Sorry to come back to this so late, but I only recently
got around to working on it. I think the discussion left
the topic incomplete.
To restate the problem: A spring is held vertically by
the upper end, and released. What is the acceleration
of the highest infinitesimal part ?
Consider a spring of mass M, that stretches by dL,
modelled as N masses separated by N-1 springs
of spring constant k.
The uppermost mass, of size M/N supports the load
(N-1)Mg/N when held in equilibrium.
When released a = [(N-1)Mg/N] / (M/N) = (N-1)g,
so that a gets very large as N is increased.
However, if we consider the time response, we
get a finite impulse.
Calculate k (the i'the spring supports the load iM/N),
and the total stretch is dL, so k = (N-1) N M g / (2 dL).
Taking the period of SHM for the topmost mass as a
measure of the time response T = 2 pi sqrt( M/(Nk))
so T =2 pi sqrt( 2 dL/ ( N (N-1) g ) )
Taking delta V = a T ~ sqrt( (N-1)/N ),
which clearly goes to 1 as N is increased.
Moreover, the momentum transfer ~ (M/N) delta V ~ 1/N
which goes to 0 as N increases.
Two more comments:
1. The same model applies to a hanging bar of material.
2. Real materials or springs are not infinitely divisible
into uniform entities.