OxfordAQA International AS and A-level Mathematics: Newton's laws: slopes, variable acceleration (9660) – Revision Notes

Quick recall for section M2.4 of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. This content sits in International A-level Unit M2 (Mechanics) and is not examined at International AS. For full explanations and longer worked examples, read the study guide. Then practise with the practice questions.

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Key results

Idea	Result
Newton’s second law	F = ma, a vector equation: resultant force = mass × acceleration
Weight on a slope at angle θ	mg sin θ along the slope (downwards), mg cos θ into the slope
Normal reaction (no other perpendicular force)	R = mg cos θ
Friction while sliding	F = μR, opposing motion
Friction when not sliding	F ≤ μR
Smooth slope, sliding	a = g sin θ, independent of mass
Rough slope, sliding down	a = g(sin θ − μ cos θ)
Rough slope, sliding up	deceleration = g(sin θ + μ cos θ)
Variable acceleration	a = dv/dt, v = dr/dt, v = ∫ a dt, r = ∫ v dt
Vectors	r = x i + y j + z k, v = dr/dt, a = d²r/dt², F = ma

Constant-acceleration equations (v = u + at, s = ut + ½at², v² = u² + 2as, s = ½(u + v)t) hold only when a is constant. Use g = 9.8 m s⁻² unless told otherwise.

Method for a slope problem

1. Draw the force diagram: weight mg, normal reaction R, friction F (opposing motion), any applied force.
2. Mark the direction of motion and the positive direction.
3. Resolve perpendicular to the slope: R = mg cos θ (adjust if an applied force has a perpendicular component).
4. If sliding, write F = μR.
5. Resolve along the slope: resultant = ma.
6. Solve for a, then use constant-acceleration equations for time, distance or speed.

Reminder: at rest or sliding? On a slope with coefficient μ, a particle stays at rest if tan θ ≤ μ. It slides if tan θ > μ. For a particle that has stopped, compare mg sin θ with the largest possible friction μ mg cos θ. Do not assume it stays put.

Reminder: up then down. Going up, friction acts down the slope, so the deceleration is g(sin θ + μ cos θ). On the way back, friction acts up the slope, so a = g(sin θ − μ cos θ). These are different values.

Method for a variable acceleration problem

1. Find a. If given a force F(t), a = F/m. If given r(t), differentiate twice.
2. Integrate to get v. Add the constant using the stated condition, usually the velocity at t = 0.
3. Integrate again to get r. Add a second constant from a stated position.
4. For vectors, do each component separately and keep the answer as a vector.
5. Check what is asked: velocity or speed, displacement or distance, vector or magnitude.

Speed and direction from v = p i + q j. Speed = √(p² + q²). Angle to i: tan⁻¹(q/p), taking care with the quadrant. In three dimensions, speed = √(p² + q² + w²) for v = p i + q j + w k.

Distance versus displacement. Find the times when v = 0. Split the integral there and add the sizes of the separate displacements.

Special conditions in vectors.

Statement	Condition
Moving parallel to i	j (and k) component of v is zero
Force acts in the j direction only	i and k components of F are zero
At rest	every component of v is zero at the same time
Passes through the origin	every component of r is zero at the same time
Magnitude of force equals N	sum of squares of components equals N²

Three dimensions. If k is vertically upwards, include −mg k in the resultant force.

Worked reminders

Smooth slope. Mass 2 kg, slope 15°, smooth: a = 9.8 sin15° = 2.54 m s⁻². The component of weight along the slope is 2(9.8) sin15° = 5.07 N.

Variable acceleration. F = 6t N acts on a 3 kg particle starting from rest at t = 0. Then a = 2t, v = t², and r = t³/3 + constant.

Force from position. r = (t³ − t) i + 3t² j with m = 3: a = 6t i + 6 j, so F = 18t i + 18 j.

Rough slope with an applied force. Mass 6 kg, slope 20°, μ = 0.3, pushed up the slope by 50 N parallel to it. R = 6(9.8) cos20° = 55.25 N, friction 0.3R = 16.58 N down the slope, weight component 20.11 N down the slope. Then 50 − 20.11 − 16.58 = 6a, so a = 2.22 m s⁻² up the slope.

Distance travelled. If v = −(t − 1)(t − 5) with x = 0 at t = 0, the velocity changes sign at t = 1. Then x(1) = −7/3 and x(5) = 25/3, so the distance is 7/3 + 32/3 = 13.

Modelling words to use

• Particle: size ignored, all forces act at a point.
• Smooth: no friction, so only the weight component and applied forces act along the slope.
• Motion under gravity: no air resistance, g constant.
• A force law F(t) is valid only over the time stated in the question.

Must-know distinctions

• F = ma uses the resultant force, not one force.
• sin θ goes with the slope direction, cos θ with the perpendicular, when θ is the angle to the horizontal.
• Limiting friction F = μR applies only when sliding or about to slide.
• Distance travelled is never less than the magnitude of the displacement.
• A definite integral from 0 to T gives the change in the quantity. An indefinite integral needs a constant.

Quick self-test

1. A particle slides down a smooth plane inclined at 40°. Find its acceleration to 3 s.f.
2. A 5 kg block rests on a plane inclined at 25°. Find the normal reaction to 3 s.f.
3. A block rests on a plane inclined at 20° with μ = 0.5. Can it remain at rest? Give a reason.
4. A particle of mass m has a = 4t − 3 and v = 2 at t = 0. Find v when t = 3.
5. A 2 kg particle starts from rest under F = (4t i + 6 j) N. Find its speed at t = 3 to 3 s.f.
6. A 3 kg particle has r = (t³ − t) i + 3t² j. Find the magnitude of the force at t = 2 to 3 s.f.
7. A block slides down a plane inclined at 30° with μ = 0.2. Find its acceleration to 3 s.f.
8. A particle has v = (t − 2)(t − 4) for t from 0 to 4, starting at x = 0. Find the distance travelled.
9. A 3 kg particle has a = (2i − j + 4t k). Find the magnitude of the force at t = 1 to 4 s.f.
10. A 4 kg particle is projected up a smooth plane inclined at 30° with speed 7.35 m s⁻¹. Find the time to come to rest and the distance travelled up the slope.

Answers

1. a = 9.8 sin40° = 6.30 m s⁻².
2. R = 5(9.8) cos25° = 44.4 N.
3. tan20° = 0.364, which is less than 0.5. The maximum friction exceeds the weight component along the slope, so yes, it can remain at rest.
4. v = 2t² − 3t + 2, so at t = 3, v = 18 − 9 + 2 = 11 (in the same units as the question uses).
5. a = (2t, 3), v = (t², 3t), at t = 3 v = (9, 9), speed = 9√2 = 12.7 m s⁻¹.
6. a = (6t, 6), F = (18t, 18), at t = 2 F = (36, 18), |F| = √1620 = 40.2 N.
7. a = 9.8(sin30° − 0.2 cos30°) = 3.20 m s⁻² down the slope.
8. x = t³/3 − 3t² + 8t. x(2) = 20/3 and x(4) = 16/3. Distance = 20/3 + (20/3 − 16/3) = 8 (same units as x).
9. F = 3a = (6, −3, 12t). At t = 1: |F| = √(36 + 9 + 144) = √189 = 13.75 N.
10. Deceleration = 9.8 sin30° = 4.9 m s⁻². t = 7.35/4.9 = 1.5 s. Distance = 7.35²/(2 × 4.9) = 5.51 m.

Where marks are usually lost

• Resolving with sin and cos the wrong way round on a slope.
• Missing a force: friction, or the applied force’s component, in the equation along the slope.
• Using the same friction direction for the upward and downward parts of a journey.
• Applying v = u + at to an acceleration that varies with t.
• Forgetting a constant of integration, or finding it from the wrong condition.
• Stopping at a vector acceleration without the factor of mass when asked for force (or dividing the wrong way).
• Giving displacement when asked for distance, because the sign change in velocity was never checked.
• Rounding too early. Keep full values until the last line, then give 3 significant figures unless told otherwise.
• Not stating the direction of an answer when it is a vector or “up/down the slope”.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, for International AS exams from May/June 2018 and International A-level exams from May/June 2019, published by OxfordAQA. These notes follow section M2.4.