# Inconsistency in applying work-energy theorem in the classical problem of a kink on an inclined wedge

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.

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.

• Why do you think there is the $cos$ factor? What's the calculation you did? Aug 25 '20 at 11:24
• @FGSUZ I applied Momentum conservation on the ball along the horizontal (There being No force on the ball along horizontal) which gives $mvcos(\theta) = mv_f$. Aug 25 '20 at 11:30
• How comes there's no horizontal force along the horizontal? What about $N_x$? Aug 25 '20 at 12:07
• 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 Aug 25 '20 at 20:35
• @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. Aug 26 '20 at 5:29

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.
$$W_C=\int F(x)\,\text dx=\int W_0\delta(x)\,\text dx=W_0$$
The work energy theorem is alright. The problem is that you are not taking $$N_x$$ into account in momentum conservation