Question

Two particles $A$ and $B$ of equal mass $M$ are moving with the same speed $v$ as shown in the figure. They collide completely inelastically and move as a single particle $C$. The angle $\theta$ that the path of $C$ makes with the $X-axis$ is given by :

• A. $\tan\theta = \frac{\sqrt{3} + \sqrt{2}}{1-\sqrt{2}} $
• B. $\tan\theta = \frac{\sqrt{3} - \sqrt{2}}{1-\sqrt{2}} $
• C. $\tan\theta = \frac{ 1 - \sqrt{2}}{\sqrt{2} (1 + \sqrt{3})} $
• D. $\tan\theta = \frac{ 1 - \sqrt{3}}{1 + \sqrt{2}} $

Answer 1

The Correct Option is A

\(2\, mv'\, sin \,\theta=\frac{mv}{\sqrt{2}}+\frac{mv\sqrt{3}}{2}\)
\(3 \,mv' \,cos\, \theta=\frac{mv}{2}-\frac{mv}{\sqrt{2}}\)
\(sin\, \theta=\frac{\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}}{\frac{1}{2}-\frac{1}{\sqrt{2}}}\)
\(=\frac{\sqrt{2}+\sqrt{3}}{1-\sqrt{2}}\)

\(\text{The Correct Option is (A):}\) \(\tan\theta = \frac{\sqrt{3} + \sqrt{2}}{1-\sqrt{2}}\)