AQA GCSE Physics Physics Paper 2 (Higher), November 2020: Question 6
24 marks · High Demand difficulty · Short Answer
Calculate the wavelength of radio waves, determine speed from a distance-time graph, and calculate resultant force and explain maximum speed for a remote-controlled car.
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Question text
06 Figure 8 shows a student playing with a remote-controlled car.
Figure 8
06.1 The remote control transmits radio waves to the car aerial.
The transmitted radio waves have a frequency of 320 MHz.
speed of radio waves = 3.0 × 108 m/s
Calculate the wavelength of the radio waves.
Give the unit.
[5 marks]
Wavelength = Unit
06.2 The car aerial is connected to an electrical circuit in the car.
Describe what happens in the electrical circuit when the car aerial absorbs radio
waves.
[2 marks]
06.3 The car produces sound waves.
Give two ways in which radio waves are different to sound waves.
[2 marks]
Figure 9 shows the distance-time graph for the first 30 seconds of the car’s motion.
Figure 9
06.4 Describe the motion of the car during the first 30 seconds.
[1 mark]
06.5 Determine the speed of the car 20 seconds after it started to move.
[4 marks]
Speed = m/s
06.6 A different car accelerated from 0.12 m/s to 0.52 m/s.
The acceleration of the car was 0.040 m/s2.
The work done to accelerate the car was 0.48 J.
*22* Calculate the resultant force needed to accelerate the car.
[6 marks]
Resultant force = N
06.7 Explain why the car has a maximum speed.
[4 marks]
Mark scheme
Show the mark scheme
Question 6
AO /
Question Answers Extra information Mark
Spec. Ref.
06.1 320 MHz = 3.2 × 108 Hz allow 320 000 000 1 AO2
3.0 × 108 = 3.2 × 108 × λ this mark may be awarded if 1 AO2
frequency is incorrectly/not
converted
3.0 × 108 this mark may be awarded if 1 AO2
λ = 8 frequency is incorrectly/not
3.2 × 10
converted
wavelength = 0.9375 allow correct calculation using 1 AO2
an incorrectly/not converted
frequency
allow an answer that rounds
to 0.94
metres or m 1 AO1
4.6.1.2
06.2 (alternating) current induced (in allow electrons vibrate / oscillate 1 AO1
the electrical circuit) (in the electrical circuit) 4.6.2.3
with the same frequency as the 1
radio wave
06.3 Any two from: 2 AO1
• (radio waves are) transverse allow sound waves are 4.6.1.1
longitudinal 4.6.1.2
allow a description of
transverse/longitudinal waves
• (radio waves) travel at a
higher speed
• (radio waves) don’t need a allow (only) radio waves travel
medium through a vacuum
• (radio waves are) allow sound waves are
electromagnetic mechanical
06.4 accelerating allow speeding up 1 AO3
4.5.6.1.4
06.5 appropriate tangent drawn 1 AO2
4.5.6.1.4
correct reading from graph for allow correct reading from their 1
change in distance and change tangent for change in distance
in time (eg 5.6 (m) and 20 (s)) and change in time
gradient of tangent shown allow correct gradient from their 1
(eg 5.6/20) tangent
0.28 (m/s) this answer only 1
allow 0.25 to 0.30 (m/s) if the
tangent is appropriate
allow 2.8 / 20 = 0.14 (m/s) for 1
mark
06.6 0.522 – 0.122 = 2 x 0.04 x s 1 AO2
4.5.2
0.522 − 0.122 1 4.5.6.1.5
s =
2 × 0.04
s = 3.2 (m) 1
0.48 = F x 3.2 this mark may be awarded if the 1
displacement is incorrectly
calculated
0.48 this mark may be awarded if the 1
F =
3.2 displacement is incorrectly
calculated
F = 0.15 (N) allow a correctly calculated F 1
using an incorrectly
calculated displacement
OR
Alternative method 1
0.52 – 0.12
t = (1)
0.04
t = 10 (s) (1)
s = 0.32 × 10 allow a correctly calculated
= 3.2 (m) (1) displacement from an incorrectly
calculated t
0.48 = F x 3.2 (1) this mark may be awarded if the
displacement is incorrectly
calculated
0.48 this mark may be awarded if the
F = (1)
3.2 displacement is incorrectly
calculated
F = 0.15 (N) (1)
allow a correctly calculated F
from incorrectly calculated
values for displacement
and / or t
OR
Alternative method 2
0.48 = (0.5 x m x 0.522) –
(0.5 x m x 0.122) (1)
0.48 = 0.1352m – 0.0072m (1)
0.48 = 0.128m (1)
m = 3.75 (1)
F = 3.75 x 0.040 (1)
allow their calculated m
F = 0.15 (N) (1)
allow correctly calculated F
using an incorrectly calculated
m
06.7 there is a maximum forward allow driving force for forward 1 AO1
force (provided by the motor) force - throughout 4.5.6.1.5
the car has a maximum
acceleration is insufficient
as the speed of the car allow friction / drag for air 1
increases air resistance resistance - throughout
increases
until air resistance is equal in allow (until) the resultant force is 1
size to forward force zero
allow forces are in equilibrium /
balanced
so the car can no longer allow the car travels at terminal 1
accelerate velocity
Total 24
How to answer it
AQA GCSE Physics Study Guide: Forces, Motion & Waves
This multi-topic question tests your ability to apply wave calculations (wave equation, unit conversions), properties of electromagnetic vs. sound waves, interpreting distance-time graphs (calculating speed via tangents), equations of motion combined with Newton's Second Law ( F = ma and work done), and explaining terminal velocity and resistive forces.
Calculating Wave Wavelength
✅ Correct Answer
Wavelength = 0.9375 m (or 0.94 m ), Unit = m or metres .
💡 Key Knowledge
- Wave equation: v = f × λ (Wave speed = frequency × wavelength).
- Rearranged: λ = v / f .
- Standard prefix: Mega (M) means ×10⁶. So 320 MHz = 320 × 10⁶ Hz or 3.2 × 10⁸ Hz .
📐 Step-by-Step Calculation
- Convert frequency: 320 MHz = 320,000,000 Hz .
- State formula: λ = v / f .
- Substitute values: λ = (3.0 × 10⁸) / (3.2 × 10⁸) .
- Calculate result: 0.9375 m .
❌ Common Errors
- Forgetting to convert MHz into Hz (losing standard form marks).
- Omitting or giving the wrong unit (must state metres or m).
Electrical Circuit Response to Radio Waves
✅ Correct Answer
An (alternating) current is induced in the electrical circuit with the same frequency as the radio wave.
💡 Key Knowledge
- Electromagnetic waves are transverse waves made of oscillating electric and magnetic fields.
- When radio waves hit a metal aerial, the oscillating electric field forces electrons in the metal to vibrate, inducing an alternating current.
🧠 Exam Technique
Use precise terminology: examiners specifically look for the terms induced current and matching frequency.
Comparing Radio Waves and Sound Waves
✅ Correct Answer (Any two)
- Radio waves are transverse, whereas sound waves are longitudinal.
- Radio waves travel at a much higher speed (speed of light vs. speed of sound in air).
- Radio waves do not need a medium (can travel through a vacuum), whereas sound waves require a medium.
- Radio waves are electromagnetic waves, whereas sound waves are mechanical.
❌ Common Errors
- Vague comparisons like "radio waves are faster" without stating properties, or mixing up transverse and longitudinal definitions.
Interpreting Motion from a Distance-Time Graph
✅ Correct Answer
The car is accelerating (or speeding up).
💡 Key Knowledge
- On a distance-time graph, a straight diagonal line represents constant speed.
- A curve getting steeper indicates that distance is increasing at a greater rate per second, meaning speed is increasing (acceleration).
Determining Speed from a Distance-Time Graph at t = 20 s
✅ Correct Answer
Speed = 0.28 m/s (Acceptable range: 0.25 to 0.30 m/s ).
🧠 Exam Technique: Tangent Method
- Draw a precise, straight tangent line touching the curve exactly at t = 20 s .
- Construct a large triangle along your tangent line to maximise accuracy.
- Read off the change in distance ( Δd ) and change in time ( Δt ).
- Calculate gradient: Speed = Δd / Δt .
❌ Common Errors
- ">
- Calculating average speed by reading a single coordinate point ( distance / time ) instead of finding the gradient of the tangent.
- Drawing too small of a tangent triangle, leading to inaccurate readings.
Calculating Resultant Force using Kinematics and Work Done
✅ Correct Answer
Resultant force = 0.15 N
💡 Key Knowledge
- SUVAT equation: v² - u² = 2as can be rearranged to find distance/displacement ( s ).
- Work done formula: Work Done (W) = Force (F) × distance (s) .
- Newton's Second Law: F = ma (Alternative method).
📐 Step-by-Step Calculation (Method 1)
- Use v² - u² = 2as : 0.52² - 0.12² = 2 × 0.04 × s .
- Simplify: 0.2704 - 0.0144 = 0.256 = 0.08s .
- Calculate distance s = 3.2 m .
- Use Work Done: W = F × s → 0.48 = F × 3.2 .
- Rearrange for force: F = 0.48 / 3.2 = 0.15 N .
❌ Common Errors
- Forgetting to square the velocity values in the SUVAT equation.
- Mixing up initial ( u ) and final ( v ) velocities.
Explaining Maximum Speed (Terminal Velocity)
✅ Correct Answer
- There is a maximum forward force provided by the motor.
- As the speed of the car increases, air resistance (drag) increases.
- Eventually, air resistance increases until it is equal in size to the forward driving force (forces are balanced / resultant force is zero).
- Therefore, the car can no longer accelerate and travels at a steady maximum speed.
🧠 Exam Technique: 4-Mark Command Structure
To secure full marks in terminal velocity questions, always follow this 4-step causal chain:
- State the initial driving force.
- Link rising speed to increasing friction/air resistance.
- State when resistive forces balance driving forces (resultant force = 0).
- Conclude that acceleration stops and maximum speed is reached.
❌ Common Errors
Saying that air resistance "stops the car" or that "speed becomes zero", rather than explaining that acceleration stops when forces balance.
Topics
Physics · P5: Forces · P6: Waves
Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Higher), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.