Question

If the kinetic energy of a body increases by 400%, what would be its percentage increase in momentum?

Answer 1

% increase in momentum
$\frac{changeinmomentum}{Initialmomentum}\times 100$
$=\frac{p_2-p_1}{p_1}\times 100=[\frac{p_2}{p_1}-1]\times 100$
$p=\frac{p_1}{\sqrt{2mE}}$
$\Rightarrow p\alpha \sqrt{E}\Rightarrow \frac{\sqrt{E_2}}{\sqrt{E_1}}$
$\therefore$ Percentage increase in momentum $=[\frac{\sqrt{E_2}}{\sqrt{E_1}}-1]\times 100$
Let $E_1=100$, then
$E_2=100+300=400$
Percentage increase in momentum
$=[\frac{\sqrt{400}}{\sqrt{100}}-1]\times 100=|\sqrt{4}-1|\times 100$
$=(2-1)\times 100=100$%
Therefore, percentage increase in momentum is 100%.