Question

A bullet of mass \[4.2\times {{10}^{-2}}kg\], moving at a speed of \[300m{{s}^{-1}}\], gets stuck into a block with a mass 9 times that of the bullet. If the block is free to move without any kind friction, the heat generated in the process will be
A. 45 cal
B. 405 cal
C. 450 cal
D. 1701 cal

Answer 1

Hint: According to the conservation of linear momentum, if the net external force \[\left( {{F}_{ext.}} \right)\] acting on a system of bodies is zero, then the momentum of the system remains constant. i.e.,
Initial momentum = final momentum \[\left( {{p}_{i}}={{p}_{f}} \right)\].
Here friction is absent so that the heat produced will be equal to the change in kinetic energy.

Complete step by step answer:
Given that:
Mass of bullet, \[m=4.2\times {{10}^{-2}}kg\]
Velocity of bullet, \[v=300m{{s}^{-1}}\]
Mass of block, \[M=9m=9\times 4.2\times {{10}^{-2}}kg\]
Let the velocity of the combined system will be, \[V\] (when bullet gets stuck inside the block)
Now on applying conservation of linear momentum i.e., \[{{p}_{i}}={{p}_{f}}\]
\[\Rightarrow mv=\left( m+M \right)V\]
On putting value of \[M\],
\[\Rightarrow mv=\left( m+9m \right)V\]
\[\Rightarrow mv=10mV\]
\[\Rightarrow V=\dfrac{v}{10}\]
As there is no any kind of friction, so heat generated will be equal to the change in kinetic energy i.e.,
\[H=\Delta \left( K.E. \right)\]
\[\Rightarrow H=\dfrac{1}{2}m{{v}^{2}}-\dfrac{1}{2}\times \left( 10m \right){{V}^{2}}\]
On putting value of \[V\],
\[\Rightarrow H=\dfrac{1}{2}m{{v}^{2}}-\dfrac{1}{2}\times \left( 10m \right){{\left( \dfrac{v}{10} \right)}^{2}}\]
\[\Rightarrow H=\dfrac{1}{2}m{{v}^{2}}-\dfrac{1}{2}\times m\times \dfrac{{{v}^{2}}}{10}\]
On solving,
\[\Rightarrow H=\dfrac{9}{20}m{{v}^{2}}\]
Now, on putting values of \[m\] and \[v\],
\[\Rightarrow H=\dfrac{\text{9}}{\text{20}}\text{ }\!\!\times\!\!\text{ 4}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{300} \right)}^{\text{2}}}\text{joule}\]
On solving,
\[\Rightarrow H=\text{1701 joule}\]
As we know that, \[\text{1 calorie = 4}\text{.2 joule}\]
So, heat generated in the process will be,
\[\Rightarrow H=\dfrac{\text{1701}}{\text{4}\text{.2}}\text{ calorie}\]
\[\Rightarrow H=\text{405 calorie}\]
Hence, the correct option is B, i.e., 405 cal.

Note: Students should understand the conservation of linear momentum i.e., \[{{p}_{i}}={{p}_{f}}\] (mathematically) and they need to know that the conservation of linear momentum can only be applied in the case where external force \[\left( {{F}_{ext.}} \right)\] is zero. Students also keep in mind that if there is absence of every kind of friction then the heat produced will be equal to the change in kinetic energy.