Two spheres (A physics olympiad problem) Browsing an archive of problems of a local physics olympiad, i stumbled upon a problem which seems not a very trivial. 
Given two identical metal spheres in vacuum, with mass $m$ and radius $R$. One sphere with a charge $Q$ and other with no charge(neutral). At initial moment, they are very far from each other. After releasing, due to the electrostatic attraction, the spheres come together and collide. Find the velocity of the spheres after the collision if the collision is:
a) perfectly elastic
b) inelastic
Ignore gravity forces and a possible spark discharge.
The part b) seems especially tricky.
Edit: 
Yes, there was probably a typo. The part b) must be simply "inelastic collision" that means kinetic energy is not conserved and the spheres do not stick. Sorry for confusion!
 A: This is an edited answer following the remark by Qmechanic pointing out that the capacitance used in the original answer assumes electrical contact between the spheres.
Here I assume that the actual contact time is very small, so no discharge occurs. On, the other hand it is assumed that the motion is slow enough that the laws of electrostatics apply. The capacitance formulas are taken from the article: "Capacitance coefficients of two spheres" by John Lekner. The same notation is used, where the charged sphere is denoted with the subscript $a$ and the other sphere with the subscript $b$. Also CGS units are used (The capacitance has units of length).
Substituting $ Q_a = Q$, $Q_b = 0$, we obtain the following formula for the effective capacitance:
$$ C_s \equiv \frac{V_a}{Q} = \frac{C_{aa}C_{bb}-C_{ab}^2}{C_{bb}},$$ 
where $s$ is the separation distance. The expressions for the capacitance matrix elements at infinite separations $s=\infty$ are given in the text eq. (8) resulting the following result:
$$ C_{\infty} = R.$$
Of course, this is just the capacitance of the charged sphere.
The capacitance matrix elements become singular when the spheres approach contact $s\to 0$, but using the asymptotic formulas (16)-(18), the effective capacitance has a well defined limit:
$$C_0 = \lim_{u \to 0}\frac{R}{2} \frac{(\ln(\frac{2}{u})-\psi(\frac{1}{2}))^2-(\ln(\frac{2}{u})+\gamma)^2}{(\ln(\frac{2}{u})-\psi(\frac{1}{2}))}= -R (\psi(\frac{1}{2})+\gamma) = 2 R \ln 2,$$
where $\gamma$ is the Euler constant and $\psi$ is the digamma function, and the following identity was used
$$\gamma +\psi(\frac{1}{2}) = -2 \ln 2.$$
By energy conservation
$$2 \frac{m v^2}{2} + \frac{Q^2}{C_0}=  \frac{Q^2}{C_{\infty}}.$$
We obtain:
$$v = \sqrt{\frac{Q^2 (2\ln 2-1)}{2m R \ln 2}},$$
which is the speed just after collision in the case of an elastic collision. Of course, the speed is zero in the case (b) of inelastic collision.
A: There is an electrostatic attraction between the two spheres even though one has a net zero charge. On the side of the uncharged sphere that faces toward the (say) positively charged sphere, there will be felt a strong electromagnetic field arising from the charged sphere. This field will attract conduction band electrons and cause them to rush to the side facing the charged sphere. As a result, the far side of the uncharged sphere will have a shortage of electrons thus will be positively charged.
