Let the velocities of particles A, B, C immediately after the impulse be $\vec{v}_1, \vec{v}_2, \vec{v}_3$. Set up a coordinate system with B at the origin and BC along the positive x-axis. The impulse is $\vec{J} = J\hat{i}$. The tension in the string is the only impulsive force on A, so its velocity $\vec{v}_1$ must be along the line BA. The direction of BA is $(\cos\alpha, -\sin\alpha)$. So, $\vec{v}_1 = v_1(\cos\alpha, -\sin\alpha)$, where $v_1$ is the speed of A. Similarly, the net impulsive force on C is along the x-axis, so $\vec{v}_3 = v_3\hat{i}$.

The total momentum of the system changes by the external impulse $\vec{J}$.

$$m_1\vec{v}_1 + m_2\vec{v}_2 + m_3\vec{v}_3 = \vec{J}$$

In component form: x-component: $m_1 v_1 \cos\alpha + m_2 v_{2x} + m_3 v_3 = J$ y-component: $-m_1 v_1 \sin\alpha + m_2 v_{2y} = 0 \implies v_{2y} = \frac{m_1}{m_2} v_1 \sin\alpha$

The inextensible strings impose kinematic constraints. The relative velocity between connected particles along the string must be zero. Along BC: $(\vec{v}_2 - \vec{v}_3) \cdot \hat{i} = 0 \implies v_{2x} = v_3$ Along AB (direction $(-\cos\alpha, \sin\alpha)$): $(\vec{v}_2 - \vec{v}_1) \cdot (-\cos\alpha, \sin\alpha) = 0 \implies \vec{v}_2 \cdot (-\cos\alpha, \sin\alpha) = \vec{v}_1 \cdot (-\cos\alpha, \sin\alpha)$

$-v_{2x}\cos\alpha + v_{2y}\sin\alpha = -v_1(\cos^2\alpha + \sin^2\alpha) = -v_1 \implies v_1 = v_{2x}\cos\alpha - v_{2y}\sin\alpha$

Substitute $v_{2y}$ and $v_3$ into the equations: The x-momentum equation becomes: $m_1 v_1 \cos\alpha + (m_2+m_3)v_{2x} = J$ The AB constraint becomes: $v_1 = v_{2x}\cos\alpha - (\frac{m_1}{m_2} v_1 \sin\alpha)\sin\alpha \implies v_{2x} = \frac{v_1}{\cos\alpha} \left(1 + \frac{m_1}{m_2}\sin^2\alpha\right)$

Substitute $v_{2x}$ into the momentum equation and solve for $v_1$:

$$m_1 v_1 \cos\alpha + (m_2+m_3) \frac{v_1}{\cos\alpha} \frac{m_2+m_1\sin^2\alpha}{m_2} = J$$ $$v_1 \left[ m_1 \cos\alpha + \frac{(m_2+m_3)(m_2+m_1\sin^2\alpha)}{m_2\cos\alpha} \right] = J$$ $$v_1 \frac{m_1 m_2 \cos^2\alpha + (m_2+m_3)(m_2+m_1\sin^2\alpha)}{m_2\cos\alpha} = J$$

The numerator simplifies to $m_2(m_1+m_2+m_3) + m_1 m_3 \sin^2\alpha$.

$$v_1 = \frac{J m_2 \cos\alpha}{m_2(m_1+m_2+m_3) + m_1 m_3 \sin^2\alpha}$$