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Question

(A) Let the mass of each particle be $M$ and their speed be $v$. After the collision,they form a single particle $C$ of mass $2M$ moving with velocity

Vedclass · Accepted Answer

$	an 	heta = \frac{\sqrt{3} + \sqrt{2}}{1 - \sqrt{2}}$

Vedclass · Answer

(A) Let the mass of each particle be $M$ and their speed be $v$. After the collision,they form a single particle $C$ of mass $2M$ moving with velocity $v'$ at an angle $	heta$ with the $X$-axis.Applying the law of conservation of linear momentum along the $X$ and $Y$ axes:Along the $X$-axis:$P_{ix} = P_{fx}$$Mv \cos(60^{\circ}) - Mv \cos(45^{\circ}) = (2M)v' \cos 	heta$$v(\frac{1}{2} - \frac{1}{\sqrt{2}}) = 2v' \cos 	heta \quad ... (i)$Along the $Y$-axis:$P_{iy} = P_{fy}$$Mv \sin(60^{\circ}) + Mv \sin(45^{\circ}) = (2M)v' \sin 	heta$$v(\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}) = 2v' \sin 	heta \quad ... (ii)$Dividing equation $(ii)$ by equation $(i)$:$	an 	heta = \frac{v(\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}})}{v(\frac{1}{2} - \frac{1}{\sqrt{2}})} = \frac{\sqrt{3} + \sqrt{2}}{1 - \sqrt{2}}$

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

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