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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="../#2.5">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.5.18^*.$ A particle of mass $m_1$ flew at a velocity $v$ on a stationary particle of mass $m_2$, which, after an elastic impact, flew at an angle $\alpha$ to the initial direction of movement of the incoming particle. Determine the velocity of a particle of mass $m_2$ after impact.
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ This phenomenon is also knowned as Compton scattering
+</p>
+<center>
+ <figure>
+ <img src="drawing1.png"
+ loading="lazy" width="320" />
+ <figcaption>
+ Particle $m_1$ flew on a particle $m_2$
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+Let's write the law of conservation of momentum:
+$$\left\{\begin{matrix}
+X\text{: }m_1v=m_1v_1\cos\beta+m_2v_2\cos\alpha \\
+Y\text{: }m_1v_1\sin\beta=m_2v_2\sin\alpha \\
+\end{matrix}\right.$$
+Let's
+$$\kappa=\frac{m_2}{m_1}$$</p><p>
+From where:
+$$v_1\sin\beta=\kappa v_2\sin\alpha$$
+$$v=v_1\cos\beta+\kappa v_2\cos\alpha$$
+We could find $\cos\beta$ from first expression:
+$$\cos\beta=\sqrt{1-\sin^2\beta}$$
+$$\cos\beta=\sqrt{1-(\kappa \frac{v_2}{v_1}\sin\alpha)^2}$$
+The law of conservation of energy:
+$$\frac{m_1v^2}{2}=\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}$$
+From where
+$$v_1^2=v^2-\kappa v_2^2$$
+From the other hand:
+
+$$(v-\kappa v_2\cos\alpha)^2=v_1^2\cos^2\beta$$
+$$(v-\kappa v_2\cos\alpha)^2=v_1^2\cdot(1-(\kappa \frac{v_2}{v_1}\sin\alpha)^2)$$
+$$(v-\kappa v_2\cos\alpha)^2=v_1^2-(\kappa v_2\sin\alpha)^2$$
+$$v^2 - 2\kappa vv_2\cos\alpha=v_1^2-\kappa^2 v_2^2$$
+From where
+$$v_2^2\kappa(1+\kappa)=2\kappa vv_2\cos\alpha$$
+$$v_2=v\frac{2\cos\alpha}{1+\kappa}$$
+And, finally
+$$\boxed{v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}}$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}$$
+ </p>
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ Yuldashev Ulugbek<br>
+ </p>
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