The tube of radius r is filled with a porous substance of density ho_0. The piston, which acts on a constant force F, moving in the pipe, compacts the substance to a density ho. 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 ho, and to the left- ho_0? At the initial moment, this boundary coincides with the surface of the piston.

Solutions of Savchenko Physics Textbook

Aliaksandr Melnichenka
October 2023

    <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>
        

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
$$\boxed{v=\sqrt{\frac{F(\rho-\rho_0)}{\pi r^2\rho\rho_0}}}$$

    <h4>Answer</h4>
    <p>
        $$v=\sqrt{F(\rho-\rho_0)/(\pi r^2\rho\rho_0)}$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 16;">Almaskhan Arsen</p>


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