Manipal Medical Manipal Medical Solved Paper-2007

  • question_answer
    A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant\[k=50\]N/m The maximum compression of the spring would be

    A)  0.15 m          

    B)  0.12 m

    C)  1.5 m            

    D)  0.5 m

    Correct Answer: A

    Solution :

     Key Idea: The kinetic energy of mass must be converted into energy stored in spring at the time mass strikes the spring. By the law of conservation of energy, kinetic energy of mass = energy stored in spring ie,        \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}k{{x}^{2}}\] \[\therefore \] \[{{x}^{2}}=\frac{m{{v}^{2}}}{k}\] \[\Rightarrow \] \[x=\sqrt{\left( \frac{m{{v}^{2}}}{k} \right)}\] \[\Rightarrow \] \[x=\sqrt{\left( \frac{0.5\times 1.5\times 1.5}{50} \right)}\] \[\therefore \] \[=0.15\,m\]


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