JEE Main & Advanced Sample Paper JEE Main Sample Paper-47

  • question_answer
    A particle strikes a smooth horizontal surface at an angle of \[45{}^\circ \] with a velocity of 100 m/sand rebounds. If the coefficient of restitution between the floor and the particle is 0.57,then the angle which the velocity of the particle after it rebounds will make with the floor is

    A) \[30{}^\circ \]

    B) \[45{}^\circ \]

    C) \[60{}^\circ \]

    D) \[90{}^\circ \]

    Correct Answer: A

    Solution :

     Along horizontal direction. \[mu\,\sin \,{{\theta }_{1}}=mv\,\,\sin \,{{\theta }_{2}}\]    ?(i) \[eu\,\,\cos \,{{\theta }_{1}}=v\,\,\cos \,{{\theta }_{2}}\]     ?(ii) Here, \[e=\frac{Velocity\,of\,separation}{Velocity\,of\,approach}\] Solving, Eqs. (i) and (ii), we get \[{{\theta }_{2}}={{\tan }^{-1}}\,\left( \frac{1}{0.57}\tan \,{{45}^{o}} \right)={{60}^{o}}\] \[\Rightarrow \]            \[\beta ={{30}^{o}}\]

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