JEE Main 2015 · 11 Apr · Shift 1 · Physics Q3

← JEE Main 2015 · 11 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 3 of 90

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 horizontal rod of length L fixed between two vertical supports, with several beads strung on it; one bead is moving.

\dfrac{mv^2}{L-nr}

\dfrac{mv^2}{L-2nr}

\dfrac{mv^2}{2(L-nr)}

zero

Answer and worked solution

Official answer: B

In elastic collisions between identical beads the velocities are exchanged, so in effect one bead moves at speed $v$ back and forth. It travels the free length $L-2nr$ (the beads take up $2nr$ ).

Each support receives an impulse $2mv$ once every round trip, $t=\dfrac{2(L-2nr)}v$ :
$F=\frac{2mv}{2(L-2nr)/v}=\frac{mv^2}{L-2nr}.$

Topics:Elastic and inelastic collisions Impulse and momentum

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JEE Main 2015 · 11 Apr · Shift 1, Physics question 3