Impulse and momentum on a system of three particles (equilateral triangle)
I found this exercise on a textbook which provided solely the description of what is happening and the answers with little explanation. The solution being provided, I am mostly interested in understanding what is happening. Consider the following system composed of three particles, each of mass $m$, connected by rigid massless rods in the form of an equilateral triangle. The system is initially at rest (particle $A$ is above particle $B$ in the initial state). An impulse $\hat{F}$ is applied to $A$, causing the particle $B$ to slide without friction on a supporting base.
The idea is to solve for the values of $\dot x$ and $\dot \theta$ velocities immediately after the impulse; and evaluate the constraint impulse $\hat{N}$.
My initial idea was to use the linear impulse, focusing in the $x$ direction first. 512
Since particle B will be sliding, the linear acceleration of the system would be $3m\ddot x$. By taking particle B as the reference point, and since the system center of mass would be engaged in an arc motion, its acceleration $x$ component (tangential) directly after the impulse would be $ml\sqrt{3}\ddot\theta$ (the radius would be $l/\sqrt{3}$).
So the impulse would be $\hat{F}=3m\dot x+ml\sqrt{3}\dot \theta$.
But the textbook solution indicates that $\hat{F}=3m\dot x+\frac{3}{2}ml\dot \theta$.
So my first problem is: where does the $\frac{3}{2}$ comes from? Or am I working this completely wrong?
Upon seeing this solution I wondered if perhaps this $\frac{3}{2}ml\ddot \theta$ tangential component of angular acceleration would be the result of the addition of an $l/2$ radius (along the vertical axis) which would be the vertical distance towards particle C from B; plus the $l$ radius which would be the vertical distance towards particle $A$; multiplied by $\ddot \theta$. Which would result in a "tangential" component along the $x$ axis of $\frac{3}{2}ml\ddot \theta$. Would this be an acceptable proposition?
My idea was then to state the angular impulse as the variation of the angular momentum, resulting in the multiplication of $\hat{F}$ by the distance between particle $B$ and the center of mass: $l/\sqrt{3}$. This way I would end up with a system of two equations and two variables and solve for $\dot x$ and $\dot \theta$.
Only, again the solution provided is rather far from my idea because it states that $\hat{F}l = \frac{3}{2}ml\dot x + 2ml^2\dot \theta$. And I see little relationship with the linear impulse equation. 101