Here is my question
A proton (mass $\text {m}$, charge $+e$) and an alpha particle (mass $4\text {m}$, charge $+2e$) approach one another with the same initial speed $\text {v}$ from an initially large distance. How close will these two particles get to one another before turning around?
The kinetic energy of the moving particles is completely transformed into electric potential energy at the point of closest approach.
$$U_e = K$$
I want to get your helps If that's possible.
On
Read up on elastic collision. When particles collide, total momentum and total kinetic energy are both conserved. Easy equations to work with. You will have two more equations to work with from this setup.
I think you will find your scenario in this article where they have worked on it. It is not as complex as it looks just read through it.
infinitylord's solution for the two equations is almost correct. The right-hand side of the equation from conservation of momentum is wrong. Come up with your own.
P.S. In reality, it all depends on the velocity. If they are traveling too fast(close to the speed of light, their mass will start to change as well then you will need more equations) since you have Hadron Collider experiment going on.
I'm a little rusty on electrodynamics, but I believe you'll need both conservation of energy and conservation of momentum. I'm going to assume this is non-relativistic, as you made no mention of that.
At their point of closest approach, you will have that $v_{\alpha} = v_{p} $
From conservaton of momentum:
$$(m_p - m_{\alpha})v = (m_p + m_{\alpha})v_c$$ So that the velocity at closest approach is: $$v_c = \frac{m_p - m_{\alpha}}{m_p + m_{\alpha}}v $$ Now, convservation of energy tells you: $$K_i + U_i = K_c + U_c $$ $U_i = 0$, as they are infinitely far apart. You have $K_i $ and $K_c $, because you know the velocities and mass at each point. Now just expand the electric potential energy and solve for the distance.