JCECE Medical JCECE Medical Solved Paper-2014

  • question_answer A mass of 0.5 kg moving with a speed of \[1.5\,\text{m}{{\text{s}}^{-1}}\]on a horizontal smooth surface, collides with a nearly weightless spring of force constant\[\text{k}\,\text{=}\,\text{50}\,\text{N}{{\text{m}}^{-1}}.\]The maximum  compression of the spring would be

    A)  0.15 m             

    B)  0.23 m

    C) 1.6 m               

    D)  0.4 m

    Correct Answer: A

    Solution :

     By the law of conservation of energy. Kinetic energy of mass = energy stored in spring \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}k{{x}^{2}}\] \[{{x}^{2}}=\frac{m{{v}^{2}}}{k}\] The maximum compression of the spring \[x=\sqrt{\frac{m{{v}^{2}}}{k}}\] \[x=\sqrt{\frac{0.5\times 1.5\times 1.5}{50}}=0.15\,m\]


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