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QJEEM20Ph30d5: JEE Main 2020 Physics - Class 11 JEEM20Ph30d5 Solution: A particle of mass m is moving along the x-axis with initial…

A particle of mass m is moving along the x-axis with initial velocity ui^u\widehat i . It collides elastically with a particle of mass 10 m at rest and then moves with half its initial kinetic energy (see figure). If sin⁡θ1=nsin⁡θ2\sin {\theta _1} = \sqrt n \sin {\theta _2} then value of n is ________. JEE Main 2020 (Online) 2nd September Evening Slot Physics - Center of Mass and Collision Question 71 English

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JEE Main 2020 (Online) 2nd September Evening Slot Physics - Center of Mass and Collision Question 71 English Explanation
12mv12=12(12mu2){1 \over 2}mv_1^2 = {1 \over 2}\left( {{1 \over 2}m{u^2}} \right)

⇒\Rightarrow v12=u22v_1^2 = {{{u^2}} \over 2}

⇒\Rightarrow v1=u2{v_1} = {u \over {\sqrt 2 }} ......(1)

Also collision is elastic : ki = kf

12mu2=12mv12+12(10m)v22{{1 \over 2}m{u^2} = {1 \over 2}mv_1^2 + {1 \over 2}\left( {10m} \right)v_2^2}

⇒\Rightarrow 12×12mu2=12(10m)v22{{1 \over 2} \times {1 \over 2}m{u^2} = {1 \over 2}\left( {10m} \right)v_2^2}

⇒\Rightarrow v22=u220{v_2^2 = {{{u^2}} \over {20}}}

⇒\Rightarrow v2=u20{v_2} = {u \over {\sqrt {20} }}

By momentum conservation along y :

m1v1sin θ\theta 1 = m2v2sin θ\theta 2

⇒\Rightarrow mv1sin θ\theta 1 = 10mv2sin θ\theta 2

⇒\Rightarrow u2{u \over {\sqrt 2 }} sin θ\theta 1 = 10 ×\times u20{u \over {\sqrt {20} }} sin θ\theta 2

⇒\Rightarrow sin θ\theta 1 = 10\sqrt {10} sin θ\theta 2

∴\therefore n = 10

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