If momentum is increased by 80%, then percentage increase in kinetic energy is

${\mathrm{P}}^2=2\mathrm{mk}$

$\begin{array}{l}\frac{{\mathrm{P}}_1^2}{{\mathrm{P}}_2^2}=\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}\\ \frac{{\mathrm{P}}_2-{\mathrm{P}}_1}{{\mathrm{P}}_1}=\times 100=80\\ \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}-1=\frac{8}{10}\\ \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\frac{18}{10}\\ \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\frac{9}{5}\\ \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{81}{25}\\ \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}-1=\frac{81-25}{25}\\ \frac{\mathrm{\Delta k}}{\mathrm{k}}=\frac{56}{25}\\ \frac{\mathrm{\Delta k}}{\mathrm{k}}\times 100=56\times 4\\ \frac{\mathrm{\Delta k}}{\mathrm{k}}=224\mathrm{\%}\end{array}$