AQA AS Level Physics Paper 2, June 2025: Question 2

12 marks · Medium difficulty · Practical Techniques & Data Analysis

Determine the centre of mass of a tapered metal beam using digital balances, including choosing appropriate measuring instruments, checking horizontal alignment, plotting and analyzing balance reading graphs, and evaluating systematic error.

Practise this question

Question

Multi-part physics practical question about finding the centre of mass of a tapered metal beam. Figure 3 shows the beam dimensions: narrow width a, wide width b, length c, and thickness d. Table 2 lists the values of a, b, c, and d. Figure 4 shows the experimental arrangement with the beam resting on two digital balances; balance 1 has a movable triangular support at distance x and rests on an adjustable jack, while balance 2 is on a fixed box. Figure 5 presents a graph of balance reading in grams against x in millimetres with a curve for m2 and points for m1. Table 3 gives two additional data points for m1. Figure 6 shows the geometry for determining distance e of the centre of mass from the wide end using the relationship 2e = c - x when m1 = m2. The final sub-questions address a procedure error where balance 1 is zeroed before placing the support.

Mark scheme

Show the mark scheme Mark scheme for question 02 consisting of parts 02.1 through 02.7 totaling 12 marks. 02.1 awards 1 mark for calipers for b, metre ruler for c, and micrometer screw gauge for d. 02.2 awards 2 marks for setting the ruler vertical with a set square, measuring height at two different points along line P, and verifying they are equal. 02.3 awards 2 marks for accurately plotting points and drawing a smooth curve of increasing gradient through all five points. 02.4 awards 2 marks for adding m1 and m2 values at a common x value from at least two pairs of values to find the beam's mass (around 1270 g). 02.5 awards 2 marks for calculating e in the range 350 to 370 mm using the intersection point where m1 = m2. 02.6 awards 1 mark for noting that values of m1 increase due to the included support mass. 02.7 awards 2 marks for stating the intersection occurs at a smaller value of x, hence e is larger.

How to answer it

Determining the Centre of Mass of a Tapered Beam

WHAT THIS QUESTION TESTS

This practical-skills question evaluates your understanding of mechanics experiments and data analysis:

  • Instrument Selection: Matching measuring instruments to dimensions based on scale and resolution (decimal places).
  • Experimental Alignment: Using geometry tools (metre ruler and set square) to ensure horizontality and reduce parallax/misalignment.
  • Graph Plotting & Line Drawing: Plotting coordinates accurately and drawing a smooth curve of increasing gradient.
  • Moments & Conservation of Mass: Recognising that the sum of balance readings gives total mass ( m₁ + m₂ = M ).
  • Deducing Centre of Mass: Interpreting intersecting curves and applying the condition m₁ = m₂ .
  • Evaluating Systematic Errors: Predicting how a zero/tare error shifts graphs and affects final calculated parameters.

Part 02.1 — Choosing Appropriate Measuring Instruments

Table 2: Dimensions, Values, and Instruments [1 mark]

✅ Correct Answer

DimensionValue / mmInstrument Used
a21.5Vernier calipers (given)
b86.0(Vernier) calipers or travelling microscope
c900(Metre) ruler
d9.74Micrometer (screw gauge) or digital calipers

Note: All three missing instruments must be correct to secure the 1 mark.

💡 Key Knowledge

  • Precision & Decimal Places: In Physics, measurements given to 1 d.p. in mm (0.1 mm resolution) correspond to standard vernier calipers.
  • Measurements given to 2 d.p. in mm (0.01 mm resolution, like d = 9.74 mm ) require a micrometer screw gauge or digital calipers.
  • Large dimensions near 1 m ( c = 900 mm ) require a metre ruler (resolution typically 1 mm).

❌ Common Errors

  • Stating “ruler” for dimension b (86.0 mm implies 0.1 mm precision, which a standard ruler cannot provide).
  • Stating standard “vernier calipers” for dimension d without specifying “digital” (standard mechanical vernier calipers read to 0.1 mm or 0.05 mm, not 0.01 mm).
Mark Breakdown: 1 mark for all three entries correct. (AO1)

Part 02.2 — Verifying Horizontality

Using a Metre Ruler and Set Square to Check Line P [2 marks]

✅ Model Response (Any 2 of 3)

  1. Place the set square flat against the bench and hold the metre ruler vertically against the perpendicular face of the set square.
  2. Measure the vertical distance (or height) between the bench surface and line P at two widely separated points along the beam.
  3. Verify that these two vertical distance measurements are equal.

🧠 Exam Technique

  • Why the set square? A metre ruler held by hand will tilt, introducing cosine errors. The set square guarantees the ruler is strictly at 90° to the bench surface.
  • Always state “at two different points” when checking whether two lines (the bench and line P) are parallel.

❌ Examiner Pitfalls

  • Saying “put the set square on the beam” — a set square alone cannot verify if a line is parallel to the ground; it only measures relative 90° corners.
  • Failing to mention measuring at two different points.
Mark Breakdown: 1 mark for using the set square to ensure ruler is vertical; 1 mark for measuring vertical height at two points and checking they are equal. (AO1)

Part 02.3 — Graph Plotting & Line Drawing

Plotting Data from Table 3 and Drawing Line for m₁ vs x [2 marks]

✅ Required Plotting & Curve

  • Point 1: Plot x = 130 mm , m₁ = 594 g accurately within ±0.5 small square.
  • Point 2: Plot x = 205 mm , m₁ = 658 g accurately within ±0.5 small square.
  • Curve: A single, continuous, smooth curve with an increasing gradient passing cleanly through all 5 points.

💡 Physics Insight: Why a Curve?

The beam is tapered (trapezoidal/wedge shaped). As the knife-edge pivot support moves towards the wide end (increasing x), the mass distribution of the overhang increases non-linearly. Hence, the rate of increase of m₁ accelerates — producing a curve that curves upwards (increasing gradient), NOT a straight line!

❌ Drawing Traps (Losing 1 or Both Marks)

  • Drawing a straight line of best fit: The points clearly curve upward; forcing a straight line forfeits the line mark.
  • Thick points / blobs: Plot marks must be sharp crosses ( × ) or fine dots with circles, under 1 mm across.
  • Discontinuous or “feathery” line: Sketching with multiple overlapping strokes loses the second mark immediately.
Mark Breakdown: 1 mark for plotting both points within minor grid square; 1 mark for a smooth curve of increasing gradient passing through all 5 points. (AO2)

Part 02.4 — Total Mass of the Beam

Determining an Accurate Value for Mass [2 marks]

📐 Step-by-Step Calculation

  1. Underlying Principle: The beam is supported only by balance 1 (reading m₁ ) and balance 2 (reading m₂ ). By vertical equilibrium:
    Total Mass M = m₁ + m₂   (at any common value of x)
  2. Sample Reading 1 (at x = 0 mm):
    From given graph: when x = 0 mm , m₂ = 760 g .
    From plotted curve: when x = 0 mm , m₁ = 510 g .
    M₁ = 760 + 510 = 1270 g .
  3. Sample Reading 2 (at x = 200 mm):
    From given graph: when x = 200 mm , m₂ = 616 g .
    From plotted curve: when x = 200 mm , m₁ = 654 g .
    M₂ = 616 + 654 = 1270 g .
  4. Average / Confirmation:
    Mass of beam ≈ 1270 g .

🧠 Exam Technique

To get full marks for an “accurate value”, the mark scheme explicitly requires reading at least two pairs of values of m₁ and m₂ at the same x, and either calculating an average or showing they yield an identical total.

❌ Common Error

Taking just a single pair of readings scores only 1 of the 2 marks. Always look for ways to reduce random error by taking multiple readings across the graph range.

Mark Breakdown: 1 mark for summing one pair of m₁ + m₂ at common x; 1 mark for determining mass from at least two pairs (or mean). Theoretical value = 1270 g (accept consistent with candidate's graph). (AO3)

Part 02.5 — Centre of Mass Position (e)

Deducing e from the Intersection of the Two Curves [2 marks]

📐 Calculation Steps

  1. Condition for Equal Readings: When m₁ = m₂ , both curves intersect on Figure 5.
  2. Read the Intersection: Locate the coordinate where the curve of m₁ crosses the line of m₂ .
    From the graph, the intersection occurs at: x ≈ 180 mm (read to half a grid square, e.g. 175 – 185 mm).
  3. Apply the Given Formula:
    Given 2e = c - x where c = 900 mm :
    2e = 900 - 180 = 720 mm
    e = 720 / 2 = 360 mm

✅ Acceptable Range

e = 350 to 370 mm
(Theoretical exact value: 360 mm. Full marks awarded if your value matches your curve's intersection point within this window).

🧠 Exam Technique

Show the substitution explicitly: write down 2e = 900 - [your x] on the answer lines so the examiner can award the first method mark even if an arithmetic slip occurs.

Mark Breakdown: 1 mark for suitable evidence of reading intersection x and using 2e = c - x ; 1 mark for final answer in range 350 – 370 mm. (AO3)

Part 02.6 — Systematic Error on the Graph

Zeroing Balance 1 Before Adding the Support [1 mark]

✅ Correct Explanation

The student zeroed the balance before placing the support on it. Therefore, balance 1 measures both the beam force AND the mass of the support.

As a result, all values of m₁ are increased by a constant amount (the graph is shifted vertically upwards; same shape/gradient, but a higher y-intercept).

❌ Common Misconception

Do NOT say that the gradient changes or that the line is steeper! The mass of the support is constant regardless of the value of x, so it is a purely additive systematic shift.

Mark Breakdown: 1 mark for stating all values of m₁ increase / line shifts up vertically because mass of support is included. (AO1)

Part 02.7 — Impact on the Calculated Value of e

Comparing the Erroneous Value of e to the Correct Value [2 marks]

📐 Logical Chain of Reasoning

  1. Effect on the curves: The curve of m₁ is shifted upwards, while the curve for m₂ remains completely unchanged.
  2. Effect on intersection: Because m₁ is higher at every point and m₂ slopes downward, the intersection point ( m₁ = m₂ ) shifts to the left (i.e. occurs at a smaller value of x).
  3. Effect on e:
    The formula is: e = (c - x) / 2
    Since x is smaller, (c - x) becomes larger.
    Therefore, the student's calculated value of e is larger than the correct value.

✅ Marking Points

  • Mark 1: Stating that the intersection of the two lines ( m₁ = m₂ ) occurs at a smaller value of x.
  • Mark 2: Concluding that e is larger (because c - x is larger).

❌ Trap to Avoid

Many students guess that because a mass is added, e must become smaller. Always write out the formula e = (c - x)/2 and track the direct mathematical effect of a smaller x on (c - x) .

Mark Breakdown: 1 mark for identifying smaller value of intersection x; 1 mark for concluding e is larger. (AO3)

Topics

Physics · Practical skills · 3.1 Measurements and their errors · 3.4 Mechanics and materials · Experimental design · Data analysis · Uncertainty and evaluation

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