question_answer

                Three masses m, 2m and 3m are moving in x-y plane with speed 3u, 2u and u respectively as shown in figure. The three masses collide at the same point at P and stick together. The velocity of resulting mass will be:   [JEE Main Online Paper ( Held On 12 Apirl  2014 )

A) \[\frac{u}{12}\left( \hat{i}+\sqrt{3}\hat{j} \right)\]                          

B) \[\frac{u}{12}\left( \hat{i}-\sqrt{3}\hat{j} \right)\]

C) \[\frac{u}{12}\left( -\hat{i}+\sqrt{3}\hat{j} \right)\]        

D) \[\frac{u}{12}\left( -\hat{i}-\sqrt{3}\hat{j} \right)\]

Correct Answer: D

Solution :

                From the law of conservation of momentum we know that, \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}+....={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}+....\] Given \[{{m}_{1}}=m,{{m}_{2}}=2m\]and\[{{m}_{3}}=3m\] and\[{{u}_{1}}=3u,{{u}_{2}}=2u\]and\[{{u}_{3}}=u\] Let the velocity when they stick \[=\vec{v}\] Then, according to question, \[m\times 3u\left( {\hat{i}} \right)+2m\times 2u\left( -\hat{i}\cos {{60}^{o}}-\hat{j}\sin {{60}^{o}} \right)\] \[+3m\times u\left( -\hat{i}\cos {{60}^{o}}+\hat{j}\sin {{60}^{o}} \right)=m(m+2m+3m)\vec{v}\]\[\Rightarrow \]\[3mu\hat{i}-4mu\frac{{\hat{i}}}{2}-4mu\left( \frac{\sqrt{3}}{2}\hat{j} \right)-3mu\frac{{\hat{i}}}{2}\] \[+3mu\left( \frac{\sqrt{3}}{2}\hat{j} \right)=6m\vec{v}\] \[\Rightarrow \]\[mu\hat{i}-\frac{3}{2}mu\hat{i}-\frac{\sqrt{3}}{2}mu\hat{j}=6m\vec{v}\] \[\Rightarrow \]\[-\frac{1}{2}mu\hat{i}-\frac{\sqrt{3}}{2}mu\hat{j}=6m\vec{v}\] \[\Rightarrow \]\[\vec{v}=\frac{u}{12}\left( -\hat{i}-\sqrt{3}j \right)\]