stated like an examiner
First law: velocity stays constant — in size and direction — unless a resultant force acts. Second law: for constant mass, resultant force equals mass times acceleration, \(F=ma\); the word resultant is not decoration, and dropping it costs the mark. Third law: if body A exerts a force on body B, body B exerts a force on A that is equal in magnitude, opposite in direction, of the same type, and acting on the other body.
That last clause dismantles the commonest wrong answer in the paper: the weight of a book and the table's normal force on it both act on the book, so they cannot be a third-law pair — and in an accelerating lift they are not even equal. The genuine pairs are book-pulls-Earth with Earth-pulls-book, and book-pushes-table with table-pushes-book. For connected bodies (a car towing a trailer), apply \(F=ma\) to the whole system to find the acceleration, then to one body alone to expose the internal force in the coupling.
MechanismMomentum and impulse — buying time
Momentum \(p=mv\) is a vector, and in any system free of external forces the total momentum is the same after an event as before — the conservation of linear momentum, which follows directly from the second and third laws. Newton actually stated his second law this way: force is the rate of change of momentum, \(F=\Delta(mv)/\Delta t\). Rearranged, \(F\Delta t=\Delta(mv)\): the product \(F\Delta t\) is the impulse, equal to the area under a force-time graph.
That rearrangement is the safety-engineering equation. The momentum change in a catch, crash or landing is fixed by the physics that came before; stretch the stopping time and the force falls in exact proportion. Crumple zones, airbags, gym mats and climbing ropes are all the same device at different scales. Collisions split into elastic (kinetic energy conserved — very rare outside particle physics and near-ideal springs) and inelastic (some kinetic energy becomes thermal energy and deformation). Momentum is conserved in both; kinetic energy is the quantity that tells them apart.
Put numbers on the drop test from the hook. Falling 4.8 m: \(v=\sqrt{2gh}=\sqrt{2\times 9.81\times 4.8}=9.7\ \text{m s}^{-1}\), so the 80 kg mass arrives with momentum \(p=80\times 9.7\approx 780\ \text{kg m s}^{-1}\). A dynamic rope stretches and brings it to rest over roughly 0.25 s: average force \(F=\Delta p/\Delta t\approx 780/0.25\approx 3.1\ \text{kN}\) — and because the force peaks well above its average, the measured maximum lands at the 8–9 kN real ropes record. A steel sling stretching a few millimetres stops the same momentum in about 5 ms: \(F\approx 780/0.005\approx 156\ \text{kN}\) — fifty times the rope's pull and almost 200 times the mass's own weight. Same momentum, different \(\Delta t\), fatal difference in \(F\).
ModelWork, energy, power — and where the joules go
Work is done when a force moves its point of application along its own line: \(W=Fs\cos\theta\), where \(\theta\) is the angle between force and displacement. Only the component along the motion counts — a force at right angles to the velocity, like the tension in a whirling string, does no work at all, which is why circular motion at constant speed costs nothing in energy.
Energy bookkeeping then runs on two accounts: kinetic energy \(E_k=\tfrac{1}{2}mv^2\) and gravitational potential energy \(\Delta E_p=mg\Delta h\). In an ideal transfer they simply swap — a dropped object gains exactly the kinetic energy its height paid for, giving \(v=\sqrt{2g\Delta h}\) with mass cancelling. In a real transfer the shortfall is the work done against friction and drag, which ends up as thermal energy; efficiency is the useful fraction, useful output divided by total input. Power is the rate of the whole business: \(P=\Delta W/\Delta t=Fv\). That last form explains top speed: a car flat out is not out of force, it is out of power — at maximum \(v\), every watt the engine makes is spent matching drag.
ModelInside the materials — density, Hooke's law and strain energy
Density \(\rho=m/V\) is the housekeeping quantity; the physics starts with deformation. Up to the limit of proportionality, extension is proportional to the applied force — Hooke's law, \(F=k\Delta L\), with stiffness \(k\) in N m\(^{-1}\). Load beyond the elastic limit and the object no longer returns to its original length: the deformation has become plastic.
But \(k\) belongs to one particular sample. To talk about the material, normalise the geometry away: tensile stress \(\sigma=F/A\) (force per unit cross-section, in pascals) and tensile strain \(\varepsilon=\Delta L/L\) (fractional extension, no units). The energy stored in a stretched sample — elastic strain energy — is the area under its force-extension graph, \(\tfrac{1}{2}F\Delta L\) while the line is straight. Unload a metal within its elastic region and it retraces the same line, handing the energy back. Rubber and nylon do not: the unloading curve sits below the loading curve, a hysteresis loop whose enclosed area is energy dissipated as heat in each cycle. That loop is why a climbing rope that has held a hard fall is retired — it comes back measurably longer and stiffer, having spent part of itself as heat and broken polymer strands to save the climber.
ModelThe Young modulus — stiffness as a property of the stuff itself
Divide stress by strain in the linear region and the sample's size cancels completely, leaving the Young modulus: \[E=\frac{\sigma}{\varepsilon}=\frac{FL}{A\,\Delta L}\] a property of the material alone, with units of pascals. Steel comes in around 200 GPa; nylon nearer 3 GPa. Under identical stress, nylon strains roughly seventy times more — the climbing-rope story compressed into one ratio.
Read a stress-strain graph the way an examiner does: the straight region's gradient is \(E\); the limit of proportionality ends the straight line; the elastic limit ends full recovery; the yield point (in metals) is where large plastic strain arrives for little extra stress; the breaking stress is the maximum stress the material survives. Keep three adjectives separate, because AQA tests the distinctions: stiff means high \(E\), strong means high breaking stress, and brittle means no plastic region — glass and cast iron snap at the top of their straight line, strong or not.
DataRequired practical 4 — the Young modulus of a wire
The apparatus is chosen by the equation. Since \(E=FL/(A\Delta L)\) and \(E\) for a metal is enormous, you need a long, thin wire — a couple of metres of about 0.3 mm diameter — so that the extension is large enough to measure: long \(L\) and small \(A\) both inflate \(\Delta L\), slashing its percentage uncertainty. Measure the original length with a tape to the millimetre; measure the diameter with a micrometer at several positions and two perpendicular orientations (wires are not perfectly round), check the micrometer's zero error, and average. Load the wire in steps with known masses, reading the extension from a marker against a millimetre scale or a vernier arrangement; a reference wire hung alongside (Searle's method) cancels any sag of the support and thermal expansion.
Plot force against extension and take the gradient of the straight region: \(E=\text{gradient}\times L/A\). The dominant uncertainty is the diameter, because it enters through \(A=\pi d^2/4\) — squaring doubles its percentage contribution. Stay inside the limit of proportionality (unload occasionally and confirm the wire returns to length), and wear eye protection: a snapping wire under tension whips.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
AQA's mark schemes reward a fixed rhythm in every calculation: equation, substitution with units, evaluation, answer with a unit and sensible significant figures — usually two or three, matching the data. On a 'show that' question, work to at least one more significant figure than the value quoted, or you have shown nothing. And remember the weighting: over 40% of A-level physics marks are AO2 application, and at least 40% demand maths at Level 2 or above, so the paper is mostly using these equations, not reciting them.
In suvat work, list u, v, a, s, t, strike out what you do not need, and fix one positive direction before the first line — sign errors in projectile questions are the single commonest way to lose a straightforward six marks. In momentum questions, say explicitly which quantity is conserved and why ('no external forces act, so total momentum is conserved') before calculating; that sentence is often a mark on its own. On the 6-mark extended responses — terminal speed is the perennial — examiners mark the causal chain, so write it as one: resultant force, therefore acceleration, therefore speed change, therefore drag change, and round again until the resultant is zero.
The required practicals resurface in Paper 3 Section A. Be ready to explain why the free-fall experiment plots \(h\) against \(t^2\) and how a delayed electromagnet release shows up as an intercept rather than corrupting the gradient, and why the wire's diameter dominates the Young modulus uncertainty — its percentage uncertainty doubles when \(d\) is squared in \(A=\pi d^2/4\). Quoting a percentage uncertainty calculation unprompted, where relevant, is the cheapest AO3 credit on the paper.