Momentum and collisions

First take right as positive. Momentum is defined as mass times signed velocity, $p=mv$. During contact the cars exert equal opposite forces, by Newton’s third law. Divide contact into sufficiently short intervals. In each interval the momentum change is approximately $F\Delta t$. The two changes are opposite at every stage, so their accumulated changes cancel and total momentum is conserved. The contact force varies: one instantaneous force cannot be multiplied by the whole contact time. An elastic collision also has equal total kinetic energy before and after contact. That is an additional collision condition, not a consequence of momentum conservation alone.

Both masses are $m$, with initial velocities $u$ and 0 and final velocities $v_1$ and $v_2$. Divide momentum conservation by $m$ and energy conservation by $m/2$ to obtain the two equations below. Kinetic energy $K=mv^2/2$ follows from work: $F\,ds=m(dv/dt)(v\,dt)=mv\,dv$. Adding from speed 0 to $v$ gives a triangle of base $v$ and height $mv$.

At least one final velocity is zero. The solution $v_2=0$, $v_1=u$ is the original state before momentum exchange. After a genuine contact and separation, the solution is $v_1=0$, $v_2=u$. With the default $u=3\ \mathrm{m/s}$, $A$ stops and $B$ moves at 3 m/s. The blue energy marker transfers during playback; mass does not.