Physics - Center of Mass Question with Solution | TestHub

Let $v_{CM}$ be the velocity of the center of mass and $\omega$ be the angular velocity about the CM. From the impulse-momentum theorem for linear motion, $J=M{v}_{CM}$, so $v_{CM}=J/M$. The angular impulse is applied at a distance $L/2$ from the CM, so $J(L/2)=I_{CM}\omega$. For a uniform rod, $I_{CM}=M{L}^2/12$. Thus, $\omega =\frac{J(L/2)}{M{L}^2/12}=\frac{6J}{ML}$. The total kinetic energy of the rod is $KE=\frac{1}{2}M{v}_{CM}^2+\frac{1}{2}{I}_{CM}{\omega }^2=\frac{1}{2}M{\left(\frac{J}{M}\right)}^2+\frac{1}{2}\left(\frac{M{L}^2}{12}\right){\left(\frac{6J}{ML}\right)}^2=\frac{J^2}{2M}+\frac{1}{2}\frac{M{L}^2}{12}\frac{36J^2}{M^2{L}^2}=\frac{J^2}{2M}+\frac{3J^2}{2M}=\frac{4J^2}{2M}=\frac{2J^2}{M}$.