AQA AS Level Physics Paper 1, November 2021: Question 3
10 marks · Medium difficulty · Short Answer
Calculate the work done, average force, and mass of a block for a dust particle impacting a spacecraft block, and compare stopping distances for different particle densities.
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Question text
03 Figure 5 shows a spacecraft travelling towards a comet.
The spacecraft has an array of blocks designed to capture small dust particles from
the comet’s tail.
Figure 5
To test the blocks before launch, a spherical dust particle P is fired at a right angle to
the surface of a fixed, stationary block.
P has a mass of 1.1 × 10−9 kg. It has a speed of 5.9 × 103 m s−1 when it hits the
surface of the block.
P comes to rest inside the block.
03.1 Calculate the work done in bringing P to rest.
[1 mark]
work done = J
03.2 P travels a distance of 2.9 cm in a straight line inside the block before coming to rest.
The resultant force on P varies as it penetrates the block.
Calculate the average force acting on P as it is brought to rest.
[2 marks]
average force = N
03.3 The block is rectangular with an area of cross-section of 8.0 cm2 and a thickness
of 3.0 cm.
*10* Figure 6 shows how the density of the block varies with depth up to its maximum
thickness.
Figure 6
Calculate the mass of the block.
[4 marks]
12 mass = kg
03.4 In another test, a spherical particle Q is fired at a right angle to the surface of an
identical block.
Q has the same mass as P and is travelling at the same speed as P when it strikes
the surface of the block.
Q is made from a less dense material than P.
*11* Compare the distance travelled by Q with that travelled by P as they are brought to
rest.
[3 marks]
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark AO
03.1 (Work done = lost KE = ½ mv2 =) 0.019 (J) 1 AO1.1b
03.2 Use of W = Fs Condone POT error in substitution 2 AO1.1a
(F =) 0.66 (N) ECF from 03.1 AO2.1b
Alternative:
Use of an appropriate suvat equation and use
of F = ma
(t = 9.8 x 10-6 s)
(a= 6.0 x 108 m s-2)
Condone POT error in substitution
No ECF from 3.1 this route
(F =) 0.66 (N)
03.3 Use of Volume = Thickness x area of cross-section Condone POT errors apart from final answer 4 AO1.1a
(V =) 8 × 10–4 × 0.03 Or (V=) 2.4 × 10–5
50+5 AO1.1b
(Average density =) = 27.5
mass Condone use of their density and volume AO1.1b
14 Use of density =
volume
AO2.1b
(mass =) 6.6 × 10–4 (kg) c.a.o
Alternative:
Use of Volume = Thickness x area of cross-section
Condone POT errors apart from final answer
(V =) 8 × 10–4 × 0.03 Or (V=) 2.4 × 10–5
mass -3 -4
Use of density = / (mass =) 1.2 x 10 or 1.2 x 10
volume
Condone use of their density and volume
1.2 ×10−3+1.2 ×10−4 (50 kg m-3 =1.2 x 10-3)
(Average mass =)
(5 kg m-3 =1.2 x 10-4)
= 6.6 × 10– 4 (kg) c.a.o
Alternative:
Condone POT errors apart from final answer
Attempts to determine the area under the graph:
Formula for area of a rectangle added to the formula for area of a triangle seen /
formula for the area of trapezium seen
(50−5) ×0.03 −2
5 × 0.03 + = 0.825 (𝑘𝑘𝑘𝑘 𝑚𝑚 )
50+5 −2
𝒐𝒐𝒐𝒐 × 0.03 = 0.825 (𝑘𝑘𝑘𝑘 𝑚𝑚 )
Multiplies their area by 8 x 10-4
Mass = 6.6 × 10–4 (kg) c.a.o
03.4 Q has a larger volume (for the same mass and KE) / Must have Q will travel a shorter distance for 3 AO3.1a
all 3 marks. AO3.1a
Q has a larger surface area (for the same mass and KE)
AO3.1a
Q will experience a greater resistive force (at any given speed)
/ Q will displace more matter per unit distance
Q will do more work per unit distance / Q will transfer more of
its kinetic energy per unit distance / Q will experience a
greater deceleration
Total 10
How to answer it
Spacecraft Dust Particle Impact & Material Testing
What this question tests
This question assesses your ability to apply the principles of energy conservation (work-done and kinetic energy), mechanics equations relating force, mass, acceleration, and distance, volume-density-mass relationships with variable distributions (interpreting graphical data), and qualitative analysis of resistive forces in materials.
Calculating Work Done
✅ Correct Answer
0.019 J (Accept 0.0186 J)
💡 Key Knowledge
- Work-energy principle: Work done in stopping a particle equals its initial kinetic energy.
- Formula: Work Done = (1/2)mv²
📐 Step-by-Step Calculation
- Identify values: m = 1.1 × 10⁻⁹ kg, v = 5.9 × 10³ m s⁻¹.
- Substitute into formula: Work Done = 0.5 × (1.1 × 10⁻⁹) × (5.9 × 10³)².
- Calculate kinetic energy: 0.5 × (1.1 × 10⁻⁹) × 34810000 = 0.0191445 J.
- Round appropriately to 2 significant figures matching input data: 0.019 J .
Calculating Average Force
✅ Correct Answer
0.66 N
🧠 Exam Technique
Always watch out for unit conversions! Distance is given in centimetres ( 2.9 cm ) and must be converted to metres before using work equations.
❌ Common Errors
Forgetting to convert cm to m (using 2.9 instead of 0.029 m) results in a power of ten error. ECF (Error Carried Forward) from part 03.1 applies if you correctly rearrange W = Fs using your previous work done value.
📐 Step-by-Step Calculation
- State formula relating work done, force, and distance: W = F × s .
- Rearrange for force: F = W / s .
- Convert distance: s = 2.9 cm = 0.029 m.
- Substitute values: F = 0.0191445 J / 0.029 m = 0.6601... N.
- Final Answer: 0.66 N .
Calculating Mass from Variable Density Graph
✅ Correct Answer
6.6 × 10⁻⁴ kg (Accept 6.6 × 10⁻⁴ to 6.63 × 10⁻⁴ kg)
💡 Key Knowledge
When density varies linearly with depth, the average density can be found by taking the mean of the minimum and maximum density values, or by finding the area under the density-depth graph and multiplying by the cross-sectional area.
📐 Step-by-Step Calculation (Mean Density Method)
- Find average density from the graph: At depth 0, density = 5 kg m⁻³. At depth 3.0 cm, density = 50 kg m⁻³. Mean density = (5 + 50) / 2 = 27.5 kg m⁻³.
- Calculate total volume of the block: V = Area × thickness = (8.0 cm²) × (3.0 cm) = 24 cm³. Convert to m³: 24 × 10⁻⁶ m³ (or 8.0 × 10⁻⁴ m² × 0.03 m = 2.4 × 10⁻⁵ m³). Note: 8.0 cm² = 8.0 × 10⁻⁴ m².
- Calculate mass using mass = density × volume : mass = 27.5 kg m⁻³ × 2.4 × 10⁻⁵ m³.
- Final Answer: 6.6 × 10⁻⁴ kg .
Comparing Stopping Distances for Different Densities
✅ Correct Answer
Particle Q will travel a shorter distance than particle P.
💡 Key Knowledge
Since particle Q is made from a less dense material than P while having the exact same mass and speed, Q must occupy a larger volume and therefore have a larger surface area/frontal profile.
🧠 Examiner Commentary
To secure all 3 marks, top-tier responses clearly linked material density to volume/surface area differences, explained the resulting increase in resistive force per unit distance, and explicitly concluded with the relative stopping distance of Q compared to P.
📐 Marking Points Breakdown
- Point 1: Q has a larger volume / larger surface area for the same mass and kinetic energy.
- Point 2: Q will experience a greater resistive force / greater deceleration / do more work per unit distance at any given speed.
- Point 3: Conclusion stating explicitly that Q travels a shorter distance.
Topics
Physics · 3.4 Mechanics and materials
Question and mark scheme from the AQA AS Level Physics examination, Paper 1, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.