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A Level · Cambridge (CIE) · 9702

Calculate density and pressure

Relate density to mass per volume and pressure to perpendicular force per area, keeping units consistent.

Section 4 · Forces, density and pressure

Forces, density and pressure · Cambridge International AS & A Level Physics 9702 · 4.3

AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 4. These original explanations and numerical practice support theory study; they do not demonstrate supervised practical competence. Full timed assessment and human subject-teacher review remain outstanding.

Before you begin

  • Use SI units, resolve perpendicular components, and distinguish force from pressure.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Which equation defines density?

What you will learn

  • Define and calculate density
  • Define and calculate pressure
  • Use consistent units for mass, volume, force and area

Density describes mass packed into a volume

Density is mass per unit volume: ρ = m/V. It is a property of the material in a specified state. The SI unit is kg m⁻³; convert dimensions to metres before calculating volume in m³.

For a regular solid, volume can be calculated from its dimensions. For an irregular object, an appropriate displacement method may estimate volume, but a written or simulated method cannot by itself demonstrate practical measurement skill.

Pressure spreads a force over an area

Pressure is the normal force per unit area: p = F/A, where F is perpendicular to the surface. Its SI unit is Pa, equivalent to N m⁻². The same force gives greater pressure when it acts over a smaller area.

If the force is angled, use its component perpendicular to the surface for normal pressure. Keep area in square metres and distinguish pressure, which is force per area, from force itself.

density ρ = mass / volume · pressure p = normal force / area

Pause and explain

What happens to pressure if the same normal force is spread over twice the area?

Put the idea to work

Worked example

A metal block has mass 2.7 kg and volume 1.0 × 10⁻³ m³. Find its density. Then find the pressure from a 120 N normal force over 0.020 m².

Show the worked solution
  1. Density = 2.7/(1.0 × 10⁻³) = 2700 kg m⁻³.
  2. Pressure = 120/0.020 = 6000 Pa.
  3. The two quantities have different definitions and units.

Answer Density 2.7 × 10³ kg m⁻³; pressure 6.0 × 10³ Pa.

Try a topic test

Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

A sample has mass 0.60 kg and volume 2.0 × 10⁻⁴ m³. Calculate its density.

Give me a hint

Divide mass by volume and preserve the powers of ten.

Compare my reasoning
  1. ρ = m/V.
  2. ρ = 0.60/(2.0 × 10⁻⁴).
  3. Write the result in kg m⁻³.

3.0 × 10³ kg m⁻³.

Look for these in your work

  • Relationship used
  • Division correct
  • SI unit stated
2 · Independent

A 500 N student stands on two shoes with a total contact area of 0.025 m². Find the average pressure on the floor.

Give me a hint

Use the total contact area of both shoes and the normal force.

Compare my reasoning
  1. p = F/A.
  2. p = 500/0.025.
  3. Report pascals or N m⁻².

2.0 × 10⁴ Pa.

Look for these in your work

  • Total force and area used
  • Pressure calculated
  • Pa unit included
3 · Transfer

A cube of side 5.0 cm has mass 0.50 kg. Find its density in kg m⁻³. State the pressure on a horizontal floor if the cube rests on one face and g = 10 N kg⁻¹.

Give me a hint

Convert centimetres to metres before finding volume and contact area.

Compare my reasoning
  1. Side = 0.050 m; volume = 0.050³ = 1.25 × 10⁻⁴ m³, so ρ = 0.50/(1.25 × 10⁻⁴) = 4000 kg m⁻³.
  2. Weight = mg = 0.50 × 10 = 5.0 N; face area = 0.050² = 0.0025 m².
  3. Pressure = 5.0/0.0025 = 2000 Pa.

Density 4.0 × 10³ kg m⁻³; average pressure on the floor 2.0 × 10³ Pa.

Look for these in your work

  • Length converted to metres
  • Density uses the full cube volume
  • Pressure uses weight over one face area

Common mistakes

  • Using cm³ with mass in kilograms without converting
  • Using total surface area instead of the contact area in a pressure problem
  • Using a force component parallel to the surface
Recall without your notes

What can you explain now?

Which equation defines density?

Compare with the explanation

ρ = m/V

Density is mass per unit volume.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

How do the units and equations show the difference between density and pressure?

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Physics

Course scope and assessment

One hundred and sixty-two original lessons cover all 12 published subject-content outcomes in Topic 1 (sections 1.1–1.4), all nine outcomes in Topic 2 (section 2.1), all 13 outcomes in Topic 3 (sections 3.1–3.3), all 13 outcomes in Topic 4 (sections 4.1–4.3), all 11 outcomes in Topic 5 (sections 5.1–5.2), all 10 outcomes in Topic 6 (sections 6.1–6.2), all 16 outcomes in Topic 7 (sections 7.1–7.5), all 12 outcomes in Topic 8 (sections 8.1–8.4), all 15 outcomes in Topic 9 (sections 9.1–9.3), all 16 outcomes in Topic 10 (sections 10.1–10.3), all 18 outcomes in Topic 11 (sections 11.1–11.2), all seven outcomes in Topic 12 (sections 12.1–12.2), all 12 outcomes in Topic 13 (sections 13.1–13.4), all eight outcomes in Topic 14 (sections 14.1–14.3), and all nine outcomes in Topic 15 (sections 15.1–15.3), plus all four outcomes in Topic 16 (sections 16.1–16.2), and all ten outcomes in Topic 17 (sections 17.1–17.3), plus all 12 outcomes in Topic 18 (sections 18.1–18.5), all nine outcomes in Topic 19 (sections 19.1–19.3), all 19 outcomes in Topic 20 (sections 20.1–20.5), and all eight outcomes in Topic 21 (sections 21.1–21.2), and all 17 outcomes in Topic 22 (sections 22.1–22.4), and all 13 outcomes in Topic 23 (sections 23.1–23.2), and all 16 outcomes in Topic 24 (sections 24.1–24.3), and all 11 outcomes in Topic 25 (sections 25.1–25.3), of the 2025–2027 syllabus. Each lesson has guided, independent and application practice with explained answers. All 25 syllabus topics now have mapped teaching and practice; the 2028–2030 changes outside Topics 23–25 still need reconciliation. Online study cannot demonstrate supervised Paper 3 laboratory competence. Paper 5 planning and analysis, full timed cumulative assessment and human subject-teacher review remain outstanding.

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