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How do you find the acceleration and tension for connected particles joined by a string over a pulley?

Analyse connected particles: model two particles joined by a light inextensible string, including over a smooth pulley, and find the common acceleration and the tension.

A CCEA GCSE Further Mathematics answer on connected particles, covering two masses joined by a light string over a smooth pulley, the common acceleration, the string tension, and the equations of motion for each particle in the Mechanics unit.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. The modelling assumptions
  3. One equation of motion per particle
  4. Solving the pair
  5. Why this matters

What this dot point is asking

Connected particles are two masses joined by a string, often passing over a pulley, and CCEA GCSE Further Mathematics asks you to find how they move together. You must model the system with a light inextensible string over a smooth pulley, write an equation of motion for each particle, and solve them simultaneously for the common acceleration and the string tension. This is an extended application of Newton's second law and a classic Mechanics question.

The modelling assumptions

The standard assumptions make the problem solvable and must be understood. A light string has negligible mass, so the tension does not vary along it; an inextensible string keeps the two particles moving with the same speed and acceleration; and a smooth pulley exerts no friction, so the tension is unchanged as the string passes over it.

One equation of motion per particle

The method is to treat each particle separately. Draw its forces, choose the direction it accelerates as positive, and apply F=maF = ma. For a hanging mass this involves its weight and the tension; for a mass on a table it involves the tension and possibly friction.

The two unknowns, the acceleration aa and the tension TT, appear in both equations, so the two equations together determine them. Because the string is inextensible, the same aa is used in both.

A useful sanity check comes from looking at the system as a whole. If you imagine the two particles as a single object, the only external force driving the motion is the difference in their weights (the part not balanced), and the total mass is the sum of the two masses. So the acceleration must equal the net driving force divided by the total mass. For two hanging masses m1m_1 and m2m_2 with m2m_2 heavier, this gives a=(m2−m1)gm1+m2a = \dfrac{(m_2 - m_1)g}{m_1 + m_2}, which is exactly what solving the two equations produces. Knowing this whole-system relationship lets you check your simultaneous-equation answer quickly, although in the exam you should still show the separate equations of motion to earn the method marks.

Solving the pair

You solve the two equations of motion simultaneously. Adding them often eliminates the tension immediately, giving the acceleration in one step.

Why this matters

Connected particles bring together everything in the Mechanics unit: Newton's second law for each mass, weight, tension, the normal reaction and friction for a mass on a rough table, and the suvat equations afterwards if the question asks how far the system moves. The technique of writing one equation per body and solving simultaneously is a transferable problem-solving method, and the modelling assumptions teach you to read a mechanics problem precisely.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA Unit 2 (style)6 marksTwo particles of mass 33 kg and 55 kg hang on the ends of a light inextensible string over a smooth pulley. Taking g=9.8g = 9.8 m/s2^2, find the acceleration of the system and the tension in the string.
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The heavier 55 kg mass descends and the 33 kg mass rises with the same acceleration aa and the string tension TT is the same throughout.

For the 55 kg mass (down positive): 5g−T=5a5g - T = 5a, so 49−T=5a49 - T = 5a.

For the 33 kg mass (up positive): T−3g=3aT - 3g = 3a, so T−29.4=3aT - 29.4 = 3a.

Add the equations to eliminate TT: 49−29.4=8a49 - 29.4 = 8a, so 19.6=8a19.6 = 8a and a=2.45a = 2.45 m/s2^2. Substitute back: T=3(2.45)+29.4=36.75T = 3(2.45) + 29.4 = 36.75 N.

Marks are for each equation of motion, eliminating TT, the acceleration, and the tension.

CCEA Unit 2 (style)6 marksA 44 kg block on a smooth horizontal table is joined by a light string over a smooth pulley at the edge to a 22 kg mass hanging freely. Taking g=9.8g = 9.8 m/s2^2, find the acceleration and the tension.
Show worked answer →

Let the acceleration be aa and the tension TT, the same throughout the string.

For the hanging 22 kg mass (down positive): 2g−T=2a2g - T = 2a, so 19.6−T=2a19.6 - T = 2a.

For the 44 kg block on the smooth table (horizontal, the only force is TT): T=4aT = 4a.

Substitute T=4aT = 4a into the first equation: 19.6−4a=2a19.6 - 4a = 2a, so 19.6=6a19.6 = 6a and a=3.27a = 3.27 m/s2^2. Then T=4(3.27)=13.1T = 4(3.27) = 13.1 N.

Marks are for both equations, the substitution, the acceleration, and the tension.

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