Dimensional Analysis & Derived Units

1. Definitions (Must Know)

• SI base quantity: a fundamental physical quantity not defined in terms of other quantities (mass, length, time, current, temperature, amount of substance).
• SI base unit: the unit for a base quantity (kg, m, s, A, K, mol).
• Derived quantity: a quantity defined using base quantities (e.g. speed, force, pressure).
• Derived unit: the unit of a derived quantity written using SI base units (e.g. N = kg m s⁻²).
• Dimension: the “type” of a quantity written in terms of base quantities (e.g. force depends on mass, length and time).
• Dimensional analysis: using SI base units (or dimensions) to check whether an equation is homogeneous.
• Homogeneous equation: an equation where every term has the same unit/dimension (so LHS and RHS match).

SI base quantities (must memorise)

Prefixes (must memorise)

• p = 10⁻¹², n = 10⁻⁹, μ = 10⁻⁶, m = 10⁻³, c = 10⁻², d = 10⁻¹, k = 10³, M = 10⁶, G = 10⁹, T = 10¹²

2. Key Ideas (What Earns Marks)

• Convert every quantity into SI base units before you compare terms or substitute into equations.
• For addition/subtraction, every term must have the same unit (e.g. you can add two energies, but not energy + momentum).
• For multiplication/division, the units combine (the result becomes a new derived unit).
• Dimensional analysis can prove an equation is wrong (not homogeneous), but it cannot prove an equation is correct (you could still be missing constants like 1/2 or 2π).

3. Detailed Explanations

A. Expressing a derived unit in base units

Workflow:

1. Write down a defining equation (e.g. F = ma, p = F/A, V = W/Q).
2. Substitute the SI units of each quantity.
3. Simplify using indices.

Mini-example (force): [F] = [m][a] = (kg)(m s⁻²) = kg m s⁻²

B. Checking if an equation is homogeneous

Workflow:

1. Write the SI unit of each term.
2. Reduce each term to SI base units.
3. If units match across all terms, the equation is homogeneous.

Mini-example ( s = ut + (1/2)at²):

• [ut] = (m s⁻¹)(s) = m
• [at²] = (m s⁻²)(s²) = m

So every term has unit m.

C. Common derived units (in base units)

4. Common Mistakes

• Treating a prefix as part of the unit (e.g. using cm inside equations without converting to m).
• Dropping indices (e.g. treating m² like m).
• Using the wrong defining equation (e.g. mixing up p = F/A and p = ρ g h without stating which pressure you mean).

5. Exam Tips

• If you are checking homogeneity, show it explicitly (e.g. “ [ut] = m and [at²] = m, so the equation is homogeneous.”).
• Memorise a few high-frequency derived-unit expansions:
  • N = kg m s⁻², J = kg m² s⁻², W = kg m² s⁻³, Pa = kg m⁻¹ s⁻²
  • C = A s, V = kg m² s⁻³ A⁻¹, Ω = kg m² s⁻³ A⁻²
• If an equation is not homogeneous, it must be wrong. If it is homogeneous, it could still be wrong (missing constants/terms).
• Next: apply these skills in Uncertainty & Error Propagation and Estimating Physical Quantities.

6. Worked Examples

Modelled example 1

Express the unit of density in SI base units

Core

Problem

Express the unit of density in SI base units.

Study the worked solution

1. Start from the defining relation

   Method

   Use ρ = m/V.

   Reason

   Base-unit derivations should begin from a quantity equation.

   Working

   [m] = kg and [V] = m³.

2. Combine the units

   Method

   Obtain kilogram per cubic metre.

   Reason

   Volume contributes three powers of length in the denominator.

   Working

   [ρ] = kg/m³ = kg m⁻³

Guided practice 2

Express the newton (N) in SI base units

About 4 min

Problem

Express the newton in SI base units.

Try this before viewing the solution

Defining equation and base unit

Hints

Hint 1: use Newton's second law

Start from F = ma.

View solution step by step

1. Expand acceleration

   Method

   Use [a] = m s⁻².

   Reason

   Acceleration is velocity change per time.

   Working

   [F] = [m][a].

2. Combine

   Method

   Obtain kg m s⁻².

   Reason

   Force is mass times acceleration.

   Working

   1 N = 1 kg m s⁻²

Common misconception 3

Express the pascal (Pa) in SI base units

Find and correct the mistake

Learner claim

A learner divides the newton by area and writes 1 Pa = 1 kg m⁻² s⁻². Diagnose the length exponent.

Try this before viewing the solution

Metre exponent

−1−2−3

Check my attempt

View solution step by step

1. Expand pressure

   Method

   Use p = F/A and 1 N = 1 kg m s⁻².

   Reason

   Both numerator and denominator length powers must be included.

   Working

   [p] = (kg m s⁻²)/m²

2. Subtract exponents

   Method

   Obtain kg m⁻¹ s⁻².

   Reason

   m¹/m² = m⁻¹.

   Working

   1 Pa = 1 kg m⁻¹ s⁻².

Examiner practice 4

Check if mgh = (1/2)mv² is homogeneous

4 marks

Examination question

Test whether mgh = (1/2)mv² is dimensionally homogeneous. State what the result can and cannot prove. [4 marks]

Try this before viewing the solution

Both sides and conclusion

View solution step by step

1. Expand the left side

   1 mark

   Method

   Combine mass, acceleration and length.

   Reason

   g has unit m s⁻².

   Working

   [mgh] = kg m² s⁻²

2. Expand the right side

   1 mark

   Method

   Square the velocity unit.

   Reason

   Numerical factors such as 1/2 are dimensionless.

   Working

   [(1/2)mv²] = kg m² s⁻²

3. Compare

   1 mark

   Method

   State the equation is homogeneous.

   Reason

   Every term has the same base unit.

   Working

   [LHS] = [RHS].

4. Limit the conclusion

   1 mark

   Method

   State that homogeneity does not prove the physics or numerical factor.

   Reason

   A dimensionally consistent equation may still omit constants or terms.

   Working

   Homogeneous: possible, not proven correct.

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Self-mark both expansions, comparison and limitation.

Mark points

1. 1 mark: Expands mgh correctly.
2. 1 mark: Expands mv² correctly.
3. 1 mark: Concludes homogeneous.
4. 1 mark: States dimensional consistency is necessary but not sufficient.

Calculate my mark

Challenge 5

Check if s = ut + (1/2)at² is homogeneous

Minimal support

Independent transfer

Test s = ut + (1/2)at² for dimensional homogeneity, showing every additive term.

Try this before viewing the solution

Units of all terms and conclusion

Hints

Hint 1: addition requires matching types

Expand [ut] and [at²] separately before comparing with [s].

View solution step by step

1. Expand the velocity term

   Method

   Obtain a length.

   Reason

   m s⁻¹ × s = m.

   Working

   [ut] = m.

2. Expand the acceleration term

   Method

   Obtain a length.

   Reason

   m s⁻² × s² = m.

   Working

   [at²] = m.

3. Compare additive terms

   Method

   Conclude homogeneous.

   Reason

   [s] = [ut] = [at²] = m.

   Working

   Every term has dimension length.

Challenge 6

Find the SI unit of p if z² = p + qt

Minimal support

Independent transfer

For the homogeneous equation z² = p + qt, t is in seconds and q has unit m s⁻¹. Find the SI unit of p and deduce the unit of z.

Try this before viewing the solution

Units of p and z

Hints

Hint 1: start inside the sum

First find [qt]; then every additive term and z² must match it.

View solution step by step

1. Find the product unit

   Method

   Cancel the time powers.

   Reason

   q is speed and t is time.

   Working

   [qt] = (m s⁻¹)(s) = m

2. Infer the additive term

   Method

   State [p] = m.

   Reason

   Quantities added together require identical dimensions.

   Working

   [p] = [qt] = m.

3. Infer the squared quantity

   Method

   State [z] = m^(1/2).

   Reason

   [z²] = m.

   Working

   [z] = square root of m = m^(1/2).

Next step

Attempt finding the dimensions of k independently.

7. Mind Stretchers

Mind stretcher 1: Find the dimensions of k if F = kv²Extension

Show Answer

Given:

• [F] = N = kg m s⁻²
• [v] = m s⁻¹ so [v²] = m²s⁻²

From F = kv²: [k] = [F]/[v²] = (kg m s⁻²)/m²s⁻² = kg m⁻¹

Mind stretcher 2: Find x and y if Δ p = kρ^x v^yExtension

Assume k is dimensionless, Δ p is pressure, ρ is density, and v is speed.

Show Answer

Units:

• [Δ p] = Pa = kg m⁻¹s⁻²
• [ρ] = kg m⁻³
• [v] = m s⁻¹

So kg m⁻¹s⁻² = (kg m⁻³)^x(m s⁻¹)^y = kg^xm^(-3x + y)s^(-y)

Compare powers:

• kg: x = 1
• s: -y = -2 ⇒ y = 2

So x = 1 and y = 2.

Mind stretcher 3: Find r and s if M = kv^rρ g^sExtension

Where k is dimensionless, M is mass, v is speed, ρ is density, and g is gravitational field strength. (r and s are dimensionless.)

Show Answer

Units:

• [M] = kg
• [v] = m s⁻¹
• [ρ] = kg m⁻³
• [g] = m s⁻²

So kg = (m s⁻¹)^r(kg m⁻³)(m s⁻²)^s = kg · m^(r-3 + s) · s^(-r-2s)

Compare powers:

• m: r-3 + s = 0
• s: -r-2s = 0 ⇒ r = -2s

Substitute r = -2s into r-3 + s = 0: -2s-3 + s = 0 ⇒ -s = 3 ⇒ s = -3 So r = -2(-3) = 6.

Mind stretcher 4: Optional (Enrichment)Extension

A. Why is it better to quote mass in kg than in N s²m⁻¹?

Mass is a base quantity with base unit kg.

Although N s²m⁻¹ simplifies to kg (because N = kg m s⁻²), you should not define a base quantity using a derived unit built from other quantities. In exams, quote mass in kg.

B. Which operations are physically meaningful if A and B have different dimensions?

Consider:

1. A + B
2. A/B
3. B - A
4. AB

Only 2 and 4 can be physically meaningful.

You cannot add or subtract quantities with different dimensions, but you can multiply or divide them (the result becomes a new derived quantity).