Explosion of 2 Carts on a Platform (Momentum)

In summary: In a closed system with no escape, momentum can only increase. The more massive the object, the less it changes velocity after each collision.
  • #1
uSee2
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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?
 

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  • #2
uSee2 said:
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.
 
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  • #3
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?
uSee2 said:
Here is where I'm sorta confused. It states that mp is really big, so shouldn't mpvp 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.

uSee2 said:
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.

uSee2 said:
What makes mpvp approach zero rather than infinity if mp is really large?
I don't see it approaching zero. Nothing asserts that.

uSee2 said:
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.
 
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  • #4
uSee2 said:
After their collision, the 2 carts are to the right of x = 0.
Yes, but you should explain why.
uSee2 said:
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.
uSee2 said:
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.
 
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  • #5
haruspex said:
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:
Screenshot 2022-12-29 214926.png


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:
Screenshot 2022-12-29 215158.png

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.
 
  • #6
uSee2 said:
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?
 
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  • #7
haruspex said:
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?
 
  • #8
uSee2 said:
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.
 
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  • #9
uSee2 said:
And since the cart itself is so massive,
You mean the platform, right?
uSee2 said:
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.
 
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  • #10
haruspex said:
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?
 
  • #11
uSee2 said:
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.
 
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  • #12
76078.jpg

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.
 
  • #13
uSee2 said:
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. :smile:
 
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  • #14
uSee2 said:
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.
 
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  • #15
haruspex said:
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)
 
  • #16
uSee2 said:
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?
 
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  • #17
uSee2 said:
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.
 
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  • #18
Halc said:
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?
 
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  • #19
haruspex said:
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:

uSee2 said:
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.

haruspex said:
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.

uSee2 said:
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:

uSee2 said:
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?
 
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  • #20
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... :wideeyed:
 
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  • #21
haruspex said:
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
 
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  • #22
Lnewqban said:
the platform has no reason to move prior to the first collision.
Nobody has suggested it does.
 
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  • #23
haruspex said:
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.
 
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  • #24
erobz said:
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.
 
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  • #25
uSee2 said:
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)
 
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  • #26
erobz said:
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} ##
 
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  • #27
Halc said:
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?
 
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  • #28
erobz said:
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.
erobz said:
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.
 
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  • #29
haruspex said:
@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.
 
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  • #30
erobz said:
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.
 
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  • #31
haruspex said:
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.
 
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  • #32
uSee2 said:
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! :smile: :smile: :smile:
 
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  • #33
erobz said:
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
 
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  • #34
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.

Carts.png


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.
Carts_B.png


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

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