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| <title>The path length of a particle of mass m is proportional to its initial momentum if the force braking the particle is proportional to its velocity (see the previous problem). Make sure of this and for a given lpha (l = lpha p), find the work of the braking force on the path x for a particle with mass m and initial momentum p.</title> | | <title>The path length of a particle of mass m is proportional to its initial momentum if the force braking the particle is proportional to its velocity (see the previous problem). Make sure of this and for a given lpha (l = lpha p), find the work of the braking force on the path x for a particle with mass m and initial momentum p.</title> |
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| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| $2.3.36^*.$ The path length of a particle of mass $m$ is proportional to its initial momentum if the force braking the particle is proportional to its velocity (see the previous problem). Make sure of this and for a given $\alpha (l = \alpha p)$, find the work of the braking force on the path $x$ for a particle with mass $m$ and initial momentum $p$. | | $2.3.36^*.$ The path length of a particle of mass $m$ is proportional to its initial momentum if the force braking the particle is proportional to its velocity (see the previous problem). Make sure of this and for a given $\alpha (l = \alpha p)$, find the work of the braking force on the path $x$ for a particle with mass $m$ and initial momentum $p$. |
| </p> | | </p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Let's make sure that $F \sim v$! | | Let's make sure that $F \sim v$! |
| $$dA = Fdx \\$$ | | $$dA = Fdx \\$$ |
| $$dE = \frac{p dp}{m} \\$$ | | $$dE = \frac{p dp}{m} \\$$ |
| Hence | | Hence |
| $$\frac{mv^2}{2} = \frac{p^2}{2m} \\$$ | | $$\frac{mv^2}{2} = \frac{p^2}{2m} \\$$ |
| Find the small increment of energy $E$ | | Find the small increment of energy $E$ |
| $$d(E) = d\left(\frac{p^2}{2m}\right) = \frac{p dp}{m} = \frac{pdp}{m} \\$$ | | $$d(E) = d\left(\frac{p^2}{2m}\right) = \frac{p dp}{m} = \frac{pdp}{m} \\$$ |
| From the law of conservation of energy, this energy has completely gone into doing work $A$: | | From the law of conservation of energy, this energy has completely gone into doing work $A$: |
| $$dA = dE \\$$ | | $$dA = dE \\$$ |
| $$Fdx = \frac{p dp}{m} \\$$ | | $$Fdx = \frac{p dp}{m} \\$$ |
| $$dx = \alpha dp \Rightarrow l = \alpha p \\$$ | | $$dx = \alpha dp \Rightarrow l = \alpha p \\$$ |
| $$\frac{dp}{dx} = \frac{1}{\alpha} \\$$ | | $$\frac{dp}{dx} = \frac{1}{\alpha} \\$$ |
| $$F = \frac{p}{m \alpha} = \frac{v}{\alpha} \Rightarrow \boxed{F \sim v}$$ | | $$F = \frac{p}{m \alpha} = \frac{v}{\alpha} \Rightarrow \boxed{F \sim v}$$ |
| That's what we had to prove! | | That's what we had to prove! |
| </p><p> | | </p><p> |
| The work of force $F$ on a small interval $dx$ is found as | | The work of force $F$ on a small interval $dx$ is found as |
| $$dA = Fdx = \frac{p^*-p}{m \alpha} dx$$ | | $$dA = Fdx = \frac{p^*-p}{m \alpha} dx$$ |
| $$dA = \frac{x}{m \alpha^2} dx - \frac{p}{m \alpha} dx$$ | | $$dA = \frac{x}{m \alpha^2} dx - \frac{p}{m \alpha} dx$$ |
| And the total work could be found as a sum of all $dA$ all along the way | | And the total work could be found as a sum of all $dA$ all along the way |
| $$\int_0^A dA = \int_0^x \frac{x dx}{m \alpha^2} - \int_0^x \frac{p dx}{m \alpha} \\$$ | | $$\int_0^A dA = \int_0^x \frac{x dx}{m \alpha^2} - \int_0^x \frac{p dx}{m \alpha} \\$$ |
| $$\boxed{A = \frac{x^2}{2m \alpha^2} - \frac{px}{m \alpha}}$$ | | $$\boxed{A = \frac{x^2}{2m \alpha^2} - \frac{px}{m \alpha}}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$A = \frac{x^2}{2m \alpha^2} - \frac{Px}{m \alpha}$$ | | $$A = \frac{x^2}{2m \alpha^2} - \frac{Px}{m \alpha}$$ |
| </p> | | </p> |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Almaskhan Arsen<br> | | Almaskhan Arsen<br> |
| </p> | | </p> |
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