AQA GCSE Combined Science: Trilogy Physics Paper 2 (Higher), 2023: Question 5

✅ Correct answer

W ∝ e

💡 Key knowledge

• For a spring obeying Hooke’s law, force is directly proportional to extension.
• Weight acts as the force stretching the spring.
• So if weight doubles, extension doubles, as long as the spring has not passed the limit of proportionality.

🧠 Exam technique

If you see “describes the relationship”, look for proportionality rather than equality unless the equation includes a constant, such as F = ke .

❌ Common errors

• Choosing W = e instead of proportionality.
• Thinking weight must be smaller than extension because the units are different. They are different physical quantities, so inequalities are not appropriate here.

📐 Calculations

1. Use Hooke’s law: F = ke
2. Convert extension into metres: 60 mm = 0.060 m
3. Substitute values: 750 = k × 0.060
4. Rearrange: k = 750 ÷ 0.060
5. Calculate: k = 12 500 N/m

Spring constant = 12 500 N/m

✅ Mark breakdown

• 1 mark for substituting into 750 = k × 0.060
• 1 mark for correct rearrangement: k = 750 / 0.060
• 1 mark for the correct answer: 12 500 N/m

💡 Key knowledge

• Spring constant unit = N/m
• Extension must be in metres when using the standard equation sheet form.
• Weight is a force, measured in newtons.

❌ Common errors

• Using 60 instead of 0.060 .
• Writing the unit as N instead of N/m .
• Mixing up spring constant and extension when rearranging.

🧠 Exam technique

In physics calculations, unit conversion is often the difference between full marks and losing the final answer mark. Write the conversion as its own step to make your method clear.

✅ Correct answer

An object that is inelastically deformed does not return to its original length when the force is removed.

💡 Key knowledge

• Elastic deformation: returns to original shape/length after the force is removed.
• Inelastic deformation: permanent change remains.
• The mark scheme allows “shape” as well as “length”.

🧠 Exam technique

For a definition question, include both parts: what happens to the object and when it happens. Here, that means permanent change after the force is removed.

❌ Common errors

• Saying only “it stretches” — that is not enough.
• Missing out the idea that the force/load has been removed.
• Confusing inelastic deformation with breaking.

📐 Calculations

1. Use the elastic potential energy equation: E = ½ke²
2. Substitute values: 1800 = ½ × 225 × e²
3. Rearrange: e² = (2 × 1800) / 225
4. Calculate: e² = 3600 / 225 = 16
5. Square root: e = 4 m

Maximum extension = 4.0 m

✅ Mark breakdown

• 1 mark for 1800 = ½ × 225 × e²
• 1 mark for correct rearrangement to find e or e²
• 1 mark for e = 4 m

💡 Key knowledge

• Elastic potential energy depends on the spring constant and the square of extension.
• Because extension is squared, you must square root at the end.
• The unit of extension is metres.

❌ Common errors

• Forgetting the ½ in the equation.
• Stopping at e² = 16 and not taking the square root.
• Giving the final unit as J or N instead of m .

🧠 Exam technique

When the unknown is squared, write each algebra step clearly. Examiners often award method marks even if the final arithmetic goes wrong later.

📐 Main calculation method

1. Use Hooke’s law: F = ke
2. Rearrange: e = F / k
3. Substitute values: e = 750 / 225
4. Calculate: e = 3.3... m
5. Compare with the distance to the ground: 30 cm = 0.30 m

The extension would be far greater than 0.30 m, so the chair would reach/rest on the ground. The spring is not suitable.

✅ Mark breakdown

• 1 mark for calculating extension: e = 750 / 225
• 1 mark for e = 3.3... m
• 1 mark for a valid conclusion consistent with the calculation: not suitable

💡 Key knowledge

• A spring can be unsuitable either because it stretches too much in use or because it exceeds safe limits.
• Here, the spring’s extension under the person’s weight is much bigger than the chair’s ground clearance.
• Since 3.3 m is also less than the 4.0 m maximum elastic extension from part 05.4, the main problem is not immediate inelastic deformation — it is that the chair would drop far too low.

🧠 Exam technique

“Evaluate” means calculate + compare + conclude. A top answer links the numbers to the real situation: the bottom of the chair is only 0.30 m above the ground, but the spring would extend by 3.3 m .

You could also use the alternative mark-scheme idea: F = ke = 225 × 0.3 = 67.5 N . That means a 0.30 m extension only supports 67.5 N , which is far less than the person’s 750 N weight, so the spring is not suitable.

❌ Common errors

• Writing only “not suitable” with no calculation.
• Forgetting to convert 30 cm to 0.30 m .
• Comparing the 1800 J directly with 750 N — these are different quantities and cannot be compared directly.
• Missing the real-world interpretation that the chair would end up on the ground.

✅ 05.1

W ∝ e

✅ 05.2

F = ke
60 mm = 0.060 m
750 = k × 0.060
k = 750 / 0.060 = 12 500 N/m

✅ 05.3

An inelastically deformed object does not return to its original length when the force is removed.

✅ 05.4

E = ½ke²
1800 = ½ × 225 × e²
e² = (2 × 1800) / 225 = 16
e = 4.0 m

✅ 05.5

e = F / k = 750 / 225 = 3.3 m
The chair is only 0.30 m above the ground, so the extension would be too great. The new spring is not suitable.

🧠 How students lost marks

• Not converting mm to m or cm to m.
• Using the wrong spring equation for the quantity asked.
• Giving a calculation without a written conclusion in the evaluation question.
• Missing key words in the definition of inelastic deformation.

💡 What strong answers did well

• Selected the correct equation immediately.
• Showed clear method lines, so method marks were secured.
• Used correct units throughout.
• Linked the final answer to the practical situation: the chair would hit the ground, so the spring is not suitable.