JEE Main & Advanced Sample Paper JEE Main Sample Paper-45

  • question_answer
    A particle of mass 1 kg is released from rest at origin to move ina conservative force field. The variation of potential energy corresponding to this conservative force with X-coordinate is as shown in the figure. Mark out the correct statement(s) for this situation.                    

    A)  Particle is moving along +ve X-axis                                    

    B)  Particle may move along+ve X-axis                                 

    C)  Particle may move along -ve X-axis

    D)  When particle crosses x =-2 m, the speed of particle is \[\sqrt{60}\,m/s\]

    Correct Answer: D

    Solution :

     Here the information about variation of field is given by along X-axis but about \[Y\] and \[Z\]-axis no information is given. On \[Y\] and \[Z\]-coordinate, U may be depending or not. If \[\frac{\partial U}{\partial y}=\frac{\partial U}{az}=0,\], then motion of particle is determined only by From graph, \[{{F}_{x}}=-\frac{\partial U}{\partial x}=-15\,N,\,-ve\] sign shows that \[{{F}_{x}}\] is along \[-ve\] X-axis. Consider two cases ; Case I \[\frac{\partial U}{\partial y}=\frac{\partial U}{\partial z}=0,\] then particle is going to move along -ve X-axis with an acceleration of \[15\,\,m/{{s}^{2}}\] (towards -ve X). Case II \[\frac{\partial U}{\partial y}\] and/or \[\frac{\partial U}{\partial z}\ne 0,\]. then particle's motion cannot be exactly determined from given information but it is for sure that in any circumstance, the particle won't move along +ve X-axis, So, particle may move in -ve X-axis. From work energy theorem \[dK=-dU\] \[\Rightarrow \]  \[\frac{m{{v}^{2}}}{2}-0=-\{[-30+{{U}_{y}}+{{U}_{z}}]\,-{{[u]}_{0,\,0,\,0}}\}\] \[v=\sqrt{60}\,m/s\] if U is not depending on y and z.

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