AQA AS Level Physics Paper 1, June 2019: Question 3

✅ Correct Answer

Weight = 1.2 × 10³ kg × 9.81 m s⁻² = 1.18 × 10⁴ N

Vertical tension component (per rope) = 3.7 × 10⁴ cos(20°) = 3.48 × 10⁴ N

Total downward force F = 2(3.48 × 10⁴) - 1.18 × 10⁴ = 5.8 × 10⁴ N (shown)

💡 Key Knowledge

• Forces in equilibrium: The clamp holds the cage stationary against opposing forces.
• Resolving vectors: Use T cos(\theta) to find the vertical component aligned with the tower.
• Multiplying by 2 accounts for the two identical symmetric elastic ropes.

🧠 Exam Technique

This is a "show that" question. You must state your intermediate values clearly (weight and vertical tension component) to secure all marking points before arriving at the target figure of 6 × 10⁴ N .

❌ Common Errors

• Using sin(20°) instead of cos(20°) by confusing the angle given with the horizontal.
• Forgetting to multiply the single rope's vertical tension component by 2.

✅ Correct Answer

Acceleration a = 50 m s⁻² (accept 48 m s⁻² if unrounded previous values are carried forward).

📐 Step-by-Step Calculation

1. Recall Newton's Second Law: F = ma .
2. Rearrange for acceleration: a = F / m .
3. Substitute values: a = (5.8 × 10⁴ N) / (1.2 × 10³ kg) .
4. Evaluate: 48.33... m s⁻² → round appropriately to 50 m s⁻² .

❌ Common Errors

Dividing mass by force instead of force by mass, leading to dimensionally incorrect fractions.

✅ Correct Answer

Total stored elastic potential energy = 4.9 × 10⁵ J (to be compared with the statement "about 5 × 10⁵ J ").

📐 Step-by-Step Calculation

1. Find stretched length of one rope using trigonometry: L = 35 / cos(20°) = 37.2 m .
2. Calculate extension ( \Delta L ): 37.2 m - 24 m = 13.2 m .
3. Apply elastic potential energy formula for one rope ( E = 0.5 F \Delta L ): 0.5 × (3.7 × 10⁴ N) × 13.2 m = 2.44 × 10⁵ J .
4. Multiply by 2 for both ropes: 2 × (2.44 × 10⁵ J) = 4.88 × 10⁵ J ≈ 4.9 × 10⁵ J .

🧠 Exam Technique

Alternative methods using E = 0.5 k (\Delta L)² are fully accepted if spring constants are derived correctly first. Always double-check geometric lengths before calculating extensions.

✅ Correct Answer

The claim is not justified. Maximum height reached above Q is 42 m (which is less than the claimed 50 m ).

💡 Key Knowledge

Conservation of Energy: Assuming no thermal losses, all elastic potential energy stored in the ropes converts entirely into gravitational potential energy ( E_{p(stored)} = mg\Delta h ).

📐 Verification Method

1. Equate energy: 4.9 × 10⁵ J = (1.2 × 10³ kg) × 9.81 m s⁻² × h .
2. Solve for h : h = (4.9 × 10⁵) / (1.2 × 10³ × 9.81) = 41.6 m ≈ 42 m .
3. Compare: 42 m < 50 m , proving the claim is false.

❌ Common Errors

Attempting to use SUVAT equations. SUVAT assumes constant acceleration, but acceleration here varies continuously as tension changes with extension. Examiners award zero credit for kinematic equations in this context.

✅ Correct Answer

Kinetic energy = 3.8 × 10⁵ J (or 3.75 × 10⁵ J depending on rounding).

📐 Step-by-Step Calculation

1. Convert speed from km h⁻¹ to m s⁻¹ : 90 / 3.6 = 25 m s⁻¹ .
2. State kinetic energy formula: E_k = 0.5 m v² .
3. Substitute values: 0.5 × (1.2 × 10³ kg) × (25 m s⁻¹)² .
4. Evaluate: 0.5 × 1200 × 625 = 3.75 × 10⁵ J ≈ 3.8 × 10⁵ J .

❌ Common Errors

Forgetting to convert km h⁻¹ into m s⁻¹ before squaring, resulting in wildly incorrect magnitudes.

✅ Correct Answer

The claim is justified because the calculated kinetic energy ( 3.8 × 10⁵ J ) is less than the total available stored elastic energy ( 4.9 × 10⁵ J ).

💡 Key Knowledge

Energy budgeting: The kinetic energy at any point plus gravitational potential energy gained cannot exceed the initial elastic potential energy provided by the ropes, minus any resistive thermal losses.

🧠 Exam Technique

This is a quick 1-mark follow-on. Look back at your figures from 03.3 ( ≈ 5 × 10⁵ J available) and 03.5 ( 3.8 × 10⁵ J required for that speed). Since required E_k is less than total energy, the speed is entirely achievable.