The quantum wavepacket spreading is underlain by the dispersiveness of the free
Schrödinger equation, which is basically a diffusion equation of sorts. For simplicity, let's consider an electron coming from outer space, and stick to one dimension, with m and ħ set equal to one—we'll reinstate them later. Also, let's not use "collapse" (reserved for observation) to denote wavepacket spreading.
In momentum space, there is no counterintuitive stuff: momenta are conserved and the wave packet profile in momentum space preserves its shape upon propagation: no forces.
Restricting attention to one dimension, the solution to the Schrödinger equation satisfying the Gaussian initial condition starting at the origin with minimal space uncertainty,
$$
u(x,0) =\int \frac{dk}{2\sqrt{\pi}}e^{ikx}e^{-(k-\bar{k})^2/4}= e^{−x^2+i\bar{k} x},
$$
is seen to be
$$\begin{align}
u(x,t) &= \frac{1}{\sqrt{1 + 2it}} e^{-\frac{1}{4}\bar{k}^2} ~ e^{-\frac{1}{1 + 2it}\left(x - \frac{i\bar{k}}{2}\right)^2}\\
&= \frac{1}{\sqrt{1 + 2it}} e^{-\frac{1}{1 + 4t^2}(x - \bar{k}t)^2}~ e^{i \frac{1}{1 + 4t^2}\left((\bar{k} + 2tx)x - \frac{1}{2}t\bar{k}^2\right)} \\
&= e^{i\bar{k}x-it\bar{k}^2/2}~~\frac{e^{-\frac{1-2it}{1 + 4t^2}(x - \bar{k}t)^2}~
}{\sqrt{1 + 2it}} ~.
\end{align} $$
The leading part is the plane wave corresponding to the "center" of the wave packet, and the trailing part contains the real exponent which demarcates the envelope, so the probability density $|u|^2\sqrt{2/\pi}$ propagates "classically" with group velocity $\bar k$, as it rapidly spreads,
$$|u(x,t)|^2 = \frac{1}{\sqrt{1+4t^2}}~e^{-\frac{2(x-\bar{k}t)^2}{1+4t^2}}~. $$
A bit too rapidly...
The width here, $\sqrt{1+4t^2} \to 2t$, after reinstating the naturalized constants, amounts to
$$\Delta x=A\sqrt{1+(\hbar t/mA^2)^2},$$ for A the initial width. If we took it to be ångströms, plugging in values for the electron mass, we see a spread to kilometers in a milisec. That is, the normalized probability envelope above has all-but dissolved to a delocalized flapjack, and the electron consists of plane wave components, which is why distant cosmic rays are modeled by plane waves. (1. Yes, multi-multi-multiple kilometers, before detection.)
The electron has not disappeared to nothingness (2. It's all there, so it will keep coming and be detected, ultimately, with the same probability), it is that its precise location has quantum-diffused to all over the place. With some probability, these electrons will come to your detector, normally modeled by plane waves, and will have mostly momenta/velocities close to $\bar k$, as posited; this distribution has not changed. The consequent spread in detection times, non-relativistically, will be
$$\Delta t = \frac{\Delta x} {\bar{v}}= \frac{t}{A\bar{k}}~. $$
Where did they come from (in x; the momenta in 3D specify the direction)? Who knows.
Quantification may be fiendishly conducive to errors by dozens of orders of magnitude, however. See Tzara 1988 for reconciliation/reassurance that the short ultra relativistic neutrino burst
of the SN1987A supernova did not violate the above standard wavepacket spreading notions, as erroneously claimed: one must compute the spreading in the wavepacket's rest frame and then transform to the terrestrial detector frame! Phew...
Numbers : Prompted by the question, let's summarize the basic numerics of the spread. Define a characteristic time
$$
\tau=A^2 \frac{m }{\hbar} , \qquad \Longrightarrow \qquad
\Delta x= A \frac{t}{\tau} .
$$
For an electron and A ~ ångström, $\tau= 10^{-16} s$, whence the above spread to a kilometer in miliseconds. But for an iron nucleus and A in microns, we have, instead, $\tau= 10^{-7}s$. This would amount to only an expansion to 10m in a second. Can you estimate the ages required to expand the probabilistic size of a quantum basketball by such a factor of 10?
