$2.2.39.$ 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.
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 $$\boxed{v=\sqrt{\frac{F(\rho-\rho_0)}{\pi r^2\rho\rho_0}}}$$
$$v=\sqrt{F(\rho-\rho_0)/(\pi r^2\rho\rho_0)}$$
Almaskhan Arsen