AQA GCSE Combined Science: Trilogy Physics Paper 2 (Higher), 2018: Question 4

10 marks · Standard Demand difficulty · Short Answer

Answer questions about spring constant, zero error in a newtonmeter, and calculate the total extension of a spring using elastic potential energy and Hooke’s law.

Practise this question

Question

The question page contains three numbered parts about newtonmeters and springs. In part 04.1, Figure 6 shows four vertical newtonmeters labelled A, B, C and D, all drawn with the same length scale but different maximum readings of 2.5 N, 5 N, 10 N and 20 N respectively; the student is asked which has the spring with the greatest spring constant and to give a reason, for 2 marks. In part 04.2, Figure 7 shows a single vertical newtonmeter labelled from 0 N at the top to 20 at the bottom, with the pointer not aligned to zero, and the student is asked to name the type of error and describe how to correct it, for 2 marks. In part 04.3, the text states that a student hangs a weight on a newtonmeter, the energy stored in the spring is 4.5 × 10 to the power minus 2 joules, then increases the weight by 2.0 N, and asks for the total extension of the spring given a spring constant of 400 N/m, for 6 marks.
Question text

04.1 Figure 6 shows four newtonmeters.

Each newtonmeter contains a spring.

Figure 6

Which newtonmeter has the spring with the greatest spring constant?

Give a reason for your answer.

[2 marks]

Newtonmeter

Reason

04.2 The newtonmeter in Figure 7 will give an error when used to make a measurement.

Figure 7

*12* Name the type of error.

Describe how this error can be corrected.

[2 marks]

Type of error

Correction

04.3 A student hangs a weight on a newtonmeter.

The energy now stored in the spring in the newtonmeter is 4.5 × 10–2 J

The student then increases the weight on the newtonmeter by 2.0 N

Calculate the total extension of the spring.

Spring constant = 400 N/m

[6 marks]

Total extension = m

Mark scheme

Show the mark scheme The mark scheme is a table listing answers, extra information, marks and specification references for questions 04.1 to 04.3, all linked to spec ref 6.5.3. For 04.1 it gives D as the correct newtonmeter and explains that it needs the greatest force to produce the same extension. For 04.2 it gives zero error, allowing systematic error, with correction by recording the offset and subtracting it from readings or adjusting the newtonmeter to zero. For 04.3 it shows the calculation using elastic energy 4.5 × 10 to the power minus 2 = 0.5 × 400 × e squared to find an initial extension of 0.015 m, then either uses F = ke to find an additional 0.005 m and adds to get 0.02 m, or finds the initial force as 6 N, adds 2 N to get 8 N, and uses Hooke’s law to get the same total extension of 0.02 m.

AO /

Question Answers Extra information Mark

Spec. Ref.

04.1 D allow 20 (N) 1 AO1

allow fourth (newtonmeter) 6.5.3

needs the greatest force to reason only scores if correct 1

extend the spring the same newtonmeter selected

amount

04.2 zero (error) allow systematic (error) 1 AO3

6.5.3

any one from:

• record the value and allow subtract 1 from all 1

subtract from readings readings

taken

• adjust the newtonmeter

to zero

AO /

Spec. Ref.

04.3 an answer of 0.02 (m) gains 6 AO2

marks 6.5.3

4.5 × 10−2 = 0.5 × 400 × e2 this mark may be awarded if the 1

standard form value is

incorrectly converted

−2 this mark may be awarded if the 1

�4.5 × 10 standard form value is

e = incorrectly converted

0.5 × 400

4.5 × 10−2

allow e2 =

0.5 × 400

e = 0.015 (m) this answer only 1

2.0 = 400 × e 1

e = 0.005 (m) this answer only 1

0.015 + 0.005 = 0.02 (m) allow their initial extension + 1

their additional extension

correctly calculated

or

4.5 × 10−2 = 0.5 × 400 × e2 (1) this mark may be awarded if the

standard form value is

incorrectly converted

this mark may be awarded if the 12

4.5 × 10−2

� standard form value is

e = (1)

0.5 × 400 incorrectly converted

4.5 × 10−2

allow e2 =

0.5 × 400

e = 0.015 (m) (1) this answer only

F = 400 × 0.015 allow an answer of 400 × their

calculated value of e

F = 6 (N) (1)

total force = 6 + 2 allow an answer that is

consistent with their calculated

8 = 400 × e (1) value of e

e = 0.02 (m) (1)

Total 10

How to answer it

Spring constant and newtonmeters

What this question tests

This question checks whether you can compare springs using the same extension, recognise a zero error, and use the spring energy/Hooke’s law relationship to calculate extension. You need to read diagrams carefully, choose the correct newtonmeter, and show clear calculation steps with units.

Part (a) 04.1 — Which newtonmeter has the greatest spring constant?

✅ Correct answer

Newtonmeter D

Reason: It needs the greatest force to extend the spring by the same amount.

💡 Key knowledge

  • The spring constant tells you how stiff a spring is.
  • A larger spring constant means a spring is harder to stretch.
  • If the same extension is produced by a bigger force, the spring constant is greater.

🧠 Exam technique

  • Use the diagram labels: D has the largest force scale, so it is the stiffest spring.
  • For the mark, the reason must match the correct choice.
  • Good wording: “needs the greatest force to extend the spring the same amount.”

❌ Common errors

  • Choosing the smallest number because it looks like the “most sensitive” meter.
  • Writing only “D” with no reason — that loses the second mark.
  • Confusing big force scale with big extension.
How marks were awarded: 1 mark for the correct newtonmeter, 1 mark for a reason linked to force and extension. The mark scheme allows “20 N” or “fourth newtonmeter” as well.

Part (b) 04.2 — Error in the newtonmeter

✅ Correct answer

Type of error: zero error

Correction: adjust the newtonmeter to zero, or record the error and subtract it from all readings.

💡 Key knowledge

  • A zero error happens when the instrument does not read zero before measuring.
  • This is a systematic error because it affects all readings by the same amount.
  • To fix it, the meter should be set to zero before use, or the offset should be corrected afterwards.

🧠 Exam technique

  • Look at the needle position when no force is applied.
  • If it is not at 0 N, the reading is shifted.
  • For full marks, name the error and describe a correction.

❌ Common errors

  • Saying “random error” — this is wrong because the offset is consistent.
  • Writing “parallax” — that is a reading error from viewing angle, not the issue shown here.
  • Forgetting to explain how to correct the error.
How marks were awarded: 1 mark for “zero error” (systematic was also allowed), and 1 mark for a correction such as “adjust to zero” or “subtract 1 from all readings.”

Part (c) 04.3 — Calculate the total extension of the spring

Given: energy stored = 4.5 × 10⁻² J, extra force = 2.0 N, spring constant = 400 N/m

📐 Calculations: step-by-step

  1. Find the initial extension from the energy stored.
    Use E = 1/2 k e²
  2. Substitute the values:
    4.5 × 10⁻² = 0.5 × 400 × e²
  3. Rearrange:
    e² = (4.5 × 10⁻²) / (0.5 × 400)
  4. Calculate:
    e² = 0.045 / 200 = 0.000225
  5. Square root:
    e = 0.015 m
  6. Find the extra extension from the extra force.
    Use F = k e
  7. Substitute:
    2.0 = 400 × e
  8. Rearrange and calculate:
    e = 2.0 / 400 = 0.005 m
  9. Add the two extensions:
    0.015 + 0.005 = 0.020 m
  10. Final answer: 0.02 m

✅ Correct answer

Total extension = 0.02 m

This is the full-mark answer expected by the mark scheme.

💡 Key knowledge

  • Elastic potential energy: E = 1/2 k e²
  • Hooke’s law: F = ke
  • Extension must be in metres.
  • Spring constant units are N/m.

🧠 Exam technique

  • Use the correct equation for each part of the question.
  • Show rearrangement clearly to pick up method marks.
  • Keep units in the working, especially N, N/m, m, and J.
  • Round the final answer appropriately: 0.02 m.

❌ Common errors

  • Using E = ke instead of E = 1/2 k e² .
  • Forgetting that the 2.0 N is an extra force added after the initial stretch.
  • Not converting correctly from joules to the equation form — the energy stays in J, but extension must be in m.
  • Missing the final addition of the two extensions.

📐 Why the answer is 0.02 m

The mark scheme accepted answers that showed the spring was first stretched by 0.015 m, then stretched a further 0.005 m, giving a total of 0.02 m. Top-level answers made both stages clear and used correct physics equations.

❌ Calculation traps the examiner was checking

  • Mixing up force and energy.
  • Using the wrong spring formula for the wrong stage.
  • Forgetting to add the second extension to the first.
  • Leaving the answer as 2 cm without showing it equals 0.02 m .
How marks were awarded: The mark scheme gives multiple method marks. It rewards: correct substitution into E = 1/2 k e² , finding e = 0.015 m , using F = ke to get 0.005 m , and then adding them to reach 0.02 m . An answer of 0.02 m could gain full credit if the working is consistent.

Quick recap: what to remember for similar questions

💡 Key knowledge

  • Greater spring constant = stiffer spring.
  • Zero error = meter does not read 0 before measuring.
  • Elastic energy uses E = 1/2 k e² .
  • Force-extension uses F = ke .

🧠 Exam technique

  • Always link your reason to the mark scheme wording.
  • Show every step in calculations.
  • Check units at the end.

❌ Common errors

  • Choosing the wrong spring from the diagram.
  • Naming the wrong type of error.
  • Forgetting to add the extra extension in part (c).

Topics

Physics · P5: Forces

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 2 (Higher), 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.