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Work Energy Theorem Application

A wooden bloc...

Question

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:

m^{2} v^{2} / 2 M^{2} g

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m^{2} v^{2} / 8 M^{2} g

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m v^{2} / 4 M g

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m v^{2} / 2 M g

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Solution

The correct option is B $m^{2} v^{2} / 8 M^{2} g$
Applying COLM
$m v + 0 = M v^{'} + m v / 2$
$M v^{'} = m v / 2$
$\Rightarrow v^{'} = \frac{m v}{2 M} = \sqrt{2 g h}$
$\Rightarrow h = \frac{m^{2} v^{2}}{8 M^{2} g}$

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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:

The correct option is B m2v2/8M2gApplying COLMmv+0=Mv′+mv/2Mv′=mv/2⇒v′=mv2M=√2gh⇒h=m2v28M2g

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:

The correct option is B $m^{2} v^{2} / 8 M^{2} g$
Applying COLM
$m v + 0 = M v^{'} + m v / 2$
$M v^{'} = m v / 2$
$\Rightarrow v^{'} = \frac{m v}{2 M} = \sqrt{2 g h}$
$\Rightarrow h = \frac{m^{2} v^{2}}{8 M^{2} g}$