AQA A-Level Physics Paper 1, June 2022: Question 3

✅ Correct Answer

brass (with supporting working shown)

💡 Key Knowledge

• Cylinder volume formula: V = π × r² × l (or equivalent using diameter: V = π × (d/2)² × l).
• Density equation: ρ = m / V.
• Mass from weight: m = W / g (using g = 9.81 m s² or 9.8 m s²).

📐 Calculation Steps

1. Find Volume of Cylinder A: V = π × (4.8 × 10² m / 2)² × 0.23 m = 4.16 × 10&sup4; m³.
2. Find Mass of A: m = 35 N / 9.81 m s² = 3.57 m kg.
3. Calculate Density of A: ρ = 3.57 kg / (4.16 × 10&sup4; m³) ≈ 8580 kg m³ ≈ 8.6 × 10³ kg m³.
4. Compare with Table 2: Matches brass (ρ = 8.6 × 10³ kg m³). Alternatively, calculate expected mass/weight of brass and compare.

❌ Common Errors

• Writing "brass" without showing calculations scores zero marks (unsupported answer rule).
• Using weight directly in the density formula instead of converting to mass first.

✅ Correct Answer

tension = 40 N (accept 40 to 40.2 N)

💡 Key Knowledge

• Frictionless pulleys mean tension throughout the rope equals the weight of A (T_rope = 35 N).
• Resolution of forces at movable pulley M: The horizontal cable balances the horizontal components of the two rope sections pulling at 55° to the horizontal.

📐 Calculation Steps

1. Identify tension in each rope branch = 35 N.
2. Calculate horizontal component for one branch: 35 × cos(55°).
3. Account for both upper and lower rope sections attached to movable pulley M: T_cable = 2 × 35 × cos(55°) = 40 N.

🧠 Exam Technique

Always double-check whether a pulley system has one or two rope branches acting on the central movable component. Missing the factor of 2 is a frequent trap.

✅ Correct Answer

As pulley M moves left, the angle of the rope segments to the horizontal decreases, increasing their horizontal components, which requires a larger tension in cable C to maintain equilibrium.

💡 Key Knowledge

• As an angle θ approaches 0°, cos(θ) approaches 1.
• Horizontal pulling force = 2T_rope cos(θ). A smaller angle increases cos(θ).

❌ Common Errors

• Vague statements like "tension in the rope increases" (the rope tension remains fixed by weight A at 35 N; it is the horizontal component and cable C tension that change).

✅ Correct Answer

moment = 8.1 N m (or 8.10 N m)

💡 Key Knowledge

• Moment = Force × Perpendicular distance from pivot (hinge) to line of action of force.
• Alternatively, resolve the force into components perpendicular and parallel to the door face.

📐 Calculation Steps

1. Method 1 (Perpendicular Distance): Perpendicular distance d_perp = 0.95 × sin(12°) = 0.198 m. Moment = 41 N × 0.198 m = 8.098 N m → 8.1 N m.
2. Method 2 (Resolve Force): Angle of cable to door normal = 90° - 12° = 78°. Perpendicular force component = 41 × cos(90° - 12°) = 41 × sin(12°) = 8.52 N. Moment = 8.52 N × 0.95 m = 8.1 N m.

❌ Common Errors

• Directly multiplying 41 N × 0.95 m without factoring in the angle (scores zero marks).

✅ Correct Answer (Any two independent changes)

1. Increase weight A: Increases rope tension, which increases the pull in cable C and increases the moment.
2. Reposition pulley R further out / increase its diameter: Increases the perpendicular distance from the hinge to cable C.
3. Decrease angle of the rope: Move P and fixed points closer together/further right to increase horizontal components and cable C tension.
4. Move attachment point D further from hinge: Increases the perpendicular distance along the door where cable C acts.

🧠 Exam Technique

When an open-ended "discuss two changes" question is asked, clearly state the modification AND explicitly link it to the mechanics principle (either increasing force/tension or increasing perpendicular distance/moment arm).