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JCECE Engineering JCECE Engineering 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 the body is projected and \[{{t}_{2}}\] is the time in second at which it reaches the ground, the change in momentum in \[kgm{{s}^{-1}}\] 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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