Solution
= Solution
{c}
The third particle is free, so
$$
\frac d{dt}(m_3q_3\times\dot q_3)=q_3\times m_3\ddot q_3=0.
$$
Thus \b[particle three's <angular momentum> is conserved].
= Solution
{c}
The third particle is free, so
$$
\frac d{dt}(m_3q_3\times\dot q_3)=q_3\times m_3\ddot q_3=0.
$$
Thus \b[particle three's <angular momentum> is conserved].