1. Given data

• Masses: $m_1 = 2m_2$
• Initial velocity of particle $A$: $\vec u_1 = (\sqrt{3}\,\hat i + \hat j)\,\text{m s}^{-1}$
• Initial velocity of particle $B$: $\vec u_2 = 0$
• Final velocity of particle $A$: $\vec V_1 = (\hat i + \sqrt{3}\,\hat j)\,\text{m s}^{-1}$

We need the angle between $\vec V_1$ and $\vec V_2$.

2. Apply conservation of linear momentum

Since no external impulse acts during collision, $m_1\vec u_1 + m_2\vec u_2 = m_1\vec V_1 + m_2\vec V_2$

Using $\vec u_2=0$ and $m_1=2m_2$: $2m_2\vec u_1 = 2m_2\vec V_1 + m_2\vec V_2$

Divide by $m_2$: $2\vec u_1 = 2\vec V_1 + \vec V_2$

So, $\vec V_2 = 2(\vec u_1 - \vec V_1)$

3. Compute $\vec V_2$

First, $\vec u_1 - \vec V_1 = (\sqrt{3}-1)\hat i + (1-\sqrt{3})\hat j$

Therefore, $\vec V_2 = 2\big[(\sqrt{3}-1)\hat i + (1-\sqrt{3})\hat j\big]$

$\vec V_2 = 2(\sqrt{3}-1)\hat i - 2(\sqrt{3}-1)\hat j$

Hence $\vec V_2$ is along the direction $\hat i - \hat j$

So its direction angle with the positive $x$-axis is $-45^\circ$

4. Direction of $\vec V_1$

$\vec V_1 = \hat i + \sqrt{3}\hat j$

Its direction angle with the positive $x$-axis is $\tan\theta_1 = \frac{\sqrt{3}}{1} = \sqrt{3}$ $\Rightarrow \theta_1 = 60^\circ$

5. Angle between $\vec V_1$ and $\vec V_2$

The directions are:

• $\vec V_1$: $60^\circ$
• $\vec V_2$: $-45^\circ$

So the angle between them is $60^\circ - (-45^\circ) = 105^\circ$

6. Check with options

• A: $105^\circ$ ✅
• B: $15^\circ$
• C: $-45^\circ$
• D: $60^\circ$

Therefore, the correct option is A.

7. Comparison with stored answer

Stored correct answer: A

This matches our derived answer.