AQA GCSE Physics Physics Paper 2 (Higher), June 2025: Question 3

7 marks · Standard Demand difficulty · Short Answer

Calculate uncertainty and error in time measurements for a falling ball and determine its acceleration using the gradient of a height against time squared graph.

Practise this question

Question

Question 3 presents an investigation of a falling gravity ball. Figure 3 shows a gravity ball held in a hand, featuring a digital stopwatch display and a release button. Subquestion 03.1 gives five time measurements (0.49 s, 0.51 s, 0.58 s, 0.56 s, 0.61 s) and asks to calculate the uncertainty. Subquestion 03.2 provides checkboxes to identify the error type: random error, systematic error, or zero error. Subquestion 03.3 asks for one reason for the error. Subquestion 03.4 displays Figure 4, a graph of height dropped from in metres (0 to 4.0) against time squared in s² (0 to 0.8), with a straight line through the origin passing through (0.7, 3.5). Students must calculate the acceleration using acceleration = 2 × (gradient of the graph).

Mark scheme

Show the mark scheme Mark scheme for Question 3 with a total of 7 marks: 03.1 awards 1 mark for calculating the range (0.61 - 0.49 = 0.12) and 1 mark for uncertainty = ± 0.06 s. 03.2 awards 1 mark for selecting 'random error'. 03.3 awards 1 mark for reasons such as varying drop height, different initial speed, or delay in pressing/releasing. 03.4 awards 3 marks: 1 mark for reading corresponding graph coordinates (e.g. 3.5 m and 0.70 s²), 1 mark for gradient = 5.0, and 1 mark for acceleration = 10.0 m/s².

How to answer it

Investigating Acceleration Due to Gravity (Free Fall)

What This Question Tests

This question evaluates practical skills in experimental mechanics: calculating experimental uncertainty from repeat measurements, identifying and explaining random errors in timing free fall, and using a linear graph (height against time squared) to extract a gradient and calculate acceleration due to gravity.

Question 03.1

Calculating Uncertainty from Repeat Data

Data: 0.49 s, 0.51 s, 0.58 s, 0.56 s, 0.61 s (2 Marks)

📐 Step-by-Step Calculation

  1. Find the range:
    Range = Max value - Min value
    Range = 0.61 - 0.49 = 0.12 s [1 mark]
  2. Divide the range by 2:
    Uncertainty = Range ÷ 2
    Uncertainty = 0.12 ÷ 2 = ± 0.06 s [1 mark]
Mark Scheme: 1 mark for finding range (0.12) or mean (0.55). 1 mark for uncertainty = ± 0.06 (s).

❌ Common Errors

  • Forgetting to divide by 2: Writing 0.12 s (the whole range) instead of half the range.
  • Calculating only the mean: Finding 0.55 s and writing that on the answer line. (The mean gets 1 working mark, but misses the answer mark).
  • Ignoring the ± sign: While the question line includes "±", always remember that uncertainty represents spread either side of the mean.
Question 03.2

Identifying Types of Experimental Error

Classifying variation in timing measurements (1 Mark)

✅ Correct Answer

[✓] Random error

The options for Systematic error and Zero error should be left unticked.

💡 Key Knowledge

  • Random errors: Cause readings to fluctuate unpredictably above and below the true value. Repeating and calculating a mean reduces their effect.
  • Systematic errors: Cause readings to differ from the true value by a consistent amount each time (e.g. zero errors on a ruler or balance).
Question 03.3

Sources of Variation in Free-Fall Experiments

Suggest one reason for the timing variation (1 Mark)

✅ Acceptable Answers (Any One)

  • The ball may have been dropped from slightly different heights each time.
  • The initial speed of the ball was different (e.g. ball was pushed downwards or held slightly instead of being released cleanly).
  • Variation or delay between pressing the start button and releasing the ball.
  • The stopwatch mechanism did not always stop immediately upon hitting the floor.

🧠 Exam Technique & Examiner Tip

Never just write "human error"!

Examiners award zero marks for the vague phrase "human error". You must clearly identify the specific mechanical or procedural cause (e.g. variation in release point or release delay).

Question 03.4

Graph Gradient & Acceleration Calculation

Equation: acceleration = 2 × (gradient of the graph) (3 Marks)

📐 Step-by-Step Solution

  1. Choose coordinates from the line of best fit:
    The line passes directly through (0, 0) and (0.70 s², 3.5 m) .
    [1 mark for reading corresponding values]
  2. Calculate the gradient:
    Gradient = Δy ÷ Δx = (3.5 - 0) ÷ (0.70 - 0) = 5.0
    [1 mark]
  3. Apply the given relationship:
    acceleration = 2 × 5.0 = 10.0 m/s² (or 10 m/s² )
    [1 mark]
Mark Scheme: MP1: Correct values read from graph. MP2: Gradient = 5.0. MP3: Acceleration = 10.0 (or 10) m/s².

❌ Common Traps

  • Forgetting the formula: Many students correctly calculate the gradient as 5.0 and stop there, forgetting to multiply by 2!
  • Inverting the gradient: Calculating Δx ÷ Δy (0.70 ÷ 3.5 = 0.2) instead of Δy ÷ Δx .
  • Physics Sanity Check: Acceleration due to gravity on Earth is around 9.8 m/s² . An answer close to 10 m/s² tells you your working is right!

Topics

Physics · P5: Forces

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.