Motion of two constrained masses

[PhMichael]

Homework Statement

The attached images shows everything. ({\bf{e}}_1 denotes the direction of X and {\bf{e}}_2 denotes the direction of Y).

Initially, the spring is force free when X_{0}=0.4 m (which yields Y_{0}=0.3 m). Also, at this instant, the velocity of B is {\bf{V}}_{B0}=-v_{0}{\bf{e}}_{2} such that the corresponding velocity of A is {\bf{V}}_{A0}=v_{0}(Y/X){\bf{e}}_{1}=0.75v_0{\bf{e}}_{1}.

I'm asked to find the minimum value of v_0 for which B hits the floor.

Homework Equations

Energy conservation yields
(\frac{1}{2}m_{B}V_B^{2}+\frac{1}{2}m_{A}V_A^{2})-(\frac{1}{2}m_{B}V_{B0}^{2}+\frac{1}{2}m_{A}V_{A0}^{2})+\frac{1}{2}k(X_{0}-L)^2=0
However, the kinematic constrains (L is constant) yields
XV_{A}+YV_{B}=0
So, when B hits the floor, Y=0 so that, at this instant, V_{A}=0. Hence,
V_B^{2}=(V_{B0}^{2}+\frac{m_{A}}{m_{B}}V_{A0}^{2})-\frac{k}{m_B}(X_{0}-L)^2=v_{0}^{2}(1+\frac{2}{3}(0.75)^{2})-\frac{100}{3}(0.4-0.5)^2
Now, my requirement is that V_B=0 when it hits the floor, so that the last equation yields
v_{0}\approx0.49 m/s
while the right answer is
v_{0}\approx0.37 m/s
Am I doing/assuming something wrong here?

I would appreciate your help!

The Attempt at a Solution

See the previous item.

 

[mfb]

While it would help if you show more steps, I agree with your result. 0.37 m/s would be the speed of A.