(a)
\(\boxed{\text{To the right of }C}\)

The disk should hit the rod as far from the pivot as possible. A collision farther from the pivot gives the disk a larger angular momentum about the pivot. Since angular momentum about the pivot is transferred to the rod-disk system, a larger distance from the pivot produces a larger final angular speed.

(b)
\(\boxed{\text{Yes}}\)

The equation \(\omega=\dfrac{m_{\text{disk}}xv_0}{I}\) shows that \(\omega\) increases as \(x\) increases. That matches the reasoning in part (a): hitting farther from the pivot gives the system more angular momentum and therefore a larger angular speed.

(c)
The equation \(\omega=\dfrac{Iv_0}{m_{\text{disk}}d^4}\) is not plausible because it predicts the wrong dependence on important physical quantities.

If \(m_{\text{disk}}\) increases, the incoming disk has more angular momentum, so the final angular speed should generally increase. But this equation has \(m_{\text{disk}}\) in the denominator, so it predicts that a larger disk mass would make \(\omega\) smaller.

Also, if the rod’s rotational inertia \(I\) is larger, the system should be harder to rotate, so \(\omega\) should decrease. But this equation has \(I\) in the numerator, so it predicts the opposite.

The units are also not consistent with angular speed. Since angular speed should have units of \(\text{s}^{-1}\), the expression \(\dfrac{Iv_0}{m_{\text{disk}}d^4}\) does not have the correct dimensions.

(d)
Use conservation of angular momentum about the pivot during the collision. The pivot can exert an external force, but it exerts no torque about the pivot point itself.

Initial angular momentum:

\(L_i=m_{\text{disk}}v_0x\)

After the disk sticks to the rod, the total rotational inertia about the pivot is

\(I_{\text{total}}=I+m_{\text{disk}}x^2\)

Final angular momentum:

\(L_f=\left(I+m_{\text{disk}}x^2\right)\omega\)

Set \(L_i=L_f\):

\(m_{\text{disk}}v_0x=\left(I+m_{\text{disk}}x^2\right)\omega\)

Solve for \(\omega\):

\(\boxed{\omega=\dfrac{m_{\text{disk}}v_0x}{I+m_{\text{disk}}x^2}}\)

(e)
\(\boxed{\text{Greater than}}\)

If the disk sticks, it keeps moving with the rod after the collision, so some angular momentum remains associated with the disk as part of the rotating rod-disk system.

If the disk bounces backward, its angular momentum about the pivot changes direction. To conserve total angular momentum about the pivot, the rod must gain more angular momentum than in the sticking case.

Since the rod gains more angular momentum in the bounce case, its postcollision angular speed is greater than when the disk sticks to the rod.