I am taking for granted that when we say that something is conserved it is understood 'in its full integrity'.

  • Energy is represented by a scalar J, and is conserved in elastic collision.

  • momentum is the product of a number (of Kg) by velocity which is a vector, and therefore has a direction: $p = m * v$. (In circular motion we say that speed is constant but velocity is constantly changing).

In an elastic collision if a ball A (m = 1) with v = p = +8 in the x axis (E = 32) hits a ball B at rest (M =3) , B will proceed ( p = 4 *3 = +12, E = 24) and A will bounce in the opposite direction (p = v = -4, E = 8). We conclude that: +12 -4 = +8, momentum is conserved (and also energy: E = 8 + 24 = 32 J)

If a perfectly elastic ball is thrown against a wall energy J is conserved, $E-i = E_f$ the value of the speed is unchanged, but the direction of the vector is reversed.

I am asking: the momentum of what is conserved? if it is the momentum of the system this is not true in the first case, if it is the momentum of the single balls this is not true in the second case.

Language of science must be precise, I hope you will not consider this as hair-splitting, but I ask: is there a general definition of 'conserved' valid for all instances? why a broad/permissive interpretation of the principle in the case of momentum? Or, most likely, where did I go wrong?


This does mean that the wall contains a momentum of 2mv (for mass m and velocity v). But note that since the mass of the wall is incredible compared to the ball, the velocity is notably imperceptible!

If the momentum of the wall were to be 2mv, the energy of the bouncing ball should be impercetibly less but yet less: $E - \epsilon < E$ and the Energy of the ball would not be conserved. That means that a perfect elastic collision against a fixed body is not possible, in the sense that no body can have a CoR = 1. right?

  • $\begingroup$ The big planet under our feet is too easy to miss. It is the huge reservoir of linear and angular momentum we constantly use without even noticing. It is the reason why the laws of conservation were not-that-easy to come up with. $\endgroup$ – safkan Jan 4 '15 at 2:34
  • $\begingroup$ If you did have CoR = 1, the collision would be elastic -- again, for conservation you need to consider the energy passed to the wall as well. Then all will be conserved. But, even with a CoR = 1 ball, you can never bounce back with the same energy -- this one is because you can never have a "fixed body". If you push it, it will move, no matter how little. $\endgroup$ – safkan Jan 4 '15 at 2:37

Momentum is conserved in magnitude and direction. So in order to analyze any situation of momentum conservation, you should always start with $$ \sum \mathbf p_{i}=\sum\mathbf p_f $$ where the subscripts denote the initial and final momenta.

As to the ball & wall, you are correct that momentum is not conserved if you are only looking at the ball. If you consider that the system includes the wall, then the momentum conservation holds. This does mean that the wall contains a momentum of $2mv$ (for mass $m$ and velocity $v$). But note that since the mass of the wall is incredible compared to the ball, the velocity is notably imperceptible!

  • $\begingroup$ The wall is not even alone -- unless you are floating in space with it -- it is connected to the planet, which is massive. ($5.972 \times 10^{24}\mathrm{kg}$..) $\endgroup$ – safkan Jan 4 '15 at 2:30
  • $\begingroup$ @safkan: You are correct. If I recall correctly, there was a comment thread here (apparently now erased) in which I brought that point up. $\endgroup$ – Kyle Kanos Jan 4 '15 at 2:34

It is the momentum of the entire system that is conserved. The fundamental reason for this is that the laws of physics are the same everywhere in space. This argument for momentum conservation is called Noether's Theorem.

So where did you go wrong in your original example? Well you assumed that the wall was completely rigid. In reality that isn't actually true. The bricks in the wall will recoil very slightly when the ball hits them. Because the wall is much more massive than the ball this tiny recoil of the bricks has enough velocity to ensure that the overall momentum is conserved (both in magnitude and direction)!


An elastic collision means that the over all kinetic energy of the entire system before and after the collision is the same.

So the ball can bounce off the wall, and the wall can recoil in such a manner that you have an elastic collision.


Physicist created momentum as the property of the system that is conserved if on the system act only internal forces. A force is defined internal in the system if, the force itself and its reaction are applied on the system.

From the previous hypotesis you can demostrate that momentum is defined as $$\vec{p} = \sum m_i\vec{\upsilon}_i$$

The last example is the easiest, infact when the first ball touches the second, A applies a force to B, for the third principle of motion, B applies a force to A so, if you consider the system of only one ball, the momentum isn't conserved because the reaction to the force isn't applied in the system, but if you consider both the balls as your system, momentum is conserved infact action and reaction belong to the system.

If a ball is thrown to the wall, in order to conserve the momentum you have to decide what is your system. Choosing only the ball won't work (momentum is a vector, change direction and you change the vector). If you choose the ball and the wall, the wall must be free to move: a possible situation would be a wall floating in the space, hit by the ball, it will start moving. But since the wall is anchored to the house and the house to the ground, there will be forces acting on them, so every force must be internal to conserve his momentum. Practically you have to consider the whole earth as your system.


protected by Qmechanic Oct 8 '16 at 15:24

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