AQA A-Level Physics: Mechanics and materials (7408) – 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 3.4, Mechanics and materials, of AQA A-level Physics (7408), following the AQA AS and A-level Physics specification (7407/7408), version 1.4, for AS and A-level exams from June 2016 onwards. They span sub-sections 3.4.1.1 to 3.4.2.2, all of which are AS and A-level content, examined on A-level Paper 1 and assumed for Paper 2; the practical questions reflect Paper 3 Section A. Sections 3.9 to 3.13 are options: you study one, assessed in Paper 3 Section B. Take g = 9.81 m s⁻².

See the study guide, revision notes, course hub and printable checklist.

Questions

1. (a) State the difference between a scalar and a vector, giving one example of each. [2] (b) A rope pulls a crate with a force of 80 N at 35° above the horizontal. Calculate the horizontal and vertical components of this force. [2]

2. (a) A driver turns a steering wheel of diameter 0.38 m by applying two equal and opposite tangential forces of 15 N at opposite ends of a diameter. Calculate the moment of this couple. [2] (b) A uniform plank 4.0 m long and of weight 300 N rests on a single support 1.5 m from its left end. A vertical force F at the left end holds the plank horizontal. Calculate F and the force from the support. [4]

3. A cyclist accelerates uniformly from rest to 8.0 m s⁻¹ in 5.0 s, travels at constant speed for 12 s, then decelerates uniformly to rest in 4.0 s. (a) Calculate the initial acceleration. [1] (b) Calculate the total distance travelled. [3] (c) Calculate the average speed for the whole journey. [1]

4. In a free-fall experiment to find g, a steel ball is released from rest by an electromagnet and timed electronically over several heights h. A graph of h against t² is a straight line of gradient 4.87 m s⁻². (a) Show how the gradient is related to g and calculate g. [2] (b) The ball is released a short, constant time after the timer starts. State the type of error this causes and explain why a graph of √h against t would remove its effect on g. [3]

5. A ball is kicked from level ground at 18 m s⁻¹ at 40° above the horizontal. Ignore air resistance. (a) Calculate the time of flight. [3] (b) Calculate the maximum height reached. [2] (c) Calculate the horizontal range. [2] (d) Describe how air resistance would change the path. [2]

6. Explain why a skydiver falling from rest reaches a terminal speed. [3]

7. A car of mass 1100 kg tows a trailer of mass 400 kg along a level road. The driving force is 2400 N. Resistive forces are 300 N on the car and 150 N on the trailer. (a) Calculate the acceleration. [2] (b) Calculate the tension in the tow bar. [2] (c) State the Newton’s third law pair of the tow bar’s pull on the trailer. [1]

8. A 0.16 kg cricket ball travelling at 25 m s⁻¹ is hit straight back at 30 m s⁻¹. Contact lasts 8.0 ms. (a) Calculate the change in momentum. [2] (b) Calculate the average force on the ball. [1] (c) The force–time graph is a triangle. Calculate the peak force. [2] (d) Suggest why a padded glove reduces the force on a fielder’s hand when catching. [1]

9. A motor lifts a 2.4 kg load 1.5 m vertically at constant speed in 4.0 s. The input power is 18 W. (a) Calculate the useful output power using P = Fv. [3] (b) Calculate the efficiency as a percentage. [1]

10. Two trolleys, of mass 0.60 kg and 0.90 kg, are held at rest against a compressed spring between them. When released, the 0.60 kg trolley moves at 1.5 m s⁻¹. (a) Calculate the velocity of the 0.90 kg trolley. [2] (b) The spring was compressed by 0.050 m. Assuming all the elastic strain energy becomes kinetic energy, calculate the spring constant. [4]

11. A cyclist and bicycle, total mass 80 kg, freewheel from rest down a road, descending 12 m vertically over 150 m of road, and reach 11 m s⁻¹. Calculate the average resistive force. [4]

12. A student uses a 2.50 m length of copper wire of diameter 0.40 mm to find the Young modulus. (a) The wire has mass 2.8 g. Calculate the density of copper. [2] (b) A load of 30 N produces an extension of 5.0 mm, within the limit of proportionality. Calculate the Young modulus. [4] (c) Calculate the elastic strain energy stored at this load. [1] (d) Explain why a long, thin wire is used and why the diameter is measured at several places. [2] (e) The load is increased until the wire extends plastically. Describe what the force–extension graph shows when the load is then removed. [2]

Answers

1. (a) A scalar has magnitude only; a vector has magnitude and direction [1]. Example: mass (scalar), force (vector) [1]. (b) Horizontal = 80 cos 35° = 65.5 N [1]; vertical = 80 sin 35° = 45.9 N [1]. Examiner insight: Both examples are usually needed for the second mark; one correct example alone is not enough.

2. (a) Moment = one force × separation = 15 × 0.38 [1] = 5.7 N m [1]. (b) Weight acts at 2.0 m, so 0.5 m right of the support [1]. Moments about support: F × 1.5 = 300 × 0.5 [1], so F = 100 N [1]. Support force = 300 + 100 = 400 N [1]. Examiner insight: Moments about the support remove its unknown force; a consistent equation about another point still earns method marks.

3. (a) a = 8.0/5.0 = 1.6 m s⁻² [1]. (b) Areas: ½ × 8.0 × 5.0 = 20 m; 8.0 × 12 = 96 m [1]; ½ × 8.0 × 4.0 = 16 m [1]. Total = 132 m [1]. (c) 132 ÷ 21 = 6.3 m s⁻¹ [1]. Examiner insight: Error carried forward is normally allowed in (c), so a wrong total in (b) can still earn the average-speed mark.

4. (a) From rest, h = ½gt², so gradient = g/2 [1]; g = 2 × 4.87 = 9.74 m s⁻² [1]. (b) Systematic error [1]. Every measured time is too long by the same delay [1]. Since √h = √(g/2) × (t − delay), the √h–t graph is still a straight line of gradient √(g/2); the delay only moves the intercept, so g from the gradient is unaffected [1]. Examiner insight: “Systematic” alone earns one mark; the explanation must link the constant offset to the intercept, not the gradient.

5. (a) u_y = 18 sin 40° = 11.57 m s⁻¹ [1]. Back to launch height: 0 = u_y t − ½gt² [1], t = 2 × 11.57/9.81 = 2.36 s [1]. (b) At the top v_y = 0: h = u_y²/(2g) [1] = 6.82 m [1]. (c) uₓ = 18 cos 40° = 13.79 m s⁻¹ [1]; range = 13.79 × 2.36 = 32.5 m [1]. (d) Range and maximum height are smaller [1]; the path is asymmetric, with a steeper descent than ascent [1]. Examiner insight: Keep unrounded values between parts; rounding t to 2.4 s gives 33.1 m, which may lose the accuracy mark.

6. At first only weight acts, so the skydiver accelerates at g [1]. Air resistance increases with speed, so the resultant force and acceleration decrease [1]. When air resistance equals weight the resultant force is zero and speed is constant [1]. Examiner insight: “Forces balance” earns credit only if the answer names which forces: air resistance and weight.

7. (a) Resultant = 2400 − 300 − 150 = 1950 N [1]; a = 1950/1500 = 1.3 m s⁻² [1]. (b) For the trailer: T − 150 = 400 × 1.3 [1], T = 670 N [1]. (c) The trailer’s pull on the tow bar (and so on the car), 670 N backwards [1]. Examiner insight: A third-law partner must act on a different body; naming the trailer’s weight or friction scores zero.

8. (a) Taking the rebound direction as positive: Δp = 0.16 × (30 − (−25)) [1] = 8.8 N s [1]. (b) F = 8.8/0.0080 = 1100 N [1]. (c) Area = ½ × 0.0080 × F_peak = 8.8 [1], so F_peak = 2200 N [1]. (d) The padding increases the stopping time, so a smaller force gives the same change in momentum [1]. Examiner insight: Using 30 − 25 for the velocity change gives 0.80 N s and loses both marks in (a); direction matters.

9. (a) v = 1.5/4.0 = 0.375 m s⁻¹ [1]; F = mg = 2.4 × 9.81 = 23.5 N [1]; P = 23.5 × 0.375 = 8.83 W [1]. (b) 8.83/18 × 100 = 49% [1]. Examiner insight: At constant speed the lifting force equals the weight; no “ma” term is needed.

10. (a) Total momentum stays zero: 0.60 × 1.5 = 0.90v [1], so v = 1.0 m s⁻¹ in the opposite direction [1]. (b) Ek = ½ × 0.60 × 1.5² + ½ × 0.90 × 1.0² [1] = 1.125 J [1]. ½k(0.050)² = 1.125 [1], so k = 900 N m⁻¹ [1]. Examiner insight: The direction is required for the second mark in (a); velocity is a vector.

11. Ep lost = 80 × 9.81 × 12 = 9418 J [1]. Ek gained = ½ × 80 × 11² = 4840 J [1]. Work against resistance = 4578 J [1]. F = 4578/150 = 31 N [1]. Examiner insight: The resistive force acts along the road, so divide by 150 m, not the 12 m height.

12. (a) A = π(0.20 × 10⁻³)² = 1.257 × 10⁻⁷ m²; V = 3.14 × 10⁻⁷ m³ [1]. ρ = 2.8 × 10⁻³/3.14 × 10⁻⁷ = 8.9 × 10³ kg m⁻³ [1]. (b) Stress = 30/1.257 × 10⁻⁷ = 2.39 × 10⁸ Pa [1]. Strain = 5.0 × 10⁻³/2.50 = 2.0 × 10⁻³ [1]. E = stress/strain [1] = 1.2 × 10¹¹ Pa [1]. (c) ½ × 30 × 5.0 × 10⁻³ = 0.075 J [1]. (d) A long, thin wire gives a large enough extension to measure precisely [1]; the wire may not be uniform, so a mean diameter reduces uncertainty in the area [1]. (e) The unloading line is straight and parallel to the loading line [1], and does not return to zero: a permanent extension remains [1]. Examiner insight: Using the diameter as the radius makes the area four times too large and E four times too small, losing the accuracy marks.

Where marks are usually lost

• Measuring a moment arm along the beam instead of perpendicular to the line of action.
• Losing the sign of a rebound velocity, so the momentum change comes out far too small.
• Rounding an intermediate time in a projectile question and missing the final accuracy mark.
• Naming two forces on the same body as a Newton’s third law pair.
• Treating terminal speed as “no forces acting” rather than zero resultant force.
• Forgetting work done against resistive forces when applying energy conservation.
• Describing the Young modulus graph method without saying the gradient of the straight section is used.

Next steps

• Revision notes
• Study guide
• Course hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA AS and A-level Physics specification (7407/7408), version 1.4, July 2026, for AS and A-level exams June 2016 onwards, published by AQA. Section 3.4 Mechanics and materials.