# Question 8d6d9

Mar 12, 2017

The kinetic energy $E$ of a body of mass $m$ and velocity $v$ is given by the following relation.

$E = \frac{1}{2} m {v}^{2}$

$\implies E = \frac{1}{2} m {v}^{2} = \frac{1}{2} {\left(m v\right)}^{2} / m = {p}^{2} / \left(2 m\right)$

$\text{where " p=mv=" linear momentum}$

Mass remaining constant Kinetic energy $E \propto {p}^{2}$

For two cases (final and initial) the relation becomes

${E}_{f} / {E}_{i} = {p}_{f}^{2} / {p}_{i}^{2}$

Given E_f=E_i+800% of E_i=9E_i

So ${p}_{f} / {p}_{I} = \sqrt{{E}_{f} / {E}_{I}} = \sqrt{9} = 3$

Hence percent increase of momentum
=(p_f-p_I)/p_ixx100%=(3p_i-p_i)/p_i xx100=200%#