# Why is momentum conserved in inelastic collisions? How is it related to momentum-impulse theory?

First of all I want to mention, that I've found many questions around this site and in other websites dealing with my question; however, I don't think they answer my question fully. So I am here to re-ask it and specify some things so that whoever will answer it can focus on my ideas and help me understand momentum better.

So here is my question: Why is momentum conserved in inelastic collisions (perfectly inelastic or just inelastic)?

So here is what I thought about in this question: I understand that elastic collisions have the K.E conserved but inelastic don't.

So in inelastic if K.E is not conserved then... Wnet=deltaKE (work energy theorm) so here that means that their is work done and a net force acting on the system.

Now from impulse momentum... Ft=deltap (isnt here F not equal zero so the change in momentum not equal zero?)

So basically my confusion comes in the fact that to have a change in momentum their must be a force and time, and the force exists because K.E changes?

I hope you will help me understand this concept more, because I cant seem to wrap my head around it. What am I doing wrong?

• en.wikipedia.org/wiki/… – lemon Mar 18 '15 at 17:18
• Momentum is always conserved in any simple collision. Energy is also always conserved, but energy can be converted to things like heat, so objects can lose kinetic energy in collisions. Even in cases where momentum may seem to disappear, like if you throw a ball straight up in the air until it stops, if you expand your system enough you'll find that the earth itself gains momentum so that total system momentum is conserved. – MonkeysUncle Mar 18 '15 at 17:31

Let's call our two colliding objects $A$ and $B$. So object $A$ and object $B$ come together, collide and ricochet away again. The collision may be elastic or inelastic.

You are quite correct that there must be a force acting during the collision, and because the force acts for some time there is an associated change in momentum. Consider just object $A$. If a force $F$ acts on $A$ for a time $t$ then the momentum of $A$ changes by the impulse $Ft$:

$$\Delta p_A = Ft$$

But remember that the force on $A$ is being exerted by $B$ during the collision. And Newton's third law tells us that the force being exerted on $A$ by $B$ must be equal and apposite to the force being exerted on $B$ by $A$. So if the force on $A$ is $F$, then the force on $B$ must be $-F$. Therefore the momentum change of $B$ is:

$$\Delta p_B = -Ft$$

The total momentum change is:

$$\Delta p_{total} = \Delta p_A + \Delta p_B = Ft - Ft = 0$$

and that's what conservation of momentum means. It means the total momentum is unchanged. The momenta of the individual objects $A$ and $B$ can and indeed do change, but the total momentum remains constant.

• OK, so it's the whole momentum that we are supposed to consider right? and another thing: inelastic collisions have friction? If so then why is momentum still conserved? – mariam.. Mar 19 '15 at 2:44
• @mariam.. the argument I've made above doesn't depend on whether the collision is elastic or inelastic. In all cases the forces on the two bodies are equal and opposite so the momentum changes must be equal and opposite. – John Rennie Mar 19 '15 at 6:13

A collision is always a continious process meaning that one shouldn't think of a force acting for a single instance of time. There's a theorem in physics called "Noether's theorem" (wikipedia) which states that if the underlying physics are invariant under certain continious transformations, there exists a conserved quantity. If that transformation is

1. translation in time
2. translation in space
3. rotation around an axis

the conserved quantity is

1. energy
2. momentum in direction of that translation
3. angular momentum in direction of that rotation axis .

So for instance if a physical process is "invariant" (see below for further clarification) under translation in any direction, the momentum in any direction is conserved.

Now, if we know that a certain quantity is conserved we can just try to skip the excat dynamics and calculate things based on our quantity, that is we take momentem "before" the collision and set it equal to momentum "after" the collision. "Before" and "after" shall mean instances of time when the forces that give rise to a momentum change of a subsystem (e.g. single particle) are still negligible so that one has movement on a straight line approximately.

Above I have used the term "invariant". What I have meant really is that the action $S=\int L \ dt$ shall be invariant under such transformations 1,2,3. In classical mechanics one has $L = T - U$ with $T$ kinetic energy and $U$ potential energy. This is part of a different forumlation of mechanics called "lagrangian mechanics". Leaving the action $S$ invariant, one can directly follow that a certain transformation leaves the laws of physics (e.g. differential equations) form invariant, so that they are (in a sense) the same for different observers seperated by that transformation. This means the qualitative behaviour of a process is the same for both observers.