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A pendulum comsists of a wooden bob of mass `m` and length `l`. A bullet of mass `m_1` is fired towards the pendulum with a speed `v_1` and it emerges from the bob with speed `v_1/3`. The bob just completes motion along a vertical circle. Then `v_1` is A. (a) `m/m_1sqrt(5gl)`B. (b) `(3m)/(2m_1)sqrt(5gl)`C. (c) `2/3(m/m_1)sqrt(5gl)`D. (d) `(m_1/m)sqrt(gl)` |
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Answer» Correct Answer - B Let velocity of m just after collision is v. Then, from conservation of momentum we have, `m_1v_1=mv+m_1v_1/3implies:. v=2/3(m_1v_1)/(m)` Now, for just completing the circle, `v=sqrt(5gl)` `:. 2/3(m_1v_1)/(m)=sqrt(5gl)` `:. v_1=(3m)/(2m_1)sqrt(5gl)` |
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