1. Physical idea

When the moving cube hits the small bump at point $O$, the point of contact acts like an instantaneous pivot during the impulsive collision.

Since the impulse acts through $O$, angular momentum about $O$ is conserved during impact.

2. Given data

• Side of cube: $a = 30\text{ cm} = 0.30\text{ m}$
• Initial speed of cube: $v = 2\text{ m s}^{-1}$

The cube is initially translating without rotation.

3. Initial angular momentum about point $O$

Before collision, the angular momentum about $O$ is due to translation of the center of mass.

$L_i = m v \times d$

where $d$ is the perpendicular distance from $O$ to the line of motion of the center of mass.

Since the cube moves horizontally and $O$ is the lower front corner, the center of mass is at height $a/2$ above $O$. Hence,

$d = \frac{a}{2}$

So,

$L_i = m v \frac{a}{2}$

4. Moment of inertia about point $O$

Immediately after collision, the cube rotates about corner $O$.

So we need moment of inertia of the cube about an axis through $O$ perpendicular to the plane of motion.

Using parallel axis theorem:

$I_O = I_{CM} + m r^2$

For a square lamina/cubical cross-section of side $a$ about its center and perpendicular to the plane,

$I_{CM} = \frac{1}{6}ma^2$

Distance of center from corner $O$:

$r = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2} = \frac{a}{\sqrt{2}}$

Thus,

$m r^2 = m\frac{a^2}{2}$

Hence,

$I_O = \frac{1}{6}ma^2 + \frac{1}{2}ma^2 = \frac{2}{3}ma^2$

5. Apply conservation of angular momentum about $O$

$L_i = L_f$

$m v \frac{a}{2} = I_O \omega$

$m v \frac{a}{2} = \frac{2}{3}ma^2\omega$

Cancel $m$ and one factor of $a$:

$v\frac{1}{2} = \frac{2}{3}a\omega$

$\omega = \frac{3v}{4a}$

Substitute $v=2$ and $a=0.30$ m:

$\omega = \frac{3\times 2}{4\times 0.30} = \frac{6}{1.2} = 5\ \text{rad/s}$

6. Option check

• A: $5.0$ ✅
• B: $6.7$ ❌
• C: $9.4$ ❌
• D: $13.3$ ❌

So the correct option is A.

7. Comparison with stored answer

Stored correct answer: A

Derived answer: A

They agree.