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A Level · Cambridge (CIE) · 9709

Distinguish towing tension from braking thrust

Find acceleration and the connecting force for a car and trailer, then decide whether a light rope or a rigid tow-bar can sustain the required force.

Mechanics · Newton’s laws and connected particles · 4.4
Exam scope and assessment notes

Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.4. Use g = 10 m s⁻² throughout these problems. Each particle has positive constant mass. Forces and acceleration are constant within each stated stage; linear motion only. Use the resultant force in F = ma. A light taut inextensible string over a smooth pulley has one tension and equal acceleration magnitudes; those constraints cease if the string becomes slack. A rope can pull but cannot push; a light rigid tow-bar may exert tension or thrust. Resistance other than the stated friction is neglected unless explicitly supplied. In our rough-contact model, sliding friction has magnitude μR and opposes motion; at rest, friction adjusts with magnitude at most μR. Check contact and motion assumptions before continuing a solution. Written practice is self-checked; human subject-teacher review remains pending.

Before you begin

  • Apply Newton’s laws to separate and combined systems.
  • Distinguish tension, thrust and Newton’s third-law pairs.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Which connecting model can transmit either tension or thrust?

What you will learn

  • Use the combined system to find acceleration and the trailer to find the connection force.
  • Interpret a negative signed pulling force as compression and distinguish rope from rod.
Only horizontal forces are shown for a 1000 kg car and 500 kg trailer moving forwards, with right positive. Backward external forces are 1750 N on the car and 250 N on the trailer. A rigid tow-bar exerts thrust 1250/3 N backward on the trailer and forward on the car, keeping acceleration −4/3 m s⁻² for both. The internal connection forces act on different bodies. A flexible rope could not supply the compression. All arrows and positions are schematic.Rigid tow-bar can transmit thrustBoth move forwards → and slow down500 kg1000 kgTow-bar250 N back1750 N backThrust 1250/3 N on each bodyTrailer: backward; car: forwarda = −4/3 m s⁻² for the rigid systemA rope cannot provide compressionSchematic arrows; use numerical balances
Only horizontal forces are shown for a 1000 kg car and 500 kg trailer moving forwards, with right positive. Backward external forces are 1750 N on the car and 250 N on the trailer. A rigid tow-bar exerts thrust 1250/3 N backward on the trailer and forward on the car, keeping acceleration −4/3 m s⁻² for both. The internal connection forces act on different bodies. A flexible rope could not supply the compression. All arrows and positions are schematic.

Internal connection forces cancel in the whole system

A 1000 kg car tows a 500 kg trailer on a horizontal road. With forward drive 3000 N, car resistance 500 N and trailer resistance 250 N, the combined resultant is 2250 N and total mass is 1500 kg. Thus a = 1.5 m s⁻² forward. On the trailer, T − 250 = 500(1.5), giving T = 1000 N. On the car, 3000 − 500 − 1000 = 1000(1.5), independently checking the result.

The connection pulls the trailer forward and the car backward with equal magnitudes. These third-law partners act on different bodies, so each separate diagram includes its own force. A light taut rope or a light rigid tow-bar can both sustain this tension state in the ideal model. Do not use the combined-system equation alone to infer the connection force; it has cancelled out there.

A rod may push, while a rope cannot

During a braking stage, suppose a 1000 kg car has 1750 N total backward external force and a 500 kg trailer has 250 N backward resistance. Treating a rigid connection as maintaining common acceleration gives a = −2000/1500 = −4/3 m s⁻² with forward positive. Let C be the signed forward connection force on the trailer. C − 250 = 500(−4/3), so C = −1250/3 N. The rod pushes the trailer backward with thrust 1250/3 N and exerts the opposite forward force on the car.

A rope cannot supply that negative pulling force. If no other constraint is stated, it becomes slack rather than maintaining the assumed shared acceleration. The car’s own backward acceleration is then −1750/1000 = −1.75, and the trailer’s is −250/500 = −0.5 m s⁻² during the stated forward motion. These separate-stage equations apply only until another contact, force change or rest event; do not describe negative tension as a working rope model.

Pause and explain

A trailer equation gives a signed forward connection force C < 0. For a rigid tow-bar, what does it mean?

Put the idea to work

Worked example

A 1000 kg car and 500 kg trailer move forwards on a horizontal road. Total backward external forces are 1750 N on the car and 250 N on the trailer. A light rigid tow-bar keeps them connected. Find a and the tow-bar force.

Show the worked solution
  1. For the combined system, −1750 − 250 = 1500a, so a = −4/3 m s⁻².
  2. For the trailer, C − 250 = 500a, giving C = −1250/3 N.
  3. This is thrust 1250/3 N backward on the trailer and forward on the car; a rope could not sustain it.

Answer a = −4/3 m s⁻²; tow-bar thrust 1250/3 N.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

A 1000 kg car tows a 500 kg trailer with a light taut connection. Forward drive is 3000 N; resistances are 500 N on the car and 250 N on the trailer. Find a and tension.

Give me a hint

Use the whole system for acceleration and the trailer alone for tension.

Compare my reasoning
  1. Combined resultant is 3000 − 500 − 250 = 2250 N, so a = 2250/1500 = 1.5 m s⁻².
  2. Trailer balance is T − 250 = 500(1.5), so T = 1000 N.
  3. Car balance gives 3000 − 500 − 1000 = 1500 N = 1000a, confirming the solution.

a = +1.5 m s⁻²; T = 1000 N.

Look for these in your work

  • Include both external resistances and both masses.
  • Check the separate car and trailer equations.
2 · Independent

A 1000 kg car and 500 kg trailer are joined by a light rigid tow-bar while moving forwards. Backward external forces are 1750 N on the car and 250 N on the trailer. Find common acceleration and the bar’s thrust.

Give me a hint

Define a signed forward connection force on the trailer and interpret its sign.

Compare my reasoning
  1. Combined balance gives a = −2000/1500 = −4/3 m s⁻².
  2. Trailer balance C − 250 = 500(−4/3) gives C = −1250/3 N.
  3. Thus the bar’s thrust is 1250/3 N backward on the trailer and forward on the car.

a = −4/3 m s⁻²; thrust 1250/3 N.

Look for these in your work

  • Retain negative acceleration during the forward braking stage.
  • Interpret compression as thrust, not as a negative physical magnitude.
3 · Transfer

Replace that braking tow-bar with a light flexible rope. The car still has 1750 N backward external force and the trailer 250 N, with masses 1000 kg and 500 kg. Test a taut-rope model and find each acceleration immediately after it slackens, before any new event.

Give me a hint

A required backward connection force on the trailer would make a rope push.

Compare my reasoning
  1. A common-acceleration calculation requires C = −1250/3 N, which is impossible for a taut rope.
  2. The rope therefore cannot maintain the shared-acceleration constraint in this model.
  3. With no rope force, car acceleration is −1750/1000 = −1.75 and trailer acceleration is −250/500 = −0.5 m s⁻² while they still move forwards.

Taut-rope model fails; after slackening, car a = −1.75 and trailer a = −0.5 m s⁻².

Look for these in your work

  • Reject compression for a flexible rope.
  • Recompute the two accelerations using separate force systems.

Common mistakes

  • Both masses contribute to the common acceleration.
  • The trailer needs its own equation to find the connecting force.
  • A negative rope tension means the taut-rope assumption fails.
Recall without your notes

What can you explain now?

Can a light rope supply the compressive force required to keep a braking trailer with its car?

Compare with the explanation

No; a rope pulls but cannot push, so that taut-rope model fails.

A light rigid tow-bar can transmit thrust; a rope requires a new slack-connection stage.

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Make it stick

Add the bodies for common acceleration, resolve one for the connection force, then check whether the connection can pull or push as required.

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Keep learning Mathematics

Course scope and assessment

312 sequenced lessons provide 936 guided, independent and application problems covering the 160 objectives in sections 1.1–1.8, 2.1–2.6, 3.1–3.9, 4.1–4.5, 5.1–5.5 and 6.1–6.5. The map lists all 38 sections across six alternative components; all eight Pure Mathematics 1 sections and all six Pure Mathematics 2 sections, from Paper 2 Algebra through Numerical solution of equations, currently have full teaching sequences. Paper 3 Algebra, section 3.1, links the six shared foundation lessons plus six lessons on partial fractions and rational binomial expansions; Paper 3 logarithms, trigonometry and numerical equations use the shared lessons, with the Paper 1 graph foundation linked into trigonometry. Paper 3 differentiation combines the shared methods with its inverse-tangent derivative lesson. Paper 3 Integration links four shared standard and identity lessons plus seven advanced lessons, covering all six section 3.5 requirements. Eight Vectors lessons cover all six section 3.7 requirements, with 2D and 3D line and solid applications. Nine Differential equations lessons cover all four section 3.8 requirements, including Paper 3 integration techniques, initial conditions, equilibrium solutions and conditional model interpretations. Eleven Complex numbers lessons cover all eight section 3.9 requirements, with seven Argand diagrams and explicit argument and boundary checks. All nine Pure Mathematics 3 sections now have teaching linked to every mapped requirement. Nine Forces and equilibrium lessons cover all seven section 4.1 requirements, with six force diagrams, signed incline-friction intervals, reaction-dependent pull and push limits and separate-body third-law checks. Nine Kinematics lessons cover the four official section 4.2 requirements, with the graph requirement mapped to three relationships. Six motion diagrams support directed quantities, graph slopes and signed areas, Paper 1 time-calculus methods, reversal distances, constant-acceleration stages and shared-clock particle meetings. Six Momentum lessons cover both section 4.3 requirements, including signed system totals, conservation assumptions, outgoing velocities, coalescence, unknown masses and successive impacts. Four diagrams show system momentum through the stipulated events, and supplied kinetic-energy checks distinguish conservation of momentum from changes in kinetic energy. Twelve Newton’s laws lessons cover all four section 4.4 requirements, with vertical support and release stages, rough horizontal and incline motion, inferred parameters, hanging and surface pulley systems, rigid tow-bar thrust and slack-string transitions. Seven force diagrams support the separate-particle balances. Nine Energy, work and power lessons cover all five section 4.5 requirements, with angle-dependent work, kinetic and reference potential energy, signed net-work balances, smooth curved slides, rough and driven slopes, ideal connected-system energy and instantaneous power on hills. Five diagrams distinguish the energy inputs, losses and valid positive-speed force model. All five Mechanics sections now have teaching and staged practice linked to their mapped requirements. Thirteen Representation of data lessons cover all five section 5.1 requirements, including choice and limitations of displays, ordered and back-to-back stems, raw summaries and box plots, unequal-width density histograms, cumulative-frequency interpolation, descriptive moment calculations, grouped estimates, coding, combining two datasets and correcting totals. Six accessible diagrams retain repeated observations and distinguish numerical spread, frequency area and estimated readings. Ten Permutations and combinations lessons cover both official section 5.2 requirements, with selection and ordered roles, repeated objects, adjacency blocks, non-adjacency gaps, leading-zero cases, labelled multi-row seating and mixed restrictions. Six accessible diagrams show outcome trees, ordered versus unordered choices, repeated copies, blocks, gaps and distinct row seats. Eleven Probability lessons cover all four section 5.3 requirements, with equiprobable elementary outcomes, selection-based probabilities, complements and disjoint regions, independence tests, restricted sample spaces, two-way tables, replacement and removal trees, reversed conditions and unknown parameters. Seven accessible diagrams show dice outcomes, random subsets, event regions, a conditioned grid, a count table and two probability trees. Twelve Discrete random variables lessons cover all three official section 5.4 requirements, with complete probability tables, moment and parameter checks, binomial assumptions, exact probabilities and moments, geometric first-success probabilities, tails, expectation and integer reliability thresholds. Six accessible diagrams distinguish outcome grouping, weighted moments, fixed trials, discrete probability bars, waiting conventions and integer tail deadlines. Twelve Normal distribution lessons cover all three official section 5.5 requirements, with continuous-area interpretation, MF19 cumulative tables, full standardisation, tail and interval regions, conditional probabilities, percentiles, one- and two-parameter inference, binomial suitability, continuity corrections and approximate integer thresholds. Seven accessible diagrams show model areas, table reading, aligned scales, intervals, quantiles, matched binomial and normal shapes and whole integer cells. All five Statistics 1 sections now have teaching and staged practice linked to their mapped requirements. Fourteen Poisson distribution lessons cover all five official section 6.1 requirements, with homogeneous event-count assumptions, exposure scaling, exact masses and tails, conditional count events, moments, parameter inference, probability duration targets, disjoint and overlapping windows, rare-count binomial approximations, matched normal approximations and continuity corrections. Nine accessible diagrams show event exposures, unit conversions, discrete bars, complete tail complements, population moments, shared windows, approximate count models and integer cells. Exact checks distinguish approximate and exact close-target decisions. 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Written practice is self-checked, not officially marked. Check the 2028–2030 syllabus separately if sitting an examination from 2028; this teaching map uses 2026–2027 requirements.

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