JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Sample Paper Topic Test - Work Energy Power

  • question_answer
    A ball of mass m hits a wall with a speed v making an angle \[\frac{5g}{14}\] with the normal. If the coefficient is e, the direction and magnitude of the velocity of ball after reflection from the wall will respectively be -

    A) \[{{\tan }^{-1}}\left( \frac{\tan \theta }{e} \right),\,v\sqrt{{{\sin }^{2}}\theta +{{e}^{2}}{{\cos }^{2}}\theta }\]

    B) \[{{\tan }^{-1}}\left( \frac{e}{\tan \theta } \right),\frac{1}{v}\sqrt{{{e}^{2}}{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }\]

    C) \[{{\tan }^{-1}}(e\tan \theta ),\frac{v}{e}\tan \theta \]

    D) \[{{\tan }^{-1}}(e\tan \alpha ),v\sqrt{{{\sin }^{2}}\theta +{{e}^{2}}}\]

    Correct Answer: A

    Solution :

    [a] Let the angle of reflection be \[\theta \]' and the magnitude of velocity after collision be v'. As there is no force parallel to the wall, the component of velocity parallel to the surface remains unchanged.
      Therefore, \[v'sin\text{ }\!\!\theta\!\!\text{  }\!\!'\!\!\text{  = v sin  }\!\!\theta\!\!\text{ }\,\,\,\,\,\,\,\,....\left( 1 \right)\]      
      As the coefficient of restitution is e, for perpendicular component of velocity
      Velocity of separation = e x velocity of approach
      \[-\left( v'\cos \text{ }\!\!\theta\!\!\text{  }\!\!'\!\!\text{  - 0} \right)=-e\left( v\cos \text{ }\!\!\theta\!\!\text{  - 0} \right)\,\,\,\,\,....\left( 2 \right)\]
      From (1) and (2)
      \[v'=v\sqrt{{{\sin }^{2}}\text{ }\!\!\theta\!\!\text{  + }{{\text{e}}^{2}}{{\cos }^{2}}\text{ }\!\!\theta\!\!\text{ }}\]
      and \[\tan \text{ }\!\!\theta\!\!\text{  }\!\!'\!\!\text{  = tan }\!\!\theta\!\!\text{ /e}\]

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