Work done and point of application I read some books and some articles regarding point of application but I could not understand what it really means
Let's suppose you have a ball and you throw it on a rigid wall now the ball comes back (neglect gravitational force). So if someone asks what is work done by the normal reaction we would say that as $W= F s \cos(\theta)$. So Newton's third law will come in and the displacement will be calculated and the answer would come but many sources claim that the $W$ done is 0 as point of application does not move.
But the point of application is part of that ball and ball moves hence work is done in the direction of normal reaction. So the normal reaction must do the work and the sources could be misleading.
So am I right in this case?
 A: 
But the point of application is part of that ball and ball moves hence work is done in the direction of normal reaction. So the normal reaction must do the work and the sources could be misleading. So am I right in this case ?

Typically, no, you would not be right. The usual simplifying assumptions in a question like this are as follows:

*

*The wall is perfectly rigid

*The ball is perfectly elastic

*The wall/earth are so much more massive than the ball that its acceleration is negligible

Assumption 3) makes it natural to use the frame where the wall is at rest. In that frame, by assumption 1) there is no motion of any point on the wall. Since a point of contact is a point where two objects are touching that means that no point of contact with the wall can move. Therefore we know that no work can be done either on or by the wall through contact forces.
Now, let’s investigate where your reasoning fails. The failure is due to assumption 2). It is true that the point of application is part of the ball and it is true that the ball moves, however because of assumption 2) the ball is not rigid. Being non-rigid means that not all parts of the ball move together. So the part of the ball that is the contact point can be stationary while the ball moves.
This is indeed what happens. The part of the ball that touches the wall is stationary, and the rest of the ball is moving, resulting in large deformations to the ball (see video below). No work is done by the wall on the ball or by the ball on the wall. Instead, the ball’s kinetic energy is briefly converted into elastic potential energy through deformation, and then back to kinetic energy as it returns to its original shape.
https://m.youtube.com/watch?v=FC8Tpi3U0H0
A: That's a good question. First of all, let's begin with a simple back of the hand calculation. To do work on a body and change its kinetic energy, you need a force to act on it. The only time there is any force on the ball is when the normal force acts on it for an instance (assuming no gravity, etc)
Also, it returns with the same speed as before (assuming ideal conditions like elastic collision and immovable wall so as to get to the essence of the question)
So, clearly, its kinetic energy didn't change and so consequently, there has not been any work done on it by the normal reaction
So what's missing here?
During the collision, the wall can only act on the ball when it is contact with it and so, the displacement of the ball when the force acted on it must have been 0 (and when the displacement did occur, there was no force since there was no contact). So, the work done was zero
