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+ <meta name="description" content="A heavy particle of mass m_1 collides with a resting light particle of mass m_2. What is the greatest angle that a heavy particle can deflect as a result of an elastic impact?">
+ <meta name="author" content="Aliaksandr Melnichenka">
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+ <h2>Solutions of Savchenko Problems in Physics</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>
+ $2.5.17^*.$ A heavy particle of mass $m_1$ collides with a resting light particle of mass $m_2$. What is the greatest angle that a heavy particle can deflect as a result of an elastic impact?
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ </p>
+<center>
+ <figure>
+ <img src="draw.png"
+ loading="lazy" width="230" />
+ <figcaption>
+ Direction of balls momentum
+ </figcaption>
+ </figure>
+</center>
+<p>
+Following from the law of conservation of energy
+$$\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}=\frac{p_0^2}{2m_1}$$
+From where
+$$p_2^2=\frac{m_2}{m_1}\left(p_0^2-p_1^2\right)\quad\text{(1)}$$
+From the cosine theorem
+$$p_2^2=p_1^2+p_0^2-2p_1p_0\cos\alpha$$
+Substituting into $\text{(1)}$:
+$$\frac{m_2}{m_1}p_0^2-\frac{m_2}{m_1}p_1^2=p_1^2+p_0^2-2p_1p_0\cos\alpha$$
+Let's regroup this into a quadratic equation
+<div class="scroll-wrapper">$$p_1^2\left(1+\frac{m_2}{m_1}\right)-2p_0p_1\cos\alpha+p_0^2\left(1-\frac{m_2}{m_1}\right)=0$$</div>
+Solving relative to $p_1$
+<div class="scroll-wrapper">$$p_1=\frac{2p_0\cos\alpha\pm \sqrt{4p_0^2\cos^2\alpha-4p_0^2\left(1-\frac{m_2^2}{m_1^2}\right)}}{2\left(1+\frac{m_2}{m_1}\right)}$$</div>
+Since $\Delta = 0$
+$$\cos^2\alpha=1-\frac{m_2^2}{m_1^2}$$
+From where we find the required ratio
+$$\frac{m_2^2}{m_1^2}=1-\cos^2\alpha=\sin^2\alpha$$
+$$\boxed{\sin\alpha=\frac{m_2}{m_1}}$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$\sin\alpha=m_2/m_1$$
+ </p>
+
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ Almaskhan Arsen<br>
+ </p>
+
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