# What happened to momentum if kinetic energy increase 3time?

Feb 21, 2018

Momentum becomes ${\left(3\right)}^{\frac{1}{2}}$ times the initial momentum given that the Mass of the object is constant.

#### Explanation:

$K {E}_{i} = \left(\frac{1}{2}\right) . m . {v}^{2}$ and ${\vec{P}}_{i} = m \vec{v}$

$K {E}_{f} = 3 K {E}_{i} = 3 \left(\frac{1}{2}\right) . m . {v}^{2}$

$\Rightarrow K {E}_{f} = \left(\frac{1}{2}\right) . m . {\left(v '\right)}^{2}$ where $v ' = {\left(3\right)}^{\frac{1}{2}} v$

$\Rightarrow {\vec{P}}_{f} = m \vec{v} ' = m {\left(3\right)}^{\frac{1}{2}} \vec{v} = {\left(3\right)}^{\frac{1}{2}} m \vec{v}$

$\therefore {\vec{P}}_{f} = {\left(3\right)}^{\frac{1}{2}} {\vec{P}}_{i}$

Feb 21, 2018

$K E = \frac{1}{2} m {v}^{2}$ and $P = m \setminus \Delta v$

#### Explanation:

So, if the kinetic energy increases 3 times (triples) that means that the velocity must have increased by $\sqrt{3}$

If you start with mass m and velocity v , as noted above $K E = \frac{1}{2} m {v}^{2}$

So, if the velocity increased by a factor of $\sqrt{3}$, the new velocity is $\sqrt{3} v$ and the new kinetic energy is $K E = \frac{1}{2} m {\left(\sqrt{3} v\right)}^{2}$
which is $K E = \frac{1}{2} m \cdot 3 \cdot {v}^{2} \to 3 K E = \frac{1}{2} m {v}^{2}$