Variation of momentum means force: $F = \dot p$. In a very non-rigorous way, we have:
$$
F = \frac{dp}{dt} \quad\Longrightarrow\quad
Fdt = dp \quad\Longrightarrow\quad
F = \frac{\Delta p}{\Delta t}
$$
The force in this case, it is likely assumed to be the gravitational force: $F = mg\sin\theta$, where $\theta$ is the angle of the ladder and the horizontal axis, and $\Delta t$ is the time taken by the collision. Hence:
$$
\sin\theta = \frac{\Delta p}{\Delta t}\frac{1}{mg}
$$
Let $A = (0, 0), B=(2a, 0), C(2a, 2a), D= (0, 2a)$
Then $M = (a, 0 ), N = (a, 2a ) $
Point $E$ is on $MN$, if the distance $NE = y$ then $E = (a, 2a - y) $
Hence, the centroid of $\triangle AEB = ( a , \dfrac{y}{3} )$
And therefore, after removing $\triangle AEB$, the centroid of the remainder is given by
$G = \dfrac { a^2 (a, a ) - a y (a, \dfrac{y}{3} ) } { 4a^2 - a y } $
Simplifying,
$G = (a, \dfrac{ 12 a^2 - y^2 }{12 a - 3 y} )$
If the y-coordinate of $G$ is equal to that of $E$ then we have
$ y = \dfrac{ 12 a^2 - y^2 }{12 a - 3 y}$
Therefore,
$ y (12 a -3 y) = 12 a^2 - y^2 $
Re-arranging,
$ 2 y^2 - 12 a y + 12 a^2 = 0 $
Hence, $ y = \dfrac{1}{4} ( 12 a - \sqrt{ 144 a^2 - 96 a^2 } ) =\dfrac{1}{4} ( 12 a - 4 \sqrt{3} a ) = (3 - \sqrt{3} )a $
From which $NE = 2a - y = 2a - (3 - \sqrt{3}) a = a (\sqrt{3} - 1) $
Best Answer
There is a simpler way : take the center of gravity of the first disk weighted by its positive area $\pi 49 a^2$ and the second one weighted by its negative weight $-\pi 9 a^2$, all being divided by the algebraic sum $\pi 49 a^2-\pi 9 a^2=\pi 40 a^2$.