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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section M2.4 (Newton’s Law of Motion) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. This content is part of International A-level Unit M2 (Mechanics) and is not examined at International AS. Use g = 9.8 m s⁻² and round answers to 3 significant figures unless told otherwise. In vector questions, i, j and k are perpendicular unit vectors, force is in newtons, time in seconds, distance in metres and velocity in m s⁻¹.

Learn the content first with the study guide and the revision notes. Course hub: /boards/oxfordaqa/a-level/mathematics/. Checklist: /checklists/oxfordaqa/a-level/mathematics/. Try the free 10-minute diagnostics to check where you stand.

Questions

1. A particle is released from rest on a smooth plane inclined at 35° to the horizontal. Find its acceleration. [3]

2. A block of mass 8 kg is pulled up a smooth plane, inclined at 20° to the horizontal, by a constant force of 60 N acting parallel to the line of greatest slope. The block starts from rest.

(a) Find the acceleration of the block. [3] (b) Find its speed after 2.5 s. [1] (c) State one modelling assumption made by treating the plane as smooth, and say how the acceleration would change if the plane were rough. [1]

3. A block slides from rest down a rough plane inclined at 25° to the horizontal. The coefficient of friction between the block and the plane is 0.2. Find the acceleration of the block, and the time taken to slide 6 m. [5]

4. A block is given an initial speed of 7 m s⁻¹ up a line of greatest slope of a rough plane at 30° to the horizontal. The coefficient of friction is 0.5. Find the distance the block travels up the plane before stopping. Show that it then slides back down, and find its speed when it returns to its starting point. [7]

5. A particle of mass 4 kg moves in a straight line under a force F = (6t² − 8) N in the direction of motion. When t = 0 its velocity is 3 m s⁻¹.

(a) Find the acceleration when t = 2. [1] (b) Find the velocity when t = 4. [3]

6. A particle of mass 1.5 kg starts from rest at the origin and moves along the x-axis under a force F = 12 sin 2t newtons in the positive x direction.

(a) Show that its velocity is 4 − 4 cos 2t. [3] (b) Find, in terms of π, its displacement from the origin when t = π/2. [2] (c) Find the maximum speed of the particle. [1]

7. A particle of mass 0.4 kg moves under a single force F = (0.8t i + (3.2 − 1.6t) j). When t = 0 its velocity is (i − 6j).

(a) Find the velocity at time t. [3] (b) Find the speed when t = 2. [1] (c) Find the two values of t when the particle is moving parallel to i. [2]

8. A particle of mass 2.5 kg has position vector r = (t³ − 3t²) i + (2t² + t) j + (4t − t³) k, relative to a fixed origin.

(a) Find the force acting on the particle at time t. [3] (b) Find the time at which the force has no i component, and the magnitude of the force at that time. [3]

9. A particle of mass 2 kg is released from rest on a smooth plane inclined at 30° to the horizontal. At time t = 0, a force P = 4t newtons begins to act on it, directed up the line of greatest slope.

(a) Show that the acceleration up the slope is (2t − 4.9) m s⁻². [3] (b) Find the positive value of t at which the particle is next at rest. [2] (c) Find the greatest distance of the particle below its starting point, measured along the slope. [3]

10. A particle of mass 2 kg moves under a force F = (12t i + (8 − 12t) j). At t = 0 it is at the origin with velocity −3i.

(a) Find its velocity at time t. [3] (b) Find the positive value of t at which the particle is moving parallel to i. [2] (c) Find the position vector of the particle at that time. [3]

Answers

1. Along the slope: mg sin35° = ma [1] a = 9.8 sin35° [1] a = 5.62 m s⁻² down the slope [1] [3] Examiner insight: Write the equation with m on both sides before cancelling. It shows the resolving step, which is where most of the credit is, even if you slip on the arithmetic.

2. (a) Up the slope: 60 − 8g sin20° = 8a [1] 8g sin20° = 26.81, so 8a = 33.19 [1] a = 4.15 m s⁻² [1] (b) v = 0 + 4.148(2.5) = 10.4 m s⁻¹ [1] (c) Particle/block treated as a point and no friction acts; with a rough surface friction acts down the slope, so the acceleration would be smaller [1] [5] Examiner insight: A modelling comment needs both parts: what is assumed and the effect on the result. Naming the term alone is rarely enough.

3. Perpendicular: R = mg cos25° [1] Friction F = μR = 0.2 mg cos25° [1] Down the slope: mg sin25° − 0.2 mg cos25° = ma [1] a = 9.8(0.4226 − 0.1813) = 2.37 m s⁻² [1] 6 = ½(2.365)t², so t = √(12/2.365) = 2.25 s [1] [5] Examiner insight: No mass is given. Writing m (or leaving it symbolic) and cancelling shows you understand why. Do not invent a numerical mass.

4. Up the slope the friction acts down: R = mg cos30° and F = 0.5 mg cos30° [1] Deceleration: ma = mg sin30° + 0.5 mg cos30°, so a = 9.8(0.5 + 0.4330) = 9.144 [1] Using v² = u² + 2as with v = 0: 0 = 49 − 2(9.144)s [1] s = 2.68 m [1] At rest: weight component mg sin30° = 4.9m, maximum friction = 0.5 mg cos30° = 4.24m. 4.9m > 4.24m, so it slides back [1] Down the slope: a = 9.8(0.5 − 0.4330) = 0.6565 m s⁻² [1] v² = 2(0.6565)(2.679) = 3.519, so v = 1.88 m s⁻¹ [1] [7] Examiner insight: The “show that” needs a stated comparison of forces (or of tan30° with μ) and a conclusion. Writing the final line “so it slides” is part of the argument.

5. (a) a = F/m = (6t² − 8)/4 = 1.5t² − 2; at t = 2, a = 4 m s⁻² [1] (b) v = ∫(1.5t² − 2) dt = 0.5t³ − 2t + c [1] v = 3 at t = 0, so c = 3 [1] At t = 4: 32 − 8 + 3 = 27 m s⁻¹ [1] [4] Examiner insight: Dividing by the mass before integrating keeps the constants simple. Substituting t = 4 without first finding c often leads to a wrong answer.

6. (a) a = 12 sin2t / 1.5 = 8 sin2t [1] v = −4 cos2t + c [1] v = 0 at t = 0, so c = 4, giving v = 4 − 4 cos2t [1] (b) x = 4t − 2 sin2t (x = 0 at t = 0) [1] At t = π/2: x = 2π − 2 sinπ = 2π m [1] (c) Greatest value of 4 − 4cos2t is when cos2t = −1: 8 m s⁻¹ [1] [6] Examiner insight: In a “show that”, each step must appear: the acceleration, the integral, and the use of the initial condition to find c.

7. (a) a = F/0.4 = 2t i + (8 − 4t) j [1] v = (t² + c1) i + (8t − 2t² + c2) j [1] Using v = i − 6j at t = 0: v = (t² + 1) i + (−2t² + 8t − 6) j [1] (b) t = 2: v = 5i + 2j, speed = √29 = 5.39 m s⁻¹ [1] (c) Set the j component to zero: −2(t − 1)(t − 3) = 0 [1] t = 1 and t = 3 [1] [6] Examiner insight: Both roots are asked for. Check the i component is non-zero at each (it is 2 and 10), otherwise the particle would be at rest rather than moving parallel to i.

8. (a) v = dr/dt = (3t² − 6t) i + (4t + 1) j + (4 − 3t²) k [1] a = (6t − 6) i + 4 j − 6t k [1] F = 2.5a = (15t − 15) i + 10 j − 15t k [1] (b) 15t − 15 = 0, so t = 1 [1] F = 10j − 15k [1] |F| = √(100 + 225) = √325 = 18.0 N [1] [6] Examiner insight: Differentiate twice before multiplying by the mass. A force written as a number rather than a vector in (a) loses the final step.

9. (a) Up the slope, taking the positive direction up: 4t − 2g sin30° = 2a [1] 2g sin30° = 9.8 [1] a = (4t − 9.8)/2 = 2t − 4.9 [1] (b) v = ∫(2t − 4.9) dt = t² − 4.9t (v = 0 at t = 0) [1] v = t(t − 4.9) = 0, so t = 4.9 s [1] (c) x = t³/3 − 2.45t² (x = 0 at t = 0) [1] v < 0 for 0 < t < 4.9, so the greatest distance below the start is at t = 4.9 [1] x(4.9) = 39.22 − 58.82 = −19.6, so the distance is 19.6 m [1] [8] Examiner insight: Justify why t = 4.9 gives the greatest distance by stating the sign of v. Choosing the time when a = 0 (t = 2.45) is a common misreading: that gives the greatest speed, not the greatest distance.

10. (a) a = F/2 = 6t i + (4 − 6t) j [1] v = (3t² + c1) i + (4t − 3t² + c2) j [1] v = −3i at t = 0: v = (3t² − 3) i + (4t − 3t²) j [1] (b) j component zero: t(4 − 3t) = 0; t > 0 gives t = 4/3 [1] (The value t = 0 is excluded by the question.) [1] (c) r = (t³ − 3t) i + (2t² − t³) j (r = 0 at t = 0) [1] At t = 4/3: i component = 64/27 − 4 = −44/27 [1] r = (−44/27) i + (32/27) j ≈ −1.63i + 1.19j [1] [8] Examiner insight: Keep exact fractions until the end. Rounding t = 4/3 to 1.33 early gives answers that differ in the third significant figure.

Where marks are usually lost

• Using cos for the component of weight along the slope instead of sin.
• Forgetting that friction reverses direction when the block turns round.
• Assuming a block stays at rest after stopping without comparing mg sin θ with μ mg cos θ.
• Using constant-acceleration equations where the acceleration is a function of t.
• Missing the constant of integration, or using one constant for both integrations.
• Dividing by the mass when you should multiply (force from acceleration) or the other way round.
• Giving only the magnitude of a vector when the question asks for a vector.
• Reporting displacement instead of distance without checking where the velocity changes sign.
• Quoting a rounded intermediate value (for example 1.33 for 4/3) and carrying it forward.

Next steps

• Review the revision notes and the study guide.
• Course hub: /boards/oxfordaqa/a-level/mathematics/ and checklist.
• Take the free 10-minute diagnostics to find your weakest topics.
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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 questions follow section M2.4.