Use energy for a taut connected particle system
Sum the two particles’ kinetic and potential changes, explain tension-work cancellation and stop the energy model when a string or contact constraint changes.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.5. Use g = 10 m s⁻² throughout these problems. Masses are positive and constant. Work by a constant force is Fd cos θ, with θ between force and displacement; scalar-product notation is not required. Kinetic energy is ½mv² and gravitational potential-energy change is mgΔh. State the system and count gravitational work or potential-energy change once, not both. Mechanical energy is conserved only under the stated loss-free model; friction and supplied resistance transfer mechanical energy to other forms. Smooth curved-slide examples use overall energy changes and assume contact throughout, rather than requiring a normal-reaction calculation. Average power is work divided by elapsed time; P = Fv applies to a force along the motion. Constant driving power generally gives variable force and acceleration; instantaneous calculations do not justify constant-acceleration formulae over a whole journey. Check string, contact and motion-state changes before continuing a model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Apply ideal taut-string constraints over a smooth pulley.
- Find gravitational and friction contributions along separate particle paths.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Two connected masses m and M have the same speed magnitude v. What is their total kinetic energy?
What you will learn
- Use the shared speed magnitude for the combined kinetic energy of ideal connected particles.
- Include friction and height changes of the correct bodies without counting internal tension work twice.
The connection links speeds and displacements
For a 4 kg particle on a rough horizontal table connected to a 2 kg hanging mass by a light taut inextensible string over a smooth pulley, the speed magnitudes are equal while the string remains taut. Starting from rest, total final K is ½(4 + 2)v² = 3v². With table coefficient 1/4, friction is 10 N. If the hanging particle falls 0.3 m, it releases 6 J potential energy and table friction removes 3 J, leaving total K = 3 J. Hence v = 1 m s⁻¹.
Tension does positive work on the table particle moving towards the pulley and equal negative work on the descending hanging particle. Under the common-tension and equal-displacement assumptions, those work terms cancel when the particle balances are added. This is a scalar work cancellation, even though the two tension force vectors do not necessarily point in opposite directions in space. It does not imply that either individual tension is zero.
Use the height change of each mass and respect events
For hanging masses 3 kg and 5 kg, with the 5 kg mass descending 0.8 m from rest, the heavier loses 40 J of potential energy and the lighter gains 24 J. The net release is 16 J, so ½(3 + 5)v² = 16 gives v = 2 m s⁻¹. Using only the descending weight as the release would omit the energy needed to raise its partner.
On a slope-and-hanging system, the slope particle’s vertical rise is its along-plane travel multiplied by sine; the hanging particle’s height change equals its vertical travel. Include friction only on the rough contact. If a mass hits the floor or the string slackens, shared-speed and tension-work assumptions can end. A floor impact may dissipate energy; do not claim mechanical-energy conservation across it without a stated impact model. Start a new interval for the unaffected particle when appropriate.
Pause and explain
A 5 kg hanging mass descends while a 3 kg mass rises by the same 0.8 m. What potential-energy release is available?
Worked example
A 4 kg rough-table particle with μ = 1/4 is linked to a 2 kg hanging mass by an ideal taut string. Released from rest, the hanging mass falls 0.3 m before any contact. Find their common speed.
Show the worked solution
- The gravitational energy release is 2 × 10 × 0.3 = 6 J.
- Table friction is 10 N and removes 10 × 0.3 = 3 J.
- Total final K is 3 J = ½(6)v², so common speed is 1 m s⁻¹.
Answer Common speed 1 m s⁻¹ before contact; combined K = 3 J.
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Test this topic →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.
A 4 kg table particle with μ = 1/4 is linked to a 2 kg hanging mass by an ideal taut string. Released from rest, the hanging mass falls 0.3 m before any contact. Find common speed using energy.
Give me a hint
Subtract table friction loss from the hanging mass’s potential release and include both kinetic energies.
Compare my reasoning
- Potential release is 6 J and friction loss is 3 J.
- Combined final K is 3 J = ½(4 + 2)v².
- Therefore v² = 1 and common speed is 1 m s⁻¹.
Common speed 1 m s⁻¹; combined K = 3 J.
Look for these in your work
- Include both moving masses in kinetic energy.
- Use the equal-displacement taut-string assumption only within the stated interval.
Masses 3 kg and 5 kg hang on an ideal taut pulley string and are released from rest. The heavier descends 0.8 m before any contact. Find their common speed by energy.
Give me a hint
The ascending mass gains potential energy and reduces the net gravitational release.
Compare my reasoning
- The heavier loses 40 J and the lighter gains 24 J, so net release is 16 J.
- Combined K is ½(8)v² = 4v².
- Thus v² = 4 and common speed is 2 m s⁻¹.
Common speed 2 m s⁻¹; net potential release 16 J.
Look for these in your work
- Use both signed height changes.
- Add both positive kinetic-energy contributions.
A 5 kg particle starts up a rough slope linked to a descending 4 kg mass by an ideal taut string. Given sin θ = 3/5, cos θ = 4/5 and μ = 1/10, the system starts from rest and travels 3 m before any boundary. Find common speed and check using Newton’s law.
Give me a hint
The hanging potential loss must supply the slope potential gain, friction loss and both kinetic energies.
Compare my reasoning
- Hanging release is 4 × 10 × 3 = 120 J; slope rise costs 5 × 10 × 3 × 3/5 = 90 J; friction 4 N removes 12 J.
- The remaining 18 J equals ½(9)v², so v² = 4 and common speed is 2 m s⁻¹.
- Newton’s equations give common a = (40 − 30 − 4)/9 = 2/3; from rest over 3 m, v² = 2 × 2/3 × 3 = 4, agreeing.
Common speed 2 m s⁻¹; total K = 18 J; Newton check a = 2/3 m s⁻².
Look for these in your work
- Count all gravitational and friction changes once.
- Check the energy result independently while the connection is still taut.
Common mistakes
- Both masses contribute to combined kinetic energy.
- Include the height change of every mass.
- Do not carry a taut-connection energy model through a floor impact or slackening event.
What can you explain now?
Does cancellation of tension work mean the two actual tension force vectors must be opposite in space?
Compare with the explanation
No; work cancels through equal tensions and opposite signed along-path work.
The two individual forces can have different spatial directions while their scalar work contributions cancel.
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Make it stick
Sum positive kinetic energies and signed height changes, cancel ideal internal tension work and end the balance at a constraint-changing event.
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