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When considering room as a system, when a ball is made to fall freely and it collides with the floor it loses speed each time it collides. The momentum hence is not conserved in this case? Can anyone elaborate on the topic, it would help me understand a great deal.

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    $\begingroup$ If your system includes the earth, momentum is conserved $\endgroup$
    – lalala
    Feb 6, 2020 at 6:54
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    $\begingroup$ Momentum is not conserved even if there is no lose of speed (perfectly elastic bouncing). $\endgroup$ Feb 6, 2020 at 7:12

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If you consider ball and earth as a system then net force should be zero but in your consideration mg act on the ball. where m is mass of ball and g is acceleration due to gravity.

If you consider a system in which net force is zero then linear momentum should be conserved. example- a ball collide to another ball horizontal ,if perfectly inelastic collision then e=0,e is coefficient of restitution.

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The conservation laws of momentum , angular momentum and energy are absolute within an inertial frame. When the earth is within the inertial frame, as in your question, the mass of the earth is so large with respect to the ball, that the momentum is taken up without the results being measurable.The loss of speed is also due to the fact that the interaction is not completely elastic, there is friction. Can get quite complicated.

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  • $\begingroup$ I do realise now that ive considered earth but isint momentum directly related to the velocity? So in inelastic collision when kinetic energy is lost, isint speed also lost? So how is momentum conserved then? $\endgroup$
    – deewhy
    Feb 6, 2020 at 17:26
  • $\begingroup$ friction on the floor dissipates energy and momentum turning it into kinetic vibrational and rotational energy and finally into hear. Momentum is $mv$ , kinetic energy s $1/2mv^2$ . when momentum is lost also kinetic energy will be lost, $\endgroup$
    – anna v
    Feb 6, 2020 at 18:18
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The complete system to be considered would include the recoil on earth as mentioned in the other answers, however if the collision were to be elastic, including the earth's recoil won't explain why the bounces actually become smaller and smaller in a non-ideal situation, meaning a real objective and not a point-like idealized particle.

The reason is really loss of energy/momentum due to non-conservative forces such as friction already mentioned. However the most relevant for this example is the shock absorption properties of the ball, and therefore the duration of the interaction. If you are thinking of a basketball vs a "fluffy" ball behave totally different, the energy is distributed differently depending on the inner structure of the material. A bounce produces deformation (breaking/elongation of some structure against a potential), heat and noise through vibrations, etc.

You can watch for example this short YouTube video to have a better intuition.

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  • $\begingroup$ I am having trouble visualising the fact that even when inelastic collision occurs, momentum is conserved. Lets not even consider earth, lets just talk of normal collision between two balls? If energy is lost then so is speed and hence the momentum not conserved? $\endgroup$
    – deewhy
    Feb 6, 2020 at 17:29
  • $\begingroup$ Yes, the thing to remember is that momentum is conserved as long as there are no external forces acting on the system, inelastic in this context means that there was some energy lost in some internal degree of freedom that is not being accounted for so one cannot attribute it formally to a force (or a direction even). Also the fact that "speed is lost" as you put it, is not precise enough, because mass and directions are also at play. $\endgroup$
    – ohneVal
    Feb 6, 2020 at 21:39
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Yeah momentum is not conserved in this case as an external impulsive force is exerted by earth on floor and hence momentum is not conserved in this case

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