The momentum of a body is increased by 50 % . The percentage increase in the kinetic energy of the body is_…

The relation of kinetic energy and momentum is given by

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

The new value of momentum is $p\text{'}=1.5p$

The new value of kinetic energy is

$\Rightarrow K\mathit{\text{'}}=\frac{{\left(1.5\mathrm{p}\right)}^2}{2\mathrm{m}}=2.25K$

Hence, the percentage increase in kinetic energy is 

$\frac{\mathit{K}\mathit{\text{'}}\mathit{-}\mathit{K}}{\mathit{K}}\times 100=\frac{2.25K-K}{K}=125\%$