1. Given data

• Rod mass: $M=4m$
• Rod length: $\ell$
• Pivoted about its centre
• A particle of mass $m$ hits one end of the rod and sticks
• Speed of particle: $v$
• Angle with rod axis: $\theta=\dfrac{\pi}{4}$

We need the angular speed just after collision.

2. Key principle: conservation of angular momentum about the pivot

During the collision, the pivot may exert an impulse, so linear momentum is not conserved. But the impulse at the pivot has zero moment about the pivot, so angular momentum about the pivot is conserved.

Thus, $L_i=L_f$

3. Initial angular momentum of the incoming mass

The particle hits the rod at one end, whose distance from the pivot is $r=\frac{\ell}{2}$

Angular momentum magnitude about the pivot is $L_i = mvr\sin\theta$

Substitute $r=\ell/2$ and $\theta=\pi/4$: $L_i = mv\left(\frac{\ell}{2}\right)\sin\frac{\pi}{4}$ $L_i = mv\left(\frac{\ell}{2}\right)\cdot \frac{1}{\sqrt{2}}$ $L_i = \frac{mv\ell}{2\sqrt{2}}$

4. Final moment of inertia of the stuck system

After collision, the rod and particle rotate together about the central pivot.

(a) Rod's moment of inertia about its centre

$I_{\text{rod}}=\frac{1}{12}M\ell^2$ Since $M=4m$, $I_{\text{rod}}=\frac{1}{12}(4m)\ell^2=\frac{m\ell^2}{3}$

(b) Particle's moment of inertia

The mass sticks at the end, distance $\ell/2$ from the pivot: $I_{\text{mass}}=m\left(\frac{\ell}{2}\right)^2=\frac{m\ell^2}{4}$

(c) Total moment of inertia

$I_f=\frac{m\ell^2}{3}+\frac{m\ell^2}{4}$ $I_f=m\ell^2\left(\frac{1}{3}+\frac{1}{4}\right)$ $I_f=m\ell^2\left(\frac{7}{12}\right)$ $I_f=\frac{7m\ell^2}{12}$

5. Apply angular momentum conservation

$L_i = I_f\omega$

So, $\frac{mv\ell}{2\sqrt{2}} = \frac{7m\ell^2}{12}\,\omega$

Cancel $m$ and one factor of $\ell$: $\frac{v}{2\sqrt{2}} = \frac{7\ell}{12}\,\omega$

Hence, $\omega = \frac{12}{7\ell}\cdot \frac{v}{2\sqrt{2}}$ $\omega = \frac{6}{7\sqrt{2}}\frac{v}{\ell}$

Now rationalize/comparably rewrite: $\frac{6}{7\sqrt{2}} = \frac{6\sqrt{2}}{14}=\frac{3\sqrt{2}}{7}$

Therefore, $\boxed{\omega=\frac{3\sqrt{2}}{7}\frac{v}{\ell}}$

6. Option matching

This matches: $\boxed{\text{A}}$

7. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.