JEE Main 2020 — Center of Mass Momentum and Collision Question with Solution

Given,

Mass of the rod, $M=0.9 \mathrm{kg}$

Mass of particle, $m=0.1 \mathrm{kg}$

Length of the rod, $l=1 \mathrm{m}$

Velocity of particle, $v=80 \mathrm{m} {\mathrm{s}}^{-1}$

No external torque acting on the system of rod and particle, so angular momentum of system about the suspension is conserved, i.e.; $L_{\mathrm{i}}=L_{\mathrm{f}}$

$\Rightarrow mvl=I\omega ...\left(1\right)$

Where, $I$ is the momentum of inertia of rod and particle system, $\omega$ is the angular velocity of system just after the collision.

$\Rightarrow mvl=\left(\frac{M{l}^2}{3}+m{l}^2\right)\omega ...\left(2\right)$

Substitute the values in equation (2)

$\Rightarrow 0.1\times 80\times 1=\left(\frac{0.9\times (1{)}^2}{3}+0.1\times (1{)}^2\right)\times \omega$

$\omega =20 \mathrm{rad} {\mathrm{s}}^{-1}$