Question

Two particles A and B of equal mass M are moving with the same speed $\upsilon$ 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 $tan\theta =\frac{\sqrt{3}+\sqrt{2}}{1-\sqrt{2}}$

Before collision

Linear momentum

alons $\times$ axis

or $P_x=P_{Ax}+P_{Bx}$

or $P_x=mv\cos {60}^{\circ }+(-mv\cos {45}^{\circ })$

$P_x=\frac{mv}{2}-\frac{mv}{\sqrt{2}}$

or $P_x=\frac{mv}{2}(1-\sqrt{2})$ __(1)

$P_y=P_{Ay}+P_{By}=mv\cos {30}^{\circ }+mv\cos {45}^{\circ }$

or $P_y=\frac{mv}{2}(\sqrt{3}+\sqrt{2})$ __(2)

After collision if speed be $V_0$ then being total mass as $2m$

we get momentum along $\times$ axis

$P_x^{\prime }=2m{v}_0\cos \theta$ __(3)

y axis $P_y^{\prime }=2m{v}_0\sin \theta$ __(4)

Conservins momentum

$P_x^{\prime }=P_x\Rightarrow 2m{v}_0\cos \theta =\frac{mv}{2}(1-\sqrt{2})$

$P_y^{\prime }=P_y\Rightarrow 2m{v}_0\sin \theta =\frac{mv}{2}(\sqrt{3}+\sqrt{2})$

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