Question

Two small identical discs, each of mass $m$, lie on a smooth horizontal plane. The discs are interconnected by a light non-deformed spring of length $l_0$ and stiffness $\kappa$. At a certain moment one of the discs is set in motion in a horizontal direction perpendicular to the spring with velocity $v_0$. If the maximum elongation of the spring in the process of motion is given as ${\epsilon }_{max}\approx \frac{xm{v}_0^2}{\kappa l_0^2}$, Find $x$

Answer 1

In the C.M. frame of the system (both the discs $+$ spring), the linear momentum of the discs are related by the relation, $\widetilde{\overrightarrow{p_1}}=-\widetilde{\overrightarrow{p_2}}$, at all moments of time.
where ${\tilde{p}}_1={\tilde{p}}_2=\tilde{p}=\mu v_{rel}$
And the total kinetic energy of the system,
$T=\frac{1}{2}\mu v_{rel}^2$
Bearing in mind that at the moment of maximum deformation of the spring, the projection of $\overrightarrow{v_{rel}}$ along the length of the spring becomes zero, i.e. $v_{rel\left(x\right)}=0$.
The conservation of mechanical energy of the considered system in the C.M. frame gives,
$\frac{1}{2}\left(\frac{m}{2}\right)v_0^2=\frac{1}{2}\kappa x^2+\frac{1}{2}\left(\frac{m}{2}\right)v_{rel\left(y\right)}^2$
Now from the conservation of angular momentum of the system about the C.M.,
$\frac{1}{2}\left(\frac{l_0}{2}\right)\left(\frac{m}{2}{v}_0\right)=2\left(\frac{l_0+x}{2}\right)\frac{m}{2}{v}_{rel\left(y\right)}$
or, $v_{rel\left(y\right)}=\frac{v_0{l}_0}{(l_0+x)}=v_0{(1+\frac{x}{l_0})}^{-1}\approx (1-\frac{x}{l_0})$, as $x\ll l_0$ (2)
Using (2) in (1), $\frac{1}{2}m{v}_0^2[1-{(1-\frac{x}{l_0})}^2]=\kappa x^2$
or, $\frac{1}{2}m{v}_0^2[1-{(1-\frac{2x}{l_0}+\frac{x^2}{l_0^2})}^2]=\kappa x^2$
or, $\frac{m{v}_0^{2}x}{l_0}\approx \kappa x^2$ [neglecting $\frac{x^2}{l_0^2}$]
As $x\ne 0$, thus $x=\frac{m{v}_0^2}{\kappa l_0}$