Two equal drops of water are falling through air with a steady velocity v. If the drops coalesced, what will be the new velocity?

Explanation

To solve the problem of finding the new velocity of the combined water drop after two equal drops coalesce, we can follow these steps:

### Step 1: Understand the initial conditions
We have two equal drops of water, each with a radius \( r \) and falling with a steady velocity \( v \).

### Step 2: Calculate the volume of the drops
The volume \( V \) of a single drop is given by the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, the total volume of the two drops is:
\[ V_{\text{total}} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

### Step 3: Determine the radius of the new drop
When the two drops coalesce, they form a larger drop with a new radius \( R \). The volume of the new drop must equal the total volume of the two smaller drops:
\[ \frac{4}{3} \pi R^3 = \frac{8}{3} \pi r^3 \]
By simplifying, we can cancel \( \frac{4}{3} \pi \) from both sides:
\[ R^3 = 2r^3 \]
Taking the cube root gives us:
\[ R = 2^{1/3} r \]

### Step 4: Relate the terminal velocity to the radius
The terminal velocity \( V_t \) of a sphere falling through a fluid is given by:
\[ V_t = \frac{2}{9} \frac{R^2 g (\rho_1 - \rho_2)}{\eta} \]
where:
- \( g \) is the acceleration due to gravity,
- \( \rho_1 \) is the density of the water,
- \( \rho_2 \) is the density of the air,
- \( \eta \) is the viscosity of the air.

### Step 5: Set up the ratio of velocities
Since the terminal velocity is proportional to the square of the radius, we can write:
\[ \frac{V}{V_t} = \frac{r^2}{R^2} \]
Substituting \( R = 2^{1/3} r \):
\[ \frac{V}{V_t} = \frac{r^2}{(2^{1/3} r)^2} = \frac{r^2}{2^{2/3} r^2} = \frac{1}{2^{2/3}} \]
Thus, we can express the new terminal velocity \( V_t \) in terms of the original velocity \( v \):
\[ V_t = v \cdot 2^{2/3} \]

### Step 6: Conclusion
The new steady velocity \( V_t \) of the larger drop after coalescence is:
\[ V_t = 2^{2/3} v \]

### Summary
When two equal drops of water coalesce, the new velocity of the resulting larger drop is \( 2^{2/3} v \).

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