Is energy conserved is this problem?

[malawi_glenn]

1 kg bead and 1000 J initial energy, quite tall ramps :)

 

[jbriggs444]

1 kg bead and 1000 J initial energy, quite tall ramps :)

Yes. When I re-ran to look at elapsed time, separation distance and total bead travel, I scaled that back to a 1 meter launch height. Expected values (over a distribution of mass ratios) for elapsed time and separation distance probably diverge (integral of an approximately harmonic function). Typical elapsed times are modest and are dominated by the final trip. For instance, about 2.5 minutes total for the first 34 one way trips and about 15 minutes additional for the last.

[In order to get meaningful figures for elapsed time, one needs for the bounces to take non-zero time. Otherwise, the distance between the ramps is zero, all the trips take place instantly and the total elapsed time is zero. So I imputed a bounce time based on twice the free fall time based on the bead's share of the impact energy at each bounce. And counted only 1 times free fall time for the launch. So I needed a reasonable launch height]

 

[Rikudo]

Will there be any deformation of the objects causing internal energies to be altered?

Umm... By "internal energies" here, do you refer to the whole system's (slide A,B and mass C)?

Just out of curiosity, which book?

Pathfinder.

 

[malawi_glenn]

Umm... By "internal energies" here, do you refer to the whole system's (slide A,B and mass C)?
Pathfinder.

The Individual objects

Never heard of pathfinder, which author and publisher?

 

[jbriggs444]

Never heard of pathfinder, which author and publisher?

Google suggests: https://www.amazon.com/dp/B072QR5CWH/?tag=pfamazon01-20

Arvind Tiwari & Sachin Singh
Pearson India, 2016 - 376 pages

 

[Orodruin]

Unfortunately it is yet another case of guessing intent from subtle detail in the wording.

Assuming the bead matches the velocity of one slide allows two slide speeds to be deduced, the one the bead last encountered and the one it is vainly chasing.
What we cannot do is say which is A and which is B. They will take it in turns to be the faster.

The question says "Find expressions for the speeds". I note that a) it says speeds, not speed, and b) it is not entirely clear that it requires us to identify which slide has which speed.
I'm fairly confident that such a solution would be more accurate than $\mathcal O(m/M)$. I tried to start with the exact solution then approximate, but I get a 3x3 transition matrix, so it's a bit messy. There's probably an easier way.

As to which is intended, I toss a coin.

The thing is, you cannot assume that the bead will have the same velocity as one of the ramps. These are extreme cases of the possible solutions. The range of solutions include the bead having any speed between these extreme solutions. You cannot know which without performing a series argument with a cutoff condition.

The assumption of assuming the bead to have the same speed as either block is therefore no more accurate than neglecting the bead momentum and energy. Even if the bead does have the same velocity as one of the blocks, you won’t know which - giving you an inherent uncertainty that goes as m/M.

 

[kuruman]

I modified my approach and assumed that the bead has a final speed that is a fraction of the final speed of the slide on the left, $v_{\text{bead}}=\beta ~V_{\text{L}}$ ($0 \leq\beta\leq 1.)$ I solved the momentum and energy conservation equations from the moment the bead starts moving at the top of the left slide until all masses are moving away from each other with final speeds $V_{\text{L}}$, $V_{\text{R}}$ and $\beta ~V_{\text{L}}.$

Then I assumed a mass ratio $\mu=m/M=10^{-4}$ and plotted the two speeds normalized to $\sqrt{2gh}$ as a function of $\beta$. The plot is shown below. The orange line is the speed of the right slide and the blue line is the speed of the right left slide. I will not post the equations because I think this homework problem is still live. However, it is fair to say that a Taylor series expansion of the difference gives a leading term $$\frac{V_{\text{R}}-V_{\text{L}}}{\sqrt{2gh}}=\frac{\mu^{3/2}}{\sqrt{2}}\beta.$$ In the limit $m<<M$, the speeds of the slides are, to a very good approximation, the same and this result is independent of $\beta$, i.e. the final speed of the bead.

I think that this holistic approach complements while I compliment the detailed work by @jbriggs444.

 

[Orodruin]

The orange line is the speed of the right slide and the blue line is the speed of the right slide.

I think one of those should be a left.

 

[haruspex]

you cannot assume that the bead will have the same velocity as one of the ramps

I see that now. In post #22 I gave a reason why I thought it would always be close to a ramp's speed, but, as I acknowledged in post #32, @jbriggs444 proved me wrong.

 

[kuruman]

I think one of those should be a left.

You're right. It's fixed, thanks.