Questions on connected particles and pulleys

Question 1

Two particles are joined by a light inextensible string over a smooth pulley. Explain why they share one acceleration and one tension.

Accepted 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.

Question 2

Explain why each particle in a connected system gets its own F = ma equation.

Accepted Answer

Newton's second law applies body by body, to the forces on that body alone. A hanging 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.

Question 3

Two particles are joined by a light inextensible string over a smooth pulley, and the two are treated as a single system. Explain why the tension does not appear in the whole-system equation.

Accepted Answer

The tension acts on both particles: it pulls one in its direction of motion and the other against its direction of motion, with the same magnitude on each because the string is light and the pulley smooth. When the two equations of motion are added, the two tension terms have opposite signs and cancel, so the whole-system equation contains only the external forces and the total mass. B1 for the tension being internal, acting on both particles with equal magnitude, B1 for the two terms cancelling in the sum. Friction on a table or a resistance is not internal, so it stays in the system equation.

Question 4

Particles of masses 5 kg and 3 kg hang from a light inextensible string over a smooth pulley, and the system is released from rest. A student calculates the tension in the string as 52 N. Without finding the acceleration, explain why this answer cannot be correct.

Accepted Answer

The weights are 5g = 49 N and 3g = 29.4 N. The 3 kg particle accelerates upwards, so the tension on it must exceed its weight: T > 29.4 N. The 5 kg particle accelerates downwards, so its weight must exceed the tension: T < 49 N. A tension of 52 N would make both particles accelerate upwards, which is impossible for a system driven only by gravity, so 52 N cannot be the tension. B1 for T above the lighter weight with the reason, B1 for T below the heavier weight with the reason, B1 for the conclusion. The true value is 2 × 5 × 3g/8 = 36.8 N, which does lie between them.

Question 5

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.

Accepted 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.

Question 6

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.

Accepted 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.

Question 7

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.

Accepted 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.

Question 8

A locomotive of mass 5000 kg pulls two trucks, of masses 3000 kg and 2000 kg, along a straight horizontal track, joined in that order by light rigid couplings. The driving force is 24000 N, and the resistances are 2000 N on the locomotive, 1200 N on the 3000 kg truck and 800 N on the 2000 kg truck. Find the acceleration of the train and the tension in each coupling.

Accepted Answer

Both coupling tensions are internal to the whole train, so the system equation ignores them: 24000 − 4000 = 10000a, giving a = 2 m s⁻². Each tension is then exposed by cutting the train at that coupling and taking everything behind the cut. The rear truck alone: T₂ − 800 = 2000 × 2, so T₂ = 4800 N. Both trucks together: T₁ − 2000 = 5000 × 2, so T₁ = 12000 N. M1 A1 for the system equation and a = 2, M1 for isolating the rear truck, A1 for T₂ = 4800 N, M1 for the two-truck equation, A1 for T₁ = 12000 N. The locomotive's own equation is a free check: 24000 − 2000 − 12000 = 10000 = 5000 × 2. The front coupling carries more than the rear because each coupling has to drag everything behind it.

Question 9

A car of mass 1200 kg tows a trailer of mass 400 kg along a straight level road by a light rigid tow bar. The car's brakes are applied, producing a braking force of 3200 N on the car only, and no other resistances act. Find the deceleration of the car and trailer, and the magnitude of the force in the tow bar, stating whether it is a tension or a thrust.

Accepted Answer

The tow bar force is internal, so for the whole system 3200 = 1600a and a = 2 m s⁻², a deceleration. The trailer has no brakes, so the only horizontal force on it comes from the tow bar, and that force must decelerate it: F = 400 × 2 = 800 N, acting backwards on the trailer. A bar that pushes backwards on the trailer, and so forwards on the car, is being compressed: the force is a thrust. M1 for the system equation, A1 for a = 2, M1 for the trailer's own equation, A1 for 800 N, B1 for thrust with the reason. Check on the car: 3200 − 800 = 2400 = 1200 × 2. A rigid bar can push as well as pull, which is why the question says bar rather than rope; a rope would go slack and the trailer would run on into the car.

Question 10

A lift of mass 600 kg carries a crate of mass 50 kg resting on its floor. The lift is raised by a vertical cable in which the tension is 7800 N. Taking g = 9.8 m s⁻², find the acceleration of the lift, and the normal reaction between the crate and the lift floor.

Accepted Answer

The lift and crate move together, so treat them as one body of mass 650 kg: 7800 − 650g = 650a, so 7800 − 6370 = 650a and a = 1430/650 = 2.2 m s⁻² upwards. For the crate alone, up positive: R − 50g = 50 × 2.2, so R = 50(9.8 + 2.2) = 600 N. M1 for the whole-system equation with the total weight, A1 for the equation, A1 for a = 2.2, M1 for the crate's own equation, A1 for 600 N. The reaction exceeds the crate's weight of 490 N because the floor has to accelerate the crate upwards as well as support it; the cable tension does not appear in the crate's equation because the cable is not attached to the crate.

Question 11

Particles of masses 2 kg and m kg, where m > 2, hang from a light inextensible string over a smooth pulley. When the system is released from rest, the 2 kg particle accelerates upwards at g/5. Find m, and the tension in the string, taking g = 9.8 m s⁻².

Accepted Answer

The heavier particle descends. Whole system: (m − 2)g = (m + 2)a with a = g/5, so 5(m − 2) = m + 2, giving 4m = 12 and m = 3. Tension from the 2 kg particle: T − 2g = 2 × g/5, so T = 2 × 1.2g = 2.4g = 23.5 N (3 s.f.). M1 for the system equation with a = g/5, A1 for m = 3, M1 for one particle's equation, A1 for 23.5 N. Check with the other particle: 3g − T = 3g/5 gives T = 2.4g as well, and 23.5 N lies between 2g = 19.6 N and 3g = 29.4 N. g cancels from the first equation, which is why m comes out as a number without a value of g being used.

Question 12

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.

Accepted 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.

Question 13

A block of mass 6 kg rests on a rough horizontal table, coefficient of friction 0.25. It is joined by a light inextensible string over a smooth pulley at the table's edge to a 4 kg mass hanging freely. The system is released from rest. Find the acceleration and the tension in the string, taking g = 9.8 m s⁻².

Accepted Answer

Nothing pulls the block up or down except its weight and the table, so R = 6g = 58.8 N and friction opposes the motion at 0.25 × 58.8 = 14.7 N. Around the string, the hanging weight drives and friction resists: 4g − 14.7 = 10a, so 24.5 = 10a and a = 2.45 m s⁻². Isolating the block, T − 14.7 = 6 × 2.45, so T = 29.4 N. B1 for the friction force 14.7 N, M1 for the system equation, A1 for a = 2.45, M1 for isolating one particle, A1 for T = 29.4 N. The hanging mass gives the same tension, 4(9.8 − 2.45) = 29.4 N. Friction belongs in the system equation because it is external to the system, unlike the tension, which cancels.

Question 14

In the system of the previous question, the 4 kg mass starts 0.9 m above the floor and the block starts 2 m from the pulley. Find the speed at which the 4 kg mass lands, and the further distance the block slides before coming to rest.

Accepted Answer

While the string is taut the acceleration is 2.45 m s⁻², so over the 0.9 m drop v² = 2 × 2.45 × 0.9 = 4.41 and v = 2.1 m s⁻¹. When the mass lands the string goes slack and the block carries on alone at 2.1 m s⁻¹, decelerated by friction only: 6a = 14.7 gives a deceleration of 2.45 m s⁻². It slides a further 4.41/(2 × 2.45) = 0.9 m, short of the 1.1 m then left to the pulley, so it stops on the table. M1 for suvat over the drop, A1 for v = 2.1 m s⁻¹, M1 for the friction-only deceleration once the string is slack, A1 for 0.9 m. The slide equalling the fall is a feature of these numbers, not a rule: here the deceleration 0.25g happens to equal the acceleration (4 − 0.25 × 6)g/10.

Question 15

A block of mass 5 kg rests on a rough horizontal table, coefficient of friction 0.5. It is joined by a light inextensible string over a smooth pulley at the edge of the table to a particle of mass 2 kg hanging freely. Show that the system remains at rest, and state the friction force acting on the block. The 2 kg particle is then replaced by one of mass 4 kg. Find the acceleration of the system and the tension in the string, taking g = 9.8 m s⁻².

Accepted Answer

R = 5g = 49 N, so the largest friction available is 0.5 × 49 = 24.5 N. At rest the string pulls the block with the hanging weight, 2g = 19.6 N, and 19.6 < 24.5, so friction can balance it and the system stays at rest, with friction taking the value needed: 19.6 N, not 24.5 N. With 4 kg hanging, 4g = 39.2 N exceeds 24.5 N, so the system moves and friction is at its limit, 24.5 N, opposing the block's motion. Whole system: 39.2 − 24.5 = 9a, so a = 1.63 m s⁻². Hanging particle: T = 4(g − a) = 4 × 8.17 = 32.7 N. M1 for the limiting friction 24.5 N, A1 for the comparison with 19.6 N, B1 for the friction being 19.6 N in equilibrium, M1 for the system equation with friction at 24.5 N, A1 for 1.63, A1 for 32.7 N. The decision whether the system moves has to come first, because friction takes its limiting value only once it does; writing 24.5 N for the friction in the 2 kg case would describe a system that accelerates the wrong way.