JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Second Law of Motion

  • question_answer
    A ball of mass 400 gm is dropped from a height of 5m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 100 newton so that it attains a vertical height of 20 m. The time for which the ball remains in contact with the bat is \[[g=10\,m/{{s}^{2}}]\]  [MP PMT 1999]

    A)             0.12s   

    B)             0.08 s

    C)             0.04 s  

    D)             12 s

    Correct Answer: A

    Solution :

                    Velocity by which the ball hits the bat \[{{v}_{1}}=\sqrt{2g{{h}_{1}}}=\sqrt{2\times 10\times 5}\]or \[\overrightarrow{{{v}_{1}}}=+10\,m/s=10\,m/s\]             velocity of rebound             \[{{v}_{2}}=\sqrt{2g{{h}_{2}}}=\sqrt{2\times 10\times 20}=20\,m/s\] or \[\overrightarrow{{{v}_{2}}}=-20\,m/s\]             \[F=m\frac{dv}{dt}=\frac{m(\overrightarrow{{{v}_{2}}}-\overrightarrow{{{v}_{1}}})}{dt}=\frac{0.4(-20-10)}{dt}=100\,N\]             by solving \[dt=0.12\,\sec \]


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