# Is this a valid way to derive conservaton of momentum?

So i’ve just been considering some basic things, and i wanted to make a quick way to show conservation of the total momentum of a system occurs under the collision of two objects.

Consider force, which is $$F = \frac {dp}{dt}$$ Where p= momentum. So, with force being the time derivative of momentum, it can be rewritten as $$F= \frac{d}{dt}(p)$$

Under a collision between two objects, in accordance with newtons third law (equal and opposite reaction), the total force in the system will equal zero.

So using the definiton of force as the time derivative of momentum, we can rewrite the total force of the system as $$\frac{d}{dt}(p_1) + \frac{d}{dt}(p_2) = 0$$ or $$\frac{d}{dt}(p_1+p_2)=0$$

Now, from basic knowledge of derivatives, the derivative of a function is zero if that function is a constant, so $$p_1+p_2$$ must have a constant value because its time derivative (total force) is zero. So the total momentum is constant, or conserved under a collision.

This may be totally wrong, but that’s why i’m asking the question here. So apologies if i sound totally stupid.

When you say

$$\vec{F} = \frac{d\vec p}{dt}$$

you have already implied conservation of momentum. I could rewrite it with "secret" subscripts and some rearranging and have:

$$\vec{F}_{\rm net,\,ext}\, dt = {d\vec p}_{\rm sys}.$$

What I mean here is that $\vec F_{\rm net,\,ext}$ is the net, external force on a system with momentum $p_{\rm sys}(t)$. The net force is outside of the system, and the impulse $\vec{F}_{\rm net,\,ext}\, dt$ it provides changes the system's momentum accordingly. It is a statement of conservation of momentum.

After all, you can always separate variables and solve this differential equation to have

$$\vec p_f = \vec p_i + \vec{F}_{\rm net,\,ext}\, \Delta t.$$

At any rate, conservation of momentum is mostly an empirical observation. Mathematically, it can be derived from Noether's theorem as a consequence of translation symmetries in position.

Your analysis of the collisions is correct. I described it in this physics SE answer.

• Thank you very much for the explanation. I haven’t really considered impluse too much, i’m still learning physics at school :/. Was what i said about p1+ p2 being constant means total momentum is conserved true? – Thatpotatoisaspy Aug 12 '18 at 2:39
• Great answer. I think it would be worth mentioning that the "mostly empirical observation" is just Newton's third law. This is what can be used to "derive" momentum conservation. i.e. you just need N3L and the definition of momentum. Of course, Noether's theorem is more fundamental, but I still think it is worth pointing out. – Aaron Stevens Aug 12 '18 at 5:56
• @Thatpotatoisaspy Yes, that is true. For a system of two particles with momenta, $p_1$ and $p_2$, the total momentum is $p_1 + p_2$. And total momentum is always conserved in absence of a net external force. – Zack Hutchens Aug 12 '18 at 11:02
• @zhutchens1 Ah, thank you for your help. – Thatpotatoisaspy Aug 12 '18 at 11:19