Solution

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

$$\boxed{u_n=\lim_{n\to\infty}\sqrt{\frac{Fl}{m\left(1+\frac{1}{n}\right)}}=\sqrt{\frac{Fl}{m}}}$$