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If an object under a constant force , say gravity. Collides with another object with no external force acting on it. Will the momentum of the system right before and after the collision be the same if the collision is instantaneous? Assume that the collision is vertical.

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  • $\begingroup$ Depends if you include the thing it's gravitating towards. If not, then it won't be conserved as you'll be seeing momentum being given to the ball seemingly for no reason. If you do include the large body that it's gravitating to then all that additional momentum the smaller falling ball is getting would sum with the gained momentum of the large body to give zero and momentum would be conserved. $\endgroup$ – user177179 Dec 20 '17 at 17:54
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The Impulse-momentum theorem is $$\Delta \vec{p}=\int \vec{F}\cdot d\vec{t},$$ where the quantity on the right is the impulse, and $\vec{F}$ is an external force (or a sum of external forces). So if you want to know when momentum is conserved, $\Delta \vec{p}=0$, you want to know when the impulse will vanish. This can happen in two different generic ways

  • $\vec{F}=0$, and the system is isolated.
  • The time period over which the collision happens is very small, so that the impulse vanishes.

In your situation $\vec{F}\neq 0$, so the system is not isolated. You're saying the collision is "instantaneous", which to me means $\Delta t \simeq 0$, so that integral vanishes. Since is vanishes, $$\Delta \vec{p}=\vec{p}_f-\vec{p}_i=0\rightarrow \vec{p}_i=\vec{p}_f,$$ i.e. the momentum right before the collision is equal to the momentum right after.

We need to be a little careful though, since there is an external force on this system. So if your "initial" state is a finite time period before the collision, or if your "final" state a finite time period after, momentum will not be conserved.

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Yes. If the collision is instantaneous (contacct time $t=0$) then the external force $F$ imparts no impulse $Ft=\Delta p$ during the collision. So momentum $p$ is conserved during the collision.

For example, object A is accelerated by gravity towards object B. Object A reaches a certain velocity $v$ just before impact. The momentum of A is then shared between A and B in some way which conserves momentum, but probably involves some loss of kinetic energy. (Or possibly there is a gain of kinetic energy if stored energy is released.) If A and B have equal mass $m$ and there is no loss (or gain) of kinetic energy then A is brought to rest instantaneously and B moves downward with velocity $v$. (This is unrealistic because B must accelerate from rest to finite velocity $v$ in no time at all, which would require an infinite force.) The total momentum just before and just after the collision is $mv$.

However if the collision is not instantaneous then A and B will move downward together while they remain in contact for a time $t$. Gravity continues to act on A, increasing the momentum of the combined object AB by $mgt$. (I assume B somehow has inertial mass but is not affected by gravity.) After the collision, when A and B have separated, the total momentum is $mv+mgt$ whereas before collision it was $mv$.

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  • $\begingroup$ So in principle it can't be since the time of collision is not exactly 0?? $\endgroup$ – Khaled Oqab 2 days ago
  • $\begingroup$ @KhaledOqab That is correct. In practice the collision lasts a short but finite time $\Delta t \approx 1ms$, during which gravity is acting so the total momentum increases by $mg\Delta t$. $\endgroup$ – sammy gerbil 2 days ago
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No momentum is not conserved conceptually in the presence of external force but for a instant in numericals we take it constant

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  • $\begingroup$ So , as the momentum instantly before and after the instant collision is conserved? $\endgroup$ – Jonathan Willianto Dec 20 '17 at 17:35
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Yes: momentum is conserved, but there is massive loss of kinetic energy. The larger the stricken object the greater the loss of energy.

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  • $\begingroup$ You would need to clarify your answer as conservation of momentum is for closed systems, on which there is no external force. $\endgroup$ – ZeroTheHero Dec 27 '17 at 21:06

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