Laws of Motion

JEE Main level, single correct, medium

50 of 210

Question 50 of 210

A large number (n)(n) of identical beads, each of mass mm and radius rr are strung on a thin smooth rigid horizontal rod of length L(L≫r)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 vv, the average force experienced by each support after a long time is (assume all collisions are elastic):

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Answer: (B) mv2L−2nr\dfrac{m{v}^{2}}{L-2nr}

Average time for 11 collision to occur on 11 wall,
=tavg=averagedistancetravelledspeed={t}_{avg}=\dfrac{averagedistancetravelled}{speed}=2(L−2nr)v=\dfrac{2\left(L-2nr\right)}{v}

⇒tavg=2(L−2nr)v\Rightarrow {t}_{avg}=\dfrac{2\left(L-2nr\right)}{v}. Change in momentum in 11 collision =2mv=2mv. Average force on each support, favg=2mvtavg=2mv22(L−2nr){f}_{avg}=\dfrac{2mv}{{t}_{avg}}=\dfrac{2m{v}^{2}}{2\left(L-2nr\right)}⇒favg=mv2L−2nr\Rightarrow {f}_{avg}=\dfrac{m{v}^{2}}{L-2nr}.

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