What is the physical intuition behind the Bragg peak? The wikipedia page says:

Energy lost by charged particles is inversely proportional to the
  square of their velocity, which explains the peak occurring just
  before the particle comes to a complete stop.

What is the physical reason why the energy loss goes inversely as the square of the velocity?
 A: See if this argument works - I am making this up on the spot so there is definitely space for argument...
Most of the interactions with the electrons will not be "head-on collisions" but rather electrostatic interactions. If we get to a certain distance of an electron, it will feel the force and undergo acceleration. If the time of the interaction is short, the distance moved will be small and the force will be approximately constant. 
At any given time, the momentum transfer will be proportional to the force:
$$\delta p = F \delta t$$
Now the force is given by the distance between the particle and the electron; while this is going to introduce all kinds of geometric terms we will ignore that for now.
Expressing time in terms of velocity and distance we get
$$\Delta p = F \frac{\Delta x}{v}$$
Energy transferred to the particle goes as momentum squared - it follows that energy lost per unit distance goes as the inverse of velocity.
One way to think about this intuitively  - the longer the particle spends near a given electron, the more time it has to exchange energy (given that all these are electrostatic interactions - not "collisions").
I found a nice derivation of the above that mostly agrees...
