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

  • question_answer
    Two balls of masses \[{{m}_{1}}\] and \[{{m}_{2}}\] are separated from each other by a powder charge placed between them. The whole system is at rest on the ground. Suddenly the powder charge explodes and masses are pushed apart. The mass \[{{m}_{1}}\] travels a distance \[{{s}_{1}}\] and stops. If the coefficients of friction between the balls and ground are same, the mass \[{{m}_{2}}\] stops after travelling the distance                           [BHU 1994]

    A)                         \[{{s}_{2}}=\frac{{{m}_{1}}}{{{m}_{2}}}{{s}_{1}}\]            

    B)                           \[{{s}_{2}}=\frac{{{m}_{2}}}{{{m}_{1}}}{{s}_{1}}\]            

    C)                         \[{{s}_{2}}=\frac{m_{1}^{2}}{m_{2}^{2}}{{s}_{1}}\]          

    D)                           \[{{s}_{2}}=\frac{m_{2}^{2}}{m_{1}^{2}}{{s}_{1}}\]

    Correct Answer: C

    Solution :

                              We know that in the given condition \[s\propto \frac{1}{{{m}^{2}}}\]             \[\therefore \] \[\frac{{{s}_{2}}}{{{s}_{1}}}={{\left( \frac{m{{}_{1}}}{{{m}_{2}}} \right)}^{2}}\] Þ \[{{s}_{2}}={{\left( \frac{m_{1}^{{}}}{m_{2}^{{}}} \right)}^{2}}\times {{s}_{1}}\]          


You need to login to perform this action.
You will be redirected in 3 sec spinner