I Solving a momentum problem where masses change after the collision

[rdemyan]

TL;DR Summary

Looking for potential methods for solving a momentum problem where the masses after the collision are different but the sums of the masses remains the same.

I'm trying to figure out what is the best way to solve the problem presented in the attached diagram which is a collision problem. The knowns are $m_1, m_2, u_1, u_2,$ and v (perhaps surprisingly). The half angle of the collision, $\beta$ is also known. The drawing shows what could potentially happen if the momentum of 1 exceeds that of 2, however, it could be the reverse or they could even be equal. I want to determine $m_b,m_f, \alpha,\phi$. The oddity here is that the masses after the collision are not the same as before the collision, BUT, the sums are, so

$$m_1 +m_2 = m_f +m_b$$

I have tried several methods, but since I am new to them, I may very well have made errors that are keeping me from coming to a solution. In particular, when I start plugging numbers into a spreadsheet, some of the reasonable initial conditions result in $m_b$ being less than zero, which is not possible. I've looked at changing to the center of mass reference frame (which seemed somewhat promising, but I wasn't able to figure out how to get the angles) and the oblique collision method (had trouble getting the additional equation which supposedly comes from the coefficient of restitution). Those are the methods I have tried, but there may be others I should look at.

I'm looking for suggestions on what method to use to solve this problem. Thanks.

EDIT: I'm not sure if this matters but this problem is always viewed as 1 and 2 going from left to right. So 2$\beta$ is never greater than 180 degrees. If it were, I would simply flip the diagram so once again 1 and 2 are traveling from left to right.

 

[haruspex]

Consider a simple head on collision. In the usual case, e is directly related to the fraction of KE remaining. What is that relationship?
I am suggesting that you could define e in this mass transfer case such that the same relationship holds.

The forces that act between two bodies in contact are friction in the plane of contact and the "normal" force, i.e. normal to the plane of contact. So ignoring friction, the change in their velocities is normal to the plane of contact. There is no change to their velocities parallel to that plane.
This means that for what you are trying to do you can reduce the collision to one dimension.

 

[rdemyan]

Consider a simple head on collision. In the usual case, e is directly related to the fraction of KE remaining. What is that relationship?
I am suggesting that you could define e in this mass transfer case such that the same relationship holds.

The forces that act between two bodies in contact are friction in the plane of contact and the "normal" force, i.e. normal to the plane of contact. So ignoring friction, the change in their velocities is normal to the plane of contact. There is no change to their velocities parallel to that plane.
This means that for what you are trying to do you can reduce the collision to one dimension.

The fraction of kinetic energy lost is proportional to $1-e^2$. But $u_1$ and $u_2$ are, in general, different. Since the masses change after the collision, it doesn't seem that straightforward to me. At collision angles of 90 degrees and less, $m_f$ will be quite a bit larger than $m_b$ and also approaching twice the mass equal to half the sum of the incoming masses. The lower the angle $\beta$, the more $m_f$ approaches $m_1+m_2$.

"There is no change to their velocities parallel to the plane". Yeah, but the velocities are different and there is mass transfer. Are you suggesting some kind of weighting technique where $v_x$ in each direction (forward and backward) is somehow weighted? Also, as I write this I'm wondering if you are suggesting that the magnitude of $v_x$ in the forward direction is equal to the magnitude of $v_x$ in the backward direction?

All I know from the problem statement is that the magnitude of $v$ in the forward direction equals the magnitude of $v$ in the backward direction. Now if the magnitudes of the x-components are equal, then the magnitudes of the y components have to be equal, which means that the magnitudes of the angles $\alpha,\phi$ are equal.

 

[haruspex]

there is mass transfer

Good point, but I would assume the transferred mass takes its momentum with it. That means it is still true that [for the body losing mass] velocities parallel to the contact plane are conserved.

 

[rdemyan]

I suspect you are right about the transferred mass taking its momentum with it. But, there isn't any way to know how much of $m_1$ at its velocity of $u_1$ gets transferred and the same with stream 2.

With regards to splitting up the bodies, my first post on this forum regarding this problem tried to use that idea. If $\phi = \pi+\alpha$, then the forward and backward directions are colinear and I can rotate the drawing such that this colinear line is on the x axis. The general idea was to look at each body individually and calculate the forward and backward amounts from that body independent of the other body. Do this for the other body as well and then $m_b, m_f$ are simply the sums of the amounts calculated from each body. Here's the link to that post.

https://www.physicsforums.com/threa...treams-impinging-at-differing-angles.1079454/

At the time I came up with this idea and analysis I felt unsure because it didn't seem to coincide with any of the classical physics types of analyses for similar collisions (albeit with no mass transfer). This morning I started looking at it again since the classical oblique impact method seemed to be a dead end. But this splitting technique does use the idea that the velocities in the x direction don't change at least if the collision is perfectly elastic.

 

[rdemyan]

Still looking for ideas on how to solve the problem mentioned in the original post. Does it seem like working in the center of mass reference frame should work?

 

[haruspex]

I suspect you are right about the transferred mass taking its momentum with it. But, there isn't any way to know how much of $m_1$ at its velocity of $u_1$ gets transferred and the same with stream 2.

(I just made a clarification to my preceding post.)
How much mass is transferred must be one of the input parameters, making the complete set:
- the initial masses and velocities
- the contact angle
- the transferred mass
(In three dimensions, the contact angle has two degrees of freedom.)
I see two ways to approach this:

1. Apply conservation of momentum and a specified reduction in KE expressed in terms of e. That should give you enough information to solve it.
2. Construct a model for the process. E.g.
   • resolve the motion of each body at contact into a velocity normal to the contact plane and one parallel to it
   • for the parallel components, the body losing mass maintains its velocity but the lost mass coalesces with the other body, conserving momentum. This means there is inevitably some loss of KE.
   • for the normal components, momentum is still conserved and KE may be conserved. Assign a coefficient of restitution, e', to this aspect.
   • whether you then define the CoR of the collision as e' or representing the overall degree of KE preservation is a matter of choice.

Interesting question: will (1) lead to an expression that breaks (e.g., sqrt of a negative value) when the transferred mass is nonzero and there is a difference in the initial parallel velocities and you set e=1? According to approach (2), that should be impossible.

 

[A.T.]

(In three dimensions, the contact angle has two degrees of freedom.)

In 3D you can use a reference frame where both incoming velocites are in the XY-plane.

Now you have to decide if the outgoinng velocites must stay the XY plane or can have Z-compononets. The outgoing Z-momenta must obviously add to zero.

 

[haruspex]

In 3D you can use a reference frame where both incoming velocites are in the XY-plane.

If the particles are small spheres, say, the trajectories may be skew lines, i.e. neither intersecting nor parallel. This results in two degrees of freedom for the contact angle.
I always consider idealisations such as point particles to be the limit of real contexts. In this case, it means that the contact angle still has two degrees of freedom even though the trajectories, in the limit, intersect at a point.

 

[rdemyan]

I see two ways to approach this:

1. Apply conservation of momentum and a specified reduction in KE expressed in terms of e. That should give you enough information to solve it.
2. Construct a model for the process. E.g.
   • resolve the motion of each body at contact into a velocity normal to the contact plane and one parallel to it
   • for the parallel components, the body losing mass maintains its velocity but the lost mass coalesces with the other body, conserving momentum. This means there is inevitably some loss of KE.
   • for the normal components, momentum is still conserved and KE may be conserved. Assign a coefficient of restitution, e', to this aspect.
   • whether you then define the CoR of the collision as e' or representing the overall degree of KE preservation is a matter of choice.

Interesting question: will (1) lead to an expression that breaks (e.g., sqrt of a negative value) when the transferred mass is nonzero and there is a difference in the initial parallel velocities and you set e=1? According to approach (2), that should be impossible.
[/QUOTE]


Let me write down the equations from what I've already done as per the link in reply #5. However, in that link I assumed that $\phi = \pi+\alpha$ and thus the outgoing streams were colinear. Let me remove that assumption. Let's look at the split of each body individually.

$$m_1u_1cos\beta = m_{f1}v_1cos\alpha - m_{b1}v_1cos\phi$$
where $m_{f1}$ and $m_{b1}$ are the splits in the forward and backward direction of the body #1 (actually a liquid stream but let's assume I can model it via particle collisions). I'm still assuming that the velocities of the forward and backward masses ($v_1$ for #1) are the same. The mass conservation for #1 is,
$$m_1 = m_{f1} + m_{b1}$$

Similarly, for #2 the equations are,
$$m_2u_2cos\beta = m_{f2}v_2cos\alpha - m_{b2}v_2cos\phi$$
$$m_2 = m_{f2} + m_{b2}$$

For now, let's keep it simple and set e = 1. When e = 1, the splits from each individual stream, hypothetically, maintain their velocity as given by that stream before the collision (i.e $v_1 = u_1$ and $v_2 = u_2$) . QUESTION: Should these equalities instead be the x-components of the velocities, for example, $v_{1x} = u_{1x}$? EDIT: But if that is the case then doesn't that mean that $\phi = \alpha$ (I've already included the negative sign in the x momentum equations for each stream above)

There's also an equation for the conservation of momentum in the y-direction or line of impact direction,

$$m_1u_1sin\beta-m_2u_2sin\beta = m_fvsin\alpha + m_bvsin\phi$$
QUESTION: I don't think I can write individual stream or particle equations for the y-direction (line of impact) like I did for the x direction. But I'm not sure I understand why I can't.
Ultimately, the amount of mass in the forward stream is,

$$m_f = m_{f1} + m_{f2}$$
and for the backward stream,

$$m_b= m_{b1} + m_{b2}$$
with the idea being that the splits are theoretically combined without any loss of energy (since I've assumed e = 1). With loss of energy, I suppose I would have to assume that this combining was included in the coefficient of restitution.

So the y momentum equation can be rewritten as,

$$m_1u_1sin\beta-m_2u_2sin\beta = (m_{f1} +m_{f2}) vsin\alpha + (m_{b1}+m_{b2})vsin\phi$$


So, these are the unknowns: $m_{f1},m_{b1},m_{f2},m_{b2}, \alpha, \phi$, which total six. I currently have five equations.

If I assume that $\phi = \pi+\phi$ so that the outgoing streams are colinear, I can already solve the system of equations.

So, I'm stuck here. Note, I am assuming that I know the value of $v$, so if I do, I'm not sure that somehow incorporating kinetic energy equations will help.

EDIT: There is still an overall x momentum balance given by,

$$m_1u_1cos\beta+m_2u_2cos\beta = (m_{f1} +m_{f2}) vcos\alpha - (m_{b1}+m_{b2})vcos\phi$$

but I'm not sure I can use this since I already used x momentum equations for each body (stream) individually.

 

[rdemyan]

UPDATE: I've come up with equations for $m_b,m_f$ and plugged them into the overall y momentum balance. Now, I'm just picking some cases on what $\alpha, \phi$ might be just to check for consistency and if any negative values are generated (for example $\phi = \pi - \alpha$ or $\phi = \pi +\alpha$ or $\phi = 0$ or $\alpha > \beta$ and what if $m_1u_1 = m_2u_2$ in which case $\alpha = \phi = 0$). So far, so good. The only time negative numbers are generated is for $m_b$ when $\alpha > \beta,$ but I already knew this from my previous work per the link posted in Reply #5. And $\alpha > \beta$ is not possible in a real world situation. As the momentum ratio increases $\alpha$ approaches $\beta$ but can never exceed it.

Just need that extra equation.

 

[haruspex]

actually a liquid stream

That seems very different from what I thought you were doing. Please post the actual problem to be solved.

the splits from each individual stream

there is a two-way mass transfer?

the splits are theoretically combined without any loss of energy

I doubt it. Coalescence always results in KE loss.

 

[rdemyan]

That seems very different from what I thought you were doing. Please post the actual problem to be solved.

there is a two-way mass transfer?

I doubt it. Coalescence always results in KE loss.

1) The drawing shown is actually pretty representative. Although I didn't realize it at the time I developed the equations in the link provided in reply #5 to this post, I think I am using the ideas developed for oblique collisions of particles (with regards to the tangential velocities). The link I posted in reply #5, shows a diagram of two equal liquid jets colliding in reply #18 of that post (the scattering angles are zero degrees in the forward direction and 180 degrees in the backward direction) and actual photographs of a liquid sheet formed from the collision in reply #12. The collision results in a stagnation point in the impingement zone around which liquid is deflected. However, the impingement zone is typically mildly turbulent, which is one of the reasons I believe that the velocity, $v$, of each outgoing stream (f and b) will be equal. Early work on this phenomenon of equal impinging jets assumed that the velocities of the outgoing streams were equal but also assumed a perfectly elastic collision (so no turbulence; streamline flow only). I am not actually looking at impinging jets, but instead something that can be modeled in two dimensions only (x and y as I've been showing on the diagrams).

2) With liquid streams you can't assume that all of the mass transfer to the larger outgoing stream (assuming $\beta$ < 90 degrees) comes from one of the streams. Even if this problem was for particles, I don't know why you would assume that the mass transfer comes from only one of the spheres.

3) When you say that coalescence always results in KE loss, it seems to me that the coefficient of restitution can handle that, can't it. A perfectly elastic collision is, of course, idealized, but, imo, the formulas shouldn't fail if I decide to set "e" to 1.

 

[haruspex]

1) The drawing shown is actually pretty representative. Although I didn't realize it at the time I developed the equations in the link provided in reply #5 to this post, I think I am using the ideas developed for oblique collisions of particles (with regards to the tangential velocities). The link I posted in reply #5, shows a diagram of two equal liquid jets colliding in reply #18 of that post (the scattering angles are zero degrees in the forward direction and 180 degrees in the backward direction) and actual photographs of a liquid sheet formed from the collision in reply #12. The collision results in a stagnation point in the impingement zone around which liquid is deflected. However, the impingement zone is typically mildly turbulent, which is one of the reasons I believe that the velocity, $v$, of each outgoing stream (f and b) will be equal. Also, the pioneers of studying this phenomenon of equal impinging jets assumed that the velocities were equal but also assumed a perfectly elastic collision (so no turbulence; streamline flow only). I am not actually looking at impinging jets, but instead something that can be modeled in two dimensions only (x and y as I've been showing on the diagrams).

ok, I had not followed that link. It explains a lot.
But it seems to me that the idea of simulating it with a collision of a single pair of particles, with two-way mass transfer, is fatally flawed. Liquid streams have cohesion, which is why the diagram in post #19 there can show just two emergent streams. It would look very different with two streams of solid particles.
You say there:
"most researchers assume that the velocity of the liquid sheet, which forms from the stagnation point in the impingement zone, is equal to the jet velocity, "
but that would violate momentum conservation, and the paper you quote does not do that.

3) When you say that coalescence always results in KE loss, it seems to me that the coefficient of restitution can handle that, can't it. A perfectly elastic collision is, of course, idealized, but, imo, the formulas shouldn't fail if I decide to set "e" to 1 at least for the initial pass.

In a multiparticle collision you can assume conservation of both momentum and energy without contradicting coalescence, but at the two particle level you can't. This strongly suggests that it would not happen at the multiparticle level either.
The cohesion within a stream, for example, will cancel out velocity components of particles normal to the stream flow without increasing the velocity of the stream.

 

[A.T.]

3) When you say that coalescence always results in KE loss, it seems to me that the coefficient of restitution can handle that, can't it. A perfectly elastic collision is, of course, idealized, but, imo, the formulas shouldn't fail if I decide to set "e" to 1.

See my previous comment on this:

If you want to mix two streams going in the same direction but at different speeds, to get a single stream moving at one uniform speed (no speed variation within the final stream), then you effectively have a completely inelastic collision during the mixing and speed homogenisation.

Like when a faster clay ball catches up to slower one, and they end up stuck to each other, moving at the same speed.

Macroscopic kinetic energy is not conserved in those cases.

 

[haruspex]

See my previous comment on this:

I believe @rdemyan is arguing that with two emergent streams, of unknown masses and velocities, the equations allow conservation of momentum and energy plus coalescence. That much appears to be true, but how it might be achieved in practice eludes me.
For example, we can imagine two elastic collisions between pairs of particles such that two particles, one emerging from each of the two collisions, happen to leave with the same velocity. But that does not really constitute a coalescence.

In the fluid flow diagrammed at post #19 of the linked thread, we could model it as two smoothly flowing streams, as though following a system of pipes. Each incoming stream, independently of the other, splits into two streams in opposite directions. With a suitable balance of flows, we can satisfy momentum conservation without requiring any net force from the imagined pipes. Bernoulli would apply, and KE conserved.
However, this does not really match two jets squirted together.

 

[rdemyan]

EDIT: The analysis presented here doesn't quite apply to impinging jets because they require a 3 dimensional analysis. What I've done here is to reduce the problem to two dimensions and the analysis does apply. So instead of two imping jets think of two impinging 2D streams only.


I've attached the diagram from reply #19 of the linked post to this reply. It is meant to be a diagram of two free liquid streams impinging upon one another. To keep it simple, assume that the streams are equal. I'm going to use the parameter terminology in the diagram attached as part of this reply and not the terminology I've been using in this thread for the diagram attached to the original post. So $m_1 = m_2 = m$ and $v_{o1} = v_{o2}= v_o$. The half impingement or collision angle is equal to $\theta$. The diagram assumes that the forward and backward streams that are formed have a velocity equal to $v$. The key to understanding the nature of the collision (100% elastic, partially elastic, or inelastic) is the amount of backflow. Let's look at the x momentum balance:

$$mv_ocos\theta +mv_ocos\theta = m_fv - m_bv$$

where $m_f$ is the mass flowrate in the forward direction of the outgoing stream (to the right) and $m_b$ is the mass flowrate in the backward direction of the outgoing stream (to the left). An inelastic collision occurs if $m_b$ is equal to zero, i.e. no liquid is ejected out the back end. All of both liquids merge much like two spheres that stick together. The equation becomes,

$$2mv_ocos\theta = m_fv$$

but mass conservation always applies so

$$m_1 + m_2 = m_f + m_b = 2m$$

Now for the inelastic case, $m_b = 0$ and $m_f = m_1 + m_2 = 2m$, so

$$v = \frac{2mv_ocos\theta}{2m} = v_ocos\theta$$

Kinetic energy is absolutely not conserved in this case and is lost as part of the collision. The coefficient of restitution for this inelastic collision is,

$$COR = \frac{v}{v_o} = \frac{v_ocos\theta}{v_o} = cos\theta$$

Now, for the case where $m_b > 0$, the first equation becomes,

$$2mv_ocos\theta = v(m_f - m_b)$$

and

$$v = \frac{2mv_ocos\theta}{(m_f - m_b)}$$

Whether the collision is inelastic, partially elastic or 100% elastic sets the difference between the outgoing mass flowrates, $m_f-m_b$. The maximum value for this difference is for an inelastic collision (because $m_b = 0$). When $m_b > 0$, the collision now is partially elastic to 100% elastic. The lowest difference in these mass flowrates (and hence the greatest outgoing stream velocity) is for the case where the collision is 100% elastic and $m_b$ achieves its greatest value. If we assume that the collision is 100% elastic, then by my definition of the COR above, $COR = 1$. This means that $v = v_o$, so,

$$v_o = \frac{2mv_ocos\theta}{(m_f - m_b)}$$

and

$$m_f - m_b = 2mcos\theta$$


as opposed to $m_f-m_b = 2m$ for the inelastic case. I want to look at the case where the incoming streams are not equal and hence have different momentums. I think you can see why I am focused on the mass flowrates of the outgoing streams. The inelastic collision case for the unequal colliding streams is simple to calculate since $m_b$ is equal to zero. But it's the elastic collision case which is challenging. At the same time I'm expecting any final solution to reduce to the equal colliding streams case discussed in this reply and also be able to handle the inelastic collision case of unequal streams.

 

[haruspex]

What I've done here is to reduce the problem to two dimensions

I understand that, but I cannot see that your particle approach is sufficiently realistic.
Much more likely is that much of the KE due to the velocity components that are initially normal to the trajectory of the common mass centre is lost.

 

[rdemyan]

I understand that, but I cannot see that your particle approach is sufficiently realistic.
Much more likely is that much of the KE due to the velocity components that are initially normal to the trajectory of the common mass centre is lost.

In my example in the previous reply, inelastic collisions lose a fair amount of energy and that is clearly shown. So, I don't understand why you are saying it is unrealistic. It turns out that if the velocity of the incoming streams is lower than the Taylor-Culick velocity (the velocity which is needed to overcome the surface tension forces), then liquid cannot eject out the back and that forces the collision to be inelastic. As the velocity increases, liquid can now eject out the back (inertial forces flowing the liquid out exceed surface tension forces trying to pull the liquid back in) and the COR increases.

I am close to getting my equations to work. Do you have a suggestion for a final equation? Kinetic energy is a scalar so it makes me a little nervous to "partition" it. With equal impinging streams, the lost energy comes from the y component velocities. It seems that the same would be true for unequal streams. Maybe I could assume that kinetic energy in the x direction is conserved. but I don't know that this will buy me anything.

 

[haruspex]

$$v = \frac{2mv_ocos\theta}{2m} = v_ocos\theta$$

Kinetic energy is absolutely not conserved in this case and is lost as part of the collision. The coefficient of restitution for this inelastic collision is,

$$COR = \frac{v}{v_o} = \frac{v_ocos\theta}{v_o} = cos\theta$$

That's not what coefficient of restitution means. The COR is a property of the two bodies involved and tells you what the result would be for a head-on collision. Oblique collisions need to be handled as I indicated, separating the motions normal to and parallel to the plane of contact.

Now, for the case where $m_b > 0$, the first equation becomes,

$$2mv_ocos\theta = v(m_f - m_b)$$

You are assuming the forward and backward jets are at the same speed. Seems unlikely, but let’s see where it leads for the unequal streams.
In the frame of reference of the COM, you have two unequal masses colliding head-on at a speed ratio fixed by the mass ratio. This turns into two other masses ejected at right angles to the original velocities, different mass ratio but with the same relationship to the new speeds.
Once you choose the new mass ratio and fraction of KE retained, the new speeds can be calculated.
If you were hoping that both the new masses and speeds could all be deduced you are out of luck. To do that you will need a more detailed model of the process.

Applying a COM frame to two streams is tricky. COM of what, exactly? Could think of it as two streams directed head-on but with their sources moving at right angles to the streams. For simplicity, we can ignore the movement of the sources and add that back later.
The fully inelastic case now leads to a static pool of water. Experience says there will be some sideways splatter. This could be modelled as a pressure build up at the stagnation point pushing out in all directions, but that will not lead to an actual "backward" jet. It takes some cohesive force to pull it back into that form, and that will mean energy loss.
So here’s a possibility: consider the initial result of the collision as being water spurting equally in all directions (if the incoming streams are flat sheets, that's all directions normal to the line of intersection of the sheets). Assume no KE loss so far. Now add in the effect of the moving sources (i.e. an equal and opposite velocity to those). Next, separate the outflows into the bundle going out the side where the sources are headed (these are the backward flows) and the bundle going out the other side. Finally, on each side, collapse the bundle into a single stream by throwing away the lateral velocities (y components) and coalescing the x components.

Note that the backward stream may turn out to have a lower speed than the moving sources. This means no backward jet would appear.

Hope that makes sense. Hard to describe without a whiteboard.

 

[rdemyan]

"That's not what coefficient of restitution means. The COR is a property of the two bodies involved and tells you what the result would be for a head-on collision. Oblique collisions need to be handled as I indicated, separating the motions normal to and parallel to the plane of contact."

The coefficient of restitution of an object bouncing off a stationary target is defined as the ratio of the object's rebound speed after collision to before collision. In a sense the collision of two "equal" streams has been modeled in this way, because the liquids combine either inelastically where there is no backflow or elastically where there is backflow.

"You are assuming the forward and backward jets are at the same speed. Seems unlikely, but let’s see where it leads for the unequal streams."

In the case of impinging liquid jets I have never seen a paper challenging the idea that the velocity of the liquid sheet formed isn't equal in backward and forward sections (assuming that the incoming jets have flat velocity profiles). Clearly, if the collision is head on where the formed sheet essentially forms a circle that has even amounts of liquid flowing in all directions, the velocity must be constant (again for flat velocity profile jets).

From what I've seen the center of mass velocity is equal to the velocity of an inelastic collision. I have looked at simulations of colliding spheres. Since the velocity and trajectory angle of the center of mass is not impacted by the coefficient of restitution, the speed and trajectory angle can be determined by simply setting "e" in the simulator to zero. The two spheres stick together and the simulator readily provided the angle and speed of the combination, which was the same as the COM.

Sorry, but I was unable to follow the rest of your reply.

 

[rdemyan]

It occurs to me that knowing the maximum velocity possible from the collision of these two unequal streams might be useful. How would I mathematically determine the theoretical maximum velocity of the two streams after impingement. I'm not sure if this means that absolutely no energy would be lost, but if I could come up with a way to determine the theoretical maximum that might be helpful.

Right now I am postulating that the theoretical maximum is,

$$u_{max} = \frac{m_1u_1+m_2u_2}{m_1+m_2}$$

If I instead use the kinetic energy equation (no energy loss) as a way to calculate this velocity,

$$u_{max} = \sqrt{\frac{m_1u_1^2 + m_2u_2^2}{m_1+m_2}}$$

However, comments by A.T. in reply#15 suggest that no kinetic energy loss can't happen, although it seemed like A.T. was talking more about an inelastic collision.

If I compare calculations of $u_{max}$ from the two equations, the value from the K.E. equation is higher than that calculated using the first equation. However, for a collision of equal incoming streams (equal meaning the same mass and velocity), both equations calculate the same result.

I'm looking for a way to derive what the maximum velocity could be of the two streams after the collision.

 

[haruspex]

The coefficient of restitution of an object bouncing off a stationary target is defined as the ratio of the object's rebound speed after collision to before collision.

Not nearly precise enough. It is "the ratio of the relative velocity of separation after a two-body collision to the relative velocity of approach before collision".
https://en.wikipedia.org/wiki/Coefficient_of_restitution
Relative velocity of separation / approach means we only use the velocity components normal to the plane of contact. This is important because it means the COR is constant for the two bodies (when colliding at the same orientations, if they're nonuniform). It does not depend on the strike angle nor the frame of reference.
Also, your definition does not work if the target moves as a result.

 

[rdemyan]

Not nearly precise enough. It is "the ratio of the relative velocity of separation after a two-body collision to the relative velocity of approach before collision".
https://en.wikipedia.org/wiki/Coefficient_of_restitution
Relative velocity of separation / approach means we only use the velocity components normal to the plane of contact. This is important because it means the COR is constant for the two bodies (when colliding at the same orientations, if they're nonuniform). It does not depend on the strike angle nor the frame of reference.
Also, your definition does not work if the target moves as a result.

Quoted from Wikepedia on coefficient of restitution:

"For an object bouncing off a stationary target, e is defined as the ratio of the object's rebound speed after the impact to that prior to impact:$$e = \frac{v}{u}$$
where

u is the speed of the object before impact
v is the speed of the rebounding object (in the opposite direction) after impact

Since the incoming sheets are equal in all ways, imo, it is possible to extend this idea. There is definitely a difference in velocity which directly results from loss of K.E. It can be called something else besides COR, but that doesn't mean it isn't a worthwhile concept.

Thanks for the time you've taken in reading and responding to my posts, but I don't believe that the purpose of solving the problem is being furthered at the moment. Thanks again.

 

[haruspex]

Quoted from Wikepedia on coefficient of restitution:

"For an object bouncing off a stationary target, e is defined as the ratio of the object's rebound speed after the impact to that prior to impact:$$e = \frac{v}{u}$$
where

u is the speed of the object before impact
v is the speed of the rebounding object (in the opposite direction) after impact

You omitted the first line of that section:
"In the case of a one-dimensional collision…"
To apply it where the contact plane is not normal to both trajectories you have to reduce it one dimension by taking only the components normal to that plane.

Also, it would appear by comparing with the earlier equation in that section that by "stationary target" they intend that it is stationary both before and after. I.e., the target is arbitrarily massive. My point is that that should have been made clearer.

Thanks for the time you've taken in reading and responding to my posts, but I don't believe that the purpose of solving the problem is being furthered at the moment. Thanks again.

I believe that in post #20 I gave you a justifiable way to proceed. Yes, it was hard to explain, but I am happy to work with you on that.

 

[rdemyan]

If you maintain that the velocities of the exiting streams are different, then your proposed method is a dead end. I already provided convincing evidence that the velocities are the same.

 

[haruspex]

If you maintain that the velocities of the exiting streams are different

I maintain that they could be different, and the only evidence for equality that I see in the thread is your statement that everyone assumes it, or maybe some tests of specific cases roughly confirm it.

Let's start with a very simple case: two equal 2D blobs (so, discs) collide head on along the y axis.
The sideways splatter is symmetric about the y axis, but the distributions of quantity of flow and speed may be functions of the angle, alpha, to the x axis.
Intuitively, I suggest that relatively little emerges near the y axis, so cohesion leads to the splatter forming two flows along the x axis. Other than that consequence, for simplicity, suppose that the volume and speed, v, are the same in all directions.
Since the speed in the x direction of a given part is $v\cos(\alpha)$, each of those flows consists of a core at speed v and successively slower outer layers.

We can transform this simple case into blobs converging from the left at angles $\pm\theta$ to the x axis by viewing them in a frame moving negatively along the x axis at speed $v\cot(\theta)$. A flow layer with x velocity $-v\cos(\alpha)$ in the original frame now has velocity $v\cot(\theta)-v\cos(\alpha)$. So a slower portion of the left moving flow is now seen as moving right, while the fastest may still be seen as moving left.
The next step is to allow each of the flows to settle down to a uniform speed by averaging over the fluid moving in that direction. It is feasible that these averages will be similar.

 

[rdemyan]

I maintain that they could be different, and the only evidence for equality that I see in the thread is your statement that everyone assumes it, or maybe some tests of specific cases roughly confirm it.

Let's start with a very simple case: two equal 2D blobs (so, discs) collide head on along the y axis.
The sideways splatter is symmetric about the y axis, but the distributions of quantity of flow and speed may be functions of the angle, alpha, to the x axis.
Intuitively, I suggest that relatively little emerges near the y axis, so cohesion leads to the splatter forming two flows along the x axis. Other than that consequence, for simplicity, suppose that the volume and speed, v, are the same in all directions.
Since the speed in the x direction of a given part is $v\cos(\alpha)$, each of those flows consists of a core at speed v and successively slower outer layers.

We can transform this simple case into blobs converging from the left at angles $\pm\theta$ to the x axis by viewing them in a frame moving negatively along the x axis at speed $v\cot(\theta)$. A flow layer with x velocity $-v\cos(\alpha)$ in the original frame now has velocity $v\cot(\theta)-v\cos(\alpha)$. So a slower portion of the left moving flow is now seen as moving right, while the fastest may still be seen as moving left.
The next step is to allow each of the flows to settle down to a uniform speed by averaging over the fluid moving in that direction. It is feasible that these averages will be similar.

And what about the real case where energy is in fact given up in the impingement zone as a result of the collision. The energy dissipation results in some level of turbulence. The turbulent eddies homogenize any gradients and therefore it is likely that liquid ejected from that zone in whichever direction has the same velocity.

 

[haruspex]

And what about the real case where energy is in fact given up in the impingement zone as a result of the collision. The energy dissipation results in some level of turbulence. The turbulent eddies homogenize any gradients and therefore it is likely that liquid ejected from that zone in whichever direction has the same velocity.

That just reduces the v in my analysis. It doesn’t change the method.

 

[rdemyan]

Yes, it reduces v but it also means that the v in the exiting streams is highly likely to be uniform for both exiting streams. If you don't assume the same v it seems unlikely to me that your method would be able to calculate $m_b, m_f, \alpha, \phi$, which are the parameters of interest. Even with the same v, I've been short one equation. It seems to me that making the velocities unequal will only exacerbate that problem.

 

[haruspex]

it also means that the v in the exiting streams is highly likely to be uniform for both exiting streams

I took v as uniform right after the collision in the initial reference frame, where the two blobs collide head on. But then we have to consider what happens as cohesion pulls the spatter into a pair of flows. It seems reasonable this mainly involves forces in the y direction, so, at first, the velocity is now seen as nonuniform across a flow. Instead, it will be fastest at the core then fall off to nearly static at the outside.
Yes, each flow would become uniform over time, but before that can happen more blobs come in. In the moving frame of reference (i.e. in the angled input streams case) the new blobs are offset in the x direction so redefine the forward and backward boundary. This is why the forward flow is the larger. Only after that should we do the velocity averaging.

If you insist on averaging the velocities within each flow before considering the moving reference frame then it is impossible to get even roughly equal exit speeds for the two flows. It just cannot happen.

 

[rdemyan]

I took v as uniform right after the collision in the initial reference frame, where the two blobs collide head on. But then we have to consider what happens as cohesion pulls the spatter into a pair of flows. It seems reasonable this mainly involves forces in the y direction, so, at first, the velocity is now seen as nonuniform across a flow. Instead, it will be fastest at the core then fall off to nearly static at the outside.
Yes, each flow would become uniform over time, but before that can happen more blobs come in. In the moving frame of reference (i.e. in the angled input streams case) the new blobs are offset in the x direction so redefine the forward and backward boundary. This is why the forward flow is the larger. Only after that should we do the velocity averaging.

If you insist on averaging the velocities within each flow before considering the moving reference frame then it is impossible to get even roughly equal exit speeds for the two flows. It just cannot happen.

Well, I'm still not really following. It sounds like your blobs are effectively spheres which can change mass. It would be interesting if you tried running some numbers, but I just don't see how you are going to have enough equations.

 

[haruspex]

Well, I'm still not really following. It sounds like your blobs are effectively spheres which can change mass. It would be interesting if you tried running some numbers, but I just don't see how you are going to have enough equations.

They are not spheres because the actual scenario is jets in the form of sheets. They are elements of the sheets, thin sections perpendicular to the planes of the sheets and parallel to their velocities. Effectively, they are 2D.

As noted, the distribution of the spatter is uncertain. It will certainly be symmetric about the y axis in the frame of reference where the jets collide head on. That is enough to make a strong case that the two exit velocities are unlikely to be equal in the frame where the collision is oblique. But it does leave open the possibility that the spatter distribution is such that the two velocities could be similar in some set-ups.
What I am suggesting as the next step is to make a simple assumption about the distribution and see what results.

Alternatively, experiment with the head-on case to see what the distribution is.

 

[haruspex]

Btw, do you understand COR now?

 

[rdemyan]

Btw, do you understand COR now?

For equal impinging streams where $m_1 = m_2 = m;u_1=u_2=u$ the kinetic energy equation is

$$EL = \frac{2mu^2}{2} - \frac{2mv^2}{2} = mu^2 - mv^2$$

where EL means kinetic energy lost. Now when no energy is lost, $v = u$.

However, as I mentioned previously I have been stating that the following works in a similar manner to the classical physics definition of the coefficient of restitution for the case of equal impinging streams,

$$COR = \frac{v}{u}$$

For an inelastic collision of the equal streams, the velocity of the stream after the collision is just equal to the x component of the velocity of the streams before the collision (which are equal). Note for an inelastic collision there is only one stream after the collision. Therefore,

$$COR = \frac{ucos\beta}{u} = cos\beta$$

Plug this into the energy lost equation

$$EL =mu^2 - mu^2cos^2\beta = mu^2(1-cos^2\beta)$$

but for this inelastic collision, $COR = cos\beta$, so

$$EL = mu^2(1-COR^2)$$

So, the fraction of energy lost, F, is

$$F = \frac{mu^2(1-COR^2)}{mu^2} = 1-COR^2$$

Isn't this exactly the result using the Physics definition of the coefficient of restitution, e, namely that the fraction of kinetic energy lost is $1-e^2$ or if you prefer, $1-COR^2$.

Now since the COR is equal to $\cos\beta$ for an inelastic collision of equal streams, the fraction of energy lost is proportional to,

$$1-COR^2 = 1-cos^2\beta = sin^2\beta$$

Thus, the energy lost comes from the kinetic energy associated with the y component of the velocity at impact, which is the velocity component along the line of impact. The velocity component along the tangent line, $v_x$ remains unaffected.

I should mention the definitions of elastic and inelastic that I use. A 100% perfectly elastic collision is where COR = 1 and from the above discussion, $v=u$. An inelastic collision is the theoretical loss of energy assuming that all of the energy associated with the y component velocity is lost and the kinetic energy associated with the velocity along the tangential line ($v_x = ucos\beta$) is not lost. An elastic collision is where only a portion of the kinetic energy associated with the y component velocity is lost and again, none of the kinetic energy associated with the tangential velocity component is lost. For an elastic collision,

$$ucos\beta < v < u$$

 

[haruspex]

For equal impinging streams where $m_1 = m_2 = m;u_1=u_2=u$ the kinetic energy equation is

$$EL = \frac{2mu^2}{2} - \frac{2mv^2}{2} = mu^2 - mv^2$$

where EL means kinetic energy lost. Now when no energy is lost, $v = u$.

However, as I mentioned previously I have been stating that the following works in a similar manner to the classical physics definition of the coefficient of restitution for the case of equal impinging streams,

$$COR = \frac{v}{u}$$

For an inelastic collision of the equal streams, the velocity of the stream after the collision is just equal to the x component of the velocity of the streams before the collision (which are equal). Note for an inelastic collision there is only one stream after the collision. Therefore,

$$COR = \frac{ucos\beta}{u} = cos\beta$$

Plug this into the energy lost equation

$$EL =mu^2 - mu^2cos^2\beta = mu^2(1-cos^2\beta)$$

but for this inelastic collision, $COR = cos\beta$, so

$$EL = mu^2(1-COR^2)$$

So, the fraction of energy lost, F, is

$$F = \frac{mu^2(1-COR^2)}{mu^2} = 1-COR^2$$

Isn't this exactly the result using the Physics definition of the coefficient of restitution, e, namely that the fraction of kinetic energy lost is $1-e^2$ or if you prefer, $1-COR^2$.

Now since the COR is equal to $\cos\beta$ for an inelastic collision of equal streams, the fraction of energy lost is proportional to,

$$1-COR^2 = 1-cos^2\beta = sin^2\beta$$

Thus, the energy lost comes from the kinetic energy associated with the y component of the velocity at impact, which is the velocity component along the line of impact. The velocity component along the tangent line, $v_x$ remains unaffected.

I should mention the definitions of elastic and inelastic that I use. A 100% perfectly elastic collision is where COR = 1 and from the above discussion, $v=u$. An inelastic collision is the theoretical loss of energy assuming that all of the energy associated with the y component velocity is lost and the kinetic energy associated with the velocity along the tangential line ($v_x = ucos\beta$) is not lost. An elastic collision is where only a portion of the kinetic energy associated with the y component velocity is lost and again, none of the kinetic energy associated with the tangential velocity component is lost. For an elastic collision,

$$ucos\beta < v < u$$

What you are effectively doing there is defining the COR in terms of the fraction of total KE lost in a specific collision, so it is not surprising that it is internally consistent with that. But it is clearly not the same as the standard physics definition of the concept.
First, it ought to be the same in all inertial reference frames. If we take a maximally inelastic head on collision there is no rebound, so COR=0. But viewed in a reference frame moving at constant velocity normal to the line of collision your definition makes it nonzero. That is not acceptable.
Secondly, it is a property of the two bodies (and their impinging surfaces) involved so if they are uniform it ought to be the same for all collisions between those same two bodies.
With your definition, it changes according to the two velocities and the obliqueness of the collision.

I already cited the Wikipedia entry that makes this clear by referring to relative velocities of approach and separation. Here are more extracts:
“0 [represents] a perfectly inelastic collision (in which the objects do not rebound at all, and end up touching”
(In your definition, the coalescence case can have nonzero e)
“The COR is a property of a pair of objects in a collision”
“e is generally treated as a dimensionless constant, independent of … relative velocities of the two objects”.

See also (5.4.4) in https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Tatum)/05:_Collisions/5.04:_Oblique_Collisions