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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

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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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