The pictures below are from the 39-4 section of The Feynman Lectures on Physics. I don't know how to prove the red line. By the conservation of momentum and energy. There are
$$
m_1u_1 + m_2 u_2 = m_1 \hat u_1 + m_2 \hat u_2 \\
m_1u_1^2 + m_2 u_2^2 = m_1 \hat u_1^2 + m_2 \hat u_2^2
$$
where $\hat u_1, \hat u_2$ are the speed vector after collision. I can't see $|\hat u_1| =|u_1|, |\hat u_2| =|u_2|$.
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1$\begingroup$ In the center-of-mass (CM) system $m_1 \vec{u}_1+m_2 \vec{u}_2=\vec{0}$ holds. $\endgroup$– HyperonCommented 2 days ago
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1 Answer
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It comes from including conservation of energy.
Let's work in the centre of momentum frame where we have:
$$ m_1 v_1 + m_2 v_2 = 0 $$
so:
$$ v_2 = -\frac{m_1}{m_2} v_1 $$
Then the initial total kinetic energy is:
$$ T_i = \frac12 m_1 v_1{}^2 + \frac12 m_2 (\frac{m_1}{m_2} v_1)^2 $$
We can use the same argument so find the final KE:
$$ T_f = \frac12 m_1 \hat{v}_1{}^2 + \frac12 m_2 (\frac{m_1}{m_2} \hat{v}_1)^2 $$
Then equating the initial and final KE we find $v_1 = \hat{v}_1$ and therefore $v_2 = \hat{v}_2$. Note that this is only true in the centre of momentum frame.
