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

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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
1596743358220.png
I can only say that:

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

Yes or not and why?
 
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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. :wink: We can't help you if we don't understand you.
 
Vanadium 50 said:
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. :wink: We can't help you if we don't understand you.

Sorry , I am spanish , my english is not very good. I already correct.
 
A13235378 said:
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) :smile:
 
berkeman said:
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) :smile:
yeah
 
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.
 
A13235378 said:
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?
 
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TSny said:
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...
 
Maybe we should wait for a problem attempt.
 
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Vanadium 50 said:
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... :smile:
 
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