AQA AS Level Physics Paper 1, November 2020: Question 2

13 marks · Medium difficulty · Short Answer

Calculate the kinetic energy, explain deceleration using Newton's laws, find the acceleration due to gravity, and deduce the effect of increased dust density on deceleration for a spacecraft entering Mars' atmosphere.

Practise this question

Question

A 4-part physics exam question about a spacecraft entering the atmosphere of Mars. Part 02.1 asks to calculate the kinetic energy of a 610 kg spacecraft traveling at 5.5 km s^-1. Part 02.2 asks to explain using Newton's laws why displacing atmospheric gas with a parachute causes a force and deceleration. Part 02.3 asks to calculate the acceleration due to gravity given mass, vertical distance, kinetic energy lost, and energy transferred to the atmosphere. Part 02.4 asks to deduce how increased dust content affecting atmospheric density will impact the deceleration.
Question text

02 A spacecraft entering the atmosphere of Mars must decelerate to land undamaged on

the surface.

Figure 1

02.1 Figure 1 shows the spacecraft of total mass 610 kg entering the atmosphere at a

speed of 5.5 km s−1.

Calculate the kinetic energy of the spacecraft as it enters the atmosphere.

Give your answer to an appropriate number of significant figures.

[3 marks]

7 kinetic energy = J

02.2 A parachute opens during the spacecraft’s descent through the atmosphere.

Figure 2 shows the parachute–spacecraft system, with the open parachute displacing

the atmospheric gas. This causes the system to decelerate.

*06* Figure 2

Explain, with reference to Newton’s laws of motion, why displacing the atmospheric

gas causes a force on the system and why this force causes the system to

decelerate.

[4 marks]

02.3 As the parachute–spacecraft system decelerates, it falls through a vertical distance

of 49 m and loses 2.2 × 105 J of kinetic energy.

During this time, 3.3 × 105 J of energy is transferred from the system to the

atmosphere.

The total mass of the system is 610 kg.

Calculate the acceleration due to gravity as it falls through this distance.

[3 marks]

acceleration due to gravity = m s−2

02.4 Dust from the surface of Mars can enter the atmosphere. This increases the density

of the atmosphere significantly.

Deduce how an increase in dust content will affect the deceleration of the system.

[3 marks]

Mark scheme

Show the mark scheme The mark scheme details answers and guidance for four parts (02.1 to 02.4). Part 02.1 awards 3 marks for using Ek = 1/2 mv^2, getting 9.2 x 10^9 J, and correct significant figures. Part 02.2 awards 4 marks for referencing momentum change, Newton's 2nd and 3rd laws, and resultant resistive forces. Part 02.3 awards 3 marks for determining energy change, using Ep = mgh, and calculating g = 3.7 m s^-2. Part 02.4 awards 3 marks for discussing increased mass/particles to displace, greater rate of momentum change, and greater resultant force leading to increased deceleration.

Question Answers Additional Comments/Guidance Mark

details

Condone POT error on 1st MP

Allow use where v where has been converted

Use of E = ½ mv2 from 5.5 km h-1

k

(Kinetic energy =) 9.2 × 109 (J)

An answer to 2 significant figures (with some working)

Significant figure mark requires evidence of AO1.1a

some relevant working.

02.1 3 AO2.1b

AO1.1b

7407_1 Version post-stand

Max 3 for why there is a force on the gas and

Why force on the gas: why there is a resistive force on the system

The gas’s momentum is changing

This require a force according to Newton’s 2nd law Must have why the system decelerates to obtain

8 all 4 marks.

Or

The reason why the resultant force causes the

The gas is being accelerated deceleration rather than the acceleration.

This require a force according to Newton’s 2nd law

Why (resistive) force on system: Allow statement that is equivalent to N1 / N2 / N3.

The gas exerts a force on the parachute (with an equal magnitude

Allow: air resistance (or drag) increases.

and opposite direction force) / there is air resistance (on the system)

Allow: there is an upward force

/ there is drag (on the system) / there is a resistive force (on the

system) AO2.1a

must have a clear action-reaction pair for this N3

(because) the Parachute exerts a force on the gas according to mark. AO2.1e

Newton’s 3rd law

02.2 4 AO2.1a

Why system decelerates: AO2.1e

The resistive force is greater than the weight so there is a resultant force

Or

The resultant force is acting in the opposite direction (to its motion). allow the resultant force is vertically upwards

acceleration in same direction as resultant force according to Or

Newton’s 2nd law

Links to violation for conditions of Newton’s 1st law

and therefore cannot continue at constant velocity.

– SICS – – JUNE 2020

Question Answers Additional Comments/Guidance Mark ID

details

1st mark: Credit an application of conservation

Attempt at determining difference = 3.3 (× 105)– 2.2 (× 105 )

of energy (allow written statement, or equation

or difference = 1.1 (x 105)

without substitution)

Ignore signs on difference and answer.

MP2 allow their energy in a substitution that is,

otherwise correct.

Condone an answer = 18.4 (m s-2) is worth 2

Use of Ep = mgh marks.

Condone mgh = ½ mv2 where rearranged to

make g subject. AO2.1f

Condone 610 x g x 49 = their energy

Alternative:

02.3 3 AO1.1a

• Attempt to use appropriate equations of

motion to determine acceleration

v2=u2 +2as rearranged to make a the AO2.1b

subject (condone use of their values for

v and u and / or g = a)

• Attempt to use W=Fs to determine the

air resistance FD (or FD= 6734(.7) (N)

(g = ) 3.7 (m s–2)

seen)

• Attempt to determine g from the

deceleration of the system

𝐹𝐷 − 𝑚𝑎

7407_1 Version post-stand 𝑔 =

𝑚

10 Must have some interaction with parachute-

spacecraft.

N/E to say there are more particles / gas / dust

/mass

More mass to displace / more particles to collide with / more

gas / dust to displace

(at any given speed)

Greater (rate of) change of momentum / More work done (per

unit distance) / Greater (resistive) force / more kinetic energy AO3.1b

02.4 transferred (per unit distance) 3 AO3.1b

AO3.1b

3rd MP for greater resultant force: allow the

idea that the difference between the drag and

weight has increased

Greater resultant force on the system (therefore greater

3rd MP

deceleration) / greater loss of velocity per second (therefore

greater deceleration) Allow clear statement that links:

• rate of change of momentum of gas /

dust to rate of change of momentum of

system

• rate of work done on gas / dust to rate

of work done by system

– SICS – – JUNE 2020

Total 13

How to answer it

Spacecraft Deceleration in the Martian Atmosphere

AQA AS Level Physics • Mechanics & Energy

What this question tests

This multi-part mechanics question evaluates your ability to apply core physical principles to a real-world space scenario. Key competencies assessed include calculating kinetic energy using standard equations, linking Newton's laws of motion (1st, 2nd, and 3rd) to fluid displacement and drag forces, applying the conservation of energy principle to gravitational potential energy, kinetic energy, and thermal transfer, and deducing proportional changes in atmospheric dynamics.

Part (0.2.1): Calculating Kinetic Energy

Calculate the kinetic energy of the spacecraft entering the atmosphere (3 marks)

✅ Correct Answer

Kinetic energy = 9.2 × 10⁹ J (to 2 significant figures)

Mark breakdown: 1 mark for correct equation (Ek = ½mv²), 1 mark for correct substitution/calculation (9.225 × 10⁹ J), 1 mark for giving the final answer to exactly 2 significant figures with working shown.

📐 Step-by-Step Calculation

  1. Identify given values: Mass m = 610 kg, Speed v = 5.5 km s⁻¹ = 5500 m s⁻¹.
  2. Recall the formula: Ek = ½mv²
  3. Substitute values: Ek = 0.5 × 610 × (5500)² = 9 227 500 000 J = 9.2275 × 10⁹ J.
  4. Apply significant figures: Since speed is given to 2 sig figs, round to 9.2 × 10⁹ J.

❌ Common Errors & Traps

  • Unit conversion failure: Forgetting to convert km s⁻¹ into m s⁻¹ (using 5.5 instead of 5500).
  • Significant figure penalties: Giving an unrounded calculator value (e.g., 9.2275 × 10⁹ J) loses the final mark, as the question explicitly asks for an appropriate number of significant figures supported by working.

Part (0.2.2): Newton's Laws and Parachute Deceleration

Explain why displacing atmospheric gas causes a force and why this causes deceleration (4 marks)

💡 Key Knowledge & Physics Principles

  • Newton's 2nd Law: Force is equal to the rate of change of momentum ( F = Δp / Δt ). Displacing gas changes its momentum.
  • Newton's 3rd Law: When the parachute exerts a forward/outward force to push the gas away, the gas exerts an equal and opposite reaction force (air resistance/drag) back on the parachute system.
  • Resultant Force & Deceleration: For the system to slow down, the upward/resistive drag force must be greater than the downward weight of the system, creating a resultant force opposing motion, which causes deceleration via Newton's 2nd law.

🧠 Exam Technique & Marking Strategy

This is a 4-mark structured explanation. To secure full marks, you must explicitly separate your logic: (1) Why there is a force on the gas (momentum change), (2) Why there is a reaction force on the system (Newton's 3rd law), and (3) Why the system decelerates (resultant force greater than weight / opposing motion).

Part (0.2.3): Gravitational Field Strength Calculation

Calculate the acceleration due to gravity (g) as the system falls through 49 m (3 marks)

✅ Correct Answer

g = 3.7 m s⁻²

Mark breakdown: 1 mark for calculating energy change (1.1 × 10⁵ J), 1 mark for using Ep = mgh (or equivalent energy balance), 1 mark for the correct final value of g.

📐 Step-by-Step Calculation

  1. Step 1: Determine net energy change.
    Initial kinetic energy lost = 2.2 × 10⁵ J.
    Energy transferred to atmosphere (thermal/work done) = 3.3 × 10⁵ J.
    Net change in mechanical energy (ΔEp) = 3.3 × 10⁵ - 2.2 × 10⁵ = 1.1 × 10⁵ J.
  2. Step 2: Relate to gravitational potential energy.
    Ep = mgh, therefore 1.1 × 10⁵ = 610 × g × 49.
  3. Step 3: Rearranging for g.
    g = (1.1 × 10⁵) / (610 × 49) = 110 000 / 29 890 = 3.68 m s⁻² ≈ 3.7 m s⁻².

❌ Common Errors & Traps

  • Energy confusion: Adding the two energy values together instead of finding their difference. Read carefully: energy is lost from kinetic, but some is also transferred to the atmosphere.
  • Expecting Earth's gravity: Students sometimes panic when they calculate an unusual value like 3.7 m s⁻² and overwrite it with 9.81 m s⁻². Trust your math—this is Mars!

Part (0.2.4): Impact of Increased Atmospheric Density

Deduce how an increase in dust content will affect the deceleration of the system (3 marks)

💡 Key Knowledge & Examiner Guidance

  • More mass/particles: Higher dust content means more particles / mass of gas to displace per unit time/distance at any given speed.
  • Greater rate of momentum change: Because there are more particles colliding with the parachute, the rate of change of momentum increases, resulting in a greater resistive force (drag).
  • Increased deceleration: A larger resistive force creates a greater resultant force opposing motion, leading to a higher rate of velocity loss (increased deceleration).

🧠 Top-Level Response Markers

Top-level responses explicitly chain the microscopic cause (more dust particles colliding) to the macroscopic mechanics formulas ( F = Δp / Δt yielding a larger force), finishing with a clear statement on how the net resultant force scales the acceleration.

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.