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(A) Let the angle of incidence be $	heta$.From the geometry of the collision shown in the figure,consider the triangle $\Delta PMO$ where $P$ is the

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$60$

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(A) Let the angle of incidence be $	heta$.From the geometry of the collision shown in the figure,consider the triangle $\Delta PMO$ where $P$ is the point of impact,$M$ is the projection of $P$ on the central line,and $O$ is the center of the sphere.The impact parameter $b = PM = 2\sqrt{3} \, cm$ and the radius $R = OP = 4 \, cm$.In $\Delta PMO$,$\sin 	heta = \frac{PM}{OP} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$.Therefore,$	heta = 60^{\circ}$.The angle of deflection $\delta$ is the angle between the initial direction of the particle and the final direction after the elastic collision.From the geometry,the angle of deflection is given by $\delta = 180^{\circ} - 2	heta$.Substituting the value of $	heta$,we get $\delta = 180^{\circ} - 2(60^{\circ}) = 180^{\circ} - 120^{\circ} = 60^{\circ}$.

$A$ small particle elastically strikes a fixed smooth sphere of $4 \, cm$ radius,with a speed of $5 \, m/s$. The impact parameter for the collision is $2\sqrt{3} \, cm$. Calculate the angle of deflection of the particle ............... $^o$

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