question_answer

A sphericia ball A of mass 4 kg, moving along a straight line strikes another spherical ball B of mass 1 kg at rest. After the collision, A and B move with velocities \[{{\upsilon }_{1}}m{{s}^{-1}}\]and \[{{\upsilon }_{2}}m{{s}^{-1}}\] respectively making angles of \[30{}^\circ \] and \[60{}^\circ \] with respect to the original direction of motion of A. The ratio\[\frac{{{\upsilon }_{1}}}{{{\upsilon }_{2}}}\] will be

A) \[\sqrt{3}/4\]                 

B) \[4/\sqrt{3}\]  

C) \[1/\sqrt{3}\]     

D)        \[\sqrt{3}\]

Correct Answer: A

Solution :

Apply the law of conservation of linear momentum along a direction perpendicular to the direction of motion (i.e. along y-axis), we get \[0+0=4{{v}_{1}}\,\sin 30{}^\circ -{{v}_{2}}\sin 60{}^\circ \] \[4{{v}_{1}}\sin 30{}^\circ ={{v}_{2}}\sin 60{}^\circ \] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{\sin 60{}^\circ }{4\sin 30{}^\circ }=\frac{\sqrt{3}}{4}\]