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Velocity From Electric Potential

  1. Jul 19, 2016 #1
    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. jcsd
  3. Jul 19, 2016 #2
    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.
     
  4. Jul 19, 2016 #3
    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?
     
  5. Jul 19, 2016 #4
    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! :biggrin:
     
  6. Jul 20, 2016 #5

    collinsmark

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    Homework Helper
    Gold Member

    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.

    [itex]m_1 \vec {v_1} + m_2 \vec {v_2} = 0. [/itex]

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

    [itex] m_1 v_1 - m_2 v_2 = 0. [/itex] [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 [itex] m_1 v_1 + m_2 v_2 = 0. [/itex] That's arguably the better approach anyway.]

    Substituting [itex] m_2 = 2m_1 [/itex] you can calculate a simple relationship between [itex] v_1 [/itex] and [itex] v_2 [/itex] (you can do that for yourself :wink:).

    Note that the energy for each particle is

    [itex] E_1 = \frac{1}{2}m_1 v_1^2, [/itex]

    [itex] E_2 = \frac{1}{2}m_2 v_2^2, [/itex]

    And the total energy is,

    [itex] E_T = E_1 + E_2, [/itex]

    if you make the appropriate substitutions, you'll find the energy ratios are 2/3 and 1/3 of the total.
     
  7. Aug 7, 2016 #6
    This does help, thank you very much :D
     
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