The shape of the lesson
By the end of the lesson, learners can
- Car and trailer: For a car towing a trailer in a straight line, you can write \(F = ma\) for the car, for the trailer, or for the whole system.
- Strings over pulleys: Over a smooth pulley the tension is the same on both sides, but the two particles move in different directions.
- When the string goes slack: If a falling particle hits the floor, the string goes slack and the tension becomes zero.
- Diagram: forces on each particle separately, with \(T\) on both ends of the string.
- Directions: each particle's own direction of motion is positive.
- Equations: \(F = ma\) for each particle (or the whole system plus one particle).
- Solve: add to eliminate \(T\), find \(a\), then \(T\).
- Slack string: \(T = 0\); continue with the velocity at that instant.
One picture every idea returns to
A light inextensible string has the same tension all along it (also over a smooth pulley), and the particles it joins have accelerations of equal size. Write \(F = ma\) for EACH particle in its own direction of motion, then solve the equations together. A tow-bar can be in tension or in thrust.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, One string, one acceleration.. A car towing a trailer, two masses over a pulley, a block dragged across a table by a falling weight: when particles are joined by a light inextensible string or a rigid rod, they share one acceleration, and Newton's second law for each particle finds both it and the tension. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Car and trailer | For a car towing a trailer in a straight line, you can write \(F = ma\) for the car, for the trailer, or for the whole system. Model the worked example Tension in a tow-rope on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Strings over pulleys | Over a smooth pulley the tension is the same on both sides, but the two particles move in different directions. Model the worked example On a rough table (Fig. 1) on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: When the string goes slack | If a falling particle hits the floor, the string goes slack and the tension becomes zero. Model the worked example Greatest height after impact on the board, one line at a time. Your turn: Particles of mass 0.6 kg and 0.4 kg hang on either side of a smooth pulley and are released from rest. Find the acceleration and the tension. | Anticipation / Reaction Guide |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 28 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in When the string goes slack, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| A rope can only pull. | A tow-bar can push, and when the car brakes hard it does. If your tension comes out negative, the connector is in thrust (for a rope, that would mean it goes slack). |
| The tension is NOT the weight of the hanging particle. | If it were, B would have no resultant force and could not accelerate. In Fig. 1, \(T = 14.4\) N, less than B's 20 N weight. |
| After the string goes slack, the rising particle does NOT stop at once; it keeps moving up and decelerates at \(g\). | Forgetting this second stage gives 1.2 m instead of 1.5 m. |
Connected particles and pulleys: worksheet
6 questions, 28 marks. Written in the style of the papers; no past paper question is reproduced.
A car of mass 1000 kg tows a trailer of mass 500 kg with a light horizontal rope on a level road. The driving force is 2000 N and there is no resistance. Find the acceleration and the tension in the rope.
A car of mass 1500 kg tows a trailer of mass 500 kg. The driving force is 3000 N; the resistances are 400 N on the car and 200 N on the trailer. Find the acceleration and the tension in the tow-rope.
Particles of mass 5 kg and 3 kg hang on either side of a smooth fixed pulley, connected by a light inextensible string. They are released from rest. Find the acceleration, the tension and the force exerted by the string on the pulley.
A block A of mass 3 kg lies on a rough horizontal table and is joined by a light inextensible string over a smooth pulley at the edge to a block B of mass 2 kg hanging freely. When released, the system accelerates at \(2.8\ \text{m s}^{-2}\). Find the coefficient of friction and the tension.
A particle A of mass 4 kg lies on a rough plane inclined at \(30^\circ\), \(\mu = 0.25\). A light inextensible string from A passes up the plane, over a smooth pulley at the top, to a particle B of mass 6 kg hanging freely. The system is released from rest. Find the acceleration and the tension.
Particles P (5 kg) and Q (3 kg) hang on either side of a smooth pulley, with P 1.2 m above the floor. They are released from rest. P hits the floor and does not rebound; Q never reaches the pulley.
(a) Find the time P takes to reach the floor.
(b) Find the greatest height of Q above its starting point.
Answer key and marking notes
- Whole: \(2000 = 1500a\), so \(a = 1.33\ \text{m s}^{-2}\).
- Trailer: \(T = 500 \times \tfrac{4}{3} = 667\) N.
- Whole: \(3000 - 600 = 2000a\), so \(a = 1.2\ \text{m s}^{-2}\).
- Trailer: \(T - 200 = 500(1.2)\), so \(T = 800\) N.
- \(50 - T = 5a\), \(T - 30 = 3a\); adding, \(a = 2.5\ \text{m s}^{-2}\).
- \(T = 37.5\) N.
- Force on pulley \(= 2T = 75\) N, vertically downwards.
- B: \(20 - T = 2(2.8)\), so \(T = 14.4\) N.
- A: \(T - \mu(30) = 3(2.8)\), so \(14.4 - 30\mu = 8.4\).
- \(\mu = 0.2\).
- A: \(R = 40\cos 30^\circ = 34.64\) N; friction \(= 8.660\) N down the plane (A moves up).
- A: \(T - 20 - 8.660 = 4a\). B: \(60 - T = 6a\).
- Add: \(31.34 = 10a\), so \(a = 3.13\ \text{m s}^{-2}\).
- \(T = 60 - 6(3.134) = 41.2\) N.
- (a) \(a = 2.5\ \text{m s}^{-2}\); \(1.2 = \tfrac{1}{2}(2.5)t^2\), so \(t^2 = 0.96\) and \(t = 0.980\) s.
- (b) Speed at impact: \(v^2 = 2(2.5)(1.2) = 6\).
- Q then moves freely under gravity: further rise \(= \dfrac{6}{20} = 0.3\) m.
- Greatest height \(= 1.2 + 0.3 = 1.5\) m.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A car of mass 1000 kg tows a trailer of mass 300 kg along a straight horizontal road using a light rope. The driving force is 2000 N, and the resistances are 200 N on the car and 100 N on the trailer. Find the acceleration and the tension in the rope.
- A\(a = 1.31\ \text{m s}^{-2}\), \(T = 492\) N
- B\(a = 1.31\ \text{m s}^{-2}\), \(T = 392\) N
- C\(a = 1.7\ \text{m s}^{-2}\), \(T = 610\) N
- D\(a = 1.54\ \text{m s}^{-2}\), \(T = 562\) N
Two particles of different masses hang on either side of a smooth fixed pulley, connected by a light inextensible string, and are released from rest. How does the tension compare with the weights?
- AIt is less than the heavier weight and more than the lighter weight
- BIt equals the heavier weight
- CIt equals the lighter weight
- DIt equals the sum of the two weights
Particles of mass 7 kg and 3 kg hang on either side of a smooth fixed pulley, connected by a light inextensible string, and are released from rest. Find the acceleration and the tension.
- A\(a = 4\ \text{m s}^{-2}\), \(T = 42\) N
- B\(a = 4\ \text{m s}^{-2}\), \(T = 70\) N
- C\(a = 4\ \text{m s}^{-2}\), \(T = 30\) N
- D\(a = 5.71\ \text{m s}^{-2}\), \(T = 47.1\) N
A block of mass 4 kg rests on a smooth horizontal table. A light inextensible string attached to it passes over a smooth pulley at the edge of the table to a particle of mass 1 kg hanging freely. The system is released from rest. Find the acceleration and the tension.
- A\(a = 2\ \text{m s}^{-2}\), \(T = 8\) N
- B\(a = 2.5\ \text{m s}^{-2}\), \(T = 10\) N
- C\(a = 2\ \text{m s}^{-2}\), \(T = 10\) N
- D\(a = 2\ \text{m s}^{-2}\), \(T = 12\) N
Block A of mass 5 kg lies on a rough horizontal table, with coefficient of friction 0.4. A light inextensible string joins A, over a smooth pulley at the edge of the table, to block B of mass 3 kg hanging freely, as shown. The system is released from rest. Find the acceleration and the tension.
- A\(a = 1.25\ \text{m s}^{-2}\), \(T = 26.3\) N
- B\(a = 3.75\ \text{m s}^{-2}\), \(T = 18.8\) N
- C\(a = 1.25\ \text{m s}^{-2}\), \(T = 30\) N
- D\(a = 2.25\ \text{m s}^{-2}\), \(T = 23.3\) N
Particle A of mass 3 kg lies on a smooth plane inclined at \(30^\circ\) to the horizontal. A light inextensible string from A runs up the plane, over a smooth pulley at the top, to particle B of mass 2 kg hanging freely, as shown. The system is released from rest. Find the acceleration and the tension.
- A\(a = 1\ \text{m s}^{-2}\), \(T = 18\) N
- B\(a = 1\ \text{m s}^{-2}\), \(T = 20\) N
- C\(a = 7\ \text{m s}^{-2}\), \(T = 6\) N
- D\(a = 4\ \text{m s}^{-2}\), \(T = 12\) N
Particles of mass 4 kg and 2 kg hang on either side of a smooth fixed pulley, connected by a light inextensible string. They are released from rest with the 4 kg particle 0.9 m above the floor, as shown. It does not rebound. Find the greatest height the 2 kg particle reaches above its starting point (it does not reach the pulley).
- A1.2 m
- B0.9 m
- C1.8 m
- D0.3 m
A car of mass 1100 kg tows a trailer of mass 500 kg with a light rigid tow-bar on a straight horizontal road. The engine is switched off and the car brakes with a force of 2400 N. The resistances are 150 N on the car and 250 N on the trailer. Find the force in the tow-bar.
- AA thrust of 625 N
- BA tension of 625 N
- CA thrust of 875 N
- DA thrust of 1125 N
Quiz key
System: \(2000 - 200 - 100 = 1300a\), so \(a = 1.31\ \text{m s}^{-2}\). Trailer: \(T - 100 = 300a\), so \(T = 100 + 392.3 = 492\) N. The tempting 392 N forgets that the rope must also overcome the trailer's resistance. 1.7 divides by the car's mass only, and 1.54 ignores both resistances.
The heavier particle accelerates downwards, so its weight must exceed \(T\); the lighter particle accelerates upwards, so \(T\) must exceed its weight. The tempting "equals the heavier weight" would leave the heavier particle with no resultant force, so it could not accelerate. The sum of the weights is (roughly) the force on the pulley's support, not the tension.
7 kg: \(70 - T = 7a\). 3 kg: \(T - 30 = 3a\). Adding: \(40 = 10a\), so \(a = 4\), and \(T = 30 + 12 = 42\) N. The tempting 70 N or 30 N sets the tension equal to one of the weights, which would stop that particle accelerating. 5.71 divides the 40 N by 7 kg only, but both particles must be accelerated.
Adding \(T = 4a\) and \(10 - T = a\) gives \(10 = 5a\), so \(a = 2\ \text{m s}^{-2}\) and \(T = 8\) N. The tempting \(T = 10\) N is the weight of the hanging particle; then it would have no resultant force. 2.5 divides by the block's mass only, and \(T = 12\) has the hanging particle's acceleration the wrong way (\(T - 10 = a\)).
A: \(T - 20 = 5a\). B: \(30 - T = 3a\). Adding: \(10 = 8a\), so \(a = 1.25\ \text{m s}^{-2}\) and \(T = 3(10 - 1.25) = 26.25 \approx 26.3\) N. The tempting 3.75, 18.8 ignores friction. \(T = 30\) N is B's weight, and 2.25 uses friction on B's weight (\(0.4 \times 30\)) instead of on A's normal reaction.
Adding the equations: \(20 - 15 = 5a\), so \(a = 1\ \text{m s}^{-2}\), and \(T = 2(10 - 1) = 18\) N. The tempting \(T = 20\) N is B's weight. \(a = 7\) has the weight component of A helping the motion instead of opposing it, and \(a = 4\) ignores the slope altogether, as if A were on a horizontal table.
\(a = \dfrac{40 - 20}{6} = \tfrac{10}{3}\ \text{m s}^{-2}\), so when the 4 kg particle lands, \(v^2 = 2 \times \tfrac{10}{3} \times 0.9 = 6\). The 2 kg particle has risen 0.9 m and then rises a further \(\dfrac{v^2}{2g} = \dfrac{6}{20} = 0.3\) m. Total \(= 1.2\) m. The tempting 0.9 m forgets that the particle keeps moving up after the string goes slack. 1.8 m uses the deceleration \(\tfrac{10}{3}\) instead of \(g\) after the string goes slack, and 0.3 m is the free stage only.
System: deceleration \(= \dfrac{2400 + 150 + 250}{1600} = 1.75\ \text{m s}^{-2}\). Trailer, forward positive: \(T - 250 = 500 \times (-1.75)\), so \(T = -625\). The negative tension means the tow-bar pushes back on the trailer: a thrust of 625 N. The tempting "tension of 625 N" ignores the sign. 875 N leaves out the trailer's own resistance, and 1125 N adds it the wrong way.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.