AQA GCSE Combined Science: Trilogy Physics Paper 2 (Foundation), June 2025: Question 1
8 marks · Low Demand difficulty · Short Answer
Identify an anomalous result, identify the direction and nature of resistive forces, and calculate change in velocity and resultant force for a diving person.
Practise this questionQuestion
Question text
01 Figure 1 shows a person about to dive from a high platform into a pool of water.
Figure 1
01.1 The person dived from the platform into the water.
The time taken between the person leaving the platform and entering the water was
measured by four people.
The measurements were:
2.1 s 1.9 s 1.3 s 2.0 s
Which measurement is anomalous?
[1 mark]
Tick ( ) one box.
2.1 s
1.9 s
1.3 s
2.0 s 3
01.2 Air resistance acted on the person during the dive.
Figure 2 shows the person moving vertically downwards.
Figure 2
Which arrow on Figure 2 shows the direction of the air resistance acting on
the person?
[1 mark]
Tick ( ) one box.
A
B
C
D 4
01.3 Air resistance is a contact force.
Which force is also a contact force?
[1 mark]
Tick ( ) one box.
Friction
Gravitational force
Magnetic force
01.4 What happened to the size of the air resistance force as the speed of the
person increased?
[1 mark]
01.5 After entering the water, the person decelerated for a time of 0.35 s.
The deceleration of the person was 60 m/s2.
Calculate the change in velocity of the person.
Use the equation:
change in velocity = deceleration × time
[2 marks]
Change in velocity =5 m/s
01.6 The mass of the person was 75 kg.
Calculate the resultant force on the person when the deceleration was 60 m/s2.
Use the equation:
resultant force = mass × deceleration
[2 marks]
Resultant force = N
Mark scheme
Show the mark scheme
Question 1
AO /
Question Answers Extra information Mark
Spec. Ref.
01.1 1.3 s 1 AO3
6.5.4.1.2
AO /
Spec. Ref.
01.2 A 1 AO2
6.5.4.1.5
AO /
Spec. Ref.
01.3 friction 1 AO1
6.5.1.2
AO /
Spec. Ref.
01.4 (air resistance) increased 1 AO1
6.5.1.2
6.5.4.1.5
AO /
Spec. Ref.
01.5 ∆v = 60 × 0.35 1 AO2
6.5.4.1.5
∆v = 21 (m/s) allow − 21 (m/s) 1
AO /
Spec. Ref.
01.6 F = 75 × 60 1 AO2
6.5.4.2.2
F = 4500 (N) allow − 4500 (N) 1
Total Question 1 8
How to answer it
Forces, Falling Objects & Deceleration Analysis
What this question tests
This question evaluates foundational understanding of forces and 1D motion under gravity and decelerating forces:
- Data analysis: Spotting an outlier/anomalous reading from a small set of experimental repeats.
- Forces in motion: Identifying the direction of resistive forces opposing downward motion.
- Classifying forces: Distinguishing between contact forces (friction) and non-contact forces (gravity, magnetism).
- Fluid resistance factors: Knowing how drag / air resistance changes with speed.
- Quantitative physics: Direct substitution into velocity change and Newton's Second Law equations with correct standard units.
Spotting an Anomalous Time Measurement
Measurements: 2.1 s, 1.9 s, 1.3 s, 2.0 s
✅ Correct Answer
1.3 s
🧠 Exam Technique
An anomaly is a result that does not fit the pattern of repeated readings. Here, three values cluster tightly between 1.9 s and 2.1 s. 1.3 s is far lower than the rest, indicating timing error or premature stopping.
Direction of Air Resistance on a Falling Object
Figure 2: Diver falling vertically downwards
✅ Correct Answer
Tick box A (arrow pointing vertically upwards).
💡 Key Knowledge
- Air resistance is a frictional drag force that always opposes the direction of motion.
- Because the diver is travelling downwards (towards C), air resistance must act vertically upwards (arrow A).
❌ Common Errors
Selecting C: Confusing the resistive force with the direction of motion or the pull of gravity (weight).
Classifying Types of Forces
Which force is also a contact force?
✅ Correct Answer
Tick box Friction.
💡 Key Knowledge
- Contact forces require physical touching: Friction, air resistance, tension, normal reaction force.
- Non-contact forces act across a field without contact: Gravitational force, magnetic force, electrostatic force.
Relationship Between Speed and Air Resistance
Effect on drag as speed increases
✅ Correct Answer
(Air resistance) increased
💡 Key Knowledge
As an object falls faster, it collides with more air particles per second and with greater impact. Therefore, drag / air resistance increases as velocity increases.
🧠 Exam Technique
Keep your answer concise. Single comparative terms like "increased" or "it got bigger" are fully credited by the mark scheme.
Calculating Change in Velocity
Given: Deceleration = 60 m/s², Time = 0.35 s
📐 Step-by-Step Calculation
- Identify equation:
change in velocity = deceleration × time - Substitute values (1 mark):
Δv = 60 × 0.35 - Calculate final answer (1 mark):
Δv = 21 m/s
❌ Common Errors & Examiner Notes
- Dividing instead of multiplying: Mistakenly calculating 60 ÷ 0.35. Always check the formula carefully when provided in the prompt.
- Signs: Deceleration implies a loss of velocity, so -21 m/s is also fully accepted.
Calculating Resultant Force
Given: Mass = 75 kg, Deceleration = 60 m/s²
📐 Step-by-Step Calculation
- Identify equation (Newton's 2nd Law):
resultant force = mass × deceleration - Substitute values (1 mark):
F = 75 × 60 - Calculate final answer (1 mark):
F = 4500 N
🧠 Exam Technique
- Units are already provided ( N ), so you only need to write the number.
- Show your working clearly: Writing 75 × 60 guarantees 1 method mark even if an arithmetic slip occurs on your calculator.
- Allowable answer: -4500 N is also accepted as a decelerating/opposing force.
Topics
Physics · P5: Forces
Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 2 (Foundation), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.