Q.The SI unit of energy is $J = \text{kg m}^2\,\text{s}^{-2}$; that of speed $v$ is $\text{m s}^{-1}$ and of acceleration $a$ is $\text{m s}^{-2}$. Which of the formulae for kinetic energy ($K$) given below can you rule out on the basis of dimensional arguments ($m$ stands for the mass of the body):

Concept: Dimensional Analysis — check whether each proposed formula has the same dimensions as energy ($[M L^2 T^{-2}]$).

Step 1: Dimensions of given quantities

$[m] = M$, $[v] = L T^{-1}$, $[a] = L T^{-2}$.

Energy $K$ has $[K] = M L^2 T^{-2}$.

Step 2: Check each option

1. $[m^2 v^3] = M^2 \cdot (L T^{-1})^3 = M^2 L^3 T^{-3}$ — not $M L^2 T^{-2}$.
2. $[\frac12 m v^2] = M \cdot (L T^{-1})^2 = M L^2 T^{-2}$ — matches.
3. $[m a] = M \cdot L T^{-2} = M L T^{-2}$ — not $M L^2 T^{-2}$. …

The key idea here is that any valid physical equation must be dimensionally consistent — the units of the left-hand side must equal the units of the right-hand side. Dimensional analysis can’t tell you if a formula is correct (it can’t distinguish between $\frac12 m v^2$ and $\frac{3}{16} m v^2$), but it can tell you if a formula is definitely wrong because its units don’t match.

We are given that energy $K$ has SI unit $\text{kg m}^2 \text{s}^{-2}$, so its dimension is:

where $M$ = mass, $L$ = length, $T$ = time.

Now we check each candidate by finding the dimension of its right-hand side and comparing.

1. Option (a): $K = m^2 v^3$

   $[m^2] = M^2$

   $[v^3] = (L T^{-1})^3 = L^3 T^{-3}$

   So $[m^2 v^3] = M^2 L^3 T^{-3}$

   Compare with $[K] = M L^2 T^{-2}$: the powers of $M$, $L$, and $T$ are all different.

   Ruled out — dimensions don’t match.

2. Option (b): $K = \frac12 m v^2$

   The constant $\frac12$ is dimensionless, so ignore it.

   $[m] = M$

   $[v^2] = (L T^{-1})^2 = L^2 T^{-2}$

   So $[\frac12 m v^2] = M L^2 T^{-2}$

   This matches $[K]$ exactly.

   Cannot be ruled out by dimensional analysis.

3. Option (c): $K = m a$

   $[m] = M$

   $[a] = L T^{-2}$

   So $[m a] = M L T^{-2}$

   Compare with $[K] = M L^2 T^{-2}$: the power of $L$ is 1 instead of 2.

   Ruled out.

4. Option (d): $K = \frac{3}{16} m v^2$

   $\frac{3}{16}$ is dimensionless.

   $[m v^2] = M L^2 T^{-2}$, same as option (b).

   Cannot be ruled out — dimensions match, even though the numerical factor differs.

5. Option (e): $K = \frac12 m v^2 + m a$

   This is a sum of two terms. For the sum to be dimensionally consistent, every term must have the same dimension as $K$. …

Method: Dimensional Analysis

Why this method?

Dimensional analysis checks whether both sides of an equation have the same kind of physical quantity (mass, length, time, etc.). If dimensions don’t match, the formula is definitely wrong — no matter what the numbers say.

Steps

Step 1 – Write the dimension of kinetic energy ($K$)

From the SI unit given:

$[K] = \text{M L}^2 \text{T}^{-2}$

(where M = mass, L = length, T = time)

Step 2 – Write dimensions of each symbol

• $[m] = \text{M}$
• $[v] = \text{L T}^{-1}$
• $[a] = \text{L T}^{-2}$

Step 3 – Check each option by substituting dimensions

(a) $K = m^2 v^3$

$[m^2 v^3] = \text{M}^2 \cdot (\text{L T}^{-1})^3 = \text{M}^2 \text{L}^3 \text{T}^{-3}$

Compare with $[K] = \text{M L}^2 \text{T}^{-2}$ → Mismatch → Rule out

(b) $K = \frac{1}{2} m v^2$

$[m v^2] = \text{M} \cdot (\text{L T}^{-1})^2 = \text{M L}^2 \text{T}^{-2}$

Matches $[K]$ exactly → Dimensionally correct (the $\frac{1}{2}$ is dimensionless)

(c) $K = m a$

$[m a] = \text{M} \cdot \text{L T}^{-2} = \text{M L T}^{-2}$

Compare: $\text{M L T}^{-2}$ vs $\text{M L}^2 \text{T}^{-2}$ → Mismatch → Rule out

(d) $K = \frac{3}{16} m v^2$ …

🧠 The Core Idea

Dimensional analysis checks whether the units (dimensions) on both sides of an equation match.

If they don’t match, the formula is definitely wrong.

If they do match, the formula could be correct (but might still be wrong for other reasons).

✓ Step 1: Write dimensions of each quantity

• Energy $K$ (SI unit: $\text{kg m}^2 \text{s}^{-2}$)

  → $[K] = M L^2 T^{-2}$

• Mass $m$ → $[m] = M$

• Speed $v$ → $[v] = L T^{-1}$

• Acceleration $a$ → $[a] = L T^{-2}$

✗ Common Mistake #1: Forgetting that constants like $\frac12$ are dimensionless

The mistake: Students think $\frac12$ or $\frac{3}{16}$ affects dimensions.

Truth: Pure numbers have no dimensions — they don’t change the dimensional formula.

✓ How to avoid: Always ignore numerical constants when doing dimensional analysis. Only look at variables and physical constants.

✗ Common Mistake #2: Not checking each term separately in sums

The mistake: In option (e), $K = \frac12 m v^2 + m a$, students check only the first term and assume the whole formula is fine.

Truth: You can only add or subtract quantities with the same dimensions.

• First term: $\frac12 m v^2$ → $M (L T^{-1})^2 = M L^2 T^{-2}$ ✓ (matches energy)
• Second term: $m a$ → $M (L T^{-2}) = M L T^{-2}$ ✗ (does not match energy)

Since the two terms have different dimensions, the sum is dimensionally invalid.

✓ How to avoid: For any formula with a sum, check every term separately. If any term has different dimensions, the whole formula is wrong.

✗ Common Mistake #3: Confusing $v^2$ with $v^3$ or $v$

The mistake: Students mis-calculate powers of $T$ when squaring or cubing speed.

Example:

• $v$ has $[v] = L T^{-1}$
• $v^2$ → $L^2 T^{-2}$
• $v^3$ → $L^3 T^{-3}$

If you accidentally treat $v^3$ as $L^3 T^{-1}$, you’ll get wrong conclusions.

✓ How to avoid: Write dimensions step-by-step:

$[v^n] = (L T^{-1})^n = L^n T^{-n}$

✗ Common Mistake #4: Thinking “it looks like the known formula” means it’s correct

The mistake: Option (b) $K = \frac12 m v^2$ is the correct formula, but students sometimes reject (d) $K = \frac{3}{16} m v^2$ because “the constant is weird.” …