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

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 :

[d] From the law of conservation of momentum we know that, \[{{\operatorname{m}}_{1}}{{u}_{1}}+{{\operatorname{m}}_{2}}{{u}_{2}}+...={{\operatorname{m}}_{1}}{{\operatorname{v}}_{1}}+{{\operatorname{m}}_{2}}{{\operatorname{v}}_{2}}+...\] Given \[{{\operatorname{m}}_{1}}=m, {{m}_{2}} =2m and {{m}_{3}} = 3m\] and \[{{\operatorname{u}}_{1}}=3u, {{u}_{2}}= 2u and {{u}_{3}}= u\] Let the velocity when they stick \[= \vec{v}\] Then, according to question, \[\operatorname{m}\times 3u \left( {\hat{i}} \right) + 2m\times 2u \left( -\hat{i}cos60{}^\circ  -\hat{j}\sin  60{}^\circ  \right)\]\[+3m\times u \left( -\hat{i}cos60{}^\circ +\hat{j}sin60{}^\circ  \right) =\left( m+2m+3m \right) \vec{v}\]\[\Rightarrow 3mu\hat{i}-4mu\frac{{\hat{i}}}{2}4mu\left( \frac{\sqrt{3}}{2}\hat{j} \right)-3mu\frac{{\hat{i}}}{2}\]                                     \[+3mu\left( \frac{\sqrt{3}}{2}\hat{j} \right)=6m\vec{v}\] \[\Rightarrow mu\hat{i}-\frac{3}{2}mu\hat{i}-\frac{\sqrt{3}}{2}mu\hat{j}=6m\vec{v}\] \[\Rightarrow -\frac{1}{2}mu\hat{i}-\frac{\sqrt{3}}{2}mu\hat{j}=6m\vec{v}\] \[\Rightarrow \vec{v}=\frac{u}{12}\left( -\hat{i}-\sqrt{3}\hat{j} \right)\]