**Derrivation of the principle of he conservation of linear momentum**

Suppose that we have an isolated system of \(n\) particles. Since we

re assuming that the system is isolated, the onyl forces acting on particles in the system must be due to other particles in the system acting on them. Suppose that any arbitrary point mass \(m_i\) exerts any arbitary force \(\vec{F}_{i,j}\) on any other arbitrary point mass \(m_j\); by Newton's third law, \(m_j\) will exert an equal-and-opposite force \(-\vec{F}_{i,j}\) on \(m_i\). Since the net force, \(\sum{\vec{F}}=\vec{F}_{i,j}+\vec{F}_{j,i}=\vec{F}_{i,j}-\vec{F}_{i,j}\), acting on the \(m_i\)-\(m_j\)-system is zero, we can write

$$\sum{\vec{F}}=\frac{d}{dt}(\vec{p}_i+\vec{p}_j)=0.$$

If the derrivative of any "something" is zero, then that something must be a constant. Thus,

$$\vec{p}=\vec{p}_i+\vec{p}_j=c.\tag{2}$$

Equation (2) tells us that, despite the fact that the masses of these two particles can, in gneral, be different, and despite the fact that they may be interacting with one another via any complicated and whcky forces altering their velocities in complicated ways, the sum \(\vec{p}_i+\vec{p}_j=m_i\vec{v}_i+m_j\vec{v}_j\), will always equal \(c\) and won'[t change (as long as our assumption that the system is isolated holds true). We showed this for any two arbitary point masses \(m_i\) and \(m_j\) in the system. If the total momentum of any pair of two point masses never changes and is always constant, then it must follow that the total momentum of the entire system never changes and is always constant. Thus, we can write

$$\vec{p}=\sum_{i=1}^{n}m_i\vec{v}_i=c.\tag{3}$$

Equation (3) is known as the conservation of momentum and applies to any system composed of \(n\) point masses. Although we won't prove it explicitly here, even if the number of point masses approaced infinite (in which case, we would be dealing with a continuous mass distribution), all of the same arguments involving Newton's laws would still apply. We could go through all the same steps and find that the total momentum is still just a constant.

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