Calculating Momentum of Propellant Gases in a Rifle Recoil

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In summary: When solving for Σmv you use the quadratic equation: Σmv=a(x-b)2+cIn this equation, a is the coefficient of restitution, or the amount of momentum that the object recaptures after it has been released, and c is the coefficient of acceleration. So in this case, the equation would say that there is a total of (-1.95)(-.0072) mv missing, or -1.95 mv that has been lost.In summary, the recoil from firing a rifle causes momentum to be lost.
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Homework Statement


The expanding gases that leave the muzzle of a rifle also contribute to the recoil. A .30 caliber bullet has mass 7.20×10−3kg and a speed of 601 m/s relative to the muzzle when fired from a rifle that has mass 3.10 kg . The loosely held rifle recoils at a speed of 1.95 m/s relative to the earth.

Find the momentum of the propellant gases in a coordinate system attached to the Earth as they leave the muzzle of the rifle.

I'm not totally sure how to set up the equations because I can't really visualize what the problem is asking?

Homework Equations

The Attempt at a Solution



m1=.0072 kg
m2=3.1 kg
v1i= 0 m/s
v1f=601 m/s
v2i = ?
v2f=-1.95 m/s
 
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  • #2
You need to apply momentum conservation here.From the perspective of the ground both the bullet and the rifle were initially at rest.Now after firing the relative velocity of the bullet is 601m/s relative to the rifle which has a relative velocity of 1.95m/s with respect to the ground.So what is the relative velocity of the bullet with respect to the ground?.Its (601-1.95)m/s.Now apply momentum conservation and see how much momentum is missing from one side.
 
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ooohffff said:
I can't really visualize what the problem is asking?
It is saying that three things move as a result of the firing of the bullet:
  • The bullet
  • The gun
  • The exhaust gases from the explosive charge
Each of these has momentum. Since they form a closed system, the sum of their momenta is...?
 
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  • #4
Daymare said:
You need to apply momentum conservation here.From the perspective of the ground both the bullet and the rifle were initially at rest.Now after firing the relative velocity of the bullet is 601m/s relative to the rifle which has a relative velocity of 1.95m/s with respect to the ground.So what is the relative velocity of the bullet with respect to the ground?.Its (601-1.95)m/s.Now apply momentum conservation and see how much momentum is missing from one side.
I may be wrong, but I interpreted ooohfff's post as seeking clarification of the question rather than hints on how to solve it.
But maybe it's the relative velocity aspect that was the block.
 
  • #5
Daymare said:
You need to apply momentum conservation here.From the perspective of the ground both the bullet and the rifle were initially at rest.Now after firing the relative velocity of the bullet is 601m/s relative to the rifle which has a relative velocity of 1.95m/s with respect to the ground.So what is the relative velocity of the bullet with respect to the ground?.Its (601-1.95)m/s.Now apply momentum conservation and see how much momentum is missing from one side.

Great, thanks...I forgot to subtract the 1.95 but fixed it. The other problem I have is that I get the right number for the missing momentum but I get the negative of it when it should be positive? I plug in mv=(.0072)(599.05)+(3.1)(-1.95)
 
  • #6
ooohffff said:
mv=(.0072)(599.05)+(3.1)(-1.95)
It is easy to confuse yourself when writing balance equations that way. Safer is the form Σmivi=0.
 

FAQ: Calculating Momentum of Propellant Gases in a Rifle Recoil

What is the "Momentum rifle problem"?

The "Momentum rifle problem" is a thought experiment in physics that involves a rifle with a large mass and a bullet with a small mass. The problem arises when the rifle and bullet are fired, and the bullet has a much higher velocity than the rifle. This results in a large momentum for the bullet, but a much smaller momentum for the rifle. The question then becomes, where does the extra momentum from the bullet go?

What is the principle of conservation of momentum?

The principle of conservation of momentum states that in a closed system, the total momentum remains constant. This means that the initial momentum of an object must be equal to the final momentum of that object and any other objects involved in the system. In the "Momentum rifle problem", the momentum of the rifle and bullet combined before firing must be equal to the momentum of the bullet after firing.

How does the "Momentum rifle problem" relate to Newton's third law of motion?

The "Momentum rifle problem" is a demonstration of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. In this case, the action is the firing of the bullet, which results in a reaction force on the rifle. The rifle moves in the opposite direction of the bullet, but with a much smaller momentum due to its larger mass.

Can the "Momentum rifle problem" be solved using the law of conservation of energy?

No, the law of conservation of energy cannot be used to solve the "Momentum rifle problem" because energy is not conserved in this scenario. When the bullet is fired, some energy is lost due to friction and air resistance, so the total kinetic energy of the system before and after firing is not equal.

What are some real-life applications of the "Momentum rifle problem"?

The "Momentum rifle problem" can be applied to understanding the recoil of firearms and the impact of projectiles on different targets. It is also used in the design and testing of rocket propulsion systems and other forms of propulsion, as well as in the study of collisions between objects with different masses.

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