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# A bullet of mass is fired from the gun, the acceleration of the bullet isA. Greater than acceleration of the gunB. Equal to the acceleration of the gunC. Equal but opposite to the acceleration of the gunD. Directly proportional to the mass of the bulletE. Inversely proportional to the force of the bullet

Last updated date: 15th Jun 2024
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Hint:Acceleration is defined as the rate of change of velocity with respect to time. Law of conservation of linear momentum states that if no external force acts on a system of constant mass, the total momentum of the system remains constant with time.

Law of Conservation of Linear Momentum:If no external force is applied on a system (called isolated) of constant mass, the total momentum of the system remains constant with time.According to Newton's second law of motion, $F = \dfrac{{dp}}{{dt}}$
In the absence of any external force, $F = 0$, then $p =$constant
Which means that for all the objects present in the isolated system,
${p_1} + {p_2} + {p_3}..... =$constant
In this case the gin and the bullet can be considered as one isolated system, where the mass of the bullet is $m$ and its velocity is$v$. And let the mass of the gun be ${m_G}$ and its velocity be ${v_G}$. Then,
${p_1} + {p_2} =$constant
$mv + {m_G}{v_G} =$constant . . . . . . . (equation 1)

Differentiating the equation 1 with respect to time,
$m\dfrac{{dv}}{{dt}} + {m_G}\dfrac{{d{v_G}}}{{dt}} = 0$
$\Rightarrow ma + {m_G}{a_G} = 0$
Where $a$denotes the acceleration of the bullet and ${a_G}$ denoted the acceleration of the gun.
$ma = - {m_G}{a_g}$ . . . . . . . . (equation 2)
In equation two, the negative sign denotes the direction of the force. Which means that the force on the bullet will be in the opposite direction of that of the gun.
From equation two,
$\dfrac{m}{{{m_G}}} = - \dfrac{{{a_G}}}{a}$ . . . . . . . . (equation 3)
The above equation tells us that the acceleration of the bullet will be inversely proportional to the mass of the bullet and will be equal to that of the gun only when the mass of gun and bullet are the same.So, options B, C and D are eliminated.

Equation two can also be written as $F = - {F_G}$ . . . . (equation 4), where $F$ is the force on the bullet and ${F_G}$ is the force on the gun. So,
$F = ma \\ a = \dfrac{F}{m} \\$
So acceleration of the bullet is directly proportional to the force on it, and hence option E is also eliminated.Looking back on equation three,
$\therefore\dfrac{a}{{{a_G}}} = - \dfrac{{{m_G}}}{m}$
It tells us that the acceleration of the bullet will depend on the ratio of mass of gun to mass of bullet. And since the ratio is far far greater than one because the mass of the gun is larger than that of the bullet.

Therefore option A is correct.

Note:Law of Conservation of Linear Momentum works only in an isolated system. If any kind of external force acts inside the system the conservation law does not work. Also the conservation law proves Newton’s Third Law of Motion. If we look at equation four closely, it tells us that the force acting on the bullet is equal in magnitude to the force action on the gun but in the opposite direction, and that is Newton’s Third Law of Motion.