# Question #db371

May 5, 2016

56.25%

#### Explanation:

Momentum is defined as mass times velocity, or $p = m v$

Kinetic energy is given by $K E = \frac{1}{2} m {v}^{2}$

We can also write KE as
$K E = \frac{1}{2} m {v}^{2} = \frac{1}{2} {\left(m v\right)}^{2} / m = \frac{1}{2} {p}^{2} / m$

So if we take the above equations as the position before the increase, then increase momentum by 25%:

$p = 1.25 m v$
$K E = \frac{1}{2} {\left(1.25 p\right)}^{2} / m = \frac{1}{2} \cdot {p}^{2} / m \cdot 1.5625$

So KE will increase by 56.25%