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EAMCET Medical EAMCET Medical Solved Paper-2006

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

    A)  40                                         

    B)  40\[\sqrt{3}\]

    C)  50\[\sqrt{3}\]                                  

    D)  60

    Correct Answer: B

    Solution :

    Momentum of the body in the horizontal direction remains unchanged throughout the motion as projectile. 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, \[{{p}_{2}}=mv\,\cos {{30}^{o}}=\frac{\sqrt{3}}{2}mv\] \[\therefore \] Change in momentum, \[\Delta p={{p}_{2}}-(-{{p}_{1}})\] \[={{p}_{2}}+{{p}_{1}}\] \[=\frac{\sqrt{3}}{2}mv+\frac{\sqrt{3}}{2}mv\] \[=\sqrt{3}mv\] Given,  m = 2 kg, v = 20 m/ s \[\therefore \]   \[\Delta p=\sqrt{3}\times 2\times 20\] \[=40\sqrt{3}kg\,m/s\]


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