Two billiard balls are rolling on a flat table . One has the velocity component `V_(x) = 1 ms^(-1) , V_(y) = sqrt(3) m^(-1)` and the other has components `V_(x) = 2 ms^(-1) and V_(y)=2 ms^(-1)` .If both the balls start moving from the same point , what is the angle between their paths ?

Explanation

To find the angle between the paths of the two billiard balls, we will follow these steps:

### Step 1: Identify the velocity components
The velocity components of the first billiard ball are:
- \( V_{x1} = 1 \, \text{m/s} \)
- \( V_{y1} = \sqrt{3} \, \text{m/s} \)

The velocity components of the second billiard ball are:
- \( V_{x2} = 2 \, \text{m/s} \)
- \( V_{y2} = 2 \, \text{m/s} \)

### Step 2: Calculate the angles of the paths
To find the angle of each ball's path relative to the x-axis, we can use the tangent function:
- For the first ball:
\[ \tan(\alpha_1) = \frac{V_{y1}}{V_{x1}} = \frac{\sqrt{3}}{1} = \sqrt{3} \]
Therefore,
\[ \alpha_1 = \tan^{-1}(\sqrt{3}) = 60^\circ \]

- For the second ball:
\[ \tan(\alpha_2) = \frac{V_{y2}}{V_{x2}} = \frac{2}{2} = 1 \]
Therefore,
\[ \alpha_2 = \tan^{-1}(1) = 45^\circ \]

### Step 3: Calculate the angle between the paths
The angle between the paths of the two balls can be found by subtracting the smaller angle from the larger angle:
\[ \text{Angle between paths} = \alpha_1 - \alpha_2 = 60^\circ - 45^\circ = 15^\circ \]

### Final Answer
Thus, the angle between the paths of the two billiard balls is \( 15^\circ \).

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