Conservation of Linear Momentum of Rigid Body

[Uchida]

Homework Statement

Consider a determined structure of three particles, each with mass m, interconnected by rigid bars of negligible mass forming an equilateral triangle on the side a.
The structure falls from a height h, as shown in the figure and the particle C hooks on the stop.

The angular velocity of the structure immediately after the impact is:

Relevant Equations

Conservation of Linear Momentum

I solved it by two methods:

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First, by conservation of linear momentum, using the vector velocities of each particle:

In the imminence of the impact, the velocity of all the three particles are the same, \vec v_0 = - \sqrt{2gh} \hat j
After que impact, C will have zero velocity, and particles A and B will have velocities with the same magnitude v_f, in the perpendicular direction of the bars connected to them. Therefore:

\vec v_A = - v_fsin(30º) \hat i - v_fcos(30º) \hat j
\vec v_B = v_fsin(30º) \hat i - v_fcos(30º) \hat j

By conservation of linear momentum:

m_A\vec v_0 + m_B\vec v_0 + m_C\vec v_0 = m_A\vec v_A + m_B\vec v_B + m_B\vec v_C

Considering m_A = m_B = m_C = m and \vec v_C = 0, we have:

3m\vec v_0 = 2m(\vec v_A + \vec v_B)

-3\sqrt{2gh} \hat j = -2v_fcos(30º) \hat j

Which give us: v_f = ||v_A|| = ||v_B|| = \sqrt{6gh}

The angular speed of the rigid body can be calculated as ω = \frac{v_t}{r}. where v_t = v_f and r = a.

Finally:
ω = \frac{\sqrt{6gh}}{a}-----------------------------------------------------

The second method, by conservation of linear momentum of the center of gravity of the rigid body.
This method is more straight forward, and the calculations give the velocity of CG after the impact equal to the velocity in the imminence of the impact. v_{f_{CG}} = v_{0_{CG}} = \sqrt{2gh}

The distance of CG from the particle C is r_{CG} = a\frac{2}{3}{cos(30º)} = \frac{a \sqrt{3}}{3}

The angular speed of the rigid body can be calculated as ω = \frac{v_{f_{CG}}}{r_{CG}}.

Finally:
ω = \frac{\sqrt{6gh}}{a}
The problem is, that the correct answer to the question is:

The "correct" answer has a 2 dividing the result I found, and there is no answer key to the result I got.

Can someone please help me find where I am wrong, or validate my methods (which would mean that the answer key to the question must be wrong).

 

[TSny]

The total linear momentum of the 3-particle system is not conserved when particle C collides with the support. The support exerts an external, impulsive force on the system that changes the linear momentum.

 

[Merlin3189]

TSny deals with the first method and the conservation of momentum issue of the second.
Another problem with the second approach is, I think, that you take $v_{CG}$ as the velocity of all the masses immediately prior to impact.
On impact, C stops and A & B do not increase their vertical speed (the only possible forces on them are upwards and sideways.) The vertical speed of the CG must be between that of C (zero) and that of A & B, hence less than $\sqrt{2gh}$, though that speed may not be relevant if momentum is changed.

The only approach that got me the official answer, was to assume C is small, having negligible moment of inertia and negligible friction with the stop bracket.

 

[Gordianus]

The angular momentum relative to point C just before colisión Is $L=2m\sqrt(2gh)a\sin(60)$ and it's conserved. Upon collision the sistem rotates around C
$2m\sqrt(2gh)a\sin(60)=2ma^2\omega$
$\omega=\frac{\sqrt(6gh)}{2a}$