AQA A-Level Maths: Mechanics — Complete Revision Guide (7357)

Mechanics is the half of applied AQA A-Level Maths (7357) that turns algebra and calculus into physics. Where statistics asks you to reason about data and probability, mechanics asks you to reason about objects: how they move, what forces act on them, when they slip, and what happens when two of them collide. It rewards candidates who can draw a clear diagram, set up a clean coordinate system, and work through a problem without skipping steps.

Mechanics appears on Paper 2 alongside pure mathematics. Paper 1, 2 and 3 are each two hours and worth 100 marks. Across the qualification, mechanics accounts for roughly a quarter of the assessment, so a candidate strong on pure but weak on mechanics is leaving a large block of marks on the table. The AQA spec covers Sections T to V.

This guide walks through ten mechanics topics: kinematics, forces and Newton's laws, variable acceleration with calculus, projectiles, moments, friction, statics, dynamics with connected particles, work-energy-power, and impulse and momentum. The last two sit at the boundary of A-Level Maths and AQA Further Maths / Physics. They are flagged as enrichment, but the LearningBro course covers them because they appear in adjacent qualifications and tighten your grip on the core topics.

Throughout, AQA uses $g = 9.8 \text{ m/s}^2$g=9.8 m/s2 unless a question specifies otherwise, and modelling assumptions — light strings, smooth pulleys, inextensible cables, particles, smooth surfaces — are assessed across every topic. You should be able to state which assumption simplifies which step of your working, and to comment on its validity when asked.

What the AQA 7357 Mechanics Specification Covers

AQA's A-Level Maths (7357) is examined through three two-hour 100-mark papers. Paper 1 is pure only. Paper 2 is pure together with mechanics. Paper 3 is pure together with statistics. Mechanics on Paper 2 typically takes up around 40-50 of the 100 marks.

The mechanics content spans Sections T (Kinematics), U (Forces and Newton's laws) and V (Moments) of the specification, plus the variable-acceleration calculus that bridges into pure. The table below lists the topics covered, the section they sit under, and a realistic estimate of how many marks each typically contributes to a Paper 2 mechanics section.

Topic	Spec Section	Typical Paper 2 mechanics marks weight
Kinematics (constant acceleration)	T	6-10 marks
Forces and Newton's laws	U	8-12 marks
Variable acceleration (calculus)	T / Pure overlap	6-10 marks
Projectiles	T / U	6-10 marks
Moments	V	4-8 marks
Friction	U	4-8 marks
Statics	U / V	4-8 marks
Dynamics with connected particles	U	6-10 marks
Work, energy and power (enrichment)	Further / Physics overlap	0-4 marks
Impulse and momentum (enrichment)	Further / Physics overlap	0-4 marks

The first eight topics are core 7357 mechanics. The final two are enrichment — they do not appear on a standard 7357 paper but they sit on AQA Further Maths and A-Level Physics. Mark weights are estimates, not guarantees for any single year.

Kinematics

Kinematics is the description of motion without reference to the forces causing it. At A-Level you work with three connected quantities — displacement $s$s, velocity $v$v and acceleration $a$a — and you describe how they change with time. The first kinematics chapter restricts itself to constant acceleration, which is the regime where the suvat equations apply.

The five suvat equations relate displacement $s$s, initial velocity $u$u, final velocity $v$v, acceleration $a$a and time $t$t:

$v = u + at$v=u+at

$s = ut + \tfrac{1}{2}at^2$s=ut+21​at2

$s = vt - \tfrac{1}{2}at^2$s=vt−21​at2

$v^2 = u^2 + 2as$v2=u2+2as

$s = \tfrac{1}{2}(u + v)t$s=21​(u+v)t

Each equation is missing one of the five variables. Choosing the right one for a question is mostly a matter of identifying which three quantities you are given and which one you want, then picking the equation that excludes the variable you are not interested in.

Vertical-motion-under-gravity questions are the standard application. AQA uses $g = 9.8 \text{ m/s}^2$g=9.8 m/s2 acting downwards, so if you take upward as positive, then $a = -9.8 \text{ m/s}^2$a=−9.8 m/s2 throughout. A common pitfall is sign-flipping mid-question. Pick a positive direction at the start and stick with it.

Velocity-time and displacement-time graphs are also examinable. The gradient of a velocity-time graph is acceleration; the area under it is displacement. The gradient of a displacement-time graph is velocity. Many candidates lose marks by reading "distance" off a velocity-time graph as if it were displacement, or by failing to account for negative areas.

For full coverage with worked suvat problems and graph practice, see the Kinematics lesson.

Forces and Newton's Laws

Newton's laws are the foundation of dynamics. Newton's first law states that an object continues at rest or in uniform motion unless a resultant force acts on it. Newton's second law gives the relationship between resultant force, mass, and acceleration:

$F = ma$F=ma

Newton's third law states that for every action there is an equal and opposite reaction. The vast majority of A-Level mechanics questions reduce to applying Newton's second law along a chosen direction, so $F = ma$F=ma is the workhorse equation of the topic.

The standard workflow is: draw a clear force diagram (sometimes called a free-body diagram), choose a coordinate system, resolve all forces into components along your axes, and apply $F = ma$F=ma along each axis. Every force on a particle near the Earth's surface includes its weight $W = mg$W=mg acting vertically downwards, plus any contact forces, tensions, friction, applied forces, or normal reactions specified by the problem.

A common pitfall is forgetting weight on a horizontal surface — even though vertical forces cancel, you still need the vertical equation to find $R$R, which feeds into friction. Another pitfall is applying $F = ma$F=ma in a non-accelerating direction: on a horizontal surface, vertical acceleration is zero, so the equation is $R - mg = 0$R−mg=0.

The standard modelling assumptions matter here. A "particle" has mass but no size. A "smooth" surface exerts no friction. A "light" string has zero mass, so tension is the same throughout. An "inextensible" string transmits motion without stretching. Each simplifies a step of your working, and AQA may ask you to state which you used and whether it is realistic.

For force-diagram practice and worked $F = ma$F=ma problems, see the Forces and Newton's Laws lesson.

Variable Acceleration

When acceleration is not constant, the suvat equations no longer apply. Instead, you use calculus to move between displacement, velocity and acceleration. This is where mechanics meets pure: differentiation and integration with respect to time replace the algebraic suvat formulae.

The core relationships are:

$v = \frac{ds}{dt} \quad\text{and}\quad a = \frac{dv}{dt} = \frac{d^2 s}{dt^2}$v=dtds​anda=dtdv​=dt2d2s​

Integrating reverses each step. Given $a$a as a function of time, integrate to find $v$v (using a known initial velocity to fix the constant of integration), then integrate again to find $s$s. Given $v$v as a function of time, differentiate to find $a$a or integrate to find $s$s.

A short worked example. A particle moves so that its velocity is $v = 3t^2 - 4t + 1$v=3t2−4t+1 at time $t$t. Its acceleration at $t = 2$t=2 is $a = \frac{dv}{dt} = 6t - 4 = 8 \text{ m/s}^2$a=dtdv​=6t−4=8 m/s2. Its displacement from the origin between $t = 0$t=0 and $t = 3$t=3 is $s = \int_0^3 (3t^2 - 4t + 1) \, dt = [t^3 - 2t^2 + t]_0^3 = 27 - 18 + 3 = 12 \text{ m}$s=∫03​(3t2−4t+1)dt=[t3−2t2+t]03​=27−18+3=12 m.

Common pitfalls include confusing displacement (the integral of velocity) with distance travelled (the integral of speed, $|v|$∣v∣, which differs whenever the particle reverses direction), and forgetting the constant of integration when working from $a$a back to $v$v or from $v$v back to $s$s. A useful check: substitute $t = 0$t=0 into your expression for velocity and verify it matches the stated initial velocity.

For variable-acceleration practice and reverse-direction questions, see the Variable Acceleration lesson.

Projectiles

A projectile is a particle moving freely under gravity, with no air resistance, after being launched with some initial velocity. The standard model assumes the only force acting after launch is weight, so horizontal acceleration is zero and vertical acceleration is $-g = -9.8 \text{ m/s}^2$−g=−9.8 m/s2 (taking upward as positive).

The crucial insight is that horizontal and vertical motion are independent. If the projectile is launched at speed $u$u at angle $\theta$θ above the horizontal, the initial velocity components are $u_x = u\cos\theta$ux​=ucosθ and $u_y = u\sin\theta$uy​=usinθ. Horizontal motion uses $x = u\cos\theta \cdot t$x=ucosθ⋅t at constant velocity. Vertical motion uses suvat with $a = -g$a=−g — so $y = u\sin\theta \cdot t - \tfrac{1}{2}g t^2$y=usinθ⋅t−21​gt2 for displacement and $v_y = u\sin\theta - gt$vy​=usinθ−gt for vertical velocity.

The standard quantities to find are the time of flight (set $y = 0$y=0 and solve), the range (substitute the time of flight into the horizontal equation), the maximum height (set $v_y = 0$vy​=0 and substitute back), and the velocity components at any specific time. A useful identity is the range formula on level ground:

$R = \frac{u^2 \sin(2\theta)}{g}$R=gu2sin(2θ)​

which shows that range is maximised at $\theta = 45°$θ=45° for any given launch speed.

A common pitfall is treating the initial velocity as a single scalar throughout the calculation rather than splitting it into components. Another is forgetting to use the negative sign for vertical acceleration consistently — once you choose upward as positive, $g$g must appear with a minus sign in the vertical suvat equations. Finally, when the projectile is launched and lands at different heights (off a cliff, for example), you cannot use the level-ground range formula; go back to suvat and solve carefully.

For trajectory sketches and worked range and height problems, see the Projectiles lesson.

Moments

The moment of a force about a point is its turning effect — the product of the force's magnitude and the perpendicular distance from the point to the line of action of the force:

$M = Fd$M=Fd

Moments are measured in newton-metres (N m). A moment is clockwise or anticlockwise depending on the direction the force would rotate the object about the chosen pivot. Conventionally, anticlockwise is taken as positive.

A rigid body is in rotational equilibrium about a point when the sum of clockwise moments about that point equals the sum of anticlockwise moments. Combined with translational equilibrium (resultant force is zero), this gives you enough equations to solve any standard A-Level statics problem with a uniform rod or beam.

The standard problem setup is a uniform rod of length $L$L and weight $W$W resting on two supports, with additional weights placed at known positions. The rod's weight acts through its centre, at $L/2$L/2 from either end. To find the reactions at the supports, take moments about one support to eliminate that reaction from the equation, solve for the other reaction, then use vertical equilibrium to find the first reaction.

A common pitfall is using the wrong distance — moments use the perpendicular distance from the pivot to the line of action of the force, not the distance to where the force is applied. For a force applied at angle $\theta$θ to a rod, the perpendicular distance is the geometric distance multiplied by $\sin\theta$sinθ. Another pitfall is forgetting to include the rod's own weight when it is described as "uniform" but not "light". If the rod is light, its weight is zero; if it is uniform, its weight acts at the midpoint and must be included.

For worked beam problems and tilting-on-the-point-of-toppling questions, see the Moments lesson.

Friction

Friction is the contact force that resists relative motion between two surfaces. At A-Level, the model assumes friction $F$F is bounded above by $\mu R$μR, where $\mu$μ is the coefficient of friction and $R$R is the normal reaction:

$F \le \mu R$F≤μR

When the particle is on the point of slipping — or actually slipping — friction takes its maximum value $F = \mu R$F=μR. When the particle is in static equilibrium and not on the point of slipping, friction takes whatever value is needed to maintain equilibrium, which may be less than $\mu R$μR.

The distinction between "in equilibrium" and "on the point of slipping" is the most commonly tested subtlety. If a question says a particle is on the point of slipping, you can replace $F$F with $\mu R$μR in your equations and solve. If a question says a particle is in equilibrium, you can only assert $F \le \mu R$F≤μR as an inequality, which often gives you a range of values for an unknown like an applied force or an angle.

On an inclined plane at angle $\theta$θ, the standard resolution gives $R = mg\cos\theta$R=mgcosθ (perpendicular to the plane) and a component of weight $mg\sin\theta$mgsinθ acting down the plane. For a particle on the point of slipping down a rough plane, equilibrium gives $\mu R = mg\sin\theta$μR=mgsinθ, leading to the classic result $\tan\theta = \mu$tanθ=μ for the angle of friction.

A common pitfall is using $F = \mu R$F=μR in static equilibrium when the particle is not on the point of slipping. Another is forgetting that $R$R on an inclined plane is $mg\cos\theta$mgcosθ, not $mg$mg. A third is reversing the direction of friction — friction always opposes motion or the tendency to motion, so you must think carefully about which way the particle is about to move before drawing the friction arrow.

For inclined-plane practice and angle-of-friction problems, see the Friction lesson.

Statics

Statics is the study of objects at rest. A particle or rigid body is in static equilibrium when the resultant force is zero and the resultant moment about any point is zero. For a particle (no size), the moment condition is automatic and only the force condition matters. For a rigid body, both must be checked.

The standard A-Level statics problems involve a particle or rigid body acted on by several forces — typically weight, normal reaction, friction, tension in a string, and any applied forces. To solve, resolve all forces into components along two perpendicular axes (usually horizontal and vertical, or parallel and perpendicular to a plane), set the sum of components in each direction to zero, and solve the resulting simultaneous equations. For rigid-body problems, also take moments about a convenient point.

The choice of pivot for taking moments is strategic. A pivot that lies on the line of action of an unknown force eliminates that force from the moment equation, which simplifies the algebra. For a ladder problem, taking moments about the foot of the ladder eliminates the friction and normal reaction at the floor, leaving an equation in only the wall reaction and the weight.

Triangle of forces and Lami's theorem are alternative tools for three-force equilibrium problems, particularly when a particle is suspended by two strings at angles. Lami's theorem states that for three concurrent forces in equilibrium, $\frac{F_1}{\sin\theta_1} = \frac{F_2}{\sin\theta_2} = \frac{F_3}{\sin\theta_3}$sinθ1​F1​​=sinθ2​F2​​=sinθ3​F3​​, where each angle is opposite the corresponding force. Resolution into components is more general; Lami's is faster when it applies.

A common pitfall is forgetting the moment condition for rigid bodies — a force balance alone is not sufficient when the object can rotate. Another is choosing an awkward pivot and then doing harder algebra than necessary. Pick a pivot on the line of action of an unknown force whenever possible.

For ladder problems and three-force equilibrium practice, see the Statics lesson.

Dynamics

Dynamics generalises Newton's second law to systems with more than one particle, typically connected by a light inextensible string passing over a smooth pulley. The two standard configurations are: two particles hanging vertically on either side of a pulley (Atwood machine), and one particle on a horizontal or inclined surface connected by a string over a pulley to a second particle hanging vertically.

The standard workflow is: treat each particle separately, draw a force diagram for each, choose a positive direction along the string for each, and write Newton's second law for each particle. The light-inextensible-string assumption means that the two particles share the same magnitude of acceleration $a$a, and the smooth-pulley assumption means that the tension $T$T is the same throughout the string. You end up with two simultaneous equations in $a$a and $T$T, which you solve.

A short worked example. Two particles of mass $3 \text{ kg}$3 kg and $5 \text{ kg}$5 kg hang on either side of a smooth pulley, connected by a light inextensible string. Taking downward as positive for the heavier particle and upward as positive for the lighter, both with the same acceleration magnitude $a$a:

$5g - T = 5a \quad \text{and} \quad T - 3g = 3a$5g−T=5aandT−3g=3a

Adding gives $2g = 8a$2g=8a, so $a = g/4 = 2.45 \text{ m/s}^2$a=g/4=2.45 m/s2. Substituting back gives $T = 3g + 3a = 3(9.8) + 3(2.45) = 36.75 \text{ N}$T=3g+3a=3(9.8)+3(2.45)=36.75 N.

A common pitfall is choosing inconsistent positive directions, so the acceleration appears with the wrong sign in one of the equations. The cleanest convention is to choose positive in the direction the system is actually accelerating — the heavier mass moves down, the lighter mass moves up, and both accelerate at $+a$+a in their chosen positive directions. Another pitfall is forgetting that tension and weight must be on the correct side of the equation: $F_{\text{net}} = ma$Fnet​=ma, where $F_{\text{net}}$Fnet​ is the resultant force in the positive direction.

For Atwood and pulley-on-incline practice, see the Dynamics lesson.

Work, Energy and Power

Work, energy and power are not on the standard 7357 specification, but they appear on AQA Further Maths and on A-Level Physics, and they sharpen your understanding of the core mechanics topics. Work done by a constant force $F$F moving a particle a distance $d$d along its line of action is $W = Fd$W=Fd. Kinetic energy is the energy of motion, $KE = \tfrac{1}{2}mv^2$KE=21​mv2. Gravitational potential energy relative to a chosen reference height is $PE = mgh$PE=mgh. Power is the rate of doing work, $P = \frac{dW}{dt}$P=dtdW​, measured in watts.

The work-energy theorem says that the work done by the resultant force on a particle equals the change in its kinetic energy. The principle of conservation of mechanical energy says that, in the absence of friction or other non-conservative forces, the sum $KE + PE$KE+PE is constant. Both let you bypass suvat or $F = ma$F=ma for problems that ask only about start and end states without needing to know the time or detailed motion in between.

A short worked example. A particle of mass $2 \text{ kg}$2 kg slides from rest down a smooth slope of height $5 \text{ m}$5 m. By conservation of energy, $mgh = \tfrac{1}{2}mv^2$mgh=21​mv2, so $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} \approx 9.9 \text{ m/s}$v=2gh​=2×9.8×5​=98​≈9.9 m/s. No suvat needed.

A common pitfall is including friction in a smooth-surface problem (or excluding it on a rough surface) and getting the wrong final speed. When friction is present, the work done against friction equals the loss of mechanical energy, and the energy equation becomes $\frac{1}{2}mu^2 + mgh_1 - F \cdot d = \frac{1}{2}mv^2 + mgh_2$21​mu2+mgh1​−F⋅d=21​mv2+mgh2​.

This topic is enrichment for 7357 candidates but is included in the LearningBro course because it strengthens conceptual understanding of the core dynamics work. For worked energy-conservation problems, see the Work, Energy and Power lesson.

Impulse and Momentum

Impulse and momentum also sit outside the standard 7357 specification but appear on AQA Further Maths and A-Level Physics. Linear momentum is $p = mv$p=mv, the product of mass and velocity. Impulse is the change in momentum, $J = \Delta p = mv - mu$J=Δp=mv−mu, and equals the integral of force over time, $J = \int F \, dt$J=∫Fdt. For a constant force, this reduces to $J = Ft$J=Ft.

The principle of conservation of linear momentum says that, in the absence of external forces, the total momentum of a closed system is constant. This is the standard tool for collision problems: two particles of masses $m_1$m1​ and $m_2$m2​ travelling at velocities $u_1$u1​ and $u_2$u2​ collide and rebound at velocities $v_1$v1​ and $v_2$v2​. Momentum conservation gives:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$m1​u1​+m2​u2​=m1​v1​+m2​v2​

For a perfectly inelastic collision the particles stick together and have a common final velocity. For elastic and partially elastic collisions, an additional equation — typically conservation of kinetic energy or Newton's experimental law for restitution — is needed to find both final velocities.

A short worked example. A particle of mass $3 \text{ kg}$3 kg moving at $4 \text{ m/s}$4 m/s collides with a stationary particle of mass $2 \text{ kg}$2 kg. They coalesce. Their common final velocity is given by $3 \times 4 + 2 \times 0 = (3 + 2)v$3×4+2×0=(3+2)v, so $v = 12/5 = 2.4 \text{ m/s}$v=12/5=2.4 m/s.

A common pitfall is mixing units or signs — momentum is a vector, so velocities in opposite directions must have opposite signs. Another is using kinetic-energy conservation for an inelastic collision, where energy is not conserved; only momentum is.

This topic is enrichment for 7357 candidates but is included in the LearningBro course because it appears in adjacent qualifications and it deepens understanding of force as the rate of change of momentum. For collision-problem practice, see the Impulse and Momentum lesson.

Common Mark-Loss Patterns Across Mechanics

Across the whole mechanics syllabus, a small set of habits accounts for a disproportionate share of lost marks. None of these are about content you do not know. They are all about content you do know, applied carelessly under exam conditions.

• Inconsistent sign conventions. Pick a positive direction at the start of every problem, write it down on your diagram, and stick with it. Sign flips mid-question are the single most common source of suvat and $F = ma$F=ma errors.
• Skipping the force diagram. A clear, labelled diagram with every force and its direction takes thirty seconds and prevents most resolution mistakes. Examiners reward clear diagrams with method marks even when the final number is wrong.
• Forgetting weight on a horizontal surface. Even when vertical motion is zero, you need the vertical force equation to find the normal reaction $R$R, which feeds into friction.
• Confusing "on the point of slipping" with "in equilibrium". Only on the point of slipping (or actually slipping) does $F = \mu R$F=μR. In static equilibrium without slipping, $F \le \mu R$F≤μR as an inequality.
• Treating a non-uniform rod as if its weight acts at the midpoint. Only uniform rods have weight at the centre. Non-uniform rods have weight at a stated or unknown position.
• Using suvat when acceleration is not constant. If the question gives acceleration as a function of time, use calculus, not suvat.
• Computing "distance" from a velocity-time graph as a signed area. Distance is the total area between the curve and the t-axis, treating areas below the axis as positive. Displacement is the net signed area.
• Forgetting the modelling assumption commentary. AQA increasingly asks "comment on the validity of the assumption that..." and these marks are easy if you have a one-sentence answer rehearsed for each standard assumption (light, smooth, inextensible, particle).
• Mixing up tension and compression in connected-particle problems. Strings can only pull (tension is positive); rods can push or pull (compression or tension).
• Not showing enough working. AQA mark schemes award method marks generously when the working is clear. A correct final answer with no working can score fewer marks than an incorrect final answer with a clean method.

A revision plan that explicitly drills these habits — not just the content — will move your grade more than another pass through the textbook.

Recommended Six-Week Revision Plan

This plan assumes about 4-5 hours per week on mechanics, with the rest of your maths revision across pure and statistics. Adjust pace as needed.

Week	Topics	Practice
1	Kinematics (suvat, vertical motion, velocity-time graphs); variable acceleration with calculus	20 suvat problems; 10 graph-interpretation problems; 10 calculus-based problems
2	Forces and Newton's laws; friction (horizontal and inclined plane)	10 force-diagram problems; 10 inclined-plane problems including angle-of-friction questions
3	Projectiles (level ground and uneven launch); moments	10 projectile problems including range and max height; 10 beam-and-pivot moment problems
4	Statics (ladder problems, three-force equilibrium); dynamics with connected particles	8 statics problems; 10 Atwood and pulley-on-incline problems
5	Enrichment: work-energy-power and impulse-momentum (skip if not on Further Maths or Physics)	10 energy-conservation problems; 10 collision and impulse problems
6	Mixed practice; AQA past-paper Section B questions; targeted review of weakest topics	One full mechanics question set per day; review marking-scheme working for any question scoring below 60%

Keep moving forward while maintaining contact with earlier topics. Do not spend three weeks on suvat and run out of time before dynamics. By the end of week 4, every core topic should have had focused contact. Week 5 is enrichment for Further Maths or Physics candidates; week 6 is consolidation using past papers.

Treat any question you got wrong as a diagnostic. Was it a content gap, a method error, or a careless arithmetic slip? Logging the cause means your next review session targets the right thing.

How LearningBro's AQA A-Level Maths Mechanics Course Helps

LearningBro's AQA A-Level Maths: Mechanics course is built around the structure of this guide. Each of the ten lessons covers one mechanics topic, in the order AQA teaches it across Sections T to V plus the variable-acceleration calculus bridge, with worked examples, practice questions and full mark-scheme-style solutions.

The course is designed to be used in two ways. As a first pass, work through the lessons in order — kinematics gives you the language of motion, Newton's laws give you the language of force, and the rest are applications of these two. As a revision tool, drop into any lesson and work the practice independently. The AI tutor gives targeted hints when you get stuck without revealing full solutions, and marks your written working with structured feedback on diagrams, sign conventions and method.

For one place to revise mechanics well with realistic Paper 2 practice, start with the AQA A-Level Maths: Mechanics course.

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