I Solving a momentum problem where masses change after the collision

[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