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NEET Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति NEET PYQ-Work Energy Power and Collision

  • 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 [NEET 2015 ]

    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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