AQA A-Level Mathematics: P: Quantities and units in mechanics (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 P: Quantities and units in mechanics (P1) 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 P in Paper 2. A calculator is required in every 7357 paper, so every question here assumes you have one. Give answers to 3 significant figures where they are not exact, unless the question says otherwise.

Revise first with the Quantities and units 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. Write down the SI unit, with its symbol, of each of these quantities: acceleration, force, moment. [3]

2. A train travels at 108 km h⁻¹. Convert this speed to m s⁻¹. [2]

3. A parcel has mass 750 g. Taking g = 9.8 m s⁻², find the weight of the parcel. [2]

4. Express the newton in terms of the SI units of mass, length and time. Hence show that a mass in kilograms multiplied by g in m s⁻² gives a quantity measured in newtons. [3]

5. A cyclist’s speed increases steadily from 18 km h⁻¹ to 27 km h⁻¹ in 4 seconds. Calculate the cyclist’s acceleration in m s⁻². [3]

6. A force of 60 N is applied at right angles to a spanner, 25 cm from the centre of a nut.

(a) Calculate the moment of the force about the centre of the nut, stating its units. [2] (b) Find the force, applied at right angles 40 cm from the centre, that has the same moment. [2]

7. A model for the air resistance R newtons on a ball moving at v m s⁻¹ is R = kv², where k is a constant.

(a) Find the SI units of k, in terms of kg, m and s. [2] (b) When v = 30, R = 54. Find k and use the model to find R when v = 20. [2]

8. A student writes the formula v = u + as for motion with constant acceleration. Use units to show that this formula cannot be correct. [3]

9. A space probe has mass 120 kg.

(a) Find its weight where g = 9.8 m s⁻². [1] (b) Find its weight at a location where g = 1.6 m s⁻². [1] (c) State the mass of the probe at the second location, and explain your answer. [2]

10. A lorry of mass 7.5 tonnes (1 tonne = 1000 kg) speeds up with constant acceleration from 36 km h⁻¹ to 63 km h⁻¹ in 15 seconds.

(a) Convert both speeds to m s⁻¹. [2] (b) Find the acceleration of the lorry. [2] (c) Find the resultant force on the lorry, giving your answer in kN. [2] (d) Taking g = 9.8 m s⁻², find the weight of the lorry in kN. [2]

11. A crate of mass 40 kg is raised vertically by a cable. The tension in the cable is 0.42 kN. Air resistance is ignored.

(a) Taking g = 9.8 m s⁻², find the weight of the crate. [1] (b) Find the resultant force on the crate and hence its acceleration. [3] (c) The crate starts from rest and the acceleration stays constant. Find its speed after 5 seconds, giving your answer in km h⁻¹. [2] (d) Repeat part (b) using g = 9.81 m s⁻², and comment on the effect of the value of g on your answer. [2]

12. A sign of mass 2.5 kg hangs from the end of a light horizontal rod fixed to a vertical wall. The sign hangs 80 cm from the wall. Take g = 9.8 m s⁻².

(a) Find the weight of the sign. [1] (b) Find the moment of the weight of the sign about the point where the rod meets the wall. [2] (c) The fixing can withstand a moment of at most 30 N m. The sign is replaced by a heavier sign of mass m kg hanging at the same point. Find the greatest possible value of m. [3] (d) Express the newton metre in terms of kg, m and s. [1]

Answers

1. acceleration: metres per second per second, m s⁻² [1] force: newton, N [1] moment: newton metre, N m [1] [3] Examiner insight: in this answer the unit name or the symbol earns the mark, but a wrong power (m s⁻¹ for acceleration) or a missing distance unit for moment (N only) earns nothing for that item.

2. 108 × 1000 ÷ 3600, or 108 ÷ 3.6 [1] = 30 m s⁻¹ [1] [2] Examiner insight: the method mark needs a correct conversion factor; multiplying by 3.6 (388.8) loses both marks.

3. mass = 750 ÷ 1000 = 0.75 kg [1] W = mg = 0.75 × 9.8 = 7.35 N [1] [2] Examiner insight: using m = 750 gives 7350, which earns nothing; the conversion to kilograms is where the method mark sits.

4. From F = ma, 1 N = 1 kg × 1 m s⁻² [1] so 1 N = 1 kg m s⁻² [1] mg has units kg × m s⁻² = kg m s⁻², which is N, as required [1] [3] Examiner insight: in a “show that” you must write the final link (kg m s⁻² = N) explicitly; stopping at kg m s⁻² leaves the last mark unearned.

5. 18 km h⁻¹ = 5 m s⁻¹ and 27 km h⁻¹ = 7.5 m s⁻¹ [1] a = (7.5 − 5) ÷ 4 [1] = 0.625 m s⁻² [1] [3] Examiner insight: working in km h⁻¹ throughout (9 ÷ 4 = 2.25) gives the wrong units altogether; convert both speeds first.

6. (a) 25 cm = 0.25 m, moment = 60 × 0.25 [1] = 15 N m [1] (b) F × 0.4 = 15 [1] F = 37.5 N [1] [4] Examiner insight: in (b) this scheme allows follow-through from your (a), but the answer must be a force in N, not a moment.

7. (a) k = R ÷ v², units N ÷ (m s⁻¹)² = kg m s⁻² ÷ m² s⁻² [1] = kg m⁻¹ [1] (b) k = 54 ÷ 30² = 0.06 [1] R = 0.06 × 20² = 24 N [1] [4] Examiner insight: in (a) writing N in base units is the key step; an answer of “N m⁻² s²” is equivalent but must be simplified if the question asks for kg, m and s.

8. v has units m s⁻¹ and u has units m s⁻¹ [1] as has units m s⁻² × m = m² s⁻² [1] m² s⁻² is not m s⁻¹, so the terms cannot be added and the formula cannot be correct [1] [3] Examiner insight: the units of each term must be shown; a bare statement that “the units don’t match”, with no working, does not earn the method marks.

9. (a) W = 120 × 9.8 = 1176 N (1180 N to 3 s.f.) [1] (b) W = 120 × 1.6 = 192 N [1] (c) 120 kg [1] Mass is the amount of matter in the probe and does not depend on g; only the weight changes [1] [4] Examiner insight: the explanation mark needs the idea that mass is independent of location or of g; “it is the same probe” alone is too vague.

10. (a) 36 ÷ 3.6 = 10 m s⁻¹ [1] 63 ÷ 3.6 = 17.5 m s⁻¹ [1] (b) a = (17.5 − 10) ÷ 15 [1] = 0.5 m s⁻² [1] (c) F = ma = 7500 × 0.5 = 3750 N [1] = 3.75 kN [1] (d) W = 7500 × 9.8 = 73 500 N [1] = 73.5 kN [1] [8] Examiner insight: using m = 7.5 gives 3.75 N in (c), a factor of 1000 out; follow-through applies to a wrong acceleration from (b), but not to an unconverted mass.

11. (a) W = 40 × 9.8 = 392 N [1] (b) tension = 0.42 kN = 420 N [1] resultant = 420 − 392 = 28 N upwards [1] a = 28 ÷ 40 = 0.7 m s⁻² upwards [1] (c) v = 0 + 0.7 × 5 = 3.5 m s⁻¹ [1] 3.5 × 3.6 = 12.6 km h⁻¹ [1] (d) W = 392.4 N, resultant = 27.6 N, a = 27.6 ÷ 40 = 0.69 m s⁻² [1] The acceleration depends on the value of g used, so the answer is only as accurate as g; a small change in g makes a noticeable change here because the resultant is the difference of two close forces [1] [8] Examiner insight: in (b) the kN must be converted before subtracting; 0.42 − 392 is a sign of an unconverted unit and earns no accuracy marks.

12. (a) W = 2.5 × 9.8 = 24.5 N [1] (b) 80 cm = 0.8 m, moment = 24.5 × 0.8 [1] = 19.6 N m [1] (c) m × 9.8 × 0.8 ≤ 30 [1] m ≤ 30 ÷ 7.84 = 3.8265… [1] greatest m = 3.83 kg (3 s.f.) [1] (d) N m = kg m s⁻² × m = kg m² s⁻² [1] [7] Examiner insight: in (c) the method mark is for forming the inequality (or the equation at the limit) with the distance in metres; using 80 for the distance gives 0.0383 kg and loses the accuracy marks.

Where marks are usually lost

• Leaving grams, centimetres, kN or km h⁻¹ unconverted inside F = ma, W = mg or a moment.
• Multiplying by 3.6 instead of dividing when converting to m s⁻¹.
• Giving the weight in kg or the mass in N.
• Writing m s⁻¹ for acceleration, or N for a moment.
• Using a different value of g from the one the question gives.
• Rounding a converted value early and carrying it forward.
• Stopping a “show that” one line short of the stated result.
• Missing units, or the direction, on a final force or acceleration.

Next steps

• Recap the rules: Quantities and units revision notes.
• Full explanations: Quantities and units study guide.
• Forces as vectors: Vectors practice questions.
• 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 3.17 P: Quantities and units in mechanics (P1), with Appendix A: mathematical notation (mechanics) and Appendix B: mathematical formulae and identities.