Added Luis's English solution of 2.2.16

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+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
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+ <h3> Statement </h3>
+
+ <p>
+ $2.2.16$
+ The radioactive core decayed into three fragments of mass $m_1$, $m_2$, and $m_3$, having mutually perpendicular velocities $v_1$, $v_2$, and $v_3$, respectively. What was the speed of the nucleus before the decay?
+ </p>
+ <h3>Solution</h3>
+ <p>
+ Let's consider the following figure
+ </p>
+ <center>
+ <figure>
+ <img src="draw.png"
+ loading="lazy" alt="2.2.16" width="200" />
+ <figcaption>
+ Velocity's analysis
+ </figcaption>
+ </figure>
+ </center>
+ <p>
+ From Momentum Conservation Law (MCL)
+ $$(m_1+m_2+m_3)~\vec{V} = m_1\vec{v_1}+m_2\vec{v_2}+m_3\vec{v_3}$$
+ Since velocity's vectors are mutually perpendicular, their directions can be fundamental for establishing the spatial coordinated system. So, let be $\hat{v_1}$, $\hat{v_2}$ and $\hat{v_3}$ the unitary vectors for each direction. Applying MCL for each direction,
+ $$(m_1+m_2+m_3)V_1 = m_1v_1 \;(1)$$
+ $$(m_1+m_2+m_3)V_2 = m_2v_2 \;(2)$$
+ $$(m_1+m_2+m_3)V_3 = m_3v_3 \;(3)$$
+ where $V_1$, $V_2$ and $V_3$ are the components of velocity $\vec{V}$,
+ $$V=\sqrt{V_1^2+V_2^2+V_3^2} \;(4)$$
+ Substituting $(1)$, $(2)$ and $(3)$ into $(4)$
+ $$\boxed{V=\frac{\sqrt{m_1^2v_1^2+m_2^2v_2^2+m_3^2v_3^2}}{m_1+m_2+m_3}}$$
+ </p>
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ BSc. Luis Daniel Fernández Quintana<br>
+ Physics Department (FCNE)<br>
+ Universidad de Oriente, Cuba<br>
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