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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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A bullet of mass mm moving with a horizontal velocity uu strikes a stationary wooden block of mass MM suspended by a string of length L=50 cmL=50 \mathrm{~cm}. The bullet emerges out of the block with speed u4\frac{u}{4}. If M=6 mM=6 \mathrm{~m}, the minimum value of uu, so that the block can complete the vertical circle, is (Take, g=10 ms−2g=10 \mathrm{~ms}^{-2} )
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Correct option: D
40 ms−140 \mathrm{~ms}^{-1}
Option D (40 ms−140 \mathrm{~ms}^{-1}) is correct because it satisfies the momentum conservation and allows the block to complete a vertical circle.

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