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JCECE Medical JCECE Medical Solved Paper-2014

  • question_answer
    A body of mass 2 kg is projected from the ground with a velocity \[20\,m{{s}^{-1}}\]at an angle \[{{30}^{o}}\] with the vertical. If\[{{t}_{1}}\] is the time in second at which of the body is projected and \[{{t}_{2}}\] is the time in second at which it reaches the ground, the change in momentum in \[\text{kgm}{{\text{s}}^{-1}}\] during the time\[({{t}_{2}}-{{t}_{1}})\]is

    A)  \[40\sqrt{2}\]

    B)  \[40\sqrt{3}\]

    C)  \[25\sqrt{3}\]

    D)  45

    Correct Answer: B

    Solution :

     Initial moment of the body in vertical upward direction \[{{p}_{1}}=mv\cos {{30}^{o}}\] \[=\frac{\sqrt{3}}{2}mv\] Final momentum of body in downward direction \[{{p}_{2}}=mv\cos {{30}^{o}}\] \[=\frac{\sqrt{3}}{2}mv\] Change in momentum \[\Delta {{p}_{2}}={{p}_{2}}-(-{{p}_{1}})\] \[={{p}_{2}}+{{p}_{1}}\] \[=\frac{\sqrt{3}}{2}mv+\frac{\sqrt{3}}{2}mv\] \[=\sqrt{3}\,mv\] \[=\sqrt{3}\times 2\times 20\] \[=40\sqrt{3}\]


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