The balloon deforms when it weakly hits the wall, as shown in the figure. The maximum deformation of the balloon x is much smaller than its radius R. Ignoring the change in excess pressure \Delta p of air in the balloon and the elasticity of the shell, estimate the time of impact with the wall. The mass of the ball m.

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Solutions of Savchenko Problems in Physics
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    <h3> Statement </h3>
    <p>
        $3.3.8^*.$ The balloon deforms when it weakly hits the wall, as shown in the figure. The maximum deformation of the balloon $x$ is much smaller than its radius $R$. Ignoring the change in excess pressure $\Delta p$ of air in the balloon and the elasticity of the shell, estimate the time of impact with the wall. The mass of the ball $m$.

    <h3>Solution</h3>
    <p>
        

$$y=R-x$$
$$r^2=R^2-(R-x)^2$$
$$S=\pi r^2=\pi (R^2-(R-x)^2)$$
Considering $x \ll R$
$$S=\pi r^2=\pi (R^2-(R-x)^2)=2\pi Rx$$
Newton's second law in differential form:
$$-F=m\ddot{x}$$
$$m\ddot{x}+F=0 \Rightarrow m\ddot{x}+2\pi Rx \cdot \Delta p=0$$
$$\ddot{x}+\frac{2\pi R\Delta p}{m}x=0$$
Recalling the differential equation of harmonic motion
$$\ddot{x}+\omega ^2x=0$$
Find the cyclic frequency $\omega$ of oscillation
$$\omega=\sqrt{\frac{2\pi R\Delta p}{m}}$$
Since we're interested in half of the full period, estimated time of impact with the wall could be found as
$$t=\frac{1}{2}\frac{2\pi}{\omega}=\frac{\pi}{\omega}$$
$$\boxed{t=\pi\sqrt{\frac{m}{2\pi R\Delta p}}}$$

    <h4>Answer</h4>
    <p>
        $$t=\pi\sqrt{m/(2\pi R\Delta p)}$$
    </p>


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