(1) m1v1 + m1v2 = (m1 + m2)* v3
(2) m1v1 * v1 + m1v2 * v2 = (m1 + m2) * v3 * v3
When rearrange for v3 :
(1) I get an answer that only works with this equation
(2) I get an answer that only works with this equation
Or solve simultaneously :
(2) / (1) I get an answer that doesn’t work for ether.
Thus I’ll come to think that I am not using the right equation, or is there some thing else I’m not doing right?
PS Currently my physics simulator can handle like charges colliding, and I want it to handle unlike charges colliding.
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Two charged particles collide
Author
I have two oppositely charged particles colliding in my physics simulator. My question is how do I determine the velocity of the resultant particle (after the particles have coalesced) what satisfies both convection of kinetic energy and momentum. I assume that there is a particle left over, and the two oppositely charged particles do not annihilate each other, and the resultant mass is just the sum of the masses.
When I Use :
Hrmm. I''m not the physics buff that I''d like to be, but I am a math buff (I''m a 3d graphics programmer, after all), and looking at the equations you put up, some stuff doesn''t click here. I don''t have my physics book here at work with me, though, so bear with me.
Ok, first off, you never use m2 on the left side. I don''t know if that was a typo or not, but being that you did it in both equations, I''d have to say that it might not be. Well, I don''t understand how that would work at all unless m1 == m2.
Secondly, just by glancing at it (and not thinking too deeply), I don''t see how those two equations could not contradict each other is when all of the v''s are 0 or 1 (or when v*v = v).
Solving 1) for v3 gives me v3 = (m1v1 + m2v2)/(m1 + m2).
Solving 2) for v3 gives me v3 = ((m1v2*v2 + m2v2*v2)/(m1 + m2))^(1/2)
The first equation, if I remember correctly, is the conservation of momentum. Guessing from your post, the 2nd is the conservation of kinetic energy, though if that is true, I don''t see how physics works at all!
Oh well, I''m going to go consult my physics book after work, and I''ll post what I come up with tomorrow. Good Luck!
-- Succinct
Ok, first off, you never use m2 on the left side. I don''t know if that was a typo or not, but being that you did it in both equations, I''d have to say that it might not be. Well, I don''t understand how that would work at all unless m1 == m2.
Secondly, just by glancing at it (and not thinking too deeply), I don''t see how those two equations could not contradict each other is when all of the v''s are 0 or 1 (or when v*v = v).
Solving 1) for v3 gives me v3 = (m1v1 + m2v2)/(m1 + m2).
Solving 2) for v3 gives me v3 = ((m1v2*v2 + m2v2*v2)/(m1 + m2))^(1/2)
The first equation, if I remember correctly, is the conservation of momentum. Guessing from your post, the 2nd is the conservation of kinetic energy, though if that is true, I don''t see how physics works at all!
Oh well, I''m going to go consult my physics book after work, and I''ll post what I come up with tomorrow. Good Luck!
-- Succinct
Two particles joining together is not an elastic collision. In fact, it is the way in which the most kinetic energy can be converted after a collision. Energy is always conserved, but KE is not. On the macro scale, it would be converted to heat. On the micro scale, some sort of radiation would be emitted.
To answer the question of where the particles go, all you need is to conserve momentum.
To answer the question of where the particles go, all you need is to conserve momentum.
Author
I knew something was iffy about those equations :/
That was a typo Succinct, and I managed to solve those equations by letting the resultant mass be unknown, which allowed me to get an anwser that satifies both equations. But after reading what Beer Hunter I realised that it was *NOT* an elastic collision and the equations I generated were invalid.
They were for when to particles collide, in a manor which conserves both kinetic energy and momentum.
The equations were :
These generated a result were the final mass was (normally) just less that the total initial mass...
That was a typo Succinct, and I managed to solve those equations by letting the resultant mass be unknown, which allowed me to get an anwser that satifies both equations. But after reading what Beer Hunter I realised that it was *NOT* an elastic collision and the equations I generated were invalid.
They were for when to particles collide, in a manor which conserves both kinetic energy and momentum.
The equations were :
v3 = (m1*v1*v1 + m2*v2*v2) / (m1*v1 + m2*v2)m3 = ((m1*v1 + m2*v2) ^ 2 )/ (m1*v1*v1 + m2*v2*v2) These generated a result were the final mass was (normally) just less that the total initial mass...
If they are charged particles, shouldn''t you take electro-static forces into account?
p_r is correct. You''ve got some potential energy (since the particles are attracting eachother from a distance) getting converted to kinetic energy as the particles speed up towards eachother.
Since they do, after all, accelerate towards eachother, electromagnetic waves will be generated. Depending on the scale of your simulation, this may or may not have an effect.
Of course, if you''re dealing with elementary particles, all bets are off and you need to refer to your quantum mechanics text.
Since they do, after all, accelerate towards eachother, electromagnetic waves will be generated. Depending on the scale of your simulation, this may or may not have an effect.
Of course, if you''re dealing with elementary particles, all bets are off and you need to refer to your quantum mechanics text.
depending on the interaction, you may have some mass being converted into energy (i.e. a gamma or two). also, particles don''t just stick together without some other sort of particle keeping them together (which will take some of the KE of the impact as well and affect somewhat the new mass). Keep in mind it has been a few years(about 5 or 6) since i''ve dealt with any of this stuff, so I only remember vague generalities.
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