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Consider the typical problems of mechanics where we had to find the velocity for some mass which reaches bottom of a wedge after metting some changes in the wedge angles (kinks). The following is a particluar type of problem. enter image description here

The ball starts sliding from the top of the incline A and reaches the ground encountering a kink at C. We have to find the velocity of the mass as it reaches the point B somewhere on the ground. Assume every surface the ball encounters is frictionless and ignore the rotational motion of the ball.

Edit: Assume that the collision of the ball with the ground at C is perfectly inelastic.

We used energy conservation from A to C and momentum conservation along the horizontal at C to state that the velocity of the ball as it reaches the ground is $\sqrt{2gh}\space cos(\theta)$.

But, work energy theorem implies that the work done by all forces on the ball must be equal to the change in the kinetic energy of the particle. What work does the impulsive normal (The kink at C) do on the ball? There being no displacement at C (C being a point), shouldn't the work done by the impulsive normal be zero? If so, shouldn't the velocity of the ball at point B be $\sqrt{2gh}$?

I am sure that the velocity at B has to be $\sqrt{2gh}\space cos(\theta)$. But, I am unable to find out why is such inconsistency? I heard Work-energy theorem is universally applicable. Any insights to such problems will be highly be appreciated. I am new to this site, so please ask for clarifications, in case of any discrepancies.

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  • $\begingroup$ Why do you think there is the $cos$ factor? What's the calculation you did? $\endgroup$
    – FGSUZ
    Aug 25 '20 at 11:24
  • $\begingroup$ @FGSUZ I applied Momentum conservation on the ball along the horizontal (There being No force on the ball along horizontal) which gives $mvcos(\theta) = mv_f$. $\endgroup$
    – UncleJohn
    Aug 25 '20 at 11:30
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    $\begingroup$ How comes there's no horizontal force along the horizontal? What about $N_x$? $\endgroup$
    – FGSUZ
    Aug 25 '20 at 12:07
  • $\begingroup$ Nothing happens at C other than a change in velocity direction, conservation of energy is enough, and the speed at B is the same than at C $\endgroup$ Aug 25 '20 at 20:35
  • $\begingroup$ @Wolphramjonny Depends entirely on the collision of the ball with the ground. I forgot the assumption that the collision of the ball with the ground is completely inelastic. Now, it has been corrected. $\endgroup$
    – UncleJohn
    Aug 26 '20 at 5:29
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If you are assuming that the vertical velocity goes to $0$ at $C$, then there has to be some work being done by the ground there. In reality nothing is rigid, and so there is a force applied for a small and finite time over a small and finite distance due to deformation of the surfaces, the nature of repulsive forces between the ball and ground, etc. The details in this case are not important, as you are just interested in the end result. You can use the work-energy theorem to determine how much work was done by this force, although further assumptions of how this force/deformation works would need to be stated to determine the distance this force was applied over.

However, we can idealize the situation using Dirac delta functions. We can say that the force does work $W_0$ by defining the force as $F(x)=W_0\delta(x)$ where $x=0$ is at point $C$. So then there work done "at $C$" is given by.

$$W_C=\int F(x)\,\text dx=\int W_0\delta(x)\,\text dx=W_0$$

Note that this is very similar to the problem where you have a ball that hits a wall with one velocity, rebounds with another velocity, and you are asked to find the impulse given to the ball. You don't really care about the actual collision process. Sure, you can model it in different ways (from a finite force applied over a small, finite time, to an infinite spike force applied over an infinitesimal amount of time), but if you already know the starting and ending conditions you don't need to do that to find the impulse. Similarly here you don't need the modeling to find the work done. Just use the work-energy theorem. Any discrepancy you think you have must actually be the work done at this kink, and this is independent of the modeling.

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The work energy theorem is alright. The problem is that you are not taking $N_x$ into account in momentum conservation

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