Physics - NLM Question with Solution | TestHub

Let $\omega$ be the angular velocity of the cube (just after the bullet strikes) about an axis passing through D. Conservation of angular momentum about this axis gives $mv\left(\frac{2\ell }{3}\right)=I\omega$ Where $I$ is the moment of intertia about the axis thruogh D . This is $\left[\frac{M{\ell }^2}{6}+\frac{M{\ell }^2}{2}\right]=\frac{2M{\ell }^2}{3}$ from the above equation $\left[\frac{M{\ell }^2}{6}+\frac{M{\ell }^2}{2}\right]=\frac{2M{\ell }^2}{3}$ The cube will topple if the center of mass is just ale to rise from $\left(\frac{1}{2}\right)to\left(\frac{1}{\sqrt{2}}\right)$. In such a cas, the rotational energy must be equated to the change off potenaial energy. Thus $\frac{1}{2}I{\omega }^2=Mg\left(\frac{\ell }{\sqrt{2}}-\frac{\ell }{2}\right)$. Using the values of $I$ and $\omega$, we get the experssion for $v$ that will just topple the cube $v=\frac{M}{m}{\left[3gl\left(\frac{\sqrt{2}-1}{2}\right)\right]}^{\frac{1}{2}}$.