
A bullet of mass 50 g is fired from a rifle of mass 2 kg and the total kinetic energy produced by the explosion is 2050 J. The kinetic energy of the bullet is.
A) 50 J
B) 2000 J
C) 2050 J
D) 587.7 J
Answer
574.5k+ views
Hint: As no external forces have been applied so the momentum of the rifle-bullet system will be conserved. From the momentum conservation, you’ll get a relationship between the velocities of the bullet and the rifle. Use it in the kinetic energy expression to get the result.
Formulae used:
Conservation of linear momentum
\[mv + MV = 0\] …………...(1)
Where,
$m$ is the mass of the bullet
$v$ is the velocity of the bullet
$M$ is the mass of the riffle
$V$ is the velocity of the riffle
Total Kinetic Energy expression
$E = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}M{V^2}$ …………….(2)
Where,
$E$ is the total kinetic energy of the system
Complete step by step answer:
Given,
Mass of the bullet, \[m = 50g = 0.05kg\].
Mass of the riffle, $M = 2kg$.
Total kinetic energy, $E = 2050J$.
To find Energy of the bullet.
Step 1
Substituting the given values of m and M in eq (1) to get the velocity relation
\[
0.05 \times v + 2 \times V = 0 \\
\Rightarrow V = - \dfrac{{0.05}}{2}v = - \dfrac{v}{{40}} \\
\] ……. (3)
(Using the relation in fraction form for easier calculation)
Step 2
Now, substitute the values of m, M and T along with the relation of eq(3) in eq(2)
$
2050 = \dfrac{1}{2} \times 0.05 \times {v^2} + \dfrac{1}{2} \times 2 \times {\left( { - \dfrac{v}{{40}}} \right)^2} \\
\Rightarrow \dfrac{{{v^2}}}{{40}} + \dfrac{{{v^2}}}{{1600}} = 2050 \\
\Rightarrow {v^2} \times \left( {\dfrac{{40 + 1}}{{1600}}} \right) = 2050 \\
\Rightarrow {v^2} = \dfrac{{2050 \times 1600}}{{41}} = 80000{(m/s)^2} \\
$
Step 3
Hence, get the kinetic energy of the bullet as $\dfrac{1}{2}m{v^2} = \dfrac{1}{2} \times 0.05 \times 80000 = 2000J$
The kinetic energy of the bullet is 2000 J. So, Option (B) is correct.
Note:
Questions like these can be solved in a tricky manner. From conservation of momentum notice that the ratio of velocities is inverse to the ratio of the masses with a negative sign (Here, \[\dfrac{m}{M} = - \dfrac{V}{v}\]). The negative sign indicates that the direction of the velocity of the bullet and rifle will be opposite in nature. Now, check the ratio of their energy.
\[\dfrac{{\tfrac{1}{2}m{v^2}}}{{\tfrac{1}{2}M{V^2}}} = \dfrac{m}{M} \times {\left( {\dfrac{v}{V}} \right)^2} = \dfrac{m}{M} \times {\left( { - \dfrac{M}{m}} \right)^2} = \dfrac{M}{m}\]
Hence, the Energy ratio is exactly opposite to their mass ratio. Here, it will be $2:0.05 = 40:1$. Since, total kinetic energy is 2050 J, so the kinetic energy of the bullet will be $2050 \times \dfrac{{40}}{{41}} = 2000J$.
Formulae used:
Conservation of linear momentum
\[mv + MV = 0\] …………...(1)
Where,
$m$ is the mass of the bullet
$v$ is the velocity of the bullet
$M$ is the mass of the riffle
$V$ is the velocity of the riffle
Total Kinetic Energy expression
$E = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}M{V^2}$ …………….(2)
Where,
$E$ is the total kinetic energy of the system
Complete step by step answer:
Given,
Mass of the bullet, \[m = 50g = 0.05kg\].
Mass of the riffle, $M = 2kg$.
Total kinetic energy, $E = 2050J$.
To find Energy of the bullet.
Step 1
Substituting the given values of m and M in eq (1) to get the velocity relation
\[
0.05 \times v + 2 \times V = 0 \\
\Rightarrow V = - \dfrac{{0.05}}{2}v = - \dfrac{v}{{40}} \\
\] ……. (3)
(Using the relation in fraction form for easier calculation)
Step 2
Now, substitute the values of m, M and T along with the relation of eq(3) in eq(2)
$
2050 = \dfrac{1}{2} \times 0.05 \times {v^2} + \dfrac{1}{2} \times 2 \times {\left( { - \dfrac{v}{{40}}} \right)^2} \\
\Rightarrow \dfrac{{{v^2}}}{{40}} + \dfrac{{{v^2}}}{{1600}} = 2050 \\
\Rightarrow {v^2} \times \left( {\dfrac{{40 + 1}}{{1600}}} \right) = 2050 \\
\Rightarrow {v^2} = \dfrac{{2050 \times 1600}}{{41}} = 80000{(m/s)^2} \\
$
Step 3
Hence, get the kinetic energy of the bullet as $\dfrac{1}{2}m{v^2} = \dfrac{1}{2} \times 0.05 \times 80000 = 2000J$
The kinetic energy of the bullet is 2000 J. So, Option (B) is correct.
Note:
Questions like these can be solved in a tricky manner. From conservation of momentum notice that the ratio of velocities is inverse to the ratio of the masses with a negative sign (Here, \[\dfrac{m}{M} = - \dfrac{V}{v}\]). The negative sign indicates that the direction of the velocity of the bullet and rifle will be opposite in nature. Now, check the ratio of their energy.
\[\dfrac{{\tfrac{1}{2}m{v^2}}}{{\tfrac{1}{2}M{V^2}}} = \dfrac{m}{M} \times {\left( {\dfrac{v}{V}} \right)^2} = \dfrac{m}{M} \times {\left( { - \dfrac{M}{m}} \right)^2} = \dfrac{M}{m}\]
Hence, the Energy ratio is exactly opposite to their mass ratio. Here, it will be $2:0.05 = 40:1$. Since, total kinetic energy is 2050 J, so the kinetic energy of the bullet will be $2050 \times \dfrac{{40}}{{41}} = 2000J$.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

