Suppose two bodies $A$ and $B$ with equal mass are kept at a distance. $A$ starts moving towards $B$ at a constant velocity, they hit, and I assume that it is a PERFECTLY INELASTIC collision here.

According to what I learnt from my book, suppose a body hits another body and inelastic collision takes place, in order to maintain conservation of momentum, the final velocity of the two bodies will be equal and they will stick together.

But I was thinking about it with respect to velocities and I came across a dilemma. Suppose the velocity of $A$ is $V_A$ and of $B$ is $V_B$. Before collision, $V_B=0$. After $A$ hits $B$, it applies a force on it, and $V_B$ gains some velocity. Since I assumed that both of them have equal mass, the acceleration of $B$ will be equal to retardation of $A$. Let the change in velocity at the instance of the collision be $x$.


relative velocity of $A$ w.r.t. B before collision$=V_A-V_B=V_A$

relative velocity of $A$ w.r.t. B at the instant of collision after instantaneous acceleration$=V_A-x-(V_B+x)=V_A-x-V_B-x = V_A-V_B-2x$

This shows that the relative velocity of $A$ w.r.t. $B$ has decreased.

But doesn't this mean that $A$ will apply less force at $B$ now?

That means,

$A$ will have less retardation, and $B$ will have less acceleration than before.

But doesn't that mean that the rate of decrease of velocity of $A$ as well as the rate of increase of velocity of $B$ will keep decreasing? Which means relative velocity of one of the bodies with respect to the other will keep decreasing till it approaches $0$, but will never be $0$?

So does this mean the two bodies never truly attain same velocity?

P.S: I referred to this post which I had asked long back before knowing about collisions based on a thought experiment


2 Answers 2


As I am not exactly sure as to your question I have given a number of possible answers.

In your previous post An overwhelming thought experiment regarding Newton's Third Law and motion of two bodies in an ideal condition it is explained that during a collision the total energy of the system of two colliding masses was the sum of kinetic energy and elastic potential energy.
In a system where the kinetic energy is not conserved internal forces do work, think of the masses oscillating with damping present, and heat and sound are produced and work is done by internal forces irreversibly breaking the bonds between atoms.
This is what would be happening in this example.

Now this might not answer your question in that you actually were asking about a variation of Zeno’s Paradox regarding Achilles and a tortoise where it appears that adding ever deceasing quantities never allows one to get to a final state?

This paradox is as follows:
Achilles, the fleet-footed hero of the Trojan War, is engaged in a race with a lowly tortoise, which has been granted a head start. Achilles’ task initially seems easy, but he has a problem. Before he can overtake the tortoise, he must first catch up with it. While Achilles is covering the gap between himself and the tortoise that existed at the start of the race, however, the tortoise creates a new gap. The new gap is smaller than the first, but it is still a finite distance that Achilles must cover to catch up with the animal. Achilles then races across the new gap. To Achilles’ frustration, while he was scampering across the second gap, the tortoise was establishing a third. The upshot is that Achilles can never overtake the tortoise. No matter how quickly Achilles closes each gap, the slow-but-steady tortoise will always open new, smaller ones and remain just ahead of the Greek hero.
I refer you to What Is the Answer to Zeno’s Paradox? for the resolution.

Another possibility is that the functions relating to the changes are of an exponential form, eg $\propto \text{e}^{-kt}$ which mathematically never reaches zero but one can say in Physics that after a time the model breaks down as in the real world the final state is reached. There are many other examples, eg reaching a terminal speed whilst falling, the complete discharge of a capacitor, etc.

  • $\begingroup$ That paradox about Achilles and the tortoise is an excellent comparison to my question. I'll check out the link you posted regarding its solution. $\endgroup$ Jan 26, 2022 at 15:14

When you think of inelastic collisions it's best to think of two lumps of clay colliding. They give the most intuitive picture.

We can move to the rest frame of the two lumps to make things easier to reason about. In the rest frame the total momentum is zero. In the rest frame your question becomes "Do the centers of mass of the two lumps keep moving together forever?". This is still a valid point because the force between the lumps does get smaller as they lose velocity. But at some point the relative velocity becomes smaller than thermal motion: the random motion of atoms and molecules gives a larger contribution to the velocity than the relative velocity from the collision. At this point you can safely say the relative velocity is zero.

You may object that you don't want to reason about thermal motion at all. You are wondering about the idealized mathematics which is a reasonable thing to ask. In this case you can model the collision in two ways. Either by a contact force, like you describe, or by modelling the collision as an instantaneous transfer of momentum (an impulse). In the first case the relative velocity only asymptotically goes to zero and the "paradox" you describe holds true. In the second case the relative velocity after a perfectly inelastic collision is defined to be zero. For an impulse collision you don't worry about the details of the collision you only worry about the outcome.

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    $\begingroup$ I see. Actually I didn't consider my situation to be real, real in the sense that the bodies will not compress or deform or anything, there are no forces or objects in the universe besides those two bodies which are the only ones capable of applying forces on each other and they can only change the state of motion of a body. I know that deformation and compression occurs in the real world, my bad for no clarifying it in the post itself. Thank you for your answer! $\endgroup$ Jan 26, 2022 at 15:17
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    $\begingroup$ @Mehmer In that case the last paragraph is till valid. That's what I meant with the mathematical model. $\endgroup$ Jan 26, 2022 at 15:51
  • $\begingroup$ Yes, I got it. Thanks $\endgroup$ Jan 26, 2022 at 15:59

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