Two masses connected by vertical rod is displaced The following question is from 2016 $F=ma$ exam. My question is with regards to one step in the reasoning of the solution.
Original question: A small ball of mass $3m$ is at rest on the ground. A second small ball of mass $m$ is positioned above the ground by a vertical massless rod of length $L$ that is also attached to the ball on the ground. The original orientation of the rod is directly vertical, and the top ball is given a small horizontal nudge. There is no friction; assume that everything happens in a single plane. Determine the horizontal displacement $x$ of the second (originally top) ball just before it hits the ground.
The given answer is $3L/4$.
I understand that the C.M. does not move horizontally, so the horizontal displacements of the balls are always in $3:1$ ratio. However, how do we prove that the heavier ball never lifts off the ground? To me it seems plausible that while the C.M. is moving down and the rod is pivoting around the C.M., the heavier ball could lose contact with the ground. Hence, it is totally plausible that when the lighter ball hits the ground, the rod is not in a horizontal position. How do I eliminate this possibility?
 A: I've deleted my previous answer as it has some flaws in logic. I will attempt to show you how this situation works below. The center of mass of the object is $L/4$ from the bottom. As the only forces acting on the object is vertical( gravity and normal reaction force at bottom) so it's center of mass will travel only vertically.
Now if we can show that there can exists a plausible normal reaction force which may vary( and that is not acting downwards, that is, it is positive) so that the end of the rod does not move vertically (after plugging in all the other accelerations of motion) then we can prove that the rod's end does not lift up.
First we find the moment of intertia about the center of mass(COM) of the rod. You'll find it is $\frac{3}{4}mL^2$. Then we take moments about COM. We will arrive at the following equation:
$$ \alpha=\frac{Nsin(\theta)}{3mL}$$
Where $N$ is the normal reaction force and $\theta$ is the angle rotated. Now the linear acceleration at bottom is given by $a=\frac{L}{4}\alpha$ so thus the vertical component of the acceleration on the bottom is given by : $a_v=asin(\theta)$( refer to diagram below).
Thus finally: $a_v=(N/12m)sin^2(\theta)$ so the vertical force on the end of the rod is given by $3ma_v$.
Now we can write an equation so that the vertical acceleration at bottom is zero and find the expression for $N$:
$$N+3ma_v-3mg=0$$
So finally: 
$$N=\frac{12mg}{3+sin^2(\theta)}$$
Here you can see that for $\theta=0-to-90$, $N$ remains positive( situation is hence plausible as there is no problem with the varying positive magnitudes of $N$). Thus the bottom of the rod does not lift up.

 Note: The horizontal component of the linear acceleration at the bottom of the rod still exists thus the bottom of the rod moves to the left from the initial position.

A: The ground doesn't push you back with a greater force
The Normal force and the force with which $3m$ mass is pushing the ground form action-reaction pairs. The Normal force has only the job to prevent the end $3m$ of rod from sinking into the ground.
Also, the rod is rotating about $CM$ and there is no other force other than Normal force that provides torque. If there would have been an other force which provided torque, then there would have been a possibility of end $3m$ lifting up.
