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I recently solved a problem in which I used the "fact" that the momentum is conserved just before and just after a collision between two (or maybe more) objects but I am not sure whether this is true or not.
This is what I've done:
Consider a given system $A.$ Then $\sum\mathbf F=\dfrac{d\mathbf p}{dt},$ where $\sum\mathbf F$ represents the sum of all external forces on $A$ and $\bf p$ represents the momentum of the system. Thus $d\mathbf p=\sum\mathbf{F}dt$ so if we only consider a case when $dt\approx0$ then $d\mathbf p\approx0.$

Is that result true?, Is my reasoning right?

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    $\begingroup$ Yes, it corrects. $\endgroup$ – qfzklm Sep 11 '15 at 7:37
  • $\begingroup$ Also, we typically assume that "ordinary" forces like gravity do not have sufficient magnitude over this short timescale, whereas friction, tension and normal reaction can have high enough magnitudes to apply impulses to objects. $\endgroup$ – aditya_stack Dec 30 '19 at 14:00
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Momentum conservation is valid always and everywhere, not just before and after a collision. The observation that dp goes to zero if dt does is trivial. Momentum cannot change on zero time, as this would require infinite force.

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This is not correct. $\Delta t$ (duration of impulse) is very short indeed, but $F_i$ (force on particle $i$) is very high. It is similar to attempting to compute $\infty\ 0$, which is not zero, but undefined. However, $\int F_i\mathrm{d}t$ obviously is defined, and it is not equal to zero if the velocity of your particle changes during the impulse. The sum of all the integrals $\sum_i\int F_i \mathrm{d}t$ that correspond to all your particles will be zero if no external force has been acting on the system and if no heat was lost during the collision(s), but not because you can set $\Delta t\to0$. The system here includes the kinetics of all particles. In other words, conservation of momentum of the particle kinetics can only be true if we assume zero heat loss. However, this is mostly not a valid assumption.

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  • $\begingroup$ Nobody said that $F\to\infty.$ Also, I obviously assume that the only forces acting on the system are the external ones (which is often assumed in basic physics)... $\endgroup$ – CIJ Sep 11 '15 at 15:51
  • $\begingroup$ If you would apply a finite force over an infinitesimal time interval, then the particle would never change velocity. A finite force implies a finite acceleration. We have $\dot{x}=\int^T_{0,T\to0}\ddot{x}\mathrm{d}t=0$ $\endgroup$ – JJM Driessen Sep 11 '15 at 16:03
  • $\begingroup$ "not defined", or indeterminate? $\endgroup$ – 299792458 Sep 11 '15 at 16:04
  • $\begingroup$ The first paragraph is misleading, or at best unclear. The second paragraph is completely incorrect. -1. $\endgroup$ – Emilio Pisanty Sep 11 '15 at 16:14
  • $\begingroup$ @The Dark Side $\infty 0$ is undefined and therefore also indeterminate. $\endgroup$ – JJM Driessen Sep 11 '15 at 16:17

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