= Solution
{c}
The <free-body diagram> shows a meridional section of the <right circular cone>. On the first particle, <gravity> acts vertically downwards, <tension> points along the <straight line> to the apex, and the <normal reaction> is perpendicular to the <surface>. The <smooth mechanical constraint> supplies no tangential <friction>. The second particle has only its downward <gravitational weight> and the upward <tension>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-iv-resit-cone-forces.png]
{title=Meridional section of the inverted cone and the forces on both particles; the reaction is normal to the cone}
Adding the <forces> on each particle and applying <Newton's second law> gives
$$
\boxed{m_1\ddot{\mathbf r}_1=-m_1g\mathbf e_z+\mathbf N+\mathbf T_1,\qquad m_2\ddot{\mathbf r}_2=-m_2g\mathbf e_z+\mathbf T_2.}
$$
At $z_1=-r_1$, a <normal vector> is $\mathbf e_r+\mathbf e_z$. Its sign depends on the supporting face of the <cone>; either sign has zero <dot product> with every permitted <velocity>. The diagram chooses the reaction towards the upper side of the <surface>.
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