1. Identifying the Conserved Quantity

The elastic cord exerts a force on the particle that is always directed towards the nail (point O). This is a central force. For any central force field, the torque about the center of force is zero.

$$\vec{\tau}_O = \vec{r} \times \vec{F} = 0 \quad (\text{since } \vec{r} \parallel \vec{F})$$

Therefore, the angular momentum of the particle about the nail O is conserved.

$$L_A = L_B$$

2. Calculating Angular Momentum

At Point A:
The particle is projected horizontally perpendicular to the cord. The distance is $l_0 = 12 \text{ cm}$.
$$L_A = m u l_0 \sin(90^\circ) = m u (12)$$

At Point B:
The velocity is $v = 24 \text{ m/s}$ at a distance $l = 10 \text{ cm}$. The angle between the velocity vector and the position vector (the cord) is $\theta = 37^\circ$.
$$L_B = m v l \sin(\theta) = m (24) (10) \sin(37^\circ)$$

3. Solving for Initial Velocity ($u$)

Equating $L_A$ and $L_B$:

$$m u (12) = m (24) (10) \sin(37^\circ)$$

We know that $\sin(37^\circ) \approx 0.6$ (or $3/5$).

$$12 u = 240 \times 0.6$$ $$12 u = 144$$ $$u = \frac{144}{12} = 12 \text{ m/s}$$