= Solution
{c}
Use $g$ as the <gravitational acceleration>, so the weights are $mg$ and $\sqrt2mg$, consistently with the requested equations. The printed introductory phrase calling $g$ a <force> magnitude omits the factor $m$.
For the <particle on a cone coupled to a hanging mass>, write $z=-r$ and use the <cylindrical coordinates> <unit vectors> $\mathbf e_r,\mathbf e_\theta,\mathbf e_z$. The <ideal string> segment from the cone particle to the hole has length $\sqrt2r$ and direction $(-\mathbf e_r+\mathbf e_z)/\sqrt2$. If $h$ is the downward length of the hanging segment and $s_0$ the fixed total length, then
$$
h=s_0-\sqrt2r=s_0+\sqrt2z,\qquad z_2=-h=-s_0-\sqrt2z.
$$
The light <ideal string> and smooth hole give the same <tension> $T$ on both sides. Applying the vertical component of <Newton's second law> to the hanging particle gives
$$
\sqrt2m\ddot z_2=T-\sqrt2mg,
\qquad\boxed{T=\sqrt2mg-2m\ddot z.}
$$
Write the <normal reaction> on the cone particle as $N(\mathbf e_r+\mathbf e_z)/\sqrt2$, allowing $N$ to be signed according to the side of the ideal constraining <surface>. The <acceleration in cylindrical coordinates> and <Newton's second law> give
$$
\begin{aligned}
m(\ddot r-r\dot\theta^2)&=\frac{N-T}{\sqrt2},\\
m(r\ddot\theta+2\dot r\dot\theta)&=0,\\
m\ddot z&=\frac{N+T}{\sqrt2}-mg.
\end{aligned}
$$
There is no azimuthal <force>. Multiplying the middle equation by $r/m$ proves
$$
\frac{d}{dt}(r^2\dot\theta)=0,\qquad
\boxed{r^2\dot\theta=av,}
$$
where the positive angular direction is chosen along the initial <velocity>. This is <conservation of angular momentum> about the vertical axis.
Subtract the radial equation from the vertical equation to eliminate the <normal reaction>. Since $\ddot r=-\ddot z$, this gives
$$
m(2\ddot z+r\dot\theta^2)=\sqrt2T-mg.
$$
Substitute the <tension> and the conserved <angular momentum>, using $r=-z$:
$$
2(1+\sqrt2)\ddot z=g-r\dot\theta^2
=g+\frac{a^2v^2}{z^3}.
$$
Multiplication by $\dot z$ now gives an exact time <derivative>:
$$
\frac{d}{dt}\left((1+\sqrt2)\dot z^2+\frac{a^2v^2}{2z^2}-gz\right)=0.
$$
Initially $z=-a$ and $\dot z=0$, because the initial <velocity> is horizontal. Therefore
$$
\boxed{(1+\sqrt2)\dot z^2+\frac{a^2v^2}{2z^2}-gz=\frac{v^2}{2}+ga.}
$$
This is also total <mechanical energy> divided by $m$, with an irrelevant constant removed: the two particles contribute $(1+\sqrt2)m\dot z^2$ to the radial and vertical <kinetic energy>, while their combined <gravitational potential energy> is $-mgz$ plus a constant.
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