= Solution
Label the initially left-hand rod $A$ and the right-hand rod $B$. At impact their centres are $(0,\ell)$ and $(0,-\ell)$, and their contact points are both the origin. Each rod has total mass $2m$ and <moment of inertia>
$$
I=m\ell^2+m\ell^2=2m\ell^2.
$$
Let the <impulse> on $B$ be $+J\mathbf e_x$, so the <impulse> on $A$ is $-J\mathbf e_x$. The impulsive form of <Newton's second law> gives
$$
\mathbf v_A=\left(V-\frac{J}{2m},0\right),\qquad
\mathbf v_B=\left(-V+\frac{J}{2m},0\right).
$$
The contact lever arms from the centres are $(0,-\ell)$ for $A$ and $(0,+\ell)$ for $B$. Both angular <impulses> are $-\ell J\mathbf e_z$. Thus, taking counterclockwise rotation as positive,
$$
\omega_A=\omega_B=-\frac{\ell J}{I}=-\frac{J}{2m\ell}.
$$
Equal and opposite linear <impulses> therefore produce equal clockwise rotations, not opposite rotations.
The restitution law uses the velocities of the contacting endpoints, including the rotational terms. For $A$ that velocity is
$$
u_A=V-\frac{J}{2m}+\ell\omega_A=V-\frac Jm,
$$
and for $B$ it is $u_B=-V+J/m$. The approach speed before impact is $2V$. The <coefficient of restitution> $e$ therefore gives
$$
u_B-u_A=-2V+\frac{2J}{m}=2eV,\qquad J=mV(1+e).
$$
Substituting into the centre and <angular velocities> yields the <endpoint collision of two freely rotating equal rods>:
$$
\boxed{\mathbf v_A=\left(\frac{1-e}{2}V,0\right),\qquad
\mathbf v_B=\left(-\frac{1-e}{2}V,0\right),\qquad
\omega_A=\omega_B=-\frac{1+e}{2\ell}V.}
$$
For the first, perfectly elastic case, $e=1$, so
$$
\boxed{\mathbf v_A=\mathbf v_B=(0,0),\qquad
\omega_A=\omega_B=-\frac V\ell.}
$$
All initial translational energy has become rotational energy. Indeed the total initial <kinetic energy> is $2mV^2$, while the final energy for general $e$ is
$$
K_{\mathrm{final}}
=2\left(m\frac{(1-e)^2V^2}{4}
+m\ell^2\frac{(1+e)^2V^2}{4\ell^2}\right)
=mV^2(1+e^2).
$$
The loss is $mV^2(1-e^2)$, zero for an <elastic collision>. Conservation of total <angular momentum> about the origin provides another check: initially it is $-4m\ell V$; after collision the centre-motion contribution is $-2m\ell V(1-e)$ and the spins contribute $-2m\ell V(1+e)$, with the same total. The initial distance $d$ determines when the collision occurs but does not enter the impulsive velocities.
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