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

Answer: (A)

$\tan\theta = \frac{\sqrt{3} + \sqrt{2}}{1-\sqrt{2}}$

Answer 2

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

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