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  • Particle A of mass moving with velocity <img alt="\left ( \sqrt{3}\hat{i}+\hat{j} \right )ms^{-1}" src="http://entrancecorner.codecogs.com/gif.latex?%5Cleft%20%28%20%5Csqr

Particle A of mass m_{1} moving with velocity \left ( \sqrt{3}\hat{i}+\hat{j} \right )ms^{-1} collides with another particle B of mass m_{2} which is at rest initially. Let \overrightarrow{V_{1}}\; \; \; and \; \; \overrightarrow{V_{2}} be the velocities of particles A and B after collision respectively. If m_{1}=2m_{2} and after collision \overrightarrow{V_{1}}=\left ( \hat{i}+\sqrt{3} \hat{j}\right )ms^{-1}, the angle between \overrightarrow{V_{1}} and \overrightarrow{V_{2}} is :
Option: 1 15^{o}
 
Option: 2 60^{o}
Option: 3 -45^{o}  
Option: 4 105^{o}

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best_answer

\begin{array}{l} \overrightarrow{\mathrm{v}}_{01}=(\sqrt{3} \hat{\mathrm{i}}+\hat{\mathrm{j}}) \mathrm{m} / \mathrm{s} \\ \overrightarrow{\mathrm{v}}_{\infty}=\overrightarrow{0} \\ \mathrm{~m}_{1}=2 \mathrm{~m}_{2} \end{array}

\begin{aligned} &\text { After collision, } \overrightarrow{\mathrm{v}}_{1}=(\hat{\mathrm{i}}+\sqrt{3} \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\\ &\overrightarrow{\mathrm{v}}_{2}=?\\ &\text { Applying conservation of linear momentum. }\\ &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{i}-\hat{j})+2(\hat{i}-\sqrt{3} \hat{j})\\ &\overrightarrow{\mathrm{v}}_{2}=2(\sqrt{3}-1)(\hat{\mathrm{i}}-\hat{\mathrm{j}}) \end{aligned}

\begin{aligned} &\text { for angle between } \overrightarrow{\mathrm{v}}_{1} \& \overrightarrow{\mathrm{v}}_{2}\\ &\cos \theta=\frac{\vec{v}_{1}, \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)}\\ &\cos \theta=\frac{1-\sqrt{3}}{2 \sqrt{2}} \Rightarrow \theta=105^{\circ} \end{aligned}

 

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Deependra Verma

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