A solid cube of wood of side 2 a and mass M is resting on a horizontal surface as shown in the figure. The cube is free to rotate about a fixed axis A B . A bullet of mass m ( < < M ) and speed v is shot horizontally at the face opposite to A B C D at a height of 4 a 3 from the surface to impart the cube and angular speed ω . It strikes the face and embeds in the cube. Then ω is close to (note: the

Step-by-step solution

Applying the parallel axis theorem, I = I cm + M a 2 , here I cm is the moment of inertia passing through centre of mass, a is the distance between two axes. The moment of inertia of solid cube about A B is I = 2 3 M a 2 + M a 2 2 I = 8 3 M a 2 The cube is free to rotate about A B axis. Thus, the angular momentum should be conserved. Angular momentum m v r = I ω , here r = 4 a 3 . So, m v × 4 a 3 = I ω ⇒ ω = m v 4 a 3 8 3 M a 2 = m v 2 M a .