$2.4.16.$ Two beads of mass $m$ each, connected to each other by a spring of stiffness $k$, are held on smooth rods rigidly fixed in the wall. The spring is stretched and its length is $l$. The distance between the free ends of the rods is equal to the length of the undeformed spring. The beads are released. How fast will the spring move in the $x$-direction after the beads jump off the rod? What will be the greatest compression strain of the spring?
Preservation of energy at the initial stage $$\frac{k\left(l-l_0\right)^2}{2} = 2\frac{mv^2}{2}$$ From where $$ v = \sqrt{\frac{k}{2m}}\left(l-l_0\right)$$ The greatest compression will be when it has nowhere to compress $(v_y=0)$ Alternative for horizontal projection $$v_x=v\cos\alpha;v_x=\text{const}$$ After substituting $$ \boxed{v_x = \sqrt{\frac{k}{2m}}\left(l-l_0\right)\cos\alpha} $$ Conservation of energy with $x_\text{max}$ deformation $$\frac{k\left(l-l_0\right)^2}{2} = 2\frac{mv_x^2}{2} + \frac{kx_\text{max}^2}{2}$$ From where $$x_\text{max}^2=\left(l-l_0\right)^2\left(1-\cos^2\alpha\right)$$ After using pythagorean trigonometric identity: $$\boxed{x_\text{max} = \left(l-l_0\right) \sin \alpha}$$
$$\begin{aligned}v_x=(l-l_0)\sqrt{k/(2m)} \cos\alpha; x=(l-l_0)\sin\alpha.\end{aligned}$$
Almaskhan Arsen