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JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Morning)

  • question_answer
      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 v towards B due to which it collides with B perfectly inelastically \[\frac{5}{6}th\] of the initial kinetic energy is lost in whole process. What is value of M/m?   [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

    A) 2                  

    B) 3

    C) 5                  

    D) 4

    Correct Answer: D

    Solution :

    \[\frac{{{P}^{2}}}{2m}\,\,-\,\,\frac{{{P}^{2}}}{2(2m+M)}\,\,=\,\,\frac{5}{6}\,\,\frac{{{P}^{2}}}{2m}\] \[\frac{1}{m}\,\,-\,\,\frac{1}{(2m+M)}\,\,=\,\,\,\frac{5}{6\,m}\] \[\frac{2m+M-m}{m(2m+M)}\,\,=\,\,\,\frac{5}{6\,m}\] \[6\left( m+M \right)=5\left( 2m+M \right)\] \[6m+6M=10m+5M\] \[4m=M\] \[\frac{m}{M}\,\,=\,\,\frac{1}{4}\] \[\Rightarrow \,\,\frac{M}{m}\,\,=\,\,4\] Option is correct.


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