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5.Work, Energy, Power and Collision
normal
A bullet of mass $m$ moving with velocity $v$ strikes a suspended wooden block of mass $M$. If the block rises to a height $h$, the initial velocity of the bullet will be
A
$\sqrt {2gh} $
B
$\frac{{(M + m)}}{m}\sqrt {2gh} $
C
$\frac{m}{{(M + m)}}\sqrt {2gh} $
D
$\frac{{(M - m)}}{m}\sqrt {2gh} $
Solution
$\mathrm{mv}+0=(\mathrm{m}+\mathrm{M}) \mathrm{V} \Rightarrow \mathrm{V}=\frac{ \mathrm{mv}}{\mathrm{M}+\mathrm{m}}$
$\because \quad V=\sqrt{2 g h}$
$\frac{\mathrm{mv}}{\mathrm{m}+\mathrm{M}}=\sqrt{2 \mathrm{gh}}$
$\mathrm{v}=\frac{(\mathrm{M}+\mathrm{m})}{\mathrm{m}} \sqrt{2 \mathrm{gh}}$
Standard 11
Physics
