 Solution

Choose the positive direction along the incoming car's velocity. Treat each car as a translating body, take $m_1,m_2>0$, and neglect external impulse, rotation and energy losses during the short elastic collision. If $v_1,v_2$ are the final signed velocities, conservation of momentum and conservation of energy giveFactoring the second relation and using the first gives $U_1+v_1=v_2$ on the collision branch. The other algebraic branch, $v_1=U_1,v_2=0$, describes no impulse and cannot describe separating bodies after contact. Solving the collision branch yieldsFor $0<R<1$ the first car rebounds and the second moves forwards. At $R=1$ the first stops and the second takes its original velocity. For $R>1$ both move forwards, with $v_2-v_1=U_1>0$, so they separate. In the limits $R\to 0$ and $R\to \infty$, the first car's velocity approaches $-U_1$ and $U_1$, respectively.

For the deformable car, take $y=0$ at first contact. The resisting force does negative work. The work-energy theorem therefore givesAt the first stop $v=0$, whence $\sqrt{L-y}=\sqrt{L}-M{V}^2/4$. The speed bound ensures that this right-hand side is positive, so squaring introduces no extraneous solution. The stopping compression isFor $0<V<2L^{1/4}/M^{1/2}$ this lies strictly between $0$ and $L$, as required before complete crushing. The coefficient in the stipulated force law is taken in its stated normalization.