Question

A block of mass of $10kg$ rests on a horizontal table. The coefficient of friction between the black and the table is $0\cdot 05$ . When hit by a bullet of mass $50g$ moving with speed u, that gets embedded in it, the block moves and comes to rest after moving a distance of $2m$ on the table. If a freely falling object were to acquire speed $\dfrac{v}{10}$ after being dropped from height h, then neglecting energy losses and taking $g=\text{ }10\text{ m/}{{\text{s}}^{2}}$ , the value of $H$ is close to:
A) $0\cdot 04\text{ km}$
B) $0\cdot 05\text{ }km$
C) $0\cdot 02\text{ }km$
D) $0\cdot 03\text{ }km$

Answer 1

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 $\begin{aligned} & m1=\text{ mass of bullet} \\\ & m2\text{= mass of black} \\\ \end{aligned}$
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
$W=P\times S$
$w=\mu k\text{ mg }s$ .