A sliding block on a large cart: Conservation of energy and momentum

Question:
A block B is placed on the cart A that has the frictionless surface and going to be sliding the height of $h$ until the frictional bottom (the shaded part). The cart has frictionless wheels and it moves due to block's falling on the slope. The masses of the block and the cart are $m$ and $M$, respectively. Find each final velocity.

Answer:
The momentum in the horizontal direction should be conserved. Let velocities be $V_A$ and $V_B$. The initial total momentum is zero since they are initially at rest. We assume that the final total momentum is the sum of these momenta. From conservation of momentum, we have
\begin{equation}
0 = MV_A+mV_B
\end{equation}
The total energies in initial and final states must be conserved. The LHS and RHS of the following equation are initial total and final total energies, respectively.
\begin{eqnarray}
 (0+mgh) + (0 + 0) &=& \left(\frac{1}{2}mV_B^2+0\right) + \left(\frac{1}{2}MV_A^2 + 0 \right) \nonumber  \\
\rightarrow \ \frac{1}{2}MV_A^2 &+& \frac{1}{2}mV_B^2 = mgh
\end{eqnarray}
Now, derive $V_B$ and $V_A$ from these equations. From (1), we have
\begin{equation}
V_A = -\frac{m}{M}V_B
\end{equation}
Plug it in (2).
\begin{eqnarray}
M \left(\frac{m}{M}\right)^2 V_B^2 + mV_B^2 &=& 2mgh  \\
\left(\frac{m}{M}+1\right)V_B^2 &=& 2gh  \\
V_B = \sqrt{\frac{2Mgh}{m+M}}
\end{eqnarray}
Then, plug (6) in (3).
\begin{equation}
V_A = -m\sqrt{\frac{2gh}{M(m+M)}}
\end{equation}

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