<h3 id="back-link"><a href="/#2.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.3.35.$ A particle of mass $m$ flies into a region where it is affected by a braking force that depends only on the distance between the particle and the boundary of the region. Find this dependence if the depth of penetration of a particle into the braking region is proportional to its initial momentum: $l = \alpha p$.
    </p>

    <h3>Solution</h3>
    <p>
        

The work done by the force $F$ when the particle moves $dl$ $$F dl = dA \quad (1)$$ From the law of conservation of energy $$\quad \frac{m u^2}{2} = \frac{p^2}{2m}$$ Considering a small increment of energy $$dE = \frac{d(p^2)}{2m} = \frac{p dp}{m} \quad (2)$$ $$\quad dA = dE \Rightarrow F dl = \frac{p dp}{m} \ $$$ \quad F = \frac{p}{m} \frac{dp}{dl} = \frac{p}{m} \frac{1}{\alpha} = \frac{1}{m} \frac{l}{\alpha} \frac{1}{\alpha} = \frac{l}{m \alpha^2} \quad (*)$$From where$$ \quad \frac{dp}{dl} = \frac{1}{\alpha} , \quad p = \frac{l}{\alpha} \$$ $$\quad \boxed{F = \frac{l}{m \alpha^2}}$$

    <h4>Answer</h4>
    <p>
        $$F={l}/{(m \alpha^2)}$$
    </p>


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