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5.Work, Energy, Power and Collision
normal
A wooden block of mass $M$ is suspended by a cord and is at rest. A bullet of mass $m,$ moving with a velocity $v$ passes through the block and comes out with a velocity $v/2$ in the same direction. If there is no loss in kinetic energy, then upto what height the block will rise
A
$m^2v^2/2M^2g$
B
$m^2v^2/8M^2g$
C
$mv^2/4Mg$
D
$mv^2/2Mg$
Solution
By $COLM$
$\mathrm{mv}+0=\frac{\mathrm{mv}}{2}+\mathrm{MV}^{\prime}$
$\mathrm{V}^{\prime}=\frac{\mathrm{mv}}{2 \mathrm{M}}$
By $COME$
$\frac{1}{2} M\left(\frac{m v}{2 M}\right)^{2} =M g H $
$H =\frac{m^{2} v^{2}}{8 M^{2} g}$
Standard 11
Physics
