I am taking for granted that when we say that something is conserved it is understood 'in its full integrity'.
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Energy is represented by a scalar J, and is conserved in elastic collision.
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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?
EDIT :
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?
Best Answer
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!