Explosion of 2 Carts on a Platform (Momentum)

[uSee2]

Homework Statement

Two small carts, Cart 1 of 5 kg and Cart 2 of 10 kg are set in contact with each other at position x = 0 on a larger platform. The platform is centered on position x = 0 and is 24 meters long. The carts and platform can roll on bearings of negligible friction and are all intially at rest.

At time t = 0, a small spring between the two carts expands, sending Cart 1 to the left with an intiial speed of 4 m/s. Both carts collide and bounce of the bumpers at the ends of the platform, which can be assumed to be perfectly elastic. When the carts collide with each other again, they stick together.

Suppose the platform's wheels are free to roll, but the platform has much more mass than the two carts. If the experiment is repeated again exactly as before, will the center of the platform be to the left, to the right, or at x = 0 when the carts collide again and stick? Explain.

Relevant Equations

$p = mv$

My Explanation:
This system is a closed system, so the center of mass velocity stays constant. It was initially at rest so the position of the center of mass is constant. After their collision, the 2 carts are to the right of x = 0. Center of mass originally was at x = 0, so the platform had to move to the left to balance it out to keep the center of mass in the same position. As such, the platform is to the left from x = 0.

I believe that my explanation above is correct, I could be wrong though. If my explanation is correct, then couldn't there be any case in which the platform moves to the right or left?

If $m_p$ is the mass of the platform, right is positive, and since momentum initially equals zero, conservation of momentum can be written like:

$p_i = 0$
$p_f = p_i = 0$
$p_f = 0= m_2v_2 - m_1v_1 \pm m_pv_p$
Since $v_1$ and masses are known it can be written as:
$0 = 10v_2 - (5)(4) \pm m_pv_p$
$0 = 10v_2 - 20 \pm m_pv_p$

Here is where I'm sorta confused. It states that $m_p$ is really big, so shouldn't $m_pv_p$ approach infinity? Even if $v_p$ gets smaller, $m_p$ would be large so it would be still large. I know that this definitely doesn't happen in real life, since if it did that would mean $10v_2 - 20$ approaches infinity in the opposite direction, so cart 2 would just fly away.

My questions:
$v_2$ has to be positive, right? It cannot be negative in the 2nd experiment as that means both carts would move to the left.
What makes $m_pv_p$ approach zero rather than infinity if $m_p$ is really large?
Also, are we allowed to assume cart 1 moves at 4 m/s in the 2nd experiment as well? If so, why?

 

[256bits]

Homework Statement:: Two small carts, Cart 1 of 5 kg and Cart 2 of 10 kg are set in contact with each other at position x = 0 on a larger platform. The platform is centered on position x = 0 and is 24 meters long. The carts and platform can roll on bearings of negligible friction and are all intially at rest.

At time t = 0, a small spring between the two carts expands, sending Cart 1 to the left with an intiial speed of 4 m/s. Both carts collide and bounce of the bumpers at the ends of the platform, which can be assumed to be perfectly elastic. When the carts collide with each other again, they stick together.

Suppose the platform's wheels are free to roll, but the platform has much more mass than the two carts. If the experiment is repeated again exactly as before, will the center of the platform be to the left, to the right, or at x = 0 when the carts collide again and stick? Explain.
Relevant Equations:: $p = mv$

After their collision, the 2 carts are to the right of x = 0.

How do you know that when the two carts collide and stick together, they are to the right of x=0?
I don't see where you have calculated v2 to know this.

The first conclusion is correct, but putting in all the necessary steps may have something to do with understanding the part where the platform is massive.

 

[Halc]

There are several unknowns: the width of the small carts (which is essentially negligible and, per the picture, nearly identical). You also don't know the mass of the platform, only that it is significantly more than that of the carts.

None of this info is needed for the questions asked. Some equations are relevant, yes, but mostly it is this event happens before/after that one, this value is positive, negative, or zero. Yes, the platform must move unless both carts hit the ends simultaneously. The question is, does it stop after the second carts does its bounce?

Here is where I'm sorta confused. It states that m_p is really big, so shouldn't m_pv_p approach infinity?

Of course not. Infinite momentum cannot appear from nowhere in a closed system. The bigger the platform mass, the less it changes velocity after each collision event. It gains whatever momentum it takes from the object colliding with it.

My questions:
v2 has to be positive, right? It cannot be negative in the 2nd experiment as that means both carts would move to the left.

Maybe it depends on how long it takes for the 2nd experiment to commence. Let's assume more or less immediately, effectively an elastic bounce of M1 and M2 instead of them sticking to each other.
It is possible for both carts to be moving left if the platform has sufficient momentum to the right, so the reasoning supplied above is insufficient on its own.

What makes mpvp approach zero rather than infinity if mp is really large?

I don't see it approaching zero. Nothing asserts that.

Also, are we allowed to assume cart 1 moves at 4 m/s in the 2nd experiment as well? If so, why?

I think you need to deduce that (or deduce otherwise). You cannot assume it.

 

[haruspex]

After their collision, the 2 carts are to the right of x = 0.

Yes, but you should explain why.

Even if vp gets smaller, mp would be large so it would be still large.

Why? No matter how large the mass is, the velocity can be made so small that the momentum is also small.

v2 has to be positive, right?

Since you have not defined v₂, nor which is the positive direction, I cannot answer that.

I don’t understand why the experiment is repeated if it is repeated exactly. That part is missing from your screenshot, so I cannot verify your statement of it.

 

[uSee2]

Yes, but you should explain why.

Why? No matter how large the mass is, the velocity can be made so small that the momentum is also small.

Since you have not defined v₂, nor which is the positive direction, I cannot answer that.

I don’t understand why the experiment is repeated if it is repeated exactly. That part is missing from your screenshot, so I cannot verify your statement of it.

Sorry everyone, I should have added more info. Here is the problem and its full parts and my thinking:
Part A:

For Part A, I correctly calculated that the collision of the carts after they bounce back is to the right of x = 0 using Kinematics. This only happens however when the platform is braked such that it does not move at all.

Part B:

The reason why I assumed the collision was to the right was because of Part A when I calculated that their collision was to the right. However I now realize that I cannot assume that without justification. I am unsure now how to justify why they collide to the right now in the 2nd experiment, since wouldn't the cart moving be affecting the movements? Cart 1 isn't moving to the left at 4 m/s anymore (since momentum is now distributed among the 2 carts and platform now), so we can't exactly determine where they collide now since no velocities are known.

 

[haruspex]

how to justify why they collide to the right now in the 2nd experiment

You could just go through the same algebra but with a mass variable for the platform.
But since that is much more massive than the carts, a qualitative explanation should suffice.
Roughly speaking, where will the lighter cart be when the heavier one hits its buffer?

 

[uSee2]

You could just go through the same algebra but with a mass variable for the platform.
But since that is much more massive than the carts, a qualitative explanation should suffice.
Roughly speaking, where will the lighter cart be when the heavier one hits its buffer?

Ohhh I see now after drawing the FBD's. Even if the platform is moving, Cart 1 still has less inertia so it accelerates more during the collision since equal forces are exerted on each cart during the collision. And since the cart itself is so massive, the cart itself barely moves. However what I am concerned about is, is the force of static friction from the wheel affect this in any way?

 

[PeroK]

Ohhh I see now after drawing the FBD's. Even if the platform is moving, Cart 1 still has less inertia so it accelerates more during the collision since equal forces are exerted on each cart during the collision. And since the cart itself is so massive, the cart itself barely moves. However what I am concerned about is, is the force of static friction from the wheel affect this in any way?

There are a number of key points that you haven't mentioned. Conservation of momentum, timings, what happens when the carts hit the buffers and what is the final outcome.

If I was just to read your answer, I wouldn't have much idea of what the question was.

Worrying about friction (actually rolling resistance) of the wagon is a minor detail.

 

[haruspex]

And since the cart itself is so massive,

You mean the platform, right?

is the force of static friction from the wheel affect this in any way

The static friction causes some of the KE to go into rotational KE of the wheel instead of linear KE of the cart or platform. But assuming the wheels are small and light, it is irrelevant.

 

[uSee2]

You mean the platform, right?

The static friction causes some of the KE to go into rotational KE of the wheel instead of linear KE of the cart or platform. But assuming the wheels are small and light, it is irrelevant.

Yep, sorry about that I did mean the platform. So since the wheels are so easy to rotate, the static friction force is incredibly small and can be neglected.

Cart 1 is moving to the left at a faster speed than Cart 2, but I'm still confused as to where they collide. Cart 1 has to travel more distance than Cart 2, but since the platform is moving as well, how is the position where they collide able to be determined?

 

[PeroK]

Cart 1 is moving to the left at a faster speed than Cart 2, but I'm still confused as to where they collide. Cart 1 has to travel more distance than Cart 2, but since the platform is moving as well, how is the position where they collide able to be determined?

Conservation of momentum.

 

[uSee2]

Is this correct? So Part B is nearly identical to Part A in terms of the collision. The platform barely moves in Part B, which means the velocities of the carts in Experiment 2 are nearly identical to the velocities of the carts in Experiment 1. This means we can still deduce that the carts still collide to the right of x = 0, since when the platform was braked they collided to the right of x = 0.

From there the center of mass has to stay constant so it can be found out that the platform is to the left of x = 0 since the carts are to the right of x = 0.

 

[berkeman]

Is this correct?

Mentor Note -- In the future, please use LaTeX to post your math here at PF. You can learn the basics quickly by clicking on the "LaTeX Guide" link below the Edit window. Thank you.

 

[haruspex]

View attachment 319572
Is this correct? So Part B is nearly identical to Part A in terms of the collision. The platform barely moves in Part B, which means the velocities of the carts in Experiment 2 are nearly identical to the velocities of the carts in Experiment 1. This means we can still deduce that the carts still collide to the right of x = 0, since when the platform was braked they collided to the right of x = 0.

From there the center of mass has to stay constant so it can be found out that the platform is to the left of x = 0 since the carts are to the right of x = 0.

Looks right, but I'd still like to see a simple explanation of whereabouts cart 1 will be when cart 2 hits its buffer.

 

[uSee2]

Looks right, but I'd still like to see a simple explanation of whereabouts cart 1 will be when cart 2 hits its buffer.

My thought process was that in Part A, when the platform was fixed to the floor, cart 1 travelled to the left at 4 m/s. Then using conservation of momentum it could be deduced that cart 2 travels to the right at 2 m/s. This can be transferred over to Part B where the cart is not fixed. Since it is so massive, it is basically fixed, so the information from Part A can be transferred over.

Since Cart 1 travels to the left faster than Cart 2 in terms of speed, Cart 1 hits its buffer first and bounces back with its exact same speed, just in an opposite direction.

Cart 1:
$t = 0$ when explosion happens
$\Delta x = v_ot + 0.5at^2$
$12 = (4)t$
Cart 1 hits its buffer at 3 seconds.

Cart 2:
$t = 0$ when explosion happens
$\Delta x = v_ot + 0.5at^2$
$12 = (2)t$
Cart 2 hits its buffer at 6 seconds.

To get where Cart 1 is when Cart 2 hits its buffer, elapsing another 3 seconds after Cart 1 bounces yields that Cart 1 travels another 12 meters, so now it is back at the origin (x=0) when Cart 2 just hits their buffer. It also tells us that they collide at t = 8 in Part A, so elapsing another 2 seconds yields that Cart 1 and Cart 2 collide to the right of x = 0. The platform moving affects it, but the platform doesn't move that much since it is just that massive.

Was my kinematics correct? (I haven't done it in a good few months so I might be wrong haha)

 

[Halc]

The platform barely moves in Part B, which means the velocities of the carts in Experiment 2 are nearly identical to the velocities of the carts in Experiment 1. This means we can still deduce that the carts still collide to the right of x = 0, since when the platform was braked they collided to the right of x = 0.

The carts colliding to the right of x=0 is not just a function of the cart velocities, which are admittedly nearly the same in both experiments. What is significantly different between parts A and B? How does that affect the collision location?

 

[haruspex]

it is back at the origin (x=0) when Cart 2 just hits their buffer

Yes, that's what I was after. And it follows from that that they will meet again well to the right.

 

[haruspex]

The carts colliding to the right of x=0 is not just a function of the cart velocities, which are admittedly nearly the same in both experiments. What is significantly different between parts A and B? How does that affect the collision location?

The reasoning in post #1 looks solid to me. Do you see a flaw?

 

[Halc]

The reasoning in post #1 looks solid to me. Do you see a flaw?

Post 1 is the OP, and I replied to that in post 3. I don't like the momentum equations expressed as differences instead of sums. In a closed system, the sum of all momentum is constant (zero in this case), not the difference. I didn't comment on that at the time since the OP seems to use positive
speeds instead of velocities. The OP expressed more questions than reasoning.

The bit of mine you quoted was from post 12 where it is explicitly stated that movement to the right is positive (velocity, not speed). In that case, the handwritten equations are wrong.
0=m2v2+m1v1+mpvp
There is no subtraction.

At no point has uSee2 done an approximate calculation of where the first collision between the carts takes place. This can be computed exactly if the platform is fixed, and approximately (in relation to the fixed answer) for the floating platform. Also, no expression of the speed of the two attached carts after the first inelastic collision has been made. Is it zero? Fast or slow? If nonzero, which direction?I cannot just blurt answers here despite the lack of actual questions posed in the description picture in the OP. The problem in the book also doesn't mention how much time passes before the carts are expelled apart the 2nd time. It is almost immediately (an elastic collision), or a significant time? I presumed (perhaps incorrectly) that the 2nd experiment expells the carts after the first inelastic collision takes place, and is not done by resetting all the objects stopped at the original position at x=0, but I may be wrong:

If the experiment is repeated again exactly as before, will the center of the platform be to the left, to the right, or at x = 0 when the carts collide again and stick?

It says if repeated exactly as before, which literally means that everything gets put back to x=0. The question, if interpreted that way, seems to ask if physics will behave differently on a 2nd try of the exact same experiment, which is a silly question. So I took it as the carts being expelled at the same velocities relative to their current mutual center of mass, which is now located somewhere other than x=0 (and is not necessarily stationary relative to the platform.

Do you see a flaw?

uSee2 in post 12 asks if the 2nd collision also takes place to the right of x=0. I can think of conditions (unspecified by the problem as stated) where this is not true, so yea, I see a flaw.

So Part B is nearly identical to Part A in terms of the collision. The platform barely moves in Part B, which means the velocities of the carts in Experiment 2 are nearly identical to the velocities of the carts in Experiment 1. This means we can still deduce that the carts still collide to the right of x = 0, since when the platform was braked they collided to the right of x = 0.

No, Part B is not nearly identical to Part A. I had asked in my reply what that difference is.
The platform barely moves. That's true. Conservation of center of mass of system demands that. But the bolded statement above does not follow from the statement immediately preceding it, so the 'this means' parts is wrong.

I'd like to use a few numbers (like where the first collision takes place approximately, and the velocity thereafter), but none of these has been supplied so far and I'm not just going to blurt it out. I'll just say that there are regions along the line where the above bolded statement is not true, so the reasoning in post 12 is not solid.

About the new post:

Cart 2 hits its buffer at 6 seconds.
...
It also tells us that they collide at t = 8 in Part A

At approximately 6 and 8 seconds respectively. The platform is massive, but not infinitely massive. Is it slightly more or less than 6 and/or 8 seconds? Where does the collision take place, near what point (value of x), and on which side of that point?
What is the end velocity of everything after the first collision?

 

[erobz]

I think cart 2 hits its bumper at approximately $t = 3 \rm{s}$ in the limit as the platform mass goes to $\infty$. If the mass of the cart is finite, it reaches the right bumper in less than 3 seconds.

Never mind. Wrong platform half distance...

 

[Lnewqban]

The reasoning in post #1 looks solid to me. Do you see a flaw?

Why is the platform moving only because the spring between the two cars simultaneously pushes on both?
In my mind, the platform has no reason to move prior to the first collision.
Would you mind commenting about that?
Thanks

 

[haruspex]

the platform has no reason to move prior to the first collision.

Nobody has suggested it does.

 

[erobz]

Nobody has suggested it does.

The OP's first few attempts seem to indicate the platform has some momentum at $t = 0$ when the separating blocks gain their momentum.

 

[haruspex]

The OP's first few attempts seem to indicate the platform has some momentum at $t = 0$ when the separating blocks gain their momentum.

Not seeing it. Please quote the relevant passages.

 

[erobz]

If $m_p$ is the mass of the platform, right is positive, and since momentum initially equals zero, conservation of momentum can be written like:

$p_i = 0$
$p_f = p_i = 0$
$p_f = 0= m_2v_2 - m_1v_1 \pm m_pv_p$
Since $v_1$ and masses are known it can be written as:
$0 = 10v_2 - (5)(4) \pm m_pv_p$
$0 = 10v_2 - 20 \pm m_pv_p$

This is what seems off to me. (I'm not sure what @Lnewqban was seeing)

 

[erobz]

This is what seems off to me. (I'm not sure what @Lnewqban was seeing)

Ignoring friction between the masses and platform taking to the left as positive ${}^+ \leftarrow$

$mv_l = -2mv_r \implies v_r = -\frac{v_l}{2}$

 

[haruspex]

So I took it as the carts being expelled at the same velocities relative to their current mutual center of mass, which is now located somewhere other than x=0

Ah, that's the key. Neither @uSee2 nor I read it that way.
I agree that would make the question a bit more interesting, but it seems pretty clear to me that the brakes are the only intended difference.
Indeed, if a different start point were intended, why also take the brakes off?

 

[haruspex]

This is what seems off to me. (I'm not sure what @Lnewqban was seeing)

@uSee2 did not define any variables. In particular it is unclear what point in time those velocities refer to. Since the verbal reasoning and conclusions drawn were correct, I did not worry much about that.

Ignoring friction between the masses and platform taking to the left as positive

$mv_l = -2mv_r \implies v_r = -\frac{v_l}{2}$

Looks fine to me, and no implication that the platform moves.

 

[erobz]

@uSee2 did not define any variables. In particular it is unclear what point in time those velocities refer to. Since the verbal reasoning and conclusions drawn were correct, I did not worry much about that.

Looks fine to me, and no implication that the platform moves.

From the values they plugged in, it seems like the initial state.

 

[haruspex]

From the values they plugged in, it seems like the initial state.

But the preceding equation starts with $p_f=$, so this is not the initial state.
So why is the velocity taken as 4m/s still? Don't know, but maybe this is on the basis that it will not have changed much. Or maybe it was not thought through carefully in making the post because that part of the question was a generic one about the product of a very small value and a very large one.

 

[erobz]

But the preceding equation starts with $p_f=$, so this is not the initial state.
So why is the velocity taken as 4m/s still? Don't know, but maybe this is on the basis that it will not have changed much. Or maybe it was not thought through carefully in making the post because that part of the question was a generic one about the product of a very small value and a very large one.

The way I did it, I only ever the considered the momentum of one of the masses and the platform before and after a given collision.

 

[Lnewqban]

If $m_p$ is the mass of the platform, right is positive, and since momentum initially equals zero, conservation of momentum can be written like:

$p_i = 0$
$p_f = p_i = 0$
$p_f = 0= m_2v_2 - m_1v_1 \pm m_pv_p$
Since $v_1$ and masses are known it can be written as:
$0 = 10v_2 - (5)(4) \pm m_pv_p$
$0 = 10v_2 - 20 \pm m_pv_p$

Then, I missunderstood the above equation.
Happy New Year!

 

[haruspex]

The way I did it, I only ever the considered the momentum of one of the masses and the platform before and after a given collision.

Using the approach in post #1, you don’t need to analyse each bounce that carefully. You just need to note that
- when the light mass hits its buffer the heavier is only half way
- the light mass bounces back with only slightly diminished speed, and the platform picks up even less
- the light mass will be nearly back to the start point when the heavy mass hits its buffer
- ergo, the masses reunite well to the right

 

[kuruman]

For whatever it's worth, here is a plot of the positions of the two cart and platform as a function of time. The parameters of the masses are 1, 2 and 80 units (1 unit = 10 kg). The corresponding initial velocities were -4 m/s, +2 m/s and 0 m/s. When the two carts stick together their common velocity is -9.6×10^-2 m/s while the platform's velocity is +3.6×10^-3 m/s.

And here is the same plot with the brake engaged. I took the mass of the platform to be equal to the mass of the Earth, 6×10²³ units.

On edit: Edited to address the issues about the plots raised in posts #35 and #36. See post #37.

 

[uSee2]

For whatever it's worth, here is a plot of the positions of the two cart and platform as a function of time. The parameters of the masses are 1, 2 and 80 units (1 unit = 10 kg). The corresponding initial velocities were -4 m/s, +2 m/s and 0 m/s. When the two carts stick together their common velocity is -9.6×10^-2 m/s while the platform's velocity is +3.6×10^-3 m/s.

View attachment 319607

And here is part B. I took the mass of the platform to be equal to the mass of the Earth, 6×10²³ units.

View attachment 319609

I'm not sure if I'm intepreting this right, but why didn't the platform move much when the red cart (by that I mean the red line) collided with its buffer? Was it just because it was so massive?

 

[haruspex]

For whatever it's worth, here is a plot of the positions of the two cart and platform as a function of time. The parameters of the masses are 1, 2 and 80 units (1 unit = 10 kg). The corresponding initial velocities were -4 m/s, +2 m/s and 0 m/s. When the two carts stick together their common velocity is -9.6×10^-2 m/s while the platform's velocity is +3.6×10^-3 m/s.

View attachment 319607

And here is part B. I took the mass of the platform to be equal to the mass of the Earth, 6×10²³ units.

View attachment 319609

I don't understand the green lines. Why do both suddenly jump from 0 to 12 when the light cart hits its buffer? And what does the mass matter in case A, with the brakes on?

 

[kuruman]

I'm not sure if I'm intepreting this right, but why didn't the platform move much when the red cart (by that I mean the red line) collided with its buffer? Was it just because it was so massive?

Please see explanation below.

I don't understand the green lines. Why do both suddenly jump from 0 to 12 when the light cart hits its buffer? And what does the mass matter in case A, with the brakes on?

I understand the green lines. They are the result of my misguided effort at multitasking: simultaneously watch football on television, celebrate the impending arrival of the new year and post here. After the first three seconds the data points for the platform shift from the position of the center to the position of the right end. I have edited the post to fix that blunder.

The first plot addresses the situation "Suppose the platform's wheels are free to roll, but the platform has much more mass than the two carts." I considered a platform mass of 80 times the mass of the lighter cart to be "much more." The second plot addresses the engaged brake situation in which case the mass of the platform is equivalent to the mass of the Earth. I added plot titles to distinguish the two.

Finally, I should note my appreciation at the clarity of thought resulting from sobriety the morning after.

 

[erobz]

Using the approach in post #1, you don’t need to analyse each bounce that carefully. You just need to note that
- when the light mass hits its buffer the heavier is only half way
- the light mass bounces back with only slightly diminished speed, and the platform picks up even less
- the light mass will be nearly back to the start point when the heavy mass hits its buffer
- ergo, the masses reunite well to the right

But that OP asks for the position of the platform relative to $x=0$ when the small masses unite, not where the position of the small impacting masses are relative to the center of the cart when they reunite. Maybe I'm under thinking it, but that seems less obvious to me.

will the center of the platform be to the left, to the right, or at x = 0 when the carts collide again and stick?

Is it the case that since both small masses are on the RHS of $x = 0$ that the center of mass of platform must be on the LHS of $x= 0$, such that the combined COM of all three masses remains at $x= 0$ for the entire duration of the collisions?

Its dawning on me that should be the case given no external forces are acting on the system.

 

[Chestermiller]

My inclination is to analyze the problem in detail. If m is either of the masses and $m_p$ is the mass of the much higher mass platform, the equations of conservation of momentum and conservation of energy for an elastic collision (multiplied by a factor of two) are: $$mv_0+m_pv_{p0}=mv_1+m_pv_{p1}\tag{1}$$and $$mv_0^2+m_pv_{p0}^2=mv_1^2+m_pv_{p1}^2\tag{2}$$Rearranging these slightly yields:$$m_1(v_1-v_0)=-m_p(v_{p1}-v_{p_0})\tag{3}$$and$$m_1(v_1^2-v_0^2)=-m_p(v_{p1}^2-v_{p_0}^2)\tag{4}$$If we divide Eqn. 3 by Eqn. 4, we obtain: $$v_1+v_0=v_{p1}+v_{p_0}\tag{5}$$Solving Eqns. 1 and 5 for the velocities after the collision yields: $$v_{1}=\frac{(m-m_p)}{(m+m_p)}v_0+\frac{2m_p}{m_p+m}v_{p0}\tag{6}$$ and$$v_{p1}=\frac{(m_p-m)}{(m_p+m)}v_{p0}+\frac{2m}{m_p+m}v_{0}\tag{7}$$Linearizing these equations with respect to (small) $\epsilon=m/m_p$ then gives: $$v_1=-(1-2\epsilon)v_0+2(1-\epsilon)v_{p0}\tag{8}$$and$$v_p=(1-2\epsilon)v_{p0}+2\epsilon v_0\tag{9}$$

The first collision occurs at t = 3 s between the left mass and the left bumper. Prior to this collision, the velocity of the left mass is -4 m/s and the velocity of the platform is zero. From the above equations, after this collision, the velocity of the platform is $v_p=-8\epsilon\ m/s$ is $v_L=(4-8\epsilon_L)\ m/s$, where $\epsilon_L=m_L/m_p$. The location of the left mass after this collision is at $$x_L=-12+(4-8\epsilon_L)(t-3)$$The location of the right bumper after this first collision is at $$x_{RB}=12-8\epsilon_L (t-3)$$The location of the right mass after this first collision is at $$x_R=6+2(t-3)$$So the collision between the right mass and the right bumper occurs when $$x_R=x_{RB}=6+2(t-3)=12-8\epsilon_L (t-3)$$at time $$t=3+\frac{3}{1+4\epsilon_L}=6-12\epsilon_L$$Prior to this collision with the right bumper, the velocity of the right mass is $v_R=2\ m/s$ and the velocity of the platform is $v_p=-8\epsilon_L\ m/s$. After this collision with the right bumper, the velocity of the right mass is $$v_R=-2(1-2\epsilon_R)-16(1-\epsilon_R)\epsilon_L$$ and the velocity of the platform is $$v_p=-8\epsilon_L(1-2\epsilon_R)+4\epsilon_R$$Linearizing with respect to the $\epsilon$s and taking into account that $\epsilon_R=2\epsilon_L$, these relationships reduce to $$v_R=-2-8\epsilon_L$$ and $$v_p=0$$
According to this, the platform stops moving after the collision of the right mass with the right bumper. According to the previous equations, the right bumper us at $$x_R=12-24\epsilon_L$$ when this collision occurs. So the center of the platform is displaced to the left $24\epsilon_L$.

 

[haruspex]

since both small masses are on the RHS of x=0 that the center of mass of platform must be on the LHS of x=0, such that the combined COM of all three masses remains at x=0 for the entire duration of the collisions?

Yes. From post #1:
"After their collision, the 2 carts are to the right of x = 0. Center of mass originally was at x = 0, so the platform had to move to the left to balance it out to keep the center of mass in the same position. As such, the platform is to the left from x = 0."

 

[Chestermiller]

To get a better idea how this all plays out in a "practical" case, I have used Eqns. 6 and 7 of post #39 to establish the details of what transpires for the specific case where the mass of the platform is 100 kg. Here are the results of the calculations:

The collision between the left mass and the left bumper occurs at 3 seconds at x = -12 m. After this collision, the velocities of the masses and the platform are $$v_L=3.619\ m/s$$$$v_R=2.0\ m/s$$and $$v_P=-0.381\ m/s$$The total momentum of the two masses and the platform remains zero.

The collision between the right mass and the right bumper occurs at 5.52 seconds at x = 11.04 m. After this collision, the velocities of the masses and the platform are $$v_L=3.619\ m/s$$$$v_R=-2.327\ m/s$$and $$v_P=0.0519\ m/s$$The total momentum of the two masses and the platform remains zero.

The final collision between the two masses occurs at 7.86 seconds at x = 5.59 m. The center of the platform at this time is at x = -0.84 m