J & K CET Engineering J and K - CET Engineering Solved Paper-2007

  • question_answer
    A stationary bomb explodes into two parts of masses in the ratio of \[1:3\]. If the heavier mass moves with a velocity \[4\text{ }m/s,\] what is the velocity of lighter part?

    A)  \[12\text{ }m{{s}^{-1}}\]opposite to heavier mass

    B)  \[12\text{ }m{{s}^{-1}}\] in the direction of heavier mass

    C)  \[\text{6 }m{{s}^{-1}}\] opposite to heavier mass

    D)  \[\text{6 }m{{s}^{-1}}\] in the direction of heavier mass

    Correct Answer: A

    Solution :

    The ratio of masses \[=1:3\] Therefore, \[{{m}_{1}}=x\,kg,\,\,\,{{m}_{2}}=3x\,\,kg\] Applying law of conservation of momentum \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}=0\] \[\Rightarrow \] \[x\times {{v}_{1}}+3x\times 4=0\] \[\Rightarrow \] \[{{v}_{1}}=-12m/s\] Therefore, velocity of lighter mass is opposite to that of heavier mass.


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