JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2020
- Hard
- MCQ
Consider a uniform rod of mass \(\mathrm{M}=4 \mathrm{m}\) and length \(\ell\) pivoted about its centre. A mass \(m\) moving with velocity \(v\) making angle \(\theta=\frac{\pi}{4}\) to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{3 \sqrt{2}}{7} \frac{v}{l}\)
Step-by-step Solution
Detailed explanation
Let angular velocity of the system after collision be \(\omega\).
By conservation of angular momentum about the hinge
\(\mathrm{m}\left(\frac{\mathrm{v}}{\sqrt{2}}\right)\left(\frac{\ell}{2}\right)=\left[\frac{4 \mathrm{m} \ell^{2}}{12}+\frac{\mathrm{m} \ell^{2}}{4}\right] \omega\)
On solving
\(\omega=\frac{3 \sqrt{2}}{7}\left(\frac{v}{l}\right)\)
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