Work Energy And Power Question 278
Question: 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
Options:
A) $ \frac{u}{12}( \hat{i}+\sqrt{3}\hat{j} ) $
B) $ \frac{u}{12}( \hat{i}-\sqrt{3}\hat{j} ) $
C) $ \frac{u}{12}( -\hat{i}+\sqrt{3}\hat{j} ) $
D) $ \frac{u}{12}( -\hat{i}-\sqrt{3}\hat{j} ) $
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Answer:
Correct Answer: D
Solution:
[d] 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 ( {\hat{i}} ) + 2m\times 2u ( -\hat{i}cos60{}^\circ -\hat{j}\sin 60{}^\circ ) $
$ +3m\times u ( -\hat{i}cos60{}^\circ +\hat{j}sin60{}^\circ ) =( m+2m+3m ) \vec{v} $
$ \Rightarrow 3mu\hat{i}-4mu\frac{{\hat{i}}}{2}4mu( \frac{\sqrt{3}}{2}\hat{j} )-3mu\frac{{\hat{i}}}{2} $
$ +3mu( \frac{\sqrt{3}}{2}\hat{j} )=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}( -\hat{i}-\sqrt{3}\hat{j} ) $
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