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(B) Let the initial velocity of the particle of mass $2m$ be $v_1$ and the final velocities of particles $2m$ and $m$ be $v_{1f}$ and $v_{2f}$ at angl

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$\frac{\pi}{6}$

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(B) Let the initial velocity of the particle of mass $2m$ be $v_1$ and the final velocities of particles $2m$ and $m$ be $v_{1f}$ and $v_{2f}$ at angles $	heta$ and $\phi$ respectively.Conservation of linear momentum along the $x$-axis:$2mv_1 = 2mv_{1f} \cos 	heta + mv_{2f} \cos \phi$  ---$(i)$Conservation of linear momentum along the $y$-axis:$0 = 2mv_{1f} \sin 	heta - mv_{2f} \sin \phi \implies 2mv_{1f} \sin 	heta = mv_{2f} \sin \phi$  ---(ii)Conservation of kinetic energy:$\frac{1}{2}(2m)v_1^2 = \frac{1}{2}(2m)v_{1f}^2 + \frac{1}{2}mv_{2f}^2 \implies 2v_1^2 = 2v_{1f}^2 + v_{2f}^2$  ---(iii)From $(i)$ and (ii),$mv_{2f} \cos \phi = 2m(v_1 - v_{1f} \cos 	heta)$ and $mv_{2f} \sin \phi = 2mv_{1f} \sin 	heta$.Squaring and adding these:$(mv_{2f})^2 = 4m^2(v_1^2 + v_{1f}^2 - 2v_1v_{1f} \cos 	heta)$.Substitute $v_{2f}^2 = 2v_1^2 - 2v_{1f}^2$ from (iii):$m^2(2v_1^2 - 2v_{1f}^2) = 4m^2(v_1^2 + v_{1f}^2 - 2v_1v_{1f} \cos 	heta)$.$v_1^2 - v_{1f}^2 = 2v_1^2 + 2v_{1f}^2 - 4v_1v_{1f} \cos 	heta$.$3v_{1f}^2 - (4v_1 \cos 	heta)v_{1f} + v_1^2 = 0$.For $v_{1f}$ to be real,the discriminant $D \geq 0$:$(4v_1 \cos 	heta)^2 - 4(3)(v_1^2) \geq 0 \implies 16v_1^2 \cos^2 	heta - 12v_1^2 \geq 0$.$\cos^2 	heta \geq \frac{12}{16} = \frac{3}{4} \implies \cos 	heta \geq \frac{\sqrt{3}}{2}$.Thus,the maximum value of $	heta$ is $\cos^{-1}\left(\frac{\sqrt{3}}{2}ight) = \frac{\pi}{6}$.

In a scattering experiment,a particle of mass $2m$ collides with another particle of mass $m$,which is initially at rest. Assuming the collision to be perfectly elastic,the maximum angular deviation $\theta$ of the heavier particle,as shown in the figure,in radians is:

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