QJEEAdv22Pha6ae: JEE Advanced 2022 Physics - Class 11 JEEAdv22Pha6ae Solution: List I describes four systems, each with two particles and in…

(I)

${v_{net}} = \sqrt 2 \omega R = \sqrt 2$

I $\to$ S

(II) ${v_A}(t = 0.13) = {{5\pi } \over {\sqrt 2 }}\cos 45\widehat i + \left[ {{{5\pi } \over {\sqrt 2 }} \times {1 \over {\sqrt 2 }} - g \times 0.1} \right]\widehat j$

$= {{5\pi } \over 2}\widehat i + \left( {{{5\pi } \over 2} - 1} \right)\widehat j$

${v_B}(t = 0.1\,\sec ) = {{ - 5\pi } \over 2}\widehat i + \left( {{{5\pi } \over 2}} \right)\widehat j$

After $t = 0.1$ , relative velocities should not change.

${v_{rel}}(t = 0.1\,\sec ) = \left| {5\pi \widehat i - \widehat j} \right|$

$= \sqrt {25{\pi ^2} + 1}$

II $\to$ T

(III) $x = {x_A} - {x_B}$

$= {x_0}\sin t - {x_0}\sin \left( {t + {\pi \over 2}} \right)$

$= \sqrt 2 {x_0}\sin \left( {t - {\pi \over 4}} \right)$

${v_{rel}} = {{dx} \over {dt}} = \sqrt 2 {x_0}\cos \left( {t - {\pi \over 4}} \right)$

$= \sqrt 2 \cos \left( {{\pi \over 3} - {\pi \over 4}} \right)$

$= \sqrt 2 \times {{\sqrt 3 + 1} \over {2\sqrt 2 }} = {{\sqrt 3 + 1} \over 2}$

III $\to$ P

(IV) ${v_{rel}} = \sqrt {{3^2} + {1^2}} = \sqrt {10}$

IV $\to$ R