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**Hint:**As the bullet is embedded in the block, the kinetic energy of the combined block and bullet system can be calculated by using the formula

$ k\cdot E\text{ = }\dfrac{1}{2}\text{ m}{{\text{v}}^{2}} $

The value of $ v $ here is calculated by applying the principle of conservation of linear momentum.

Next, as the system stops at a distance, so the kinetic energy loss due to friction be made equal to the work done by using the initial condition given in the question.From here, the value of $ h $ can be calculated.

**Complete step-by-step solution**

From conservation of linear momentum, we know that:

$ {{m}_{1}}\text{ v = }\left( {{m}_{2}}+{{m}_{1}} \right)\text{ v} $

here $ \begin{align}

& m1=\text{ mass of bullet} \\

& m2\text{= mass of black} \\

\end{align} $

kinetic energy of combined black and bullet system is given by:

$ E\text{ = }\dfrac{1}{2}\text{ }\left( {{m}_{1}}+{{m}_{2}} \right)\text{ }{{\left( v \right)}^{2}} $

New, from equation 1 , we get

$ \text{v = }\dfrac{{{m}_{1}}v}{\left( {{m}_{1}}+{{m}_{2}} \right)} $

Putting this value in equation 2, we get:

$ \text{E = }\dfrac{1}{2}\text{ }\left( {{m}_{1}}+{{m}_{2}} \right)\text{ }\dfrac{{{m}_{1}}^{2}{{v}^{2}}}{{{\left( {{m}_{1}}+{{m}_{2}} \right)}^{2}}\text{ }} $

$ E\text{ = }\dfrac{{{\left( {{m}_{1}}v \right)}^{2}}}{2\left( {{m}_{1}}+{{m}_{2}} \right)} $

As the system stops at a distance on the table, so

$ K\cdot E $ loss due to friction = work done

So, $ H\left( {{m}_{1}}+{{m}_{2}} \right)\text{ }{{g}^{\And }}\text{ }\dfrac{{{\left( {{m}_{1}}v \right)}^{2}}}{2\left( {{m}_{1}}+{{m}_{2}} \right)} $

As the freely falling object acquires speed $ \left( \dfrac{v}{10} \right) $ when it falls a distance h

Therefore, $ {{v}^{2}}-{{u}^{2}}=2gh $

$ u=0,\text{ v = }\dfrac{v}{10} $

So, $ {{\left( \dfrac{v}{10} \right)}^{2}}=2gh $

$ {{v}^{2}}=200gh $

Putting this value in equation 4, we get

$ H\left( {{m}_{1}}+{{m}_{2}} \right)gs=\dfrac{{{\left( {{m}_{1}} \right)}^{2}}\left( 200gh \right)}{2\left( {{m}_{1}}+{{m}_{2}} \right)} $

$ H=\dfrac{2m{{g}^{\And }}{{\left( {{m}_{1}}+{{m}_{2}} \right)}^{2}}}{200g\text{ m}{{\text{1}}^{2}}} $

= $ \dfrac{2\times 0\cdot 05\times {{\left( 10\cdot 5 \right)}^{2}}}{100\times {{\left( 0\cdot 5 \right)}^{2}}} $

= $ 40\cdot 4m $

$ H=0\cdot 04km $ .

**Note**

Consider that a body is lying on a Horizontal surface. If R is the normal reaction and F is the limiting static friction, then

$ F\propto r $

$ F=\mu r $

On a horizontal surface, $ \mu k $ is the coefficient of kinetic friction between the two surfaces in contact. The force of friction between the block and horizontal surface is given by:

$ F=\mu r $ R

= $ \mu r\text{ }mg $

To move the block without acceleration, the force required will be just equal to the force of friction

$ P=F=\mu rmg $

If $ s $ is the distance moved, then work done is given by

$ W=P\times S $

$ w=\mu k\text{ mg }s $ .

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