AQA GCSE Physics Physics Paper 2 (Higher), June 2022: Question 8

12 marks · High Demand difficulty · Extended Answer

Explain the equilibrium of forces on a submerged diving brick at the bottom of a pool, calculate the water density from the force on its top surface, and determine the force at a greater depth to 3 significant figures.

Practise this question

Question

Question 8 features two diagrams. Figure 14 shows a rectangular diving brick resting on the bottom of a pool with dimensions: length 25 cm, width 10 cm, and height 10 cm. Part 08.1 asks to explain why the forces on the brick make it stationary (3 marks). Part 08.2 states the top surface is 2.50 m deep with a downward force of 637 N from the water, g = 9.8 N/kg, and asks to calculate the water's density (6 marks). Figure 15 depicts a very deep swimming pool with the brick at a depth of 49.9 m. Part 08.3 gives that at 2.50 m depth the force was 618 N, and asks to determine the force at 49.9 m depth to 3 significant figures (3 marks).
Question text

08 Diving bricks sink to the bottom of a swimming pool.

Figure 14 shows a diving brick.

Figure 14

Swimmers practise diving to the bottom of the swimming pool to pick up the

diving brick.

08.1 Explain why the forces on the brick at the bottom of the pool cause the brick to be

stationary.

[3 marks]

08.2 When the brick from Figure 14 is at the bottom of the pool, the top surface of the brick

is 2.50 m below the surface of the water.

The force acting on the top surface of the brick due to the weight of the water is

637 N.

gravitational field strength = 9.8 N/kg

Calculate the density of the water in the swimming pool.

Use the Physics Equations Sheet.

[6 marks]

30Density of water = kg/m3

08.3 Professional divers are trained in a very deep swimming pool.

The density of the water in this pool is not the same as the density of the water in

Question 08.2

The diving brick was dropped into the very deep swimming pool.

When the brick was at a depth of 2.50 m, the force due to the weight of the water on

the top surface of the brick was 618 N.

Figure 15 shows the diving brick at the bottom of the very deep swimming pool.

Figure 15

Determine the force due to the weight of the water on the top surface of the brick

in Figure 15.

Use the Physics Equations Sheet.

Give your answer to 3 significant figures.

[3 marks]

Force (3 significant figures) = N

Mark scheme

Show the mark scheme Mark scheme for Question 8. Part 08.1 awards 3 marks: upthrust acts upwards, normal contact force acts upwards, and weight equals upthrust plus normal contact force (or resultant force equals zero). Part 08.2 awards 6 marks for: Area = 0.25 × 0.10 = 0.025 m²; P = F/A = 637 / 0.025 = 25 480 Pa; substitution into P = hρg (25 480 = 2.5 × ρ × 9.8); rearranging for ρ = 25 480 / (9.8 × 2.5); giving ρ = 1040 kg/m³. An alternative method calculating water volume and mass is also provided. Part 08.3 awards 3 marks: F = 618 × (49.9 / 2.5) = 12 335.28 N, rounded to 12 300 N (3 sig figs).

Question 8

AO /

Question Answers Extra information Mark

Spec. Ref.

08.1 upthrust acts (upwards on the 1 AO1

brick) 4.5.1.2

4.5.5.1.2

normal contact force acts 1

upwards (on the brick)

weight is equal to upthrust plus allow resultant force is equal to 1

normal contact force zero only if all three forces are

given

AO /

Spec. Ref.

08.2 A = 0.25 × 0.10 = 0.025 (m2) 1 AO2

4.5.5.1.1

4.5.5.1.2

637 allow correct substitution of 1

P = incorrectly calculated value of A

0.025

P = 25 480 (Pa) allow correct calculation using 1

an incorrectly calculated value

of A

to gain further marks, P = F/A

or an incorrect rearrangement of

P = F/A must have been used

with the given data

25 480 = 2.5 × ρ × 9.8 allow correct substitution of 1

incorrectly calculated value of P

25 480 allow correct rearrangement 1

ρ = using an incorrectly calculated

9.8 × 2.5

value of P

allow use of h = 2.6 (m)

ρ = 1040 (kg/m3) allow correct calculation using 1

an incorrectly calculated value

of P

allow use of h = 2.6 (m)

Alternative method

A = 0.25 × 0.10 = 0.025 (m2) 1

volume of water column allow use of an incorrectly 1

(V) = 0.025 × 2.5 calculated value of A

V = 0.0625 (m3) allow use of an incorrectly 1

calculated value of A

637 1

m (= )= 65 (kg)

9.8

65 allow use of an incorrectly 1

ρ = calculated value of V

0.0625

allow use of an incorrectly 1

3 calculated value of V

ρ = 1040 (kg/m )

AO / 25

Spec. Ref.

08.3 49.9 allow calculation of density = 1 AO3

F = 618 × 1008.979 (kg/m3) 4.5.5.1.1

2.5

4.5.5.1.2

F = 12 335.28 1

F = 12 300 (N) 1

allow correct rounding of an

incorrectly calculated value of F

allow max of 2 marks if 50 m is

used

Total Question 8 12

How to answer it

Forces, Equilibrium, and Fluid Pressure (Submerged Brick)

📋 What this question tests

This question assesses core concepts from AQA GCSE Physics Topic 5 (Forces):

  • Identifying all forces acting on a submerged object resting on a solid surface (equilibrium of three forces).
  • Converting dimensions from centimetres to metres and calculating surface area.
  • Applying the fluid pressure formula: p = h × ρ × g combined with p = F ÷ A .
  • Using direct proportionality or density substitution to determine force at depth, rounded to 3 significant figures.
Question 08.1 (3 Marks)

Equilibrium of Forces on a Sunken Brick

Explain why the forces on the brick at the bottom of the pool cause the brick to be stationary.

✅ Model Answer (Mark Scheme)

  • Upthrust acts upwards on the brick. [1 mark]
  • Normal contact force acts upwards on the brick (from the pool floor). [1 mark]
  • Weight acting downwards is equal to upthrust plus normal contact force (meaning resultant force is zero). [1 mark]
Examiner note: Stating "resultant force = 0" only scores the third mark if you have explicitly named all three forces!

💡 Key Physics Concepts

Newton's First Law: For an object to remain stationary, the resultant force must be zero.

When an object is resting on the bottom of a pool, there are three vertical forces:

  • Downward: Weight ( W )
  • Upward: Upthrust ( U ) from water displaced
  • Upward: Normal contact force ( N ) from the pool floor
  • Equilibrium: W = U + N

❌ Common Errors

  • Forgetting normal contact force: Saying "weight equals upthrust" is incorrect because diving bricks sink; its weight is greater than upthrust alone.
  • Vague answers: Just stating "forces are balanced" without naming which forces are acting.
  • Confusing water pressure with a distinct force: Pressure creates upthrust (difference in water pressure between top and bottom).

🧠 Exam Technique

Whenever an exam question asks why a resting submerged object is stationary:

  1. Name every upward force.
  2. Name every downward force.
  3. Explicitly state that the sum of upward forces equals the sum of downward forces (resultant force = 0).
Question 08.2 (6 Marks)

Calculating the Density of Swimming Pool Water

Calculate the density of the water given depth = 2.50 m, force on top surface = 637 N, and g = 9.8 N/kg.

📐 Step-by-Step Calculation

  1. Convert dimensions to metres and calculate area of top surface:
    Length = 25 cm = 0.25 m , Width = 10 cm = 0.10 m
    A = 0.25 m × 0.10 m = 0.025 m² [1 mark]
  2. Calculate pressure on the top surface:
    p = F ÷ A
    p = 637 ÷ 0.025 = 25 480 Pa [2 marks: 1 for substitution, 1 for calculation]
  3. Set up the liquid column pressure formula:
    p = h × ρ × g
    25 480 = 2.50 × ρ × 9.8 [1 mark]
  4. Rearrange for density (ρ):
    ρ = 25 480 ÷ (2.50 × 9.8) = 25 480 ÷ 24.5 [1 mark]
  5. Calculate final value:
    ρ = 1040 kg/m³ [1 mark]

💡 Alternative Method (Volume & Mass of Column)

  • Area = 0.025 m²
  • Volume of water column: V = 0.025 × 2.50 = 0.0625 m³
  • Mass of column: m = W ÷ g = 637 ÷ 9.8 = 65 kg
  • Density: ρ = m ÷ V = 65 ÷ 0.0625 = 1040 kg/m³

❌ Common Traps to Avoid

  • Unit conversion error: Calculating area in cm² ( 250 cm² ) without converting to m² ( 0.025 m² ). Dividing 250 by 100 instead of 10 000 is a frequent mistake.
  • Wrong face: Using the 10 cm × 10 cm end instead of the top surface (25 cm × 10 cm).
  • Arithmetic order: In the rearrangement 25 480 ÷ (2.5 × 9.8) , typing 25 480 ÷ 2.5 × 9.8 into a calculator gives an incorrect value of 100 000! Always use brackets.
Question 08.3 (3 Marks)

Proportional Scaling & Significant Figures at Depth

At depth 2.50 m, force is 618 N. Find force at depth 49.9 m to 3 significant figures.

📐 Step-by-Step Calculation

Because F = p × A = (h × ρ × g) × A , and surface area A , gravity g , and density ρ remain constant, force is directly proportional to depth ( F ∝ h ).

  1. Set up ratio equation:
    F₂ = F₁ × (h₂ ÷ h₁)
    F = 618 × (49.9 ÷ 2.50) [1 mark]
  2. Calculate unrounded value:
    F = 618 × 19.96 = 12 335.28 N [1 mark]
  3. Round to 3 significant figures:
    Third digit is 3, followed by 3 → round down.
    Force = 12 300 N [1 mark]

💡 Alternative Full Calculation Route

You can also work out the new density first:

ρ = F ÷ (A × h × g) = 618 ÷ (0.025 × 2.50 × 9.8) = 1008.98 kg/m³

Then recalculate force at 49.9 m:

F = h × ρ × g × A = 49.9 × 1008.98 × 9.8 × 0.025 = 12 335.28 N → 12 300 N

❌ Common Errors

  • Using 50 m instead of 49.9 m: Reading the diagram carelessly loses 1 mark (capped at 2 marks max).
  • Significant figures penalty: Leaving the answer as 12 335 N or 12 335.28 N loses the final mark. Always check the required precision!
  • Using density from 08.2: The question explicitly says: "The density of the water in this pool is not the same as the density in Question 08.2." You must use the new value ( 618 N at 2.50 m ).

Topics

Physics · P5: Forces

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Higher), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.