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JEE Main & Advanced Sample Paper JEE Main Sample Paper-11

  • question_answer
    Consider the following statements \[{{\text{S}}_{\text{1}}}\]: Non zero work has to be done on a moving particle to change its momentum. \[{{\text{S}}_{2}}\]: To change momentum of a particle a non zero net force should act on it. \[{{\text{S}}_{3}}\]: Two particles undergo rectilinear motion along different straight lines. Then the centre of mass of system of given two particles also always moves along a straight line. \[{{\text{S}}_{4}}\]: If direction of net momentum of a system of particles (having nonzero net momentum) is fixed, the centre of mass of given system moves along a straight line. State, in order, whether \[{{\text{S}}_{\text{1}}}\text{,}{{\text{S}}_{\text{2}}}\text{,}{{\text{S}}_{\text{3}}}\text{,}{{\text{S}}_{\text{4}}}\] are true or false.

    A)  TTFT                     

    B)  FTFT

    C)  FTTT                     

    D)  FFTF

    Correct Answer: B

    Solution :

     \[{{\text{S}}_{\text{1}}}\]: No work is done by net force, it only changes direction of momentum of particle. Hence \[{{\text{S}}_{\text{1}}}\] is false. \[{{\text{S}}_{2}}\]: True by definition. \[{{\text{S}}_{3}}\]: Nothing is said about acceleration of both particles. Hence angle between velocity and acceleration of centre of mass may not be zero. Consequently centre of mass may not move along a straight line. Hence \[{{\text{S}}_{3}}\]is false. \[{{\text{S}}_{4}}\]: \[{{\vec{v}}_{cm}}=\frac{{{m}_{1}}{{{\vec{v}}}_{1}}+{{m}_{2}}{{{\vec{v}}}_{2}}+.....+{{m}_{n}}{{{\vec{v}}}_{n}}}{{{m}_{1}}+{{m}_{2}}+....+{{m}_{n}}}=\frac{{{{\vec{F}}}_{net}}}{({{m}_{1}}+{{m}_{2}}+....+{{m}_{n}})}\]Direction at \[{{\vec{P}}_{net}}\]is fixed so \[{{\vec{V}}_{cm}}\]is also constant in the direction. So path of CM will be straight line.


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