Three masses m, 2 m and 3 m are moving in x-y plane with speed 3 u, 2 u 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:

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$\frac{u}{12} (\hat{i} + \sqrt{3} \hat{j})$

$\frac{u}{12} (- \hat{i} - \sqrt{3} \hat{j})$

$\frac{u}{12} (\hat{i} - \sqrt{3} \hat{j})$

$\frac{u}{12} (- \hat{i} + \sqrt{3} \hat{j})$

answer is D.

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Detailed Solution

From the law of conservation of momentum, we know that,

$m_{1} u_{1} + m_{2} u_{2} + \dots . = m_{1} v_{1} + m_{2} v_{2} + \dots$

$Given m_{1} = m, m_{2} = 2 m and m_{3} = 3 m and u_{1} = 3 u, u_{2} = 2 u and u_{3} = u$

$Let the velocity when they stick = \vec{v}$

$Then, according to question,$

$\begin{array}{l} m \times 3 u (\hat{i}) + 2 m \times 2 u (- \hat{i} \cos 60 ° - \hat{j} \sin 60 °) + 3 m \times u (- \hat{i} \cos 60 ° + \hat{j} \sin 60 °) = (m + 2 m + 3 m) \vec{v} \\ \Rightarrow 3 m u \hat{i} - 4 m u \frac{\hat{i}}{2} - 4 m u (\frac{\sqrt{3}}{2} \hat{j}) - 3 m u \frac{\hat{i}}{2} + 3 m u (\frac{\sqrt{3}}{2} \hat{j}) = 6 m \vec{v} \\ \Rightarrow m u \hat{i} - \frac{3}{2} m u \hat{i} - \frac{\sqrt{3}}{2} m \hat{j} = 6 m \vec{v} \\ \Rightarrow - \frac{1}{2} m u \hat{i} - \frac{\sqrt{3}}{2} m u \hat{j} = 6 m \vec{v} \\ \Rightarrow \vec{v} = \frac{u}{12} (- \hat{i} - \sqrt{3} \hat{j}) \end{array}$