Getting back to the old things #2

Apart from horizontal reaction forces keeping the rod in place, the only forces acting on the rod are gravity and the normal reaction force $N$ from the wedge. Equally, the only force acting on the wedge that has a horizontal component is the normal reaction $N$ from the rod.

Since the angle of incline of the wedge is $45^\circ$ , the distance fallen by the rod and the horizontal distance moved by the wedge are always the same, and so the downwards acceleration of the rod is equal to the horizontal acceleration $a$ of the wedge. Thus we have the equations $ma \; = \; mg - \tfrac{1}{\sqrt{2}}N \hspace{2cm} ma \; = \; \tfrac{1}{\sqrt{2}}N$ which together imply that $a = \tfrac12g$ . Thus, when the wedge has moved a distance $x$ , is has speed $\sqrt{2ax} \; = \; \sqrt{gx}$ , which means that $a = \boxed{1}$ .