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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Afternoon)

  • question_answer
              A body of mass \[{{m}_{1}}\] moving with an unknown velocity of \[{{\text{v}}_{1}}\hat{i},\]undergoes a collinear collision with a body of mass\[{{m}_{2}}\]moving with a velocity \[{{\text{v}}_{2}}\hat{i}.\]After collision, \[{{m}_{1}}\]and \[{{m}_{2}}\]move with velocities of \[{{\text{v}}_{3}}\hat{i}\]and \[{{\text{v}}_{4}}\hat{i},\]respectively. If \[{{m}_{2}}=0.5{{m}_{1}}\]and \[{{\text{v}}_{3}}=0.5{{\text{v}}_{1}},\]then \[{{\text{v}}_{1}}\]is :-                                                             [JEE Main 8-4-2019 Afternoon]

    A) \[{{\text{v}}_{4}}-\frac{{{\text{v}}_{2}}}{4}\]                

    B) \[{{\text{v}}_{4}}-\frac{{{\text{v}}_{2}}}{2}\]

    C) \[{{\text{v}}_{4}}-{{\text{v}}_{2}}\]

    D)   \[{{\text{v}}_{4}}\text{+}{{\text{v}}_{2}}\]

    Correct Answer: C

    Solution :

    Applying linear momentum conservation \[{{m}_{1}}{{\text{v}}_{1}}\hat{i}+{{m}_{2}}{{\text{v}}_{2}}\hat{i}={{m}_{1}}{{\text{v}}_{3}}\hat{i}+{{m}_{2}}{{\text{v}}_{4}}\hat{i}\] \[{{m}_{1}}{{\text{v}}_{1}}+0.5\,\,{{m}_{1}}{{\text{v}}_{2}}={{m}_{1}}\text{(0}\text{.5}{{\text{v}}_{1}})+0.5\,\,{{m}_{1}}{{\text{v}}_{4}}\] \[0.5\,\,{{m}_{1}}{{\text{v}}_{2}}=\text{0}\text{.5}\,\,{{\text{m}}_{1}}({{\text{v}}_{4}}-{{\text{v}}_{2}})\] \[{{\text{v}}_{1}}\text{=}{{\text{v}}_{4}}-{{\text{v}}_{2}}\]   


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