AQA A-Level Mathematics: R: Forces and Newton's laws (7357) – 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 cover Section R: Forces and Newton’s laws (R1 to R6) of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards. The specification places Section R in Paper 2. A calculator is required in every 7357 paper, so every question here is calculator allowed. Take g = 9.8 m s⁻² unless a question says otherwise, and give non-exact answers to 3 significant figures.

Revise first with the Forces and Newton’s laws study guide and the revision notes. Course hub: /boards/aqa/a-level/mathematics/. Checklist: /checklists/aqa/a-level/mathematics/. Not sure where you stand? Try the free 10-minute diagnostics.

Questions

1. A particle moves in a straight line with constant velocity. State what this tells you about the forces acting on it, and name the law you have used. [2]

2. Two forces, (4i + 7j) N and (−1.5i − 2j) N, are the only forces acting on a particle of mass 2.5 kg. Find the acceleration of the particle, and its magnitude. [3]

3. A rover has mass 180 kg. At one location g = 9.78 m s⁻², and at another g = 9.83 m s⁻². Find the weight of the rover at each location, and state what happens to its mass. [3]

4. A parachutist of mass 85 kg falls vertically. At one instant the air resistance is 620 N.

(a) Find the acceleration of the parachutist at that instant. [3] (b) Later the parachutist falls at constant speed. Find the air resistance then, giving a reason. [2]

5. Particles P, of mass 5 kg, and Q, of mass 3 kg, are attached to the ends of a light inextensible string that passes over a fixed smooth pulley. Both hang vertically and are released from rest.

(a) Find the acceleration of the particles and the tension in the string. [4] (b) Find the speed of P after it has fallen 1.2 m, assuming Q does not reach the pulley. [2] (c) Find the force exerted by the string on the pulley. [2]

6. A crate of mass 30 kg rests on rough horizontal ground. The coefficient of friction is 0.45.

(a) A horizontal force of 100 N is applied. Show that the crate does not move, and state the friction force. [3] (b) The horizontal force is increased to 180 N. Find the acceleration of the crate. [2]

7. A particle of mass 6 kg is held in equilibrium on a smooth plane inclined at 25° to the horizontal by a horizontal force of P N, in the vertical plane through a line of greatest slope.

(a) Find P. [3] (b) Find the normal reaction of the plane on the particle. [2]

8. Four horizontal forces act on a particle in equilibrium: 8 N in the direction of i; 10 N at 60° anticlockwise from i; P N in the direction of −j; and Q N in the direction of −i. Find P and Q. [4]

9. A car of mass 1100 kg tows a trailer of mass 500 kg up a straight road inclined at α to the horizontal, where sin α = 1/20, using a light rigid tow bar parallel to the road. Resistances are 400 N on the car and 250 N on the trailer. The acceleration is 0.5 m s⁻².

(a) Find the driving force of the car. [3] (b) Find the tension in the tow bar. [2] (c) When the speed is 12 m s⁻¹, the tow bar breaks. The resistance on the trailer stays the same. Find how far the trailer travels up the road before it stops. [4]

10. A block of mass 3 kg is on a rough plane inclined at 35° to the horizontal. A force of 40 N, acting up a line of greatest slope, pulls the block up the plane with acceleration 1.2 m s⁻².

(a) Find the coefficient of friction between the block and the plane. [5] (b) The 40 N force is removed when the block is moving up the plane at 2.5 m s⁻¹. Find how much further the block travels up the plane. [3] (c) Determine whether the block stays at rest once it has stopped. [2]

11. Particle A, of mass 4 kg, is on a rough horizontal table, with coefficient of friction 0.3. A light inextensible string joins A, over a smooth pulley at the edge of the table, to particle B, of mass 3 kg, which hangs 0.9 m above the floor. The system is released from rest, and A does not reach the pulley.

(a) Find the acceleration of the system and the tension in the string. [5] (b) Find the speed of B when it hits the floor. [2] (c) B does not rebound. Find the further distance A travels before it comes to rest. [3]

12. A box of mass 15 kg rests on the floor of a lift. The lift is moving upwards and slowing down at 0.8 m s⁻². Find the force exerted by the box on the floor, and name the law you use. [3]

Answers

1. The resultant force is zero (the forces balance) [1]. Newton’s first law [1] [2] Examiner insight: “no forces act” earns nothing; the forces have zero resultant.

2. Resultant F = (2.5i + 5j) N [1] a = F ÷ 2.5 = (i + 2j) m s⁻² [1] |a| = √(1² + 2²) = √5 = 2.24 m s⁻² [1] [3] Examiner insight: add the force vectors before dividing by the mass; adding their magnitudes loses every mark after the first.

3. 180 × 9.78 = 1760.4 = 1760 N [1] 180 × 9.83 = 1769.4 = 1770 N [1] The mass stays 180 kg; only the weight changes with g [1] [3] Examiner insight: the mass comment is a separate mark; a weight in kg loses it.

4. (a) Weight = 85 × 9.8 = 833 N [1] Down positive: 833 − 620 = 85a [1] a = 213 ÷ 85 = 2.51 m s⁻² downwards [1] (b) Constant speed, so resultant force is zero (first law) [1] Air resistance = weight = 833 N [1] [5] Examiner insight: in (b) the reason is credited separately from the value; quote zero resultant or Newton’s first law.

5. (a) P: 5g − T = 5a [1] Q: T − 3g = 3a [1] Add: 2g = 8a, so a = 2.45 m s⁻² [1] T = 3(9.8 + 2.45) = 36.75 = 36.8 N [1] (b) v² = 0 + 2 × 2.45 × 1.2 = 5.88 [1] v = 2.42 m s⁻¹ [1] (c) Both sections of string pull down on the pulley: 2T = 2 × 36.75 = 73.5 N [1] vertically downwards [1] [8] Examiner insight: treating the system as one body gives a, but T needs one particle’s own equation; write a separate equation for each particle.

6. (a) R = 30 × 9.8 = 294 N; maximum friction = 0.45 × 294 = 132.3 N [1] 100 N < 132.3 N, so the crate does not move [1] Friction = 100 N [1] (b) 180 − 132.3 = 30a [1] a = 47.7 ÷ 30 = 1.59 m s⁻² [1] [5] Examiner insight: for a “show that”, compare the applied force with μR explicitly; stating 132.3 N as the friction in (a) loses the final mark.

7. (a) Resolve along the plane: P cos 25° = 6g sin 25° [1] P = 6 × 9.8 × tan 25° [1] = 27.4 N [1] (b) Perpendicular: R = 6g cos 25° + P sin 25° [1] = 53.29 + 11.59 = 64.9 N [1] [5] Examiner insight: the horizontal force has a component into the plane, so leaving out the P term gives 53.3 N, which is wrong.

8. i: 8 + 10 cos 60° − Q = 0 [1] Q = 13 [1] j: 10 sin 60° − P = 0 [1] P = 5√3 = 8.66 [1] [4] Examiner insight: each resolving equation is a method mark in its own right, so state which direction you are resolving in; 5√3 and 8.66 both answer the question.

9. (a) Whole system, up the slope: D − 400 − 250 − 1600g × (1/20) = 1600 × 0.5 [1] D − 650 − 784 = 800 [1] D = 2234 N [1] (b) Trailer: T − 250 − 500g × (1/20) = 500 × 0.5 [1] T = 250 + 245 + 250 = 745 N [1] (c) Trailer alone, down the slope: 250 + 245 = 500d [1] d = 0.99 m s⁻² (deceleration) [1] 0 = 12² − 2 × 0.99 × s [1] s = 144 ÷ 1.98 = 72.7 m [1] [9] Examiner insight: in (c) the tension term must go; keeping 745 N in the trailer’s equation after the bar breaks loses the method mark.

10. (a) R = 3g cos 35° = 24.08 N [1] Up the plane: 40 − F − 3g sin 35° = 3 × 1.2 [1] F = 40 − 16.86 − 3.6 = 19.54 N [1] μ = F ÷ R = 19.54 ÷ 24.08 [1] = 0.811 [1] (b) Friction now acts down the plane with the weight component: 16.86 + 19.54 = 3d, d = 12.13 m s⁻² [1] 0 = 2.5² − 2 × 12.13 × s [1] s = 0.258 m [1] (c) Weight component down the plane 16.9 N < maximum friction 19.5 N [1] so the block stays at rest [1] [10] Examiner insight: carry unrounded R and F into μ, or the third figure can drift.

11. (a) R = 4g = 39.2 N, so friction on A = 0.3 × 39.2 = 11.76 N [1] A: T − 11.76 = 4a [1] B: 3g − T = 3a [1] Add: 29.4 − 11.76 = 7a, a = 2.52 m s⁻² [1] T = 3(9.8 − 2.52) = 21.8 N [1] (b) v² = 2 × 2.52 × 0.9 = 4.536 [1] v = 2.13 m s⁻¹ [1] (c) String slack, so only friction acts on A: deceleration = 0.3g = 2.94 m s⁻² [1] 0 = 4.536 − 2 × 2.94 × s [1] s = 0.771 m [1] [10] Examiner insight: (c) is a new stage; reusing a = 2.52, or keeping T in A’s equation, loses the marks after (b).

12. Up positive, a = −0.8: R − 15g = 15 × (−0.8) [1] R = 15 × 9.0 = 135 N, the floor’s force on the box [1] By Newton’s third law the box pushes down on the floor with 135 N [1] [3] Examiner insight: “slowing down while moving up” means the acceleration is downwards; using +0.8 gives 159 N and loses the accuracy mark.

Where marks are usually lost

• Writing R = mg on a slope (question 10), or when another force has a component perpendicular to the surface (Q7).
• Taking friction as μR when the body is not moving or about to move (Q6(a)).
• Getting the direction of friction wrong once the motion changes, as in question 10(b).
• Keeping the tension in an equation after a string goes slack or a tow bar breaks.
• Losing the sign of a deceleration in a lift or vertical-motion question.
• Rounding R or F early and carrying the rounded value into μ or a distance.

Next steps

• Condensed recall: Forces and Newton’s laws revision notes.
• Full explanations: Forces and Newton’s laws study guide.
• Force vectors: Vectors practice questions.
• Related practice: Kinematics and Moments.
• Course hub: /boards/aqa/a-level/mathematics/. Checklist: /checklists/aqa/a-level/mathematics/.
• All free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018, for A-level exams June 2018 onwards: Section 3.19 R: Forces and Newton’s laws (R1 to R6), with Appendix B: mathematical formulae and identities.