The momentum of a body is increased by 50 % . The percentage...

The relation of kinetic energy and momentum is given by

$K = \dfrac{{p}^{2}}{2m}$

The new value of momentum is $p ' = 1 . 5p$

The new value of kinetic energy is

$⇒ K ' = \dfrac{{\left(1 . 5p\right)}^{2}}{2m} = 2 . 25K$

Hence, the percentage increase in kinetic energy is

$\dfrac{K ' - K}{K} \times 100 = \dfrac{2 . 25K - K}{K} = 125$