AQA AS Level Physics Paper 1, June 2023: Question 5

17 marks · Medium difficulty · Extended Answer

Calculate various physical quantities related to a robotic helicopter on Mars, including mass flow rate, mass of the helicopter, battery energy usage, flight time extension, vertical motion parameters, and analyze forces using Newton's laws.

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Question text

05 Figure 7 shows a robotic helicopter that is used on Mars. The helicopter is powered

by a battery. Before each flight, the battery is charged by a solar panel.

Figure 7

Figure 8 shows the helicopter hovering at a constant height above the surface of

Mars. The rotor blades move a column of atmospheric gas vertically downwards at a

velocity of 17.2 m s−1. The diameter of this column is 1.2 m.

Figure 8

05.1 The gas moved by the rotor blades has a density of 0.020 kg m−3.

Show that the helicopter moves approximately 0.4 kg of gas every second.

[3 marks]

The movement of the gas creates an upward force on the helicopter. This upward

force enables the helicopter to hover at a constant height.

The gravitational field strength on Mars is 3.72 N kg−1.

05.2 Calculate the mass of the helicopter.

[3 marks]

18 mass = kg

05.3 The battery stores 0.035 kW h of energy before a flight.

The flight lasts for 39 s.

The battery has a power output of 340 W during the flight.

Determine the percentage of the initial energy stored in the battery that is transferred

*17* during the flight.

[2 marks]

percentage = %

05.4 The helicopter has a maximum flight time of a few minutes due to the limited amount

of energy stored in the battery. The battery accounts for about 15% of the helicopter’s

mass.

A student suggests that adding another identical battery that doubles the energy

available to the helicopter would double its flight time.

Deduce without calculation whether the student’s suggestion is correct.

[3 marks]

Figure 9 shows a simplified side view of the helicopter moving vertically upwards with

a speed of 0.55 m s−1.

At the instant shown, the helicopter is at a height h and the blades stop rotating.

Figure 9

The gravitational field strength on Mars is 3.72 N kg−1.

The weight of the helicopter is the only force acting on it when the blades stop

rotating. Drag forces on the helicopter are negligible as it rises to a maximum height

and then falls back to the surface.

05.5 Calculate the time taken for the helicopter to reach its maximum height from the

instant the blades stop rotating.

[2 marks]

time = s

05.6 When the helicopter makes contact with the surface it has a velocity of 2.2 m s−1.

Calculate h.

[2 marks]

h = m

05.7 A student suggests that the acceleration of the helicopter is constant from the instant

the blades stop rotating until the helicopter makes contact with the surface.

Discuss this suggestion with reference to an appropriate Newton’s law of motion.

*19* [2 marks]

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidance Mark AO

05.1 2 3 -1 774𝜋𝜋 3 AO1.1a

πd (Volume per second =) 19.45 (m s ) =

(Use of volume (per sec) =) × 17.2 125

22 ×

πd 9𝜋𝜋 AO2.1b

× 17.2 = × 17.2

4 25

Use of ρ = Substitutes their volume (per second) and

𝑚𝑚

density where would be subject. Do not

𝑡𝑡

award MP2 if 2 errors are made in substitution.

–1

(mass per second =) 0.389 (kg s ) Answer seen to at least 2 sf.

Calculator display = 0.3890548342

05.2 m Possible ECF from 05.1 where their m rounds 3 AO1.1a

Use of F = × v or (F =) 6.69 N or 6.708 (N) or 6.88 (N)

to 0.4 kg.

OR 2 ×

AO2.1b

Use of W=mg

W = 3.72m seen or 3.72m as the subject of a

OR force equation.

statement:

Do not allow 3.72 x 0.4 as use of W=mg

Upward force = weight

Applies condition for equilibrium by setting F = mg

OR 19

6.69 = 3.72 m or 6.708 = 3.72 m or 6.88 = 3.72 m

Accept answer correctly rounded to at least 2 sf.

(m =) 1.80 (kg)

𝑚𝑚

F= 6.88 N where = 0.4

𝑡𝑡

m=1.85 kg or 1.8 kg

05.3 Use of E = Pt (E =) 340 × 39 or 13260 (J) 2 AO1.1a

OR AO2.1b

(0.035 kWh =) 35 × 3600 or 126000 (J)

converts kWh to J Alternative MP1 converts to any of the

following units of energy.

• 0.34 (kW) x 0.0108 (h) or 0.00368 (kWh)

• 0.035 kWh = 35 (Wh)

13 221

• 340 (𝑊𝑊) × (h) or (𝑊𝑊ℎ) 𝑜𝑜𝑜𝑜 3.683 (Wh)

1200 60

Or equivalent e.g W mins

Do not accept incorrect unit.

Do not accept incorrect subject.

MP2

Do not allow answers obtained using incorrect

126000

power ( )

340 340

such as 126000

𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑡𝑡 𝑝𝑝𝑖𝑖𝑝𝑝𝑖𝑖𝑖𝑖

(=) 11% 39

Accept answer correctly rounded to at least 2 sf.

Calculator display = 10.5238

05.4 Incorrect: 3 3 ×

• this will increase weight OR helicopter must provide a Do not accept increase in resistive forces or 21

AO3.1b

greater lift OR (more mass therefore) greater GPE (for increase in drag for increase in weight.

same height) OR (more mass therefore) greater KE (for

same speed) OR idea that more energy is required. Must state that it is incorrect for all 3

• the helicopter must displace more (atmospheric) gas (every marks.

second to produce greater lift force) OR blades must spin Maximum of 2 marks for suggestions that more

faster than doubles flight time.

• the helicopter must do more work every second (so will

Accept lift or thrust or upward force.

transfer stored energy at a greater rate) OR the helicopter

needs more power to fly

A maximum of 1 mark for MP3 and MP1

where only mark seen is : idea that more

energy is required.

Incorrect:

MP2 can be scored independent of this.

• this will increase weight

• atmosphere is too thin and can’t displace sufficient mass of

gas per second OR blades can’t spin fast enough

• can’t get off ground due to insufficient lift force

05.5 Use of an appropriate equation of motion: 2 AO1.1a

AO2.1b

v = u + at By correct substitution including signs or

correct rearrangement to make t subject.

(t =) 0.15 (s) Accept answer correctly rounded to at least 2 sf.

22 Calculator display = 0.14784946236559

05.6 MP1 (Downward journey) 2 AO1.1a

Use of v2 = u2 + 2as Allow s = 0.65m AO2.1b

2.22 = 02 + 2× 3.72 × s

12 1 2

Use of v = u + at and s = ut + at ECF 2.2 = 0 +3.72 t and s = 0 + 3.72 t ECF

11 m× 3.72× ∆h = m2.2

mg∆h = mv2 – mu2 2

MP1 (Upward journey)

Allow s=0.041 m obtained from

s = 0.55× 0.15 − 3.72× 0.15 ECF

OR 23

02 = 0.552 − 2× 3.72 × s

Check possible ECF for t from 05.5 used in

calculation.

Condone sign suppression in MP1 where answer of

0.65 m or 0.041 m or 0.6(1) m is seen.

(h =) 0.61 (m) ECF

Accept answer correctly rounded to at least 2 sf.

Calculator display = 0.60987903225806

05.7 Student is correct: 2 AO1.1a

MP1 statement that the object is in freefall.

Where (resultant) force is mentioned must be AO2.1a

identified as weight.

Weight is the only force acting on the helicopter.

OR Where acceleration is quoted must have correct

unit.

Acceleration = (-)3.72 ms-2

Accept F=ma as a statement of Newton’s 2nd law.

Due to Newton’s 2nd law , the acceleration acts in the same

direction as the weight (which is always downwards). MP2 Accept no mention of force being weight

where mass is included their answer, for e.g.:

Due to Newton’s 2nd law the acceleration is

OR constant because the force and mass are constant.

Neutral for statements that refer to deceleration /

Due to Newton’s 2nd law , the acceleration is constant acceleration.

because the (mass and) weight are constant

Do not accept arguments based on drag or air

resistance affecting the motion of the helicopter.

Zero marks for statement that indicates the

acceleration varies.

Must state that student is correct or that the

acceleration is constant to gain 2 marks.

Total 17

How to answer it

Mars Robotic Helicopter Physics Analysis

AQA AS Level Physics • Mechanics & Energy

What this question tests

This multi-part application question tests your mastery of mechanics, momentum, dynamics, and energy transfer. You will need to combine volume flow rates with density to find mass flow, apply Newton's second and third laws of motion, analyse power and energy efficiency, and use kinematic equations of motion under constant gravitational field strength.

Question 05.1 — Mass Flow Rate of Gas

Show that the helicopter moves approximately 0.4 kg of gas every second. [3 marks]

📐 Step-by-Step Calculation

Step 1: Find the cross-sectional area of the gas cylinder.
A = πd² / 4 = π(1.2)² / 4 = 1.131 m²
Step 2: Calculate volume of gas moved per second.
Volume/s = A × v = 1.131 × 17.2 = 19.45 m³ s⁻¹
Step 3: Calculate mass per second using density (ρ = m/V).
mass/s = density × volume/s = 0.020 × 19.45 = 0.389 kg s⁻¹ (≈ 0.4 kg s⁻¹)

❌ Common Errors & Examiner Pitfalls

Using the radius instead of diameter ( r = 0.6 m ) without squaring correctly, or confusing diameter and radius in the area formula.
Forgetting that a "show that" question requires intermediate values to be stated clearly to at least 2 significant figures before rounding.

Mark breakdown: 1 mark for volume per second equation/substitution, 1 mark for applying density, 1 mark for final evaluation yielding 0.389 kg s⁻¹ or better.

Question 05.2 — Mass of the Helicopter

Calculate the mass of the helicopter. [3 marks]

✅ Correct Answer

mass = 1.80 kg (or 1.8 kg)

🧠 Exam Technique & Physics Link

For a hovering helicopter, upward force equals downward weight (Newton's First Law / Equilibrium). Upward force equals rate of change of momentum: F = (m/t) × v .

📐 Calculation Steps

Upward force F = (0.389 kg s⁻¹) × (17.2 m s⁻¹) = 6.69 N (or using unrounded 0.4 gives ~6.88 N).
Equilibrium condition: Upward Force = Weight → F = mg
m = F / g = 6.69 / 3.72 = 1.80 kg (Accept 1.85 kg if using raw 0.4 kg s⁻¹).

Mark breakdown: 1 mark for force expression (F = (m/t)v or W=mg), 1 mark for equating forces (F = mg), 1 mark for correct numerical evaluation.

Question 05.3 — Energy Transfer Percentage

Determine the percentage of the initial energy stored in the battery that is transferred during the flight. [2 marks]

✅ Correct Answer

percentage = 11% (Accept 10.5% or 11% based on rounding)

📐 Calculation Steps

Step 1: Calculate energy transferred during flight (E = Pt).
E = 340 W × 39 s = 13260 J
Step 2: Convert initial battery storage from kW h to Joules.
0.035 kW h = 0.035 × 1000 W × 3600 s = 126,000 J
Step 3: Find percentage.
(13260 / 126000) × 100% = 10.52% → 11%

Mark breakdown: 1 mark for energy transferred or correct unit conversion of kW h to Joules, 1 mark for calculating the correct percentage.

Question 05.4 — Battery Addition Evaluation

Deduce without calculation whether the student's suggestion is correct. [3 marks]

✅ Correct Answer: The student is INCORRECT

Adding another identical battery will not double the flight time because doubling the battery increases the total mass of the helicopter.

💡 Key Knowledge & Examiner Guidance

Battery accounts for 15% of the total mass. Adding an identical battery increases total mass and therefore increases the helicopter's weight.
A heavier helicopter requires greater lift, meaning it must displace more gas per second or eject gas at higher velocity, consuming stored energy at a faster rate.
To gain all 3 marks, you must explicitly state that the student is incorrect and link increased weight to increased energy demands.

Mark breakdown: 1 mark for stating student is incorrect, 1 mark for noting the increase in weight/mass, 1 mark for explaining that greater weight requires more energy/lift per second, preventing doubled flight time.

Question 05.5 — Time to Maximum Height

Calculate the time taken for the helicopter to reach its maximum height from the instant the blades stop rotating. [2 marks]

✅ Correct Answer

time = 0.15 s

🧠 Exam Technique

At maximum height, final velocity v = 0 m s⁻¹ . Initial upward velocity u = 0.55 m s⁻¹ . Acceleration is due to gravity on Mars: a = -3.72 m s⁻² .

📐 Calculation Steps

Use kinematic formula: v = u + at
0 = 0.55 + (-3.72)t
t = 0.55 / 3.72 = 0.1478 s → 0.15 s

Mark breakdown: 1 mark for selecting/using correct kinematic equation with proper signs, 1 mark for correct time evaluation to at least 2 sf.

Question 05.6 — Calculating Height h

Calculate h. [2 marks]

✅ Correct Answer

h = 0.61 m (Accept 0.60 m to 0.65 m depending on path taken)

💡 Method Options

You can solve this using either suvat equations for the total journey, or by calculating upward displacement and downward displacement separately and combining them, or via energy conservation ( ΔE_k = ΔE_p ).

📐 Calculation Steps (Using v² = u² + 2as for whole trip)

Final velocity just before hitting surface: v = 2.2 m s⁻¹
Initial velocity when blades stop: u = 0.55 m s⁻¹ (upwards, treat downwards as positive or use vector signs carefully)
Using v² = u² + 2as (taking downwards as positive):
(2.2)² = (-0.55)² + 2(3.72)s
4.84 = 0.3025 + 7.44s → 4.5375 = 7.44s
s = 0.61 m

Mark breakdown: 1 mark for correct equation and proper substitution including signs, 1 mark for correct displacement h. ECF allowed from 05.5.

Question 05.7 — Acceleration and Newton's Laws

Discuss this suggestion with reference to an appropriate Newton's law of motion. [2 marks]

✅ Correct Answer: The student is CORRECT

The acceleration of the helicopter is constant once the blades stop rotating.

💡 Scientific Justification

Freefall condition: With rotor blades stopped and drag forces considered negligible, weight is the only force acting on the helicopter.
Newton's Second Law ( F = ma ): Since the resultant force equals weight (mg) and mass remains constant, the acceleration is constant and equal to the gravitational field strength ( g = 3.72 m s⁻² downwards).

Mark breakdown: 1 mark for stating student is correct and identifying weight as the only force (or stating acceleration is constant at 3.72 m s⁻²), 1 mark for citing Newton's 2nd law linking constant force and constant mass to constant acceleration.

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.