6.2.1. A vertically oriented construction consisting of two identical masses connected by a spring (see figure) is subjected to an impact, as a result of which the upper mass begins to move with an initial velocity $V_{1}$ at an angle $\beta$ to the horizontal, and the lower mass - with a velocity $V_{2}$ at an angle $\alpha$ to the horizontal. To what maximum height relative to its initial position will point $A$ - the center of the spring - rise?
# Solution.

Thesis 1. The center of mass will always be located at point $A$. Let the position vectors of the points with masses be $r_{1}\left(x_{1} ; y_{1}\right), r_{2}\left(x_{2} ; y_{2}\right)$. Then the position vector of the center of mass is determined by the relation:

$r_{c}=\frac{m r_{1}+m r_{2}}{m+m}=\frac{r_{1}+r_{2}}{2}(1)$

That is, the center of mass will always be at point $A$.

Thesis 2. Newton's second law for the motion of the center of mass (only external forces relative to the system of two bodies on a spring are acting):

$2 m \bar{a}=2 m \bar{g} \Rightarrow \bar{a}=\bar{g}$

The motion of the center of mass is uniformly accelerated: $\bar{r}_{c}(t)=\bar{r}_{c 0}+\bar{V}_{c 0} t+\frac{\bar{g} t^{2}}{2}$

Thesis 3. The initial velocity is determined from (1):

$\bar{V}_{c 0}=\dot{r}_{c 0}=\frac{V_{10}+V_{20}}{2}$

Thesis 4. We find the displacement of the center of mass in the projection on the vertical direction $y$:

$y_{c}-y_{c 0}=V_{c 0} t+g t^{2} / 2$,

where $V_{c 0}=\left(V_{1} \sin \beta+V_{2} \sin \alpha\right) / 2$.

Then we are looking for the maximum value of the function

$F(t)=t\left(V_{1} \sin \beta+V_{2} \sin \alpha\right) / 2+g t^{2} / 2$,

which is $F_{\max }=\frac{1}{2 g}\left(\frac{V_{1} \sin \beta+V_{2} \sin \alpha}{2}\right)^{2}$

Answer: $\frac{1}{2 g}\left(\frac{V_{1} \sin \beta+V_{2} \sin \alpha}{2}\right)^{2}$