OCR A Level Mathematics A: Two-dimensional equilibrium, connected particles and pulleys (H240) – 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 exam’s 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 follow outcomes 3.03l to 3.03q of the OCR A Level Mathematics A specification (H240, Version 3.1, August 2026, for first assessment in 2018). Only 3.03n is Stage 1 (common with AS); the rest is Stage 2, so it is A Level only. Read the study guide first and keep the revision notes to hand. The course page is OCR A Level Mathematics, the printable list is the checklist, and the free diagnostics are a good warm-up. Use g = 9.8 m s⁻². All surfaces are smooth, strings are light and inextensible, and pulleys are smooth, unless stated.

Questions

1. Forces of 13 N and 8 N act at a point. The angle between them is 70°. Find the magnitude of the resultant and the angle it makes with the 13 N force. [4]

2. A particle is in equilibrium under the forces (pi + 4j) N, (−3i + qj) N and (2i − 7j) N. (a) Find p and q. [2] (b) The particle has mass 0.5 kg. The force (2i − 7j) N is removed. Find the magnitude of the acceleration. [2]

3. A 2.4 kg particle is suspended in equilibrium by two light strings, one on each side, from a horizontal ceiling. The strings are inclined at 25° and 60° to the ceiling. Find both tensions. [5]

4. A particle of mass 3.5 kg is held at rest on a smooth plane inclined at 35° to the horizontal by a force P. P points up the slope along a steepest line, at 15° above the plane’s surface. Find P and the normal reaction. [4]

5. A van of mass 1300 kg pulls a trailer of mass 700 kg along a level straight road, joined by a light horizontal coupling. The van’s driving force is 3600 N. Resistances of 450 N on the van and 250 N on the trailer oppose the motion. (a) Find the acceleration. [2] (b) Find the tension in the coupling. [2] (c) Use Newton’s third law to state the force the coupling exerts on the van, and confirm it by considering the van alone. [2]

6. A light inextensible string passes over a smooth fixed pulley. It carries a particle of mass 8 kg on one end and a particle of mass 6 kg on the other. Both hang freely and are released from rest. (a) Show that the acceleration is 1.4 m s⁻². [3] (b) Find the tension in the string. [1] (c) State the role of each assumption: light string, inextensible string, smooth pulley. [3]

7. A particle A of mass 3 kg is on a horizontal table. Strings from A pass over smooth pulleys at opposite edges of the table to particles B (2 kg) and C (1.2 kg), each hanging freely. The system is released from rest. (a) Find the acceleration of the system. [4] (b) Find the tension in each string. [2] (c) State the direction in which A moves. [1]

8. A particle A of mass 5 kg lies on a plane inclined at 40° to the horizontal. A string parallel to the plane passes over a smooth pulley at the top of the plane and carries a particle B of mass 4 kg hanging freely. The system is released from rest. (a) Show that A moves up the plane, and find the acceleration. [4] (b) Find the normal reaction on A. [1] (c) B has descended 0.9 m. Find the speed of the particles. [2] (d) The string now breaks. Find the further distance A travels up the plane before coming to rest. [2]

9. A 3 kg particle moves under the force (18t i + (24 − 6t) j) N, where t is the time in seconds. At t = 0 it is at rest at position vector (4i + 2j) m. (a) Find an expression for its velocity at time t. [3] (b) Find its position vector when t = 3. [2] (c) Find its speed when t = 2. [2]

10. Crates A (4 kg) and B (6 kg) rest on a horizontal floor, joined by a light horizontal rope, with B further to the right. A force of 50 N pulls B to the right at 35° above the horizontal. (a) Find the acceleration. [2] (b) Find the tension in the rope. [2] (c) Find the normal reaction on each crate. [3]

11. A particle A of mass 2.5 kg lies on a plane inclined at 30° to the horizontal. A string parallel to the plane passes over a smooth pulley at the top of the plane and carries a particle B of mass 3.5 kg hanging freely. (a) Find the acceleration when the system is released. [3] (b) Find the tension. [1] (c) Find the size and direction of the force that the string applies to the pulley. [4]

Answers

1. Component along the 13 N force: 13 + 8 cos 70° = 15.74 [1]. Component perpendicular to it: 8 sin 70° = 7.52 [1]. Magnitude √(15.74² + 7.52²) = 17.4 N [1]. Angle = tan⁻¹(7.52 / 15.74) = 25.5° from the 13 N force [1]. [4] Examiner insight: Say which force the angle is measured from.

2. (a) i-components: p − 3 + 2 = 0, so p = 1 [1]. j-components: 4 + q − 7 = 0, so q = 3 [1]. (b) The remaining forces now have resultant (−2i + 7j) N [1]. |F| = √53, so a = √53 / 0.5 = 14.6 m s⁻² [1]. [4] Examiner insight: Part (b) is quicker if you see that the remaining resultant is minus the removed force.

3. Weight = 2.4 × 9.8 = 23.52 N [1]. Horizontal: T₁ cos 25° = T₂ cos 60° [1]. Vertical: T₁ sin 25° + T₂ sin 60° = 23.52 [1]. Then T₁ = 0.5517 T₂, so 1.0992 T₂ = 23.52, giving T₂ = 21.4 N [1]. T₁ = 11.8 N [1]. [5] Examiner insight: Write the weight and both equations before substituting, so your method is visible even if the arithmetic slips.

4. Along the plane: P cos 15° = 3.5g sin 35° [1]. P = 19.67 / 0.9659 = 20.4 N [1]. Perpendicular to the plane: R + P sin 15° = 3.5g cos 35° [1]. R = 28.10 − 5.27 = 22.8 N [1]. [4] Examiner insight: A force that lifts the particle reduces R. Check the sign of the P sin 15° term against your diagram.

5. (a) Whole system: 3600 − 450 − 250 = (1300 + 700)a [1]. a = 1.45 m s⁻² [1]. (b) Trailer: T − 250 = 700 × 1.45 [1]. T = 1265 N [1]. (c) The coupling pulls the van backwards with 1265 N, equal and opposite to the force on the trailer [1]. Van: 3600 − 450 − 1265 = 1885 = 1300 × 1.45, which agrees [1]. [6] Examiner insight: The internal force drops out of the whole-system equation. Find a first, then separate the bodies.

6. (a) 8 kg particle: 8g − T = 8a [1]. 6 kg particle: T − 6g = 6a [1]. Adding: 2g = 14a, so a = 19.6 / 14 = 1.4 m s⁻² [1]. (b) T = 8(9.8 − 1.4) = 67.2 N [1]. (c) Light string: the tension is the same along its whole length (and the string needs no equation of its own) [1]. Inextensible: both particles move with the same speed and acceleration size [1]. Smooth pulley: no friction, so the tension is the same on both sides [1]. [7] Examiner insight: In a “show that” question, include every step to the printed value. In part (c), link each assumption to its effect on the equations.

7. (a) B (down): 2g − T₁ = 2a [1]. C (up): T₂ − 1.2g = 1.2a [1]. A: T₁ − T₂ = 3a [1]. Adding: 0.8g = 6.2a, so a = 1.26 m s⁻² [1]. (b) T₁ = 2(g − a) = 17.1 N [1]. T₂ = 1.2(g + a) = 13.3 N [1]. (c) A moves towards B’s side of the table, since B is heavier [1]. [7] Examiner insight: Three particles need three equations. The two tensions differ, so label them T₁ and T₂.

8. (a) B: 4g − T = 4a [1]. A: T − 5g sin 40° = 5a [1]. Adding: a = (39.2 − 31.50) / 9 = 0.856 m s⁻² [1]. This is positive, so A moves up the plane (4g is greater than 5g sin 40°) [1]. (b) R = 5g cos 40° = 37.5 N [1]. (c) v² = 2 × 0.8559 × 0.9 [1]. v = 1.24 m s⁻¹ [1]. (d) After the string breaks A decelerates at g sin 40° = 6.30 m s⁻² [1]. Distance = 1.2412² / (2 × 6.299) = 0.122 m [1]. [9] Examiner insight: Carry unrounded values from part (a) into (c) and (d), so rounding does not shift your third significant figure.

9. (a) a = F/m = 6t i + (8 − 2t) j [1]. Integrating: v = 3t² i + (8t − t²) j + c [1]. At rest when t = 0, so c = 0: v = 3t² i + (8t − t²) j [1]. (b) r = (t³ + 4) i + (4t² − t³/3 + 2) j, using the initial position [1]. At t = 3: r = 31i + 29j [1]. (c) At t = 2, v = 12i + 12j [1]. Speed = 12√2 = 17.0 m s⁻¹ [1]. [7] Examiner insight: Each integration needs its own constant, found from the starting condition. Do not leave out the initial position in part (b).

10. (a) Whole system, horizontal: 50 cos 35° = 10a [1]. a = 4.10 m s⁻² [1]. (b) A alone: T = 4a [1], so T = 4 × 4.0958 = 16.4 N [1]. (c) A: R = 4g = 39.2 N [1]. B: R + 50 sin 35° = 6g [1], so R = 58.8 − 28.68 = 30.1 N [1]. [7] Examiner insight: A pull above the horizontal lowers the reaction on the crate it acts on. Resolve vertically for each crate separately.

11. (a) B: 3.5g − T = 3.5a [1]. A: T − 2.5g sin 30° = 2.5a [1]. Adding: a = (34.3 − 12.25) / 6 = 3.675 m s⁻², so 3.68 m s⁻² [1]. (b) T = 3.5(9.8 − 3.675) = 21.4 N [1]. (c) The two string parts each pull the pulley with 21.44 N. Their directions are down the slope (30° below horizontal) and vertically down, so the angle between them is 60° [1]. Magnitude = 2 × 21.4375 × cos 30° = 37.1 N [1]. The resultant lies along the bisector of the 60° angle [1], so it points 60° below the horizontal, towards the side of the plane [1]. [8] Examiner insight: Draw both string parts leaving the pulley and mark the angle between them before using 2T cos(half-angle).

Where marks are usually lost

• Giving a resultant angle without naming the line it is measured from.
• Mixing up sin and cos when resolving along and perpendicular to a plane.
• Using the same tension for strings joined to different particles in a three-body system.
• Ignoring an angled force when finding the normal reaction.
• Treating the force on a pulley as T or 2T when the strings are not parallel.
• Forgetting constants of integration.
• Rounding a or T early and reusing the rounded value.
• Quoting Newton’s third law without naming the two bodies involved.

Next steps

• Re-read the revision notes and redo any question you lost marks on.
• Go back to the study guide for the method.
• Tick off topics on the printable checklist or see the course page.
• Take the free diagnostics to find the next weak area.
• Book a free trial class if you want to work through these questions with a tutor.

Official syllabus

OCR, A Level Mathematics A (H240), Version 3.1 (August 2026), first assessment 2018: section 3.03, Forces and Newton’s Laws.