Complete Worked Paper

#CODEX: Complete Worked Paper — WME01 January 2025 (Mechanics M1)

Provenance (2026-09-29). Series used: January 2025, WME01/01 (Pearson Edexcel International Advanced Level, Mechanics M1). All three primary documents were fetched fresh this pass and read in full:

• Mark scheme — wme01-01-rms-20250306.pdf, downloaded directly from qualifications.pearson.com/content/dam/pdf/International-Advanced-Level/Mathematics/2018/Exam-materials/ (HTTP 200, confirmed genuine via file; Publications Code WME01_01_2501_MS).
• Examiner report — wme01-01-pef-20250306.pdf, same source and confirmation (WME01_01_2501_ER).
• Question paper — a direct fetch of the likely qualifications.pearson.com question-paper filename returned an HTTP redirect rather than the PDF itself this pass (unlike the mark scheme and examiner report, which both returned the file directly); consistent with, though not confirmed as, Pearson's live question papers being access-restricted for a period after sitting. This was instead read from the Physics & Maths Tutor mirror (pmt.physicsandmathstutor.com/download/Maths/A-level/Papers/Edexcel-IAL/Mechanics/M1/QP/January%202025%20QP.pdf, HTTP 200), which reproduces the genuine Pearson paper watermarked "PMT" on every page, carrying the real cover page (Friday 10 January 2025, Paper reference WME01/01, Question Paper Log Number P76197A) and the same Q1(b) Clarification Notice already on file in this repo's own content/veridian/question-bank/wme01.ts ("acts in the direction of..." corrected to "acts in a direction that is parallel to..."). That exact match — found independently, not copied from the question-bank file — is itself the cross-check that this is the same genuine paper, not a mislabelled mirror.

All three PDFs converted with pdftotext -layout. Every number, mark code and final answer below was independently derived by hand from the question paper's own data and checked against the real mark scheme's stated intermediate values and final answer before being written up — none was copied from the scheme and presented as if derived. As an end-to-end validity check, this document re-sums the tariff independently in §2 and again in §5; both sums reconcile exactly against the paper's own printed total of 75. Question wording is paraphrased throughout — the real question paper's prose is Pearson's copyrighted text — and only short (under-15-word) instructional phrases are quoted verbatim, with attribution. Numbers, equations and mark codes are not copyrightable expression and are reproduced exactly, since exactness there is the entire point of an exam-technique document.

#1. Why this series, and not another

research/veridian/WME01-verified-facts.md §5 lists 17 WME01 series with at least one genuine primary document reviewed by this course, of wildly uneven depth — most contribute one or two deep-mined questions, not a full paper. Two later documents in this same folder, forge-exam-technique-guide.md and signal-frequency-and-boundaries.md, each independently mined both the January 2025 and June 2025 mark schemes question-by-question, and named January 2025 as "the first place in this course with a full, question-by-question mark breakdown" — but a mark breakdown (topic tags, tariffs, trap digests) is not the same artefact as a complete set of worked solutions, and neither document builds one.

January 2025 is the better-evidenced of the two candidates for that harder job, on three independent grounds checked directly this pass, not assumed from the prior documents' own claims:

1. All three primary documents are independently confirmed genuine and fully legible — question paper, mark scheme, and examiner report — covering every one of the paper's 7 questions. June 2025's own mark scheme was confirmed genuine by a prior pass but explicitly "not deep-mined for topic content," and its examiner report's existence was never checked at all (facts bank §5, §8).
2. This repo already has independent, from-the-real-question-paper confirmation of two of this series' questions — content/veridian/question-bank/wme01.ts quotes Q1(b) and Q7(c) directly from WME01_01_2501 with source: "question-paper" — so this pick is corroborated by a second, earlier, independent read of the same primary source, not resting on this pass's fetch alone.
3. Internal arithmetic consistency. Re-deriving every mark from scratch (§4 below) and summing the tariff gives 8+11+12+10+7+12+15 = 75 marks across 7 questions — matching both the mark scheme's own per-question totals and the real question paper's own printed "TOTAL FOR PAPER IS 75 MARKS." A paper this internally consistent, checked end to end by an independent hand, is the honest bar for calling a series "well-evidenced enough to work start to finish" — not a claim that January 2025 is somehow a "typical" WME01 paper.

#2. Paper anatomy (recap only)

7 questions, 75 marks, tariff 8, 11, 12, 10, 7, 12, 15 (independently re-summed: 8+11+12+ 10+7+12+15 = 75, matching the paper's own printed total). 1 hour 30 minutes. Calculator permitted (Pearson's standard IAL restriction: no symbolic algebra, differentiation or integration, no stored formulae). $g = 9.8\text{ m s}^{-2}$ throughout, printed on the paper's own front cover, with final answers to 2 or 3 significant figures. This document does not repeat the wider policy discussion (the M/A/B/DM coding system, the general rounding and diagram conventions) already covered in forge-exam-technique-guide.md §§2–3 — see that file for it; nothing here mechanically restates that ground.

#3. What this document adds against the rest of the course

Checked directly (grep across every .ts file in content/veridian/curriculum/wme01/, content/veridian/question-bank/wme01.ts, and every .md file in content/veridian/technique-hub/wme01/) before writing a line below:

Five of seven questions (1(a), 2, 3, 4, 5, 6, 7(a)–(b) — six, in fact, counting 7(a)–(b) as one entry) are new, full worked content for this course; the remaining two parts (1(b), 7(c)) already have a deeper, dedicated marked-solution object elsewhere and are deliberately kept short here, with a pointer, rather than restated.

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#4. The paper, question by question

#Question 1 (8 marks) — resultant of three forces, then a direction constraint

Spec 2.1, 4.1. A particle of mass $2.5$ kg on a smooth horizontal plane is acted on by three horizontal forces $\mathbf F_1=(6\mathbf i+8\mathbf j)$ N, $\mathbf F_2=(-16\mathbf i+2\mathbf j)$ N, $\mathbf F_3=(-2\mathbf i+8\mathbf j)$ N.

(a) Find the magnitude of the acceleration.

Resultant: $\mathbf F_1+\mathbf F_2+\mathbf F_3 = (6-16-2)\mathbf i+(8+2+8)\mathbf j = -12\mathbf i+18\mathbf j$ N.

Newton's second law: $\mathbf a = \mathbf F/m = (-12\mathbf i+18\mathbf j)/2.5$.

Magnitude: $|\mathbf F| = \sqrt{12^2+18^2} = \sqrt{468} = 6\sqrt{13}$ N, so $|\mathbf a| = 6\sqrt{13}/2.5 = 12\sqrt{13}/5 \approx 8.65\text{ m s}^{-2}$.

Marks: M1 the three forces summed (the $\mathbf i$'s and $\mathbf j$'s need not be collected separately for this mark); M1 N2L applied, in either vector or scalar form, to the candidate's own resultant; M1 Pythagoras used to extract a magnitude — credited whether it is applied to the resultant force or to the acceleration, i.e. before or after dividing by the mass; A1 the final value, $8.65\text{ m s}^{-2}$ or better (exact value $\tfrac{12\sqrt{13}}5 \approx 8.653$).

Examiner's report: the most common way to lose the final mark, despite an otherwise correct method, was leaving the answer "as a vector instead of finding the magnitude" — the question asks for a magnitude, a scalar, and a fully-worked vector is not itself the answer. A smaller number of candidates added an unstated force — the paraphrase is exact: some added a weight component $2.5g$ in the $\mathbf j$ direction that the question's own "smooth horizontal plane" setup never puts there. A separate slip, purely transcriptional: "a number of candidates misread 18 for 8 at various stages."

No full worked treatment of this real question exists elsewhere in this repo — newtons-second-law-in-vector-form.ts teaches the general vector-N2L technique on its own, different, VERIDIAN-original forces.

(b) A fourth force $\mathbf F_4=(p\mathbf i+p\mathbf j)$ N is added; the resultant of all four now acts in a direction parallel to $(7\mathbf i+2\mathbf j)$. Find $p$.

(Full step-by-step marked solution, including the real examiner-report quotation on the "parallel to" misreading and a reconstructed common wrong path, already lives at content/veridian/question-bank/wme01.ts, entry wme01-jan25-q1b-vector-resultant-direction — not duplicated here.)

In brief: adding $\mathbf F_4$ to part (a)'s resultant gives $(-12+p)\mathbf i+(18+p)\mathbf j$. "Parallel to $(7\mathbf i+2\mathbf j)$" fixes the ratio of components, not their exact value: $\dfrac{-12+p}{7}=\dfrac{18+p}{2}$, so $2(-12+p)=7(18+p) \implies -150=5p \implies p=-30$.

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#Question 2 (11 marks) — a speed–time graph with two unknown times

Spec 3.1. Fixed points $A,B,C$ lie on a straight horizontal road. A motorbike passes $A$ at $t=0$ with speed $5\text{ m s}^{-1}$, accelerates uniformly to speed $V$ at $B$ (taking time $T_1$), then decelerates uniformly to rest at $C$ (taking time $T_2$). $AB=132$ m, $BC=136$ m.

(a) Sketch the speed–time graph. (3 marks, all B marks)

A single "kink" trapezium: start at $(0,5)$ — not the origin, since the bike is already moving at $5\text{ m s}^{-1}$ when it passes $A$ — rising in a straight line to $(T_1,V)$, then falling in a straight line to $(T_1+T_2,\,0)$, finishing on the time axis.

Marks: B1 the shape itself, starting on the speed axis above the origin and finishing on the time axis, with no extra continuous line anywhere; B1 the time axis correctly labelled at $T_1$ and $T_1+T_2$; B1 the speed axis correctly labelled at $5$ and $V$.

Examiner's report: "a few candidates added a solid vertical line and lost a mark" — drawing a vertical jump in a graph that should show two sloped segments. Separately, and more costly: a "special case" note records that many candidates treated the total time from $A$ to $C$ as if it were $T_2$ on its own, consistently throughout — this alone lost the two time-axis and labelling marks even where the resulting arithmetic, applied consistently, still earned follow-through credit elsewhere.

(b) Find, in terms of $V$: (i) $T_1$, (ii) $T_2$. (4 marks)

Area under each segment equals the distance covered (a trapezium for stage 1, a triangle for stage 2):

$$132 = \frac{(5+V)}{2}T_1 \implies T_1 = \frac{264}{5+V}$$ $$136 = \frac{V}{2}T_2 \implies T_2 = \frac{272}{V}$$

Marks: M1 a complete method for $T_1$ using the distance $132$; A1 the correct expression $T_1=264/(5+V)$; M1 a complete method for $T_2$ using the distance $136$; A1 the correct expression $T_2=272/V$.

(c) Given the total time $A$ to $C$ is 28 s, find $V$. (2 marks)

$$\frac{264}{5+V}+\frac{272}{V}=28$$

Clearing denominators: $264V+272(5+V) = 28V(5+V) \implies 536V+1360 = 140V+28V^2 \implies 28V^2-396V-1360=0 \implies 7V^2-99V-340=0$.

Discriminant $=99^2+4(7)(340) = 9801+9520=19321=139^2$, so $V = \dfrac{99+139}{14}=\dfrac{238}{14}=17$ (the negative root, $-20/7$, is rejected — a motorbike's speed cannot be negative).

Marks: M1 the two expressions from (b) combined via the total-time condition into an equation in $V$ alone; A1 $V=17$, having correctly rejected the negative root.

(d) Find the deceleration. (2 marks)

$T_2 = 272/17 = 16$ s. Deceleration $= V/T_2 = 17/16 = 1.0625 \approx 1.06\text{ m s}^{-2}$.

Marks: M1 a complete method using the candidate's own $V$ and $T_2$ (or the graph's gradient directly); A1 the correct, positive value, $1.06\text{ m s}^{-2}$ or better.

Examiner's report: "a substantial number gave a negative answer and lost a mark" — reporting the signed rate of change of a strictly decelerating motion instead of the unsigned deceleration the question actually asks for. A small number failed to reject the quadratic's negative root in part (c) and lost that mark instead. This is a live, real-paper instance of the single most cross-series-confirmed trap in this course's whole WME01 archive (reported speed/deceleration with the wrong sign) — see momentum-impulse-sign-convention.ts for the dedicated treatment of that general pattern on a different question.

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#Question 3 (12 marks) — impulse, then friction, then a collision

Spec 3.1, 4.3, 4.4. Particles $P$ (mass $3m$ kg) and $Q$ (mass $m$ kg) rest $1.4$ m apart on a rough horizontal surface. An impulse of magnitude $\lambda$ N s is applied to $P$ towards $Q$, giving $P$ speed $5\text{ m s}^{-1}$ immediately afterwards.

(a) Find $\lambda$ in terms of $m$. (2 marks)

Impulse $=$ change in momentum: $\lambda = 3m(5-0) = 15m$.

Marks: M1 the impulse–momentum relation used with mass $3m$ and speed $5$; A1 the correct answer, $\lambda=15m$.

Examiner's report: "occasionally either 3 or m was missing and $-15m$ was also seen" — a sign slip inconsistent with an impulse applied in the direction of travel, which must be positive on the same convention $P$'s own resulting speed is positive on.

(b) Given $P$'s speed immediately before it reaches $Q$ (1.4 m away) is $2.5\text{ m s}^{-1}$, find the coefficient of friction $\mu$. (7 marks)

$suvat$: $v^2=u^2+2as \implies 2.5^2 = 5^2+2a(1.4) \implies a = \dfrac{6.25-25}{2.8} = -\dfrac{375}{56} \approx -6.696\text{ m s}^{-2}$.

Vertically, $P$ is on a horizontal surface with no vertical acceleration, so $R=3mg$. Friction (opposing $P$'s motion, since it is decelerating) is at its limiting value, $F=\mu R = 3\mu mg$. By Newton's second law along the direction of travel:

$$-3\mu mg = 3ma \implies \mu = -\frac{a}{g} = \frac{375/56}{9.8} \approx 0.683$$

Marks: M1 a complete $suvat$ method reaching an equation in $a$ only, using $u=5$ (not $0$); A1 the correct unsimplified equation; B1 $R=3mg$, stated or clearly implied; M1 $F=\mu R$ used for the candidate's own $R$; M1 a correct N2L equation in $m$, $\mu$ and $a$, with $3m$ (or $m$) present consistently on both sides; A1 the correct unsimplified equation; A1 the final value, $\mu\approx0.683$ (or $0.68$).

Examiner's report: the dominant error was using $u=0$ instead of $u=5$ in the $suvat$ step — treating $P$ as starting from rest at this point, rather than continuing from the speed it already had. A separate, specific dimensional slip the report names directly: "some candidates... used $3m$ on one side of the equation and $m$ on the other" — exactly the kind of structurally-broken equation forge-exam-technique-guide.md §2 already documents as scoring no method credit at all, regardless of the arithmetic that follows, since a resolution or N2L equation must use the same mass consistently throughout.

(c) $Q$'s speed immediately after the collision is $2.1\text{ m s}^{-1}$. Find $P$'s speed immediately after. (3 marks)

Conservation of linear momentum: $3m(2.5)+m(0) = 3m(v)+m(2.1)$, taking $P$'s original direction as positive throughout:

$$7.5m = 3mv+2.1m \implies v = \frac{7.5-2.1}{3} = 1.8$$

$P$'s speed after the collision is $1.8\text{ m s}^{-1}$ — less than its speed before ($2.5$), as physically expected once some momentum has transferred to $Q$.

Marks: M1 a dimensionally correct CLM equation in $v$ alone, correct number of terms, correct pairing of masses and speeds; A1 the correct unsimplified equation; A1 the correct, positive speed.

Examiner's report: "a few [used] $v=5$ instead of $2.5$" for $P$'s pre-collision speed — an easy slip given the same particle carries two different named speeds across the question (the launch speed from part (a), and the reduced impact speed found in part (b)). The report also notes a small minority equated the two particles' impulses directly rather than using conservation of momentum — a valid alternative route, since one particle's impulse is, by Newton's third law, the negative of the other's.

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#Question 4 (10 marks) — a suspended rod, a moved particle, and a tilting point

Spec 6.1. A uniform rod $AB$, length $3$ m, weight $W$ N, hangs horizontally from two light vertical ropes: one at $A$, one at $C$, where $AC=2.1$ m (so $CB=0.9$ m). With the rope at $C$ under tension $350$ N, the rod is in equilibrium.

(a) Show that $W=490$. (3 marks)

The rod's own weight acts at its centre of mass — the midpoint of a uniform rod — $1.5$ m from $A$, which is $2.1-1.5=0.6$ m on the $A$-side of $C$. Taking moments about $A$ (which removes the unknown tension at $A$ from the equation):

$$M(A):\quad 350\times2.1 = W\times1.5 \implies W = \frac{735}{1.5} = 490 \text{ N (as required)}$$

Marks: M1 a complete, dimensionally-correct moments equation in $W$ alone (about $A$, or an equivalent two-equation route eliminating the tension at $A$); A1 the correct unsimplified equation; A1* the given answer reached with the working shown in full — this is a "show that" mark (cso, correct solution only): no credit for writing the target value down without the moments equation that reaches it.

Examiner's report: well answered, with the large majority correctly choosing to take moments about $A$; a minority used two equations (a vertical resolve plus moments about $B$) instead, "mostly successful, but errors were more likely with the algebra." Because the answer is given, the report separately notes "the occasional fudge" — working adjusted after the fact to land on 490 rather than followed through honestly.

(b) A particle $P$ of weight $210$ N is attached at distance $d$ metres from $A$; the tension at $C$ rises to $600$ N; the rod remains horizontal. Find $d$. (3 marks)

Taking moments about $A$ again (the same elimination trick, and the question's own "distance from $A$" phrasing is a direct hint to reuse it):

$$M(A):\quad 600\times2.1 = 490\times1.5+210\times d \implies 1260 = 735+210d \implies d=2.5$$

Marks: M1 a complete equation in $d$ alone (about $A$, correct number of terms, dimensionally correct); A1 the correct unsimplified equation; A1 $d=2.5$.

Examiner's report: the majority again took moments about $A$ and reached a correct solution directly; those who instead resolved vertically and took moments about a different point "often mak[ing] errors with the distances," specifically the two different ways of expressing the remaining length from $P$ to the far end (e.g. $1.5-d$ or $2.1-d$, which are not the same quantity and are easy to conflate under time pressure).

(c) $P$ is removed; a particle $Q$ of weight $X$ N is attached at $B$ instead. The rod remains horizontal and is now on the point of tilting. Find $X$. (4 marks)

Adding weight at $B$ — beyond $C$, away from $A$ — tends to lift the rod at $A$. Since a rope can only pull down, not push, the rope at $A$ is the one that goes slack first: tension at $A$ is zero, and the rod pivots about $C$. Taking moments about $C$ (distance from $C$ to the rod's centre of mass is $2.1-1.5=0.6$ m; distance from $C$ to $B$ is $CB=0.9$ m):

$$M(C):\quad 490\times0.6 = X\times0.9 \implies X = \frac{294}{0.9} = \frac{980}{3}\approx326.7 \text{ N}$$

Marks: B1 tension at $A$ is zero, stated or clearly implied by the following moments equation; M1 a complete, dimensionally-correct moments equation in $X$ alone about $C$ (scores nothing if the tension at $A$ is never treated as zero); A1 the correct unsimplified equation; A1 the final answer, $980/3$ or $326.7$ N (or better).

Examiner's report: many candidates "did not understand the significance of" the phrase "on the point of tilting" — this part was the least successful of the three. Most successful candidates correctly took moments about $C$; the majority of unsuccessful ones took moments about $A$ instead but — the report's own wording — "with the tension at $C$ still 600N or the particle $P$ still attached," carrying forward a force from an earlier part of the same question into a scenario where it no longer exists (the $600$ N tension belonged to part (b)'s different loading; $P$ was explicitly removed before part (c) even begins). This is the exact real-paper anchor for the "stale force carried forward" failure mode named, from a single quoted fragment, in codex-deep-dive-moments-rods-on-two-pivots.md's Case E — that document explicitly declined to reconstruct this question beyond the fragment; the full solution above is the first time it has been worked end to end anywhere in this course. A very small number of candidates additionally "used $X$ as a mass" — substituting a force, in newtons, into a place a mass, in kilograms, belonged.

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#Question 5 (7 marks) — position vectors, velocity, and a distance to a fixed target

Spec 2.1, 2.2. A ball $B$ is modelled as a particle moving with constant velocity across a horizontal surface. Its position vector is $(-2\mathbf i+5\mathbf j)$ m at $t=1$ s and $(10\mathbf i-3\mathbf j)$ m at $t=9$ s.

(a) Find the velocity of the ball. (3 marks)

$$\mathbf v = \frac{(10\mathbf i-3\mathbf j)-(-2\mathbf i+5\mathbf j)}{9-1} = \frac{12\mathbf i-8\mathbf j}{8} = 1.5\mathbf i-\mathbf j \ \left(\text{m s}^{-1}\right)$$

Marks: M1 displacement over time attempted, as a genuine difference of the two given position vectors; A1 the correct unsimplified expression; A1 the correct velocity, $1.5\mathbf i-\mathbf j$.

Examiner's report: generally well done, with most candidates correctly using a time difference of $8$ (not $9$, the second time value alone). A "significant number" found the correct velocity but then reported its speed, $\sqrt{1.5^2+1^2}=\sqrt{3.25}\approx1.80\text{ m s}^{-1}$ — the mark scheme applies no penalty for this, but the report flags it as a real example of needing to read exactly which quantity (velocity vs speed) a part is asking for, since the very same substitution error costs a mark elsewhere on this same paper (Q2(d)). A smaller number attempted a non-vector method entirely and could score nothing, since the question's data is only usable in $\mathbf i,\mathbf j$ form.

(b) The target's position vector is $(13\mathbf i-5\mathbf j)$ m. Find the distance of $B$ from the target at $t=7$ s. (4 marks)

Position at $t=7$: starting from the $t=1$ position and adding $6\mathbf v$ (six seconds further on):

$$(-2\mathbf i+5\mathbf j)+6(1.5\mathbf i-\mathbf j) = (-2+9)\mathbf i+(5-6)\mathbf j = 7\mathbf i-\mathbf j$$

Distance to the target: difference $= (13-7)\mathbf i+(-5-(-1))\mathbf j = 6\mathbf i-4\mathbf j$, magnitude $=\sqrt{6^2+4^2}=\sqrt{52}=2\sqrt{13}\approx7.21$ m.

Marks: M1 a complete method using the candidate's own velocity to find $B$'s position at $t=7$ (from either given position, adding the correct multiple of $\mathbf v$); A1 the correct unsimplified position, which reduces to $7\mathbf i-\mathbf j$; M1 Pythagoras used to find the distance between the candidate's own position and the target — available only if a genuine position vector for $B$ at $t=7$ was actually found first; A1 the final answer, $2\sqrt{13}$ or $7.21$ m (or better).

Examiner's report: found more challenging than part (a). Successful candidates used one of two equally valid routes to $B$'s position at $t=7$ — from the $t=1$ point (adding $6\mathbf v$) or from a freshly-found $t=0$ point (adding $7\mathbf v$) — with "a few" working backward from the $t=9$ position instead. The report is explicit that the distance-finding method mark depends on that position genuinely being found first — the mark scheme calls it available only "if a sensible method had been used to find the position" first. Pythagoras applied to the wrong pair of vectors (for instance, straight from the two given positions, ignoring $t=7$ entirely) does not earn it.

This is the exact real-paper source for the position-vector-then-distance-to-a-fixed-point technique resultant-forces-and-bearings.ts teaches on its own VERIDIAN-original numbers (fixed point $A=(17\mathbf i+10\mathbf j)$, not this question's $(13\mathbf i-5\mathbf j)$) — see that file for the deeper trap-taxonomy; this is the first time the real $B$/target numbers above have been carried through a finished, checked answer.

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#Question 6 (12 marks) — limiting friction on an incline, then a string that breaks

Spec 4.4, 5.2, 5.3. A particle $P$ of mass $m$ is held on a rough plane inclined at angle $\alpha<45°$ by a light string, itself inclined at angle $\alpha$ to the plane (in the vertical plane containing the line of greatest slope). With tension $0.75mg$, $P$ is on the point of moving up the plane.

(a) Find an expression for the friction force on $P$, in terms of $m$, $g$, $\alpha$. (3 marks)

Resolving along the plane (friction acts down the slope, opposing the impending upward motion):

$$0.75mg\cos\alpha = mg\sin\alpha+F \implies F = 0.75mg\cos\alpha-mg\sin\alpha$$

Marks: M1 resolution along the plane attempted, correct number of terms, every resolved term genuinely carrying a $\sin$ or $\cos$ factor; A1 the unsimplified expression, allowing at most one sign/trig error; A1 the fully correct expression above.

Examiner's report: generally well answered, though a few candidates, in the report's words, "did not appear to understand what was required" and never wrote down an expression for the friction at all — a real instance of a question asking for an unevaluated symbolic expression, rather than a numeric answer, catching candidates expecting the latter.

(b) Given $\mu=\tfrac12$, show that $\tan\alpha=\tfrac25$. (6 marks)

Resolving perpendicular to the plane — the string's tension has a component pulling $P$ away from the surface, reducing the normal reaction below $mg\cos\alpha$:

$$R = mg\cos\alpha-0.75mg\sin\alpha$$

At the point of moving, friction is at its limiting value, $F=\mu R = \tfrac12R$. Substituting both expressions:

$$0.75mg\cos\alpha-mg\sin\alpha = \frac12\left(mg\cos\alpha-0.75mg\sin\alpha\right)$$ $$0.75mg\cos\alpha-mg\sin\alpha = 0.5mg\cos\alpha-0.375mg\sin\alpha$$ $$0.25mg\cos\alpha = 0.625mg\sin\alpha \implies \tan\alpha=\frac{0.25}{0.625}=\frac25 \text{ (as required)}$$

Marks: M1 resolution perpendicular to the plane attempted; A1 the correct unsimplified equation for $R$; B1 $F=\tfrac12R$ used; M1 $F$ and $R$ eliminated to give one equation in $mg$ and $\alpha$ alone; A1 the correct equation; A1* the given result reached by a complete method — specifically, collecting the $\sin\alpha$ terms and the $\cos\alpha$ terms (or dividing through by $\cos\alpha$ to collect $\tan\alpha$ terms) before asserting the final line, since this is a "show that" mark and a jump straight to $\tfrac25$ earns nothing on its own.

Examiner's report: the reaction $R$ was generally found correctly, but the most common error was using $R=mg\cos\alpha$ — "ignoring the fact that the tension" also affects the reaction — treating this as an ordinary incline problem and forgetting the string's own perpendicular pull. Separately, some candidates "used $0.75$ instead of $0.75mg$" for the tension, or left it as the unresolved symbol $T$ — a bare number, or an unresolved symbol, substituted where a force with its own $mg$ factor belonged, breaking the equation's dimensional correctness exactly as the general M-mark rule in forge-exam-technique-guide.md §2 describes. As this is a "show that" question, the report stresses candidates needed to make clear use of $\tan\alpha=\sin\alpha/\cos\alpha$ and reach the given answer explicitly, not merely assert it.

(c) The string breaks. Determine whether $P$ remains at rest, justifying your answer. (3 marks)

With the string gone, the normal reaction reverts to $R'=mg\cos\alpha$, and the only force with a component down the slope is gravity, $mg\sin\alpha$. Using $\tan\alpha=\tfrac25$ (so, in a right triangle with opposite $2$, adjacent $5$, hypotenuse $\sqrt{29}$: $\sin\alpha=2/\sqrt{29}$, $\cos\alpha=5/\sqrt{29}$):

$$\text{Maximum available friction} = \mu R' = \frac12mg\cdot\frac{5}{\sqrt{29}} = \frac{5mg}{2\sqrt{29}} = \frac{5\sqrt{29}}{58}mg \approx 0.464mg$$ $$\text{Weight component down the slope} = mg\sin\alpha = \frac{2mg}{\sqrt{29}} = \frac{2\sqrt{29}}{29}mg \approx 0.371mg$$

Since $0.464mg>0.371mg$, the maximum friction available exceeds the force gravity supplies down the slope, so $P$ remains at rest.

Marks: M1 both the new maximum friction and the weight component down the slope found (condoning sign/trig confusion); A1 both expressions correct, with $\alpha$ substituted and an explicit numeric or algebraic comparison made (an inequality or a difference — words alone are not sufficient); A1 the correct conclusion, backed by that comparison.

Examiner's report: "a good discriminator," and "sometimes omitted" entirely. The single dominant way to score zero here despite full understanding elsewhere: many candidates "wrote an 'essay' with no calculations" to explain their conclusion — on a mechanics paper, a correct qualitative argument earns nothing without the two numbers it depends on, actually computed and compared. A separate group kept the part (a)/(b) friction expression with its tension term still attached, never removing it once the string had broken. The report also names the specific misconception behind the final lost mark: some candidates plainly did not understand limiting friction, believing friction could actively push the box up the plane — but friction is a resistive force; it can only oppose an object's actual or impending motion, never drive it.

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#Question 7 (15 marks) — connected particles over a pulley, then free motion

Spec 4.1, 4.2, 3.1. A light inextensible string connects particle $A$ (mass $2m$) to particle $B$ (mass $3m$) over a small smooth fixed pulley; both hang with the string's straight parts vertical, both held $0.5$ m above the ground. Released from rest, $B$ descends.

(a) Find the tension in the string, in terms of $m$ and $g$. (5 marks)

Equation of motion for each particle (each using its own direction of motion as positive, linked by the shared tension $T$ and shared acceleration $a$):

$$\text{For }A\text{ (rising):}\quad T-2mg=2ma \qquad \text{For }B\text{ (falling):}\quad 3mg-T=3ma$$

Adding the two equations eliminates $T$:

$$3mg-2mg = 5ma \implies a=\frac g5$$

Substituting back into $A$'s equation:

$$T = 2mg+2ma = 2mg+2m\left(\frac g5\right) = \frac{12mg}5 = 2.4mg$$

Marks: M1 a correct equation of motion for either particle or for the system as a whole, correct number of terms; A1 the correct equation; M1 a second equation of motion (for the other particle, or the system, whichever was not used first); A1 the correct second equation; A1 the final answer for $T$ — which must be expressed as a multiple of $mg$, not a decimal, since the masses are only ever given symbolically.

Examiner's report: the whole-system equation was "rarely seen or used" — nearly everyone built two separate per-particle equations instead, which is equally valid but requires an extra elimination step. Real, specific errors named: using $m$ instead of $2m$ or $3m$ for one particle's mass; omitting $g$ from a weight term while including it in the $ma$ term (or the reverse); and, most tellingly, "$T-2mg=3mg-T$" — an equation built from real, correctly-signed forces with no acceleration term anywhere in it, which (if solved) describes a system in equilibrium, not one actually accelerating. This is exactly the general M-mark rule forge-exam-technique-guide.md §2 already documents (a resolution or N2L equation must be dimensionally the right kind of statement, not merely built from correct-looking terms) shown live on this real question.

(b) Find the speed of $B$ the instant it reaches the ground. (4 marks)

$$v^2 = u^2+2as = 0^2+2\left(\frac g5\right)(0.5) = \frac{9.8}{5} = 1.96 \implies v = 1.4\text{ m s}^{-1}$$

Marks: B1 the acceleration, $g/5$, carried through from part (a); M1 a complete $suvat$ method using that acceleration; A1 the correct equation in $v$; A1 the final value, $v=1.4$ (exact — the report explicitly notes $7/5$ left unevaluated as a decimal does not itself earn this last mark).

Examiner's report: candidates who found the correct acceleration in part (a) "often went on to gain full marks" here — but "a common error (scoring no marks) was to use $g$ for the acceleration" instead of the smaller, already-found system value $g/5$ — reaching for the generic constant instead of the specific number the question's own earlier part just produced.

(c) $A$ does not reach the pulley and $B$ does not rebound. Find the total time from release to the instant $A$ first reaches a height of $1.06$ m above the ground. (6 marks)

(Full step-by-step marked solution, including the real examiner-report quotation on the "two parts to the motion" trap and a reconstructed common wrong path, already lives at content/veridian/question-bank/wme01.ts, entry wme01-jan25-q7c-two-stage-pulley-kinematics — not duplicated here.)

In brief: while the string is taut (stage 1), $A$ rises with $B$ at the shared acceleration $g/5$, covering the same $0.5$ m $B$ falls, reaching $A$'s starting height $+0.5=1.0$ m at speed $1.4\text{ m s}^{-1}$ after $t_1=0.5/0.7=5/7$ s. The instant $B$ lands, the string goes slack and $A$ — now a free particle — continues upward under gravity alone, needing a further $0.06$ m (not the full $0.56$ m) to reach $1.06$ m: $0.06=1.4t_2-4.9t_2^2$, giving $t_2=(5-\sqrt{10})/35 \approx0.0525$ s (the quadratic's other root, $\approx0.233$ s, is rejected — it would place $A$ back below the height it had already reached). Total time $=5/7+0.0525\approx0.767$ s (0.77 to 2 s.f.).

Examiner's report: "a large number failed to realise that there are two parts to the motion" and made no progress, using just one equation for the entire journey — the identical structural error connected-particles-pulleys-and-lifts.ts already teaches, on its own VERIDIAN-original numbers, as the "pulley-then-free-motion" trap, here confirmed as this real question's own anchor. Among those who did split the motion correctly, common wrong distances for the second stage were $0.56$ m (the total rise, not the remainder) or even $1.06$ m (the target height, mistaken for a distance).

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#5. Closing provenance note

Total marks re-confirmed by direct sum of the seven worked questions above: $8+11+12+10+7+12+ 15=75$, matching both the real question paper's own printed total and the real mark scheme's per-question tally independently — the same reconciliation already claimed as this series' selection justification in §1, now shown against the finished document rather than assumed in advance of writing it.

Every intermediate and final numeric value above was derived independently from the question's own given data, then checked against the real mark scheme's stated value before being written up — none was assumed or reverse-engineered from the answer alone. Two parts of this paper (Q1(b), Q7(c)) already carry a fuller, dedicated marked-solution treatment elsewhere in this course (content/veridian/question-bank/wme01.ts) and are deliberately kept to a condensed answer plus a pointer here, rather than restated — consistent with this document's own stated job of being genuinely additive, not a third copy of material this course already owns. No question, mark, or examiner-report figure in this document was invented, and no likelihood or grade-boundary claim is made anywhere above.