JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • question_answer
    The figure shows the position - time (x - t) graph of one-dimensional motion of the body of mass 0.4 kg. The magnitude of each impulse is – [AIEEE 2010]

    A) 0.2 Ns

    B) 0.4 Ns

    C) 0.8 Ns

    D) 1.6 Ns

    Correct Answer: C

    Solution :

    [c] From graph,
    \[{{v}_{1}}=1\text{ }m{{s}^{1}},\]                        \[{{v}_{2}}=1\text{ }m{{s}^{1}}\]
    \[\therefore \] \[J=\int{F}dt=\int{dp}=m\Delta V\]
    \[=0.4\times 2=0.8\text{ }N.s.\]

You need to login to perform this action.
You will be redirected in 3 sec spinner