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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2011

  • question_answer
    A body at rest splits into three parts of mass m, m and 4m respectively. The two equal masses fly off perpendicular to each other and each with speed of v. The speed of 4m will be

    A) \[\frac{v}{2\sqrt{2}}\]   

    B)                        \[\frac{v}{\sqrt{2}}\]

    C)  \[\frac{v}{2}\]                  

    D)         \[\sqrt{2\,v}\]

    Correct Answer: A

    Solution :

    From conservation of momentum                 \[mv\sqrt{2}=4m{{v}_{1}}\]                 \[{{v}_{1}}=\frac{v\sqrt{2}}{4}\]                 \[{{v}_{1}}=\frac{v}{2\sqrt{2}}\]


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