
What is the total momentum of the bullet and the gun before firing?
Answer
466.8k+ views
Hint: To answer this question, we will be using the concept of conservation of momentum. And we know that conservation of momentum says that the total momentum of a system of objects remains constant during any interaction, if no external force acts on the system.
Complete step by step answer:
When a bullet is shot from a gun, both the bullet and the gun are initially at rest, with zero total momentum. When a bullet is fired, it gains forward motion. The cannon obtains a backward momentum as a result of momentum conservation. With a forward velocity of $v$, a bullet with mass $m$ is discharged. The mass $M$ gun achieves a rearward velocity $V$. The overall momentum before firing is zero, and the total momentum after firing is also zero.
Let us assume, $m$ be the mass of the bullet and $M$ be the mass of the gun. If $v$ is the velocity of the bullet and $V$ is the velocity of the gun, after the firing then, we can say that,
$0 = mv + MV \\
\Rightarrow V = - \dfrac{{mv}}{M} $
Here, $V$ is the recoil velocity of the gun. And here the mass of the bullet is very less as compared to the gun i.e., $m \ll M$ . And the backward velocity of the gun is very small i.e., $v \ll V$ . And here the negative sign says that the gun moves in the direction to that of the bullet.
Note: We must have a thorough comprehension of all three Newton's laws of motion in order to answer these types of problems. We must understand the ideas of forces and momentum, as well as how to calculate them. We must be careful with the sign conventions because these are all vector quantities.
Complete step by step answer:
When a bullet is shot from a gun, both the bullet and the gun are initially at rest, with zero total momentum. When a bullet is fired, it gains forward motion. The cannon obtains a backward momentum as a result of momentum conservation. With a forward velocity of $v$, a bullet with mass $m$ is discharged. The mass $M$ gun achieves a rearward velocity $V$. The overall momentum before firing is zero, and the total momentum after firing is also zero.
Let us assume, $m$ be the mass of the bullet and $M$ be the mass of the gun. If $v$ is the velocity of the bullet and $V$ is the velocity of the gun, after the firing then, we can say that,
$0 = mv + MV \\
\Rightarrow V = - \dfrac{{mv}}{M} $
Here, $V$ is the recoil velocity of the gun. And here the mass of the bullet is very less as compared to the gun i.e., $m \ll M$ . And the backward velocity of the gun is very small i.e., $v \ll V$ . And here the negative sign says that the gun moves in the direction to that of the bullet.
Note: We must have a thorough comprehension of all three Newton's laws of motion in order to answer these types of problems. We must understand the ideas of forces and momentum, as well as how to calculate them. We must be careful with the sign conventions because these are all vector quantities.
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