A ball of density ais released from rest from the centre of an accelerating sealed cubical vessel completely filled with a liquid of density `rho`. If the trough moves with a horizontal acceleration `veca_(0)` find the initial acceleration of the ball.

Explanation

To solve the problem of finding the initial acceleration of a ball of density \( \sigma \) released from rest in an accelerating sealed cubical vessel filled with a liquid of density \( \rho \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Ball
When the ball is released, it experiences two main forces:
1. The gravitational force acting downwards, which is \( F_g = mg = V \sigma g \), where \( V \) is the volume of the ball.
2. The buoyant force acting upwards due to the liquid, which is \( F_b = V \rho g \).

### Step 2: Set Up the Free Body Diagram
We can analyze the forces in both the horizontal and vertical directions. The vessel is accelerating horizontally with an acceleration \( \vec{a_0} \).

### Step 3: Analyze the Horizontal Motion
In the horizontal direction, the ball will experience an effective acceleration due to the acceleration of the vessel. The net horizontal force on the ball can be expressed as:
\[ F_{x} = m a_{x} = V \rho a_{0} \]
where \( a_{x} \) is the horizontal acceleration of the ball.

### Step 4: Analyze the Vertical Motion
In the vertical direction, the net force acting on the ball is:
\[ F_{y} = F_b - F_g = V \rho g - V \sigma g \]
This can be simplified to:
\[ F_{y} = V ( \rho - \sigma ) g \]

### Step 5: Apply Newton's Second Law
Using Newton's second law in both directions, we can write:
1. For horizontal motion:
\[ m a_{x} = V \rho a_{0} \]
Thus,
\[ a_{x} = \frac{V \rho a_{0}}{m} \]
Since \( m = V \sigma \), we have:
\[ a_{x} = \frac{\rho a_{0}}{\sigma} \]

2. For vertical motion:
\[ m a_{y} = V (\rho - \sigma) g \]
Thus,
\[ a_{y} = \frac{V (\rho - \sigma) g}{m} \]
Again substituting \( m = V \sigma \):
\[ a_{y} = \frac{(\rho - \sigma) g}{\sigma} \]

### Step 6: Combine the Accelerations
The net acceleration \( \vec{a} \) of the ball can be expressed as a vector sum of the horizontal and vertical components:
\[ \vec{a} = a_{x} \hat{i} + a_{y} \hat{j} \]
Substituting the values we found:
\[ \vec{a} = \left( \frac{\rho a_{0}}{\sigma} \right) \hat{i} + \left( \frac{(\rho - \sigma) g}{\sigma} \right) \hat{j} \]

### Final Result
Thus, the initial acceleration of the ball is given by:
\[ \vec{a} = \left( \frac{\rho a_{0}}{\sigma} \right) \hat{i} + \left( \frac{(\rho - \sigma) g}{\sigma} \right) \hat{j} \]