# Homework Help: Velocity From Electric Potential

1. Jul 19, 2016

### PurelyPhysical

1. The problem statement, all variables and given/known data

http://imgur.com/UsKsaOn

2. Relevant equations

Why is the answer in joules multiplied by 2/3 and 1/2? I can follow the rest of the problem.

3. The attempt at a solution

I get the same solution as my teacher if I use the constants 2/3 and 1/2. I just don't understand where they are coming from.

2. Jul 19, 2016

### Divya Shyam Singh

I would want to correct the information you have given by looking at the solved equation.
The ratio of distribution of engergy for particle 1 and particle 2 is 2:1
So the energy of particle 1 will be: 2/(2+1)= 2/3 times the total energy
Energy of particle 2: 1/(1+2)= 1/3 times the total energy.
This "total energy" is basically the change in the electric potential as the particles move from being 0.1m apart to 0.2m apart.

3. Jul 19, 2016

### PurelyPhysical

Thank you very much! This clears it up for me. It never occurred to me to look at ratios that way. Is there a name for what this particular concept is in math?

4. Jul 19, 2016

### Divya Shyam Singh

Ummm....not really. It is generally covered in the topic of ratios and proportions.
I will explain a bit more here for you to have a better understanding:
Suppose a line segment of length A units is to be divided into a ratio of x:y
Then the length of each unit can be calculated as:
Ax/(x+y)
Similarly, the length of the other part will be:
Ay/(x+y)

To check we can easily see that the sum of each of the part of the line segment should be equal to the total length of the line segment.
so
Ax/(x+y) + Ay(x+y)= A

Hope it helped!

5. Jul 20, 2016

### collinsmark

The energy ratios can be calculated by invoking conservation of momentum.

The total momentum of the system starts out as 0 and ends as 0.

$m_1 \vec {v_1} + m_2 \vec {v_2} = 0.$

Placing the particles on a line and noting that they move in opposite directions, we can get rid of the vector notation.

$m_1 v_1 - m_2 v_2 = 0.$ [Edit: I'm just using the magnitudes of the velocities in this equation. If you'd rather allow negative velocities to indicate direction, then use $m_1 v_1 + m_2 v_2 = 0.$ That's arguably the better approach anyway.]

Substituting $m_2 = 2m_1$ you can calculate a simple relationship between $v_1$ and $v_2$ (you can do that for yourself ).

Note that the energy for each particle is

$E_1 = \frac{1}{2}m_1 v_1^2,$

$E_2 = \frac{1}{2}m_2 v_2^2,$

And the total energy is,

$E_T = E_1 + E_2,$

if you make the appropriate substitutions, you'll find the energy ratios are 2/3 and 1/3 of the total.

6. Aug 7, 2016

### PurelyPhysical

This does help, thank you very much :D