AQA GCSE Physics Physics Paper 2 (Higher), June 2025: Question 2

✅ Correct Answer

Force applied to the spring (or weight of the masses).

Also allowed: Mass hung from the spring.

💡 Key Knowledge

The independent variable is the variable that the experimenter deliberately changes or selects values for. In this investigation, weights are added incrementally to vary the applied force.

❌ Common Errors

Confusing independent and dependent variables. Writing "length of the spring" or "extension" scores 0 marks because those are the dependent variables (what is measured).

✅ Accepted Pairs (1 mark for risk + 1 mark for linked precaution)

• Risk: Stand tipping over and hitting/falling on the student.
  Precaution: Clamp the base of the stand to the desk (or place a heavy G-clamp / mass on the base).
• Risk: Heavy masses falling off the hanger and hurting feet/hands.
  Precaution: Work standing up, keep feet away, or clamp masses securely.
• Risk: Spring snapping / breaking loose and hitting eyes/face.
  Precaution: Wear safety goggles / glasses, or limit the maximum load added.

🧠 Exam Technique

A "risk of harm" requires both an event and the resulting injury (e.g. "masses falling off and hitting feet"). Just stating "the stand might fall" is an hazard, not a full risk of harm.

❌ Common Errors

Vague precautions like "be careful" or "handle with care" never gain credit in AQA practical questions. Always suggest an active laboratory control (e.g. G-clamp, eye protection).

✅ Any Two Improvements (1 mark each)

• Clamp the ruler in place (close to the spring) instead of holding it by hand.
• Ensure the ruler is vertical (e.g. using a set square against the bench).
• Attach a horizontal pointer (fiducial marker) to the bottom of the spring to read the ruler level directly.
• Use a ruler with a higher resolution (e.g. millimetre increments).
• Move ruler so it is directly alongside and in line with the spring.

💡 Parallax & Human Error

In Figure 1, the student holds the ruler at an angle and away from the spring. Holding a ruler introduces hand tremors and tilt, while reading from an angle causes parallax error.

✅ Correct Answer (Any one)

• The unstretched (initial) length of the spring is not zero (it has an initial length of ~3.6 cm at 0 N).
• The spring length is not directly proportional to the force (length was plotted rather than extension).

❌ Critical Distinction: Length vs Extension

A graph of extension against force passes through (0,0) because at 0 N there is 0 extension. However, this graph plots total length, so the y-intercept represents the original length of the spring before any load was applied.

✅ Mark Scheme Breakdown

• Mark 1: The spring is inelastically deformed (or has exceeded its elastic limit).
• Mark 2: The extension is no longer directly proportional to the force applied (it has passed the limit of proportionality / no longer obeys Hooke's Law).

💡 Elastic vs Inelastic Deformation

Elastic deformation: The spring returns to its original length when the force is removed. Force and extension are directly proportional.

Inelastic (plastic) deformation: The spring is permanently stretched. Adding further force produces much larger extensions, causing the line on a length–force graph to curve upwards steeply.

✅ Selected Option

force = spring constant × extension ☑️ (First box)

In standard symbol form: F = k × e

🧠 Physics Equations Sheet

This formula is directly given on the AQA GCSE Physics equation sheet. Always take a second to check the sheet to guarantee you don't pick an inverse or divided variant by accident!

❌ Calculation Traps

• Rearrangement blunder: Doing 0.064 / 4.0 = 0.016 instead of F / e . Always ensure you divide force by extension!
• Overlooking units: If extension had been in centimetres, you would first divide by 100 to get metres. Here it was already given in metres ( 0.064 m ).

Mark Summary:
• 1 mark for correct substitution: 4.0 = k × 0.064
• 1 mark for correct rearrangement: k = 4.0 / 0.064
• 1 mark for correct final answer: 62.5 (or 63 )