JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Critical Thinking

  • question_answer
    Two blocks A and B each of mass m are connected by a massless spring of natural length L and spring constant K. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length as shown in figure. A third identical block C also of mass m moves on the floor with a speed v along the line joining A and B and collides with A. Then                                                            [IIT-JEE 1993]  

    A)            The kinetic energy of the A-B system at maximum compression of the spring is zero

    B)            The kinetic energy of the A-B system at maximum compression of the spring is \[m{{v}^{2}}/4\]

    C)            The maximum compression of the spring is \[v\sqrt{m/K}\]

    D)            The maximum compression of the spring is \[v\sqrt{m/2K}\]

    Correct Answer: B

    Solution :

     Let the velocity acquired by A and B be V, then \[mv=mV+mV\Rightarrow V=\frac{v}{2}\] Also \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{V}^{2}}+\frac{1}{2}m{{V}^{2}}+\frac{1}{2}k{{x}^{2}}\]                    Where x is the maximum compression of the spring. On solving the above equations, we get \[x=v{{\left( \frac{m}{2k} \right)}^{1/2}}\]                    At maximum compression, kinetic energy of the            A ? B system \[=\frac{1}{2}m{{V}^{2}}+\frac{1}{2}m{{V}^{2}}=m{{V}^{2}}=\frac{m{{v}^{2}}}{4}\]


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