Multiple Balls Bouncing

(https://youtu.be/2UHS883_P60)

The bigger ball transfers some of its momentum to the smaller ball. Since mass of the bigger ball is more than the mass of smaller ball, momentum produced due to velocity of the bigger ball will result in more velocity in the smaller ball. ( When momentum is constant, mass is inversely proportional to velocity.)

When put in normal conditions a ball that has been dropped would gather up kinetic energy on the way down and when it hits the ground it will lose some of it's energy to the ground in order to bounce back up and complete it's given task. However, no matter how many times the ball bounces it will never be able to reach it's original height of drop or to put it another way, it will never have enough energy to get back up to where it started, hence the law of conservation of energy.

But now since a smaller ball, let's call it ball b , is placed on top of the larger or normal sized ball, let's call that ball a , the energy that will be given at the point of impact to the ground, will not only be to the ground but to the ball b on top of it as well. When the larger ball hits the ground it doesn't just bounce back up like a sheet of metal, but instead it stretches and pushes itself upwards due to the amount of elasticity in it's material. And when this stretching happens in normal conditions, the ball only touches the air molecules around it, which are so small that they have a negligible affect on the ball's performance and it bounces nearer to the original drop point.

But in the case with ball b , the ball a has another surface on which it can transfer most of it's remaining elastic energy, or in other words, it's jumping power. Since the ball b has a lot less area than the ball a for air resistance to slow it down, ball b actually goes much higher than the point of drop due to it's lack of area, mass and volume to keep it below the drop point. Also the fact that it now has the energy of an object 3x it's size, so that's another, shorter, reason why it goes off shooting into space.

Say balls have radius and mass $r, m$ and $R, M$ , small and big ball correspondingly. Based on energy conservation, both balls will reach the floor falling a total height $h = 10 - 2R$ counting from their center of mass and speed positive downwards

$v_o = \sqrt{2gh}$

Now the bigger ball will collide first with the ground and since the collision is elastic it will bounce back with reverse speed $-v_0$ . Immediately afterwards it will collide with the smaller ball , so we take momentum and energy conservation.

$m v_0 -M v_0 = m v' + M V'$

$\frac{1}{2}m v_0^2 +\frac{1}{2}M v_0^2 = \frac{1}{2}m v'^2 + \frac{1}{2}M V'^2$

Solve the first w.r.t. $V'$ , replace in the second and bring the second to standard quadratic form w.r.t. $v'$ . Result is

$V' = \frac{m-M}{M}v_0 - \frac{m}{M}v'$

$v'^2 - 2\frac{m-M}{m+M}v_0 v' + \frac{m-3M}{m+M}v_0^2 = 0$

Of the two roots only one makes sense since the other one would imply $V' = -v_0$ and $v' = v_0$ so that the balls pass through each other. The solutions are

$V' = -\frac{M-3m}{m+M}v_0$

$v' = -\frac{3M-m}{m+M}v_0$

Note that they are both upward and $|v'| > |V'|$ so it makes physical sense. Now $\frac{3M-m}{m+M} > 1 \hspace{2ex} \forall M > m$ so that $|v'| > |v_0|$ . By energy conservation the small ball will rise to a higher height $h'$ than $h$ and the large ball will rise to a lower height. Specifically

$h' = \left(\frac{3M-m}{m+M}\right)^2 h$

If one wanted to argue without exact calculation one could say that in two-body collisions the lighter object always gains speed and the heavier object always loses speed. Therefore the lighter ball will have greater speed then before the collision and due to energy conservation reach a greater height.

The trap here is that you would think that the ball cannot bounce higher than the point it starts from because the potential energy has to be equal to the kinetic energy but in this case the mesurement is 10 meters from the ground not the point from which the small ball bounces of from. (the top of the big ball) On top of that, the bigger ball will transfer some of its energy to the smaller one

Girl placed a hamster on the ball :)

we know that: $distant+power>distant$

If it wasn't for the assumption that the collisions are perfectly elastic, with no conversion from kinetic energy to heat, which a normal ball would leave on the ground after impact, you would not be able to tell exactly without more information. In this case with negligible air resistance, and with all the collisions being perfectly elastic, the small ball in this example, ought to bounce approximately at least as high as it's starting point returning from the first bounce, because only a negligible amount of opposing forces are there to oppose it.

Since The Ball jumps lower, when a double bounce happens, the velocity is distributed to the top, the momentum also distributes to the top, that's why the smaller ball goes higher than before.

The potential energy of the two balls combined must be the same before letting go and when the balls 'float' in the air the second time. Because the small ball is higher than the large one after the first bounce, its potential energy must be higher (it has to compensate for the loss of height of the large ball) and therefore it's higher.