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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Second Law of Motion

  • question_answer
    Two trolleys of mass m and 3m are connected by a spring. They were compressed and released once, they move off in opposite direction and comes to rest after covering distances \[{{S}_{1}}\]and \[{{S}_{2}}\] respectively. Assuming the coefficient of friction to be uniform, the ratio of distances \[{{S}_{1}}:{{S}_{2}}\] is [EAMCET (Engg.) 1995]          

    A)                              1 : 9                

    B)                         1 : 3            

    C)                         3 : 1                

    D)                         9 : 1                

    Correct Answer: D

    Solution :

                                When trolley are released then they posses same linear momentum but in opposite direction. Kinetic energy acquired by any trolley will dissipate against friction.             \[\therefore \] \[\mu \,mg\,s=\frac{{{p}^{2}}}{2m}\] Þ \[s\propto \frac{1}{{{m}^{2}}}\] [As P and u are constants]              Þ \[\frac{{{s}_{1}}}{{{s}_{2}}}={{\left( \frac{{{m}_{2}}}{{{m}_{1}}} \right)}^{2}}={{\left( \frac{3}{1} \right)}^{2}}=\frac{9}{1}\]


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