Problem source

A wedge of mass $M$ rests on a smooth horizontal surface. The face of the
wedge is a smooth plane inclined at an angle $\alpha$ to the horizontal.
A particle of mass $m$ slides down the face of the wedge, starting from rest.
At a later time $t$, the speed $V$ of the wedge, the speed $v$ of the particle
and the angle $\beta$ of the velocity of the particle below the horizontal
are as shown in the diagram.
\begin{center}
\psset{xunit=0.55cm,yunit=0.55cm,algebraic=true,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-3.96,-2.9)(9.6,6.78) \psline(0,0)(0,6) \psline(8,0)(0,6) \psline(8,0)(0,0) \psline{->}(1.13,2.31)(-1.98,2.31) \rput[tl](6.8,0.52){$\alpha$} \rput[tl](3.05,2.02){$v$} \psline(3.31,1.38)(4.47,-2.55) \rput[tl](4.08,-0.14){$\beta$} \rput[tl](-2.83,2.54){$V$} \psline{->}(2.41,4.53)(3.03,2.24) \begin{scriptsize} \psdots[dotsize=10pt 0,dotstyle=*](2.41,4.53) \end{scriptsize} \end{pspicture*}
\par\end{center}
\noindent Let $y$ be the vertical distance
descended by the particle. Derive the following results, stating in \textbf{(ii)}
and \textbf{(iii)} the mechanical principles you use:
\begin{questionparts}
\item $V\sin\alpha=v\sin(\beta-\alpha)$;
\item $\tan\beta=(1+m/M)\tan\alpha$;
\item $2gy=v^2(M+m\cos^2\beta)/M$.
\end{questionparts}
Write down a differential equation for $y$ and hence show that
$$y={gMt^2\sin^2\beta \over 2\,(M+m\cos^2\beta)}.$$