Bouncy balls put on top of each other

Since all collisions between balls are elastic, the coefficient of restitution is $1$ . This means $v_{sep} = v_{app}$ , where $v_{app}$ is the velocity of approach between two adjacent balls just before collision and $v_{sep}$ is the velocity of separation between those two balls just after collision.

Let's assume upward direction as positive and downward as negative.

Just before ball 1 collides with the ground, all the balls will have the same velocity. Let this velocity be $v_0$ downwards, that is $-v_0$ . Just after colliding with the ground, velocity of ball 1 will become $v_0$ upwards, that is $+ v_0$ . Let's call it $v_1$ .

Now, ball 1 will collide with ball 2. We know that velocity of ball 2 is $-v_0$ just before collision. Also, velocity of ball 1 is $v_1$ just before collision. Since $M_1 \gg M_2$ , we can assume velocity of ball 1 won't change at all, and the only change in velocity will be observed in ball 2, which has much less mass than ball 1. Therefore, velocity of ball 1 will remain $v_1$ just after collision as well. It's given that the velocity of ball 2 after collision is $v_2$ . The collision is again elastic, so we can still use $v_{sep} = v_{app}$ ,

$v_{app} = v_1 - (-v_0$ )

$v_{sep} = v_2 - v_1$

From the above equation we get

$v_2 - v_1 = v_1 + v_0 \implies v_2 = 2 v_1 + v_0$

By observing further, by induction, we get the following equation for $v_n$ for $n\geq 1$

$v_n = 2v_{n-1} + v_0$

But, we know $v_1 = v_0$ . For $n=2$ , we get $v_2 = 2v_0 + v_0\ = 3v_0$ . For $n = 3$ , $v_3 = 6v_0 + v_0 = 7v_0$

So, again, by observation we get a simplified general equation for $v_n$ , that is

$v_n = (2^{n} -1) v_0$

Now, we simply put $n=20$ and $n=5$ to get

$\dfrac{v_{20}}{v_5} = \dfrac{(2^{20}-1)v_0}{(2^{5}-1)v_0}$

$\dfrac{v_{20}}{v_5} = \dfrac{1048575}{31}$

Finally we get $\boxed{\dfrac{v_{20}}{v_5} = 33825}$