# Center of Mass and Linear Momentum of pellet gun

• michaelmayhem
In summary, a pellet gun fires ten 2.0 g pellets per second with a speed of 500 m/s. The pellets are stopped by a rigid wall. The magnitude of the momentum of each pellet is found using the formula P=mv, and the kinetic energy is found using the formula K=1/2mv^2. The magnitude of the average force on the wall from the stream of pellets can be calculated by multiplying the force of each pellet by the number of pellets hitting the wall per second. This is different from the average force calculated in (c) because it is based on a different time interval. To solve this problem, one can use the impulse-momentum theorem and the relationship between impulse, force, and change
michaelmayhem
1. A pellet gun fires ten 2.0 g pellets per second with a speed of 500 m/s. The pellets are stopped by a rigid wall. What are (a) the magnitude of the momentum of each pellet, (b) the kinetic energy of each pellet, and (c) the magnitude of the average force on the wall from the stream of pellets? (d) If each pellet is in contact with the wall for 0.60 ms, what is the magnitude of the average force on the wall from each pellet during contact? (e) Why is this average force so different from the average force calculated in (c)?

2. Can someone please explain to me step by step how I might go about solving this problem? I am stuck. Thanks. FYI this isn't really homework due tomorrow but I though I would try and get ahead.

3. I was going to post my work here but I'm sure it's completely wrong.

Welcome to PF Michael. It's a good thing that you're trying to get ahead; now that you got stuck you have plenty of time to really understand the problem without the pressure of time.

a) and b) are rather easy, basically it's just plugging given numbers into the correct formula. Can you write down the formula for momentum and kinetic energy (you should know it by heart)?

Actually for c) I think you cannot really calculate the number, unless I am missing something. For d) you can again use an appropriate formula, which relates the average force to the change in momentum... can you find that formula?

well K=1/2mv^2
and P=mv
I'm not really sure how to incorporate these into the equation.
I'm not really sure about the formula that relates average force to the change in momentum.

CompuChip said:
Actually for c) I think you cannot really calculate the number, unless I am missing something.
The only difference between finding the average force in c) versus the average force in d) is the time interval over which you are averaging. For c) consider a time interval of seconds.

For michaelmayhem: Look up the impulse-momentum theorem.

Oh yes the impulse is equal to the change in momentum. I can find the impulse by multiplying the force times the change in time. The momentum is simply P=mv.

Doc Al said:
The only difference between finding the average force in c) versus the average force in d) is the time interval over which you are averaging. For c) consider a time interval of seconds.
Yes, I thought something like that too, but to really calculate anything quantitatively you would need the number (e.g. the distance from the wall or the time-of-flight). Assuming that t ~ seconds will only give you an estimate.

michaelmayhem said:
Oh yes the impulse is equal to the change in momentum. I can find the impulse by multiplying the force times the change in time. The momentum is simply P=mv.

Very well, that's the results I was hinting at.

CompuChip said:
Yes, I thought something like that too, but to really calculate anything quantitatively you would need the number (e.g. the distance from the wall or the time-of-flight). Assuming that t ~ seconds will only give you an estimate.
You are given that 10 bullets, of known momentum, are fired every second--thus 10 bullets hit the wall every second. That's all you need to calculate the average force.

Doc Al said:
You are given that 10 bullets, of known momentum, are fired every second--thus 10 bullets hit the wall every second. That's all you need to calculate the average force.

OK that was kind of stupid... completely missed that piece of info.

Ok thank you. I'll give it another go.

## What is the center of mass of a pellet gun?

The center of mass of a pellet gun is the point at which the mass of the gun is evenly distributed in all directions. It is the balance point of the gun and can be found by balancing the gun on a pencil or by using mathematical calculations.

## Why is the center of mass important in a pellet gun?

The center of mass is important in a pellet gun because it determines how the gun will behave when fired. If the center of mass is not aligned with the barrel, the gun may have difficulty shooting accurately. Additionally, knowing the center of mass can help determine how the gun will recoil and how it should be held for optimal stability.

## How is the center of mass calculated for a pellet gun?

The center of mass for a pellet gun can be calculated by finding the weighted average of all the mass points in the gun. This can be done by dividing the gun into smaller sections and finding the center of mass for each section, then combining them to find the overall center of mass.

## What is linear momentum in relation to a pellet gun?

Linear momentum is a measure of the motion of an object in a straight line. In a pellet gun, it refers to the velocity of the pellet as it is fired from the gun. The more momentum the pellet has, the farther it will travel and the more force it will have upon impact.

## How does the center of mass affect the linear momentum of a pellet gun?

The location of the center of mass can affect the linear momentum of a pellet gun. If the center of mass is not aligned with the barrel, the gun may have difficulty achieving high velocities and therefore lower linear momentum. Additionally, if the center of mass is too far forward or too far back, it can cause the gun to recoil in an undesirable way, affecting the linear momentum of the pellet.

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