# Question 13b3d

Mar 6, 2014

The kinetic energy increases by 125 %.

#### Explanation:

Let initial momentum be $p$ and kinetic energy be $K$.

$p = m v$

If $p$ increases by 50% (0.50$p$)

the new momentum $p ' = p + 0.50 p = 1.50 p$

The initial $K = \frac{1}{2} m {v}^{2} = {\left(m v\right)}^{2} / \left(2 m\right) = {p}^{2} / \left(2 m\right)$

The new kinetic energy $K ' = {\left(p '\right)}^{2} / \left(2 m\right)$

So $\frac{K '}{K} = \frac{p {'}^{2}}{p} ^ 2$

$\frac{K '}{K} = {\left(1.50 p\right)}^{2} / {p}^{2} = {1.50}^{2} / 1 = 2.25$

$K ' = 2.25 K$

% change = (K' – K)/K × 100 % = (2.25 K – K)/K × 100 % = 125 %#

% increase in kinetic energy = 125%