How can I show that in an elastic collision i can set $| vec{v_2} - vec{v_1} | = | vec{v_2}' - vec{v_1}' | $?

Question

I am getting stuck in a really easy problem in Statistical Mechanics that involves elastic collisions, it is really very shameful that I am getting stuck at such a simple thing, but from:
$$|vec{v_1}|^2 +|vec{v_2}|^2 = |vec{u_1}|^2 +|vec{u_2}|^2$$
and $$vec{v_1}+vec{v_2} = vec{u_1} + vec{u_2}$$
How can I get $$|vec{v_2}-vec{v_1}|=|vec{u_2}-vec{u_1}|$$
I tried completing the square in the first equation like:
$$vec{v_1}cdotvec{v_1} +vec{v_2}cdotvec{v_2} -2vec{v_1}cdotvec{v_2}= (vec{v_2}-vec{v_1})cdot(vec{v_2}-vec{v_1})=|vec{v_2}-vec{v_1}|^2= vec{u_1}cdotvec{u_1} +vec{u_2}cdotvec{u_2} -2vec{v_1}cdotvec{v_2}$$
and then using the second equation to get:
$$=vec{u_1}cdotvec{u_1} +vec{u_2}cdotvec{u_2} -2vec{v_1}cdot(vec{u_1}+vec{u_2}-vec{v_1})$$
but I cannot seem to be able to simplify this to $$vec{u_1}cdotvec{u_1} +vec{u_2}cdotvec{u_2} -2vec{u_1}cdotvec{u_2} = |vec{u_2}-vec{u_1}|^2$$
Can someone help me with this? I am sure it is quite simple, but since I am stuck I am losing way too much time on this.

Brain Stroke Patient · Accepted Answer

I'm typing in mobile so I'm ignoring all the vector signs. The problem is to show that
$$a^2 + b^2 = c^2 + d^2$$
and
$$a + b = c + d$$
Gives you
$$|a - b| = |c-d|$$
From the second eq you get by squaring both sides
$$a^2 + b^2 + 2ab = c^2 + d^2 + 2cd$$
Using the first equation you then get
$$ab = cd$$
Now subtract $2ab= 2cd$ on both sides of the first equation and you get
$$(a-b)^2 = (c-d)^2$$
Which is the required answer. To get the vector equivalent just replace the regular product with the dot product.

mike stone · Answer

Use the reduced mass identity to write the total energy as
$$
frac 12 m_1 |{bf v}_1|^2+ frac 12 m_2 |{bf v}_2|^2=
frac 12 frac{m_1m_2}{m_1+m_2} |{bf v}_1-{bf v}_2|^2+ frac 12 (m_1+m_2)|{bf v}_{CofM}|^2.
$$
Then, if energy is conserved (definition of elastic) we must have that $|{bf v}_1-{bf v}_2|$ is the same before and after.

How can I show that in an elastic collision i can set $| vec{v_2} - vec{v_1} | = | vec{v_2}' - vec{v_1}' | $?

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