1. For a body of constant mass $m$, momentum and kinetic energy are related by

$p = mv$

and

$K = \frac{1}{2}mv^2$

Eliminating $v$, we get

$K = \frac{p^2}{2m}$

So, for constant mass,

$K \propto p^2$

2. The momentum is increased by $20\%$, so the new momentum is

$p' = 1.2p$

3. Hence the new kinetic energy is

$K' = \frac{{p'}^2}{2m} = \frac{(1.2p)^2}{2m} = 1.44\frac{p^2}{2m} = 1.44K$

4. Therefore, the increase in kinetic energy is

$K' - K = 1.44K - K = 0.44K$

So the percentage increase is

$0.44 \times 100 = 44\%$

5. Checking options:

• A: $36\%$ ❌
• B: $40\%$ ❌
• C: $44\%$ ✅
• D: $48\%$ ❌

Therefore, the correct answer is C: $44\%$.