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NEET AIPMT SOLVED PAPER SCREENING 2012

  • question_answer
    A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity\[4\,\text{m}{{\text{s}}^{\text{-1}}}\].It collides with a horizontal spring of force constant\[\text{200}\,\text{N}{{\text{m}}^{\text{-1}}}\]. The maximum compression produced in the spring will be

    A) 0.5 m

    B)                        0.6 m    

    C)                        0.7 m    

    D)        0.2 m

    Correct Answer: B

    Solution :

    Loss in KE = Gain in spring energy \[\frac{1}{2}m{{v}^{2}}\left[ 1+\frac{{{K}^{2}}}{{{R}^{2}}} \right]=\frac{1}{2}k\,x_{\max }^{2}\] where k is the force constant. Given, v = 4 m/s, m = 3 kg, k = 200 N/m For solid cylinder, \[\frac{{{K}^{2}}}{{{R}^{2}}}=\frac{1}{2}\] \[\therefore \]\[\frac{1}{2}\times 3\times {{(4)}^{2}}\left[ 1+\frac{1}{2} \right]=\frac{1}{2}\times 200\times x_{\max }^{2}\] The maximum compression in the spring                 \[x_{\max }^{{}}=0.\,6\text{m}\]


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