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Question:

A large number (n) of identical beads, each of mass m and radius r are strung on a thin smooth rigid horizontal rod of length L (L >> r) and are at rest at random positions. The rod is mounted between two rigid supports. If one of the beads is now given a speed v, the average force experienced by each support after a long time is (assume all collisions are elastic)

mv²/L−nr

mv²/2(L−nr)

zero

mv²/L−nr

Solution:

Total length available for collision is L−nr
Assuming the collision be between beads be elastic, each bead comes to rest after collision and the collided beads moves with speed v.
Average time for a collision is
tavg = (L−nr)/v
since a bead will collide with all other beads before returning to the same state (position and velocity) from where it started.
The collision with a support will transfer momentum mv to the support.
The average force experienced by each support is
F = ΔP/ΔT = mv/tavg = mv/(L−nr)/v = mv²/L−nr