| ### Statement | | ### Statement |
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| $2.5.38^*.$ 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. |
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| ### Solution | | ### Solution |
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| Let's consider 1st and 2nd collision **First collision** : From the law of conservation of energy | | Let's consider 1st and 2nd collision **First collision** : From the law of conservation of energy |
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| $$ | | $$ |
| v_1^2=2a_1l | | v_1^2=2a_1l |
| $$ | | $$ |
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| Considering Newton's 2nd law ($a_1=\frac{F}{m}$) | | Considering Newton's 2nd law ($a_1=\frac{F}{m}$) |
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| $$ | | $$ |
| v_1^2=\frac{2Fl}{m} | | v_1^2=\frac{2Fl}{m} |
| $$ | | $$ |
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| Where $v_1$ is the velocity before the collision Law of conservation of momentum | | Where $v_1$ is the velocity before the collision Law of conservation of momentum |
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| $$ | | $$ |
| mv_1=2mv_1' | | mv_1=2mv_1' |
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| From the law of conservation of energy | | From the law of conservation of energy |
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| $$ | | $$ |
| v_2^2=v_1^2+2a_2l | | v_2^2=v_1^2+2a_2l |
| $$ | | $$ |
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| Likewise, considering $a_2=\frac{F}{2m}$: | | Likewise, considering $a_2=\frac{F}{2m}$: |
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| $$ | | $$ |
| v_2^2=\frac{Fl}{2m}+\frac{Fl}{m}=\frac{3Fl}{2m} | | v_2^2=\frac{Fl}{2m}+\frac{Fl}{m}=\frac{3Fl}{2m} |
| $$ | | $$ |
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| $$ | | $$ |
| v_2=\sqrt{\frac{3Fl}{2m}}\quad\text{(2)} | | v_2=\sqrt{\frac{3Fl}{2m}}\quad\text{(2)} |
| $$ | | $$ |
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| Where $v_2$ is the velocity after the collision 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 Thus leading to the following recurrence relation | | Where $v_2$ is the velocity after the collision 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 Thus leading to the following recurrence relation |
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| $$ | | $$ |
| \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)} |
| $$ | | $$ |
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| Where $v_n$ is the velocity before the $n^\text{th}$ collision Law of conservation of momentum of the $n^\text{th}$ collision | | Where $v_n$ is the velocity before the $n^\text{th}$ collision Law of conservation of momentum of the $n^\text{th}$ collision |
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| $$ | | $$ |
| v_nmn=u_nm(n+1) | | v_nmn=u_nm(n+1) |
| $$ | | $$ |
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| $$ | | $$ |
| {u_n=\frac{1}{1+\frac{1}{n}}v_n} | | {u_n=\frac{1}{1+\frac{1}{n}}v_n} |
| $$ | | $$ |
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| Substituting into the expression $\text{(3)}$: | | Substituting into the expression $\text{(3)}$: |
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| $$ | | $$ |
| \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)}}} |
| $$ | | $$ |
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| When $n\to\infty$, $\frac{1}{n}\to0$: | | When $n\to\infty$, $\frac{1}{n}\to0$: |
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| $$ | | $$ |
| \lim_{n\to\infty}\frac{1}{n}=0 | | \lim_{n\to\infty}\frac{1}{n}=0 |
| $$ | | $$ |
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| 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 |
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| $$ | | $$ |
| \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}}} |
| $$ | | $$ |
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| #### Answer | | #### Answer |
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| $$ | | $$ |
| v_n=\sqrt{\frac{Fl}{m}(1+1/n)} | | v_n=\sqrt{\frac{Fl}{m}(1+1/n)} |
| $$ | | $$ |
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| $$ | | $$ |
| u_n=\sqrt{\frac{Fl}{m(1+1/n)}} | | u_n=\sqrt{\frac{Fl}{m(1+1/n)}} |
| $$ | | $$ |
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| $$ | | $$ |
| v_n\to\sqrt{\frac{Fl}{m}}\text{ with }n\to\infty. | | v_n\to\sqrt{\frac{Fl}{m}}\text{ with }n\to\infty. |
| $$ | | $$ |
