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.
Solutions of Savchenko Problems in Physics
Aliaksandr Melnichenka
October 2023
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<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.
For problem $2.5.38^*$
<h3>Solution</h3>
<p>
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
$$\boxed{u_n=\lim_{n\to\infty}\sqrt{\frac{Fl}{m\left(1+\frac{1}{n}\right)}}=\sqrt{\frac{Fl}{m}}}$$
