Rebuild connected motion after a string goes slack
Calculate the motion up to a floor impact, carry the unaffected particle’s velocity into a new stage and stop when a new force or static state applies.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.4. Use g = 10 m s⁻² throughout these problems. Each particle has positive constant mass. Forces and acceleration are constant within each stated stage; linear motion only. Use the resultant force in F = ma. A light taut inextensible string over a smooth pulley has one tension and equal acceleration magnitudes; those constraints cease if the string becomes slack. A rope can pull but cannot push; a light rigid tow-bar may exert tension or thrust. Resistance other than the stated friction is neglected unless explicitly supplied. In our rough-contact model, sliding friction has magnitude μR and opposes motion; at rest, friction adjusts with magnitude at most μR. Check contact and motion assumptions before continuing a solution. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Solve a table-and-hanging system.
- Carry final velocity between constant-acceleration stages.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which quantity is carried into the table particle’s slack-string stage under the stated no-jump model?
What you will learn
- Identify the event that ends the taut-string common-acceleration stage.
- Recalculate the unaffected particle’s forces and motion after slackening.
A boundary event ends the original constraint
A 4 kg particle on a rough horizontal table with μ = 1/4 is linked to a 2 kg hanging mass. Released from rest, it starts moving because the hanging weight 20 N exceeds the 10 N static limit. During taut motion, a = (20 − 10)/6 = 5/3 m s⁻². If the hanging particle falls 0.3 m before hitting the floor, both particles reach speed v with v² = 2(5/3)(0.3) = 1, so v = 1 m s⁻¹ at that event.
Suppose the hanging particle then remains on the floor, the string slackens and there is no instantaneous velocity change of the table particle. The table particle retains speed 1 m s⁻¹ towards the pulley but no longer has string tension. Its only horizontal force is friction 10 N away from the pulley, giving acceleration −10/4 = −2.5 m s⁻² with towards the pulley positive. Equal accelerations and the previous tension no longer apply.
Use a new force diagram and a local clock
From v = 1 under a = −2.5, the table particle takes 0.4 s to stop and moves a further 1²/(2 × 2.5) = 0.2 m. Its total distance is 0.3 + 0.2 = 0.5 m. These numbers use the stated no-impact velocity change for this particle, not a conservation assumption across the hanging particle’s floor impact. There must be enough table length and enough slack for this next stage to occur as specified.
Once the table particle stops with no applied horizontal drive, the needed static friction is zero; it does not reverse merely because the old sliding equation had negative acceleration. On a smooth table, loss of tension instead gives zero acceleration and the unaffected particle continues at constant velocity until another stated event. Describe the new interval and force assumptions so a multi-stage answer does not imply that a slack string still constrains the two particles.
Pause and explain
After the rough-table particle stops with no horizontal drive and slack string, what happens to friction?
Worked example
A 4 kg table particle with μ = 1/4 is linked to a 2 kg hanging mass and released from rest. The hanging mass falls 0.3 m to the floor, stays there and the string goes slack; the table particle has no instantaneous velocity change and sufficient table space. Find its further and total travel before stopping.
Show the worked solution
- Before impact, a = 5/3 and reached speed satisfies v² = 2(5/3)(0.3) = 1.
- After slackening, table friction is 10 N and acceleration is −2.5. Further stopping distance is 1/5 = 0.2 m.
- Total table travel is 0.3 + 0.2 = 0.5 m; time after slackening to rest is 1/2.5 = 0.4 s.
Answer Further travel 0.2 m; total table travel 0.5 m; stops 0.4 s after slackening.
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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 rough-table particle with μ = 1/4 is linked to a 2 kg hanging mass, released from rest. The hanging mass falls 0.3 m, stays on the floor and the string slackens without a table-particle velocity jump. Find the reached speed and further stopping distance, with enough table space.
Give me a hint
Separate the taut acceleration from the friction-only deceleration after slackening.
Compare my reasoning
- Taut acceleration is (20 − 10)/6 = 5/3 m s⁻², giving reached speed √(2 × 5/3 × 0.3) = 1 m s⁻¹.
- After slackening, table acceleration is −10/4 = −2.5 m s⁻².
- Further distance is 1²/(2 × 2.5) = 0.2 m; total travel is 0.5 m.
Reached speed 1 m s⁻¹; further distance 0.2 m; total 0.5 m.
Look for these in your work
- End the shared-acceleration model at the floor event.
- Use the unchanged reached speed in the new stopping calculation.
A 3 kg rough-table particle with μ = 1/5 is linked to a 2 kg hanging mass and released from rest. The hanging mass falls 5/7 m and then stays on the floor, making the string slack without a table-particle velocity jump. Find reached speed, further stopping time and distance before any other event.
Give me a hint
The table friction is 6 N in both moving stages, but tension disappears in the second.
Compare my reasoning
- During taut motion, a = (20 − 6)/5 = 14/5, so v² = 2(14/5)(5/7) = 4 and v = 2 m s⁻¹.
- After slackening, a = −6/3 = −2 m s⁻², so stopping time is 2/2 = 1 s.
- Further distance is 2²/(2 × 2) = 1 m; total table travel is 5/7 + 1 = 12/7 m.
Reached speed 2 m s⁻¹; further time 1 s; further distance 1 m; total 12/7 m.
Look for these in your work
- Calculate both stages from their own resultants.
- Stop the friction-only motion at zero velocity and then check rest.
A 3 kg particle on a smooth horizontal table is linked to a 2 kg hanging mass and released from rest. After the hanging mass falls 0.5 m it stays on the floor and the string slackens without a table-particle velocity jump. With sufficient table space, find the table particle’s speed and additional distance during the next 2 s.
Give me a hint
On the smooth table, slackening removes its only horizontal force.
Compare my reasoning
- Before the event, common acceleration is 20/5 = 4 m s⁻² and v² = 2 × 4 × 0.5 = 4, so speed is 2 m s⁻¹.
- After slackening, the horizontal resultant is zero and acceleration is zero.
- It continues towards the pulley at 2 m s⁻¹, moving an additional 4 m in the next 2 s; total travel is 4.5 m.
Speed remains 2 m s⁻¹; additional distance 4 m; total table travel 4.5 m.
Look for these in your work
- Apply the smooth-contact assumption after the tension vanishes.
- Do not confuse zero acceleration with stopping.
Common mistakes
- The taut-string constraint ends when the string slackens.
- Do not reset the unaffected particle’s velocity at the stated release event.
- At final rest, replace sliding friction with the required static balance.
What can you explain now?
On a smooth table, what happens to a moving particle when the connecting string becomes slack and no other horizontal force acts?
Compare with the explanation
It continues at its reached constant velocity until another event.
Zero resultant after slackening means zero acceleration, not an instant stop.
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Make it stick
Locate the constraint-changing event, carry the permitted velocity forward and write a new Newton equation for the next stage.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.