AQA AS Level Physics Paper 1, June 2023: Question 4

💡 Key Knowledge

• When an object is in equilibrium, the resultant force in any direction is zero.
• To find tension T , resolve the two angled tension forces horizontally to balance the pulling force F .
• Horizontal component of one side of string = T cos(75°) . Since there are two sides, total horizontal force is 2T cos(75°) .

✅ Correct Answer

T = 310 N (Accept 309 N to 310 N depending on rounding; calculator value is 309.096... N).

📐 Step-by-Step Calculation

1. Equilibrium condition: F = 2T cos(75°)
2. Rearrange for T: T = F / (2 cos(75°))
3. Substitute values: T = 160 / (2 cos(75°))
4. Evaluate: T = 309.096 N → round to 2 significant figures: 310 N .

❌ Common Errors

• Forgetting the factor of 2: Omitting the fact that the string pulls from both above and below point P loses the method mark.
• Trig confusion: Mixing up sin and cos by referencing the vertical angle instead of the horizontal angle provided.

🧠 Exam Technique

Read graph values with extreme care. For s = 0.22 m , locate 0.22 on the horizontal axis of Figure 6, track up to the curve, and read the corresponding force value F on the vertical axis.

✅ Correct Answer

Initial horizontal acceleration = 9900 m s⁻² (Acceptable range: 9900 to 10000 m s⁻² ).

📐 Step-by-Step Calculation

1. Read F from graph: At s = 0.22 m , F = 208 N (accepted range 208 N to 210 N).
2. State equation: F = ma → a = F / m
3. Convert mass to SI: m = 21 g = 21 × 10⁻³ kg
4. Calculate acceleration: a = 208 / (21 × 10⁻³) = 9904.76... m s⁻² → 9900 m s⁻² (2 s.f.).

❌ Common Errors

• Failing to convert grams to kilograms ( 21 g used as 21 kg ), resulting in a power-of-ten error.
• Writing an unacceptable 1 significant figure answer like 1 × 10⁴ m s⁻² .

💡 Key Knowledge

The work done in pulling the bow string—which equals the energy stored in the bow—is represented by the area under the force-displacement ( F-s ) graph.

✅ Correct Answer

sᵣ = 0.385 m (Acceptable range between 0.38 m and 0.385 m ).

🧠 Exam Technique

Since the curve isn't a standard geometric shape, use square-counting on the grid. Determine the energy represented by one small square, then count squares backwards from the peak until you accumulate 64 J of energy at displacement sᵣ .

❌ Common Errors

• Using the simple formula W = Fs (assuming constant force). This is incorrect because force varies continuously with distance as shown by the curve.

💡 Key Knowledge

• Efficiency formula: Efficiency = (Useful output energy / Total input energy)
• Kinetic energy transferred to the arrow: Eₖ = 0.5 × m × v²
• Velocity unit conversion: 190 km h⁻¹ must be converted to m s⁻¹ by dividing by 3.6 .

✅ Correct Answer

m = 0.038 kg (or 38 g ).

📐 Step-by-Step Calculation

1. Find useful energy transferred to arrow: Eₖ = 0.82 × 64 J = 52.48 J
2. Convert velocity: v = 190 / 3.6 = 52.778 m s⁻¹
3. Rearrange kinetic energy equation for mass: m = 2 Eₖ / v²
4. Substitute and solve: m = (2 × 52.48) / (52.778)² = 104.96 / 2785.5 = 0.03768 kg → round to 2 s.f.: 0.038 kg .

❌ Common Errors

• Forgetting to apply the efficiency factor (using 64 J directly instead of 52.48 J).
• Forgetting to square the velocity term in the kinetic energy equation ( v instead of v² ).