Solve Lava Bubble Layer Problem: Speed & Direction

[DivGradCurl]

I've got the answer to this problem, but I didn't find it clear enough as an explanation on how to backsolve it. I was wondering if you guys could help me with it.

Thanks!

Here it goes...

Answer: "1.5 cm/s downward (the bubbles rise but layers descend)."

Problem: "Some solidified lava contains a pattern of horizontal bubble layers separated vertically with few intermediate bubbles. (Researchers must slice open solidified lava to see these bubbles.) Apparently, as the lava was cooling, bubbles rising from the botton of the lava separated into these layers and then were locked into place when the lava solidified. Similar layering of bubbles has been studied in certain creamy stouts poured fresh from tap into a clear glass. The rising bubbles quickly become sorted into layers. The bubbles trapped within a layer rise at speed v1; the free bubbles between the layers rise at a greater speed v2. Bubbles breaking free from the top of one layer rise to join the botton of the next layer. Assume that the rate at which a layer loses weight at its top is (dy/dt)=v2 and the rate at which it gains height at its botton is (dy/dt)=v2. Also assume that v2=2.0*v1=1.0 cm/s. What are the speed and direction of the motion of the layer's center of mass?"

 

[DivGradCurl]

I need some guidance throught the main steps; I'm not looking for somebody to do my stuff!

 

[M98Ranger]

To solve this problem, we can use the concept of conservation of mass. We know that the total amount of lava and bubbles in the system remains constant. Therefore, the rate at which the layers lose weight at the top must be equal to the rate at which they gain weight at the bottom. This can be represented mathematically as:

v1 * A = v2 * A

where v1 is the speed of the bubbles within the layer, v2 is the speed of the free bubbles between the layers, and A is the cross-sectional area of the layer.

We also know that the bubbles within the layer rise at a speed of v1 and the free bubbles between the layers rise at a speed of v2. This means that the net velocity of the layer is the difference between these two speeds, which is v2 - v1.

Now, to find the speed and direction of the motion of the layer's center of mass, we need to use the formula for velocity:

v = ∆x/∆t

where v is the velocity, ∆x is the change in position, and ∆t is the change in time.

In this case, we can consider the change in position (∆x) to be the height of the layer (h) and the change in time (∆t) to be 1 second. Therefore, we can rewrite the formula as:

v = h/1

Since the layer is moving downward (as indicated by the bubbles rising but layers descending), the velocity will be negative. So, we can write the final equation as:

v = -h/1

Now, we can substitute the values we know into this equation:

- (v2 - v1) = -h/1

- (2.0*v1 - v1) = -h/1

- (v1) = -h/1

Therefore, the speed of the layer's center of mass is equal to v1, which is 1.0 cm/s, and the direction is downward. This means that the bubbles within the layer are rising at a speed of 1.0 cm/s while the layers themselves are descending at the same speed.