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JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Critical Thinking

  • question_answer
    Two particles of masses \[{{m}_{1}}\] and \[{{m}_{2}}\] in projectile motion have velocities \[{{\vec{v}}_{1}}\] and \[{{\vec{v}}_{2}}\] respectively at time t = 0. They collide at time \[{{t}_{0}}\]. Their velocities become \[{{\vec{v}}_{1}}'\] and \[{{\vec{v}}_{2}}'\] at time \[2{{t}_{0}}\] while still moving in air. The value of \[|({{m}_{1}}\overrightarrow{{{v}_{1}}}'\,+{{m}_{2}}\overrightarrow{{{v}_{2}}}')-({{m}_{1}}\overrightarrow{{{v}_{1}}}\,+{{m}_{2}}\overrightarrow{{{v}_{2}}})\]| is           [IIT-JEE Screening 2001]

    A)             Zero    

    B)             \[({{m}_{1}}+{{m}_{2}})g{{t}_{0}}\]

    C)             \[2({{m}_{1}}+{{m}_{2}})g{{t}_{0}}\]

    D)               \[\frac{1}{2}({{m}_{1}}+{{m}_{2}})g{{t}_{0}}\]

    Correct Answer: C

    Solution :

                    The momentum of the two-particle system, at t = 0 is \[{{\vec{P}}_{i}}={{m}_{1}}{{\vec{v}}_{1}}+{{m}_{2}}{{\vec{v}}_{2}}\] Collision between the two does not affect the total momentum of the system. A constant external force \[({{m}_{1}}+{{m}_{2}})g\] acts on the system. The impulse given by this force, in time t = 0 to \[t=2{{t}_{0}}\] is \[({{m}_{1}}+{{m}_{2}})g\times 2{{t}_{0}}\] \ |Change in momentum in this interval             \[=\,|{{m}_{1}}\vec{v}{{'}_{1}}+{{m}_{2}}\vec{v}{{'}_{2}}-({{m}_{1}}{{\vec{v}}_{1}}+{{m}_{2}}{{\vec{v}}_{2}})|\,=2({{m}_{1}}+{{m}_{2}})g{{t}_{0}}\]


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