Physics problem about two different liquids

Question

There are two immiscible layers of different liquids in a bowl, the densities and the thicknesses are ρ1, ρ2, h1, h2, respectively. Then, a small ball is placed to sink from the surface of the liquid into the bowl. The ball reaches the bottom of the bowl at the moment when its velocity becomes zero. What is the density of the ball?

Answer 1

The less dense fluid will be in the top layer. Since the ball eventually reaches the bottom of the bowl, the ball must be less dense than the upper layer. Also, since the ball's velocity is zero at the bottom, the second layer's buoyant force must be greater than the weight of the ball so that the net force on the ball is in the up direction.

The net force on the ball in the upper layer is equal to the ball's weight minus the buoyant force on the ball. This net force results in an acceleration of the ball through the fluid so that it has a terminal velocity at the bottom of the upper fluid given by V_term² = 0 + 2 x a1 x h1 where a1 is the calculated acceleration obtained by

Net force / mass of the ball. The mass of the ball is not known but it cancels out in the following calculations.

When the ball enters the bottom layer, it experiences a net force equal to the weight of the ball minus the buoyant force on the ball, resulting in a negative acceleration ending with the ball having a zero final velocity. The equation to set up is 0 = Vterm² + 2 x a2 x h2 where a2 is now the negative acceleration due to the negative net force.

The algebra involved is very messy, so I can't show it here, but I got an equation for the ball's density of

ρ_ball = (ρ1 x h1 - ρ2 x h2) / (h1 - h2).

Try to derive it yourself. (I hope it's right!)