# Work done on a rigid body in a collision

1. May 13, 2014

### lightarrow

Sorry if this question has already been considered in this Forum (I'm sure it is but I can't find it now).
A rigid ball slide without friction in an horizontal plane and collide elastically with a wall, the plane of which is perpendicular to the ball's velocity.
During the collision the ball loses its kinetic energy Ek, in the firs part, then re-acquires it in the second part.
How do I write the work-kinetic energy theorem Wext = ΔEk, since the contact point of the ball and wall doesn't move during the process?

Last edited: May 13, 2014
2. May 13, 2014

### Staff: Mentor

Do you mean the work - kinetic energy theorem? (What does L stand for in your equation?)

Realize that the ball must compress at least a little as it rebounds off the wall.

3. May 13, 2014

### lightarrow

I presume, I don't know the exact name in english language.
L stands (oops, I wasn't thinking in english! ) for W = work done on the system by the external forces. I have edited the post.
Yes. This means that I cannot consider the ball as a rigid body and so I cannot apply the theorem?

--
lightarrow

4. May 13, 2014

### Staff: Mentor

Realize that no real work is done by the external force (to a good approximation) since the wall doesn't move. All energy changes are internal to the ball.

5. May 13, 2014

### lightarrow

Ok, so the theorem cannot be applied here?

6. May 13, 2014

### Staff: Mentor

What is it that you are trying to calculate?

7. May 13, 2014

### mattt

This is an interesting problem and you can learn many different things depending on the level of "rigour" you want to apply. In fact I think I am going to ask you some more questions about this example to make you think about it more deeply.

First of all, in Newtonian Mechanics (which this problem belongs to), for any system of point-particles (and rigid bodies and non-rigid bodies in general are a particular case of this, because you can always consider a large body as being a huge collection of tiny-"point"-particles), if the total exterior force (the sum of all exterior forces applied on every point particle that make up our system) is zero AND the interior forces (the force each particle exerts upon other point particle) are newtonian (i.e. $$\vec{F}_{ij} = - \vec{F}_{ji}$$ ), then the total linear momentum vector of the system (with respect to an inertial frame) is conserved.

That is a mathematical theorem. What do you think about this theorem and your example? Is the total linear momentum vector of the system conserved?

Last edited: May 13, 2014
8. May 13, 2014

### lightarrow

Certainly not since a non zero total force is applied on the system (= ball) during the collision. But the total energy of the ball + the wall is conserved. If we assume that the wall is so rigid that we can neglect its vibrations and elastic potential energy and any sort of energy variations (included of course its internal energy), then "who" changes the ball's kinetic energy during the collision?
Probably is exactly the theorem you cited that makes me (incorrectly) think there shoud have been an external work on the ball, while instead only internal forces makes work, and change the energy, of the ball. But it's counterintuitive, isnt'it?

9. May 13, 2014

### lightarrow

I'm trying to understand which is the "domain of validity" (don't know if we can say so in english) of the work-kinetic energy theorem and, if the theorem cannot be applied here, if it's possible to find a generalization which it is.

10. May 13, 2014

### mattt

Well, first of all, what you call "the system" and "the rest of the Universe" is a decision to make. I mean, the same "problem" may be modeled in two (or more) different ways (depending on what you call "the system" and "the rest"), and they produce (if done correctly) exactly equivalent results (what in one model is an "external force" is "internal" in the other model, and such...).

In your example, if we imagine we have only the ball and the wall (and nothing else, no Earth :-) ,no floor, nothing :-) ) and we call the system = ball + wall, then it is an isolated system, and so in this model, there would be no external forces....so the total linear momentum vector $$\vec{P}$$ would be conserved (because of the mathematical theorem I cited earlier).

But the linear momentum vector of the ball (the sum of all linear momentum vectors of all "point particles" the make up the ball, if you want), does change obviously, so the linear momentum vector of the wall (the sum of all linear momentum vectors of all "point particles" that make up the wall) also changes, so that the total linear momentum vector (ball + wall, and always with respect to some inertial frame) is constant in time.

Another equivalent way of modeling this problem would be to consider the system = "the ball only", and the wall would be "the rest of the Universe (that is important for this problem)".

In this case, the total linear momentum vector of the system does change in time (the linear momentum vector of the ball changes), and this change would be the result of the effect the wall has on the ball, modeled as an external force.

Obviously they are two different but equivalent ways of modeling your problem.

What happens with the other theorem, the one you are interested in?

It is another mathematical theorem which says the following:

"In Newtonian Mechanics, if you have a system of point particles (again, just like the previous theorem, dealing with large rigid or non-rigid bodies is a particular case) and all the forces are differentiable vectorial functions of time (this also is necessary in the previous theorem, which I forgot to mention), then for any two time instants t0 and t1:

$$\sum_{i} \int_{t_0}^{t_1}\vec{F_i}(t)\cdot\vec{v_i}(t)dt = \sum_{i}(\frac{1}{2}m_i v_i(t_1)^2) - \sum_{i}(\frac{1}{2}m_i v_i(t_0)^2)$$

with respect to any inertial frame. ($$\vec{F_i}(t)$$ is the resultant or total force acting on the point-particle of mass m_i at time t)

Imagine your ball is made of many many tiny particles and that the force between one such particle and any other such particle in the ball, at least for those close enough, are "cohesive forces", for example imagine any two such particles of the ball (that are close enough) are connected by springs and you can think exactly the same about the wall, made of tiny "almost point-like" particles, connected by springs (this is indeed more realistic than thinking about both the ball and the wall as rigid bodies).

More so, if in this model (the internal forces between the particles that make up the ball, and also between the particles that make up the wall, being spring-like) we consider those "spring-like" internal forces as "ideal", then we have that all the forces are conservative, so the total (kinetic + elastic potential) energy of the whole system (of all particles that make up both the ball and the wall) is constant in time.

If they are more like "real" springs, then there will be energy dissipation after the interaction.

Imagine the wall is at rest (it is an inertial frame at the beginning) and the ball is coming to hit it (and in this model there are only ball + wall, nothing else :-) ).

Suppose (in the case of "real" springs) at time t_0 the ball is still coming (at constant speed), not "touching" the wall still, then at time t_1 the ball is "touching the wall, ball and wall deforming in the process and some time after", and at a later time t_2 both the ball and the wall have regained their original shape again and both going away from each other.

In this model:

Do you think the total kinetic energy of the system (ball + wall) has changed between t_0 and t_1?

And what about between t_1 and t_2?

And between t_0 and t_2?

Try to answer these same three questions in the same model but with "ideal" spring-like internal forces.

Note: in reality you don't need to model this problem this way, it can be much simpler depending on what do they want you to do. But I wanted to make you think about it this way so you can appreciate the power and generality of those few fundamental mathematical theorems that make up Newtonian Mechanics.

Newtonian Mechanics is fascinating because with just very few concepts (point particles and forces, and euclidean space and time), two laws

$$\vec{F}=m.\vec{a}$$

and

$$\vec{F_{ij}}=-\vec{F_{ji}}$$

and another law for gravitation

$$\vec{F}=-G\frac{Mm}{d^2}\vec{u_r}$$

and some Vector Algebra and Vector Calculus, you can virtually model successfully any system of macroscopic massive bodies with velocities negligible with respect to the speed of light, and with moderate densities.

11. May 13, 2014

### Staff: Mentor

Again, what do you want to do with the theorem?

Here's something that I wrote in another thread that may clarify how the "work"-energy theorem, when thought of as an application of Newton's 2nd law, may be applied to deformable bodies.

12. May 14, 2014

### lightarrow

I don't have clear which are its exact assumptions and hypothesis, simply this.
The problem I see reasoning with the CM (centre of mass), if the system is not rigid, is that in our case the external force acting on the ball (constraint reaction by the wall) does make work on the ball during the collision because the CM does move during the process, if we assume the wall rigid and the ball deforming elastically; if instead I see forces acting on point masses I have to conclude the the external force, which acts only on the ball's particles coming in contact with it, does not make work on the ball. Even if I model the ball as a massless spring attached to a point mass (in the opposite side with respect to the wall) the question remain.

13. May 14, 2014

### Staff: Mentor

You must distinguish between "real" work in the thermodynamic sense, which is the force times the displacement of the point of application. In this example, the real work done is zero and no energy is transmitted from the wall to the ball.

Multiplying the net force times the displacement of the center of mass looks like a work term, but it is really not. Or rather it should be called something like "center of mass work". It is a consequence of Newton's 2nd law. You can always perform this calculation, regardless of whether "real" work is done.

Given this, can you restate your question?

14. May 14, 2014

### lightarrow

Of course, defining the system is extremely important in physics in general.
Ok, this formulation is more precise than what I knew.
If your question on kinetic energy (KE) is intended as:
KE(t_1) - KE(t_0)
KE(t_2) - KE(t_1)
KE(t_2) - KE(t_0)

and at time t_1 the ball and wall ARE deformed (I don't have it perfectly clear because you say "deforming") then my answer is:

1) Yes.
2) Yes
3) No.

--
lightarrow

15. May 14, 2014

### lightarrow

--
lightarrow

16. May 14, 2014

### mattt

From t_0 to t_1 what happens is that part of the kinetic energy (of all the particles that make up the ball) starts to be converted into elastic potential energy (of the particles that make up both the ball and the wall, as ball and wall start to deform).

So the total kinetic energy of the system (ball + wall) at t_0 is greater that the total kinetic energy of the system at t_1 (because the system, ball + wall, at t_1 has now some total elastic potential energy).

It happens that:

$$\sum_i(\frac{1}{2}m_iv_i^2(t_1))-\sum_i(\frac{1}{2}m_iv_i^2(t_0)) = \sum_i\int_{t_0}^{t_1}\vec{F_i}(t)\cdot\vec{v_i}(t)dt$$

(the previous theorem), and the second member of this equation is what physicists usually call "work".

What are those $$\vec{F_i}(t)$$? "Who" exert them?

It is not about "who". It is just the following:

Put your attention in any one particle that make up the ball. At first this particle is moving with URM (uniform rectilinear motion) so at first (before the ball and the wall start to interact) we have $$\vec{F_i}(t) = \vec{0}$$.

From certain moment on, $$\vec{v_i}(t)$$ is no longer constant in time. From this moment on, we have that $$\vec{F_i}(t)\not = \vec{0}$$.

You can think exactly the same about any one particle that make up the wall.

If the internal forces are "ideal-spring-like", then any particle that make up the ball (and this is also true for any one particle that make up the wall) will never again have $$\vec{v_i}(t)$$ constant in time, because once the ball (and the wall) start to deform, they will be "oscillating" forever. In the case of "real-spring-like" internal forces, after some time (after the ball-wall interaction) all particles will have $$\vec{v_i}(t)$$ constant in time again, because in this case there will be energy dissipation.

In the case of "ideal-spring-like" internal forces, from t_1 to t_2 (or better said, after the ball-wall interaction has ended) the total energy of the system (kinetic plus elastic potential) is constant in time (and equal to the total kinetic energy at the beginning which was the total energy then too), but whereas before the interaction this total energy was kinetic energy only, after the interaction this same constant total energy is the sum of two forever varying (oscillating) energies: kinetic and elastic potential.

In the case of "real-spring-like" internal forces, some time after the interaction it comes a time when the total energy of the system is again only kinetic energy, but less than the total energy at the beginning (which was also only kinetic energy then too).

The theorem applies every time you have a system of point-particles with forces that are differentiable vector function of time (and it is stated relative to an inertial frame).

17. May 14, 2014

### lightarrow

Ok.
I outline the system as a spring attached to a point mass (or to a rigid ball) of mass m which collides with the wall at velocity v (as I wrote in the post you quoted). The spring has negligible mass with respect to m; the wall is rigid (in case I post a picture but I think you understood what I mean).

So the system considered is: (spring+ball).

Can I apply/use the work-kinetic energy theorem to this system to compute the system's kinetic energy variation (or the external force) between two arbitrary instants of time during the collision?
If no, why; if yes, why and how you write it .

18. May 14, 2014

### lightarrow

The system I considered in my last post:

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19. May 14, 2014

### Vibhor

Let us consider the system to be a horizontally moving elastic ball colliding with a vertical wall with a speed v.At t=0,the ball comes in contact with the wall and at t=t1 it is at maximum compression. Then applying work kinetic energy theorem ,we can write this as ∫F.dx = 0 - (1/2)mv2 .dx is the displacement of the CM of the ball and F is the force exerted by the ball .Here work done by the wall will be negative.

20. May 14, 2014

### mattt

Of course, as I told you earlier, you can make a much simpler mathematical model which is still quite good an approximation (depending on what you want, of course).

If your model is a point particle, a rigid wall (that will not move because the wall + Earth mass is so much greater than the particle mass), an a "spring-like" interaction between wall and point-particle, then you can use just this very simple mathematical model:

The system is just one point-particle of mass "m" and the interaction is just $$\vec{F}(x)=-kx\vec{i}$$ (valid for $$x\in [-a,0]$$ being $$a$$ the difference between the spring length and the spring length when it is maximally compressed)

which is conservative.

Then if t_0 is the moment of time when the spring (of your picture) touch the wall, t_1 when it is maximally compressed, and t_2 when it again leaves the wall, then:

$$\int_{t_0}^{t_1}\vec{F}(t)\cdot\vec{v}(t)dt = 0 - \frac{1}{2}mv(t_0)^2$$

And also:

$$\int_{t_1}^{t_2}\vec{F}(t)\cdot\vec{v}(t)dt = \frac{1}{2}mv(t_2)^2 - 0$$

Being $$v(t_2) = v(t_0)$$ and $$\vec{v}(t_2) = - \vec{v}(t_0)$$

And obviously

$$\int_{t_0}^{t_1}\vec{F}(t)\cdot\vec{v}(t)dt = - \int_{t_1}^{t_2}\vec{F}(t)\cdot\vec{v}(t)dt$$

And $$\int_{t_0}^{t_2}\vec{F}(t)\cdot\vec{v}(t)dt = 0$$