Two balls are rolling on a flat smooth table. One ball has velocity components `sqrt(3) hat(j)` and `hat(i)` while the other has components `2 hat(i)` and `2 hat(j)`. If both start moving simultaneously from the same point, the angle between their paths is-

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Explanation

To find the angle between the paths of the two balls, we can use the dot product of their velocity vectors. Here’s a step-by-step solution:

### Step 1: Define the velocity vectors
Let the velocity vector of ball 1 be:
\[ \mathbf{v_1} = \hat{i} + \sqrt{3} \hat{j} \]
Let the velocity vector of ball 2 be:
\[ \mathbf{v_2} = 2\hat{i} + 2\hat{j} \]

### Step 2: Calculate the magnitudes of the velocity vectors
The magnitude of \(\mathbf{v_1}\) is calculated as follows:
\[ |\mathbf{v_1}| = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

The magnitude of \(\mathbf{v_2}\) is calculated as:
\[ |\mathbf{v_2}| = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

### Step 3: Calculate the dot product of the velocity vectors
The dot product \(\mathbf{v_1} \cdot \mathbf{v_2}\) is given by:
\[ \mathbf{v_1} \cdot \mathbf{v_2} = (1)(2) + (\sqrt{3})(2) = 2 + 2\sqrt{3} \]

### Step 4: Use the dot product to find the cosine of the angle
Using the formula for the dot product:
\[ \mathbf{v_1} \cdot \mathbf{v_2} = |\mathbf{v_1}| |\mathbf{v_2}| \cos \theta \]
Substituting the values we found:
\[ 2 + 2\sqrt{3} = (2)(2\sqrt{2}) \cos \theta \]
This simplifies to:
\[ 2 + 2\sqrt{3} = 4\sqrt{2} \cos \theta \]

### Step 5: Solve for \(\cos \theta\)
Rearranging gives:
\[ \cos \theta = \frac{2 + 2\sqrt{3}}{4\sqrt{2}} = \frac{1 + \sqrt{3}}{2\sqrt{2}} \]

### Step 6: Find the angle \(\theta\)
To find \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1 + \sqrt{3}}{2\sqrt{2}}\right) \]
Calculating this gives:
\[ \theta \approx 15^\circ \]

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

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