TITLE: two identical point charges can't collide
QUESTION [6 upvotes]: I've convinced myself intuitively that if you place two massless classical particles with the same charge in $\mathbb{R}^n$, with arbitrary initial velocities and (distinct) positions, they will never collide. However, I'm have a heck of a time trying to prove it, and would appreciate some help.
Formally, consider $q_1, q_2: \mathbb{R} \rightarrow \mathbb{R}^n$ satisfying $$\ddot{q_i} = \frac{1}{\|q_i - q_j\|^3} (q_i - q_j)$$With $q_1(0) \neq q_2(0)$. The claim is that $q_1(t) \neq q_2(t)$ for all $t > 0$. 
So my questions are (i) is this true? (ii) what happens if we replace the exponent 3 in the denominator with say $\alpha > 0$ ?
N.B. The question's already a bit long, but I'd be happy to post my thoughts so far. 
Edit All the answers were very helpful, thanks so much everyone!

REPLY [2 votes]: This is as good a place as any to mention how you can derive conservation of energy from scratch, starting with nothing but a differential equation for $\ddot q_i$... as long as your force term is conservative, that is, it is the negative gradient of a scalar "potential energy" function. (I'll deal only with a one-particle system, but you can handle multiple particles simply by packing in all the position variables $q_1, q_2, \ldots, q_m$ into a single vector in $\mathbb R^{mn}$.)
Consider $\ddot q=f(q)$ where $f$ is conservative, i.e. $f(q) = -\frac{\mathrm d}{\mathrm dq} U(q)$ for some scalar-valued potential $U$. Introduce a momentum variable $p=\dot q$ so that $\dot p = f(q).$ Observe that $p = \frac{\mathrm d}{\mathrm dp} T(p)$ where $T(p) = \frac12\lVert p\rVert^2$. Define the energy function $H(q,p) = U(q) + T(p)$, and observe that
$$\dot H(q,p) = \frac{\partial H}{\partial q}\cdot\dot q + \frac{\partial H}{\partial p}\cdot\dot p = -\dot p\cdot\dot q + \dot q\cdot\dot p = 0,$$
so $H$ is constant over time for any solution.
For your problem, your $U(q_1,q_2)$ will depend only on $\lVert q_1-q_2\rVert$, and you'll want to check whether $H$ is infinite in a colliding configuration. I think you need $\alpha>1$ for that to happen.