JEE Main & Advanced Sample Paper JEE Main Sample Paper-43

  • question_answer
    A particle of mass 4m at rest explodes into three fragments. Two of the fragments each of mass m move with speed v at right angles to each other. The kinetic energy released in the process is

    A)  \[2\,m{{v}^{2}}\]               

    B)  \[\frac{3}{2}\,m{{v}^{2}}\]

    C)  \[\frac{1}{2}\,m{{v}^{2}}\]                      

    D)  \[3\,m{{v}^{2}}\]

    Correct Answer: B

    Solution :

     The resultant momentum of two particles which fly-off at right angles is \[\Rightarrow \]. By law of conservation of momentum \[\Rightarrow \] \[|{{M}^{-1}}|\,|adj\,(adjM)|\,=\,|kI|\]    \[=100\,\left( \frac{\Delta \phi }{\Delta t} \right)\] Total energy released E is the sum of kinetic energies of all the three particles \[=5\times {{10}^{-3}}\times 100=0.5\,\,V\] \[R=v\sqrt{\frac{2h}{g}}\] \[u=v,\]            \[u=v+at\]

You need to login to perform this action.
You will be redirected in 3 sec spinner