OCR A Level Mathematics A: Newton's laws, weight and equilibrium in one dimension (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 OCR A Level Mathematics A (H240), Version 3.1 (August 2026), first assessment 2018, section 3.03, items 3.03a to 3.03k. Questions 8 and 9 use 3.03e and are A Level only. Use g = 9.8 m s⁻². Give answers to 3 significant figures unless exact. Read the study guide and revision notes first. Course hub: OCR A Level Mathematics. Printable checklist.

Questions

1. A parcel of mass 6.5 kg rests on a horizontal shelf.

(a) Find its weight. [1] (b) State the normal reaction on the parcel and the Newton’s third law partner of that reaction. [2] (c) Explain why the weight would be different on another planet while the mass is not. [1]

2. A trolley of mass 15 kg is on a smooth horizontal track. A force of 80 N pushes it forwards and a force of 35 N acts backwards. It starts from rest. Find its acceleration and its speed after 6 s. [3]

3. A stone of mass 0.5 kg is dropped from rest. Air resistance is a constant 0.9 N.

(a) Find the acceleration of the stone. [2] (b) Find its speed after falling 5 m. [2]

4. A passenger of mass 64 kg stands on the floor of a lift.

(a) Find the force from the floor when the lift accelerates upwards at 0.9 m s⁻². [3] (b) Find the force from the floor when the lift accelerates downwards at 2.1 m s⁻². [2]

5. A lift of mass 250 kg carries a passenger of mass 80 kg. The lift cable has tension T and the lift accelerates upwards at 0.5 m s⁻².

(a) Find T. [3] (b) Find the force the passenger exerts on the floor of the lift. [2] (c) State why the lift and passenger can be treated as one particle in part (a). [1]

6. Only two forces, (12i − 9j) N and (−2i − 15j) N, act on a particle of mass 2 kg.

(a) Find the acceleration as a vector. [2] (b) Find its magnitude and the angle it makes with i. [2]

7. A 3 kg particle is acted on by (2i + 5j) N and by a second force F N. Its acceleration is (4i − j) m s⁻².

(a) Find F in vector form. [3] (b) Find the magnitude of F. [1]

8. (A Level only) A body of mass 6 kg is pulled along a smooth horizontal surface by a rope at 40° above the horizontal with tension 45 N.

(a) Find the acceleration. [2] (b) Find the normal reaction. [3] (c) The tension is increased, the angle stays at 40°. Find the tension at which contact is lost. [2]

9. (A Level only) A body of mass 4 kg slides down a line of greatest slope of a smooth plane inclined at 24° to the horizontal.

(a) Find its acceleration. [3] (b) Find the normal reaction. [2]

10. A lift of total mass 600 kg moves between two floors. It accelerates from rest at 1.5 m s⁻² for 2 s, travels at constant speed for 6 s, then decelerates uniformly at 1 m s⁻² to rest. The cable is light and inextensible.

(a) Find the tension during the first 2 s. [2] (b) Find the tension during the constant-speed stage. [1] (c) Find the time taken to decelerate, and the tension during that stage. [3] (d) Find the total distance travelled. [4]

11. A particle of mass 1.8 kg is raised vertically on a light inextensible string. The string breaks if the tension exceeds 36 N.

(a) Find the greatest upward acceleration the string allows. [3] (b) The particle moves upwards at 6 m s⁻¹ when the string is cut. Find the greatest height it reaches above that point, and the time for it to return to that point. [5] (c) State one way in which the string model is unrealistic. [1]

Answers

1. (a) W = 6.5 × 9.8 = 63.7 N [1] (b) At rest, R = W, so R = 63.7 N upwards [1]. The partner is the force of the parcel on the shelf, 63.7 N downwards [1]. (c) W = mg and g depends on location, but the mass does not change [1]. Examiner insight: In (b) name the body each force acts on. “Weight” alone is not the third-law partner of R, because both act on the parcel.

2. Resultant = 80 − 35 = 45 N forwards [1]. 45 = 15a, so a = 3 m s⁻² [1]. v = 0 + 3 × 6 = 18 m s⁻¹ [1]. Examiner insight: Writing the resultant as a single signed expression makes your method easy to follow and easy to check.

3. (a) Downwards positive: 0.5 × 9.8 − 0.9 = 0.5a [1]; 4.9 − 0.9 = 4, so a = 8 m s⁻² [1]. (b) v² = 0 + 2 × 8 × 5 = 80 [1]; v = 8.94 m s⁻¹ [1]. Examiner insight: Part (b) needs the acceleration from (a), not g. Using 9.8 here changes the answer.

4. (a) Upwards positive: R − 64 × 9.8 = 64 × 0.9 [1]; R = 627.2 + 57.6 [1]; R = 685 N [1]. (b) Downwards positive: 64 × 9.8 − R = 64 × 2.1 [1]; R = 627.2 − 134.4 = 493 N [1]. Examiner insight: State your positive direction. The two parts have opposite sign patterns, and a clear convention shows why.

5. (a) Total mass 330 kg [1]. T − 330 × 9.8 = 330 × 0.5 [1]; T = 3399 N [1]. (b) Passenger: R − 80 × 9.8 = 80 × 0.5 [1]; R = 824 N, so by the third law the passenger pushes the floor with 824 N downwards [1]. (c) There is no relative motion between them, so they share the same acceleration [1]. Examiner insight: Use the third law explicitly in (b). The force on the passenger and the force on the floor are equal but act on different bodies.

6. (a) Resultant = (10i − 24j) N [1]; a = (10i − 24j) ÷ 2 = (5i − 12j) m s⁻² [1]. (b) |a| = √(25 + 144) = 13 m s⁻² [1]; tan⁻¹(12 ÷ 5) gives 67.4° below i [1]. Examiner insight: Divide the resultant, not one force, by the mass. Say which side of i the acceleration points; a bare 67.4° does not say.

7. (a) ma = 3(4i − j) = 12i − 3j [1]; F = ma − (2i + 5j) [1]; F = (10i − 8j) N [1]. (b) |F| = √(100 + 64) = √164 = 12.8 N [1]. Examiner insight: Subtract the known force from the resultant, not from the acceleration. Check by adding F back.

8. (a) 45 cos 40° = 6a [1]; a = 5.75 m s⁻² [1]. (b) R + 45 sin 40° = 6 × 9.8 [1]; 45 sin 40° = 28.93 [1]; R = 29.9 N [1]. (c) Contact is lost when R = 0 [1]; T sin 40° = 58.8, so T = 91.5 N [1]. Examiner insight: Resolve the pull both ways. A common slip is to take R = 58.8, which ignores the lifting effect of the rope.

9. (a) Along the slope: 4a = 4 × 9.8 × sin 24° [1]; mass cancels [1]; a = 3.99 m s⁻² [1]. (b) Perpendicular to the slope there is no acceleration: R = 4 × 9.8 × cos 24° [1]; R = 35.8 N [1]. Examiner insight: Show which component of the weight you use. Swapping sin and cos is the usual error.

10. (a) T − 600 × 9.8 = 600 × 1.5 [1]; T = 6780 N [1]. (b) Constant speed gives zero resultant, so T = 5880 N [1]. (c) Speed reached = 1.5 × 2 = 3 m s⁻¹, so time = 3 ÷ 1 = 3 s [1]; 5880 − T = 600 × 1 [1]; T = 5280 N [1]. (d) First stage: ½ × 1.5 × 2² = 3 m [1]. Middle: 3 × 6 = 18 m [1]. Last: 3² ÷ (2 × 1) = 4.5 m [1]. Total 25.5 m [1]. Examiner insight: Multi-stage questions reward a clear table of stages. The tension in stage three is below the weight because the acceleration is downward.

11. (a) T − 1.8 × 9.8 = 1.8a [1]; 36 − 17.64 = 18.36 [1]; a = 10.2 m s⁻² [1]. (b) After cutting, only weight acts, so a = 9.8 m s⁻² downwards [1]. 0 = 6² − 2 × 9.8 × h [1]; h = 1.84 m [1]. Time up = 6 ÷ 9.8, so total time = 2 × 6 ÷ 9.8 [1]; 1.22 s [1]. (c) A real string has mass and stretches (or can only be approximately light and inextensible) [1]. Examiner insight: Cutting the string removes the tension but not the velocity. Use 6 m s⁻¹ as the starting speed with acceleration downwards.

Where marks are usually lost

• Leaving the weight out of a vertical equation of motion.
• Using R = mg for a body in an accelerating lift.
• Dividing a single force, not the resultant, by the mass.
• Taking the wrong sign for acceleration when the positive direction is not stated.
• Treating weight and R as a third-law pair.
• Swapping sin and cos when resolving along or across a slope.
• Rounding g to 10 or rounding early, so the final digits are wrong.
• Leaving out the model statement when asked, such as light string or no relative motion.

Next steps

• Review the revision notes and the study guide.
• Course hub: OCR A Level Mathematics and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

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