AQA GCSE Physics Physics Paper 1 (Higher), November 2021: Question 8
8 marks · Standard Demand difficulty · Short Answer
Explain closed systems, calculate the maximum speed of a toy car descending a track using conservation of energy, and determine the kinetic energy needed to complete a vertical loop.
Practise this questionQuestion
Question text
08 Figure 11 shows a toy car in different positions on a racing track.
Figure 11
08.1 The toy car and racing track can be modelled as a closed system.
Why can the toy car and racing track be considered ‘a closed system’?
[1 mark]
Tick ( ) one box.
The racing track and the car both have gravitational potential energy.
The racing track and the car are always in contact with each other.
The total energy of the racing track and the car is constant.29
08.2 The car is released from rest at position A and accelerates due to gravity down the
track to position B.
mass of toy car = 0.040 kg
vertical height between position A and position B = 90 cm
gravitational field strength = 9.8 N/kg
Calculate the maximum possible speed of the toy car when it reaches position B.
[5 marks]
Speed = m/s
Figure 11 is repeated below.
Figure 11
*0298.*3 At position C the car’s gravitational potential energy is 0.20 J greater than at
position B.
How much kinetic energy does the car need at position B to complete the loop of
the track?
Give a reason for your answer.
[2 marks]
Tick ( ) one box.
Less than 0.20 J
Exactly 0.20 J
More than 0.20 J
Reason
Mark scheme
Show the mark scheme
Question 8
AO /
Question Answers Extra information Mark
Spec. Ref.
08.1 the total energy of the racing 1 AO1
track and the car is constant. 4.1.2.1
08.2 Ep = 0.040 × 9.8 × 0.90 allow a correct substitution of an 1 AO2
incorrectly/not converted value 4.1.1.1
of h 4.1.1.2
Ep = 0.3528 (J) this answer only 1
0.3528 = 0.5 × 0.040 × v2 allow a correct substitution of a 1
calculated Ep
2 0.3528 allow a correct rearrangement 1
v = using a calculated E
0.5 × 0.040 p
allow an answer consistent with 1
v = 4.2 (m/s)
their calculated Ep
more than 0.20 J 1
08.3 AO3
4.1.1.1
(because) the car needs to be this mark is dependent on 1
moving at the top of the loop scoring the first mark
or
(because) the car needs to be
moving to complete the loop
or
not all Ek at B will be transferred
to Ep at C
allow energy dissipated to the
surroundings
Total 8
How to answer it
Energy Conservation & Toy Car on a Track
- Closed systems: Definition of a closed system in terms of total energy conservation.
- Gravitational Potential Energy (Ep): Calculating change in potential energy using Ep = m × g × h and converting distance units correctly ( cm → m ).
- Kinetic Energy (Ek): Calculating speed using Ek = 0.5 × m × v² by equating lost potential energy to gained kinetic energy.
- Energy transfers & Real-world mechanics: Explaining why extra energy is needed to clear a vertical loop.
Closed System Definition
Why can the toy car and track be considered a 'closed system'?
✅ Correct Answer
"The total energy of the racing track and the car is constant."
Tick the 3rd box.
💡 Key Knowledge
A closed system is an isolated system where:
- No energy is transferred into or out of the system.
- The total energy remains constant (Conservation of Energy).
- Energy can be transferred between different stores within the system, but the sum never changes.
❌ Common Errors
- Choosing "The racing track and the car are always in contact" — physical contact does not define an energetic closed system.
- Confusing a closed system with a stationary or frictionless system.
🧠 Exam Technique
Whenever GCSE questions mention a closed system, always look immediately for statements stating that total energy is constant or that no net energy transfer takes place across the boundary.
Calculating Maximum Speed at the Bottom of the Track
Calculate the maximum possible speed of the toy car at position B when released from rest at position A.
📐 Step-by-Step Calculation
Mass, m = 0.040 kg (standard SI unit ✅)
Gravitational field strength, g = 9.8 N/kg ✅
Height, h = 90 cm = 90 / 100 = 0.90 m (MUST convert cm to m!)
Ep = m × g × h
Ep = 0.040 × 9.8 × 0.90
Ep = 0.3528 J
For maximum speed, assume 100% efficient transfer (no air resistance or friction):
Ek = 0.5 × m × v²
0.3528 = 0.5 × 0.040 × v²
0.3528 = 0.020 × v²
v² = 0.3528 / (0.5 × 0.040) = 0.3528 / 0.020 = 17.64
v = √17.64 = 4.2 m/s
✅ Final Answer
Speed = 4.2 m/s
Full 5 marks awarded for this correct final value even if some intermediate steps are omitted, but setting working out clearly ensures method marks.
❌ Common Errors & Traps
- Unit trap: Using 90 instead of 0.90 gives v = 42 m/s (loses conversion mark).
- Forgetting to square root: Leaving the answer as 17.64 instead of calculating √17.64 .
- Formula confusion: Forgetting the factor of 0.5 in kinetic energy.
🧠 Examiner Insight
This is a standard 5-mark synoptic calculation testing two core equations together. Notice how the mark scheme awards each discrete step:
- Correct substitution into Ep = mgh (with or without cm conversion) [1 mark]
- Correct calculated value of Ep = 0.3528 J [1 mark]
- Equating energy value to 0.5 × m × v² [1 mark]
- Correct rearrangement for v² [1 mark]
- Correct square root evaluation yielding 4.2 m/s [1 mark]
Energy Required to Complete the Loop
At position C, the car's Ep is 0.20 J greater than at B. How much kinetic energy does it need at B to complete the loop?
✅ Correct Answer
Box to tick: More than 0.20 J
Reason (any one of):
- The car must still be moving at the top of the loop (position C).
- Not all kinetic energy at B is converted into gravitational potential energy at C (some must remain as kinetic energy).
- Some energy is dissipated to the surroundings (due to friction/air resistance).
💡 Physical Explanation
If the car had exactly 0.20 J of kinetic energy at B, all of it would convert into potential energy at C, leaving Ek = 0 at the peak.
With zero speed at the highest point of the loop, the car would simply fall off the track rather than completing the circular path!
🧠 Exam Technique
The second mark is dependent on scoring the first mark. If you choose "Exactly 0.20 J", you score 0 marks even if your reasoning mentions friction. Always ensure your reasoning directly explains why more energy is required.
❌ Common Errors
- Ticking "Exactly 0.20 J" by assuming ideal conservation without considering that stationary objects at the top fall vertically.
- Writing vague reasons like "because it goes fast" without referring to motion at the top or energy dissipation.
Topics
Physics · P1: Energy
Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Higher), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.