Question

A cubical block of side $a$ is moving with velocity $v$ on a horizontal smooth plane as shown. It hits a ridge at point O. The angular speed of the block after is hits $O$ is

• A. $\frac{3v}{4a}$
• B. $\frac{3v}{2a}$
• C. $\frac{\sqrt{3}v}{\sqrt{2}a}$
• D. zero

Answer 1

Answer: (A)

Angular momentum of block w.r.t.
$O$ before collision with $O=Mv\frac{a}{2}$
On collision, the block will rotate about the side passing through $O$.
Now its angular momentum $=I\omega$
By law of conservation of angular momentum
$Mv\frac{a}{2}=I\omega$
$\Rightarrow Mv\frac{a}{2}=\left(\frac{M{a}^2}{6}+\frac{M{a}^2}{2}\right)\omega$
$\Rightarrow \omega =\frac{3v}{4a}$
where $I$ is moment of inertia of the block about the axis perpendicular to the plane passing through $O$.