Question

If the momentum of a body is increased by 50%, what would be the percentage increase in its kinetic energy?

Answer 1

Percentage increase in K.E,
$\left(E\right)=\left[\frac{E_2-E_1}{E_1}\right]\times 100=[\frac{E_2}{E_1}-1]\times 100$
But $E\alpha p^2\Rightarrow \frac{E_2}{E_1}=\frac{p_2^2}{p_1^2}$
$\therefore$ % increase in K.E. $=[\frac{p_2^2}{p_1^2}-1]\times 100$
Let $p_1=100$, then
$p_2=100+50=150$% increase in
$K.E.=[\frac{(150{)}^2}{(100{)}^2}-1]\times 100=[\frac{225}{100}-1]\times 100$
$=125$%
Therefore, the percentage increase in its kinetic energy is 125%