Spring coupled platforms & conservation of momentum - can it be solved with freshman physics?

Question

This question came up as an exercise in a first year undergraduate course I was a TA for. It turned out to be a lot more difficult (impossible?) than anticipated...

Two platforms of mass $M_1$ and $M_2$ ($M_1\neq M_2$) are connected by a spring of constant $k$, and are initially at rest with the spring unstretched, sitting on a frictionless surface. A man of mass $m$ stands on one platform and begins to run, always with a constant speed $v$ measured relative to the platform he is running on. What is the maximum speed reached by the other platform, relative to the ground?

My intuition is telling me that I need to know something about how the man gets from rest to his constant speed (instantaneously? very slowly? with some smooth acceleration?) to solve this, but I haven't been able to prove to myself that this is required. If this is a requirement, I think the most reasonable assumption would be that he reaches his full speed instantaneously.

What I'm most interested in is whether this problem can be tackled with a typical freshman's toolbox - simple arguments around conservation of energy/momentum, no/very limited differential equations, basic calculus. I can see an easy way to get an upper bound on the maximum speed from energy/momentum considerations, but I don't see a way to check if this speed is ever reached.

Answer 1

You can ignore the spring to get the stead state solution using conservation of linear momentum.

With the spring, you will need the natural frequency of the system and its mode shapes to find the maximum velocity. If the man starts with an impulse to $M_1$ to get him from $0$ to $v$ in a small amount of time then the problem is someone hitting the two mass system with a hammer and it can be solved (albeit not on freshman level).

I was able to solve it by fitting the following general solution to the two differential equations

$$ x_1(t) = X_1 \sin(\omega t+\varphi)-X_1 \omega t \cos\varphi-X_1 \sin\varphi-\frac{m v}{M_1} t \\ x_2(t) = X_2 \sin(\omega t+\varphi)-X_2 \omega t \cos\varphi-X_2 \sin\varphi $$

and solving for $X_1$, $X_2$, $\varphi$, $\omega$. It helps to start with $X_1=-\frac{M_2}{M_1+m} X_2$.

The above fits the initial conditions at rest, but with $v_1 =-\frac{m v}{M_1}$.