Assume initial direction of motion to be positive. Also assume that `u_1 > u_2` (so that collision may take place). Conservation of momentum gives:
`m_1u_1+m_2u_2=m_1v_1+m_2v_2` `...eq^n (i)`
`=> m_1(u_1-v_1)=m_2(v_2-u_2)` `...eq^n (ii)`
In case of perfectly elastic collision, total kinetic energy also remains conserved i.e. KE will be same before and after the collision:
`\frac1 2 m_1u_1^2+\frac1 2m_2u_2^2=\frac1 2 m_1v_1^2+\frac1 2 m_2v_2^2` `...eq^n (iii)`
`=> m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2)` `...eq^n (iv)`
Dividing `eq^n(iv)` by `eq^n(ii)`, we get:
`u_1+v_1=u_2+v_2` `...eq^n (v)`
`=> (u_1-u_2)=(v_1-v_2)` `...eq^n (vi)`
Thus, in 1-D perfectly elastic collision, "velocity of approach" before collision is equal to the "velocity of recession" after collision.
Now, let's multiply `eq^n (vi)` by `m_2` and subtract it from `eq^n (ii)`:
`(m_1-m_2)u_1+2m_2u_2=(m_1+m_2)v_1` `...eq^n (vii)`
`v_1 = (\frac(m_1-m_2) (m_1+m_2)) u_1 + (\frac(2m_2) (m_1+m_2)) u_2` `...eq^n (viii)`
Similarly, multiplying `eq^n(vi)` by `m_1` and adding it to `eq^n(i)`:
`2m_1u_1+(m_2-m_1)u_2=(m_1+m_2)v_2` `...eq^n (ix)`
`v_2 = (\frac(2m_1) (m_1+m_2)) u_1 + (\frac(m_2-m_1) (m_1+m_2)) u_2` `...eq^n (x)`
This is the derivation of final speeds of both objects after collision. `eq^n(viii)` and `eq^n(x)` give the final speed of `m_1` and `m_2` respectively. Care should be taken before applying these equations that these are valid only in case of perfectly elastic collisions.