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Edexcel · GCSE · Physics · Revision Notes

Forces and matter

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Quick answer

Forces and matter examines how materials respond to forces and how pressure works in different contexts. Master the key equations: F = ke for springs (up to the limit of proportionality), Ee = ½ke² for elastic energy, p = F/A for solid pressure, and p = hρg for liquid pressure. Remember that elastic deformation is reversible while plastic deformation is permanent. Density determines whether objects float, and atmospheric pressure decreases with altitude. Always convert units to standard form (metres, kilograms, newtons) before calculating.

What you'll learn

This topic explores how forces affect materials and fluids. You'll learn to calculate spring constants, understand elastic and inelastic deformation, and work with pressure in solids, liquids and gases. These principles apply to everything from car suspension systems to hydraulic brakes and scuba diving equipment.

Key terms and definitions

Elastic deformation — when a material returns to its original shape after the force is removed

Inelastic (plastic) deformation — permanent deformation where the material does not return to its original shape after the force is removed

Limit of proportionality — the point beyond which the extension is no longer directly proportional to the force applied

Spring constant (k) — the force required to extend or compress a spring by one metre, measured in newtons per metre (N/m)

Pressure — the force per unit area acting perpendicular to a surface, measured in pascals (Pa) or N/m²

Upthrust — the upward force exerted by a fluid on an object immersed in it

Density — mass per unit volume of a substance, measured in kg/m³

Atmospheric pressure — the pressure exerted by the weight of the atmosphere, approximately 100,000 Pa at sea level

Core concepts

Density and its applications

Density describes how much mass is packed into a given volume. The equation is:

density (kg/m³) = mass (kg) ÷ volume (m³)

Or using symbols: ρ = m ÷ V

Different materials have characteristic densities. Steel has a density of approximately 7800 kg/m³, while water has a density of 1000 kg/m³. Air at sea level has a much lower density of about 1.2 kg/m³.

To find the density of a regular solid:

  • Measure its mass using a balance
  • Calculate volume from its dimensions (e.g., for a cuboid: length × width × height)
  • Divide mass by volume

For irregular solids, use the displacement method:

  • Measure the object's mass
  • Submerge it in a displacement can or measuring cylinder
  • Measure the volume of water displaced (this equals the object's volume)
  • Calculate density using the formula

Objects float or sink depending on their density relative to the fluid. An object floats if its average density is less than the fluid's density. This explains why steel ships float despite steel being denser than water—the ship's overall density (including air spaces) is less than water.

Forces and elasticity

When forces are applied to objects, they can stretch, compress or bend. Springs and elastic materials demonstrate these effects clearly.

Hooke's Law states that the extension of an elastic object is directly proportional to the force applied, provided the limit of proportionality is not exceeded.

The equation is:

force (N) = spring constant (N/m) × extension (m)

Or: F = k × e

The extension is the increase in length from the original length:

extension = stretched length − original length

Key points about Hooke's Law:

  • It only applies up to the limit of proportionality
  • Beyond this limit, the relationship becomes non-linear
  • The elastic limit is the point beyond which permanent deformation occurs
  • A stiffer spring has a larger spring constant

A force-extension graph for a spring shows:

  • A straight line through the origin (linear region) where Hooke's Law applies
  • The gradient equals the spring constant
  • Beyond the limit of proportionality, the line curves
  • If the elastic limit is exceeded, the spring won't return to its original length when the force is removed

Multiple springs in combination behave predictably:

  • Springs in series (end-to-end): extension doubles, spring constant halves
  • Springs in parallel (side-by-side): extension halves, combined spring constant doubles

Elastic potential energy

When an elastic object is stretched or compressed, work is done and energy is stored as elastic potential energy. This energy can be calculated when the limit of proportionality is not exceeded:

elastic potential energy (J) = 0.5 × spring constant (N/m) × extension² (m²)

Or: Ee = ½ × k × e²

On a force-extension graph, the elastic potential energy equals the area under the line up to that extension. For a linear relationship (triangle), this confirms the ½ke² formula.

This principle applies to:

  • Springs in mechanical devices
  • Elastic cords in bungee jumping
  • Bow and arrows (storing energy when drawn)
  • Catapults and slingshots

Pressure in solids

Pressure measures how concentrated a force is over an area:

pressure (Pa) = force normal to surface (N) ÷ area (m²)

Or: p = F ÷ A

One pascal (Pa) equals one newton per square metre. Pressure is also measured in N/m² or N/cm².

The force must act perpendicular (normal) to the surface. If you push at an angle, only the perpendicular component creates pressure.

Applications of pressure in solids:

Reducing pressure:

  • Wide tyres on tractors spread weight over larger areas, preventing sinking in mud
  • Skis have large surface areas to reduce pressure on snow
  • Foundations spread building weight over large areas

Increasing pressure:

  • Knife blades are thin to concentrate force on small areas
  • Drawing pins have sharp points to penetrate surfaces easily
  • Ice skates have narrow blades creating high pressure that melts ice slightly

Pressure in liquids

Liquids exert pressure on submerged objects. This pressure:

  • Acts equally in all directions at a given depth
  • Increases with depth
  • Increases with liquid density
  • Is independent of the container's shape or surface area

The equation for pressure in a liquid column is:

pressure (Pa) = height of column (m) × density (kg/m³) × gravitational field strength (N/kg)

Or: p = h × ρ × g

Where g = 10 N/kg on Earth (or 9.8 N/kg for more precise calculations).

This explains why:

  • Dams are thicker at the bottom (greater pressure at depth)
  • Submarine hulls must withstand enormous pressures at depth
  • Divers experience increasing pressure as they descend
  • Water pressure at the bottom of a tall thin tube equals that at the bottom of a wide tank of the same depth

The total pressure at depth also includes atmospheric pressure acting on the liquid surface:

total pressure = atmospheric pressure + pressure due to liquid column

Atmospheric pressure and upthrust

The Earth's atmosphere creates pressure due to the weight of air molecules above. At sea level, atmospheric pressure is approximately 101,000 Pa (101 kPa).

Atmospheric pressure:

  • Decreases with altitude (less air above)
  • Varies with weather conditions
  • Acts equally in all directions

Upthrust is the upward force exerted by fluids on immersed objects. It occurs because pressure increases with depth, so the upward force on the bottom of an object exceeds the downward force on top.

Archimedes' principle states: The upthrust on an object equals the weight of fluid displaced

For an object to float:

  • Upthrust must equal or exceed the object's weight
  • This occurs when the object's average density ≤ fluid density

For a floating object: upthrust = weight of object

For a sinking object: upthrust < weight of object

This explains why:

  • Helium balloons rise (helium density < air density)
  • Hot air balloons float (heated air expands, reducing density)
  • Ships with cargo can adjust buoyancy using ballast
  • Submarines control depth by adjusting their average density

Worked examples

Example 1: Calculating spring constant

Question: A spring extends from 8.0 cm to 14.0 cm when a force of 12 N is applied. Calculate the spring constant. [3 marks]

Solution:

Original length = 8.0 cm = 0.08 m

Extended length = 14.0 cm = 0.14 m

Extension = 0.14 − 0.08 = 0.06 m [1 mark]

Using F = k × e

Rearranging: k = F ÷ e [1 mark]

k = 12 ÷ 0.06 = 200 N/m [1 mark]

Example 2: Pressure in liquids

Question: A diver descends to a depth of 25 m in seawater. Seawater has a density of 1030 kg/m³. Calculate the pressure due to the seawater at this depth. (g = 10 N/kg) [3 marks]

Solution:

Using p = h × ρ × g [1 mark]

p = 25 × 1030 × 10 [1 mark]

p = 257,500 Pa (or 258 kPa or 0.258 MPa) [1 mark]

Example 3: Elastic potential energy

Question: A spring with spring constant 150 N/m is compressed by 0.04 m. Calculate the elastic potential energy stored in the spring. [3 marks]

Solution:

Using Ee = ½ × k × e² [1 mark]

Ee = 0.5 × 150 × (0.04)² [1 mark]

Ee = 0.5 × 150 × 0.0016 = 0.12 J [1 mark]

Common mistakes and how to avoid them

  • Forgetting to convert units — Always convert centimetres to metres, grams to kilograms, and cm² to m² before substituting into equations. Extension of 5 cm must become 0.05 m.

  • Confusing extension with stretched length — Extension is the increase in length, not the total length. Always subtract original length from stretched length.

  • Using the wrong rearrangement — When calculating spring constant from F = ke, you must divide force by extension (k = F/e), not multiply. Practice triangle methods or algebraic rearrangement.

  • Forgetting that pressure acts in all directions — In liquids, pressure doesn't just act downwards. At any point, it acts equally in all directions—sideways, upwards and downwards.

  • Not recognizing when Hooke's Law doesn't apply — If a question states the elastic limit has been exceeded, or shows a curved force-extension graph, you cannot use F = ke. The relationship is no longer linear.

  • Mixing up density and pressure equations — Density is ρ = m/V (mass divided by volume). Pressure in liquids is p = hρg (height times density times g). Check which quantity the question asks for.

Exam technique for "Forces and matter"

  • Show all working clearly — Even if your final answer is wrong, you can gain marks for correct method. Write the equation, substitute values with units, then calculate.

  • Identify command words — "Calculate" requires numerical working and units. "Explain" needs reasons using physics principles. "Describe" requires observations without necessarily explaining causes.

  • Use correct units — Spring constant is N/m, pressure is Pa or N/m², density is kg/m³. Writing the wrong unit loses marks even with correct numbers.

  • For 3-mark calculations — Typically: 1 mark for correct equation/method, 1 mark for correct substitution, 1 mark for correct answer with unit. Show each step separately.

Quick revision summary

Forces and matter examines how materials respond to forces and how pressure works in different contexts. Master the key equations: F = ke for springs (up to the limit of proportionality), Ee = ½ke² for elastic energy, p = F/A for solid pressure, and p = hρg for liquid pressure. Remember that elastic deformation is reversible while plastic deformation is permanent. Density determines whether objects float, and atmospheric pressure decreases with altitude. Always convert units to standard form (metres, kilograms, newtons) before calculating.

Forces and matter: common questions

What do you need to know about Forces and matter for Edexcel GCSE Physics?

Forces and matter examines how materials respond to forces and how pressure works in different contexts. Master the key equations: F = ke for springs (up to the limit of proportionality), Ee = ½ke² for elastic energy, p = F/A for solid pressure, and p = hρg for liquid pressure. Remember that elastic deformation is reversible while plastic deformation is permanent. Density determines whether objects float, and atmospheric pressure decreases with altitude. Always convert units to standard form (metres, kilograms, newtons) before calculating.

What are the most common mistakes in Forces and matter?

Forgetting to convert units: Always convert centimetres to metres, grams to kilograms, and cm² to m² before substituting into equations. Extension of 5 cm must become 0.05 m. Confusing extension with stretched length: Extension is the increase in length, not the total length. Always subtract original length from stretched length. Using the wrong rearrangement: When calculating spring constant from F = ke, you must divide force by extension (k = F/e), not multiply. Practice triangle methods or algebraic rearrangement.

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