Collision with a damped spring I am struggling to grasp what happens in the scenario below.
Say a ball, of radius $r$, is rolling on a flat horizontal plane with linear velocity, $V$, and angular velocity, $\omega$. It then collides with a massless spring, attached to a stationary trolley. The spring has a constant damping factor $c$ and stiffness $k$. 
My question is how would you compute the linear velocities of the trolley and ball after collision and how does this collision affect the angular velocity of the ball? 
I know that the restoring force of the spring is given by $$F = -kx - c\nu $$ and I think that $\nu$ is the speed of the object touching the end of the spring. 
You can then set up a second order ODE as 
$$mx^{''} + cx^{'} + kx = 0 .$$ 
Now if I solve this equation for $x(t)$ I could just differentiate it to find the linear velocity of the ball, couldn't I? Then, once I have the linear velocity I can write, assuming no slipping, the angular velocity as $\omega = \nu/r$.  For the trolley at the other end would the ODE be $mx^{''} - cx^{'} - kx = 0 $ or am I going down the wrong path here?
My sticking point here is that I am not clear on the physics for a damped spring. When the ball collides with the spring it must compress and the spring exerts the restoring force =−− as it is a damped spring. This restoring force must retard the motion of the ball. But at what point does the trolley start to accelerate? Is it when the spring is fully compressed or does it start to move instantly when the ball hits the spring attached to the trolley?
Actually thinking about this further if the trolley doesn't instantly accelerate, the restoring force initially can't contain − because the spring is only being compressed. Therefore is there two parts to this collision? Firstly initial compression where there is =− acting on the ball until maximum compression has been reached. Then the damped oscillation where the force =−−$_\text{diff}$ acts on both trolley and ball where $_\text{diff}$ is the relative velocity between the trolley and ball?
 A: $v=\dot x$ is the rate of change of extension of the spring, which equals the difference in velocities between its 2 ends. I would assume that the spring is massless.
As soon as the ball touches one end of the spring the compression force starts to act on the trolley at the other end. The spring is not attached to the ball so the ball loses contact again when the spring reaches its natural length again after compression - ie when $x=0$. In effect the ball and trolley oscillate on the ends of the spring for one half cycle.
I would draw separate free body diagrams for the ball and trolley. The same spring compression force acts on each, but there is also unknown and variable static friction force on the ball which is just sufficient to ensure that there is no slipping. 
Your approach should work but as @Archimaredes suggests using conservation of energy and momentum would be easier. The fraction of KE lost per half cycle of oscillation can be calculated from the Q factor for the spring-mass system, using the reduced mass $\mu=m_1m_2/(m_1+m_2)$ in the definition of Q. Angular momentum of the ball is not conserved because of the friction force, but there is no slipping so friction does no work on the ball and the rotational motion is determined by the linear motion. However there is an external force here so linear momentum is not conserved either. I am not convinced that this simpler method will give the correct result in this case, but if there were no external force (friction) it would be correct.
A: Your approach is correct. You recognized there is only one degree of freemdom, and you have chosen $x$ to be it. The rotation $\theta$ of the ball is dependent on $x$ given a no-slip condition.
With a simple free body diagram, you formulate the equation of motions. The forces acting on the ball are the spring force and the friction to the ground.

$$ \begin{aligned}
  m \ddot{x} & = -F_{\rm s} + F_{\rm f} \\
  I \ddot{\theta} & = -r\, F_{\rm f} 
\end{aligned} $$
Now use $\ddot{x} = r\, \ddot{\theta}$ to solve the problem, but making it all in terms of $x$ and its derivatives.
