A body falling on an inclined surface Say, an object is falling down with the initial speed v on an inclined surface:

Gravity is acting, so immediately before the impact, the object will have the velocity of v+gΔt:

Now, let's assume that there is no friction and no inertia. Also, let's assume that the body is absolutely inelastic: it won't bounce, just smack against the surface and immediately roll down without friction.
Now, it is my understanding that the velocity of the ball down the slope will be equal to (v+gΔt)×sinα. In other words, a sinα portion of its vertical velocity will be converted into down-the-slope diagonal speed.

(v+gΔt)×sinα is its initial speed upon contact: the ball will also accelerate at gsinα due to gravity. As for the cosα portion of the pre-contact speed: it's gonna be absorbed by the impact, and the corresponding portion of kinetic energy will be converted into thermal energy, or whatever.
Am I correct so far?
Now, my understanding is that a portion of velocity has to be absorbed in the first place because on inclined surfaces, the vertical and horizontal speeds are interdependent: the slower an object moves horizontally, the slower it descends (and vice versa). In other words, the horizontal speed acts as a restrictor to how fast an object can descend: since it can only get so much horizontal speed upon impact, its vertical speed has to be capped. Reduction of vertical speed means that some energy must be dissipated.
What follows is that if an object had some horizontal speed beforehand (for example, it was thrown at an angle), it will allow it go move down the slope faster, preserve more vertical speed, and reduce the impact:

To calculate the down-the-slope speed upon impact and the amount of energy dissipated, I would just rotate the image and treat the combined horizontal+vertical velocity as if it was "falling" speed:

From here, I can apply sinα and cosα for the new angle and find out that indeed down-the-slope speed would be much higher compared to after straigh-down falling.
My first question is: is all of the above correct, and is there anything that I'm missing?
And now to my second question. Suppose we run the very first scenario again (straight-down falling), but this time our object will have significan inertial mass. According to my calculations, its initial down-the-slope speed upon impact should be (v+gΔt)×sinα. However, it shouldn't be able to generate that speed instantaneously because of inertia: the body has been falling down with pure vertical speed, but down-the-slope speed has a horizontal component that can't just pop up out of nowhere.
So, what exactly is going to happen? More kinetic energy will be dissipated into heat because the initial down-the-slope speed will be even slower than (v+gΔt)×sinα? And how much slower? Is there a way to calculate the dependency between the amount of dissipated energy/down-the-slope initial speed and the object's mass?
What makes me even more confused is that technically, an object that has any mass whatsoever requires SOME time to acquire any speed at all. Which means that momentarily, the object's speed upon impact must be completely reduced to zero. And yet that's not what we see with real objects: a falling something will move down the slope faster than if just put on the slope and let to be accelerated by gravity.
Maybe the horizontal inertia gets cancelled somehow by mg×sinα in a way that I just don't see?
 A: This is best understood by introducing a concept of momentum and impulse. The impulse-momentum theorem says
$$\boxed{\vec{p}_0 + \vec{J} = \vec{p}_1} \tag 1$$
where $\vec{p} = m\vec{v}$ is the momentum of the system (object plus ramp), and $\vec{J}$ is the impulse:
$$\boxed{\vec{J} = \vec{F}_\text{net} \Delta t} \qquad \text{or} \qquad \boxed{\vec{J} = \int_{t_1}^{t_2} \vec{F}_\text{net} dt} \tag 2$$
What this means is that at the moment of impact there was some huge (but finite) force between the object and the ramp that was acting in a finite time $\Delta t$, and this force has changed the momentum of the system. Since momentum of the ramp before and after the impact remained the same (its velocity is zero), the falling object lost some of its kinetic energy to other (non-recoverable) forms of energy during the collision. In general, in an inelastic collision the kinetic energy of the system is not preserved!
In order to quantify object's velocity after the impact, you would need to know properties of the collision, i.e. the collision force profile over time. Since this force is (usually) much larger than other forces acting on the object, the collision force dominates and other forces such as gravitational or friction forces can be neglected in the calculation of impulse $\vec{J}$. See below for an example of a force during an inelastic collision.

Figure 1. Example of a force during an inelastic collision (Source here)
In an elastic collision, the kinetic energy of the system is preserved. However, for a collision to be elastic, relative velocities of the two objects before and after the collision must be the same in magnitude and opposite in direction. Since ramp is at rest all the time, this means the ball should bounce vertically after the collision. Your example mandates the ball to keep moving on the ramp without bouncing, hence the collision is perfectly inelastic meaning that all energy is converted into heat and deformation of objects. The ball keeps moving on the ramps from zero velocity.
A: I believe your troubles will all disappear if you allow one of two things: either (1) impact is instantaneous and impact forces are infinite (of course, real forces aren’t infinite, but this model can still be a useful way to think about things) or (2) the impact occurs over some short but finite time.
In either case, assuming no friction, so that the force on the ball is perpendicular to the ramp surface, momentum in the along-ramp direction will be conserved—the velocity component in the downhill direction will not change during the impact.
So, the downhill velocity doesn’t “come from nowhere”; it was there already. And the sudden change in the velocity component perpendicular to the ramp is brought about by a large force acting for a short time, or, if you like, for conceptual purposes, an infinite force applied at a single point in time.
In the above explanation I have chosen, for simplicity, to decompose the velocity of the ball just prior to impact into components parallel and perpendicular to the ramp. The impact leaves the former unchanged and reduced the latter to zero. If you choose instead to decompose it into vertical and horizontal components, then you find that the impact force, being at an angle, changes both components of the velocity—the vertical component of the impact force slows (but does not stop) the ball’s rate of vertical descent, and the horizontal component of the impact force starts it moving to the left.
EDIT One further possible source of confusion is the idea that in a “completely inelastic collision” the two bodies must momentarily come to rest with respect to each other. All that is implied by “inelastic” is that no energy is stored in deformation of the bodies during the collision and then given back. So there is no “bouncing off”, but in the frictionless case there can be relative motion along the direction of the interface, and unless the impact occurs normal to the surface, there will be.
A: The understanding shown in the first part of your question, about picking a frame in which you have rotated the ramp, is fine as far as it goes, but if you do rotate the ramp in that way, you need to account for the change when you consider the acceleration due to gravity. Also it makes no sense to assume 'no inertia', when you question is all about how a body moves under the influence of forces!
The mistake you have made in the second part of your question is to assume that if the object did come to a stop on impacting the ramp that the only force accelerating it subsequently would be the component of gravity down the ramp. That would be true if the collision was entirely inelastic- the body would come to a complete stop and then start to accelerate down the ramp in exactly the same way that an object placed on the ramp would accelerate. If the collision wasn't elastic, then the body would still come to a momentary halt, but its acceleration would then be the result of two forces- firstly an elastic restoring force which would cause it to rebound at an angle from the ramp and secondly the force of gravity acting down the slope of the ramp.
