Question

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 1

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 $2050 \times \dfrac{{40}}{{41}} = 2000J$.