Dimensional analysis — check homogeneity of equation v² = u² + 2as

Question

Using dimensional analysis, check whether the equation $v^2 = u^2 + 2as$ is dimensionally consistent. State the principle of homogeneity and explain what dimensional analysis can and cannot tell us about a physical equation.

(NCERT Class 11, Units and Measurements)

Solution — Step by Step

State the principle of homogeneity

The principle of homogeneity of dimensions states: a physical equation is valid only if the dimensions of every term on both sides are identical. You cannot add or equate quantities with different dimensions.

Write the dimensions of each quantity

$v$ (final velocity): $[\text{LT}^{-1}]$
$u$ (initial velocity): $[\text{LT}^{-1}]$
$a$ (acceleration): $[\text{LT}^{-2}]$
$s$ (displacement): $[\text{L}]$

Check each term

LHS: $[v^2] = [\text{LT}^{-1}]^2 = [\text{L}^2\text{T}^{-2}]$

RHS term 1: $[u^2] = [\text{LT}^{-1}]^2 = [\text{L}^2\text{T}^{-2}]$

RHS term 2: $[2as] = [as] = [\text{LT}^{-2}][\text{L}] = [\text{L}^2\text{T}^{-2}]$

(The constant 2 is dimensionless and doesn't affect the dimensions.)

Conclude

All three terms have the same dimensions: $[\text{L}^2\text{T}^{-2}]$ .

$[\text{LHS}] = [\text{RHS term 1}] = [\text{RHS term 2}] = [\text{L}^2\text{T}^{-2}]$

The equation is dimensionally consistent. ✓

Why This Works

Dimensions are like the "DNA" of physical quantities. Just as you can't add apples and oranges, you can't add metres and seconds. Every valid physical equation must respect this.

However, dimensional consistency is a necessary but not sufficient condition. The equation $v^2 = u^2 + 3as$ is also dimensionally consistent (the factor 3 is dimensionless), but it's physically wrong. Dimensional analysis can:

Detect definitely wrong equations (if dimensions don't match, the equation is wrong)
Derive relations between quantities (up to a dimensionless constant)
Convert units between systems

It cannot:

Verify the exact numerical coefficients (2, $\pi$ , etc.)
Distinguish between quantities with the same dimensions (work and torque both have $[\text{ML}^2\text{T}^{-2}]$ )
Handle trigonometric, exponential, or logarithmic functions (their arguments must be dimensionless)

Alternative Method

A quicker check: in the equation $v^2 = u^2 + 2as$ , notice that $v$ and $u$ are both velocities, so $v^2$ and $u^2$ automatically have the same dimensions. We only need to verify that $2as$ also has the dimension of velocity squared:

$[as] = \frac{[\text{velocity}]}{[\text{time}]} \times [\text{distance}] = \frac{[\text{L/T}]}{[\text{T}]} \times [\text{L}] = [\text{L}^2/\text{T}^2] = [\text{velocity}]^2$ ✓

💡 Expert Tip

For JEE, dimensional analysis is most useful for quick elimination in MCQs. If only one option has the correct dimensions, pick it without solving the full problem. This can save 2-3 minutes per question. Always check dimensions as a sanity check after solving any physics problem.

Common Mistake

⚠️ Common Mistake

Students sometimes argue that a dimensionally consistent equation must be correct. Consider $v^2 = u^2 + 4as$ — it's dimensionally fine but physically wrong (the coefficient should be 2, not 4). Dimensional analysis is a necessary check, not a sufficient proof. It catches blatant errors (like adding force and velocity) but can't verify numerical constants.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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