JEE Main 2024 — Center of Mass Momentum and Collision Question with Solution

Impulse $J=0.2 \mathrm{N} \mathrm{s}$

$\Rightarrow \mathrm{J}=\int Fdt=0.2 \mathrm{N} \mathrm{s}$

Now, angular impulse $\left(\overrightarrow{M}\right)$ will be

$M_c=\int \tau dt$

$=\int F\frac{L}{2}dt$

$=\frac{L}{2}\int Fdt=\frac{L}{2}\times \mathrm{J}$

$=\frac{0.3}{2}\times 0.2$

$=0.03$

Moment of inertia of the rod about the centre of mass, $I_{\mathrm{cm}}=\frac{M{L}^2}{12}=\frac{2\times (0.3{)}^2}{12}=\frac{0.09}{6}$

Angular impulse will be equal to the change in angular momentum.

$M=I_{\mathrm{cm}}\left({\omega }_f-{\omega }_i\right)$

$\Rightarrow 0.03=\frac{0.09}{6}\left({\omega }_f\right)$

$\Rightarrow {\omega }_f=2 \mathrm{rad} {\mathrm{s}}^{-1}$

For angular displacement, we can write $\theta =\omega t$

$\Rightarrow t=\frac{\mathrm{\theta }}{\mathrm{\omega }}=\frac{\mathrm{\pi }}{2\times 2}=\frac{\mathrm{\pi }}{4} \mathrm{s}$.

Therefore, $x=4$.