Added 2.5.36

astrosander edited Sep 30, 2024, 3:49 PM

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		+	<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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		+	<i><b>knowledge must be free</b></i>
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		+	<h3 id="back-link"><a href="../../#2.5">$\leftarrow$Back</a></h3>
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		+	<h3> Statement </h3>
		+	<p>
		+	$2.5.36.$ A bullet of mass $m_1$ having an initial velocity $v$ penetrates a lead ball of mass $m_2$ suspended on a thread and flies out of it at half speed. How much of the bullet's kinetic energy was converted to heat?
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="2.5.36_1.png"
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		+	<p>
		+
		+	After the collision, the velocity of the lead ball $u$ could be found from the preservation of momentum
		+	$$m_1 v = m_2 u + \frac{m_1v}{2}$$
		+	From here the velocity of the lead ball $u$ is
		+	$$u = v \frac{m_1}{2m_2}$$
		+	Considering the energy $Q$ converted to heat, the conservation of energy could be written as
		+	$$\frac{m_1v^2}{2} = \frac{m_1(v/2)^2}{2} + \frac{m_2u^2}{2} + Q \quad(1)$$
		+	Where the initial kinetic energy
		+	$$E_K = \frac{m_1v^2}{2}\quad(2)$$
		+	From $(1)$, the amount of heat released
		+	$$Q = \frac{m_1v^2}{8} + \frac{m_1^2v^2}{2m_2} - \frac{m_1v^2}{2}\quad(3)$$
		+	Deviding $(2)$ and $(3)$ equations, we obtain
		+	$$\boxed{\frac{E_K}{Q} = \frac{3- {m_1}/{m_2}}{4}}$$
		+	</p>
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	Aliaksandr Melnichenka<br>
		+	</p>
		+
		+	<h4>Answer</h4>
		+	<p>
		+	$$Q/K=(3-m_1/m_2)/4$$
		+	</p>
		+
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			<meta name="author" content="Aliaksandr Melnichenka">
			<meta name="date" content="2023-10" scheme="YYYY-MM">
			<meta property="og:title" content="A bullet of mass m_1 having an initial velocity v penetrates a lead ball of mass m_2 suspended on a thread and flies out of it at half speed. How much of the bullet's kinetic energy was converted to heat?">
			<meta property="og:image" content="img/logo.png">
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			<div id="logo">
			<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
			</div>
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			<i><b>knowledge must be free</b></i>
			</p>
			</header>

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

			<h3> Statement </h3>
			<p>
			$2.5.36.$ A bullet of mass $m_1$ having an initial velocity $v$ penetrates a lead ball of mass $m_2$ suspended on a thread and flies out of it at half speed. How much of the bullet's kinetic energy was converted to heat?
			</p>

			<h3>Solution</h3>
			<p>

			</p>
			<center>
			<figure>
			<img src="2.5.36_1.png"
			loading="lazy" width="280" />
			</figure>
			</center>
			<p>

			After the collision, the velocity of the lead ball $u$ could be found from the preservation of momentum
			$$m_1 v = m_2 u + \frac{m_1v}{2}$$
			From here the velocity of the lead ball $u$ is
			$$u = v \frac{m_1}{2m_2}$$
			Considering the energy $Q$ converted to heat, the conservation of energy could be written as
			$$\frac{m_1v^2}{2} = \frac{m_1(v/2)^2}{2} + \frac{m_2u^2}{2} + Q \quad(1)$$
			Where the initial kinetic energy
			$$E_K = \frac{m_1v^2}{2}\quad(2)$$
			From $(1)$, the amount of heat released
			$$Q = \frac{m_1v^2}{8} + \frac{m_1^2v^2}{2m_2} - \frac{m_1v^2}{2}\quad(3)$$
			Deviding $(2)$ and $(3)$ equations, we obtain
			$$\boxed{\frac{E_K}{Q} = \frac{3- {m_1}/{m_2}}{4}}$$
			</p>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			Aliaksandr Melnichenka<br>
			</p>

			<h4>Answer</h4>
			<p>
			$$Q/K=(3-m_1/m_2)/4$$
			</p>


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