each time $t$, such that there exist $i$ and $j$, such that $x_i(t)=x_j(t)$. It is problematic because such times $t$ include not only each time of a collision but also each time at which some particles are stuck together. Thus if for example two particles $x_i$ and $x_j$ start with the same position and velocity, then we may lose uniqueness at an arbitrary time $t>0$. Therefore we will enforce that the once stuck trajectories cannot separate. We will do this by replacing equation (\[cs2\]) with (\[cs3\]), which does not distinguish trajectories that once stuck together. Hereinafter we consider (C-S) model defined by (\[cs1\]) and (\[cs3\]). For this model we still do not have uniqueness but the times at which we lose it are restricted only to the times of collisions, which as we will prove occur in some sense rarely.\
Thus our problem and it’s solution is defined as follows.

\[sol\] Let $\{T_n\}_{n\in{\mathbb N}}\cup \{0\}$ be the set of all times of collision of some particles and for each $n$ let $0<T_n\leq T_{n+1}$. For $n\geq -1$, on each interval $[T_n,T_{n+1}]$ (we assume that $T_{-1}=0$) we consider the problem $$\begin{aligned}
\label{cs}
\left\{
\begin{array}{lll}
\frac{dx_i}{dt}=v_i,\\
\frac{dv_i}{dt}=\frac{1}{N}\sum_{k\in B_i(T_n)}(v^n_k-v^n_i)\psi_n(|x_k^n-x^n_i|),
\end{array}
\right.\end{aligned}$$ for $t\in[T_n,T_{n+1}]$, with 