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+ <meta name="description" content="A thin rod of mass m and length l lies on a smooth horizontal surface. A plasticine ball of mass m with a velocity v perpendicular to the rod hits one of its ends and sticks to it. How much heat will be released from such an impact?">
+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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+ <div id = "logo">
+ <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
+ </div>
+ <p class="author">
+ Solutions of Savchenko Problems in Physics <br>
+ <i><b>knowledge must be free</b></i>
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+ </header>
+
+ <h3 id="back-link"><a href="../../#2.7">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.7.41.$ A thin rod of mass $m$ and length $l$ lies on a smooth horizontal surface. A plasticine ball of mass $m$ with a velocity $v$ perpendicular to the rod hits one of its ends and sticks to it. How much heat will be released from such an impact?
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ </p>
+<center>
+ <figure>
+ <img src="2.7.41_1.png"
+ loading="lazy" width="230" />
+ <figcaption>
+ Collision of a ball with a rod
+ </figcaption>
+ </figure>
+</center>
+<p>
+After collision they will move as a unit, rotating around the centre of mass, which is at a distance $l/4$ from the contact point of the balls
+
+</p>
+<center>
+ <figure>
+ <img src="2.7.41_2.png"
+ loading="lazy" width="230" />
+ <figcaption>
+ Rotations around the centre of mass
+ </figcaption>
+ </figure>
+</center>
+<p>
+Conservation of angular momentum
+$$mv\cdot\frac{L}{4}=m\frac{\omega L}{4}\cdot\frac{L}{4}+I\omega\quad(1)$$
+Determening the moment of inertia, using parallel axis theorem
+$$I=\frac{1}{2}mL^2+m\left(\frac{L}{4}\right)^2=\frac{7}{48}mL^2\quad(2)$$
+From the $(1)$ and $(2)$
+$$v=\frac{5}{6}\omega L\Rightarrow \boxed{\omega=\frac{6v}{5L}}$$
+Also, in addition to rotational motion, translational motion will be observed. From the law of conservation of momentum
+$$mv=2mu\Rightarrow u=v/2$$
+$u$ — centre of mass velocity</p><p>
+THe initial energy was as a kinetic energy
+$$E_0=\frac{mv^2}{2}$$
+The total energy after the collision:
+$$E_f=\frac{2mu^2}{2}+\frac{I\omega ^2}{2}+\frac{m(\omega\cdot L/u)^2}{2}$$
+Conservation of energy
+$$E_0=E_f+Q$$
+Where, $Q$ — the released energy
+$$Q=E_0-E_f$$
+<div class="scroll-wrapper">
+$$Q=\frac{mv^2}{2}-\frac{mv^2}{2\cdot2}-\frac{7mL^2}{48}\cdot \left(\frac{6v}{5L}\right)^2\cdot \frac{1}{2}+\frac{m\cdot(6v/5L\cdot L/4)^2}{2}$$
+ </div>
+After the mathematical transformations
+$$\boxed{Q=\frac{mv^2}{10}}$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$Q=1/10\cdot mv^2$$
+ </p>
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ Almaskhan Arsen<br>
+ </p>
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