AQA GCSE Combined Science: Trilogy Physics Paper 2 (Foundation), November 2020: Question 4

9 marks · Low Demand difficulty · Short Answer

Determine bicycle speed from a distance-time graph, identify a non-contact force, calculate work done with units, and complete sentences regarding energy changes and friction.

Practise this question

Question

Question 04 contains four parts based on a bicycle ride. Figure 5 shows a distance-time graph plotted from 0 to 50 seconds on the horizontal axis and 0 to 250 metres on the vertical axis, showing a straight line from (0, 0) to (50, 250). Part 04.1 asks to determine the speed of the bicycle from the gradient in m/s for 2 marks. Part 04.2 asks which force is a non-contact force from options: air resistance, friction, gravitational force, or normal contact force for 1 mark. Part 04.3 asks to calculate work done given a distance of 250 m and horizontal force of 30 N using the equation work done = force × distance, and select the unit from J, kg, or m for 3 marks. Part 04.4 asks to complete three sentences about forces and energy stores using words from a box: chemical, frictional, kinetic, magnetic, and tension, for 3 marks.
Question text

04 Figure 5 shows a distance-time graph for 50 seconds of a bicycle ride.

Figure 5

04.1 The gradient of the distance-time graph gives the speed of the bicycle.

Determine the speed of the bicycle.

[2 marks]

13 Speed = m/s

04.2 Which force acting on the moving bicycle is a non-contact force?

[1 mark]

Tick ( ) one box.

Air resistance

Friction

Gravitational force

Normal contact force

04.3 The bicycle travels a distance of 250 m

The bicycle exerts a constant horizontal force of 30 N on the ground.

Calculate the work done.

Use the equation:

work done = force × distance

Choose the unit from the box.

[3 marks]

J kg m

Work done =14 Unit

04.4 The bicycle travels at a constant speed.

Complete the sentences.

Choose answers from the box.

[3 marks]

*13* chemical frictional kinetic

magnetic tension

As the bicycle moves, work is done against forces.

There is no change in the cyclist’s store of energy.

There is a decrease in the cyclist’s store of energy.

Mark scheme

Show the mark scheme Mark scheme for Question 04 showing: 04.1 requires gradient = (250 - 0)/(50 - 0) for 1 mark (AO2) and speed = 5.0 m/s for 1 mark (AO2). 04.2 answers 'gravitational force' for 1 mark (AO1). 04.3 shows W = 30 × 250 for 1 mark (AO2), W = 7500 for 1 mark (AO2), and unit J for 1 mark (AO1). 04.4 awards 1 mark each (AO1) for 'frictional', 'kinetic', and 'chemical'. Total marks = 9.

AO /

Question Answers Extra information Mark

Spec. Ref.

04.1 250 - 0 allow any correct pair of values 1 AO2

gradient =

50 - 0 substituted 6.5.4.1.4

speed = 5.0 m/s allow 5 (m/) 1

allow the use of s = v t

04.2 gravitational force 1 AO1

6.5.1.2

04.3 W = 30 × 250 1 AO2

W = 7500 1 AO2

J 1 AO1

6.5.2

04.4 frictional 1 AO1

6.5.2

kinetic 1

chemical 1

Total 9

How to answer it

Bicycle Motion: Distance-Time Graphs, Forces & Work Done

📋 What this question tests

This question covers fundamental concepts in AQA GCSE Combined Science Physics (Forces topic):

  • Distance-time graphs: Interpreting a linear graph and calculating speed from the gradient (rise ÷ run).
  • Types of forces: Distinguishing between contact forces (friction, normal force, air resistance) and non-contact forces (gravity, magnetism, electrostatic).
  • Work done equation: Applying work done = force × distance and identifying standard scientific units (Joules).
  • Energy stores and transfers: Identifying chemical, kinetic, and frictional energy interactions when an object moves at a constant speed.
Part 04.1 (2 marks)

Calculating Speed from a Distance-Time Graph

Determining the gradient of a straight-line graph

📐 Step-by-Step Calculation

  1. Identify coordinates: The line starts at (0, 0) and ends clearly at (50 s, 250 m).
  2. Find the gradient:
    gradient = change in y ÷ change in x
    gradient = (250 − 0) ÷ (50 − 0) = 250 ÷ 50
  3. Calculate value:
    Speed = 5.0 m/s (or 5 m/s )

✅ Mark Scheme Breakdown

  • [1 mark]: Correct substitution into gradient formula, e.g. 250 / 50 or using any matching coordinates along the line (e.g. 100 / 20 ).
  • [1 mark]: Final value of 5.0 (or 5 ).

🧠 Exam Technique

Always pick points that are easy to read from the grid. Using the full line from (0, 0) to the very end point (50, 250) gives the most reliable values and avoids misreading small grid squares.

❌ Common Errors

Inverting the fraction (doing 50 ÷ 250 = 0.2 ). Remember that on a distance-time graph, distance (m) is on the vertical y-axis and time (s) is on the horizontal x-axis: speed = distance ÷ time .

Part 04.2 (1 mark)

Contact vs Non-Contact Forces

Identifying non-contact forces

✅ Correct Answer

Tick (✔) the box next to: Gravitational force

💡 Key Knowledge

  • Non-contact forces: Act without physical contact between objects. The three main examples at GCSE are gravity, magnetic force, and electrostatic force.
  • Contact forces: Require objects to physically touch. Examples: air resistance, friction, normal contact force, and tension.

❌ Common Errors

Confusing air resistance as a non-contact force. Even though air is an invisible gas, the bicycle must physically collide with air particles to experience resistance, making it a contact force.

Part 04.3 (3 marks)

Work Done Calculation and Units

Applying W = F × s

📐 Step-by-Step Calculation

  1. Identify values from the question:
    Force, F = 30 N
    Distance, s = 250 m
  2. Substitute into the equation:
    work done = 30 × 250
  3. Calculate answer:
    work done = 7500
  4. Select the correct unit:
    Work done is an energy transfer, measured in Joules ( J ).

✅ Mark Scheme Breakdown

  • [1 mark]: Correct substitution: 30 × 250
  • [1 mark]: Calculation value: 7500
  • [1 mark]: Unit: J

❌ Common Errors

  • Selecting kg (unit of mass) or m (unit of distance) instead of J .
  • Dividing force by distance instead of multiplying. The question provides the formula, so follow it directly: force × distance .
Part 04.4 (3 marks)

Energy Stores and Transfers

Completing sentences using the word box

✅ Correct Sentence Completions

  • "As the bicycle moves, work is done against frictional forces." [1 mark]
  • "There is no change in the cyclist’s kinetic store of energy." [1 mark]
  • "There is a decrease in the cyclist’s chemical store of energy." [1 mark]

💡 Why These Answers Are Correct

  • Frictional forces: The tyres and moving components push against ground friction and air drag.
  • Kinetic store: The prompt explicitly states the bicycle travels at a constant speed. Since kinetic energy depends on speed ( Ek = 0.5 × m × v² ), constant speed means no change in kinetic energy!
  • Chemical store: The cyclist's muscles burn glucose from food to produce pedalling force, steadily depleting their chemical energy store.

🧠 Exam Technique

Always re-read the context sentence: "The bicycle travels at a constant speed." This is your clue that kinetic energy cannot be changing, immediately solving the second blank.

Topics

Physics · P1: Energy · P5: Forces

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 2 (Foundation), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.