JEE Main & Advanced Sample Paper JEE Main Sample Paper-22

  • question_answer
    A ball strikes a horizontal floor at \[45{}^\circ \], 25 % of its kinetic energy is lost in collision, find the coefficient of restitution.

    A)  \[\frac{1}{2}\]                         

    B)   \[\frac{1}{\sqrt{2}}\]

    C)  \[\frac{1}{2\sqrt{2}}\]              

    D)  \[\frac{1}{4}\]

    Correct Answer: B

    Solution :

    Fraction of KE lost in collision \[=\frac{\frac{1}{2}m{{u}^{2}}-\frac{1}{2}m{{v}^{2}}}{\frac{1}{2}m{{u}^{2}}}\] \[=1-{{\left( \frac{v}{u} \right)}^{2}}=\frac{1}{4}\] (given) \[v=u\sqrt{\frac{3}{4}}\] The ball strikes at \[{{45}^{0}}\] component of velocity parallel to wall \[(u\cos {{45}^{0}})\] will not change while component of velocity normal to wall will change. \[v=u\,\sqrt{\frac{3}{4}}\,={{\left[ {{\left( \frac{u}{\sqrt{2}} \right)}^{2}}+\,{{\left( \frac{eu}{\sqrt{2}} \right)}^{2}} \right]}^{1/2}}\] \[=u\,{{\left[ \frac{1}{2}\,+\frac{{{e}^{2}}}{2} \right]}^{1/2}}\]on solving \[e=\frac{1}{\sqrt{2}}\]


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