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OCR A Level Further Mathematics B (MEI) Paper 1 June 2026 Mark Scheme – Mechanics Major (Y421/01)

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Official OCR A Level Further Mathematics B (MEI) Paper 1 (Y421/01) Mechanics Major mark scheme – June 2026 series. Complete marking instructions with detailed solutions for all 12 questions. Section A (shorter questions): projectile motion (time of flight T = 2u sinθ/g, range R = u cosθ × T, vertical displacement formula), Newton’s laws and friction (limiting friction F_max = μR, static vs kinetic friction, inclined plane tanθ = μ at slipping), work-energy principle (KE = ½mv², work done by gravity mgh, work done by friction Fd, change in KE), circular motion (centripetal acceleration a = v²/r, centripetal force F = mv²/r, conical pendulum resolving vertically and horizontally, tanθ = v²/rg, tension T = mg/cosθ), momentum and impulse (impulse = change in momentum m(v-u), conservation of linear momentum m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, coefficient of restitution e = relative speed after / relative speed before), simple harmonic motion (x = Acos(ωt), v² = ω²(A² – x²), v_max = ωA, period T = 2π/ω), rotational motion (moment of inertia of rod about centre I_cm = 1/12 ML², parallel axis theorem I = I_cm + Mh², torque τ = Fr, angular acceleration α = τ/I). Section B (longer questions): connected particles with variable forces (Newton’s second law equations, simultaneous solution for acceleration), elastic strings/springs and energy (Hooke’s law T = λx/l, equilibrium mg = λx/l, elastic potential energy EPE = λx²/2l, energy conservation mg(h+x) = λx²/2l for maximum extension), motion in a vertical circle (forces at top mg + T = mv²/r, minimum speed for complete circle v_min = √(rg), energy conservation between bottom and top, tension calculation), differential equations in mechanics (air resistance m dv/dt = mg – kv, terminal velocity v_T = mg/k, separation of variables to derive v = v_T(1 – e^{-kt/m})), statics of rigid bodies (moments about a point, uniform rod weight at centre, equilibrium of forces, tilting condition R_A = 0). Includes mark allocation per part (method marks M, accuracy marks A, independent marks B, follow-through FT), grade boundaries (A* 80%+, A 70-79%), and examiner guidance (use of g = 9.8 m/s², significant figures, calculator use). Perfect for OCR A Level Further Mathematics revision, mechanics major exam preparation, teacher marking, and student self-assessment.

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1|Page


# OCR A LEVEL FURTHER MATHEMATICS B
(MEI)
## PAPER 1 JUNE 2026 MARK SCHEME (Y421/01:
MECHANICS MAJOR)




## MARKING INSTRUCTIONS


### General Principles


1. **Mark strictly to the mark scheme** – Every mark has a clear
descriptor.


2. **Follow-through marks (FT)** – When a candidate makes an error
in a previous part and uses that incorrect value in subsequent parts,
marks may be awarded for correct method applied to their incorrect
value. Look for the **"FT"** annotation in the scheme.


3. **Method marks (M marks)** – Awarded for demonstrating a correct
method, even if the final answer is wrong. Method marks are
independent unless marked as `dep*` (dependent on a previous mark).

,2|Page


4. **Accuracy marks (A marks)** – Awarded only for completely
correct answers, following from correct working.


5. **B marks** – Awarded for independent marks, such as stating a
formula or writing an equation.


6. **Crossed-out work** – If a clear alternative is provided, the crossed-
out response is ignored. If no alternative is given, mark the crossed-out
work if legible.


7. **Contradictory responses** – No mark awarded.


8. **Use of g** – Unless specified otherwise in the question, use g = 9.8
m s⁻².


---


## PAPER STRUCTURE


| Section | Format | Marks |
|---------|--------|-------|
| **Section A** | Shorter questions with minimal reading and
interpretation | ~60 marks |

, 3|Page


| **Section B** | Longer questions with more problem-solving | ~60
marks |
| **Total** | | **120 marks** |


All questions are compulsory.


---


# SECTION A: SHORTER QUESTIONS


## Question 1 (Projectile Motion)


| Part | Answer | Marks | Guidance |
|------|--------|-------|----------|
| (a) | Horizontal velocity = \( u \cos \theta \), Vertical velocity = \( u \sin
\theta - gt \) | B1 | Both components stated |
| (b) | Time of flight = \( \frac{2u \sin \theta}{g} \) | M1 | Using \( s = ut
+ \frac{1}{2}at^2 \) with vertical displacement = 0 |
| | \( T = \frac{2 \times 28 \times \sin 35^\circ}{9.8} \) | A1 | Substitution
correct |
| | \( T = \frac{56 \times 0.574}{9.8} = \frac{32.14}{9.8} = 3.28 \) s
(3.28 to 3.3) | A1 | Answer to 3 s.f. |
| (c) | Range = \( u \cos \theta \times T \) | M1 | Horizontal velocity ×
time |

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