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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="../#2.2">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.2.39.$ The tube of radius $r$ is filled with a porous substance of density $\rho_0$. The piston, which acts on a constant force $F$, moving in the pipe, compacts the substance to a density $\rho$. At what speed does the piston move if the compaction of the substance occurs in a jump, i.e. the interface moves at a certain speed in the pipe, to the right of which the density of the substance is $\rho$, and to the left-$\rho_0$? At the initial moment, this boundary coincides with the surface of the piston.
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+</p>
+<center>
+ <figure>
+ <img src="draw.png"
+ loading="lazy" width="280" />
+ <figcaption>
+ Geometry of the problem
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+From the drawing
+$$\Delta L=L_0-L$$
+During the time $\Delta t$ the piston with velocity $v$ has moved by the amount $\Delta l$
+$$\Delta L=v\Delta t$$
+Given a constant cross-sectional area $S=\pi r^2$, the expression from above can be rewritten as
+$$\frac{m}{\rho_0S}-\frac{m}{\rho S}=v\Delta t$$
+$$\frac{m}{\Delta t}=Sv\frac{\rho\rho_0}{\rho-\rho_0}$$
+Because the mass remains constant ($m=\text{const}$):
+$$m=\rho_0SL_0=\rho SL$$
+From where
+$$\frac{\rho}{\rho_0}=\frac{L}{L_0}$$
+Representing the force by a small change in momentum
+$$F_\text{average}=\frac{mv_\text{system}}{\Delta t}$$
+Substituting the value of $\frac{m}{\Delta t}$:
+$$F_\text{average}=Sv\frac{\rho\rho_0}{\rho-\rho_0}v_\text{system}$$
+$$v_\text{system}=\frac{0+v}{2}=\frac{v}{2}$$
+$$F_\text{average}=\frac{0+F}{2}=\frac{F}{2}$$
+Substitute into the formula for $F_\text{average}$:
+$$\frac{F}{2}=Sv\frac{\rho\rho_0}{\rho-\rho_0}\frac{v}{2}$$
+From where
+$$\boxed{v=\sqrt{\frac{F(\rho-\rho_0)}{\pi r^2\rho\rho_0}}}$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$v=\sqrt{F(\rho-\rho_0)/(\pi r^2\rho\rho_0)}$$
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