AQA GCSE Physics Physics Paper 1 (Higher), June 2023: Question 4

11 marks · Standard Demand difficulty · Short Answer

Calculate the vertical height change and time taken for energy transfer in a hydroelectric system, and explain how solar power improves reliability.

Practise this question

Question

Question 4 features three parts concerning electricity generation for a remote village. Part 04.1 asks for the mean change in vertical height of 2,500,000 kg of water with a GPE change of 367.5 MJ and g = 9.8 N/kg. Part 04.2 asks for the time in standard form for a 3.0 kW generator to transfer 2.16 × 10^7 J of energy. Part 04.3 provides a line graph showing generator power output varying throughout the months from a high of around 5 kW in January and December to a low under 1 kW in June and July, asking why adding solar power improves reliability.
Question text

04 A remote village in the UK uses a hydroelectric generator to provide electricity.

04.1 In one day, 2 500 000 kg of water passes through the hydroelectric generator.

The change in gravitational potential energy of the water is 367.5 MJ.

gravitational field strength = 9.8 N/kg

Calculate the mean change in vertical height of the water as it moves through the

hydroelectric generator.

Use the Physics Equations Sheet.

[4 marks]

Mean change in vertical height =11 m

04.2 The generator transfers 3.0 kW of electrical power.

Calculate the time taken for the generator to transfer 2.16 × 107 J of energy.

Use the Physics Equations Sheet.

Give your answer in standard form.

[5 marks]

Time taken (in standard form) =12 s

04.3 Figure 5 shows how the power output of the generator varied during one year.

Figure 5

A solar power system is installed in the remote village in addition to the

hydroelectric generator.

Explain why this improves the reliability of the electricity supply to the village.

Use information from Figure 5.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 4: 04.1 awards 4 marks for converting 367.5 MJ to 367,500,000 J, substituting into Ep = mgh, rearranging for h, and obtaining h = 15 m. 04.2 awards 5 marks for converting 3.0 kW to 3000 W, substituting into P = E / t, rearranging for t, calculating 7200 s, and writing the final answer as 7.2 × 10^3 s in standard form. 04.3 awards 2 marks for stating that hydroelectric output is lower in summer when solar output is greater, resulting in less variation in total power output.

Question 4

AO /

Question Answers Extra information Mark

Spec. Ref.

04.1 Ep = 367 500 000 (J) 1 AO2

4.1.1.2

367 500 000 = 2 500 000 × 9.8 allow a correct substitution using 1

× h an incorrectly/not converted

value of Ep

367 500 000 allow a correct rearrangement 1

h = using an incorrectly/not

2 500 000 × 9.8

converted value of Ep

h = 15 (m) allow an answer consistent with 1

their value of Ep

AO /

Spec. Ref.

04.2 3 kW = 3000 W 1 AO2

4.1.1.4

2.16 × 107 all subsequent marks can score 1

3000 = using an incorrectly / not

t

converted value of P

2.16 × 107

t =

3000

t = 7200 (s) 1

t = 7.2 × 103 (s) allow an answer given in 1

standard form– from a calculationHYSICS– –

using data given in the question

AO /

Spec. Ref.

04.3 allow reference to specific AO3

months eg April to September 4.1.3

in the summer the power output allow power output of 1

from the hydroelectric generator hydroelectric generator depends

is lower but the solar power on rainfall and power output of

output would be greater solar power system depends on

12 light intensity

so less variation in total power allow electricity supply for total 1

output (which improves the power output

reliability of the supply)

Total Question 4 11

How to answer it

Energy, Power & Hydroelectric Systems

WHAT THIS QUESTION TESTS

This question assesses fundamental physics calculation skills and data interpretation from Energy (Specification 4.1.1 & 4.1.3):

  • Unit Conversions: Converting MegaJoules (MJ) to Joules (J) and kilowatts (kW) to Watts (W).
  • Formula Recall & Application: Gravitational Potential Energy ( Ep = m g h ) and Electrical Power ( P = E / t ).
  • Algebraic Rearrangement: Changing the subject of an equation accurately.
  • Standard Form: Expressing numerical values in scientific notation ( A × 10n ).
  • Graph Evaluation: Analysing seasonal energy trends to explain reliability in renewable energy systems.
PART 04.1 • 4 MARKS

Gravitational Potential Energy & Vertical Height

Calculating vertical drop from change in gravitational potential energy

📐 Step-by-Step Calculation

  1. Convert energy to standard SI units (Joules):
    Mega means 106 (or 1 000 000).
    367.5 MJ = 367.5 × 106 J = 367 500 000 J
    awarded for correct conversion of Ep to Joules.
  2. Select and substitute into the equation:
    Ep = m × g × h
    367 500 000 = 2 500 000 × 9.8 × h
    awarded for correct substitution of values.
  3. Rearrange to make height (h) the subject:
    h = 367 500 000 / (2 500 000 × 9.8)
    h = 367 500 000 / 24 500 000
    awarded for correct algebraic rearrangement.
  4. Calculate final value:
    h = 15 m
    awarded for the correct final answer.

❌ Common Errors

  • Forgetting prefix conversion: Using 367.5 instead of 367 500 000 gives 1.5 × 10-5 m , losing the first mark.
  • Bracket errors on calculator: Typing 367500000 ÷ 2500000 × 9.8 multiplies by 9.8 instead of dividing by it! Always bracket the denominator: ÷ (2 500 000 × 9.8) .

🧠 Exam Technique

  • Error Carried Forward (ECF): If you fail to convert MJ to J but substitute correctly and rearrange properly, you can still gain 3 out of 4 marks! Always show every single step clearly.
  • Check your answer makes physical sense: A 15 m drop for a hydroelectric dam is realistic. 0.000015 m is not!
PART 04.2 • 5 MARKS

Power, Energy Transfer & Standard Form

Calculating time taken and expressing the result in scientific notation

📐 Step-by-Step Calculation

  1. Convert power to standard SI units (Watts):
    Kilo (k) means 1000.
    3.0 kW = 3000 W
    awarded for converting kW to W.
  2. Select and substitute into the power equation:
    Power = Energy / Time (P = E / t)
    3000 = (2.16 × 107) / t
    awarded for correct substitution.
  3. Rearrange for time (t):
    Multiply by t , then divide by 3000 :
    t = (2.16 × 107) / 3000
    awarded for correct rearrangement.
  4. Calculate the numerical time:
    t = 21 600 000 / 3000 = 7200 s
    awarded for calculating 7200 seconds.
  5. Convert final answer to standard form:
    Move the decimal point 3 places to the left:
    t = 7.2 × 103 s
    awarded specifically for standard form format.

❌ Common Errors

  • Rearrangement trap: Incorrectly calculating t = P × E or t = P / E . Remember, if P = E / t , then t = E / P .
  • Ignoring standard form: Leaving the answer as 7200 forfeits the 5th mark. The question explicitly states: "Give your answer in standard form."
  • Invalid standard form: Writing 72 × 102 or 0.72 × 104 . The first number must be between 1 and 10!

🧠 Exam Technique

  • Highlight command words: Underline requirements like "standard form" when you first read the prompt so you don't forget at the end.
  • All subsequent marks apply: Even if you forgot to convert 3 kW to 3000 W and used 3, you could still earn up to 4 marks via ECF if you rearranged and gave your answer in standard form.
PART 04.3 • 2 MARKS

Renewable Energy Reliability & Seasonal Variation

Explaining why combining solar power with hydroelectric power improves system reliability

✅ Model Answer

In the summer (or from April to September), the power output from the hydroelectric generator is low, but the output from the solar power system will be high / greater.

This means there is less variation in the total power output (or the combined electricity supply remains constant throughout the year), improving overall reliability.

• Mark 1: Comparing seasons/months – hydroelectric output is lower in summer, but solar output is higher (or linking hydro to rainfall and solar to light intensity).
• Mark 2: Concluding that this results in less variation / a more constant total power output throughout the year.

💡 Key Knowledge

  • Hydroelectric output: Relies on rainfall collecting in elevated reservoirs. In the UK, rainfall is lowest during late spring and summer (May–August), causing hydro output to drop (to nearly 0.3 kW in July).
  • Solar output: Depends on sunlight intensity and daylight hours, which peak during UK summer months.
  • Complementary supply: Combining systems whose peaks match each other's troughs smooths out total generation.

❌ Common Misconceptions

  • Vague answers: Stating simply "solar works when hydro doesn't" without specifying the summer/seasonal context or citing data from Figure 5.
  • Missing the 'reliability' link: Forgetting to explain that the total supply has less variation across the year. Always address both halves of the comparison!

Topics

Physics · P1: Energy

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Higher), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.