A uniform bar of length 12 cm and mass 20 m lies on a smooth… | JEE Main 2026 (22 Jan, Shift 2)

Q14JEE Main 2026 · 22 JAN · SHIFT 2Difficult

Angular Momentum & its Conservation

A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses $m$ and $2m$ are moving in opposite directions with same speed of $v$ and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency $\omega$ . The ratio of $v$ and $\omega$ is:

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Answer: (a)

From the figure the dashed middle line is the bar centre: $2m$ hits 2 cm left of it, $m$ hits 4 cm right of it; both give the same sense of rotation.
Centre of mass of the system: $\frac{2m(-2) + m(4)}{23m} = 0$ , so it stays at the bar centre.
$L = 2m(v)(2) + m(v)(4) = 8mv$ (cm units).
$I = \frac{1}{12}(20m)(12)^2 + 2m(2)^2 + m(4)^2 = 240m + 8m + 16m = 264m$
$\omega = \frac{8mv}{264m} = \frac{v}{33} \Rightarrow \frac{v}{\omega} = 33$ (cm)
So the answer is (a) 33.

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Angular Momentum & its Conservation question from JEE Main 2026 (22 Jan, Shift 2)

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