Two balls are moving on the same smooth horizontal plane. Their velocity components along one adge of the square plane are `10sqrt(3)m//s` & `20 m//s` respectively. Their velocity components along a perpenficular edge are `30 m//s` & `20 m//s`. Find the angle between their directions of motion.

Explanation

To find the angle between the directions of motion of the two balls, we can follow these steps:

### Step 1: Identify the velocity components
Let’s denote the velocity components of the first ball as:
- \( V_{1x} = 10\sqrt{3} \, \text{m/s} \) (along one edge)
- \( V_{1y} = 30 \, \text{m/s} \) (along the perpendicular edge)

For the second ball, the velocity components are:
- \( V_{2x} = 20 \, \text{m/s} \) (along one edge)
- \( V_{2y} = 20 \, \text{m/s} \) (along the perpendicular edge)

### Step 2: Calculate the angles of motion
To find the angle of motion for each ball, we can use the tangent function, which relates the opposite side to the adjacent side in a right triangle.

For the first ball:
\[ \tan(\alpha) = \frac{V_{1y}}{V_{1x}} = \frac{30}{10\sqrt{3}} = \frac{30}{10 \cdot 1.732} = \frac{30}{17.32} \approx 1.732 \]
Thus, \( \alpha = \tan^{-1}(1.732) \). We know that \( \tan(60^\circ) = \sqrt{3} \), so:
\[ \alpha = 60^\circ \]

For the second ball:
\[ \tan(\beta) = \frac{V_{2y}}{V_{2x}} = \frac{20}{20} = 1 \]
Thus, \( \beta = \tan^{-1}(1) \). We know that \( \tan(45^\circ) = 1 \), so:
\[ \beta = 45^\circ \]

### Step 3: Find the angle between the two directions of motion
The angle \( \theta \) between the two directions of motion is given by:
\[ \theta = \alpha - \beta = 60^\circ - 45^\circ = 15^\circ \]

### Final Answer
The angle between the directions of motion of the two balls is \( 15^\circ \).

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