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NEET NEET SOLVED PAPER 2015 (C)

  • question_answer
    Two particles of masses \[{{m}_{1}},{{m}_{2}}\] move with initial velocities \[{{u}_{1}}\] and \[{{u}_{2}}\]. On collision, one of the particles get excited to higher level, after absorbing energy \[\varepsilon \]. If final velocities of particles be \[{{v}_{1}}\] and \[{{v}_{2}},\] then we must have                                                                                                                                                                    

    A)  \[m_{1}^{2}{{u}_{1}}+m_{2}^{2}{{u}_{2}}-\varepsilon =m_{1}^{2}{{v}_{1}}+m_{2}^{2}{{v}_{2}}\]

    B)  \[\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}-\varepsilon \]

    C)  \[\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}-\varepsilon =\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}\]

    D)  \[\frac{1}{2}m_{1}^{2}u_{1}^{2}+\frac{1}{2}m_{2}^{2}u_{2}^{2}+\varepsilon =\frac{1}{2}m_{1}^{2}v_{1}^{2}+\frac{1}{2}m_{2}^{2}v_{2}^{2}\]

    Correct Answer: C

    Solution :

    Total initial energy = \[\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}\] Since, after collision one particle absorb energy \[\varepsilon \]. \[\therefore \] Total final energy = \[\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{1}^{2}+\varepsilon \] From conservation of energy, \[\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}+\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}+\varepsilon \] \[\Rightarrow \,\,\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}-\varepsilon \]                       \[=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}\]


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