= Solution
Let $\mathbf w=\mathbf v_1-\mathbf v_2$ be the <relative velocity> and $\mathbf V=(M_1\mathbf v_1+M_2\mathbf v_2)/(M_1+M_2)$ the <centre-of-mass velocity>. Inverting gives
$$
\mathbf v_1=\mathbf V+\frac{M_2}{M_1+M_2}\mathbf w,\qquad
\mathbf v_2=\mathbf V-\frac{M_1}{M_1+M_2}\mathbf w.
$$
Substitute into the sum of the two <kinetic energies>. The cross terms cancel, while the coefficient of $|\mathbf w|^2/2$ is the <reduced mass> $\mu=M_1M_2/(M_1+M_2)$. Hence
$$
T=\frac12(M_1+M_2)|\mathbf V|^2+\frac12\mu|\mathbf w|^2.
$$
The first term is translation of the <centre of mass>; the second is the <kinetic energy> measured in the <centre-of-momentum frame>. For an isolated <perfectly inelastic collision>, the bodies finish with a common <velocity>, while their total <momentum> and thus $\mathbf V$ are unchanged. The <kinetic energy> converted to internal forms is therefore
$$
\boxed{T_i-T_f=\frac12\mu v^2,}
$$
where $v$ is their initial relative <speed>. Total <energy> remains conserved; it is the translational-plus-relative <kinetic energy> that decreases.
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