Creative Beading

collisions..external force?

Beads of equal mass are strung at equal distances on a long horizontal wire. The beads are initially at rest but can move without friction.
—>F
—O—O—O—-O—O—–…….
One of the beads is continuously accelerated (towards the right by a constant force F. what are he speeds of the accelerated bead and the front of the ‘shock wave’, after a long time, if the collisions of the beads are:
(i) completely inelastic,
(ii) perfectly elastic?

(i) The acceleration of the first bead will be F/m, where m is the bead’s mass. But after colliding (inelastically) with the second bead, the acceleration will be F/2m. After the third bead, the acceleration will be F/3m.
Whether the resulting velocity converges to a specific value or diverges to infinity depends, I think, on the relative values of F and m. But I’ve long since forgotten whatever I knew about series.

(ii) The acceleration of the first bead is still F/m, but when it strikes bead 2 it transfers ALL of its energy to it (i believe, since “perfectly elastic”). It stops, but begins accelerating again at F/m. Meanwhile bead 2, traveling at v = at = Ft/m (where t is the amount of time it took to reach bead 1 to reach bead 2), strikes bead 3, transferring all of it’s Ek to it. It stops. It is struck by bead 1, which has once again reached v = Ft/m. Meanwhile, bead 3 has struck bead 4, imparting velocity Ft/m to it. … bead 4 strikes bead 5… 3 strikes 4… 2 strikes 3… 1 strikes 2…
So if after “a long time” there are still beads that have not been hit, it seems to me that the velocity of the shock wave is = Ft/m with F, t and m as defined above.
This might be better expressed in terms of the inter-bead spacing x, obtained via
x = ½at² = ½(F/m)t²
t = sqrt(2mx/F)
so
v = F•sqrt(2mx/F)/m = sqrt(2xF/m)
imo

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