= Solution
{c}
The <Lagrangian> is $L=\sum_i m_i|\dot q_i|^2/2-f(r)$. Away from a collision, its equations of motion are
$$
m_1\ddot q_1=-\frac{f'(r)}r(q_1-q_2)=:F,\qquad
m_2\ddot q_2=-F,\qquad m_3\ddot q_3=0.
$$
Summing forces proves conservation of total <linear momentum>. The total <mechanical energy> $E=\sum_i m_i|\dot q_i|^2/2+f(r)$ is also conserved: its derivative is $F\cdot(\dot q_1-\dot q_2)+f'(r)\dot r=0$, because $\dot r=(q_1-q_2)\cdot(\dot q_1-\dot q_2)/r$.
For the first proposed <angular momentum>,
$$
\frac d{dt}(m_1q_1\times\dot q_1)=q_1\times F
=\frac{f'(r)}r q_1\times q_2,
$$
which need not vanish. Thus \b[particle one's <angular momentum> about the fixed origin is not generally conserved]. Its force is central relative to particle two, not necessarily relative to that origin.
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