Cambridge AS & A Level Further Mathematics 9231/32 is the May/June 2026 Paper 3 (Further Mechanics), variant 2. It lasts 1 hour 30 minutes, is worth 50 marks, and you answer all seven questions. It counts towards the full A Level and is one of the two applied options for AS Further Maths. Every question was worth 5 to 9 marks. The biggest were Question 3 on motion against a resistance proportional to v² and Question 5 on a vertical circle where the string goes slack (9 marks each). This guide goes through the paper question by question with the final answers and what the official mark scheme gave marks for. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of all four 9231 papers from this series, see A Level Further Maths 9231 May/June 2026 past papers: what came up and how it was marked.

9231/32 at a glance

Question	Topic	Marks
1	Centre of mass of a hemisphere (radius 3r) on a cylinder (radius r, height x); angle when suspended from the rim	7
2	A particle on a string fixed at two points, moving in a horizontal circle: both tensions	5
3	Driving force 8 N, resistance 0.5v² N: speed after 0.8 m, then t in terms of v	9
4	Oblique impact with a smooth barrier: e from a kinetic energy ratio, then a second barrier at 120°	8
5	Vertical circle on a string: tension in terms of θ, greatest height after the string goes slack	9
6	Elastic string with a falling particle: v² as a function of x, maximum extension, maximum speed	6
7	Projectile that strikes an inclined plane at right angles: find u and θ	6
	Total	50

Question-by-question walkthrough

The mark scheme uses the standard Cambridge mark types. M marks are for a correct method applied to this problem (not lost for slips), A marks are for accurate answers and need the M mark before them, and B marks stand alone. A DM (or DB) mark depends on an earlier mark flagged with an asterisk, such as M1* or B1*: lose the starred mark and the dependent one goes too. FT (follow through) lets you score with your own earlier wrong value. AG means the answer is given on the paper, so every step has to be shown. The cover asks for non-exact answers to 3 significant figures, or 1 decimal place for angles in degrees, and g = 10 m s⁻².

Question 1: hemisphere and cylinder (7 marks)

A solid hemisphere of radius 3r sits on a solid cylinder of radius r and height x, on the same axis and made of the same material.

• (a) Centre of mass from the base of the cylinder (4 marks): the volumes are πr²x (cylinder) and ⅔π(3r)³ = 18πr³ (hemisphere). Their centres of mass are x/2 and x + 9r/8 from the base of the cylinder (the hemisphere’s centre of mass is ⅜ of its radius from its flat face). Taking moments:
  • x̄(r²x + 18r³) = r²x × ½x + 18r³(x + 9r/8)
  • x̄ = (2x² + 72rx + 81r²) / (4x + 72r), or an equivalent simplified form.
  • Marking: M1 for a dimensionally correct moments equation with three terms, B1 for at least two terms correct, A1 for all correct, then A1 for the simplified answer.
• (b) Angle when hung from A on the rim of the hemisphere’s base, with x = r (3 marks): hanging freely, the centre of mass is vertically below A. With x = r, x̄ = 155r/76, so the centre of mass is 155r/76 − r = 79r/76 from the hemisphere’s flat face. Then tan θ = (79r/76) ÷ 3r = 79/228, giving θ = 19.1°.
  • Marking: M1* for tan θ = (x̄ − x)/3r or an equivalent ratio, DM1 for substituting your x̄ correctly in terms of r (or x) only, A1 for 19.1° (or 0.334 radians). A negative angle as the final answer lost the A mark.

Question 2: two strings and a horizontal circle (5 marks)

A string of length 28 m is tied to A and to B directly below it. A 7 kg particle P is fixed on the string with AP = 15 m and moves at 4 rad s⁻¹ in a horizontal circle whose centre is 9 m below A.

• Geometry: the radius is √(15² − 9²) = 12 m, so BP = 13 m sits 5 m below the plane of the circle. AP makes an angle with sine 3/5 with the horizontal; BP makes one with sine 5/13.
• Resolve:
  • vertically: ⅗T_A = (5/13)T_B + 7g
  • horizontally (towards the centre): ⅘T_A + (12/13)T_B = 7 × 12 × 4² = 1344
• Answer: T_A = 675 N and T_B = 871 N.
• Marking: B1 for the radial force 7 × 12 × 4², M1 for an attempt at Newton’s second law in either direction (correct number of terms, sin and cos substituted), A1 for each equation, then A1 for both tensions. Angles of about 37° and 23° were accepted in place of the exact ratios.

Question 3: resistance proportional to v² (9 marks)

A 6.4 kg particle starts from rest with a driving force of 8 N and a resistance of 0.5v² N.

• (a) Speed after 0.8 m (5 marks): use a = v dv/dx, because the question links speed to distance.
  • 8 − 0.5v² = 6.4v dv/dx
  • x = ∫ 6.4v/(8 − 0.5v²) dv = −6.4 ln(8 − 0.5v²) + c
  • x = 0 when v = 0 gives c = 6.4 ln 8
  • at x = 0.8: 8 − 0.5v² = 8e^(−1/8), so v = 4√(1 − e^(−1/8)) = 1.37 m s⁻¹
  • Marking: B1* for Newton’s second law with v dv/dx, M1* for separating variables and reaching a log form, A1 for the correct integral, DM1 for using the initial condition, then A1 (WWW, without wrong working). Substituting v = 0 into ln(0.5v² − 8), the log of a negative number, scored DM0.
• (b) t in terms of v (4 marks): this time a = dv/dt.
  • 8 − 0.5v² = 6.4 dv/dt, so t = ∫ 12.8/(16 − v²) dv
  • partial fractions give t = 1.6 ln((4 + v)/(4 − v)), and t = 0 when v = 0 gives c = 0
  • Marking: B1 for the equation with dv/dt, M1 for separating and integrating to two log terms (3.2 artanh(v/4) was also accepted), A1 for the integral and A1 for showing c = 0. Writing the answer in x instead of v capped the part at 2 marks.

Question 4: two smooth barriers (8 marks)

P hits a smooth barrier at an angle α to it, with tan α = 12/5, and leaves with 4/13 of its original kinetic energy.

• (a) Coefficient of restitution (4 marks): sin α = 12/13 and cos α = 5/13. After impact, the component perpendicular to the barrier is (12/13)eu and the component along it stays (5/13)u. Then:
  • ½mv² = (4/13) × ½mu², so v² = (4/13)u²
  • (12eu/13)² + (5u/13)² = (4/13)u², so 144e² + 25 = 52
  • e = √3/4 ≈ 0.433
  • Marking: B1 for both velocity components, B1* for the kinetic energy equation, DM1 for combining the components into an equation in e, A1 for the answer.
• (b) Angle after hitting the second barrier at 120° to the first (4 marks): P leaves the first barrier at angle β with tan β = e × 12/5 = 3√3/5 (β ≈ 46.1°). The two barriers and P’s path make a triangle with a 120° angle at the corner, so P meets the second barrier at 60° − β, where tan(60° − β) = √3/7 (about 13.9°). The same restitution rule then gives tan θ = e tan(60° − β) = 3/28, so θ = 6.12° (the question asked for 3 significant figures).
  • Marking: B1*FT for tan β from your e (if 0 < e < 1), DB1 for the angle with the second barrier, M1 for both component equations for the second impact, A1 for 6.12 (6.11 also accepted).

Question 5: vertical circle (9 marks)

A particle hangs at A on a string of length a and is projected horizontally with speed 2√(ag).

• (a) Tension after turning through θ (4 marks):
  • energy: ½m(4ag) = ½mv² + mga(1 − cos θ), so v² = 2ag + 2ag cos θ
  • Newton’s second law along the string: T − mg cos θ = mv²/a
  • T = mg(3cos θ + 2)
  • Marking: M1* for a three-term energy equation, B1 for the radial equation, DM1 for combining them, A1 for the answer.
• (b) Greatest height above A (5 marks):
  • the string goes slack when T = 0, so cos θ = −2/3, and then v² = 2ag/3
  • the particle is at height a(1 − cos θ) = 5a/3 above A and becomes a projectile
  • its vertical velocity component is v sin θ, with sin θ = √5/3, so (v sin θ)² = 10ag/27
  • extra height = (10ag/27) ÷ 2g = 5a/27
  • greatest height = 5a/3 + 5a/27 = 50a/27 (about 1.85a)
  • Marking: M1* for setting your tension to zero, then three DM1 marks in a chain: v² at the slack point, the vertical component v sin θ, and v² = u² + 2as for the extra height. The final A1 depended on full marks in part (a), or on a fully correct restart in (b).

Question 6: elastic string (6 marks)

A 1.5 kg particle on an elastic string (natural length 2.5 m, modulus 15 N) is released from rest at the fixed end O.

• (a) Show that v² = −4(x² − 5x − 12.5) (3 marks, AG): when the extension is x the particle has fallen 2.5 + x. Energy:
  • 1.5g(2.5 + x) = ½ × 1.5v² + 15x²/(2 × 2.5)
  • 37.5 + 15x = 0.75v² + 3x², so v² = 50 + 20x − 4x²
  • Marking: B1 for the elastic energy, M1 for a three-term energy equation, A1 for reaching the printed result with at least one line of working. Any error seen lost the A1. A second route, Newton’s second law with v dv/dx then using v² = 50 at x = 0, was marked the same way.
• (b) Maximum extension (1 mark): v = 0 when x² − 5x − 12.5 = 0, so E = 5(1 + √3)/2 ≈ 6.83. If the negative root appeared, it had to be rejected.
• (c) Maximum speed (2 marks): the speed is greatest when the acceleration is zero, at x = 2.5 (where the tension 6x equals the weight 15 N). Then v² = 75, so v = 5√3 ≈ 8.66 m s⁻¹. B1 for x = 2.5 and B1 for the speed.

Question 7: projectile onto an inclined plane (6 marks)

O and A are 10 m apart on level ground. A plane rises from A at angle α, with tan α = 4/3, and B is 5 m up it. A particle projected from O hits B moving at right angles to the plane.

• Distances: B is 10 + 3 = 13 m horizontally and 4 m vertically from O.
• Equations:
  • 13 = (u cos θ)t and 4 = (u sin θ)t − 5t²
  • at B the velocity is perpendicular to the plane, so (u sin θ − gt) / (u cos θ) = −3/4
• Solve: eliminating u and θ gives 20t² = 55, so t = √11/2 ≈ 1.66 s. Then u cos θ = 26/√11 and u sin θ = 71/(2√11), so tan θ = 71/52.
• Answer: θ = 53.8° and u = 13.3 m s⁻¹.
• Marking: B1* for both distances, DM1 for s = ut + ½at² horizontally and vertically, B1 for the perpendicular condition, M1 for eliminating down to one unknown, then A1 for each answer. The scheme also credited two routes through the trajectory equation, each worth the same 6 marks.

Where the marks are on 9231/32

• Circular motion carried 14 of the 50 marks: the horizontal circle in Question 2 (5) and the vertical circle in Question 5 (9).
• Variable-force motion carried 9 marks: Question 3, with one part in v dv/dx and one in dv/dt.
• Oblique collisions carried 8 marks: Question 4.
• Centre of mass and equilibrium carried 7 marks: Question 1.
• Elastic strings and energy carried 6 marks: Question 6.
• Projectiles carried 6 marks: Question 7.

Energy equations opened Questions 5(a) and 6(a), and Newton’s second law set up the working in Questions 2, 3 and 5. Question 6(a) was the only printed answer, so it was the only part where any error cost the accuracy mark outright.

How to self-mark 9231/32 with the mark scheme

1. Check the set-up equation first. Most questions open with a B1 or M1 for Newton’s second law or an energy equation. In Question 3 the B1 needed v dv/dx in part (a) and dv/dt in part (b), so the wrong form of acceleration loses the first mark.
2. Follow the dependency chain. Starred marks control the marks after them. In Question 5(b), three DM1 marks all depend on setting T = 0, so if that step is missing, the rest of the part scores nothing.
3. Credit “their” values where the scheme allows. Question 4(b) follows through your e from part (a), and Question 1(b) uses your x̄, as long as it is dimensionally correct.
4. Be strict on the show-that. In Question 6(a) the A1 needs a valid line of working from the energy equation to the printed result, with no error anywhere.
5. Check accuracy and units. Answers to 3 significant figures (1.37 m s⁻¹, 675 N, 871 N, 13.3 m s⁻¹) and angles to 1 decimal place (19.1°, 53.8°), except Question 4(b), which asked for 3 significant figures (6.12°). Exact forms such as 4√(1 − e^(−1/8)) and 5√3 were accepted too.
6. Look for capped or rejected answers. In Question 3(b) an answer in x rather than v scored at most 2 of 4. In Question 6(b) a negative root left unrejected cost the mark.
7. Log each lost mark as a wrong model (forces or energy), an integration or algebra slip, a missed dependency, or rounding, and revise in that order.

Sit 9231/32 as a timed mock

The paper gives 90 minutes for 50 marks, about 1.8 minutes per mark.

Question	Marks	Time
Question 1 (centre of mass)	7	12 minutes
Question 2 (horizontal circle)	5	8 minutes
Question 3 (variable resistance)	9	16 minutes
Question 4 (two barriers)	8	14 minutes
Question 5 (vertical circle)	9	16 minutes
Question 6 (elastic string)	6	8 minutes
Question 7 (inclined plane projectile)	6	12 minutes
Check, especially signs and rounding	–	4 minutes

Question 7 gets extra time for its 6 marks because it has three unknowns and a perpendicular condition to set up. Question 6 can be quicker, since part (a) gives you the result to work towards.

Other May/June 2026 9231 papers

• 9231 Paper 12 (Further Pure Mathematics 1)
• 9231 Paper 22 (Further Pure Mathematics 2)
• 9231 Paper 42 (Further Probability & Statistics)

Practise and download

• Download 9231/32 and its mark scheme on the Learning Portal →
• All 9231 May/June 2026 papers: overview and marking patterns
• Try a free A Level Further Maths quiz

Need help with a 9231 topic? Book a free trial lesson with an A Level Further Maths tutor →

Frequently asked questions

What topics were on 9231 Paper 32 May/June 2026? +

Seven questions: the centre of mass of a hemisphere joined to a cylinder, a particle on two strings in a horizontal circle, motion against a resistance of 0.5v², an oblique impact with two barriers at 120°, a vertical circle where the string goes slack, an elastic string, and a projectile that hits an inclined plane at right angles.

How long is 9231/32 and how many marks is it? +

It lasts 1 hour 30 minutes and is out of 50 marks. All seven questions are compulsory. Questions 3 and 5 are the biggest at 9 marks each, and you use g = 10 m s⁻² throughout.

What was the coefficient of restitution in 9231/32 Question 4? +

e = √3/4, about 0.433. After the impact the components are 12eu/13 perpendicular to the barrier and 5u/13 along it, and setting the new kinetic energy to 4/13 of the old gives 144e² = 27. The angle in part (b) was 6.12°.

How do you find the maximum height in 9231/32 Question 5(b)? +

Set the tension mg(3cos θ + 2) to zero, so cos θ = −2/3 and v² = 2ag/3 when the string goes slack. The particle is then a projectile. Its vertical speed component gives an extra 5a/27, so the greatest height above A is 5a/3 + 5a/27 = 50a/27.

Is 9231 Paper 3 needed for AS Further Maths? +

The AS route is Paper 1 plus either Paper 3 (Further Mechanics) or Paper 4 (Further Probability & Statistics). The full A Level takes all four papers, so Paper 3 is needed for the full A Level and is one of the two options for AS.

Where can I download 9231/32 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge AS & A Level Further Mathematics 9231 past papers page.

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