Question

If kinetic energy of a body is increased by 300%, then percentage change in momentum will be

• A. $100\%$
• B. $150\%$
• C. $265\%$
• D. $73.2\%$

Answer 1

The correct option is B $150\%$

$KE=\frac{1}{2}m{v}^2=\frac{1}{2}\frac{m^2{v}^2}{m}(\because \text{momentum},p=mv)=\frac{1}{2}\frac{p^2}{m}$
so, $p=\sqrt{2\left(KE\right)m}$
$\Rightarrow \frac{\Delta p}{p}=\frac{1}{2}\frac{\Delta KE}{KE}$

given, $\frac{\Delta KE}{KE}\times 100=300$ %= $3$
$\frac{\Delta KE}{KE}=3$
Therefore, $\frac{\Delta P}{P}=\frac{1}{2}\times 3=1.5$
$\Rightarrow \frac{\Delta P}{P}\times 100=150$ %