HookThe stopping-distance table that hasn't changed since the 1940s
Every learner driver in Britain memorises the same table. The Highway Code says a car doing 30 mph needs about 23 metres to stop — 9 metres of thinking distance while the driver reacts, then 14 metres of braking distance while the brakes do their work. At 70 mph the total is 96 metres: 21 metres of thinking, 75 of braking. Those figures were fixed by stopping tests on the drum-braked cars of the 1940s and have barely been touched since — modern tyres and discs stop you sooner, but the Code was never rewritten around them.
Look closer and the table is doing physics in public. From 30 to 70 mph the speed rises by a factor of about 2.3. Thinking distance rises by the same factor — 9 m to 21 m — because it is simply speed × reaction time. But braking distance jumps from 14 m to 75 m, a factor of 5.4, which is almost exactly 2.3 squared. Hidden in a government booklet is the entire argument of P5: forces are vectors, a resultant force changes motion (Newton), braking is work done against friction that empties the car's kinetic store — and that store grows with the square of speed. Learn the machinery and the table stops being a list to memorise and becomes something you can rebuild from scratch.
ModelScalars, vectors and the two families of force
A scalar has magnitude only: speed, distance, mass, time, energy. A vector has magnitude and direction: displacement, velocity, acceleration, force, weight, momentum. On paper a vector is an arrow — the length shows the size, the direction shows the direction — and that little convention is worth marks every time a question says 'represent' or 'draw'.
A force is a push or pull that acts on an object because it interacts with something else. Contact forces need the objects touching: friction, air resistance, tension in a rope, the normal contact force from a surface. Non-contact forces act at a distance: gravitational, electrostatic and magnetic. Every force, either kind, is a vector.
Gravity supplies the non-contact force you live with. Mass (kg) is the amount of matter and never changes; weight (N) is the force of gravity on that mass: \(W = mg\), where \(g\) is the gravitational field strength — about \(9.8\) N/kg at the Earth's surface. Weight acts at a single point called the centre of mass, and it is measured with a calibrated spring balance, a newtonmeter. Because \(g\) is a constant at a given place, weight and mass are directly proportional — double one and you double the other.
NASA's Perseverance rover has a mass of 1,025 kg. On Earth its weight is \[W = mg = 1025 \times 9.8 = 10{,}045\ \text{N} \approx 10\ \text{kN}.\] On Mars, where \(g = 3.7\) N/kg, the same rover weighs \[W = 1025 \times 3.7 \approx 3{,}800\ \text{N}.\] The mass has not changed by a single gram — only the field pulling on it has. 'Mass is constant, weight depends on \(g\)' is a one-line mark in almost every Forces paper.
ModelResultant forces and the free-body diagram
Real objects rarely feel one force at a time, so physics compresses them: the resultant force is the single force that would have the same effect as all the forces acting together. Along a straight line the arithmetic is easy — add forces in the same direction, subtract opposing ones. A car with 7,000 N of driving force and 5,500 N of drag and friction has a resultant of 1,500 N forwards. If the resultant is zero the forces are balanced and the object is in equilibrium: it either stays still or, crucially, keeps moving exactly as it was.
The tool for seeing this is the free-body diagram: the object reduced to a point or box, with every force on it drawn as a labelled arrow of sensible relative length, acting at the object. Weight down, normal contact force up, thrust forwards, resistance backwards — nothing that acts on other objects is allowed in.
Higher tier adds two skills. A single force can be resolved into two components at right angles — usually horizontal and vertical — which each behave independently; and when two forces act at an angle to each other, you find the resultant with an accurate scale vector diagram, drawing the arrows tip-to-tail and measuring the closing side. No trigonometry is demanded: a ruler, a protractor and a stated scale earn the marks.
MechanismWork done — and why springs store it
When a force moves something, energy is transferred, and physics calls that transfer work done: \(W = Fs\), force multiplied by the distance moved along the line of action of the force. One joule is one newton-metre — the two units are the same thing. Push with 50 N for 10 m and you have shifted 500 J between stores, whatever the object. Work done against friction has a signature you can feel: the energy ends up in the thermal store, so surfaces that rub get hot.
Springs turn work into storage. First, an honesty rule the examiners test: stretching, compressing or bending needs more than one force — pull a spring at only one end and it simply accelerates. Deformation is elastic if the object returns to its original shape when the forces are removed, and inelastic if it doesn't. Within the elastic region, extension is directly proportional to force, \(F = ke\), where \(k\) is the spring constant in N/m — a stiffness rating. The proportionality holds only up to the limit of proportionality; beyond it the force–extension graph curves and the neat linear rule fails.
The work done stretching a spring (within the linear region) sits in its elastic potential store: \(E_e = \tfrac{1}{2}ke^2\). Note the square — it will matter in the required practical, and it echoes the kinetic-energy square from the Highway Code table.
A car's brakes apply a total resistive force of 6.0 kN while the car travels 25 m before stopping. Work done: \[W = Fs = 6000 \times 25 = 150{,}000\ \text{J} = 150\ \text{kJ}.\] That 150 kJ leaves the car's kinetic store and lands in the thermal store of the brake discs and pads — which is exactly why brakes glow cherry-red on a long mountain descent, and why 'work done against friction causes a rise in temperature' is a stock mark in this topic.
DataRequired practical 18 — force and extension of a spring
The method is disarmingly simple; the marks are in the discipline. Hang a spring from a clamp stand (secured to the bench with a G-clamp or counterweight so the whole rig cannot topple), fix a metre rule vertically beside it, and record the spring's natural length before anything is added. Add slotted masses one at a time — each 100 g adds a weight of about 0.98 N, near enough 1.0 N — and after each one record the new length. Extension = stretched length − natural length, every time. The independent variable is the force; the dependent variable is the extension; you control the spring itself and keep adding masses gently so the spring never bounces.
Plot force against extension. Within the linear region the points make a straight line through the origin, and the gradient of that line is the spring constant \(k\). The classic errors are all measurement-craft: read the rule at eye level to dodge parallax; measure to a fixed point on the spring (a pointer or marker helps); wait for the masses to hang still; and never overload the spring — past the limit of proportionality the deformation turns inelastic and every later reading is corrupted, because the natural length itself has changed. Wear eye protection: a spring that slips off its hook under load is a projectile.
Sample results: natural length 4.0 cm. With 2.0 N applied the length is 12.0 cm, so \(e = 8.0\ \text{cm} = 0.080\ \text{m}\) and \[k = \frac{F}{e} = \frac{2.0}{0.080} = 25\ \text{N/m}.\] With 4.0 N the length is 20.0 cm, \(e = 0.16\) m, and \(4.0 \div 0.16 = 25\) N/m again — the constant ratio is your evidence the spring is still in its linear region. The energy stored at 4.0 N is \[E_e = \tfrac{1}{2}ke^2 = \tfrac{1}{2} \times 25 \times 0.16^2 = 0.32\ \text{J}.\] The lethal slip is leaving extension in centimetres: \(\tfrac{1}{2} \times 25 \times 16^2\) gives an answer 10,000 times too big. Convert to metres before you square.
ModelDescribing motion — graphs and the equations
Motion vocabulary comes in scalar–vector pairs. Distance (how far, scalar) pairs with displacement (how far in a straight line from start to finish, and in which direction — vector). Speed pairs with velocity. For steady speed, \(s = vt\). The spec expects typical values: walking about 1.5 m/s, running about 3 m/s, cycling about 6 m/s, and sound in air 330 m/s.
Graphs carry the story. On a distance–time graph the gradient is the speed: flat means stationary, straight means steady, curved means changing speed (Higher tier reads the speed at an instant from the gradient of a tangent). On a velocity–time graph the gradient is the acceleration, \(a = \Delta v / t\), and — Higher tier — the area under the line is the distance travelled. One axis label changes everything, so read it before you read the shape.
When acceleration is uniform, one equation covers motion without time appearing at all: \(v^2 - u^2 = 2as\) (given on the equations sheet). Near the Earth's surface anything falling freely accelerates at about \(9.8\ \text{m/s}^2\). But a real object falling through air doesn't accelerate forever: as it speeds up, air resistance grows until it balances weight, the resultant force hits zero and the object continues at a steady terminal velocity — a preview of Newton's first law.
A car joins a motorway slip road at \(u = 13\) m/s and reaches \(v = 27\) m/s over \(s = 200\) m. Its acceleration: \[a = \frac{v^2 - u^2}{2s} = \frac{27^2 - 13^2}{2 \times 200} = \frac{729 - 169}{400} = 1.4\ \text{m/s}^2.\] Square both speeds before subtracting — the single most common slip is calculating \((27-13)^2\) instead, which gives 0.49 m/s² and loses every mark after the first.
MechanismNewton's three laws
First law: if the resultant force on an object is zero, a stationary object stays stationary and a moving object keeps moving at the same speed in the same direction. Motion does not need a force to continue — only a change of motion does. A car cruising at a steady 70 mph has a driving force exactly balancing resistance, not 'no forces'. Higher tier names the idea behind this reluctance to change: inertia.
Second law: acceleration is proportional to resultant force and inversely proportional to mass — \(F = ma\), with \(F\) the resultant. Higher tier defines inertial mass as the ratio \(F/a\): a measure of how hard an object is to accelerate. The spec also wants order-of-magnitude sense here — a family car's everyday acceleration is a few m/s², so the resultant forces involved are a few thousand newtons.
Third law: when two objects interact, they exert forces on each other that are equal in size, opposite in direction and of the same type. The pair act on different objects — that is the clause that keeps the law from contradicting everything else. The Earth pulls you down; you pull the Earth up with an identical gravitational force. It is not the same thing as the normal contact force balancing your weight, which is two different forces acting on one object.
A 1,400 kg car accelerates from rest to 27 m/s (about 60 mph) in 9.0 s. Acceleration: \(a = \Delta v/t = 27 \div 9.0 = 3.0\ \text{m/s}^2\). Resultant force: \[F = ma = 1400 \times 3.0 = 4{,}200\ \text{N}.\] If air resistance and friction total 1,100 N at these speeds, the engine must actually supply \(4200 + 1100 = 5{,}300\) N — because \(F = ma\) uses the resultant, not the driving force. Spotting that distinction is routinely the final mark of a 4-marker.
DataRequired practical 19 — how acceleration depends on force and mass
The apparatus: a trolley on a bench or track, pulled by a string that passes over a pulley to a hanging stack of masses; light gates (or a motion sensor) time the trolley through known points and give its acceleration. Two separate investigations hide inside one rig. To test acceleration against force, change the accelerating force — but here is the design point examiners love — keep the total accelerating mass constant by transferring slotted masses from the trolley onto the hanger, never adding new ones. The falling masses are part of the accelerated system, so if you just piled more on, you would be changing force and mass at once and the test would not be fair. To test acceleration against mass, keep the hanging force fixed and load extra mass onto the trolley.
The expected results are Newton's second law in graph form: acceleration is directly proportional to force (a straight line through the origin, gradient \(1/m\)) and inversely proportional to mass. The main systematic error is friction: it eats part of the applied force, so every measured acceleration comes out low and the graph misses the origin. Compensate by tilting the runway slightly until the trolley, nudged, rolls at a steady speed before you start. Light gates beat stopwatches because they remove human reaction time — a random error of a few tenths of a second in timings that only last about a second.
CaseStopping a car — and what momentum adds
Now rebuild the Highway Code table. Stopping distance = thinking distance + braking distance. Thinking distance is speed × reaction time, and typical human reaction times run from 0.2 s to 0.9 s — the Code's 9 m at 30 mph (13.4 m/s) implies about 0.67 s. Tiredness, alcohol, drugs and phone distraction all stretch reaction time, and because the relationship is linear, a doubled reaction time doubles the thinking distance at any speed.
Braking distance answers to energy. The brakes apply a friction force that does work, transferring the kinetic store \(\tfrac{1}{2}mv^2\) into the thermal store of the discs — and because that store grows with \(v^2\), braking distance grows with the square of speed: the table's 14 m at 30 mph becoming 75 m at 70 mph. Wet or icy roads and worn tyres or brakes cut the available friction force, so the same energy takes more distance to remove. And braking very hard from high speed demands a huge deceleration, hence (by \(F = ma\)) a huge force: brakes can overheat and the driver can lose control — the physics behind motorway pile-ups.
Higher tier finishes with momentum: \(p = mv\), measured in kg m/s, a vector pointing the way the object moves. In a closed system — no external forces — the total momentum before an event equals the total momentum after it: the conservation of momentum. It is the bookkeeping rule of every collision, from snooker balls to shunted cars.
Higher tier. A 1,200 kg car travelling at 20 m/s runs into the back of a stationary 800 kg car, and the two lock together. Momentum before: \[p = 1200 \times 20 + 800 \times 0 = 24{,}000\ \text{kg m/s}.\] The wreckage has mass 2,000 kg, so \[v = \frac{24{,}000}{2000} = 12\ \text{m/s}\] in the original direction of travel. State the closed-system assumption and give the direction — momentum is a vector, and both are creditworthy points.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
P5 lives on Physics Paper 2, and it is the most calculation-dense topic in Combined Science — physics questions carry the highest mathematical demand of the three sciences, so expect multi-step work. Since the 2025 series you are given a full equations sheet, which moves the marks from recall to selection and execution: pick the equation whose quantities you actually have, substitute in SI units before rearranging, and give the answer with a unit and a sensible number of significant figures. The two recurring unit traps are centimetres left unconverted in spring calculations (extension must be in metres before you square it in ½ke²) and speeds squared incorrectly in v² − u² = 2as — square first, subtract second.
AQA weights the papers 40% AO1 (recall), 40% AO2 (application) and 20% AO3 (analysis and evaluation), and at least 15% of all marks touch practical work — so RP18 and RP19 are not optional extras. Be ready to name independent, dependent and control variables, and to argue error direction: friction makes measured accelerations too low in RP19 (hence the tilted, friction-compensated runway), and exceeding the limit of proportionality in RP18 corrupts every subsequent reading because the spring is permanently deformed. 'How would you improve the accuracy?' wants light gates over stopwatches, with the reason — removing human reaction time.
On extended answers, discipline of language is what levels the response. Newton's third law statements need all three clauses — equal size, opposite direction, acting on different objects and of the same type. Stopping-distance questions want the split stated first (thinking + braking), then each factor tied to the right half: reaction-time factors stretch thinking distance linearly, road and vehicle condition stretch braking distance, and speed stretches both — linearly for one, quadratically for the other. Higher-tier momentum answers earn their final marks for the phrase 'in a closed system, total momentum before = total momentum after' and for giving the direction of the answer, because momentum is a vector.