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Connected particles and pulleys questions
Two objects tied together are one system when you want the acceleration and two separate objects when you want the tension. Knowing which view to take, and when, decides the question.
6 original questions · 24 marks · the connected particles and pulleys notes · Mechanics
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Two particles are joined by a light inextensible string over a smooth pulley. Explain why they share one acceleration and one tension.
Worked answer
Inextensible means the string cannot stretch, so its ends move together and the two accelerations match in magnitude. Light string and smooth pulley mean nothing along the way absorbs force, so the tension is the same at both ends. B1 for the shared acceleration, B1 for the shared tension. A rough pulley would break the second half of that, since friction at the pulley leaves the two sides pulling with different tensions.Explain why each particle in a connected system gets its own F = ma equation.
Worked answer
Newton's second law applies body by body, to the forces on that body alone. The 7 kg mass feels its own weight and the tension and nothing else. B1 for the law applying to one body at a time, B1 for the forces being those on that body alone. Writing one equation per particle, then combining them, is the method from start to finish.Masses of 7 kg and 3 kg hang from a light string over a smooth pulley. Write the two equations of motion and find the acceleration and the tension, taking g = 9.8 m s⁻².
Worked answer
Take each mass's own direction of motion as positive. Heavy side: 7g − T = 7a. Light side: T − 3g = 3a. Adding the pair removes T and gives 4g = 10a, so a = 2g/5 = 3.92 m s⁻². Substituting back, T = 3(g + a) = 3 × 13.72 = 41.2 N. M1 A1 for the pair of equations, M1 for eliminating T, A1 for a = 3.92, A1 for T = 41.2. Check it in the other equation as well: 7(9.8 − 3.92) = 41.2 N, which agrees. The tension also has to sit between 3g = 29.4 N and 7g = 68.6 N. If T came out equal to either weight, that mass would not be accelerating at all.A 1000 kg car tows a 250 kg trailer with driving force 2000 N, against resistances of 400 N on the car and 100 N on the trailer. Find the acceleration and the tension in the towbar.
Worked answer
Treat the whole system first, because the towbar tension is internal to it and cancels: 2000 − 500 = 1250a, so a = 1.2 m s⁻². Now take the trailer on its own: T − 100 = 250 × 1.2, so T = 400 N. M1 A1 for the system equation, A1 for a = 1.2, M1 for the trailer equation, A1 for T = 400. The car's own equation is a free check, 2000 − 400 − 400 = 1000 × 1.2. Choosing the system equation first is what makes a come out so cleanly, and the smaller body then exposes T with the least arithmetic.A 4 kg block on a smooth table is joined by a string over a smooth pulley at the table's edge to a 2 kg mass hanging freely. Find the acceleration and the tension.
Worked answer
Hanging mass: 2g − T = 2a. Block: T = 4a, since the table is smooth and the string is horizontal. Adding gives 2g = 6a, so a = g/3 = 3.27 m s⁻² and T = 4a = 13.1 N. M1 A1 for the two equations, A1 for a = 3.27, A1 for T = 13.1. Only the hanging weight drives the system, but it has to move all 6 kg, which is why the block's mass appears in the denominator even though gravity never pulls it sideways.The 7 kg and 3 kg masses hang from a light string over a smooth pulley, with the 7 kg mass 1.4 m above the floor and the 3 kg mass on the floor. The system is released from rest. Find the speed at which the 7 kg mass strikes the floor, and the greatest height above the floor reached by the 3 kg mass.
Worked answer
While the string is taut the acceleration is 3.92 m s⁻², as before. Over the 1.4 m drop, v2 = 0 + 2 × 3.92 × 1.4 = 10.976, so v = 3.31 m s⁻¹. The 3 kg mass is by then 1.4 m up and moving upwards at the same 3.31 m s⁻¹. The moment the 7 kg mass lands the string goes slack, so the 3 kg mass becomes a free particle under gravity alone. It rises a further v2/(2g) = 10.976/19.6 = 0.56 m, reaching 1.4 + 0.56 = 1.96 m above the floor. M1 A1 for v = 3.31 over the drop, B1 for the string going slack, M1 for the free-flight stage, A1 for the extra 0.56 m, A1 for 1.96 m. The acceleration changes at the instant of landing, from 3.92 m s⁻² upwards to 9.8 m s⁻² downwards, so the motion has to be split into two suvat stages. Running a single suvat through the whole thing assumes constant acceleration where there is none, so its numbers are wrong from the landing onward.
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