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QJEEM20Phb2c3: JEE Main 2020 Physics - Class 11 JEEM20Phb2c3 Solution: A particle of mass m with an initial velocity collides…

A particle of mass m with an initial velocity ui^u\widehat i collides perfectly elastically with a mass 3 m at rest. It moves with a velocity vj^v\widehat j after collision, then, v is given by :

Multiple Choice Options

  • A v=23uv = \sqrt {{2 \over 3}} u
  • B v=u3v = {u \over {\sqrt 3 }}
  • C v=u2v = {u \over {\sqrt 2 }}
  • D v=16uv = {1 \over {\sqrt 6 }}u

Step-by-Step Verified Solution

JEE Main 2020 (Online) 2nd September Morning Slot Physics - Center of Mass and Collision Question 73 English Explanation

From momentum conservation

m(u) i^\widehat i + 3m(0) = mv j^\widehat j + 3m v′→\overrightarrow {v'}

⇒\Rightarrow 3m v′→\overrightarrow {v'} = m(u) i^\widehat i - mv j^\widehat j

⇒\Rightarrow v′→=ui^−vj^3\overrightarrow {v'} = {{u\widehat i - v\widehat j} \over 3}

∴\therefore ∣v′→∣=u2+v23\left| {\overrightarrow {v'} } \right| = {{\sqrt {{u^2} + {v^2}} } \over 3}

⇒\Rightarrow ∣v′→∣2=u2+v29{\left| {\overrightarrow {v'} } \right|^2} = {{{u^2} + {v^2}} \over 9} ......(1)

As collision is perfectely elastic hence

KEi = KEj

⇒\Rightarrow 12mu2+12(3m)×02=12mv2+123m(v′)2{1 \over 2}m{u^2} + {1 \over 2}\left( {3m} \right) \times {0^2} = {1 \over 2}m{v^2} + {1 \over 2}3m{\left( {v'} \right)^2}

⇒\Rightarrow u2 = v2 + 3 (v′)2{\left( {v'} \right)^2}

⇒\Rightarrow u2 = v2 + 3 (u2+v29)\left( {{{{u^2} + {v^2}} \over 9}} \right)

⇒\Rightarrow 3u2 = 3v2 + u2 + v2

⇒\Rightarrow 2u2 = 4v2

⇒\Rightarrow v=u2v = {u \over {\sqrt 2 }}

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