Distance of closest approach

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I solved this in the method based on the following assumptions.1)For closest distance their velocity along the line joining them must be zero.2)Momentum conservation along the x,y directions.3)Energy conservation.4)Conservation of angular momentum about the centre of mass.This yielded $100x^2-25x-4=0$ .where $(x=D/5L)$ .So the answer came out to be $89$ . Although conservation of momentum along x,y is quite apparent better method would be to conserve it along radial and tangential directions.

As usual, a two-body problem like this (i.e. with a potential depending on the relative position of the particles) can be cast as a problem of a single particle subject to a potential. One needs just to choose the centre-of-mass frame, in which the Lagrangian is the following: $\mathcal{L} = \frac{1}{2}\mu(\dot r^2 + r^2\dot\phi^2) - \frac{e^2}{2\pi\epsilon_0r}\;,$ where $r$ and $\phi$ are the polar coordinates of the relative position between the $\alpha$ -particle and the proton and $\mu \equiv \frac{mm_\alpha}{m + m_\alpha} = \frac{4}{5}m\;$ is the reduced mass.

Now, it is not difficult obtain the Euler-Lagrange equations of motion and to integrate them once, obtaining: $E = \frac{2}{5}m\dot r^2 + \frac{5l^2}{8mr^2} + \frac{e^2}{2\pi\epsilon_0r}\;,$ where $E$ is the energy and $l \equiv \mu r^2\dot\phi\;,$ is the angular momentum. Both $E$ and $l$ are constants of the motion.

Now we use the data that the problem provide. The energy is given by the sum of the kinetic energies of the particles at infinity: $E = \frac{2}{5}m\dot r^2 = \frac{2}{5}m4v^2 = \frac{2e^2}{5\pi\epsilon_0L}\;,$ where recall that the relative velocity is twice $v$ .

Similarly, the angular momentum is given by $l = \mu L 2v = \frac{8}{5}mLv\;.$

Now, we substitute these result in the equation we found above for the energy. The closest approach distance is given by the condition $\dot r = 0$ , since this is a turning point. In other words, the relative position decreases from infinity to $D$ as the particles approach and then starts to increase again.

Therefore, we have, after some algebra: $\frac{D^2}{L^2} - \frac{5D}{4L} - 1 =0\;.$ In order for the $x$ defined in the problem to appear, we must multiply by 64 and add 25. This leaves the result: $\boxed{x = 89}$