JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Graphical Questions

  • question_answer
    In the figure given below, the position-time graph of a particle of mass 0.1 Kg is shown. The impulse at \[t=2\sec \] is                                                                 [AIIMS 2005]

    A)                 0.2\[kg\,m{{\sec }^{-1}}\]

    B)                    \[-0.2kg\,m{{\sec }^{-1}}\]

    C)                    \[0.1kg\,m{{\sec }^{-1}}\]

    D)                                 \[-0.4kg\,m{{\sec }^{-1}}\]

    Correct Answer: B

    Solution :

                    Velocity  between \[t=0\] and \[t=2\sec \]                 Þ \[{{v}_{i}}=\frac{dx}{dt}=\frac{4}{2}=2\ m/s\]                 Velocity at t\[=2\sec \], \[{{v}_{f}}=0\]                 Impulse = Change in momentum \[=m({{v}_{f}}-{{v}_{i}})\]                 \[=0.1(0-2)\] \[=-0.2\ kg\ m\ {{\sec }^{-1}}\]


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