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Q. A large block of wood of mass $M =5.99 \,kg$ is hanging from two long massless cords. A bullet of mass $m =10 \,g$ is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their centre of mass rising a vertical distance $h =9.8 \,cm$ before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is : (Take $g=9.8\, ms ^{-2}$ )
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JEE MainJEE Main 2021Work, Energy and Power

Solution:

From energy conservation,
[after bullet gets embedded till the system comes momentarily at rest]
$( M + m ) g h =\frac{1}{2}( M + m ) v _{1}^{2}$
$\left[ v _{1}\right.$ is velocity after collision $]$
$\therefore v _{1}=\sqrt{2 gh }$
Applying momentum conservation, (just before and just after collision)
$mv =( M + m ) v _{1} $
$v =\left(\frac{ M + m }{ m }\right) v _{1}$
$=\frac{6}{10 \times 10^{-3}} \times \sqrt{2 \times 9.8 \times 9.8 \times 10^{-2}}$
$\approx 831.55 m / s$