Formula and Method Bank

Provenance: written 2026-09-28, built directly from content/veridian/technique-hub/wme01/forge-exam-technique-guide.md (completed the same day) and research/veridian/WME01-verified-facts.md (17 series, Oct 2018–Jun 2025, including the Jan 2025/Jun 2025 deep-read both documents already share). Cross-checked against the nine real lesson files in content/veridian/curriculum/wme01/ — paper-anatomy.ts, resultant-forces-and-bearings.ts, newtons-second-law-in-vector-form.ts, kinematics-constant-acceleration-two-stage-motion.ts, momentum-impulse-sign-convention.ts, connected-particles-pulleys-and-lifts.ts, equilibrium-of-a-particle-coplanar-forces.ts, friction-equilibrium-and-motion-on-a-rough-plane.ts, moments-rods-on-two-pivots.ts, modelling-assumptions-and-show-that-discipline.ts — specifically to confirm no formula, convention, or worked scenario below duplicates a full worked exemplar those files already own. No new primary-source PDF was fetched for this document; every fact traces to the FORGE guide or the facts bank, both already checked against the actual Pearson spec/formula-booklet/mark-scheme/examiner-report PDFs in prior passes. This document's own job is to organise that already-verified material into a formula masterlist (Part 1) and a short worked-method layer (Part 2) distinct from both.

#CODEX · WME01 Formula & Identity Masterlist and Worked-Method Reference

#Pearson Edexcel IAL Mechanics 1 (WME01/01)

This is the third, and structurally different, layer of this course's WME01 material. content/veridian/curriculum/wme01/ teaches each topic with full worked chains, MCQs, and trap taxonomies. forge-exam-technique-guide.md teaches the paper's shape — timing, mark-coding, the self-mark checklist. This document does neither: it is the thing a candidate should be able to reconstruct from memory in the first ninety seconds of the exam — every real formula the spec actually requires, tagged with whether it is printed in the formula booklet or has to already be in your head — followed by short, method-only skeletons for the question shapes that recur across the eleven real series this course has verified in detail. Nothing here is a full worked exemplar; every numeric illustration below is short, VERIDIAN-original, and deliberately distinct from every scenario already used in the curriculum files or the FORGE guide.

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#PART 1 — Formula & Identity Masterlist

#1.0 What the formula booklet actually gives M1 — the fact every other row in this part depends on

Verbatim, the entire Mechanics M1 section of the real exam formula booklet (Mathematical Formulae and Statistical Tables, Issue 2, January 2021, p.13):

"There are no formulae given for M1 in addition to those candidates are expected to know." "Candidates sitting M1 may also require those formulae listed under Pure Mathematics P1 and P2."

That is the whole entry. No diagram, no equation. The practical consequence for this masterlist: every M1-specific relationship below — not only the three the spec's own wording singles out — has to come from memory, because there is nothing M1-shaped anywhere in the exam room to look it up on. The spec's own text (M1.2, point 3) explicitly names three formula groups as things that must therefore be memorised: Momentum = mv, Impulse = mv − mu, and the five constant-acceleration (suvat) equations. The rest of the relationships this masterlist lists (F = ma, weight = mg, moment = force × distance, F = μR, F ≤ μR) aren't separately flagged in that spec sentence — almost certainly because they're basic definitional content taught directly inside the spec's own numbered items, not because a booklet gives them. They are tagged "not given — memorise" below by direct entailment from the empty M1 booklet entry, not because the spec repeats the same explicit warning for each one; that distinction is marked honestly in the table so it isn't presented as more directly sourced than it is.

The one genuine exception is the cosine rule, a² = b² + c² − 2bc cos A — a Pure Mathematics P1/P2 result the booklet's own contents page extends to M1 candidates, and confirmed as a live, not merely theoretical, M1 technique (Jun 2024 Q2's examiner report discusses candidates using it on a resultant-of-two-forces question). Basic right-angle trigonometry and Pythagoras' theorem are prerequisite general mathematics, not a booklet-specific M1 entry either way — assumed throughout, never flagged as "given" or "must memorise" for M1 specifically.

#1.1 The g-value convention — the single most consequential number on this paper

• Use g = 9.8, not g = 10 and not g = 9.81. It is printed on the front of the question paper itself.
• Round final numerical answers to 2 or 3 significant figures. Using g = 9.81 is penalised once per complete question; over- or under-accuracy is penalised once per complete question; premature rounding mid-working is penalised every single time it happens, even more than once within the same question.
• "Show that" questions built on an exact fraction (e.g. sin α = 3/5, cos α = 4/5) must be carried through in that exact fractional form — a decimal substitution anywhere in the chain, even one that rounds correctly, breaks the given-answer mark on a "show that" line (verified directly in June 2025's mark scheme, per the FORGE guide §2).
• Units are not usually a penalised accuracy error if omitted or wrong — write them anyway as habit, but don't spend exam time agonising over a missing "N" or "m s⁻¹."

(Full sourcing for all of the above: FORGE guide §2 "Units" and "The rounding convention," and facts bank §3.)

#1.2 Mathematical models in mechanics (spec 1.1) — vocabulary, not a formula

Spec 1.1's own guidance names the modelling vocabulary a candidate is expected to know the consequence of, not just the word:

This table is definitional scaffolding only — the technique for using it (rereading a question's own stated model before naming a credit-worthy modification) is modelling-assumptions-and-show-that-discipline.ts's job, not repeated here.

#1.3 Vectors in mechanics (spec 2.1–2.2)

• Resolving a force into components, angle θ measured from a stated reference direction: the component along the reference direction is F cos θ; the component perpendicular to it is F sin θ. Every resolved term must actually carry a sin or cos factor — a bare force with the angle only mentioned in words earns nothing (the mark scheme's own general-principles rule: an incomplete resolution is a method error).
• Magnitude of a vector xi + yj: |v| = √(x² + y²) — Pythagoras, not a booklet entry, assumed general mathematics.
• Bearing → component conversion — the single most-confirmed vectors-specific trap in the archive. A bearing is measured clockwise from North, which swaps the usual sin/cos roles: for a force/vector of magnitude F on bearing θ, the East (i) component is F sin θ and the North (j) component is F cos θ — sin goes with the i-component here, the reverse of the "angle-from-the-x-axis" convention most students default to. Getting this backwards is the exact error a real examiner report names directly (Jun 2024 Q7: "a significant number showed a lack of knowledge of bearings", producing the wrong-swapped answer).
• Resultant of two forces from a common point, angle θ between their lines of action: R² = P² + Q² + 2PQ cos θ (the parallelogram law). The triangle-of-forces/cosine-rule route is algebraically the same result reached by drawing the forces tip-to-tail and using the interior angle of that triangle (which is 180° − θ, not θ itself) — the real, confirmed trap here (Jun 2024 Q2) is substituting the wrong one of these two angles into the cosine-rule formula, even when the triangle itself is drawn correctly.
• Lami's theorem, a real accepted alternative for three concurrent coplanar forces in equilibrium (confirmed independently in Jan 2022 Q1 and Jan 2023 Q6's examiner reports, though named nowhere in the spec's own text): for forces P, Q, R with angles α, β, γ respectively opposite each force, P/sin α = Q/sin β = R/sin γ.
• Resultant of several forces at once: sum every i-component, sum every j-component separately, then take the magnitude/direction of the combined (ΣFx)i + (ΣFy)j — the direct generalisation of the two-force case, and the method that scales to any number of forces without needing a triangle at all.
• Position vector under constant velocity (spec 2.2's own stated case): r(t) = r₀ + t v, where r₀ is the position vector at t = 0 and v the constant velocity vector. This is the vector-mechanics content spec 2.2 actually tests — not two-dimensional projectile motion under gravity (horizontal constant velocity combined with vertical constant acceleration). That classic "projectile" scenario is genuinely absent from this spec's M1 unit content as transcribed in the facts bank (§1) — spec item 3 restricts kinematics to "motion in a straight line," and every real vector-kinematics question the facts bank has verified (Jan 2025 Q5, Jun 2024 Q7) uses r(t) = r₀ + tv under constant velocity, never a vector suvat equation combining a nonzero acceleration with a nonzero starting velocity in two dimensions at once. Flagged explicitly here as a correction to an assumption the task brief for this document made — see "Genuine gaps flagged" below.

#1.4 Kinematics of a particle moving in a straight line (spec 3.1)

The five constant-acceleration (suvat) equations, all five must be memorised (spec-named, not in the booklet):

Choosing which one to use is almost always a matter of which four of the five quantities (s, u, v, a, t) the question actually supplies — three known, one unknown, one equation left over that never needed the missing quantity in the first place.

Graph conventions (spec 3.1's own guidance names displacement-time, velocity-time, speed-time and acceleration-time graphs as all potentially examinable; the archive this course has verified in detail confirms live, separately-marked use of velocity-time and speed-time graphs specifically — displacement-time and acceleration-time graphs are spec-permitted but not independently confirmed as tested in the series reviewed, flagged honestly rather than assumed equally common):

• On a velocity-time graph, the gradient is the acceleration: a = (v − u)/t, which rearranges directly to v = u + at.
• The area under the graph is the displacement — a trapezium with parallel sides u and v and height t has area ½(u+v)t, which is exactly the fifth suvat equation, arrived at with no algebra at all.
• The shape must start where the object's own motion actually starts, not automatically at the origin (real B-mark condition, both 2025 series); a continuous line must never be drawn where the real motion has a corner or a gap (also a real, separately-confirmed B0 condition in both 2025 series). Full self-mark detail: FORGE guide §2 and §3.

#1.5 Dynamics of a particle moving in a straight line or plane (spec 4.1–4.4)

• Newton's second law, scalar form: F = ma, where F is the resultant force, not any single applied force.
• Newton's second law, vector form (spec 4.1's own stated case, "as a vector of the form ai + bj"): F₁ + F₂ + ... = ma, where the resultant and the mass × acceleration vector are equated component-by-component — the i-components give one scalar equation, the j-components give a second, independent scalar equation. A vector equation is not one equation; treating it as one (solving only the i-component, say, and assuming the j-component "matches") is a confirmed real trap (newtons-second-law-in-vector-form.ts's own trap taxonomy).
• Connected particles — the equation of motion is not itself a single named formula, it's an identity you build: for each particle, resolve (resultant force in that particle's own direction of motion) = (that particle's mass) × (the shared acceleration), with each particle's own direction of motion taken as positive for that particle — this is the real, standard, mark-scheme-endorsed convention (confirmed directly in the January 2025 mark scheme), not an inconsistency to fix. See §2.8 below for the single most important worked-method entry in this document, addressing exactly the error this identity most commonly loses marks to.
• Momentum: momentum = mv (spec-named, must memorise).
• Impulse: impulse = mv − mu (spec-named, must memorise) — the change in momentum a force produces, where u is the velocity immediately before and v immediately after.
• Conservation of momentum, two particles colliding directly: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ — total momentum before equals total momentum after. Not separately quoted in the spec's own "must memorise" sentence the way momentum/impulse are named individually, but it is spec-named content (4.3) and the M1 booklet gives it no more than it gives anything else, so it is tagged "not given — memorise" here as a direct inference, flagged as such rather than presented as a literal spec quote. Every velocity in this equation is signed — the strongest, most cross-series-confirmed trap in the entire WME01 archive is reporting a signed velocity where the question's own wording asked for an unsigned speed (three independent series: Jan 2020, Oct 2022, Jun 2024; full treatment in momentum-impulse-sign-convention.ts).
• Friction, moving case: F = μR, where R is the normal reaction and μ the coefficient of friction (spec 4.4's own guidance: "An understanding of F = μR when a particle is moving").

#1.6 Statics of a particle (spec 5.1–5.3)

• Weight: W = mg — a force, in newtons, not to be confused with mass in kilograms. A real, confirmed examiner-report trap: candidates writing 5g where the question already supplied a force of 5 N (weight-mass confusion, verified in Jan 2022 Q1's report).
• Equilibrium of a particle under coplanar forces: resolve in two independent directions (almost always horizontal and vertical, or along/perpendicular to an incline) and set each resultant to zero — ΣFₓ = 0 and ΣF_y = 0. Every term that needs resolving must carry its sin or cos factor; every force actually present in the system must appear in at least one of the two equations (an incomplete resolution, e.g. 5 = F cos 30° where a second force's own component was also needed, is a confirmed real error, Jan 2022 Q1).
• Friction, equilibrium/limiting case: F ≤ μR — friction adjusts up to, but never exceeds, μR, and only takes the value μR exactly at the point of limiting equilibrium (about to slide). Below that, F is whatever value resolving requires to keep the particle in balance, found from the equilibrium equations themselves, not from μR.

#1.7 Moments (spec 6.1)

• Moment of a force about a point: moment = force × perpendicular distance from that point to the force's own line of action. If a force acts exactly at the point moments are being taken about, its perpendicular distance from that point is zero, so its moment there is zero regardless of the force's own (possibly unknown) magnitude — the entire mechanism behind "choose the pivot that eliminates an unknown reaction."
• Equilibrium condition for a rigid body under coplanar parallel forces (spec 6.1's own restriction — "coplanar parallel forces," almost always all-vertical: weights and reactions): sum of clockwise moments about any point = sum of anticlockwise moments about that same point, equivalently net moment = 0, together with ΣF = 0 in the direction the forces act (for an all-vertical system, Σ(upward reactions) = Σ(downward weights/loads)). A moments question with two unknown reactions is normally solved most efficiently with one moments equation (about either support, eliminating that support's own reaction) plus the vertical-resolve equation — reaching for a second moments equation is the less efficient, more error-prone route in that specific case (three independent series converge on this exact rule: Jan 2023, Jun 2024, Oct 2023). The one confirmed exception: if the rod's own weight or centre-of-mass position is also unknown (not just the reactions), a single moments equation cannot separate two unknowns from one line, and a second, genuinely independent moments equation (about the other support) is the correct, credited method (Jan 2023 Q4).

#1.8 Summary masterlist — every result above, one table

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#PART 2 — Worked-Method Reference

Each entry below is a method skeleton, not a full worked exemplar: the recognisable question shape, the ordered steps a credited method actually takes, and one short, self-contained numeric check — never a multi-part narrated solution with a diagram, MCQ, or trap taxonomy of its own. Every scenario and number here is VERIDIAN-original and distinct from every worked chain, marked solution, or MCQ already in the curriculum files or the FORGE guide, checked directly against those files before writing this section.

#2.1 Resultant of two or more forces (spec 2.1)

Recognise it from: two or more forces given either as vectors, as magnitude-and-angle pairs, or as magnitude-and-bearing pairs, asked for a single combined force (magnitude and/or direction).

Method skeleton:

1. Convert every force to i/j component form (§1.3 — watch the bearing swap if bearings are given).
2. Sum all i-components, sum all j-components, separately.
3. Magnitude: Pythagoras on the two sums. Direction: tan⁻¹(|j-sum| / |i-sum|), adjusted for the correct quadrant.
4. If the question explicitly wants a triangle-of-forces/cosine-rule/Lami's-theorem route instead (or offers it as a faster method for exactly two forces), identify which angle the formula needs — the angle between the forces themselves, or the triangle's own interior angle — before substituting; conflating the two is the single most-confirmed error on this exact question type (Jun 2024 Q2).

Short check: forces P = (6i − 2j) N and Q = (−1i + 5j) N. Resultant = 5i + 3j N. Magnitude = √(25+9) = √34 ≈ 5.83 N.

#2.2 Bearing-to-component conversion (spec 2.1–2.2)

Recognise it from: any force or velocity given as "magnitude on a bearing of θ°."

Method skeleton: East (i) component = F sin θ; North (j) component = F cos θ. Sketch a small compass triangle before substituting if there's any doubt — the swap trap costs the whole method mark, not just accuracy, because a wrongly-paired component still "looks like" a resolved term.

Short check: a force of 14 N on a bearing of 065°. Components: East = 14 sin 65° ≈ 12.69 N, North = 14 cos 65° ≈ 5.92 N — not the reverse.

#2.3 Position vectors and relative motion under constant velocity (spec 2.2)

Recognise it from: an initial position vector, a constant velocity vector, and a question asking for position, distance travelled, or distance to a fixed point at a later time — not a scenario combining a starting velocity with a constant acceleration in two dimensions (§1.3's flagged non-coverage of 2D projectile motion).

Method skeleton:

1. Substitute the given t into r(t) = r₀ + tv to get the position vector at that instant.
2. If the question asks for distance to a fixed point A, compute r(t) − A first, then take the magnitude of that difference — never the magnitude of r(t) alone (a confirmed real trap: computing distance from the origin instead of from the stated point).
3. If the question asks for a minimum distance between two moving particles instead, that needs calculus on |r_A(t) − r_B(t)|² or a "components equal → no real roots" argument — a genuinely harder variant this document does not expand on, since it is single-series-confirmed and already flagged as such in resultant-forces-and-bearings.ts's own header.

Short check: r₀ = (2i − j) m, v = (3i + 4j) m/s, fixed point A = (20i + 15j) m. At t = 4: r(4) = (2+12)i + (−1+16)j = 14i + 15j. r(4) − A = −6i + 0j. Distance = 6 m.

#2.4 Choosing the right suvat equation (spec 3.1)

Recognise it from: a single, uninterrupted phase of constant-acceleration motion in a straight line, with exactly one of s, u, v, a, t unknown.

Method skeleton: list the five quantities, mark the three given and the one asked for; the equation that doesn't need the fifth (unlisted) quantity is the one to use — never introduce it by trying to find it first unless the question requires it for a later part.

Short check: u = 4 m/s, a = 1.5 m/s², t = 6 s, find s. No v needed → s = ut + ½at² = 4(6) + ½(1.5)(36) = 24 + 27 = 51 m.

#2.5 Two-stage motion (spec 3.1)

Recognise it from: any journey with a described change partway through — a different acceleration, a force removed or added, a collision with the ground, a parachute opening.

Method skeleton: resolve stage 1 completely and independently first, using only stage 1's own u/a/t values; carry forward only the single handover value the second stage actually needs (almost always v at the boundary, sometimes also a position); resolve stage 2 entirely on its own terms, with its own (possibly different) u, a, and t. Never reuse the same time variable for two stages that start at different clock-times, and never assume stage 2 must end at rest — it may instead be cut off by a stated distance or time boundary while the particle is still moving (both confirmed, independent real traps; full treatment: kinematics-constant-acceleration-two-stage-motion.ts).

Short check: stage 1 — from rest, a = 2 m/s² for t = 6 s → v = 12 m/s, s = 36 m. Stage 2 — decelerates at 3 m/s² from that 12 m/s to rest → t = 4 s, s = 24 m. Total distance = 60 m, total time = 10 s.

#2.6 The velocity-time graph — construction and the B-marks it actually carries (spec 3.1)

Recognise it from: any question that either supplies a v-t graph to read from, or explicitly asks for one to be sketched or drawn before further parts can be answered.

Method skeleton:

1. Identify every phase's own u, v, a, t (some as unknowns).
2. Draw the shape starting exactly where the motion starts (not automatically the origin), with a straight-line segment for each constant-acceleration phase and a sharp corner, never a smooth curve or an extra vertical line, at every phase boundary.
3. Label both axes with every known numeric value; leave unknowns as clearly marked variables.
4. Use the total area (sum of the trapezium/triangle/rectangle regions) to form an equation in the remaining unknown(s) — this is exactly the fifth suvat equation, applied region by region.

Short check: a particle accelerates from rest to 20 m/s in T seconds, travels at a constant 20 m/s for 15 s, then decelerates to rest over a further 4 s. Total distance covered is 500 m. Area equation: ½T(20) + 15(20) + ½(4)(20) = 500 → 10T + 300 + 40 = 500 → 10T = 160 → T = 16 s.

#2.7 Newton's second law in vector form (spec 4.1)

Recognise it from: two or more forces given in i/j form, a mass, and either the resultant/acceleration to find, or an unknown coefficient inside one of the forces to solve for.

Method skeleton:

1. Sum the given forces to form the resultant (component-wise).
2. Equate to ma — this produces two scalar equations (i-component, j-component), not one.
3. If an unknown appears in both equations at once (a genuinely coupled pair, not one unknown isolated per equation), solve them simultaneously — do not stop at "equated the coefficients" and assume the job is done.
4. If the question gives a direction rather than the resultant itself, remember a direction vector is not the resultant — it only tells you the resultant is some scalar multiple of that vector, which typically requires eliminating that unknown scalar (λ) to reach an equation in the real unknown alone. This exact technique, and the real mark-scheme wording behind it, is already built out in newtons-second-law-in-vector-form.ts's Jan 2025-derived worked chain — not repeated here.

Short check: mass 4 kg, F₁ = (6i − 2j) N, F₂ = (−2i + 5j) N. Resultant = (4i + 3j) N. a = resultant / m = (1i + 0.75j) m/s², magnitude = √(1² + 0.75²) = √1.5625 = 1.25 m/s² exactly. (A confirmed real trap worth restating here even though it belongs to a different question family: a fully correct method that stops at the vector a without taking its magnitude, when the question asked for a speed or magnitude of acceleration, still loses the final accuracy mark — Jan 2025 Q1(a).)

#2.8 Connected particles — and the dimensional-completeness / missing-ma-term error, addressed directly

Recognise it from: two particles joined by a string (over a pulley, or in a straight-line two-string system), moving with a shared magnitude of acceleration.

Method skeleton:

1. Draw a separate force diagram for each particle — the single most examiner-endorsed way to avoid mixing up which tension or which weight belongs to which body.
2. Write one equation of motion per particle, each resolved in that particle's own direction of motion, taken as positive for that particle alone. This is not an inconsistency; it is the real, credited convention.
3. Every equation must be dimensionally complete: every term is a genuine force (weight, tension, or friction), and the right-hand side is that particle's own mass times the shared acceleration — never omit the ma/Ma term, and never let two equations quietly assume zero acceleration by cancelling it away.
4. Solve the two equations simultaneously for the two unknowns (typically T and a) — usually fastest by adding the two equations to eliminate T first.

The error this entry exists to address, worked through with fresh numbers. Particle X, mass 5 kg, hangs from one end of a string over a smooth fixed pulley; particle Y, mass 9 kg, hangs from the other end. Released from rest, Y falls and X rises.

Correct, dimensionally complete equations — X rising is X's own positive direction; Y falling is Y's own positive direction:

• For X: T − 5g = 5a
• For Y: 9g − T = 9a

Adding eliminates T: 9g − 5g = 14a → 4(9.8) = 14a → a = 2.8 m/s². Back-substituting: T = 5(9.8 + 2.8) = 5(12.6) = 63 N. Checked independently from Y's own equation: T = 9(9.8 − 2.8) = 9(7) = 63 N — consistent.

The real, confirmed error (this is the exact structural pattern the FORGE guide's self-review pass identified from January 2025's examiner report, reproduced here with this entry's own fresh numbers rather than the real question's): a candidate writes

T − 5g = 9g − T

Every term present is a genuine force, correctly assigned to the right side of the string — but there is no a anywhere in this equation. Solved, it gives 2T = 14g → T = 7g = 68.6 N: a plausible-looking number, but wrong (the real, correct tension is 63 N), and — more importantly for the M mark — this equation cannot produce a value for acceleration at all, because acceleration was never a term in it. On a paper where the first thing nearly every connected-particles question asks for is the system's acceleration, an equation that is structurally incapable of yielding one is worth zero M marks regardless of what number the T-line happens to produce. The general marking-guidance rule this illustrates: an M mark in Mechanics requires the resulting equation to be dimensionally correct, not just "attempted" — every term must genuinely be the right kind of quantity for the equation it sits in, and a shared acceleration term is exactly as load-bearing as a mass or a length. Self-check before moving on from any connected-particles equation: does each side of each equation contain a mass-times-acceleration term, and does a completely separate line exist (or a clear plan to derive one) for the shared value of a? If the answer to either is no, the equation is not yet a real equation of motion, however tidy it looks.

#2.9 Momentum, impulse, and conservation of momentum (spec 4.3)

Recognise it from: one particle given an impulse (find the resulting velocity or the impulse itself), or two particles colliding directly (find an unknown velocity after, using conservation).

Method skeleton:

1. Fix a positive direction before writing anything down.
2. For a single-particle impulse question: impulse = m(v − u), solve directly for whichever of v, u, or the impulse itself is unknown.
3. For a collision: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, substitute every known signed velocity, solve for the one unknown.
4. Before writing the final answer, re-read the exact word the question used — speed (unsigned, take the magnitude) or velocity (signed, keep the sign as found). This single check is the highest-value line in this entire document by frequency of real marks lost.

Short check: particle M, mass 7 kg, moving at 6 m/s, collides with particle N, mass 2 kg, at rest. After the collision N moves at 3.5 m/s (same direction as M's original motion). Conservation: 7(6) + 2(0) = 7v_M + 2(3.5) → 42 = 7v_M + 7 → v_M = 5 m/s. Impulse on N = 2(3.5 − 0) = 7 N s.

#2.10 Equilibrium of a particle under coplanar forces (spec 5.1–5.2)

Recognise it from: a particle held motionless by three or more forces (weight plus at least two others — a string, a horizontal push, a second string), asked for an unknown force magnitude or angle.

Method skeleton:

1. Draw every force acting on the particle, including weight.
2. Resolve in two convenient, independent directions — usually horizontal/vertical, but along/perpendicular to a stated string or plane if that eliminates an unknown faster.
3. Set each resultant to zero; solve the resulting pair of equations.
4. Sense-check: does every equation actually contain every force that has a component in that direction? An omitted term is a method error, not just an accuracy slip.

Short check: a particle of weight 8 N hangs in equilibrium, held by a string at 40° above the horizontal (tension T) and a horizontal force H. Vertical: T sin 40° = 8 → T ≈ 12.4 N. Horizontal: H = T cos 40° ≈ 12.4 × 0.766 ≈ 9.5 N.

#2.11 Friction — equilibrium, limiting equilibrium, and motion (spec 4.4/5.3)

Recognise it from: a rough surface (horizontal or inclined), asked either to show a value of friction, find a coefficient, or decide whether an object moves at all.

Method skeleton:

1. Resolve perpendicular to the surface first to find R — include every force with a perpendicular component, not just weight (a string tension or an applied force at an angle to the plane both contribute to R if they aren't parallel to it).
2. Resolve along the surface. If the object is in equilibrium, friction is whatever value balances the equation (F ≤ μR, not automatically equal to it). If the object is moving or is asked "is it on the point of moving," compare the driving force component to the maximum available friction, μR, as an explicit numeric comparison — a written conclusion with no numbers behind it earns nothing on this paper (a confirmed real trap, Jan 2025 Q6(c)).
3. If limiting equilibrium is stated or implied ("on the point of sliding"), set F = μR exactly and solve.

Short check — does it move?: a 4 kg block rests on a rough plane inclined so that sin α = 7/25, cos α = 24/25, with μ = 0.3. Driving force down the slope: mg sin α = 4(9.8)(7/25) ≈ 10.98 N. Maximum available friction: μ mg cos α = 0.3(4)(9.8)(24/25) ≈ 11.29 N. Since 10.98 N < 11.29 N, the block remains in equilibrium — friction adjusts down to exactly 10.98 N, comfortably under its own ceiling.

#2.12 Moments — a rod on two supports (spec 6.1)

Recognise it from: a rod or beam resting horizontally on two supports (or held by two strings), with an added load, asked for one or both reactions, or for the load/position at which the rod is on the point of tilting about one support.

Method skeleton:

1. Identify the rod's own weight and where it acts (centre, if uniform; a stated point, if not).
2. If only the two reactions are unknown: resolve vertically (ΣF = 0) and take one moments equation about either support (eliminating that support's own reaction) — this two-equation combination is the efficient, most-credited route.
3. If the rod's own weight or centre-of-mass position is also unknown, one moments equation cannot separate two unknowns on its own; take a second, genuinely independent moments equation about the other support instead.
4. For a "point of tilting about support X" question, set the reaction at the other support to zero and take moments about X alone — a single equation, one unknown, because the zeroed reaction was never in it.

Short check: a uniform rod AB, length 8 m, weight 50 N, rests on supports at C (1 m from A) and D (7 m from A — i.e. 1 m from B). A load W is placed at B. Find W for the rod to be on the point of tilting about D (so the reaction at C is zero). Taking moments about D: the 50 N weight acts 3 m to the left of D (anticlockwise moment 50 × 3 = 150 N m); the load W acts 1 m to the right of D (clockwise moment W × 1). Equilibrium: W = 150 N.

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#Genuine gaps flagged, not papered over

• This document's own task brief anticipated a "projectiles" method entry for WME01. That entry does not appear above, and its absence is a finding, not an oversight: the real M1 spec's kinematics content (item 3) is restricted to "motion in a straight line," and every real vector-kinematics question independently verified for this course (Jan 2025 Q5, Jun 2024 Q7) uses constant-velocity position vectors, r(t) = r₀ + tv, never a two-dimensional suvat combination of a starting velocity with gravity acting in a different direction. Two-dimensional projectile motion under gravity is standard content on Pearson's M2 unit, not this one. If a genuine WME01 projectile-style question exists in a series outside the 17 this course's facts bank has reviewed, this correction would need revisiting — but nothing in the archive checked supports building a projectile-motion method entry for this paper, so none was invented.
• Whether the sine rule itself sits in the P1/P2 formula-booklet pages available to M1 candidates was not independently confirmed. The facts bank confirms the cosine rule's presence and live use directly; its own listing of what else the P1/P2 extension covers (mensuration, series, logs, binomial, trapezium rule) does not mention the sine rule either way. Lami's theorem is confirmed as an examiner-accepted alternative method but is named nowhere in the spec's own text, and there is no evidence it is a booklet entry at all. Table row 21 above is marked unverified rather than guessed at.
• The "not given — memorise (inferred)" tag on F = ma, W = mg, F = μR, F ≤ μR, and the moments formula/equilibrium condition is this document's own entailment from the empty M1 booklet section, not a direct spec quote the way momentum/impulse/suvat are. The spec's own explicit "must be memorised" sentence names only the latter three formula groups. This is flagged in Part 1 itself as well as here, so the distinction between "the spec says so directly" and "this follows necessarily from the booklet's own empty M1 entry" isn't lost in the summary table.
• No mark-scheme evidence reviewed for this course names a specific mark lost for an incomplete or malformed connected-particles equation missing its ma term on a real paper — the FORGE guide's own citation for this pattern (January 2025's examiner report) describes the error qualitatively ("T − 2mg = 3mg − T," no numeric mark-loss stated beyond the general M0 consequence the general marking guidance implies) rather than quoting an exact mark line for that specific wrong-working instance. §2.8 above states the M-mark consequence as a direct, correctly-sourced entailment of the general marking-guidance rule (an M mark requires a dimensionally correct equation), not as a quoted mark-scheme line for that exact error.
• This document did not attempt a full per-question, per-mark spec-topic weighting table (which sub-topic carries how many of the 75 marks, series by series) — the facts bank itself flags this as out of scope for the research pass this document is built from (§8), and it remains out of scope here.