= Solution
For constant particle masses define the <centre of mass> by $M\boldsymbol X=\sum_i m_i\boldsymbol r_i$, with $M=\sum_i m_i$. Summing <Newton's second law> over all particles gives
$$
M\ddot{\boldsymbol X}=\sum_i\boldsymbol F_i^e+
\sum_{i<j}(\boldsymbol F_{ij}+\boldsymbol F_{ji}).
$$
Each pair of internal <forces> cancels by <Newton's third law>. Hence
$$
\boxed{M\ddot{\boldsymbol X}=\boldsymbol F^e,\qquad
\boldsymbol F^e=\sum_i\boldsymbol F_i^e.}
$$
Only the action-reaction condition is needed for this translation law; centrality of the internal <forces> is needed for the <angular momentum> result in part (ii).
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