Hide a solution

**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*

See a solution

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
$\frac{l_{1}}{l_{2}}=\frac{M}{m}$, there is center of mass,
$\upsilon_{ц}$ is its speed (fig.2). Piece stiffness a snua of
length $l_{1}$ is equal to

*Fig. 2*

$$k_{1}=k\frac{l_{1}+l_{2}}{l_{1}}=k\left(1+\frac{m}{M}\right),$$

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