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QJEEM17Phb776: JEE Main 2017 Physics - Class 11 JEEM17Phb776 Solution: Two particles A and B of equal mass M are moving with the same…

Two particles A and B of equal mass M are moving with the same speed υ\upsilon as shown in the figure. They collide completely inelastically and move as a single particle C. The angle θ\theta that the path of C makes with the X-axis is given by :

JEE Main 2017 (Online) 9th April Morning Slot Physics - Center of Mass and Collision Question 103 English

Multiple Choice Options

  • A tan θ\theta = 3+21−2{{\sqrt 3 + \sqrt 2 } \over {1 - \sqrt 2 }}
  • B tan θ\theta = 3−21−2{{\sqrt 3 - \sqrt 2 } \over {1 - \sqrt 2 }}
  • C tan θ\theta = 1−22(1+3){{1 - \sqrt 2 } \over {\sqrt 2 \left( {1 + \sqrt 3 } \right)}}
  • D tan θ\theta = 1−31+2{{1 - \sqrt 3 } \over {1 + \sqrt 2 }}

Step-by-Step Verified Solution

JEE Main 2017 (Online) 9th April Morning Slot Physics - Center of Mass and Collision Question 103 English Explanation

Using conservation of linear moments,

Along X-axis,

2MV'cos θ\theta = MVsin30o −- MVsin45o . . . . . (1)

Along Y - axis,

2Mv' sin θ\theta = Mv cos30o + Mv cos45o . . . . . (2)

Dividing (2) by (1), we get,

sin⁡θcos⁡θ{{\sin \theta } \over {\cos \theta }} = cos⁡30∘+cos⁡45∘sin⁡30∘−sin⁡45∘{{\cos {{30}^ \circ } + \cos {{45}^ \circ }} \over {\sin {{30}^ \circ } - \sin {{45}^ \circ }}}

= 32+1212−12{{{{\sqrt 3 } \over 2} + {1 \over {\sqrt 2 }}} \over {{1 \over 2} - {1 \over {\sqrt 2 }}}}

∴   \therefore\,\,\, tan θ\theta = 3+21−2{{\sqrt 3 + \sqrt 2 } \over {1 - \sqrt 2 }}

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