Derive an expression for centre of mass of a two particles system.

Explanation

Consider a system of two particles `P_1 and P_2` having masses `m_1 and m_2` . Let `vecr_1 and vecr_2` be their position vectors at any instant t, with respect to origin O.
The velocity and acceleration of two particles are
`vecv_1=(dvecr_1)/(dt) and veca_1=(dvecv_1)/(dt)=(d^2vecr_1)/(dt^2)`
`vecv_2=(dvecr_2)/(dt) and veca_2=(dvecv_2)/(dt)=(d^2vecr_2)/(dt^2)" "...(i)`
Total force acting on `P_1 and P_2`
Where `F_1 and F_2` are external forces acting on particles 1 and 2 respectively.

`F_(12)` : Internal force acts on particle 1 due to 2.
`F_(21)` : Internal force acts on particle 2 due to 1.
Now, according to Newton.s `2^(nd)` law of motion
`m_1veca_1=vecF_(P_1)=vecF_(12)+vecF_1 " "..(ii)`
`m_2veca_2=vecF_(P_2)=vecF_(21)+vecF_2 " "..(iii)`
On-adding equations (ii) and (iii), we have
`m_1veca_1+m_2veca_2= (vecF_(12)+vecF_1)+(vecF_(21)+vecF_2)`
`m_1veca_1+m_2veca_2= (vecF_(12)+vecF_(21))+(vecF_(1)+vecF_2)`
As `vecF_(12) = -vecF_(21)` [According to Newton.s `3^(rd)` law]
Therefore `m_1veca_1+m_2veca_2 =vecF_1+vecF_2=vecF`
So `m_1veca_1+m_2veca_2=vecF" "...(iv)`
Total mass of system `M= m_1+m_2`
Let us suppose the force `vecF` acting on system of mass M produces acceleration `veca_(c.m)` , then
`Mveca_(c.m.)=vecF" "...(v)`
Equating (iv) and (v) , we have
`m_1veca_1+m_2veca_2=Mveca_(cm)`
`veca_(cm)=(m_1a_1+m_2a_2)/M=(m_1a_1+m_2a_2)/(m_1+m_2)" "[:. M =m_1+m_2]`
`(d^2vecR_(cm))/(dt^2)=(m_1(d^2vecr_1)/(dt^2)+m_2(d^2vecr_2)/(dt^2))/(m_1+m_2)`
`d^2/(dt^2)[vecR_(cm)]=(d^2)/(dt^2)[(m_1vecr_1+m_2vecr_2)/(m_1+m_2)] " ".....(vi)`
On integrating equation (iv) , twicely w.r.t. time , we have
`vecR_(cm)= (m_1vecr_1+m_2vecr_2)/(m_1+m_2)`
This is the expression for centre of mass of two particle system.