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+ <title>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.</title>
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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../#2.5">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <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.
+
+</p>
+<center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" width="280" />
+ <figcaption>
+ For problem $2.5.38^*$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ Let's consider 1st and 2nd collision
+<br>
+<b>First collision</b>:
+<br>
+From the law of conservation of energy
+$$v_1^2=2a_1l$$
+Considering Newton's 2nd law
+$$v_1^2=\frac{2Fl}{m}$$
+Where $v_1$ is the velocity before the collision
+<br>
+Law of conservation of momentum
+$$mv_1=2mv_1'$$
+$$v_1=\sqrt{\frac{Fl}{2m}}\quad\text{(1)}$$
+<b>Second collision</b>:<br>
+From the law of conservation of energy
+$$v_2^2=v_1^2+2a_2l$$
+Likewise, considering $a_2=\frac{F}{2m}$:
+$$v_2^2=\frac{Fl}{2m}+\frac{Fl}{m}=\frac{3Fl}{2m}$$
+$$v_2=\sqrt{\frac{3Fl}{2m}}\quad\text{(2)}$$
+Where $v_2$ is the velocity after the collision
+<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
+<br>
+Thus leading to the following recurrence relation
+$$\boxed{v_n=\sqrt{\frac{El}{m}\left( 1+ \frac{1}{n} \right)}}\quad\text{(3)}$$
+Where $v_n$ is the velocity before the $n^\text{th}$ collision
+<br>
+Law of conservation of momentum of the $n^\text{th}$ collision
+$$v_nmn=u_nm(n+1)$$
+$${u_n=\frac{1}{1+\frac{1}{n}}v_n}$$
+Substituting into the expression $\text{(3)}$:
+$$\boxed{u_n=\sqrt{\frac{Fl}{m\left(1+\frac{1}{n}\right)}}}$$
+When $n\to\infty$, $\frac{1}{n}\to0$:
+$$\lim_{n\to\infty}\frac{1}{n}=0$$
+Whence it follows that the velocity $u_n$ after $n^\text{th}$ collision, where $n\to\infty$, will be equal to
+$$\boxed{u_n=\lim_{n\to\infty}\sqrt{\frac{Fl}{m\left(1+\frac{1}{n}\right)}}=\sqrt{\frac{Fl}{m}}}$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$v_n=\sqrt{\frac{Fl}{m}(1+1/n)}$$
+$$u_n=\sqrt{\frac{Fl}{m(1+1/n)}}$$
+$$v_n\to\sqrt{\frac{Fl}{m}}\text{ with }n\to\infty.$$
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