neilparker62 said:
if the string has "negligible" mass and is perfectly elastic, the implication is that the instantaneous collision is only between the two balls
Don't confuse elasticity and extensibility.
In reality, all strings and solid bodies are partly elastic and at least a bit extensible/compressible. That means they have two spring 'constants', ##k_{stress}>k_{relax}>0##. Common idealisations are:
- Perfectly elastic: ##k_{relax}\rightarrow k_{stress}##
- Perfectly inelastic: ##k_{relax}\rightarrow 0##
- Inextensible/incompressible: ##k_{stress}\rightarrow\infty##
Idealised solutions are only valid as the limits of realistic solutions. (Feel free to quote this as Haruspex' Principle, or
Principia Haruspicis.)
When more than one such idealisation is applied, ambiguities can arise wrt the relative rates of the limits. In the current problem we have the spring constants for the ball collision and for the string. You have assumed ##k_{stress,string}/k_{stress,balls}\rightarrow 0##.
Consider a much simpler problem: a light elastic spring sits on a rigid mass, which sits on another light elastic spring, which sits on firm ground. A second equal rigid mass falls on top at speed v.
If ##k_{upper}<<k_{lower}## then most of the energy gets stored in the upper spring and the lower mass barely moves. Having a much higher frequency of oscillation, it rebounds immediately. The lower spring might as well be rigid.
If ##k_{upper}>>k_{lower}## then we have your bungee cord model. The upper mass comes almost to a stop and the lower mass moves down at speed v, then bounces back up. In the meantime, the upper mass is in free fall, and the value of ##\frac{g^2m}{v^2k_{lower}}## will be critical to the subsequent motion. If it is very small we can ignore gravity; the upper mass is stationary while the lower rebounds and, on second collision, transfers all the KE back to the upper mass, producing the same result as above.
In between, we could consider ##k_{upper}=k_{lower}##, which produces a compound oscillation involving the Golden Ratio. To figure out what happens, we need to find the first time at which one of the springs returns to its relaxed length, thereby losing contact with the mass above it. E.g. for the lower spring I got ##(\lambda-1)\sin(\omega't)=\sin(\omega t)##, where ##\lambda=\frac{\sqrt 5-1}2## and ##\omega'=-\frac {\omega}{\lambda^2}##. That could be solved numerically for ##\alpha=\omega t##.
In terms of the post #1 problem, we have to make some assumption about the ratio of the two spring constants.
Taking the string to have a much higher constant seems reasonable (##k_{string}/k_{balls}\rightarrow \infty##), but it still not clear to me what result that produces. Most likely it does allow the combination of KE conservation with a resulting purely horizontal velocity for the suspended mass, but I am uncertain.
If instead we take ##k_{string}/k_{balls}\rightarrow 0## ("bungee" model), we have to worry about a second/concurrent collision. For a given angle of incidence, and again ignoring gravity, that will depend on the value of ##\frac{v^2m}{r^2k_{string}}##, where r is the radius of the ball. If sufficiently large, your original claim applies, and since we are taking ##k_{string}\rightarrow 0##, that will be true.
Finally, if we take ##k_{string}=k_{balls}=k## and ignore gravity, simulation would seem to be the only reasonable prospect. Varying ##\frac{v^2m}{r^2k}## would need to be tried.
There's a dissertation in this.
and refer you to my Insights article: