Can mv^2 /2 + mu^2 / 2 + kq^2 / L = kq^2 /X? Find Out!

[A13235378]

Homework Statement

Two identical particles with velocities u and v, which form angles α and β with the straight line that joins them, are at a distance l from each other. The charge of each particle is q. Determine the mass of the particles, knowing that the minimum distance between the two is X. Despise the gravitational interaction.

Relevant Equations

mv^2 /2
kq^2 /d

I can only say that:

mv^2 /2 + mu^2 / 2 + kq^2 / L = kq^2 /X

Yes or not and why?

 

[Vanadium 50]

Whatever translation software you are using isn't very good. "knowing that the minimum distance between the two is worth" sounds like a critical part of the problem, and I have no idea what that means. As for "Despise the gravitational interaction", I already do. We can't help you if we don't understand you.

 

[A13235378]

Whatever translation software you are using isn't very good. "knowing that the minimum distance between the two is worth" sounds like a critical part of the problem, and I have no idea what that means. As for "Despise the gravitational interaction", I already do. We can't help you if we don't understand you.

Sorry , I am spanish , my english is not very good. I already correct.

 

[berkeman]

Sorry , I am spanish , my english is not very good. I already correct.

And "Despise the gravitational interaction" means to ignore the gravitational attraction between the masses, right?

(I also edited your thread title for you to make it a bit clearer)

 

[A13235378]

And "Despise the gravitational interaction" means to ignore the gravitational attraction between the masses, right?

(I also edited your thread title for you to make it a bit clearer)

yeah

 

[TSny]

Are you familiar with reducing a two-body problem with a central force to a one-body problem with a reduced mass $\mu$? For example, see this .

It seems to me that you will be able to find only the reduced mass. You won't be able to find the individual masses of the two particles.

 

[haruspex]

mv^2 /2 + mu^2 / 2 + kq^2 / L = kq^2 /X

That would be true if they were stationary at closest approach. Will that be so?

 

[hutchphd]

It seems to me that you will be able to find only the reduced mass. You won't be able to find the individual masses of the two particles.

It says they are identical...

 

[Vanadium 50]

Maybe we should wait for a problem attempt.

 

[berkeman]

Maybe we should wait for a problem attempt.

But I was about to post my Excel simulation as a cross-check for the OP to check their work. Oh, wait...