Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Section B / 7

The original diagram has a horizontal bottle neck, so weight does not enter the axial force balance. Gauge pressure acts on the cork's end area and supplies outward force $F_g=\pi a^2{P}_g$. The radial contact force is the contact pressure times the cylindrical contact area, $N=2\pi a(l-x)P_c$. The limiting friction is $\mu N$, so outward sliding occurs whenThe assumption of equal static and kinetic friction coefficients removes a separate force drop when slipping starts.

Initially the pressure is marginal at $x=0$, so $\pi a^2{P}_g=2\pi al\mu P_c$. During extraction the contact length decreases, while the pressure force stays constant. The net force is therefore $F(x)=\pi a^2{P}_{g}x/l$. The cork's mass is $m_c=\rho \pi a^{2}l$, giving the pressure-driven cork ejection equationFor $x(0)=0$ and $\dot{x}(0)=v_0>0$, solving the constant-coefficient equation givesThe equilibrium is unstable: as the cork moves out, friction decreases and the unbalanced force grows.

Multiplying the equation of motion by $\dot{x}$ and integrating gives ${\dot{x}}^2=v_0^2+b^2{x}^2$. The contact ends at $x=l$, soEquivalently, the net work is ${\int }_0^{l}F(x)dx=\pi a^2{P}_{g}l/2$. Exactly zero initial velocity would leave the ideal marginal configuration stationary. The exit time $b^{-1}\mathrm{arsinh}(bl/v_0)$ diverges as $v_0\to 0^{+}$, even though the limiting exit speed is finite.

The density is $500\mathrm{kg}{\mathrm{m}}^{-3}$, so $v_{\mathrm{exit}}\simeq \sqrt{400000/500}=28.3\mathrm{m}{\mathrm{s}}^{-1}$. For an ideal upward launch at this speed, conservation of mechanical energy gives height above the exitThis neglects air resistance and the small gravitational work during extraction compared with the launch energy. Actual gas pressure need not stay constant, so it is an idealized estimate.

Adding a stopper that doubles the moving mass, while preserving the same pressured area and contact friction, leaves the extraction work unchanged. The squared exit speed and the ballistic height are therefore halved: