Imagining zero orbital angular momentum for s-orbitals Orbital Angular momentum of a s-orbital is always zero. One can easily imagine why this is so: QM says $\hat{p}=-i\hbar \nabla_{r}$, and since the s-wave functions are radially symmetric, the momentum $\bar{p}$ always points radially out, i.e in the direction of the concerned $\bar{r}$. Hence $\hat{L}=\hat{r}\times\hat{p}=0$.
Now, 1) Is my argument sensible ? 2) If, yes electrons in s-orbital appear to move only in radial direction and any other place they want to be, they have to magically (Quantum mechanically) appear at some other point. What is it in the structure of QM which prohibits the electron from straight forward collision into the nucleus ? Because it appears there is a finite probability of it whizzing through a radius vector $=-\bar{r}$ straight into $\bar{r}=0$!  What am I missing here ?
 A: Any time we try to invoke the classical picture of a pointlike electron to help "picture" a quantum state, we are asking for trouble. It just doesn't work. Quantum physics is more fundamental. Classical physics — when it works — is derived from quantum physics as an approximation. The picture of a particle as a tightly-localized thing moving around in space works well enough under the right circumstances, but an electron in an s-orbital is not one of those circumstances.
An s-orbital in a single-electron atom is an example of a stationary state. Ignoring the atom's center-of-mass motion, this is a state that does not change in time at all. It is completely static. The electron is not moving. It is spread out in space according to the wavefunction 
$$
\psi(\mathbf{x})\propto \exp\left(-\frac{|\mathbf{x}|}{a}\right),
$$
where $a$ is a constant. It simply doesn't have a sharply-defined location, and it doesn't have a sharply-defined linear momentum, either. 
What the electron does have, in this particular case, is a sharply defined energy. (This statement again ignores the center-of-mass dynamics of the whole atom.) States with sharply defined energy don't have any moving parts. They are completely stationary, and this is related to the fact that the energy operator (the Hamiltonian) is the operator that generates translations in time.
As stated in the question, the s-orbital also has a sharply defined angular momentum. (Again, ignoring the center-of-mass dynamics of the atom.) In terms of components, the angular momentum operators are 
$$
  L_{jk}\propto i(x_j\nabla_k-x_k\nabla_j).
$$
The identity $L_{jk}|\mathbf{x}|=0$ implies
$$
  L_{jk}\exp\left(-\frac{|\mathbf{x}|}{a}\right) = 0,
$$
so the s-orbital has zero angular momentum. The classical picture of angular momentum being expressed in terms of the cross-product of a sharply-defined location with a sharply-defined momentum does not apply in quantum physics, because a particle cannot have both of these qualities in quantum physics (the "uncertainty principle"). 
For the same reason, saying that the momentum operator always points radially out in the s-orbital is incorrect. The momentum operator isn't associated with any location. A particle cannot have both a sharply-defined location and (along the same direction) a sharply-defined momentum. If we pick a point $\mathbf{x}$ in space, in a coordinate system centered on the nucleus, and assume that the electron's momentum has a sharply-defined direction proportional to $\mathbf{x}$, then the uncertainty principle implies the electron has a completely undefined location transverse to that direction. The idea of a "radially directed momentum" assumes that the electron's transverse location and transverse momentum can both be sharply defined; but they can't.

What is it in the structure of QM which prohibits the electron from straight forward collision into the nucleus?

In QM, the s-orbital is a stationary state in which the electron is spread out in space. The electron does have some "presence" within the tiny space occupied by the nucleus, and in this sense, "collision" with the nucleus is not prevented in QM. The s-orbital solution takes all of this into account, and the result is a stationary state.
Trying to make sense of these statements in terms of a classical picture of the electron does not work, and it's not necessary. Quantum theory gives us a new way to think about things. It takes a while to get used to, but it is in excellent agreement with all of our experience, including all of our experience with ordinary macroscopic objects that might as well have sharply-defined locations and momenta, because the limits imposed by the uncertainty principle are utterly negligible for macroscopic objects. They are not negligible at all for atoms.
A: Yes, $s$ electrons have a contact density on the nucleus. This is responsible for hyperfine effects that are seen in nuclear magnetic resonance spectroscopy (NMR) and in the Mößbauer effect. No "collisions" though.
In neutron physics, there are "collisions". Thermal neutrons have low momentum and their angular momentum can be smaller than $\hbar$ when whizzing past a nucleus. In some nuclei like cadmium this leads to cross sections that much larger than the typical size of a nucleus: $s$-wave capture.
For bound states, one can draw elliptical Sommerfeld orbits.
A: I’m not sure what you mean when stating $s$-wave functions are radially symmetric: the angular part of the orbital is spherically symmetric and there is no “direction” to a wavefunction, i.e, the momentum does not point in any direction no more than there is a momentum direction in a 1d wavefunction.  
In fact, the probability density for a solution of the time-independent Schrodinger equation is time-independent, and is a scalar quantity without a direction. 
Actually, $\hat p_r=-i\hbar \nabla_r$ is not quite correct either as you might gather from answers to this question: the problem comes (in part) from the radial variable, which must be strictly non-negative.
So in short, your argument is not very sensible because of the difficulties with the radial $\hat p_r$.  It does make sense that something that is spherically symmetric like the probability density of an $s$-wavefunction have $0$ average angular momentum since the angular momentum about any direction would be averaged out by the momentum about the opposite direction: by going to Cartesian coordinates $\langle x\rangle=\langle y\rangle=\langle z\rangle=0$ by spherical symmetry so that any $\hat L_k$ also averages out to $0$.
