OxfordAQA International AS and A-level Mathematics: Forces, Newton's laws and momentum (9660) – 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 sections M1.3 and M1.4 of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International AS, Unit PSM1. Take g = 9.8 m/s² and treat strings as light and inextensible and pulleys as smooth unless told otherwise. Notes: study guide, revision notes, course hub, printable checklist and free diagnostics.

Questions

1. A parcel has mass 7.5 kg.

(a) Find its weight. [1] (b) It is lifted vertically by a cable with tension 90 N. Find its acceleration. [2]

2. A sledge of mass 40 kg is pulled along rough horizontal ground by a horizontal rope with tension 100 N. The coefficient of friction between sledge and ground is 0.15. Find the sledge’s acceleration. [4]

3. A lift of mass 600 kg carries a load of mass 90 kg. The cable tension is 7107 N and the lift accelerates upwards.

(a) Find the acceleration. [3] (b) Find the normal reaction between the load and the floor of the lift. [2]

4. A crate of mass 20 kg rests on horizontal ground.

(a) State the magnitude and direction of the normal reaction on the crate. [1] (b) Name the force that forms the Newton’s third law pair with this normal reaction, and say what body it acts on. [2]

5. A car of mass 1350 kg pulls a caravan of mass 450 kg along a straight horizontal road. A light rigid bar joins them. Constant resistances are 560 N on the car and 240 N on the caravan.

(a) The engine force is 2960 N. Find the acceleration and the tension in the bar. [5] (b) The engine is switched off and a braking force of 1360 N acts on the car. Find the force in the bar, stating whether it is a tension or a thrust. [4]

6. A light inextensible string passes over a smooth fixed pulley. A 0.9 kg particle is attached to one end and a 0.6 kg particle to the other. Both hang freely and the system is released from rest, with the 0.9 kg particle 1.2 m above the floor.

(a) Find the acceleration and the tension in the string. [4] (b) Find the speed of the 0.9 kg particle as it hits the floor. [2] (c) The 0.6 kg particle keeps moving upwards and the string goes slack. Find the further height it rises. [2]

7. A block P of mass 6 kg rests on a rough horizontal table, with μ = 0.5. A light string joins P, over a smooth pulley at the table edge, to a block Q of mass 8 kg that hangs below the pulley. Both are let go together.

(a) Find the acceleration of the particles and the tension. [5] (b) Q hits the floor after falling 1.4 m. P does not reach the pulley. Find how far P slides after Q lands. [4]

8. A particle of mass 6 kg moving at 3 m/s collides with a particle of mass 4 kg moving in the opposite direction at 2 m/s. They join and move together.

(a) Find their common velocity, stating its direction. [3] (b) Find the magnitude of the impulse exerted on the 4 kg particle. [2]

9. A ball of mass 0.15 kg hits a smooth fixed wall at right angles at 20 m/s and rebounds at 14 m/s. The contact lasts 0.015 s.

(a) Find the impulse on the ball. [2] (b) Find the average force, assuming it is constant. [2]

10. A sphere of mass 0.4 kg is released from rest in a tank of oil and falls 2.4 m in 2 s with constant acceleration. Find the resistance, assuming it is constant. [4]

11. A truck of mass 3000 kg moves at 4 m/s along a straight horizontal track and couples to a stationary wagon of mass 1000 kg.

(a) Find the speed immediately after coupling. [2] (b) A constant total resistance of 600 N then acts on the coupled vehicles. Find the distance they travel before stopping. [3] (c) The coupling takes 0.5 s. Find the average force on the wagon in that time. [2]

Answers

1. (a) W = 7.5 × 9.8 = 73.5 N [1] (b) 90 − 73.5 = 7.5a [1]; a = 2.2 m/s² upwards [1] Examiner insight: Write the equation with the force terms and the direction stated; a bare number gives the reader nothing to credit if it is wrong.

2. R = 40 × 9.8 = 392 N [1]; F = 0.15 × 392 = 58.8 N [1]; 100 − 58.8 = 40a [1]; a = 1.03 m/s² [1] Examiner insight: Show R before F. Friction is μR, and many errors come from using the weight without saying why R equals it.

3. (a) Total mass 690 kg; weight 690 × 9.8 = 6762 N [1]; 7107 − 6762 = 690a [1]; a = 0.5 m/s² [1] (b) R − 90 × 9.8 = 90 × 0.5 [1]; R = 927 N [1] Examiner insight: For (b), apply the second law to the load alone. Using the whole lift would find the tension again, not R.

4. (a) 196 N upwards [1] (b) The crate’s push on the floor [1], of the same magnitude and downwards, acting on the floor [1] Examiner insight: A third-law pair acts on two different bodies. Naming the weight as the partner is a typical error, so say what body each force acts on.

5. (a) Whole system: 2960 − 560 − 240 = 1800a [1]; a = 1.2 m/s² [1]. Caravan: T − 240 = 450 × 1.2 [1]; T = 780 [1]; tension 780 N [1] (b) Whole system: 1360 + 560 + 240 = 1800a, so a = 1.2 m/s² deceleration [1]. Caravan, forces backwards: 240 + X = 450 × 1.2 [1]; X = 300 [1]; the bar is in thrust of 300 N [1] Examiner insight: Say which body each equation is for. A negative tension is a good prompt to state that the bar is in thrust, and the words matter as much as the sign.

6. (a) 0.9g − T = 0.9a and T − 0.6g = 0.6a [1]; add: 0.3g = 1.5a [1]; a = 1.96 m/s² [1]; T = 0.6 × (9.8 + 1.96) = 7.056 N [1] (b) v² = 2 × 1.96 × 1.2 = 4.704 [1]; v = 2.17 m/s (3 s.f.) [1] (c) The 0.6 kg particle moves freely with deceleration g: 0 = 4.704 − 2 × 9.8 × h [1]; h = 0.24 m [1] Examiner insight: In (c), explain that the string is slack so only gravity acts. Using a = 1.96 again would give a different, wrong answer.

7. (a) R = 6 × 9.8 = 58.8 N, F = 0.5 × 58.8 = 29.4 N [1]; P: T − 29.4 = 6a [1]; Q: 78.4 − T = 8a [1]; add: 49 = 14a, a = 3.5 m/s² [1]; T = 50.4 N [1] (b) v² = 2 × 3.5 × 1.4 = 9.8 [1]; after landing the string is slack and P decelerates at μg = 4.9 m/s² [1]; 0 = 9.8 − 2 × 4.9 × s [1]; s = 1 m [1] Examiner insight: Write down both particle equations before adding. In part (b), state that friction alone acts on P once the string is slack.

8. (a) Taking the direction of the 6 kg particle as positive: 6 × 3 − 4 × 2 = 10v [1]; 10 = 10v [1]; v = 1 m/s in the direction the 6 kg particle was moving [1] (b) Impulse on the 4 kg particle = 4 × (1 − (−2)) [1] = 12 N s [1] Examiner insight: The opposite directions need opposite signs. Stating the direction of the answer as well as its size is the safer habit.

9. (a) Take towards the wall as positive: impulse = 0.15 × (−14 − 20) [1] = −5.1, so 5.1 N s away from the wall [1] (b) F = 5.1 / 0.015 [1] = 340 N [1] Examiner insight: Use −14 for the rebound. Subtracting the speeds (20 − 14) instead of adding their sizes is the common slip.

10. s = ½at² gives 2.4 = ½ × a × 4 [1]; a = 1.2 m/s² [1]; 0.4 × 9.8 − D = 0.4 × 1.2 [1]; D = 3.44 N [1] Examiner insight: Finding the acceleration first, as a separate labelled step, keeps the method visible even if the final subtraction slips.

11. (a) 3000 × 4 = 4000v [1]; v = 3 m/s [1] (b) Deceleration a = 600 / 4000 = 0.15 m/s² [1]; 0 = 3² − 2 × 0.15 × s [1]; s = 30 m [1] (c) Impulse on the wagon = 1000 × 3 = 3000 N s [1]; force = 3000 / 0.5 = 6000 N [1] Examiner insight: Use the combined mass for the deceleration in (b) and the wagon’s mass alone in (c). Say which body each calculation is about.

Where marks are usually lost

• Using R = mg without checking whether another vertical force acts.
• Taking g as 10 or 9.81 when the question says 9.8.
• Using one sign convention for both particles in a pulley problem, so the tension appears in the wrong sign.
• Forgetting that the string goes slack when the hanging particle lands.
• Mixing mass and weight, for example writing 0.9 where 0.9g is needed.
• Dropping the sign of a velocity in a momentum equation.
• Calling a force in a rod a tension without checking its sign.
• Giving an impulse without a direction, or a force without units.
• Rounding intermediate values before the final line.

Next steps

• Revise with the revision notes and re-read the study guide.
• See the course hub and the printable checklist.
• Try all free 10-minute diagnostics at /diagnostics/.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA: Unit PSM1, sections M1.3 and M1.4.