AQA AS Level Physics Paper 2, June 2024: Question 1

12 marks · Hard difficulty · Practical Techniques & Data Analysis

Analyze an experiment using an oscillating hacksaw blade attached to an accelerating trolley on a ramp to determine acceleration, uncertainties, and acceleration due to gravity.

Practise this question

Question

A series of diagrams and questions investigating the motion of a trolley down a ramp using an oscillating hacksaw blade with a pen attached to draw sinusoidal traces on graph paper. Figure 1 and 2 show the side and top views of the oscillating blade. Figure 3, 4a-c, and 5 show the setup of the trolley with graph paper and the resulting trace with points P and Q marking the start and end of the oscillations used to calculate time and displacement.
Question text

01 A hacksaw blade is a thin flexible strip of metal.

Figure 1 shows a blade clamped between two blocks above a horizontal bench.

A pen is attached to the free end of the blade.

Figure 1

The free end of the blade is displaced and released.

The blade oscillates in a horizontal plane as shown in Figure 2.

Figure 2

The time for each oscillation is T.

01.1 Table 1 shows repeated measurements of 60T.

Table 1

Measurements of 60T / s

25.20 25.05 24.97 25.10

Show that T is about 0.42 s. 4

[1 mark]

Figure 3 shows a trolley placed on a ramp that is inclined at a small angle to the

bench.

A piece of graph paper is fixed to the upper surface of the trolley.

The blade and pen are positioned so that the tip of the pen rests on the graph paper.

The dashed line shows the rest position of the pen.

Figure 3

The free end of the blade is displaced as shown in Figure 4a.

The blade and the trolley are then both released at the same moment.

The blade oscillates horizontally.

The pen remains in contact with the graph paper as the trolley moves.

Figures 4b and 4c show the trolley as it moves down the ramp with uniform

acceleration.

Figure 5 shows the graph paper.

Points P and Q mark the start and end of the continuous line drawn by the pen after

the trolley is released.

Figure 5

TPQ is the time for the pen to draw the line from P to Q.

s is the displacement of the trolley during TPQ.

01.2 Determine TPQ.

Assume that the time for each full oscillation of the blade is 0.42 s.

[2 marks]

TPQ = s

01.3 Determine s.

The scale of the graph paper is shown on Figure 5.

[1 mark]

s = m

01.4 Determine the acceleration a of the trolley.

[2 marks]

8 a = m s−2

01.5 A teacher suggests that the absolute uncertainty in s is ±2 mm.

Explain why this is a valid suggestion.

[2 marks]

01.6 The percentage uncertainty in TPQ is 0.46%.

Determine the percentage uncertainty in your result for a.

[2 marks]

percentage uncertainty9 = %

01.7 Figure 6 is a diagram drawn by a student to explain why the trolley accelerates.

The diagram is incomplete because the student has ignored the friction forces

involved.

Figure 6

Using Figure 6 it can be shown that:

a

g =

sin

where a is the acceleration of the trolley.

The student determines g using this equation.

State and explain how the student’s value of g compares with 9.81 m s−2.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme detailing numerical answers, acceptable ranges for time periods and cycles, kinematic calculations for acceleration, absolute uncertainty reasoning based on graph grid resolution, combination of percentage uncertainties, and comparison of experimental acceleration due to gravity with standard values taking friction into account.

Question Answers Additional Comments/Guidance Mark AO

01.1 Expect to see 25.08 (mean average) divided by 60, 1 AO2

calculates, using all 4 values, a mean of 0.418 (s)

or 100.32 (sum) divided by 240 in working

Expect TPQ = 1.15, 1.16 or 1.2 (s)

01.2 2.75 cycles (between P and Q) 1 2 2 × AO3

2 Allow use of >2 sf TPQ that rounds to 0.42 (s)

TPQ = 0.42 × their number of cycles 2 2 Their number of cycles must be between 2.5

and 3

01.3 0.170 (m) Condone 2 sf value on answer line if working shows 1 AO2

a 3 sf value or “170 mm” seen or “20 mm” used e.g.

‘8.5 × 20 mm’.

Expect to see a = 0.24, 0.25 or 0.26 (m s−2)

01.4 correct use of an appropriate equation of motion 1 2 2 × AO2

correct evaluation of their a 2

2 × their s

1 Expect a = 2 OR

their (TPQ)

their s 2 × their mean v

meanv = AND a =

their TPQ their TPQ

Expect mean v = 0.14 or 0.15 (m s−1)

1 Allow s in mm

01.5 links (absolute) uncertainty of 1 mm for one reading to the 1 Condone ‘uncertainty in a single reading is half 2 2 × AO1

resolution of 2 mm of the graph paper 1 a grid division’

idea that s is based on two readings so (absolute) 2 Allow ‘s is based on two readings so uncertainty

uncertainties in each reading are added 2 in s is double the uncertainty of each reading’

01.6 0.002 Expect to see % uncertainty in a = 2.1 2 2 × AO3

× 100 7

𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑠𝑠 1 Expect % uncertainty in s = 1.2. Calculator value

OR is 1.17647.

2 × 0.46 or 0.92 seen 1 1 Allow values in mm

2 Allow 1 or 2 sf values only

% uncertainty in a = (their % uncertainty in s) + 0.92 2

01.7 resultant force should be lower 1 1 Default interpretation of “a” is the experimental 2 2 × AO3

value (from 01.4) unless otherwise defined.

1 Allow idea that experimental value of a would be

larger in absence of friction.

1 Credit algebraic expression that includes friction

(F): ma = mgsinθ – F. Condone missing “m”.

g 9.81 m s−2 2 is contingent on 1

(student’s value of) is less (than ) 2

Total 12

How to answer it

Hacksaw Blade & Trolley Acceleration Study Guide

AQA AS Level Physics — Mechanics & Uncertainty Analysis

What this question tests

This multi-step mechanics question evaluates your ability to process experimental timing data, interpret graphical traces, apply equations of uniform acceleration, propagate absolute and percentage uncertainties, and analyse real-world discrepancies using Newtonian mechanics and free-body force diagrams.

Part 01.1: Mean Time Calculation

Show that T is about 0.42 s (1 mark)

✅ Correct Answer

Find the mean of the 4 given values for 60T and divide by 60 (or find the total sum and divide by 240).

Mean 60T = (25.20 + 25.05 + 24.97 + 25.10) / 4 = 25.08 s

T = 25.08 / 60 = 0.418 s (~0.42 s)

🧠 Exam Technique

Always show your intermediate sum or mean average clearly before dividing by the multiplier. Examiners look for explicit evidence that all four values were utilised.

Mark scheme note: 1 mark awarded for calculating a mean of 0.418 s using all 4 values.

Part 01.2: Determining Time Interval TPQ

Determine TPQ assuming each full oscillation is 0.42 s (2 marks)

✅ Correct Answer

TPQ = 1.15 s, 1.16 s, or 1.2 s (corresponding to 2.75 complete cycles)

📐 Calculation Steps

  1. Count cycles: Count the number of waves/oscillations between point P and point Q on Figure 5. There are 2.75 cycles.
  2. Multiply: TPQ = 2.75 × 0.42 s = 1.155 s (rounds to 1.16 s or 1.2 s).

❌ Common Errors

Miscounting cycles by missing the fractional start/end fractions (P and Q start/end mid-oscillation, giving 0.75 of a wave plus 2 full waves).

Mark scheme note: 1 mark for identifying 2.75 cycles between P and Q; 1 mark for multiplying by 0.42 s.

Part 01.3: Determining Displacement s

Determine s from the graph paper scale (1 mark)

✅ Correct Answer

s = 0.170 m (or 170 mm)

💡 Key Knowledge

Use the provided 20 mm grid scale reference box on Figure 5 to measure the linear displacement between points P and Q along the dashed centre line.

Mark scheme note: 1 mark for 0.170 m (condone 2 sf values if working clearly supports it).

Part 01.4: Calculating Acceleration a

Determine the acceleration a of the trolley (2 marks)

✅ Correct Answer

a = 0.24 to 0.26 m s⁻² (using s = 0.17 m and TPQ = 1.16 s)

📐 Calculation Steps

  1. Select kinematic equation: s = (1/2) a t² (since initial velocity u = 0).
  2. Rearrange for acceleration: a = 2s / (TPQ)²
  3. Substitute values: a = (2 × 0.170) / (1.16)² = 0.252... m s⁻²

❌ Common Errors

Forgetting to multiply displacement by 2 or failing to square the time interval in the denominator.

Mark scheme note: 1 mark for correct kinematic equation selection; 1 mark for correct numerical evaluation of a.

Part 01.5: Validating Absolute Uncertainty

Explain why an absolute uncertainty in s of ±2 mm is valid (2 marks)

✅ Correct Answer

1. The resolution of the graph paper grid is 1 mm, meaning a single reading has an uncertainty of ±1 mm (or half a grid division).

2. Because displacement s is determined from two separate readings (start point P and end point Q), their absolute uncertainties must be added together (1 mm + 1 mm = 2 mm).

🧠 Exam Technique

When asked to "explain" an uncertainty value, always explicitly state instrument resolution rules and how combining multiple measurements doubles the absolute uncertainty.

Mark scheme note: 1 mark for linking 1 mm uncertainty to graph paper resolution; 1 mark for stating s relies on two separate readings so absolute uncertainties add.

Part 01.6: Uncertainty Propagation

Determine the percentage uncertainty in your result for a (2 marks)

✅ Correct Answer

Percentage uncertainty in a ≈ 2.1% (Accept 2.1% to 2.2% depending on rounded intermediate values)

📐 Calculation Steps

  1. Find % uncertainty in s: (2 mm / 170 mm) × 100 = 1.176%
  2. Since a = 2s / T² and T has a given % uncertainty of 0.46%, use power rules for uncertainties:
    % uncertainty in (T²) = 2 × (% uncertainty in T) = 2 × 0.46% = 0.92%
  3. Add percentage uncertainties together:
    % uncertainty in a = (% uncertainty in s) + (% uncertainty in T²) = 1.176% + 0.92% = 2.096% → 2.1%
Mark scheme note: 1 mark for calculating % uncertainty in s or correctly doubling T's % uncertainty (0.92%); 1 mark for summing them to find final % uncertainty in a.

Part 01.7: Comparing Experimental g

State and explain how the student's value of g compares with 9.81 m s⁻² (2 marks)

✅ Correct Answer

The student's experimental value of g will be less than 9.81 m s⁻².

💡 Key Knowledge & Explanation

Friction was ignored in the student's derivation. Friction acts as a retarding force opposing motion down the slope, meaning the net accelerating force is smaller than the theoretical component ( mg sin(θ) - F_friction ). Consequently, measured acceleration a is lower, resulting in a lower calculated value of g.

Mark scheme note: 1 mark for recognising that the resultant force / experimental acceleration should be lower due to friction; 1 mark (contingent) for stating g is less than 9.81 m s⁻².

Topics

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

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.