Velocity From Electric Potential

In summary, the conversation discusses the concept of energy distribution between two particles and the use of ratios to calculate the energy of each particle. The energy ratios can be calculated using conservation of momentum and the total energy can be calculated by combining the energy of each particle. The concept of ratios and proportions is also mentioned in relation to this problem.
  • #1
PurelyPhysical
31
0

Homework Statement



http://imgur.com/UsKsaOn

Homework Equations



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

The Attempt at a Solution


[/B]
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.
 
Physics news on Phys.org
  • #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.
 
  • Like
Likes PurelyPhysical
  • #3
Divya Shyam Singh said:
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.

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
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:
 
  • Like
Likes PurelyPhysical
  • #5
PurelyPhysical said:
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?
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.
 
  • #6
Divya Shyam Singh said:
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:

This does help, thank you very much :D
 

1. What is velocity from electric potential?

Velocity from electric potential is the measure of the speed at which charged particles move in an electric field. It is directly related to the electric potential, which is the amount of electric potential energy per unit charge at a specific point in the electric field.

2. How is velocity from electric potential calculated?

Velocity from electric potential can be calculated using the formula v = √(2qV/m), where v is the velocity, q is the charge of the particle, V is the electric potential, and m is the mass of the particle.

3. Can velocity from electric potential be negative?

Yes, velocity from electric potential can be negative. A negative velocity indicates that the charged particle is moving in the direction opposite to the direction of the electric field.

4. What factors affect velocity from electric potential?

The velocity from electric potential is affected by the strength of the electric field, the charge and mass of the particle, and the distance between the charged particles.

5. How is velocity from electric potential related to kinetic energy?

Velocity from electric potential and kinetic energy are closely related. As the velocity increases, the kinetic energy of the charged particle also increases, and vice versa. This relationship is described by the equation K = ½mv², where K is the kinetic energy, m is the mass of the particle, and v is the velocity.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
162
  • Introductory Physics Homework Help
Replies
2
Views
365
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
2
Replies
64
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
349
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top