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JEE Main & Advanced Physics Rotational Motion Question Bank Mock Test - Centre of Mass, Conservation of Linear Momentum and Collisions

  • question_answer
    The velocities of two particles \[A\] and \[B\] of same mass are \[{{\vec{V}}_{A}}=a\vec{i}\,and\,{{\vec{V}}_{B}}=b\hat{j}\] where a and b are constants. The acceleration of particle\[A\] is \[(2a\hat{i}+4b\hat{j})\] and acceleration of particle \[B\] is \[(a\hat{i}-4b\hat{j})\] (in \[m/{{s}^{2}}\]). The centre of mass of two particle will move in

    A) straight line       

    B) parabola

    C) ellipse  

    D) circle

    Correct Answer: A

    Solution :

    [a] \[{{\vec{V}}_{com}}=\frac{{{m}_{1}}{{{\vec{V}}}_{1}}+{{m}_{2}}{{{\vec{V}}}_{2}}}{{{m}_{1}}+{{m}_{2}}}=\frac{{{{\vec{V}}}_{1}}+{{{\vec{V}}}_{2}}}{2}=\frac{a\hat{i}+b\hat{j}}{2}\] \[{{\vec{a}}_{com}}=\frac{{{a}_{1}}+{{a}_{2}}}{2}=\frac{3}{2}(a\hat{i}+b\hat{j})\] \[{{\vec{V}}_{com}}\] is parallel to \[{{\vec{a}}_{com}}\]. Hence path will be a straight line.


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