MGIMS WARDHA MGIMS WARDHA Solved Paper-2014

  • question_answer
    Two particles of masses m^ and 7712 in projectile motion have velocities\[{{\upsilon }_{1}}\]and\[{{\upsilon }_{2}}({{\upsilon }_{1}}<{{\upsilon }_{2}})\]respectively at\[t=0\]. They collide at time\[{{t}_{0}}\]Their velocities become\[\overline{\upsilon }{{}_{1}}\]and\[\overline{\upsilon }{{}_{2}}\]at time\[2{{t}_{0}}\]while still moving in it. The value of \[|({{m}_{1}}\overline{\upsilon }+{{m}_{2}}\overline{\upsilon }{{}_{2}})-({{m}_{1}}{{\overline{\upsilon }}_{1}}+{{m}_{2}}{{\overline{\upsilon }}_{2}})|\]is

    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: A

    Solution :

                    By law of conservation of linear momentum, the required value is 0.


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