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In an inelastic collision where mechanical energy is not conserved, momentum is still conserved. Should the Earth or the ground be considered as part of the system in order to account for the conservation of momentum? Since there is friction in a collision, is it appropriate to include the ground as part of the system to analyze the conservation of momentum?

Are air particles considered to be a part of this isolated system?

To clarify I am specifically talking about two vertically-moving objects colliding.

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    $\begingroup$ Kinetic energy never has to be conserved in any system. Total energy does. In an inelastic collision, kinetic energy is converted to strain energy and thermal energy $\endgroup$
    – RC_23
    May 30 at 18:08
  • $\begingroup$ I meant mechanical energy instead of kinetic. Edited the question just now. But concerning the conservation of momentum, is the ground/Earth seen as a component in our system? $\endgroup$
    – Authentic Melody
    May 30 at 18:13
  • $\begingroup$ It depends on the exact situation. Are any objects moving vertically or hitting the ground? Newton's 2nd Law (in integrated impulse form) says $\int Fdt = \Delta p$, i.e. a force can add or remove momentum from the system. This is why an object skidding against the ground will slow down and stop, as its momentum is transferred away by friction $\endgroup$
    – RC_23
    May 30 at 18:21
  • $\begingroup$ "Since there is friction in a collision" - that's not necessarily the case, friction need not play any role at all in a collision, like in the case of two birds colliding mid-flight, or in the case of an object being dropped straight down onto the floor. $\endgroup$
    – Nuclear Hoagie
    May 30 at 19:43

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Momentum is conserved in an isolated system. An isolated system has no (important) net external forces acting on it. You get to choose what constitutes the system, and therefore what forces are external and whether the system is isolated.

See Clarification about what makes a system isolated for more.

Example: You drop a rock which hits the earth in an inelastic collision.

  1. You take the rock as your system. The rock and earth exert gravitational and collision forces on each other. The forces on the rock come from outside the system. They are external forces. Momentum is not conserved. First it increases and then decreases back to its starting value. The forces on the earth are forces on a different system.

  2. You the the rock and the earth as your system. The rock and the earth exert the same forces as before. Now they are all forces one part of the system exerts on another part. They are internal forces. The sum of such forces must add to $0$. While different parts of the system gain and lose momentum, the total momentum of the rock + earth stays constant.

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  • $\begingroup$ But is it appropriate to consider the system above isolated? $\endgroup$
    – Authentic Melody
    May 30 at 18:38
  • $\begingroup$ Would it be better to say no net external forces? $\endgroup$
    – Bob D
    May 30 at 18:46
  • $\begingroup$ @BobD - You are right. A system can be considered isolated if the external forces total to $0$. But see the link for why I said important forces. $\endgroup$
    – mmesser314
    May 30 at 20:50
  • $\begingroup$ Are air particles considered to be a part of this isolated system? $\endgroup$
    – Authentic Melody
    May 31 at 21:10
  • $\begingroup$ Your choice. But forces from air such as buoyancy and wind resistance are often so small that they are ignored in problems like this. Air pressure is very big, but balanced, and so can usually be ignored. $\endgroup$
    – mmesser314
    May 31 at 21:38
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Should the Earth or the ground be considered as part of the system in order to account for the conservation of momentum?

If you are talking about an inelastic collision between an object and the ground, then the ground must be considered part of the system for conservation of momentum.

Hope this helps.

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Your question is too vague to answer.

If we're working in 3 dimensions, the ground breaks translational invariance an momentum normal to it is no longer conserved (e.g: bouncing).

If you include the ground as part of your system, then said bounce will impart momentum to the ground such that total momentum is conserved. Of course, this requires no energy, as $KE=p^2/2M$, and $M$ is enormous.

If you exclude the ground, but say it has gravity: then momentum is again, no longer conserved, as $\vec g$ breaks translational invariance.

If you decide to include the Earth, then everything is conserved again...but you have utterly failed to simplify the problem.

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