JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Self Evaluation Test - Work, Energy and Power

  • question_answer
    Two small bodies of masses 'm' and '2m' are placed in a fixed smooth horizontal circular hollow tube of mean radius 'r' as shown. The mass 'm' is moving with speed 'u' and the mass '2m' is stationary. After their first collision, the time elapsed for next collision is [ coefficient of restitution e = 1/2 ]

    A) \[\frac{2\pi r}{u}\]

    B) \[\frac{4\pi r}{u}\]

    C) \[\frac{3\pi r}{u}\]

    D) \[\frac{12\pi r}{u}\]

    Correct Answer: B

    Solution :

    [b] If just after collision, relative velocity = v then \[\frac{v}{u}=\frac{1}{2}\therefore {{\omega }_{\operatorname{rel}}}=\frac{v}{r}=\frac{u}{2r}\] \[\therefore \] time between 1st and 2nd collision, \[t=\frac{2\pi }{{{\omega }_{\operatorname{rel}}}}\] \[=\frac{4\pi r}{u}\]


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