Connected particles — pulleys, pegs, lifts, and cars with trailers

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

Connected particles: one shared acceleration, one equation per particle

Spec item 4.2 covers "simple applications including the motion of two connected particles," and its own guidance names three distinct problem shapes worth learning as three separate pictures rather than one blurred idea: "(i) the motion of two connected particles moving in a straight line or under gravity when the forces on each particle are constant; problems involving smooth fixed pulleys and/or pegs may be set (ii) motion under a force which changes from one fixed value to another, e.g. a particle hitting the ground (iii) motion directly up or down a smooth or rough inclined plane." Two connectors recur across all three: a light inextensible STRING, which can only pull, and a light rigid ROD or tow-bar, which can push as well as pull. Both rely on the spec's own modelling vocabulary from item 1.1, which is worth being fluent in before any of this makes sense: particle, rigid body, rod (light, uniform, non-uniform), inextensible string, smooth and rough surface, light smooth pulley, peg.

Whatever the picture, the method is the same three-step routine, and it works because of two separate physical facts, not one. First: the connection is INEXTENSIBLE, so however the system moves, every connected particle covers the same distance in the same time as every other one — they share the same speed at every instant, and therefore the same acceleration, in MAGNITUDE. Not necessarily in labelled direction: over a pulley, one particle can be rising while the other falls. 'The same acceleration' means the same number, $a$ , once each particle's own direction of motion is taken as that particle's own positive direction. Second: the connector is LIGHT (massless), so Newton's second law applied to the connector itself — mass zero — forces the net force on it to be exactly zero, whatever it's doing. That is the reason a single length of string carries the same tension throughout its own length, and a rigid rod carries the same thrust or tension throughout its own length. Put the two facts together and the method is: draw every particle separately, choose that particle's own direction of motion as positive, write $F = ma$ for it alone using every force actually touching it, and solve the resulting equations — one per particle — in whichever order leaves the fewest unknowns first.

There is a genuine shortcut, and it's worth knowing exactly what it can and cannot do. Add every particle's own equation together and every internal tension or thrust cancels — it appears once with a $+$ sign and once with a $-$ sign across the whole system, because it pulls two connected particles toward each other. What's left is a single whole-system equation: (driving force) $-$ (resisting force) $=$ (total mass of every connected particle) $\times a$ . That's the fastest route to the common acceleration whenever no other unknown is wanted — but it can never hand you an internal tension or a reaction force between two particles, because those are exactly the forces that cancelled out of it. A question wanting both the acceleration and an internal force always needs at least one single-particle equation too, however tidy the system equation looks on its own.

Mechanism

Why the acceleration is shared but the tension isn't — unless it's the same string

Two physical facts do all the work here, and it matters that they're different facts. Inextensibility is a length constraint: however the system moves, the total length of a taut string (or the fixed geometry of a rigid rod) cannot change, which forces every particle joined by it to move exactly the same distance, in exactly the same time, as every other particle attached to the same connector — so their speeds match at every instant, and therefore so do their accelerations, in magnitude. That argument never mentions force. Masslessness is what produces the shared tension, and it only reaches as far as a single connector's own length: apply Newton's second law to the string (or rod) itself, treating it as a particle of mass zero, and the net force on it must be zero however it's accelerating — $F = ma$ with $m = 0$ forces $F = 0$ , regardless of $a$ . For a straight length of string this means the pull at one end must exactly balance the pull at the other, so the tension is the same number everywhere along that one string. A smooth pulley or peg doesn't break this argument, it just bends the string's direction without adding any resisting force along it — spec 1.1's own modelling vocabulary calls this a 'light smooth pulley' for exactly that reason — so the tension on the two sides of a single string over a single smooth pulley is still one number. But the argument was always about ONE connector's own length. A second, physically separate string runs its own independent zero-net-force argument along its own length, with no reason at all to produce the same number as the first string's — the two tensions are related only through whichever particle both strings happen to touch, via that particle's own equation of motion, and nothing shortcuts that. This is precisely the distinction a real examiner report's advice targets when it recommends drawing a separate diagram for each connected particle: one diagram per particle keeps each connector's own tension attached to the string it actually belongs to, instead of letting a single symbol T quietly do service for two different physical quantities.

Diagram — Two particles connected over a smooth pulley at the top of a rough inclined plane

x-axis: along the ground, left to right · y-axis: height above the ground

Particle B, on the rough plane: Mass m_B, resting on the plane's rough surface, connected to a string running up the line of greatest slope to the pulley at the top. Free to slide up or down the slope; no motion perpendicular to the plane, since it never leaves the surface.
Particle A, hanging freely: Mass m_A, hanging vertically once the string passes over the pulley at the top of the plane. Moves only vertically, sharing the same speed and acceleration magnitude as B at every instant, because the string is inextensible and the pulley is smooth.
Smooth fixed pulley at the top of the plane: Changes the string's direction from 'up the slope' to 'straight down' without adding any force along the string itself: a smooth peg or pulley exerts no friction on the string passing over it, so one continuous string carries the same tension T on both sides of it.

Tension T — on B, up the slope; on A, upward: A string can only pull, never push, so T always acts along the string, away from the particle it's pulling toward the pulley. One tension here, because it is one continuous string over one smooth pulley — this is NOT the two-tension case covered later in this lesson, which needs a second, separate string.
Weight component along the plane — m_B g sinθ, down the slope: Only the part of B's weight that lies along the slope drives motion along it; the rest is balanced by the normal reaction. Resolving is unavoidable on an incline, because 'weight' itself points straight down, neither along the slope nor perpendicular to it.
Normal reaction R = m_B g cosθ — perpendicular to the plane: Balances the perpendicular component of B's weight exactly, since B has no acceleration perpendicular to the plane. R feeds directly into the friction force below, because friction depends on how hard the two surfaces are pressed together, not on B's weight alone.
Friction F = μR — direction set by B's actual motion, not assumed: Friction always opposes relative motion, never assists it. Its direction has to be decided from which way B is actually moving (or about to move), never drawn first out of habit — get this backwards and every term downstream keeps whatever sign was chosen, silently.

Common error: Drawing friction acting in the same direction as B's motion, or fixing its direction on the diagram before working out which way B actually slides.

Correct: Decide B's direction of motion first — from the physics of the whole system, or from what the question states — then draw friction directly opposing that direction, every time, on every incline question.

examiner-report · Oct 2022 · Q3

Marked, line by line

Particle A, of mass 1 kg, hangs freely from a light inextensible string. The string passes over a smooth pulley fixed at the top of a rough plane inclined at angle θ to the horizontal, where sinθ = 3/5 and cosθ = 4/5, and is attached to particle B, of mass 3 kg, resting on the plane. The coefficient of friction between B and the plane is 0.25. The system is released from rest with the string taut. (a) Show that B slides down the plane rather than remaining in equilibrium. (b) Find the acceleration of the system. (c) Find the tension in the string. (d) Find the speed of B after it has moved 2 m down the plane. Take g = 9.8 m/s². (VERIDIAN-original question — not a reproduction of any past-paper question; structured after the real WME01 incline-and-pulley scenario type, spec 4.2(i) and (iii) together, with independently chosen numbers.)

11 marks available

(a) — 3 marks

01
$R = m_Bg\cos\theta = 3 \times 9.8 \times 0.8 = 23.52$ N, so $F_{max} = \mu R = 0.25 \times 23.52 = 5.88$ N.
Method mark for finding the normal reaction from perpendicular equilibrium and the maximum available friction from it — necessary before any claim about whether the system moves can be justified, not assumed from the picture.
M1
02
Compare the driving forces WITHOUT friction: B's weight component down the plane is $m_Bg\sin\theta = 3 \times 9.8 \times 0.6 = 17.64$ N, against A's weight of $m_Ag = 1 \times 9.8 = 9.8$ N pulling the other way through the string. Net driving force, before friction: $17.64 - 9.8 = 7.84$ N.
Method mark: compares the net force that would exist with no friction at all against the maximum friction available — the only valid way to justify a direction of motion, rather than assert it.
M1
03
$7.84 > 5.88$ , so friction cannot hold the system in equilibrium: B does slide down the plane, and A rises.
Accuracy/conclusion mark for the correct comparison and the correct conclusion. Reversing the comparison, or stating the direction from the picture without the numerical check, does not earn this mark even where the stated direction happens to be right.
A1

(b) — 4 marks

101
With B confirmed sliding down, friction takes its moving value $F = \mu R = 5.88$ N, up the plane (opposing B's motion). Equation of motion for B, down the plane positive: $m_Bg\sin\theta - T - F = m_Ba$ , i.e. $17.64 - T - 5.88 = 3a$ .
Method mark: B's own equation, correct number of terms — driving weight component, tension opposing it, friction opposing the motion, mass times a — and dimensionally consistent throughout.
M1
102
Equation of motion for A, upward positive (A's own direction of motion): $T - m_Ag = m_Aa$ , i.e. $T - 9.8 = a$ .
Method mark: A's own equation, independently correct — no friction term, since A is in free vertical motion.
M1
103
Eliminate T between the two equations: $17.64 - 5.88 - (9.8 + a) = 3a$ .
Dependent method mark for correctly combining the two single-particle equations to eliminate T — dependent on both method marks above already being earned.
dM1
104
$1.96 = 4a$ , so $a = 0.49$ m/s² (3 s.f.).
Accuracy mark, correct answer only.
A1

(c) — 2 marks

201
Substitute the value of a from part (b) into A's equation: $T = 9.8 + 0.49$ .
Follow-through method mark for substituting a value of a — correct or carried from an earlier slip — into a correctly-formed single-particle equation to find T.
M1 ft
202
$T = 10.29$ N (3 s.f.).
Accuracy mark, correct answer only — and positive, confirming the string is in tension throughout, exactly as the model assumed.
A1

(d) — 2 marks

301
Use $v^2 = u^2 + 2as$ with $u = 0$ (released from rest), $a = 0.49$ , $s = 2$ : $v^2 = 0 + 2 \times 0.49 \times 2 = 1.96$ .
Method mark for selecting and correctly substituting into the suvat equation matching the three known quantities and the one wanted. None of the five constant-acceleration equations appear in the M1 formula-booklet entry — the booklet's own M1 section states there are no formulae for M1 beyond what candidates are expected to know, and suvat is one of the results the spec itself lists as required knowledge, not a reference-sheet lookup.
M1
302
$v = 1.4$ m/s (exact, since $1.96 = 1.4^2$ ) — B's own speed, and by the string's inextensibility, A's speed too, at that same instant.
Accuracy mark, correct answer only.
A1

In your own words

In one sentence: why does a system with two separate strings over two separate pulleys generally have two different tensions, even though every particle in it shares the same acceleration magnitude?

Car and trailer: a connector that can push as well as pull

A car towing a trailer is connected by a tow-bar, not a string — the spec's 'rigid rod' (light, per spec 1.1's own vocabulary), which changes one thing about the method above: a rod can carry a THRUST (pushing its two ends apart) as well as a tension (pulling them together), where a string can only ever pull. The equations are written exactly the same way — one $F = ma$ per particle, sharing one acceleration — but the sign of the connector's own force at the end of the calculation now means something a pulley-and-string question never has to ask: is the tow-bar being stretched, or squeezed?

This matters concretely whenever the driving force changes value mid-question. Spec 4.2(ii)'s own wording is a force which changes from one fixed value to another, e.g. a particle hitting the ground — and a car's engine force is a textbook case of exactly that: constant while driving, then suddenly zero (or reversed, under braking) the instant the engine cuts or the brakes are applied. Solve for the tow-bar force again with the new driving value and its sign can flip: what was a tension while the car actively pulled the trailer forward can become a thrust once the car itself starts relying on the trailer to help slow the combined system, or vice versa.

One real trap survives across BOTH phases of a question like this, and it has nothing to do with tension versus thrust. If the road is inclined, every equation of motion — the whole-system one, and each single-particle one — needs a weight component resolved along the slope, in addition to whatever driving force or resistance is stated directly. A real examiner report on exactly this scenario type records candidates dropping those weight-component terms and still reaching a numerically plausible driving force, because the two omissions happened to cancel — which earned no credit at all, since a mark scheme checks the equation's own terms, not just its final number.

Same question, every valid method

A car of mass 900 kg tows a trailer of mass 300 kg up a straight road inclined at angle θ to the horizontal, where sinθ = 0.1, connected by a light rigid tow-bar. Constant resistances to motion of 100 N (car) and 50 N (trailer) act down the slope on each. The system accelerates at 0.2 m/s² up the slope under a driving force D from the car's engine. Take g = 9.8 m/s². Find D and the force in the tow-bar. (VERIDIAN-original question — not a reproduction of any past-paper question; structured after the real Oct 2019 WME01 Q3 car-and-trailer scenario type, with independently chosen numbers.)

2 valid methods · every one reaches D = 1566 N; tow-bar force = 404 N (tension). · 5 marks available

01
Whole system, resolving along the slope, positive up the slope: $D - R_c - R_t - (M_c + M_t)g\sin\theta = (M_c + M_t)a$ .
Method mark: whole-system equation of motion, correct number of terms — the general principles for mechanics marking require an equation to have the correct number of terms and be dimensionally consistent before it earns credit.
M1
02
$D - 100 - 50 - 1200 \times 9.8 \times 0.1 = 1200 \times 0.2$ , i.e. $D - 1326 = 240$ .
Accuracy mark for correct substitution, including BOTH weight components combined via the total mass. Omitting that weight term is the exact omission a real Oct 2019 examiner report records on this scenario type — not a resistance term, specifically the weight components.
A1
03
$D = 1566$ N.
Accuracy mark, correct answer only for D.
A1
04
Isolate the trailer alone, now that a is known: $T - R_t - M_tg\sin\theta = M_ta$ , i.e. $T - 50 - 300 \times 9.8 \times 0.1 = 300 \times 0.2$ .
Method mark for a single-particle equation, applied once a is already known — this is the efficient move examiner reports repeatedly flag: solve whichever equation has fewer unknowns first, then reuse the result.
M1
05
$T - 50 - 294 = 60$ , so $T = 404$ N.
Accuracy mark, correct answer only for the tow-bar force. Positive, so it's a tension — the tow-bar is pulling the trailer forward.
A1

Always the faster route once the whole-system equation delivers the acceleration (or, as here, confirms a given one is consistent with an unknown driving force) directly. It also isolates the arithmetic: get D wrong somewhere and the trailer step below still earns its own method mark, since it depends only on the given a, never on the value of D at all.

Worked, in full

The driving force is suddenly removed — the new deceleration and the new tow-bar force (spec 4.2(ii): motion under a force which changes from one fixed value to another; VERIDIAN-original scenario, continuing the car-and-trailer numbers above)

01
Read the change precisely. The same car and trailer, same masses, same slope, same resistances — only D changes, from 1566 N to 0. Nothing about the geometry or the resistances is assumed to change with it, so every other term in the whole-system equation carries over unchanged.
Earns: Nothing gradeable yet — but stating this explicitly is exactly what the 'omission of the weight components... fortuitously led to a correct value' failure mode from earlier warns against, from the other direction: skipping past which terms genuinely change and which don't is how a mark is lost even when a final number happens to look plausible.
02
Whole-system equation, D now zero: $-R_c - R_t - (M_c + M_t)g\sin\theta = (M_c + M_t)a_2$ , so $-100 - 50 - 1176 = 1200a_2$ , giving $-1326 = 1200a_2$ .
Earns: M1 — the same whole-system method as before, re-applied with the one term that changed set to its new value.
03
$a_2 = -1.105$ m/s² (3 s.f.) — negative, meaning the system decelerates. It is still moving up the slope at the instant D is removed, just slowing down under gravity and resistance alone, not reversing direction.
Earns: A1 — correct value, and correct interpretation of the sign as deceleration rather than a reversal of motion.
04
Isolate the trailer to find the new tow-bar force F (same equation shape as before, new $a_2$ ): $F - R_t - M_tg\sin\theta = M_ta_2$ , so $F - 50 - 294 = 300 \times (-1.105)$ , giving $F - 344 = -331.5$ .
Earns: M1 — single-particle equation, correctly re-used with the new acceleration.
05
$F = 12.5$ N — small, but still positive: the tow-bar is still in tension, still pulling the trailer forward, even with no engine force at all. That isn't a coincidence: the trailer's own resistance-plus-weight per unit mass, $(50 + 294)/300 \approx 1.15$ m/s², is slightly larger than the car's, $(100 + 882)/900 \approx 1.09$ m/s² — left alone, the trailer would naturally decelerate faster than the car, so the tow-bar has to keep pulling it along to hold both at the one shared deceleration $a_2$ .
Earns: A1 — correct final value. A different pair of masses or resistances can flip this result to a negative (thrust) tow-bar force instead; the sign isn't fixed by which phase of the question you're in, only by the actual numbers on each side of the connector.

Lift with two occupants: one whole-system equation, one equation per person

A lift carrying passengers is a connected-particles system even though nothing about it looks like a pulley or a slope. The cable is one connector, always in tension (cables can't push, so a lift cable behaves exactly like the string in the pulley scenarios above); each occupant is connected to the lift not by a string but by ordinary contact with the floor, through a normal reaction force. That reaction is a genuine Newton's-third-law pair: the floor pushes up on each occupant with some force N, and each occupant pushes down on the floor with the same force N in return — which means N belongs in TWO different equations, once upward in that occupant's own equation of motion, and once downward in the lift's own equation of motion, whenever the lift is isolated on its own.

The whole-system shortcut still applies, and it's the fast route to the cable tension: treat the lift and every occupant inside it as one combined mass, since each occupant's floor contact force is exactly the kind of internal force that cancels out of a whole-system equation — it appears once pushing up on an occupant and once pushing down on the lift, and adding every particle's own equation together cancels both. What the shortcut CANNOT give you is any single occupant's own floor reaction; for that, isolate that one occupant alone, the same move as isolating any single particle in any other connected system.

Two real, independently-confirmed traps sit either side of this shortcut. Using the lift's own mass alone in the whole-system equation understates the total mass the cable actually has to support and accelerate — the occupants' weight doesn't stop mattering just because they aren't part of the lift's own structure. Isolating the lift on its own and forgetting the occupants' reaction forces makes the same mistake from the opposite equation: the lift's own weight is not the only downward force on it once people are standing inside it. Both routes, worked through with real numbers below, land on exactly the same wrong total — the clearest evidence that they are one underlying error, not two.

Marked, line by line

A lift cage of mass 400 kg carries two people, of mass 70 kg and 50 kg, standing on its floor. The lift accelerates upward from rest at 0.5 m/s². (a) Find the tension in the cable. (b) Find the force the floor exerts on the 70 kg person. (c) Find the force the floor exerts on the 50 kg person. Take g = 9.8 m/s². (VERIDIAN-original question — not a reproduction of any past-paper question; structured after the real WME01 lift-with-two-occupants scenario type, confirmed independently in two series, Oct 2022 Q4 and Jan 2023 Q7, with independently chosen numbers.)

7 marks available

(a) — 3 marks

01
Treat the lift and both occupants as one system, total mass $400 + 70 + 50 = 520$ kg, since the floor's internal contact forces cancel out of a whole-system equation.
Method mark for a whole-system equation using the FULL combined mass, not the lift's own mass alone. A real examiner report on this exact scenario type records candidates using the mass of the lift alone instead of the whole-system mass — that substitution alone is enough to lose this mark, whatever happens afterward.
M1
02
$T - 520g = 520 \times 0.5$ , so $T - 5096 = 260$ .
Accuracy mark for correct substitution: 520g = 5096 N (weight of the whole system) and 520 × 0.5 = 260 N (mass times a, whole system).
A1
03
$T = 5356$ N (3 s.f.).
Accuracy mark, correct answer only.
A1

(b) — 2 marks

101
Isolate the 70 kg person alone. The only vertical forces on them are their own weight, 70g downward, and the normal reaction $N_A$ from the lift floor, upward — the cable tension never touches them directly, only through the floor.
Method mark for a correctly-isolated single-particle equation: exactly two forces, weight and the floor's own reaction, nothing else.
M1
102
$N_A - 70g = 70 \times 0.5$ , so $N_A = 70(9.8 + 0.5) = 70 \times 10.3 = 721$ N (3 s.f.).
Accuracy mark, correct answer only. Note $N_A > 70g$ : the accelerating floor pushes up harder than it would at rest — the everyday 'heavier in an accelerating lift' feeling is exactly this number.
A1

(c) — 2 marks

201
Same method, the 50 kg person's own equation: $N_B - 50g = 50 \times 0.5$ .
Follow-through method mark: identical structure to part (b), applied to the second person's own mass.
M1 ft
202
$N_B = 50 \times 10.3 = 515$ N (3 s.f.).
Accuracy mark, correct answer only.
A1

Complete it yourself

Complete the chain — the two tensions in a three-particle, two-pulley system

01
Q (mass 4 kg) rests on a rough horizontal table (μ = 0.5). A light inextensible string attached to Q's left side passes over a smooth pulley at the table's left edge, hanging down to particle P (mass 1 kg). A SECOND, separate light inextensible string attached to Q's right side passes over a smooth pulley at the table's right edge, hanging down to particle R (mass 5 kg). Both strings are taut; the system is released from rest. Two separate strings mean two separate tensions to expect from the outset — $T_1$ in the P–Q string, $T_2$ in the Q–R string — and a real examiner report on exactly this kind of two-tension setup names the fix directly: drawing a separate diagram for each of P and Q would have helped some candidates set up these equations correctly. Draw all three particles separately before writing anything down.
02
R (5 kg) outweighs P (1 kg), so expect R to fall, dragging Q to the right and lifting P upward — set each particle's own positive direction to match its own actual motion. Start with P: upward positive, $T_1 - m_Pg = m_Pa$ , so $T_1 - 9.8 = a$ (using $m_P = 1$ ).

Named traps

car-trailer-weight-components-dropped: Verified on a real car-and-trailer question on an incline: "the most common mistake was omission of the weight components which, if done consistently, fortuitously led to a correct value of D but this received no credit due to the missing terms in the two equations" (Oct 2019, Q3). The word "fortuitously" is doing real work — dropping both weight components in a consistent way can make two errors cancel and still produce a plausible-looking final number, which is exactly why a mark scheme built from M/A/B codes checks that every term that should be in an equation is actually there, independent of whether the final number happens to come out right.
peg-pulley-unfamiliar-two-tension-terms: Verified on a real two-tension peg/pulley question: "many seemed unfamiliar with this type of situation involving two tensions" (Jun 2024, Q3), with the specific error named as "including an extra 3mg term in their system equation of motion or for missing the 3mg term when giving the equation for Q." The fix the same report names directly: "drawing a separate diagram for each of P and Q would have helped some candidates to set up these equations correctly."
peg-pulley-given-tension-substituted-wrong-place: Verified on the same question: "some responses showed confusion between the T value given and the T value to be found, resulting in the substitution of T = 3mg into the wrong place" (Jun 2024, Q3). Two different unknowns sharing the letter T — or a value of T handed to you partway through a multi-part question, then needed again for something new — is a labelling trap as much as a mechanics one: rename a value the moment it stops being unknown, rather than letting one symbol carry two different meanings.
lift-mass-of-lift-alone: Verified: candidates "using the mass of the lift alone... instead of the whole-system" mass when writing the equation of motion for a lift carrying occupants (Jan 2023, Q7).
lift-occupant-reactions-omitted: Verified, the same family of error seen from the other equation: "many candidates attempted an equation of motion for the lift but omitted the reactions from the two occupants" (Oct 2022, Q4). Confirmed as a repeat Pearson question type across two separate series, not a one-off — and, worked through with real numbers, both versions of the mistake collapse onto exactly the same missing terms (see the marked solution above).
incline-friction-direction-reversed: Verified, the single most consistently reported error on this exact rough-incline scenario type: "the most common error was having the frictional force acting down rather than up the plane" (Oct 2022, Q3). Friction opposes actual (or impending) relative motion — never gravity, never "the picture" — so its direction is the one force on an incline diagram that has to be reasoned out LAST, after the direction of motion is already settled, not drawn first out of habit.
incline-changed-acceleration-not-recognised: Verified on a two-part incline question where the force condition changed between parts: "a substantial number did not recognise that the system was subject to a different acceleration" in the second part (Jun 2024, Q6). An acceleration found in an earlier part of a question is not a constant carried forward automatically — any change to a force in the system (a different driving force, friction switching on or off, a changed angle) generally means a genuinely new equation of motion and a genuinely new value of a, exactly the discipline spec 4.2(ii)'s own guidance — a force which changes from one fixed value to another — is testing directly.

Retrieval — with feedback on every choice

Question 1

2 marks

A car (mass 800 kg) tows a trailer (mass 200 kg) along a horizontal road with driving force D. Resistances totalling 150 N act on the whole system. The system accelerates at 0.5 m/s². Which route gives D fastest, using the fewest unknowns?

Question 2

2 marks

A parcel Q of mass 6 kg rests on a smooth horizontal table. Q is attached by a string over a pulley at one edge to a hanging particle P, and by a SEPARATE string over a pulley at the opposite edge to a hanging particle R. The tension in the P–Q string is found to be 20 N. Can you conclude anything about the tension in the Q–R string without further working?

Question 3

2 marks

A particle is held at rest on a rough plane inclined at 30° by a string running up the line of greatest slope to a fixed point at the top. The string is cut, and the particle immediately begins to slide down. At the instant just after the string is cut, in which direction does friction act, and does its value change from the instant just before?

Question 4

2 marks

A lift of mass 500 kg carries one passenger of mass 80 kg and accelerates upward at 0.4 m/s². Which is a correctly-formed equation for finding the cable tension T?

Question 5

2 marks

A car and trailer, connected by a rigid tow-bar, are decelerating because the driving force has been removed. Solving the trailer's equation of motion gives a NEGATIVE value for the tow-bar force, where positive was defined as tension (pulling the trailer forward). What does this negative value mean?

Reference — not a study method, a lookup

One string, one smooth pulley/peg: same tension both sides, same |a| for every connected particle.
Two separate strings in one system: two different tensions — draw each particle separately.
Incline: resolve along the slope (mg sinθ) and perpendicular (R = mg cosθ); friction = μR, opposing actual motion.
Whole-system equation: driving − resisting = (total mass) × a. Fast for a or D; isolate one particle for an internal force.
Lift with occupants: whole-system mass for the cable tension; each occupant's own equation for their own floor reaction.
g = 9.8; round final numerical answers to 2 or 3 s.f.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme, examiner report, or the specification in this lesson was independently verified against the primary Pearson document (WME01-verified-facts.md, itself checked against the downloaded PDFs) before being included here — none of it was carried over from prior course material, since no prior WME01 material exists anywhere in this course. Every question in this lesson — prequestion, marked solutions, method comparison, worked chains, chain drill, and MCQs alike — is VERIDIAN-original: no scenario, number, or wording reproduces a real Pearson question. The research bank confirms four genuine scenario families and multiple genuinely quoted student errors for this spec point, but supplies no complete numeric scenario for any of them; the one real numeric fragment that does survive, T = 3mg from a Jun 2024 report on a differently-numbered version of the two-tension peg/pulley question, is quoted directly in the trap taxonomy and chain drill above rather than folded into this lesson's own worked numbers. Every mass, angle, coefficient of friction, and force value below was chosen and hand-checked against independent arithmetic, not transcribed from a real mark scheme; the mark-scheme CONVENTIONS attached to them (what M, A, B, and dM mean, the g = 9.8-and-round convention, the specific mechanics-marking principles about weight components and resolution terms) are verbatim-verified from real WME01 mark schemes via WME01-verified-facts.md §3.

Question 12 marks

A car (mass 800 kg) tows a trailer (mass 200 kg) along a horizontal road with driving force D. Resistances totalling 150 N act on the whole system. The system accelerates at 0.5 m/s². Which route gives D fastest, using the fewest unknowns?

The whole-system equation, $D - 150 = 1000 \times 0.5$ , since D is the only unknown in it once the total mass and total resistance are known.
Correct. Every internal tow-bar force cancels out of the whole-system equation, leaving D as the single unknown against a known total mass and total resistance.
BThe car's equation alone, $D - T - (\text{car's share of resistance}) = 800 \times 0.5$ , solved directly.
This equation has TWO unknowns, D and T — it can't be solved alone, whatever the car's share of the resistance is assumed to be.
CThe trailer's equation alone, $T - (\text{trailer's share of resistance}) = 200 \times 0.5$ , reading D off it.
The trailer's own equation doesn't contain D at all — D only acts on the car. This equation gives T, not D, however it's solved.
DBoth particles' equations solved simultaneously — there's no way to find D without that.
T does need both equations (or the shortcut used with one isolated particle) — but D specifically comes straight out of the whole-system equation on its own, faster than setting up two equations and combining them.

Traps tested: Single particle equation with two unknowns treated as solvable · Force attributed to the wrong particles equation · Unnecessarily treats simultaneous solving as the only route

Question 22 marks

A parcel Q of mass 6 kg rests on a smooth horizontal table. Q is attached by a string over a pulley at one edge to a hanging particle P, and by a SEPARATE string over a pulley at the opposite edge to a hanging particle R. The tension in the P–Q string is found to be 20 N. Can you conclude anything about the tension in the Q–R string without further working?

No — the two strings are physically separate, so their tensions are two independent unknowns until Q's own equation of motion (and P's and R's) is actually solved.
Correct. Nothing links two different strings' tensions except the shared particle's own equation of motion, which hasn't been used yet in this question.
BYes — it must also be 20 N, since Q is a single particle and both strings pull on the same object.
A single particle can have several different forces of different sizes acting on it at once. Being attached to the same particle guarantees nothing about two of those forces being equal.
CYes — the smooth pulleys guarantee both tensions are equal to each other.
Smoothness only equalises tension across a single pulley for the SAME string; it never links two physically different strings.
DYes, but only if P and R have equal masses.
Even equal masses for P and R don't force the two tensions to match — Q also has its own mass and inertia feeding into the system, and (on a rough table) its own friction feeding asymmetrically into each string's own equation.

Traps tested: Single particle forces assumed equal · Smooth pulley property applied across different strings · Equal masses assumed to force equal tensions

Question 32 marks

A particle is held at rest on a rough plane inclined at 30° by a string running up the line of greatest slope to a fixed point at the top. The string is cut, and the particle immediately begins to slide down. At the instant just after the string is cut, in which direction does friction act, and does its value change from the instant just before?

Friction still acts up the plane, opposing the new downward sliding — but its value can change: held at rest it needed only an equilibrium value up to μR, whereas once sliding it takes the fixed value μR exactly.
Correct. Direction is unchanged because the particle would slide down in both cases; the value changes because F = μR only holds once the particle is actually moving — before that, static friction supplies only whatever is needed to prevent motion, up to the limiting value.
BFriction acts down the plane once the particle starts moving, since it now moves with gravity instead of against it.
Friction opposes actual motion, full stop. Sliding down the slope still gets friction acting up the slope, whatever role gravity is playing.
CFriction stays exactly the same value and direction throughout, since μ and R haven't changed.
R (from perpendicular equilibrium) is indeed unchanged, but the VALUE of friction is not always μR — only once the particle is actually moving is friction forced to equal μR exactly. Held at rest by the string, it could have been any value up to that limit, not necessarily the maximum.
DFriction's direction can't be determined without knowing the coefficient of friction.
Direction depends only on which way the particle is (or would be) moving; the coefficient decides the MAGNITUDE, not the direction.

Traps tested: Friction assumed to always act downward · Static friction assumed always at the limiting value · Friction direction confused with friction magnitude

Question 42 marks

A lift of mass 500 kg carries one passenger of mass 80 kg and accelerates upward at 0.4 m/s². Which is a correctly-formed equation for finding the cable tension T?

$T - 580g = 580 \times 0.4$ , using the combined mass of the lift and the passenger together.
Correct — the cable has to support and accelerate both the lift and everyone inside it.
B $T - 500g = 500 \times 0.4$ , using the mass of the lift alone.
This is exactly the confirmed real trap: the passenger's weight doesn't stop mattering to the cable just because they aren't structurally part of the lift. Leaving their mass out understates both the weight term and the mass-times-a term.
C $T - 500g - N = 500 \times 0.4$ , where N is the passenger's reaction on the lift floor, without a separate equation for the passenger to find N.
This equation's shape (lift alone, including the reaction) is legitimate in principle, but N is now a new unknown with nothing to solve it — without the passenger's own equation, this can't actually be used to find T.
D $T - 80g = 80 \times 0.4$ , using the passenger's mass alone.
This is the passenger's OWN equation, which finds the floor's reaction on the passenger — not an equation for T at all, since T never touches the passenger directly.

Traps tested: Whole system mass understated by excluding an occupant · Reaction force included but not solved for · Force attributed to the wrong particles equation

Question 52 marks

A car and trailer, connected by a rigid tow-bar, are decelerating because the driving force has been removed. Solving the trailer's equation of motion gives a NEGATIVE value for the tow-bar force, where positive was defined as tension (pulling the trailer forward). What does this negative value mean?

The tow-bar is in thrust, pushing the trailer forward rather than pulling it — which a rigid rod can do, unlike a string.
Correct. A rigid connector is precisely the case where a negative (compressive) result is physically meaningful, not a sign of a mistake.
BThe calculation must contain an error, since a tow-bar force can never be negative.
A negative result flags an error for a STRING (which can't push, so a negative tension would mean it's gone slack) — but a rigid rod genuinely can push, so a negative sign here is a real, meaningful answer, not evidence of a mistake.
CThe trailer has become detached from the car at this instant.
A compressive (thrust) internal force implies nothing about the connector failing — a rigid rod under thrust is still fully connected and doing exactly its job.
DThe negative sign means the trailer is now moving backward.
The sign describes the INTERNAL FORCE convention (tension vs thrust in the tow-bar) — a separate question from which way the trailer is actually travelling. It can still be moving forward while decelerating, with the tow-bar in thrust.

Traps tested: Negative result assumed to always be an error · Compressive internal force mistaken for detachment · Internal force sign confused with velocity direction

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Oct 2022 · Q3 — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WME01.

Connected particles — pulleys, pegs, lifts, and cars with trailers

Connected particles: one shared acceleration, one equation per particle

Why the acceleration is shared but the tension isn't — unless it's the same string

Car and trailer: a connector that can push as well as pull

The driving force is suddenly removed — the new deceleration and the new tow-bar force (spec 4.2(ii): motion under a force which changes from one fixed value to another; VERIDIAN-original scenario, continuing the car-and-trailer numbers above)

Lift with two occupants: one whole-system equation, one equation per person

Complete the chain — the two tensions in a three-particle, two-pulley system

Deep dive

Practice this for real