Average time for 1 collision to occur on 1 wall,
=tavg=speedaveragedistancetravelled=v2(L−2nr)
⇒tavg=v2(L−2nr). Change in momentum in 1 collision =2mv. Average force on each support, favg=tavg2mv=2(L−2nr)2mv2⇒favg=L−2nrmv2.
JEE Main 2015 — Physics Mechanics
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 (see figure). 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):

Held on 11 Apr 2015 · Verified 6 Jul 2026.
L−nrmv2
L−2nrmv2
2(L−nr)mv2
Zero
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