JEE Main 2026 Jan 22 Shift 2, Physics Q4: Angular Momentum, Its Conservation and Angular Impulse

JEE Main2026Jan 22, Shift 2Physics

A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses $m$ and $2 m$ are moving in opposite directions with same speed $v$ 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 $ω$ . The ratio of $v$ and $ω$ is :

A
33
B
$288$
C
66
D
32

Solution

Conservation of angular momentum about the bar's centre:

$L_{i} = 2 m \cdot v \cdot 2 + m \cdot v \cdot 4 = 8 m v$

Final moment of inertia about COM: $I = \frac{( 20 m ) ( 12 ) ^{2}}{12} + m (4)^{2} + 2 m (2)^{2} = 240 m + 16 m + 8 m = 264 m$ cm $^{2}$ .

$L_{f} = I ω = 264 m ω$ . Equating: $8 m v = 264 m ω \Rightarrow \frac{v}{ω} = \frac{264}{8} = 33$ .

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JEE Main 2026 Jan 22 Shift 2, Physics Q4: Angular Momentum, Its Conservation and Angular Impulse

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