Question

A shell of mass 200 g is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The velocity of the shell is:
\[(A)102{\text{ }}m{s^{ - 1}}\]
\[(B)82{\text{ }}m{s^{ - 1}}\;\]
\[(C)41{\text{ }}m{s^{ - 1}}\]
\[(D)120{\text{ }}m{s^{ - 1}}\]

Answer 1

Hint: Find a relation between the velocities of the gun and the shell after the explosion from the law of conservation of momentum. Note that the initial velocity of the gun is zero.
Find another relation between these velocities by applying the law of conservation of energy since the final energy is given.
From these two relations find the velocity of the shell as well as the gun.

Formula used:
If ${m_1}$ is the mass of the shell and ${m_2}$ is the mass of the gun and they gained velocities ${v_1}$ and ${v_2}$respectively after the explosion, from the law of conservation of momentum we can write,
${m_1}{v_1} = {m_2}{v_2}$
If $E$ is the final energy, From the law of conservation of energy, we can write
$\dfrac{1}{2}{m_1}{v_1}^2 + \dfrac{1}{2}{m_2}{v_2}^2 = E$

Complete step by step answer:
Let, ${m_1}$ is the mass of the shell and ${m_2}$ is the mass of the gun. After the explosion, if the velocity of the gun is ${v_1}$ and the velocity of the shell is ${v_2}$, then applying the law of the conservation of momentum we may write,
${m_1}{v_1} = {m_2}{v_2}............(1)$ [ since the initial velocity of the shell the gun is zero]
Given, ${m_1} = 200gm = 0.2kg$
${m_2} = 4kg$
Putting the values of ${m_1}$ and ${m_2}$ we get From eq. (1)
$\Rightarrow 0.2{v_1} = 4{v_2}$
$ \Rightarrow {v_2} = \dfrac{{0.2{v_1}}}{4}$
$ \Rightarrow {v_2} = \dfrac{{{v_1}}}{{20}}..............(2)$
They generate $E$ amount of energy after the explosion,
So, From the law of conservation of energy, we can write
$\Rightarrow \dfrac{1}{2}{m_1}{v_1}^2 + \dfrac{1}{2}{m_2}{v_2}^2 = E$
Now putting the value of ${v_2}$ in terms of ${v_1}$ from eq. (2)
$\Rightarrow {m_1}{v_1}^2 + {m_2}\dfrac{{{v_1}^2}}{{400}} = 2E$
Given, ${m_1} = 0.2kg$, ${m_2} = 4kg$
And, $E = 1.05KJ = 1050J$
$ \Rightarrow 0.2 \times {v_1}^2 + 4 \times \dfrac{{{v_1}^2}}{{400}} = 1050$
On dividing the terms and we get,
\[ \Rightarrow 0.2{v_1}^2 + 0.01{v_1}^2 = 2 \times 1050\]
Let us add the term and we get,
\[ \Rightarrow 0.21{v_1}^2 = 2100\]
On dividing the term and we get,
\[ \Rightarrow {v_1}^2 = \dfrac{{2100}}{{0.21}}\]
Hence we get,
\[ \Rightarrow {v_1}^2 = 10000\]
Taking square on both sides, we get
\[ \Rightarrow {v_1} = 100\]
So the velocity of the shell is \[ \Rightarrow {v_1} = 100m/s\].

Hence the right answer is in option $(A)$.

Note: The law of conservation of momentum states that the total momentum remains the same for a whole system. The momentum is the product of the mass and the velocity of an object.
The law of conservation of energy states that the total energy remains the same for a whole system.