How does conservation of momentum work here? Imagine 2 oppositely charged particles attracted to each other, say an electron and a nucleus, assume their force of attraction is very large compared with any external applied force we're about to apply here:
If some particle with mass $m$ that has a velocity $ \vec v$ towards the electron and about to collide with it, the electric force between the electron and nucleus is very large so the electron's momentum is not changed, but the particle with mass $m$ loses energy and thus loses momentum, where did that lost energy go? Sorry if it's too basic, I'm trying to retrieve some classical mechanics basics.
 A: There is some confusion here:

say an electron and a nucleus,

electrons and nuclei are quantum mechanical entities, not classical mechanical.
An electron, attracted to the nucleus falls  into  a quantum mechanical bound state, as the electron and the proton forming hydrogen here, a neutral entity. 
Thus 

If some particle with mass m that has a velocity v⃗  towards the electron

The scatter has to happen with the whole neutral atom, that has a mass M , not with individual constituents, by construction of the atom.

and about to collide with it, 

One has a scattering problem, m hitting atom.

the electric force between the electron and nucleus is very large so the electron's momentum is not changed,

forget it , this is wrong, the scatter will happen with the whole atom.

but the particle with mass m loses energy and thus loses momentum, where did that lost energy go? 

It is transferred to a motion of the whole atom, a two body scatter, conserving energy and momentum.
Now classically the problem is the same : if you have a +charge lump close to a -charged lump, they will be attracted and become one neutral body, if the charges are equal, or a charged one with the difference between the charges. The mass m hitting the neutral body will transfer part of its energy to the neutral body as kinetic energy.
A: You're using classical mechanics to model a very small system. Classical mechanics breaks down on very small scales. That doesn't mean we can't use classical mechanics to model small systems. We just have to be very careful with them. It's a case of Pablo Picasso's quip "Learn the rules like a pro, so you can break them like an artist."
The answer to "Where does the lost energy go?" is "It depends". One answer is "into kinetic energy". Think of your atom an elastic ball. We can do this because we assume the nucleus-electron "force of attraction is very large compared with any external applied force we're about to apply here". If a particle collides with your elastic ball then your elastic ball will go shooting off in some direction. In this case, the electron's momentum is not changed in the atom's reference frame but the electron's momentum is changed in our laboratory reference frame. Thus, the entire atom acquires kinetic energy. Less than 0.1% of the atom's kinetic energy comes from the atom's electrons and more than 99.9% of the atom's kinetic energy comes from the nucleus's new momentum.
Quantum mechanically, there are many other places for the energy to go, like knocking the electron into an excited orbital. I'm skipping that here to keep things simple.
I think your misunderstanding comes from overlooking two things. The first mistake is forgetting that wherever the electron goes so does the nucleus. (If it doesn't then you don't have an atom anymore.) The second mistake is treating the electron's momentum as insignificant. It's not. If the particle with mass $m$ loses energy from a collision with an electron (and no new particles are released in the collusion) then all of the energy lost by the incident particle is transfered to the electron. Yes, the electron has small mass but the electron receives a cooresponding LARGE velocity, sufficient to counteract the small mass. If this energy is significant then the electron now has significant momentum.
A: When the particle will hit the electron if the coulombs force is very strong, the particle will retain its speed in opposite dirn(like a ball hitting a wall) and the electron will start to oscillate in SHM . hence the net energy will remain constant.
for your question the energy lost = energy to start SHM for electron
