Q67 / 75 · JEE Main 2026 (22 January Shift 2) · Physics
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 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 :
- A66
- Correct answerB33
- C32
- D$2 \sqrt{88}$
Solution
Linear momentum conservation: $mv - 2mv = 23m \cdot v_{cm}$, giving $v_{cm} = -\frac{v}{23}$.
Angular momentum about center of mass from the collision geometry:
Taking collision points 2 cm and 4 cm from the 12 cm bar's center at different positions, the angular momentum is $L = m \cdot v \cdot r_1 + (-2m) \cdot v \cdot r_2$.
Calculating moment of inertia: $I = \frac{1}{12}(20m)(0.12)^2 + m(0.02)^2 + 2m(0.04)^2 = 0.0276m$
The angular frequency $\omega$ is obtained from $\omega = \frac{L}{I}$, giving ratio $\frac{v}{\omega} = 33$.
Angular momentum about center of mass from the collision geometry:
Taking collision points 2 cm and 4 cm from the 12 cm bar's center at different positions, the angular momentum is $L = m \cdot v \cdot r_1 + (-2m) \cdot v \cdot r_2$.
Calculating moment of inertia: $I = \frac{1}{12}(20m)(0.12)^2 + m(0.02)^2 + 2m(0.04)^2 = 0.0276m$
The angular frequency $\omega$ is obtained from $\omega = \frac{L}{I}$, giving ratio $\frac{v}{\omega} = 33$.
Hence, the correct answer is option: 'B'.