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

This guide covers specification section M2.4 (Newton’s Law of Motion) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International A-level (A2) Unit M2 (Mechanics), which is part of the International A-level only and is not examined at International AS. Unit M2 is a written paper of 1 hour 30 minutes, 80 marks, with a calculator allowed. The same unit also contains M2.2 (kinematics with calculus) and M2.3 (statics and forces), which you will use throughout.

Course hub: /boards/oxfordaqa/a-level/mathematics/. Printable checklist: /checklists/oxfordaqa/a-level/mathematics/. Companion pages: revision notes and practice questions. Not sure where your gaps are? Try the free 10-minute diagnostics.

What you must be able to do

Reference	What the specification asks	Where in this guide
M2.4	Apply Newton’s laws to linear motion with constant acceleration, including motion up or down an inclined plane	Sections 1 and 2
M2.4	Apply Newton’s laws where the acceleration is variable; problems may be in one, two or three dimensions and may need integration or differentiation	Sections 3 to 5
M2.2	Position, velocity and acceleration vectors by differentiating or integrating with respect to t	Sections 3 to 5
M2.3	Friction with F ≤ μR, resolving forces	Section 2
M2.1	Comment on assumptions such as particle, smooth surface, motion under gravity	Section 6

1. Newton’s second law for constant acceleration

For a particle of mass m (kg) acted on by forces with resultant F (N), the acceleration a (m s⁻²) satisfies

F = ma     (resultant force = mass × acceleration)

In one dimension, choose a positive direction and write “resultant in the positive direction = ma”. If the resultant is zero the acceleration is zero (Newton’s first law). When a is constant you may also use the constant-acceleration equations:

v = u + at     s = ut + ½at²     v² = u² + 2as     s = ½(u + v)t

Do not use these when a changes with time. That is the subject of sections 3 to 5.

Method: draw a force diagram, mark the direction of motion, write one equation of motion for each direction, then solve.

2. Motion on an inclined plane

A particle of mass m on a plane at angle θ to the horizontal has weight mg acting vertically downwards. Resolve it along and perpendicular to the plane:

component down the slope = mg sin θ
component into the slope = mg cos θ

There is no acceleration perpendicular to the plane, so the normal reaction is R = mg cos θ if no other force has a component perpendicular to the plane. Along the plane you apply F = ma.

When the particle slides, friction opposes the motion and is limiting, so F = μR. If a particle is not sliding, friction only needs to satisfy F ≤ μR. Take g = 9.8 m s⁻² unless told otherwise.

Worked example 1 (smooth plane). A particle of mass 3.5 kg is released from rest on a smooth plane inclined at 32° to the horizontal. Find its acceleration, the time it takes to travel 4.5 m, and its speed after 4.5 m.

Down the plane: 3.5g sin32° = 3.5a
a = 9.8 sin32° = 5.19 m s⁻²
s = ½at²: 4.5 = ½(5.193)t²  →  t = 1.32 s
v² = 2as = 2(5.193)(4.5)  →  v = 6.84 m s⁻¹

The mass cancels, so on a smooth slope the acceleration does not depend on mass. a = 5.19 m s⁻², t = 1.32 s, v = 6.84 m s⁻¹.

Worked example 2 (rough plane, applied force). A block of mass 6 kg is pushed up a rough plane inclined at 20° by a force of 50 N acting parallel to the line of greatest slope. The coefficient of friction is 0.3. Find the acceleration.

Perpendicular: R = 6g cos20° = 55.25 N
Friction: F = μR = 0.3(55.25) = 16.58 N, acting down the slope
Weight component: 6g sin20° = 20.11 N, down the slope
Up the slope: 50 − 20.11 − 16.58 = 6a
6a = 13.31  →  a = 2.22 m s⁻²

Acceleration 2.22 m s⁻² up the slope.

Worked example 3 (finding μ from observed motion). A 2.4 kg block, starting at rest on a rough 30° incline, slides 3 m in 2 s. Find μ.

s = ½at²: 3 = ½a(4)  →  a = 1.5 m s⁻²
Perpendicular: R = 2.4g cos30°
Down the slope: 2.4g sin30° − μ(2.4g cos30°) = 2.4(1.5)
Divide by 2.4: 4.9 − μ(8.487) = 1.5
μ = 3.4 / 8.487 = 0.401

μ = 0.401. A useful check: a particle on a rough slope can only start from rest and slide down if tan θ > μ. Here tan30° = 0.577, which is larger than 0.401, so it does slide.

3. Variable acceleration in one dimension

When the force depends on time, F(t), the acceleration is a(t) = F(t)/m. Then:

a = dv/dt     v = dx/dt     v = ∫ a dt     x = ∫ v dt

Each integration gives a constant. Use the given conditions (for example, the velocity at t = 0) to find it. Always write the constant before you substitute.

Worked example 4. A particle of mass 2 kg moves on the x-axis under a single force F = (12 − 4t) N in the positive x direction. At t = 0 the particle is at the origin with velocity −5 m s⁻¹. Find (a) its velocity at time t, (b) the times when it is instantaneously at rest, (c) the total distance travelled between t = 0 and t = 5.

(a) a = (12 − 4t)/2 = 6 − 2t
    v = 6t − t² + c;  v = −5 when t = 0, so c = −5
    v = −t² + 6t − 5 = −(t − 1)(t − 5)
(b) v = 0 at t = 1 and t = 5
(c) x = −t³/3 + 3t² − 5t   (x = 0 at t = 0)
    x(1) = −1/3 + 3 − 5 = −7/3
    x(5) = −125/3 + 75 − 25 = 25/3

The velocity is negative for 0 < t < 1 and positive for 1 < t < 5, so the particle turns round at t = 1. Distance = |x(1)| + (x(5) − x(1)) = 7/3 + 32/3 = 13 m. Displacement alone, 25/3 m, would be the wrong answer to a “distance travelled” question.

4. Variable acceleration in two dimensions

Forces and motion are written as vectors in i and j (or as column vectors; the specification allows either). Each component obeys Newton’s law separately: F = ma means Fx = max and Fy = may. Integrating a vector integrates each component and adds a constant vector, found from the initial velocity or position.

Worked example 5. A particle of mass 0.5 kg moves under a force F = (6t i + 4 j) N, with t in seconds. At t = 0 it is at the origin with velocity (2i − 3j) m s⁻¹. Find its velocity and position at t = 1, and its speed and direction at that time.

a = F/0.5 = 12t i + 8 j
v = (6t² + c1) i + (8t + c2) j;  at t = 0 v = 2i − 3j
v = (6t² + 2) i + (8t − 3) j
r = (2t³ + 2t) i + (4t² − 3t) j        (r = 0 at t = 0)
t = 1: v = 8i + 5j,  r = 4i + j
speed = √(8² + 5²) = √89 = 9.43 m s⁻¹
angle above the i direction: tan⁻¹(5/8) = 32.0°

v = (8i + 5j) m s⁻¹, r = (4i + j) m, speed 9.43 m s⁻¹ at 32.0° to i.

5. Variable acceleration in three dimensions, and using differentiation

The method is identical with a third component k. If k is vertical, the weight mg acts in the −k direction and must be included in the resultant force.

Sometimes you are given the position vector r(t) and asked for the force. Then you differentiate twice: v = dr/dt, a = dv/dt, and F = ma.

Worked example 6. A particle of mass 2 kg has position vector r = 2t² i + (t³ − 6t) j + 4t k metres at time t seconds. Find (a) the force acting on it at time t, (b) the value of t when the magnitude of the force is 10 N, (c) its speed when it is moving parallel to the plane containing i and k.

(a) v = 4t i + (3t² − 6) j + 4 k
    a = 4 i + 6t j
    F = 2a = (8 i + 12t j) N
(b) |F|² = 64 + 144t² = 100  →  t² = 1/4  →  t = 0.5 (t > 0)
(c) Moving parallel to the i–k plane means the j component of v is zero:
    3t² − 6 = 0  →  t = √2
    v = 4√2 i + 4 k,  speed = √(32 + 16) = √48 = 4√3 = 6.93 m s⁻¹

(a) F = 8i + 12t j N, (b) t = 0.5 s, (c) 6.93 m s⁻¹.

6. Modelling comments

The specification expects you to comment on terms such as particle, smooth surface, light, inextensible string and motion under gravity. For this unit:

• Particle: the object’s size and rotation are ignored, so all forces act at one point.
• Smooth surface: friction is zero, so the only forces along the plane are the weight component and any applied force.
• Motion under gravity: air resistance is ignored and g is taken as constant.
• A force given as a function of t is a model; it only holds for the time interval in which the model is stated to apply.

Common errors

• Using mg cos θ down the slope and mg sin θ into it. If the angle is between the plane and the horizontal, sin goes along the plane.
• Putting friction in the wrong direction. It opposes motion, or the tendency to move.
• Using constant-acceleration equations when a depends on t.
• Omitting a constant of integration, or applying the t = 0 condition to the wrong integral: the constant for r differs from the one for v.
• Reporting displacement when asked for distance. Split the motion where v changes sign.
• Dropping the vector structure: an answer to “find the velocity” should be a vector, not a number.
• Forgetting the weight term in a three-dimensional problem where k is vertical.

Next steps

Condense this into the revision notes, then test yourself with the practice questions. Return to the course hub or the checklist to see what else is in Unit M2, and use the diagnostics to find weaker topics.

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. This guide follows section M2.4 and draws on M2.1 to M2.3.