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

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 $ .