Connected Particles and Pulleys

This lesson covers problems involving particles connected by strings passing over pulleys, as required by the Edexcel 9MA0 specification. These problems combine Newton's Second Law with the constraints imposed by an inextensible string.

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Key Assumptions

Before solving pulley problems, state the standard assumptions:

• The string is light (massless) — so the tension is the same throughout the string.
• The string is inextensible — so both particles have the same magnitude of acceleration.
• The pulley is smooth — so the tension is the same on both sides of the pulley.
• Particles are modelled as point masses.

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Atwood's Machine (Vertical Pulley)

Two particles of masses m₁ and m₂ (where m₁ > m₂) are connected by a light inextensible string passing over a smooth fixed pulley.

The heavier mass m₁ accelerates downwards, and the lighter mass m₂ accelerates upwards, both with the same acceleration a.

Setting Up the Equations

For m₁ (taking downwards as positive for this particle): m₁g - T = m₁a ... (1)

For m₂ (taking upwards as positive for this particle): T - m₂g = m₂a ... (2)

Solving

Add equations (1) and (2): m₁g - m₂g = (m₁ + m₂)a a = (m₁ - m₂)g / (m₁ + m₂)

Substitute back to find T: T = 2m₁m₂g / (m₁ + m₂)

Example: m₁ = 5 kg, m₂ = 3 kg, g = 9.8 m s⁻².

a = (5 - 3)(9.8) / (5 + 3) = 19.6 / 8 = 2.45 m s⁻² T = 2(5)(3)(9.8) / (5 + 3) = 294 / 8 = 36.75 N

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Particle on a Table with Hanging Particle

A particle A (mass m₁) rests on a smooth horizontal table. A light inextensible string passes over a smooth pulley at the edge of the table and connects to a hanging particle B (mass m₂).

Setting Up the Equations

For A (horizontally, in the direction of motion): T = m₁a ... (1)

For B (vertically, taking downwards as positive): m₂g - T = m₂a ... (2)

Solving

Add (1) and (2): m₂g = (m₁ + m₂)a a = m₂g / (m₁ + m₂)

Substitute back: T = m₁m₂g / (m₁ + m₂)

Example: m₁ = 4 kg (on table), m₂ = 6 kg (hanging), g = 9.8 m s⁻².

a = 6(9.8) / (4 + 6) = 58.8 / 10 = 5.88 m s⁻² T = 4(6)(9.8) / (4 + 6) = 235.2 / 10 = 23.52 N

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Inclined Plane with Pulley

Particle A (mass m₁) is on a smooth plane inclined at angle θ. A string passes over a pulley at the top of the plane and connects to a hanging particle B (mass m₂).

Assume B moves downwards and A moves up the plane.

Setting Up the Equations

For A (up the plane is positive): T - m₁g sin θ = m₁a ... (1)

For B (downwards is positive): m₂g - T = m₂a ... (2)

Solving

Add (1) and (2): m₂g - m₁g sin θ = (m₁ + m₂)a a = (m₂ - m₁ sin θ)g / (m₁ + m₂)

Exam Tip: If a comes out negative, it means the assumed direction of motion was wrong — A actually slides down the plane and B rises.

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What Happens When the String Breaks or Goes Slack

If the string breaks at some point during the motion:

1. Before the break: Use the connected-particle model to find velocity at the moment of the break.
2. After the break: Each particle moves independently. The tension becomes zero, and you apply Newton's Second Law to each particle separately.

Example: In an Atwood's machine, after the string breaks at time t₀, m₁ falls freely under gravity (a = g downward) and m₂ decelerates then falls (a = g downward, starting from whatever velocity it had at t₀).

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Multi-Stage Problems

Some problems involve two stages:

1. Stage 1: The system moves under the connected-particle model until something changes (e.g. a particle hits the ground).
2. Stage 2: The remaining particle continues with changed forces.

Use SUVAT at the end of Stage 1 to find the velocity, then use this as the initial velocity for Stage 2.

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Three Connected Particles

If three particles are connected in a line (e.g. A—B—C on a surface), there are two different tensions in the two sections of string. Apply Newton's Second Law to each particle and to the whole system.

For the whole system: resultant external force = total mass x acceleration. For individual particles: this gives you the internal tensions.

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Summary

• For pulley problems, apply Newton's Second Law to each particle separately.
• Light, inextensible string and smooth pulley means: same tension throughout, same magnitude of acceleration.
• Use the whole system to find acceleration; individual particles to find tensions.
• If a negative acceleration results, the direction of motion is opposite to what was assumed.
• When strings break or particles hit the ground, split the problem into stages.

A-Level Deep Dive: Connected Particles and Pulleys

Spec mapping

Edexcel 9MA0-03 specification, Paper 3 — Statistics and Mechanics, Section 10 (Forces and Newton's laws), sub-strand 10.3 covers newton's second law for motion in a straight line for bodies of constant mass moving under the action of constant forces; extension to situations where forces need to be resolved (restricted to two dimensions); connected particles, including problems involving smooth pulleys (refer to the official specification document for exact wording). This sits alongside section 8 (Kinematics — SUVAT) and section 11 (Moments) and is examined exclusively on Paper 3, where mechanics shares the paper with statistics. Connected-particle systems are the canonical Newton-II problem on 9MA0: they appear nearly every series, usually as an 8–11 mark question combining setup, equations, solution and a final assumption-criticism part. The Edexcel formula booklet provides no help here — the modelling assumptions (light, inextensible, smooth) and the equations of motion must be reconstructed from first principles each time.

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Worked example with full mark scheme

Question (8 marks):

Two particles $A$A and $B$B, of masses $3\,\text{kg}$3kg and $5\,\text{kg}$5kg respectively, are connected by a light inextensible string. The string passes over a small smooth pulley fixed at the edge of a smooth horizontal table. Particle $A$A rests on the table; particle $B$B hangs vertically below the pulley. The system is released from rest with the string taut.

(a) Find the acceleration of the system and the tension in the string. (6)

(b) State one modelling assumption you have used and explain how the answer would change if it were not made. (2)

Solution with mark scheme:

(a) Step 1 — draw a force diagram and choose positive directions.

For $A$A (on the table), positive direction is horizontally toward the pulley. The horizontal force on $A$A is the tension $T$T. (The weight $3g$3g and the normal reaction $R$R are vertical and balance, since the table is horizontal.)

For $B$B (hanging), positive direction is downward. The forces are weight $5g$5g down and tension $T$T up.

M1 — resolving forces and identifying $T$T as the same tension on both particles. This is only valid because the string is light and inextensible and the pulley is smooth — students who introduce $T_1$T1​ on one side and $T_2$T2​ on the other lose this M1.

Step 2 — Newton's second law for each particle.

For $A$A:

$T = 3a$T=3a

For $B$B:

$5g - T = 5a$5g−T=5a

M1 A1 — one mark for each correct equation of motion. Take $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2.

Step 3 — solve simultaneously.

Add the two equations to eliminate $T$T:

$5g = 8a \implies a = \dfrac{5g}{8} = \dfrac{5 \cdot 9.8}{8} = 6.125\,\text{m s}^{-2}$5g=8a⟹a=85g​=85⋅9.8​=6.125m s−2

M1 A1 — method mark for valid simultaneous solution and accuracy mark for the correct numerical acceleration. Acceptable answer: $a = 6.125\,\text{m s}^{-2}$a=6.125m s−2 or $\tfrac{5g}{8}\,\text{m s}^{-2}$85g​m s−2 in exact form.

Substitute back into $T = 3a$T=3a:

$T = 3 \cdot 6.125 = 18.375\,\text{N}$T=3⋅6.125=18.375N

A1 — final accuracy mark for tension to 3 sf: $T \approx 18.4\,\text{N}$T≈18.4N (or $\tfrac{15g}{8}\,\text{N}$815g​N exact).

(b) Assumption discussion.

B1 — naming a valid assumption: e.g. "the string is inextensible, so both particles have the same magnitude of acceleration".

B1 — explaining the consequence: e.g. "if the string stretched, $A$A and $B$B would have different accelerations, and the single value of $a$a used in both equations of motion would no longer be valid". Equivalent answers naming "light string" (so tension is constant along the string), "smooth pulley" (so tension is the same on both sides), or "smooth table" (so no friction on $A$A) score full marks if the consequence is stated.

Total: 8 marks (M3 A3 B2).

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Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): Two particles $P$P (mass $2\,\text{kg}$2kg) and $Q$Q (mass $4\,\text{kg}$4kg) are connected by a light inextensible string passing over a smooth fixed pulley. The particles hang freely on either side and the system is released from rest. Take $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2.

(a) Find the acceleration of $Q$Q and the tension in the string. (4)

(b) Find the speed of $Q$Q after it has descended $0.5\,\text{m}$0.5m, assuming neither particle has reached the pulley or the ground. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO3.1b — modelling) — equations of motion: $T - 2g = 2a$T−2g=2a for $P$P (upward positive) and $4g - T = 4a$4g−T=4a for $Q$Q (downward positive).
• M1 (AO1.1b — procedure) — adding to eliminate $T$T: $2g = 6a$2g=6a.
• A1 (AO1.1b) — $a = \tfrac{g}{3} = \tfrac{9.8}{3} \approx 3.27\,\text{m s}^{-2}$a=3g​=39.8​≈3.27m s−2.
• A1 (AO1.1b) — $T = 2(g + a) = 2g + 2a = 2(9.8) + 2(3.27) \approx 26.1\,\text{N}$T=2(g+a)=2g+2a=2(9.8)+2(3.27)≈26.1N (exact $\tfrac{8g}{3}$38g​).

(b)

• M1 (AO3.1a — selecting model) — using SUVAT $v^2 = u^2 + 2as$v2=u2+2as with $u = 0$u=0, $s = 0.5$s=0.5, $a \approx 3.27$a≈3.27 from (a).
• A1 (AO1.1b) — $v^2 = 2(3.27)(0.5) = 3.27$v2=2(3.27)(0.5)=3.27, so $v \approx 1.81\,\text{m s}^{-1}$v≈1.81m s−1.

Total: 6 marks split AO1 = 4, AO3 = 2. AO3 marks recognise that the candidate has selected an appropriate kinematic equation given a constant acceleration found from a Newton-II model — the synoptic Mechanics skill.

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Synoptic links

Connects to:

1. Section 10.1 — Newton's laws of motion: the entire topic is an application of $F = ma$F=ma to two particles linked by a constraint. The constraint (inextensible string) is what forces both particles to share the same acceleration magnitude, reducing two unknowns to one.
2. Section 10.4 — Friction: if the table in the worked example were rough, the equation for $A$A becomes $T - \mu R = 3a$T−μR=3a where $R = 3g$R=3g is the normal reaction. The pulley algebra is identical; only the force budget on the table-side particle changes.
3. Section 10.5 — Modelling assumptions: "light, inextensible, smooth" is the canonical triple for pulley problems. Each assumption disables a complication: light removes string mass from the Newton-II analysis; inextensible equalises acceleration magnitudes; smooth equalises tensions on both sides of the pulley.
4. Section 8 — Kinematics (SUVAT): once $a$a is known and constant, SUVAT gives velocity, displacement and time in the subsequent motion — up until a string goes slack or a particle hits the floor, at which point the system enters a new phase with different forces.
5. Section 9 — Vectors: when the pulley redirects motion through $90°$90° (table-and-pulley setup) the displacement of $A$A is horizontal while $B$B's is vertical. The string constraint links the speeds (equal magnitudes) — a vector idea even when the algebra looks scalar.

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Mark-scheme literacy

Connected-particle questions on 9MA0-03 split AO marks roughly:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Correct equations of motion, accurate simultaneous solution, substitution back for tension, SUVAT follow-up
AO2 (reasoning / interpretation)	10–20%	Justifying that tension is the same throughout the string, signs consistent across both equations, units stated
AO3 (problem-solving / modelling)	25–35%	Setting up the model from a worded scenario, criticising assumptions, deciding when phase-change occurs (string slack, pulley reached)

Examiner-rewarded phrasing: "taking downward as positive for the heavier particle"; "since the string is inextensible, both particles have the same acceleration magnitude $a$a"; "since the pulley is smooth, the tension $T$T is the same throughout"; "after the string becomes slack, the only force on $B$B is gravity, so it decelerates at $g$g". Phrases that lose marks: "force = mass times acceleration" without identifying which forces act on which particle; using $T_1$T1​ and $T_2$T2​ without justification; mixing positive directions between the two particles' equations without flagging it.

A specific Edexcel pattern: questions often add a final part "after $B$B hits the ground, find the further distance $A$A travels" — this is a phase change, and the new equation of motion for $A$A alone has only friction (or nothing, if smooth) acting. Read carefully when the problem moves into a second phase.

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Grade-band model answers

3-mark question

Question: Two particles, of masses $1\,\text{kg}$1kg and $2\,\text{kg}$2kg, are connected by a light inextensible string over a smooth pulley and released from rest. Find the acceleration. Take $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2.

Grade C response (~150 words):