Particle A of mass

Solution:

(4) Before collision,

Velocity of particle $A, u_{1}=(\sqrt{3} \hat{i}+\hat{j}) \mathrm{m} / \mathrm{s}$

Velocity of particle $B, u_{2}=0$

After collision,

Velocity of particle $A, v_{1}=(\hat{i}+\sqrt{3} \hat{j})$

Velocity of particle $B, v_{2}=0$

Using principal of conservation of angular momentum

$m_{1} \vec{u}_{1}+m_{2} \vec{u}_{2}=m_{1} \vec{v}_{1}+m_{2} \vec{v}_{2}$

$\Rightarrow 2 m_{2}(\sqrt{3} \hat{i}+\hat{j})+m_{2} \times 0=2 m_{2}(\hat{i}+\sqrt{3} \hat{j})+m_{2} \times \vec{v}_{2}$

$\Rightarrow 2 \sqrt{3} \hat{i}+2 \hat{j}=2 \hat{i}+2 \sqrt{3} \hat{j}+\vec{v}_{2}$

$\Rightarrow \vec{v}_{2}=(\sqrt{3}-1) \hat{i}-(\sqrt{3}-1) \hat{j} \Rightarrow \vec{v}_{1}=\hat{i}+\sqrt{3} \hat{j}$

For angle between $\vec{v}_{1}$ and $\vec{v}_{2}$,

$\cos \theta=\frac{\vec{v}_{1} \cdot \vec{v}_{2}}{\vec{v}_{1} \vec{v}_{2}}=\frac{2(\sqrt{3}-1)(1-\sqrt{3})}{2 \times 2 \sqrt{2}(\sqrt{3}-1)}=\frac{1-\sqrt{3}}{2 \sqrt{2}}$

$\Rightarrow \theta=105^{\circ}$

Angle between $\vec{v}_{1}$ and $\vec{v}_{2}$ is $105^{\circ}$