# Homework Help: Double KE, what happens to velocity

1. Oct 21, 2004

### justinbaker

if you double the KE, by what factor does the speed increase? and vice versa

i am a little lost

2. Oct 21, 2004

### Pyrrhus

The speed will increase by $\sqrt(2)$

Use the equation

$$K = \frac{1}{2}mv^2$$

If the speed increases twice

$$K = \frac{1}{2}m(2v)^2$$

$$K = 4 (\frac{1}{2}mv^2)$$

$$K = 4K$$

Kinetic energy will increase four times.

Now if the kinetic energy increases four times

$$v = \sqrt{\frac{2K}{m}}$$

$$v =\sqrt{\frac{(4)2K}{m}}$$

$$v = \sqrt{4} \sqrt{\frac{2K}{m}}$$

$$v = 2 \sqrt{\frac{2K}{m}}$$

$$v = 2v$$

The speed will increase two times.

Last edited: Oct 22, 2004
3. Oct 22, 2004

### robphy

To clarify Cyclovenom's reply (i.e. to avoid statements like $$K = 4K$$, which implies that $$K=0$$), try this.

The relation between kinetic energy and speed is
$$K=\frac{1}{2}mv^2$$.

To double the speed, think $$v_\text{new}=2 v_\text{old}$$.
Since you're probably assuming that the mass is unchanged, $$m_\text{new}=m_\text{old}$$.

Then,
\begin{align*} K_\text{new}&=\frac{1}{2}m_\text{new}v_\text{new}^2\\ &=\frac{1}{2}(m_\text{old})({\color{red}2}v_\text{old})^2\\ &=\frac{1}{2}(m_\text{old}){\color{red}2^2}(v_\text{old})^2\\ &={\color{red}4}\frac{1}{2}(m_\text{old})(v_\text{old})^2\\ &={\color{red}4} K_\text{old}\\ \end{align*}