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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 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../#2.5">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../#2.5">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.5.38^*.$ In one straight line on a smooth horizontal plane with equal intervals there are bars of mass $m$ each. A constant horizontal force $F$ is applied to the first of the bars. Determine the speed of the bars before and immediately after the nth impact. Consider the speed limit value for $n$ tending to infinity, if the width of the gaps between the bars is $l$. The blows of the bars are absolutely inelastic. | | $2.5.38^*.$ In one straight line on a smooth horizontal plane with equal intervals there are bars of mass $m$ each. A constant horizontal force $F$ is applied to the first of the bars. Determine the speed of the bars before and immediately after the nth impact. Consider the speed limit value for $n$ tending to infinity, if the width of the gaps between the bars is $l$. The blows of the bars are absolutely inelastic. |
| | | |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="280" /> | | loading="lazy" width="280" /> |
| <figcaption> | | <figcaption> |
| For problem $2.5.38^*$ | | For problem $2.5.38^*$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Let's consider 1st and 2nd collision | | Let's consider 1st and 2nd collision |
| <br> | | <br> |
| <b>First collision</b>: | | <b>First collision</b>: |
| <br> | | <br> |
| From the law of conservation of energy | | From the law of conservation of energy |
| $$v_1^2=2a_1l$$ | | $$v_1^2=2a_1l$$ |
| Considering Newton's 2nd law ($a_1=\frac{F}{m}$) | | Considering Newton's 2nd law ($a_1=\frac{F}{m}$) |
| $$v_1^2=\frac{2Fl}{m}$$ | | $$v_1^2=\frac{2Fl}{m}$$ |
| Where $v_1$ is the velocity before the collision | | Where $v_1$ is the velocity before the collision |
| <br> | | <br> |
| Law of conservation of momentum | | Law of conservation of momentum |
| $$mv_1=2mv_1'$$ | | $$mv_1=2mv_1'$$ |
| $$v_1=\sqrt{\frac{Fl}{2m}}\quad\text{(1)}$$ | | $$v_1=\sqrt{\frac{Fl}{2m}}\quad\text{(1)}$$ |
| <b>Second collision</b>:<br> | | <b>Second collision</b>:<br> |
| From the law of conservation of energy | | From the law of conservation of energy |
| $$v_2^2=v_1^2+2a_2l$$ | | $$v_2^2=v_1^2+2a_2l$$ |
| Likewise, considering $a_2=\frac{F}{2m}$: | | Likewise, considering $a_2=\frac{F}{2m}$: |
| $$v_2^2=\frac{Fl}{2m}+\frac{Fl}{m}=\frac{3Fl}{2m}$$ | | $$v_2^2=\frac{Fl}{2m}+\frac{Fl}{m}=\frac{3Fl}{2m}$$ |
| $$v_2=\sqrt{\frac{3Fl}{2m}}\quad\text{(2)}$$ | | $$v_2=\sqrt{\frac{3Fl}{2m}}\quad\text{(2)}$$ |
| Where $v_2$ is the velocity after the collision | | Where $v_2$ is the velocity after the collision |
| <br> | | <br> |
| From $v_1$ and $v_2$, we can see that the velocity index is the same as the coefficient in front of the mass and $\text{index}+1$ at the top | | From $v_1$ and $v_2$, we can see that the velocity index is the same as the coefficient in front of the mass and $\text{index}+1$ at the top |
| <br> | | <br> |
| Thus leading to the following recurrence relation | | Thus leading to the following recurrence relation |
| $$\boxed{v_n=\sqrt{\frac{Fl}{m}\left( 1+ \frac{1}{n} \right)}}\quad\text{(3)}$$ | | $$\boxed{v_n=\sqrt{\frac{Fl}{m}\left( 1+ \frac{1}{n} \right)}}\quad\text{(3)}$$ |
| Where $v_n$ is the velocity before the $n^\text{th}$ collision | | Where $v_n$ is the velocity before the $n^\text{th}$ collision |
| <br> | | <br> |
| Law of conservation of momentum of the $n^\text{th}$ collision | | Law of conservation of momentum of the $n^\text{th}$ collision |
| $$v_nmn=u_nm(n+1)$$ | | $$v_nmn=u_nm(n+1)$$ |
| $${u_n=\frac{1}{1+\frac{1}{n}}v_n}$$ | | $${u_n=\frac{1}{1+\frac{1}{n}}v_n}$$ |
| Substituting into the expression $\text{(3)}$: | | Substituting into the expression $\text{(3)}$: |
| $$\boxed{u_n=\sqrt{\frac{Fl}{m\left(1+\frac{1}{n}\right)}}}$$ | | $$\boxed{u_n=\sqrt{\frac{Fl}{m\left(1+\frac{1}{n}\right)}}}$$ |
| When $n\to\infty$, $\frac{1}{n}\to0$: | | When $n\to\infty$, $\frac{1}{n}\to0$: |
| $$\lim_{n\to\infty}\frac{1}{n}=0$$ | | $$\lim_{n\to\infty}\frac{1}{n}=0$$ |
| Whence it follows that the velocity $u_n$ after $n^\text{th}$ collision, where $n\to\infty$, will be equal to | | Whence it follows that the velocity $u_n$ after $n^\text{th}$ collision, where $n\to\infty$, will be equal to |
| $$\boxed{u_n=\lim_{n\to\infty}\sqrt{\frac{Fl}{m\left(1+\frac{1}{n}\right)}}=\sqrt{\frac{Fl}{m}}}$$ | | $$\boxed{u_n=\lim_{n\to\infty}\sqrt{\frac{Fl}{m\left(1+\frac{1}{n}\right)}}=\sqrt{\frac{Fl}{m}}}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$v_n=\sqrt{\frac{Fl}{m}(1+1/n)}$$ | | $$v_n=\sqrt{\frac{Fl}{m}(1+1/n)}$$ |
| $$u_n=\sqrt{\frac{Fl}{m(1+1/n)}}$$ | | $$u_n=\sqrt{\frac{Fl}{m(1+1/n)}}$$ |
| $$v_n\to\sqrt{\frac{Fl}{m}}\text{ with }n\to\infty.$$ | | $$v_n\to\sqrt{\frac{Fl}{m}}\text{ with }n\to\infty.$$ |
| </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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