Particle $\mathrm{A}$ of mass $\mathrm{m}_{1}$ moving with velocity $(\sqrt{3} \hat{\mathrm{i}}+\hat{\mathrm{j}}) \mathrm{ms}^{-1}$ collides with another particle B of mass $\mathrm{m}_{2}$ which is at rest initially. Let $\overrightarrow{\mathrm{V}}_{1}$ and $\overrightarrow{\mathrm{V}}_{2}$ be the velocities of particles $\mathrm{A}$ and $\mathrm{B}$ after collision respectively. If $m_{1}=2 m_{2}$ and after collision $\overrightarrow{\mathrm{V}}_{1}=(\hat{\mathrm{i}}+\sqrt{3} \hat{\mathrm{j}}) \mathrm{ms}^{-1}$, the angle between $\vec{V}_{1}$ and $\vec{V}_{2}$ is :
Correct Option: , 4
$\overrightarrow{\mathrm{v}}_{01}=(\sqrt{3} \hat{\mathrm{i}}+\hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}$
$\overrightarrow{\mathrm{v}}_{02}=\overrightarrow{0}$
$\mathrm{m}_{1}=2 \mathrm{~m}_{2}$
After collision, $\overrightarrow{\mathrm{v}}_{1}=(\hat{\mathrm{i}}+\sqrt{3} \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}$
$\overrightarrow{\mathrm{v}}_{2}=?$
Applying conservation of linear momentum,
$\mathrm{m}_{1} \overrightarrow{\mathrm{v}}_{01}+\mathrm{m}_{2} \overrightarrow{\mathrm{v}}_{02}=\mathrm{m}_{1} \overrightarrow{\mathrm{v}}_{1}+\mathrm{m}_{2} \overrightarrow{\mathrm{v}}_{2}$
$2 \mathrm{~m}_{2}(\sqrt{3} \hat{\mathrm{i}}+\hat{\mathrm{j}})+0=2 \mathrm{~m}_{2}(\hat{\mathrm{i}}+\sqrt{3} \hat{\mathrm{j}})+\mathrm{m}_{2} \overrightarrow{\mathrm{v}}_{2}$
$\overrightarrow{\mathrm{v}}_{2}=2(\sqrt{3} \hat{\mathrm{i}}+\hat{\mathrm{j}})-2(\hat{\mathrm{i}}+\sqrt{3} \hat{\mathrm{j}})$
$=2(\sqrt{3} \hat{\mathrm{i}}-\hat{\mathrm{j}})+2(\hat{\mathrm{i}}-\sqrt{3} \hat{\mathrm{j}})$
$\overrightarrow{\mathrm{v}}_{2}=2(\sqrt{3}-1)(\hat{\mathrm{i}}-\hat{\mathrm{j}})$
for angle between $\vec{v}_{1} \& \vec{v}_{2}$,
$\cos \theta=\frac{\overrightarrow{\mathrm{v}}_{1}, \overrightarrow{\mathrm{v}}_{2}}{\overrightarrow{\mathrm{v}}_{1} \overrightarrow{\mathrm{v}}_{2}}=\frac{2(\sqrt{3}-1)(1-\sqrt{3})}{2 \times 2 \sqrt{2}(\sqrt{3}-1)}$
$\cos \theta=\frac{1-\sqrt{3}}{2 \sqrt{2}} \Rightarrow \theta=105^{\circ}$
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