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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 1

  • question_answer
    A circus acrobat of mass M leaps straight up with initial velocity \[{{V}_{0}}\] from a trampoline. As he rises up, he takes a trained monkey of mass m hanging from a branch at a height h above the trampoline. What is the maximum height attained by the pair (from the branch)?

    A) \[\frac{M}{M+m}\left( \frac{V_{0}^{2}}{2g}-h \right)\]

    B) \[{{\left( \frac{M}{M+m} \right)}^{2}}\left( \frac{V_{0}^{2}}{2g}-h \right)\]

    C) \[\left( \frac{m}{M+m} \right)\left( \frac{V_{0}^{2}}{2g}-h \right)\]

    D) \[{{\left( \frac{m}{M+m} \right)}^{2}}\left( \frac{V_{0}^{2}}{2g}-h \right)\]

    Correct Answer: B

    Solution :

    [b] velocity of acrobat it height his \[V=\sqrt{V_{0}^{2}-2gh}\] Let's say now both (acrobat + monkey) leave the branch velocity is V?.             \[MV=(m+M)V'\]             \[V'=\left( \frac{M}{m+M} \right)\sqrt{V_{0}^{2}-2gh}\] for maximum height from branch \[\frac{{{(V')}^{2}}}{2g}={{\left( \frac{M}{m+M} \right)}^{2}}\left( \frac{V_{0}^{2}}{2g}-h \right)\]


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