m1 x v1 = m2 x v2 both are the same formula m1u1 + m2u2 = m1v1 + m2v2
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This cant be fully derived but a part of it can be.
In an elastic collision kinetic energy is conserved, so
1/2*m1*u1^2 + 1/2*m2*u2^2 = 1/2*m1*v1^2 + 1/2*m2*v2^2
m1*u1^2 + *m2*u2^2 = *m1*v1^2 + *m2*v2^2
m1*u1^2 - *m2*v1^2 = *m2*v2^2 - *m2*u2^2
m1(u1+v1)(u1-v1)=m2(v2+u2)(v2-u2) ---------- equation 1
NOW,
Accordinbg to conservation of linear momentum,
m1u1 + m2u2 = m1v1 + m2v2
m1(u1-v1) = m2(v2-u2) ---------- equation 2
dividing equation 1 by 2 we get ,
u1 + v1 = u2 + v2
u1-u2 = v2-v1
v2-v1/u1-u2 = 1
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