Join vertical force stages with constant acceleration motion
Turn cable forces into acceleration, calculate a stopping point and change the model when a cable is released.
Section 4.4 · Newton’s laws of motion
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
- Use weight and tension in a vertical balance.
- Apply v = u + at and v² = u² + 2as with signed quantities.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A 4 kg particle is pulled up by 48 N. With g = 10, what is the upward resultant?
What you will learn
- Combine a vertical Newton equation with constant-acceleration formulae.
- Carry the reached velocity into a released free-motion stage.
Find acceleration before choosing a motion formula
A 4 kg particle lifted by a constant upward cable force 48 N has weight 40 N. With up positive, 48 − 40 = 4a, so a = +2 m s⁻² while that force acts. Starting from rest for 3 s gives v = 6 m s⁻¹ and upward displacement 9 m. The cable force is 48 N, not the resultant 8 N; substituting 48/4 directly as acceleration would omit weight.
If a 5 kg particle moving upward at 6 m s⁻¹ has cable tension 40 N, the resultant is downward: a = (40 − 50)/5 = −2. It comes momentarily to rest after 3 s and rises 9 m. Momentary rest is a kinematic event, not proof of force equilibrium; if the same positive tension continues, downward motion follows. A cable that would require negative tension needs a different, slack-string model.
Release changes forces without resetting velocity
Releasing the 4 kg particle after its 3 s lifting stage removes cable tension. With no resistance, the next stage has a = −10 m s⁻² but starts with u = +6 m s⁻¹. It rises a further 6²/(2 × 10) = 1.8 m before stopping at the highest point, reached 0.6 s after release. The release itself is stipulated to produce no instantaneous velocity change.
Use a local time origin and the reached height for each new stage. While the particle is free, acceleration remains downwards at the highest point where velocity is zero. Continue only over the interval requested and before any stated contact or impact. If a model includes air resistance, use its specified force rather than silently inserting a new resistance or assuming the free-fall acceleration.
Pause and explain
Immediately after release at upward velocity 6 m s⁻¹, which next-stage data apply without resistance?
Worked example
A 4 kg particle starts from rest under an upward 48 N cable force for 3 s, then is released without an instantaneous velocity change. Neglect resistance. Find its maximum rise from the start.
Show the worked solution
- During lifting, a = (48 − 40)/4 = 2 m s⁻²; after 3 s, v = 6 and rise is 9 m.
- After release, a = −10 and u = 6, so the extra rise is 6²/20 = 1.8 m.
- Total maximum rise is 9 + 1.8 = 10.8 m; it occurs 3 + 0.6 = 3.6 s after starting.
Answer Maximum rise 10.8 m at total time 3.6 s.
Try a topic test
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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 3 kg particle starts from rest, pulled upward by a constant 36 N for 2 s. Find acceleration, velocity and rise.
Give me a hint
Use the net vertical force before the motion formulae.
Compare my reasoning
- Weight is 30 N, so a = (36 − 30)/3 = 2 m s⁻².
- Velocity is v = 0 + 2 × 2 = 4 m s⁻¹ upwards.
- Rise is s = ½ × 2 × 2² = 4 m.
a = +2 m s⁻²; v = +4 m s⁻¹; rise 4 m.
Look for these in your work
- Include weight in the force equation.
- Use the stated 2 s duration with the derived constant acceleration.
A 5 kg particle moves upward at 6 m s⁻¹ with constant upward cable tension 40 N. Find the time and extra rise to its first instant of rest.
Give me a hint
The upward-positive acceleration is negative because tension is below weight.
Compare my reasoning
- Acceleration is a = (40 − 50)/5 = −2 m s⁻².
- Set v = 0: 0 = 6 − 2t, so t = 3 s.
- Using v² = u² + 2as gives 0 = 36 − 4s, so rise is 9 m.
a = −2 m s⁻²; stopping time 3 s; rise 9 m.
Look for these in your work
- Keep tension positive while the resultant is negative.
- Stop at the requested first zero velocity, not at assumed equilibrium.
A 2 kg particle starts at rest under upward cable force 30 N for 2 s. The cable is then released without a velocity jump. Neglect resistance. Find the greatest height above its starting level and total time to that point.
Give me a hint
Use separate accelerations and carry the speed reached after the first 2 s.
Compare my reasoning
- During the pull, a = (30 − 20)/2 = 5, so after 2 s velocity is 10 m s⁻¹ and rise is 10 m.
- After release, a = −10; the extra time is 10/10 = 1 s and extra rise is 10²/20 = 5 m.
- Total maximum height is 15 m, reached after 3 s; gravitational acceleration is still downward there.
Greatest rise 15 m; total time 3 s.
Look for these in your work
- Change acceleration when tension disappears.
- Add displacements and elapsed times from both correctly joined stages.
Common mistakes
- Subtract weight from an upward cable force.
- Carry velocity into the release stage.
- At maximum height, velocity is zero but free acceleration remains downward.
What can you explain now?
Does a released particle acquire zero velocity immediately when its cable force vanishes?
Compare with the explanation
No; the reached velocity continues at release under the stated model.
The changed resultant determines the subsequent acceleration, not an instantaneous reset of velocity.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Force balance gives each stage’s acceleration; carry its final velocity into the next force model.
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.