AQA AS Level Physics Paper 1, June 2018: Question 5

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidelines Mark

(use of = M/V)

M = 4.0 × 10–6 × 920 = 3.68 × 10–3 (kg)

–3 –2 1

05.1 weight = 3.68 × 10 × 9.81 = 3.6 × 10 (N) ecf for second mark

1 sig.fig. -1 mark

05.2 –3 –6 3 ecf 5.1 from mass calculation 1

V = 3.68 × 10 /1000 = 3.7 (3.68) × 10 m

THREE FROM:

any mass divided by 7800 Ignore mass value if awarding first mark 1

V × 7800 + (4.0 × 10−6 –V) × 920 = 3.9 × 10−3

05.3 1

6880 V = 3.9 × 10−3 −3.68 × 10−3 1

V = 3.2 × 10−8 m3 (MAX 3)

Total 6

How to answer it

Density, Weight and Archimedes Principle Study Guide

What this question tests

This multi-part problem assesses your mastery of density equations (rho = M / V), the relationship between mass, gravitational field strength and weight (W = mg), and the application of Archimedes' Principle (where an object floating in a fluid displaces an equal weight of that fluid). It also tests advanced composite-object modelling where multiple materials share a fixed total volume.

Part 05.1 [2 marks]

Calculating the Weight of an Ice Cube

💡 Key Knowledge

Density formula: rho = M / V rearranged to M = rho * V .
Weight formula: W = m * g , where g = 9.81 m s⁻² (standard AQA acceleration due to gravity).

📐 Step-by-Step Calculation

Find mass: M = (920 kg m⁻³) * (4.0 × 10⁻⁶ m³) = 3.68 × 10⁻³ kg .
Find weight: W = (3.68 × 10⁻³ kg) * (9.81 m s⁻²) = 3.61 × 10⁻² N (or 3.6 × 10⁻² N to 2 sig fig).

✅ Correct Answer

Weight = 3.6 × 10⁻² N (or 3.61 × 10⁻² N)

❌ Common Errors & Examiner Tips

Forgetting g: Multiplying volume by density gives mass in kg, not weight in Newtons! You must multiply by 9.81.
Significant figures: The data is given to 2 sig figs. Giving an answer to 1 sig fig loses a mark (-1 penalty in mark scheme). Stick to 2 or 3 sig figs.

Mark breakdown: 1 mark for correct mass calculation, 1 mark for correct weight calculation using g = 9.81. Error carried forward (ecf) applies if mass was wrong.

Part 05.2 [1 mark]

Volume of Water Displaced

🧠 Exam Technique

Read the stem carefully: "The weight of water displaced is equal to the weight of the ice cube." Use this principle directly to bridge back from weight to mass, and then to volume of water.

📐 Step-by-Step Calculation

Mass of water displaced = Weight / g = 3.68 × 10⁻³ kg (which is numerically equal to the ice mass).
Volume = Mass / density of water = (3.68 × 10⁻³ kg) / (1000 kg m⁻³) = 3.7 × 10⁻⁶ m³ (or 3.68 × 10⁻⁶ m³).

✅ Correct Answer

Volume = 3.7 × 10⁻⁶ m³

Mark breakdown: 1 mark for correct use of mass / density of water. Full ecf allowed from part 05.1 mass values.

Part 05.3 [3 marks]

Composite Object: Ice Cube Containing Iron

💡 Key Knowledge

When an object is composed of two materials, their volumes add up to the total volume ( V(ice) + V(iron) = V(total) ), and their individual masses add up to the total mass displaced.

📐 Step-by-Step Calculation

Let the volume of the iron be V . Therefore, the volume of the ice is (4.0 × 10⁻⁶ - V) .
Set up the total mass equation from the displaced water mass (3.9 × 10⁻³ kg):
V(7800) + (4.0 × 10⁻⁶ - V)(920) = 3.9 × 10⁻³
Expand and simplify terms:
6880 V = 3.9 × 10⁻³ - 3.68 × 10⁻³
6880 V = 2.2 × 10⁻⁴
Solve for V :
V = 3.2 × 10⁻⁸ m³

✅ Correct Answer

Volume of iron = 3.2 × 10⁻⁸ m³

❌ Common Errors & Examiner Tips

This is a high-level discriminating question. Top-level candidates successfully set up simultaneous mass/volume expressions using algebraic substitution. A common error is mixing up densities or failing to subtract the iron volume from the total volume to find the remaining ice volume.

Mark breakdown: (MAX 3 marks) Awarded for structural algebra steps: 1 mark for any mass divided by 7800, 1 mark for correct substitution into total mass/density balance equation, 1 mark for final evaluated volume 3.2 × 10⁻⁸ m³ .

Topics

Physics · 3.4 Mechanics and materials