Added alternative solution to 2.5.18

astrosander edited Sep 8, 2024, 5:56 AM

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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>
		@@ -55,7 +55,7 @@
			<center>
			<figure>
			<img src="draw.png"
		−	loading="lazy" width="230" />
		+	loading="lazy" width="280" />
			<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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	<body style="">		<body style="">
	<header style="text-align:center;">		<header style="text-align:center;">
	<h2>Solutions of Savchenko Problems in Physics</h2>		<h2>Solutions of Savchenko Problems in Physics</h2>
	<p class="author">		<p class="author">
	Aliaksandr Melnichenka <br/>		Aliaksandr Melnichenka <br/>
	October 2023		October 2023
	</p>		</p>
	</header>		</header>

	<h3 id="back-link"><a href="../#2.5">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../#2.5">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<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?		$2.5.17^*.$ 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>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	</p>		</p>
	@@ -55,7 +55,7 @@
	<center>		<center>
	<figure>		<figure>
	<img src="draw.png"		<img src="draw.png"
	loading="lazy" width="230" />		loading="lazy" width="280" />
	<figcaption>		<figcaption>
	Direction of balls momentum		Direction of balls momentum
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	Following from the law of conservation of energy		Following from the law of conservation of energy
	$$\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}=\frac{p_0^2}{2m_1}$$		$$\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}=\frac{p_0^2}{2m_1}$$
	From where		From where
	$$p_2^2=\frac{m_2}{m_1}\left(p_0^2-p_1^2\right)\quad\text{(1)}$$		$$p_2^2=\frac{m_2}{m_1}\left(p_0^2-p_1^2\right)\quad\text{(1)}$$
	From the cosine theorem		From the cosine theorem
	$$p_2^2=p_1^2+p_0^2-2p_1p_0\cos\alpha$$		$$p_2^2=p_1^2+p_0^2-2p_1p_0\cos\alpha$$
	Substituting into $\text{(1)}$:		Substituting into $\text{(1)}$:
	$$\frac{m_2}{m_1}p_0^2-\frac{m_2}{m_1}p_1^2=p_1^2+p_0^2-2p_1p_0\cos\alpha$$		$$\frac{m_2}{m_1}p_0^2-\frac{m_2}{m_1}p_1^2=p_1^2+p_0^2-2p_1p_0\cos\alpha$$
	Let's regroup this into a quadratic equation		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>		<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 $p_1$		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>		<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 $\Delta = 0$		Since $\Delta = 0$
	$$\cos^2\alpha=1-\frac{m_2^2}{m_1^2}$$		$$\cos^2\alpha=1-\frac{m_2^2}{m_1^2}$$
	From where we find the required ratio		From where we find the required ratio
	$$\frac{m_2^2}{m_1^2}=1-\cos^2\alpha=\sin^2\alpha$$		$$\frac{m_2^2}{m_1^2}=1-\cos^2\alpha=\sin^2\alpha$$
	$$\boxed{\sin\alpha=\frac{m_2}{m_1}}$$		$$\boxed{\sin\alpha=\frac{m_2}{m_1}}$$
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$\sin\alpha=m_2/m_1$$		$$\sin\alpha=m_2/m_1$$
	</p>		</p>

	<p style="text-align: right; font-style: italic; font-size: 14;">		<p style="text-align: right; font-style: italic; font-size: 14;">
	Almaskhan Arsen<br>		Almaskhan Arsen<br>
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