JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    In a collinear collision, a particle with an initial speed \[{{v}_{o}}\] strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles/after collision, is:                 [JEE Main Online 08-04-2018]

    A)  \[\frac{{{v}_{o}}}{2}\]                               

    B)  \[\frac{{{v}_{o}}}{\sqrt{2}}\]

    C)  \[\frac{{{v}_{o}}}{4}\]                               

    D)  \[\sqrt{2}{{v}_{o}}\]

    Correct Answer: D

    Solution :

    From C.O.L.M. \[m{{v}_{0}}=m{{v}_{1}}+m{{v}_{2}}\]   ?(1) And \[K.{{E}_{f}}=\frac{3}{2}K.E{{.}_{i}}\] \[\frac{1}{2}mv_{1}^{2}+\frac{1}{2}mv_{2}^{2}=\frac{3}{2}\left( \frac{1}{2}mv_{0}^{2} \right)\] \[v_{1}^{2}+v_{2}^{2}=\frac{3}{2}V_{0}^{2}\]                     ?(2) Solving equation (1) and (2) we get \[{{v}_{rel}}={{v}_{2}}-{{v}_{1}}=\sqrt{2}{{v}_{0}}\]


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