JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
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)
- A \(\frac{{m{v^2}}}{{2\left( {L - nr} \right)}}\)
- B \(\frac{{m{v^2}}}{{L - 2nr}}\)
- C \(\frac{{m{v^2}}}{{L - nr}}\)
- D zero
Answer & Solution
Correct Answer
(B) \(\frac{{m{v^2}}}{{L - 2nr}}\)
Step-by-step Solution
Detailed explanation
Space between the supports rm for motion of beads is \(L - 2nr\) Average force experinced by each support \(F = \frac{{2mv}}{{\frac{{2\left( {L - 2nr} \right)}}{V}}} =\frac{{m{V^2}}}{{L - 2nr}}\)
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