Problem F2066 from magazine "Quant" (2008 №1)

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Ф22066. Carts with masses $m = 1\,kg$ and $M = 2\,kg$ are connected by a light elastic cord $L = 0.3 \,m$ long. At first, the carts are motionless, and the cord almost pulled. A light cart is given a speed $\upsilon_{0} = 2\,m/s$ in direction
Fig. 1
the cord connecting them (Fig. 1). How long will it take for the carts to hit about each other? Cord stiffness $k = 20\,H/m$.
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Center of mass of bogies all time moves with the speed $$u_{ц} = \upsilon\frac{m}{M + m}.$$ Let's move to this frame of reference. The point of the cord that divides the length into with respect to
Fig. 2
$\frac{l_{1}}{l_{2}}=\frac{M}{m}$, there is center of mass, $\upsilon_{ц}$ is its speed (fig.2). Piece stiffness a snua of length $l_{1}$ is equal to
and cord remains stretched for a period of time $$\tau_{1}=\frac{1}{2}T_{1}=\frac{\pi}{\omega_{1}}=\frac{\pi}{\sqrt{\frac{k_{ 1}}{m}}}=\pi\sqrt{\frac{mM}{k(M + m)}}.$$ After that, the cord no longer affects the movement of the system (before impact bodies - already for sure!), bodies go towards each other with speeds $\upsilon_{1}$ and $\upsilon_{2}$ , which can be found from the equations (we are still in a system related to the center of mass) describing laws of conservation of momentum and energy: $$ \begin{cases} mu_{1}=Mu_{2},\\ \frac{mu^2_{1}}{2} + \frac{Mu^2_{2}}{2} = \frac{(m(\upsilon_{0} - u_{c})^2}{2} + \frac{Mu^2_{ц}}{2}. \end{cases}$$ You can, of course, solve this system, but you can and remember that the relative velocity of bodies after an absolutely elastic frontal impact (and this is our case) remains unchanged, i.e. equal $\upsilon_{0}$, and the length of the cord at the moment when it becomes loose, is equal to $l$. Then the total time to impact will be equal to $$t=\tau+\frac{1}{\upsilon_{0}}=\pi\sqrt{\frac{mM}{k(M+m)}}+\frac{1}{\upsilon_{0} }=0.7\,c.$$
P. Aleksandrov