$2.5.38^*.$ 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.
Let's consider 1st and 2nd collision
First collision:
From the law of conservation of energy
$$v_1^2=2a_1l$$
Considering Newton's 2nd law ($a_1=\frac{F}{m}$)
$$v_1^2=\frac{2Fl}{m}$$
Where $v_1$ is the velocity before the collision
Law of conservation of momentum
$$mv_1=2mv_1'$$
$$v_1=\sqrt{\frac{Fl}{2m}}\quad\text{(1)}$$
Second collision:
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
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
$$\boxed{v_n=\sqrt{\frac{Fl}{m}\left( 1+ \frac{1}{n} \right)}}\quad\text{(3)}$$
Where $v_n$ is the velocity before the $n^\text{th}$ collision
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
$$\boxed{u_n=\lim_{n\to\infty}\sqrt{\frac{Fl}{m\left(1+\frac{1}{n}\right)}}=\sqrt{\frac{Fl}{m}}}$$
$$v_n=\sqrt{\frac{Fl}{m}(1+1/n)}$$ $$u_n=\sqrt{\frac{Fl}{m(1+1/n)}}$$ $$v_n\to\sqrt{\frac{Fl}{m}}\text{ with }n\to\infty.$$
Almaskhan Arsen