Resolve Newton equations along a smooth inclined plane
Separate the normal and downslope weight components, find acceleration and connect smooth-slope force balances to stopping and return motion.
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
- Resolve gravitational weight relative to an inclined plane.
- Apply constant-acceleration formulae on a linear axis.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For a slope at angle θ to horizontal, which weight component acts down the plane?
What you will learn
- Use mg sin θ along the plane and mg cos θ perpendicular to it.
- Combine a smooth-slope acceleration with signed motion calculations.
Choose axes parallel and perpendicular to the slope
For a smooth plane inclined at θ above horizontal, weight has component mg sin θ down the plane and mg cos θ into it. If no other force has a normal component, R = mg cos θ. Smooth means no friction in this model; it does not mean no normal reaction. Taking up the plane as positive for a freely moving particle gives −mg sin θ = ma, so a = −g sin θ.
The mass cancels in this gravity-only acceleration, but not in the reaction. On a 30° smooth slope, a = −5 m s⁻² with up positive. An applied force P parallel up the plane gives P − mg sin θ = ma. A 3 kg particle with sin θ = 3/5 and P = 24 N has a = (24 − 18)/3 = 2 m s⁻² upwards; with cos θ = 4/5 its reaction is 24 N.
The acceleration need not reverse at the turning point
A particle projected up a smooth 30° slope at 10 m s⁻¹ has constant a = −5. It first stops after 2 s and rises 10 m along the plane. Gravity’s downslope component still acts there, so it returns with negative velocity; at its original point v = −10 m s⁻¹ and total time is 4 s. Use distance along the plane rather than vertical height in the along-plane motion equation.
A new applied force, loss of contact or impact changes the model and may require a new stage. An angled applied force also changes the normal equation; the simple R = mg cos θ relation assumes every other applied force is parallel to the slope. Do not use a friction term on a surface specified to be smooth, or import air resistance that the question has not supplied.
Pause and explain
A gravity-only particle on a smooth 30° slope has what acceleration with up positive?
Worked example
A 3 kg particle is pulled up a smooth slope with sin θ = 3/5 and cos θ = 4/5 by a 24 N force parallel to the plane. Find reaction and acceleration.
Show the worked solution
- Perpendicular balance gives R = 3 × 10 × 4/5 = 24 N.
- Downslope weight component is 3 × 10 × 3/5 = 18 N.
- Upward balance gives 24 − 18 = 3a, so a = +2 m s⁻².
Answer R = 24 N; a = +2 m s⁻² up the plane.
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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 particle is released from rest on a smooth slope with sin θ = 3/5 and cos θ = 4/5. Find reaction, downslope acceleration and distance after 2 s before any boundary is reached.
Give me a hint
Resolve weight in both directions; take down the slope positive.
Compare my reasoning
- Reaction is 40(4/5) = 32 N.
- Along the plane, 40(3/5) = 4a, giving a = 6 m s⁻² downwards.
- Distance in 2 s is ½ × 6 × 2² = 12 m along the slope.
R = 32 N; downslope a = 6 m s⁻²; distance 12 m.
Look for these in your work
- Use the correct two weight components.
- Calculate distance along the plane rather than vertical drop.
A 3 kg particle is pulled up a smooth slope by a parallel 24 N force. Given sin θ = 3/5 and cos θ = 4/5, find reaction and acceleration.
Give me a hint
The parallel pull does not change the perpendicular weight balance.
Compare my reasoning
- R = 30(4/5) = 24 N.
- The downslope weight component is 30(3/5) = 18 N.
- The along-plane resultant is 6 N up, so a = 6/3 = 2 m s⁻² up.
R = 24 N; a = +2 m s⁻² up the plane.
Look for these in your work
- Omit friction on the stated smooth surface.
- Divide the net along-plane force by mass.
A particle is projected up a long smooth 30° plane at 10 m s⁻¹. With up positive, find time and distance to first rest and its return time and velocity at the starting point.
Give me a hint
Keep a = −5 through the turning point; use the positive return-time root.
Compare my reasoning
- Acceleration is −10 sin 30° = −5 m s⁻²; 0 = 10 − 5t gives first rest at 2 s.
- The greatest along-plane rise is 10²/(2 × 5) = 10 m.
- Set s = 10t − 2.5t² = 0 and exclude t = 0: return at 4 s with v = 10 − 5 × 4 = −10 m s⁻¹.
First rest at 2 s after 10 m; return at 4 s with v = −10 m s⁻¹.
Look for these in your work
- Retain the same gravitational direction on both legs.
- Exclude the initial-time root from the return event.
Common mistakes
- The along-plane weight component uses sine for this angle definition.
- Smooth removes friction, not weight or reaction.
- Use along-plane displacement in the along-plane motion equations.
What can you explain now?
For a smooth slope, does gravity’s downslope acceleration vanish when upward velocity first reaches zero?
Compare with the explanation
No; it remains −g sin θ with up positive.
Momentary rest at a turning point does not cancel the gravitational component.
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Make it stick
Resolve weight into sine along and cosine normal, then use the signed along-plane acceleration through each valid stage.
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