AQA GCSE Physics Physics Paper 1 (Higher), June 2025: Question 9

📐 Step-by-Step Calculation

1. Convert height to metres:
   h = 70 cm = 0.70 m [1 mark]
2. State and substitute into the GPE equation:
   E_p = m × g × h
   0.343 = m × 9.8 × 0.70 [1 mark]
3. Rearrange to solve for mass (m):
   m = 0.343 / (9.8 × 0.70) = 0.343 / 6.86 [1 mark]
4. Calculate final answer:
   m = 0.050 kg (or 0.05 kg ) [1 mark]

✅ Correct Answer

Mass = 0.050 kg

🧠 Exam Technique

• Always check the diagram for hidden values! The vertical lift height ( 70 cm ) is labeled on Figure 11.
• "Each second" links energy to power, but because time is 1 s, the energy transferred in 1 s is simply 0.343 J , giving the mass pumped in 1 s directly.

❌ Common Errors to Avoid

• Unit Trap: Forgetting to divide 70 cm by 100, which results in m = 0.0005 kg (loses conversion and accuracy marks).
• BODMAS error on calculator: Typing 0.343 / 9.8 × 0.70 without brackets gives 0.0245 kg. You must divide by the entire denominator: 0.343 / (9.8 × 0.70) .

📐 Step-by-Step Calculation

1. Substitute into power equation P = I²R :
   4.86 = I² × 6.0 [1 mark]
2. Rearrange to find I² :
   I² = 4.86 / 6.0 = 0.81 [1 mark]
3. Calculate current ( I ):
   I = √0.81 = 0.90 A [1 mark]
4. Convert time from minutes to seconds:
   t = 30 × 60 = 1800 s [1 mark]
5. Substitute into charge equation Q = I × t :
   Q = 0.90 × 1800 [1 mark]
6. Calculate final charge:
   Q = 1620 C (or 1600 C ) [1 mark]

✅ Correct Answer

Charge flow = 1620 C (or 1.62 × 10³ C)

💡 Key Knowledge

This is an unstructured multi-step problem requiring two linked equations from the Physics Equations Sheet:

• Power = (current)² × resistance ( P = I²R )
• Charge flow = current × time ( Q = I t )

❌ Common Errors to Avoid

• Forgetting the square root: Leaving current as 0.81 A instead of taking the square root ( √0.81 = 0.90 A ).
• Time in minutes: Calculating 0.90 × 30 = 27 C . Time in physics equations must always be in seconds (s)!

✅ Model Answer

Effect: The rate of water flow will increase / a greater mass (or volume) of water will be pumped each second. [1 mark]

Explanation: Because a greater proportion of the input power is converted into useful power output (or less energy/power is wasted). [1 mark]

💡 Key Knowledge

• Efficiency = useful power output / total power input
• If total power input remains constant and efficiency increases, useful power output must increase.
• Useful work done here is lifting water: higher useful power = more gravitational work done per second = more water lifted per second.

🧠 Exam Technique: "Explain" Questions

Always structure your answer into two distinct parts:

1. What happens: State the physical change to the flow clearly (more water / higher volume flow rate).
2. Why it happens: Connect it back to the scientific definition of efficiency (greater useful output / less wasted energy).

❌ Examiner Watch-Outs

• Insufficient response: Saying "the water moves faster" is ignored by the mark scheme. The question asks about flow rate (mass or volume pumped per unit time).
• Vague statements: Writing "it works better" or "it has more power" gets 0 marks. You must specify useful power or less wasted energy.