01 · Height at the event
A starts 1 m above the ground. Find its height when B lands after falling 2 m.
Hint
Add A’s rise to its initial height.
Worked solution
A is 3 m above ground.
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Motion when a hanging particle reaches the ground
Join taut-string motion to free flight or resistance-only motion after a hanging particle lands, finding extra height, stopping distance and when the string becomes taut again.
Before you startPulley equations, vertical motion and staged SUVAT.
01 / A landing event can remove the string tension
The other particle may keep moving even though the connection becomes slack.
Redraw the forces immediately after the event.
Assume the descending particle settles on the ground. If the other particle continues toward the pulley, the geometric path needed by the string shortens, so the string becomes slack and tension is zero. The other particle retains its velocity in this ideal model; the forces determine its new acceleration.
Follow the event, then reset the clockExplore
A (3 kg) and B (5 kg) hang over a fixed smooth pulley on a taut light inextensible string. Released from rest, B descends 2 m and settles on the ground. A starts 1 m above ground and has enough space above it. After B lands, A continues upward under gravity while the string is slack. Select an event; no motion after the string becomes taut again is modelled.
02 / Find the state just before the landing
The old equations remain valid only up to the event.
A is 3 kg and B is 5 kg. B falls 2 m from rest while A rises; g = 9.8.Worked example
49 − T = 5a and T − 29.4 = 3a
Compatible downward/upward coordinates.
a = 2.45 m/s²; T = 36.75 N
Both bodies share speed magnitude while taut.
v² = 2 × 2.45 × 2 = 9.8
Speed just before landing is √9.8 ≈ 3.13 m/s.
t₁ = √(4/2.45) ≈ 1.28 s
A has risen 2 m when B reaches the ground.
Watch: landing ends the shared-acceleration stage
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Velocity at the event
State A’s velocity immediately after B lands under the slack-string model.
Hint
The string slackens and supplies no stopping impulse to A.
Worked solution
A continues upward at √9.8 m/s. Its velocity does not instantly become zero.
03 / Start free flight with the event velocity
For A, tension is now zero and acceleration is g downward.
03 · Extra height
Find A’s further rise after B lands.
Hint
Use u = √9.8, v = 0 and downward acceleration magnitude 9.8.
Worked solution
Extra height = u²/(2g) = 9.8/19.6 = 0.5 m.
04 · Greatest height
Find A’s greatest height above ground.
Hint
Include initial height, taut-stage rise and extra rise.
Worked solution
1 + 2 + 0.5 = 3.5 m.
05 · Time to the top
Find the time from B’s landing until A reaches greatest height.
Hint
Use 0 = u − gτ.
Worked solution
τ = √9.8/9.8 = 1/√9.8 ≈ 0.319 s. Total time from release is t₁ + τ ≈ 1.60 s.
04 / The string becomes taut again at a geometric event
Do not extend free flight beyond that event without a new model.
06 · Return to event height
How long after B lands does A first return to height 3 m?
Hint
Set its displacement relative to landing height to zero and exclude τ = 0.
Worked solution
0 = uτ − ½gτ² gives τ = 2u/g = 2/√9.8 ≈ 0.639 s.
07 · Just before tautness
Find A’s velocity just before the string becomes taut again in question 6.
Hint
Use v = u − gτ while the string remains slack.
Worked solution
v = −u = −√9.8 m/s, taking upward positive. A is descending at about 3.13 m/s.
08 · Why stop there?
Can that same free-flight equation determine A’s velocity immediately after the string becomes taut?
Hint
Tautening can produce an impulsive interaction.
Worked solution
No. The free-flight calculation gives the state just before tautness. The subsequent state needs an additional impact or string-tautening model; it is not justified by the previous equations.
05 / On a table, only the remaining horizontal forces act
A grounded hanger no longer supplies tension through a slack string.
A 3 kg table particle and a 2 kg hanger have acceleration 3 m/s² from rest. Resistance on A is 4.6 N. The hanger falls 1.5 m and settles on the ground.Worked example
Speed at landing = √(2 × 3 × 1.5) = 3 m/s
A has the same speed while the string is taut.
After landing: −4.6 = 3a₂
Assume enough table remains and resistance stays constant while A moves toward the pulley.
a₂ = −23/15 m/s²
The shared acceleration 3 m/s² no longer applies.
09 · Time to stop
For this table example, find the time from landing until A stops.
Hint
Use 0 = 3 − (23/15)τ.
Worked solution
τ = 45/23 s ≈ 1.96 s.
10 · Further table distance
Find A’s extra distance before stopping.
Hint
Use 0 = 3² + 2a₂s.
Worked solution
s = 135/46 m ≈ 2.93 m. This requires at least that much table before another boundary.
11 · Total distance
How far does A travel from the original release to rest in this model?
Hint
Add the taut-stage distance and the extra distance.
Worked solution
1.5 + 135/46 = 102/23 m ≈ 4.43 m.
06 / Without resistance a table particle does not stop by itself
A boundary may end the model before any stopping event.
12 · Smooth table after landing
If a table particle is moving at 3 m/s when its hanger lands and the table is smooth, what happens while the string stays slack?
Hint
The remaining horizontal resultant is zero.
Worked solution
It continues at 3 m/s until another event, such as reaching the pulley or table boundary. Zero force does not mean zero velocity.
13 · Compare time origins
What is the difference between t₁ and τ in these calculations?
Hint
Each clock belongs to a stage.
Worked solution
t₁ is elapsed time from release to landing. τ is time since landing. A final event’s total time is t₁ plus the appropriate second-stage duration.
07 / State the landing and clearance assumptions
Do not silently invent a rebound or an impact law.
14 · A rebounding hanger
If the descending particle rebounds from the ground, can the same slack-string timeline be assumed automatically?
Hint
The new position of the grounded end changes the string geometry.
Worked solution
No. Its post-impact motion depends on the rebound model, and the string may become taut at a different time. Rebuild the geometry and force stages from the specified impact information.
The worked hanging example assumes B remains on the ground, A does not hit the pulley and the string remains in the stated fixed geometry. State the event that ends each stage; check that it occurs before any other boundary.
08 / Carry velocity across a slackening event, then change the forces
Use a new clock and stop at the next event.
Find the velocity and position at landing with the taut-string equations. With tension zero, use gravity for a free hanging particle or the remaining table resistance for a horizontal particle. Distinguish extra travel, total travel and height above ground.
Section 1 of 8 · A landing event can remove the string tension