Question

A solid cube of wood of side $2a$ and mass $M$ is resting on a horizontal surface as shown in the figure. The cube is free to rotate about a fixed axis $AB$. A bullet of mass $m(m<<M)$ and speed $v$ is shot horizontally at the face opposite to $ABCD$ at a height of $4a/3$ from the surface to impart the cube an angular speed $\omega$. It strikes the face and embeds in the cube. Then $\omega$ is close to (note: the moment of inertia of the cube about an axis perpendicular to the face and passing through the centre of mass is $2M{a}^2/3)$.

• A. $Mv/ma$
• B. $Mv/2ma$
• C. $mv/Ma$
• D. $mv/2Ma$

Answer 1

The correct option is D $mv/2Ma$

Angular momentum about the axis AB should be conserved.

So,

$mv\times \frac{4a}{3}=I\times \omega$

Using parallel axis theorem $I=\frac{2M{a}^2}{3}+M\times (\frac{2a}{\sqrt{2}}{)}^2$

$\therefore \omega =\frac{mv}{2Ma}$