Question

From a rifle of mass \[4{\text{ }}kg\], a bullet of mass \[50{\text{ }}g\] is fired with an initial velocity of \[35{\text{ }}m{s^{ - 1}}\]. Calculate the initial recoil velocity of the rifle.

Answer 1

Hint: To address this query, we will first assume the mass of the rifle and the mass of the bullet, as well as the gun's Recoil velocity with a specified term, and then calculate the overall initial momentum of the rifle and bullet system, as well as the combined momentum of both after firing. Finally, using the rule of conservation of momentum, we will determine the gun's recoil velocity.

Complete step by step answer:
Let us consider mass of the given rifle be ${m_1}$ and mass of the bullet be ${m_2}$.
Now, it is given to us in the question that;
Mass of the rifle $({m_1})$ = $4\,kg$.
Mass of the bullet $({m_2})$ = $50\,g = 0.5\,kg$.
Let us consider the Recoil velocity of the gun is = ${v_1}$.
And the bullet is fired with an initial velocity of $({v_2})$ = $35\,m{s^{ - 1}}$.
So, here for the given question we need to find out the initial recoil velocity of the gun.
The rifle is initially at rest. As a result, its initial velocity \[\left( v \right)\] is zero.
The rifle's and bullet system's overall initial momentum,
$\left( {{m_1} + {m_2}} \right) \times v = 0$

After firing, the combined momentum of the rifle and bullet system
${m_1}{v_1} + {m_2}{v_2} \\
\Rightarrow \left( {4 \times {v_1}} \right) + \left( {0.05 \times 35} \right) \\
\Rightarrow 4{v_1} + 1.75 $
According to the law of momentum conservation,
Total momentum after the firing = Total momentum before the firing
$4{v_1} + 1.75 = 0 \\
\Rightarrow {v_1} = \dfrac{{ - 1.75}}{4} \\
\therefore {v_1} = - 0.4375m{s^{ - 1}} \\ $
Therefore, the gun recoils backwards with a velocity of \[0.4375{\text{ }}m{s^{ - 1}}\] , as indicated by the negative sign.

Note: In this question we have applied the conservation of linear momentum inorder to find the recoil velocity of the gun because there was no external force. In some cases there might be some external forces acting on the system then we cannot apply conservation of momentum. Students must take care of this thing while dealing with these types of numericals related to momentum conservation.