Question 203: Centre of Mass and Rotational Motion

Question

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 $$	heta $$ that the path of C makes with the X-axis is given by :

Accepted Answer

Using conservation of linear moments,
Along X-axis,
2MV'cos$$	heta $$ = MVsin30
o
$$-$$ MVsin45
o
. . . . . (1)
Along Y - axis,
2Mv' sin$$	heta $$ = Mv cos30
o
+  Mv cos45
o
. . . . . (2)
Dividing (2) by (1), we get,
$${{\sin 	heta } \over {\cos 	heta }}$$ = $${{\cos {{30}^ \circ } + \cos {{45}^ \circ }} \over {\sin {{30}^ \circ } - \sin {{45}^ \circ }}}$$
= $${{{{\sqrt 3 } \over 2} + {1 \over {\sqrt 2 }}} \over {{1 \over 2} - {1 \over {\sqrt 2 }}}}$$
$$	herefore\,\,\,$$ tan$$	heta $$ = $${{\sqrt 3  + \sqrt 2 } \over {1 - \sqrt 2 }}$$

Answer

tan $$	heta $$ = $${{\sqrt 3  + \sqrt 2 } \over {1 - \sqrt 2 }}$$

Answer

tan $$	heta $$ = $${{\sqrt 3  - \sqrt 2 } \over {1 - \sqrt 2 }}$$

Answer

tan $$	heta $$ = $${{1 - \sqrt 2 } \over {\sqrt 2 \left( {1 + \sqrt 3 } ight)}}$$

Answer

tan $$	heta $$ = $${{1 - \sqrt 3 } \over {1 + \sqrt 2 }}$$

Centre of Mass and Rotational Motion

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 :

Choose the correct answer: