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. $\frac{3 v}{4 a}$
• B. $\frac{3 v}{2 a}$
• 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 $O=M v \frac{a}{2}$
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
$M v \frac{a}{2}=I \omega$
$\Rightarrow M v \frac{a}{2}=\left(\frac{M a^{2}}{6}+\frac{M a^{2}}{2}\right) \omega$
$\Rightarrow \omega=\frac{3 v}{4 a}$
where $I$ is moment of inertia of the block about the axis perpendicular to the plane passing through $O$.