Question

1. imagine that you are out turkey hunting for thanksgiving. upon firing your gun, the bullet (2.6 grams) gains a forward velocity 300 $\frac{m}{s}$.

a. what type of \collision\ is occurring when the bullet leaves the gun?

b. how much forward momentum has the bullet gained? be sure to convert the bullets mass to kilograms.

c. how much backward momentum will the rifle gain?

d. for a standard 3.0 kg rifle, how quickly will it move backward?

e. the rifle moving backward is called recoil which can lead to shoulder bruises for inexperienced hunters. if the rifle is hugged tightly to the shoulder, the mass of the hunter can be added to the rifle’s mass. the combined mass of the hunter and rifle is 83 kg. how quickly will the hunter-rifle combination move backward when the rifle is fired?

Explanation:

Part a

When a bullet is fired from a gun, the internal explosion (propellant burning) causes the bullet and gun to separate, and momentum is conserved. This is a perfectly inelastic collision in reverse (or an explosion - a type of collision where the objects separate after being together, and momentum is conserved, kinetic energy increases). The key is that initially, the bullet and gun are at rest (combined mass), and then they separate with the bullet moving forward and the gun (rifle) moving backward. So the type of "collision" is an explosion (or a perfectly inelastic collision in reverse, but explosion is more accurate for the firing of a gun as energy is released to separate them).

Step1: Recall the formula for momentum

The formula for momentum is $p = mv$, where $p$ is momentum, $m$ is mass, and $v$ is velocity. First, we need to convert the mass of the bullet from grams to kilograms.
Given the mass of the bullet $m = 2.6$ grams. Since $1$ kilogram $= 1000$ grams, we convert grams to kilograms by dividing by $1000$. So $m=\frac{2.6}{1000}=0.0026$ kg. The velocity of the bullet $v = 300$ m/s.

Step2: Calculate the momentum

Using the formula $p = mv$, substitute $m = 0.0026$ kg and $v = 300$ m/s.
$p=(0.0026\space kg)\times(300\space m/s)= 0.78\space kg\cdot m/s$

According to the law of conservation of momentum, the total initial momentum of the system (bullet + rifle) is zero (since they are at rest initially). After firing, the total final momentum should also be zero. So the momentum of the bullet forward and the momentum of the rifle backward must be equal in magnitude (and opposite in direction) to satisfy $p_{bullet}+p_{rifle}=0$. Therefore, the backward momentum of the rifle is equal in magnitude to the forward momentum of the bullet.

Answer:

Explosion (or a type of collision where the objects (bullet and rifle) were initially at rest together and then separate, conserving momentum, often referred to as an explosion in the context of gun firing)

Part b