JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति JEE PYQ - Work Energy Power and Collision

  • question_answer
    The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant k and compresses it by length L. The maximum momentum of the block after collision is           [AIEEE 2005]

    A) \[\sqrt{MK}L\]

    B) \[\frac{k{{L}^{2}}}{2M}\]  

    C) zero

    D) \[\frac{M{{L}^{2}}}{k}\]

    Correct Answer: A

    Solution :

    [a] Momentum would be maximum when KE would be maximum and this is the case when total elastic PE is converted into KE.
    According to conservation of energy,
    ( for spring)
                              
     


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