Particle A of mass m_{1} moving with velocity | vedclass

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Question

(D) Initial velocity of particle $A$: $\vec{u}_{1} = (\sqrt{3} \hat{i} + \hat{j}) \, m/s$.Initial velocity of particle $B$: $\vec{u}_{2} = 0$.Given $m

Vedclass · Accepted Answer

$105$

Vedclass · Answer

(D) Initial velocity of particle $A$: $\vec{u}_{1} = (\sqrt{3} \hat{i} + \hat{j}) \, m/s$.Initial velocity of particle $B$: $\vec{u}_{2} = 0$.Given $m_{1} = 2m_{2}$.Final velocity of particle $A$: $\vec{V}_{1} = (\hat{i} + \sqrt{3} \hat{j}) \, m/s$.By the law of conservation of linear momentum:$m_{1} \vec{u}_{1} + m_{2} \vec{u}_{2} = m_{1} \vec{V}_{1} + m_{2} \vec{V}_{2}$$2m_{2}(\sqrt{3} \hat{i} + \hat{j}) + 0 = 2m_{2}(\hat{i} + \sqrt{3} \hat{j}) + m_{2} \vec{V}_{2}$Dividing by $m_{2}$:$2(\sqrt{3} \hat{i} + \hat{j}) = 2(\hat{i} + \sqrt{3} \hat{j}) + \vec{V}_{2}$$\vec{V}_{2} = 2(\sqrt{3} \hat{i} + \hat{j} - \hat{i} - \sqrt{3} \hat{j}) = 2[(\sqrt{3}-1) \hat{i} + (1-\sqrt{3}) \hat{j}] = 2(\sqrt{3}-1) (\hat{i} - \hat{j})$.Now,the angle $	heta$ between $\vec{V}_{1}$ and $\vec{V}_{2}$ is given by $\cos 	heta = \frac{\vec{V}_{1} \cdot \vec{V}_{2}}{|\vec{V}_{1}| |\vec{V}_{2}|}$.$\vec{V}_{1} \cdot \vec{V}_{2} = (\hat{i} + \sqrt{3} \hat{j}) \cdot [2(\sqrt{3}-1) (\hat{i} - \hat{j})] = 2(\sqrt{3}-1) (1 - \sqrt{3}) = -2(\sqrt{3}-1)^{2}$.$|\vec{V}_{1}| = \sqrt{1^{2} + (\sqrt{3})^{2}} = 2$.$|\vec{V}_{2}| = 2(\sqrt{3}-1) \sqrt{1^{2} + (-1)^{2}} = 2\sqrt{2}(\sqrt{3}-1)$.$\cos 	heta = \frac{-2(\sqrt{3}-1)^{2}}{2 \cdot 2\sqrt{2}(\sqrt{3}-1)} = \frac{-(\sqrt{3}-1)}{2\sqrt{2}} = \frac{1-\sqrt{3}}{2\sqrt{2}}$.Since $\cos 105^{\circ} = \cos(60^{\circ} + 45^{\circ}) = \cos 60^{\circ} \cos 45^{\circ} - \sin 60^{\circ} \sin 45^{\circ} = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1-\sqrt{3}}{2\sqrt{2}}$,therefore $	heta = 105^{\circ}$.

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