JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 6)
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 :
_9th_April_Morning_Slot_en_6_1.png)
tan $$\theta $$ = $${{\sqrt 3 + \sqrt 2 } \over {1 - \sqrt 2 }}$$
tan $$\theta $$ = $${{\sqrt 3 - \sqrt 2 } \over {1 - \sqrt 2 }}$$
tan $$\theta $$ = $${{1 - \sqrt 2 } \over {\sqrt 2 \left( {1 + \sqrt 3 } \right)}}$$
tan $$\theta $$ = $${{1 - \sqrt 3 } \over {1 + \sqrt 2 }}$$
Explanation
_9th_April_Morning_Slot_en_6_2.png)
Using conservation of linear moments,
Along X-axis,
2MV'cos$$\theta $$ = MVsin30o $$-$$ MVsin45o . . . . . (1)
Along Y - axis,
2Mv' sin$$\theta $$ = Mv cos30o + Mv cos45o . . . . . (2)
Dividing (2) by (1), we get,
$${{\sin \theta } \over {\cos \theta }}$$ = $${{\cos {{30}^ \circ } + \cos {{45}^ \circ }} \over {\sin {{30}^ \circ } - \sin {{45}^ \circ }}}$$
= $${{{{\sqrt 3 } \over 2} + {1 \over {\sqrt 2 }}} \over {{1 \over 2} - {1 \over {\sqrt 2 }}}}$$
$$\therefore\,\,\,$$ tan$$\theta $$ = $${{\sqrt 3 + \sqrt 2 } \over {1 - \sqrt 2 }}$$
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