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QJEEM19Phe6d5: JEE Main 2019 Physics - Class 11 JEEM19Phe6d5 Solution: Three blocks A, B and C are lying on a smooth horizontal…

Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m while C has mass M, Block A is given an initial speed υ\upsilon towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically 56{5 \over 6} th of the initial kinetic energy is lost in whole process. What is value of M/m ?

JEE Main 2019 (Online) 9th January Morning Slot Physics - Center of Mass and Collision Question 99 English

Multiple Choice Options

  • A 5
  • B 2
  • C 4
  • D 3

Step-by-Step Verified Solution

As in elastic or in elastic clollision, momentum is conserved.

∴\therefore    Pi = Pf

Pi = Initial momentum

Pf = Final Momentum

      mv = (2m + m) Vf

⇒\Rightarrow    Vf = mv2m+M{{mv} \over {2m + M}}

Here due to collision 56{5 \over 6} th of kinetic energy is lost.

∴\therefore    Remaining kinetic energy,

      Kf = 16{1 \over 6} Ki

⇒\Rightarrow     12{1 \over 2} (2m + M) ×\times (mv)2(2m+M)2{{{{\left( {mv} \right)}^2}} \over {{{\left( {2m + M} \right)}^2}}} = 16×12mv2{1 \over 6} \times {1 \over 2}m{v^2}

⇒\Rightarrow     m2v22m+M{{{m^2}{v^2}} \over {2m + M}} = 16mv2{1 \over 6}m{v^2}

⇒\Rightarrow     m2m+M{m \over {2m + M}} = 16{1 \over 6}

⇒\Rightarrow    6m = 2m + M

⇒\Rightarrow    M = 4m

⇒\Rightarrow     Mm{M \over m} = 4

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