In the vessel are two immiscible liquids with densities ρ1 and ρ2 and layers of thickness h1 and h2 respectively. From the surface of liquid is lowered a streamlined object which reaches the bottom right at the moment when its speed becomes zero. What is the density of the material from which the object is made?

Solution. Posted on Saturday, January 3, 2015.

The amount of gravitational potential energy is equal to the work done by the resistance forces.

Thereby, m*g*(h1+h2)=w1+w2 , where w1 – work of the resistance force in the first liquid, w2 – work of the resistance force in the second liquid.

The resistance force is the buoyancy force. So, F1=ρ1*g*V and F2=ρ2*g*V, where V is the volume of the object, F1 is the resistance force in the first liquid and F2 is the resistance force in the second liquid.

The total work done by the resistance forces is w1+w2=ρ1*g*V*h1+ρ2*g*V*h2 .

Let’s substitute it into the first equation, m*g*(h1+h2)=ρ1*g*V*h1+ρ2*g*V*h2 .

We know that m=ρ*V, where ρ is the density of the material from which the object is made.

Thereby, the last equation will be rewritten as

ρ*V*g*(h1+h2)=ρ1*g*V*h1+ρ2*g*V*h2.

Finally, ρ=(ρ1*h1+ρ2*h2)/(h1+h2) .