# Gun-Projectile Elastic Collision

• Calvin Pitts
In summary, the conversation discusses the scenario of a projectile being fired from a gun held in rigid support and another scenario where the gun is free to recoil. The mass of the projectile and gun are given in both cases, and it is hinted that the kinetic energy of the system is the same in both scenarios. The conversation covers the concept of conservation of momentum and the importance of considering the total momentum of the system before and after the shot. The final equation for the momentum after the collision is given as m1v1 = -m2v2.
Calvin Pitts
1. Homework Statement

A projectile of 45kg has a muzzle speed of 656.6 m/s when fired horizontally from a gun held in rigid support (no recoil). What will be the muzzle speed (relative to the ground) of the same projectile when fired from a gun that is free to recoil? The mass of the gun is 6600 kg. (Hint: The kinetic energy of the gun-projectile system is the same in both cases.

## Homework Equations

(Pictures below)

3. The Attempt at a Solution

Calvin Pitts said:
View attachment 206793 View attachment 206794 1. Homework Statement
A projectile of 45kg has a muzzle speed of 656.6 m/s when fired horizontally from a gun held in rigid support (no recoil). What will be the muzzle speed (relative to the ground) of the same projectile when fired from a gun that is free to recoil? The mass of the gun is 6600 kg. (Hint: The kinetic energy of the gun-projectile system is the same in both cases.

## Homework Equations

(Pictures below)View attachment 206793
View attachment 206794
3. The Attempt at a Solution
View attachment 206795
I'm not really sure if I'm on the right track or not...

One way to approach this is that the total momentum created by firing the weapon will be the same, whether there is recoil or not. The first situation is simpler and you can find out the value of this momentum

scottdave said:
One way to approach this is that the total momentum created by firing the weapon will be the same, whether there is recoil or not. The first situation is simpler and you can find out the value of this momentum
So would that be as simple as adding the momentums together and calling it the momentum of the system?

Yes, that will give you a second equation. Then you have 2 equations and two unknowns.

So that's where I'm at now...(ignore that first line at the top)

It looks to me like you are trying to conserve the momenta between the two scenarios... this is not what conservation of momentum means.

In the recoil-less scenario, momentum is not conserved because outside forces are exerted (to hold the gun in place). We don't need to find any variables in this case though, because all the information is given!

In the case with recoil, all forces are taken to be absent, so then the momentum of the gun+bullet will be the same before and after firing.

scottdave
I feel like that's similar to the very last line, p = p'. I didn't think of it in that extent, though.

I'm getting really confused with this system of equations. Maybe I've just had too much calculus and I've forgotten my algebra, but I keep getting some type of proof, like v(projectile) = v(projectile).

Calvin Pitts said:
I feel like that's similar to the very last line, p = p'. I didn't think of it in that extent, though.
In that last line you are equating the net momentum of two different systems... why should they be equal??

Conservation of momentum is about comparing a single system at two different times (provided no outside forces act on the system... because that is the definition of force; the rate of change of momentum).

All the algebra in the world won't help if the physics isn't right.

scottdave and jbriggs444
Now I'm really confused. Haha I may need someone to explicitly tell me what to do in this case cause I'm just getting super thrown off at this point :( I have no idea how I've been passing this class[emoji23]

Calvin Pitts said:
Now I'm really confused. Haha I may need someone to explicitly tell me what to do in this case cause I'm just getting super thrown off at this point :( I have no idea how I've been passing this class[emoji23]
The mechanical energy supplied by the explosion of the gun-powder in the gun is the same in both scenarios. The momentum is conserved during the shot, but the system is the earth+gun+projectile in the first case, and the gun+projectile in the second case. You can use conservation of momentum in the second case (total momentum is the same before and after the shot) , and you get the energy of the system from the muzzle speed in the first case.

berkeman
ehild said:
The mechanical energy supplied by the explosion of the gun-powder in the gun is the same in both scenarios. The momentum is conserved during the shot, but the system is the earth+gun+projectile in the first case, and the gun+projectile in the second case. You can use conservation of momentum in the second case (total momentum is the same before and after the shot) , and you get the energy of the system from the muzzle speed in the first case.

So far so good?

ehild said:
You can use conservation of momentum in the second case (total momentum is the same before and after the shot) , and you get the energy of the system from the muzzle speed in the first case.
So wait, if the total momentum is the same before and after, wouldn't it just be zero? Because I thought the momentum before the shot would be zero because nothing is moving.

Calvin Pitts said:
View attachment 206849

So far so good?
Yes. Now you write the relation between v1 and v2 using conservation of momentum. And yes, the total momentum is zero before the shot.

Calvin Pitts said:
So wait, if the total momentum is the same before and after, wouldn't it just be zero? Because I thought the momentum before the shot would be zero because nothing is moving.
Correct, the sum before and after is zero. But remember that momentum is a vector quantity, having both magnitude and direction...

EDIT -- beaten out by ehild

ehild said:
Yes. Now you write the relation between v1 and v2 using conservation of momentum. And yes, the total momentum is zero before the shot.

So then for the momentum after the collision, I would have an equation looking like:
m(sub1)v(sub1) = -m(sub2)v(sub2)

?

Calvin Pitts said:
So then for the momentum after the collision, I would have an equation looking like:
m(sub1)v(sub1) = -m(sub2)v(sub2)

?
Yes. m1v1= -m2v2.
(Use x2 and x2 for subscript and superscript)

So long story short, I think I finally figured out the answer (with much pain and agony). From that last equation you put in, I solved for v1, plugged that equation in the kinetic energy equation and solved for the new velocity of the projectile, which came out to be 655.7 m/s. Seems like a valid answer. I definitely won't try to type out the work because the app doesn't support the math functions, but should I post a picture of my ugly work? Haha just thinking for future people who might stumble across this thread or for myself to see later.

Just post a picture of your work, ugly or not, it is useful.

ehild said:
Just post a picture of your work, ugly or not, it is useful.

I'm rewriting it now so it's a little cleaner :) I also rounded off a bit for that first time. I'm simplifying the equation in terms of variables right now so I can just go straight through with no rounding. I came up with 654.4 m/s.

Thanks to everyone for the help. I really like this little physics community :)

berkeman
Calvin Pitts said:
View attachment 206851

I'm rewriting it now so it's a little cleaner :) I also rounded off a bit for that first time. I'm simplifying the equation in terms of variables right now so I can just go straight through with no rounding. I came up with 654.4 m/s.
The result is correct, although I can not follow your notations. Sooner or later you learn to type in the Maths.

berkeman and Calvin Pitts
ehild said:
The result is correct, although I can not follow your notations. Sooner or later you learn to type in the Maths.

There's only so much math I can type. Haha I use the app which doesn't allow you to use a keyboard or click function. I have to type anything and everything. :(

Hiero said:
In that last line you are equating the net momentum of two different systems... why should they be equal??

Conservation of momentum is about comparing a single system at two different times (provided no outside forces act on the system... because that is the definition of force; the rate of change of momentum).

All the algebra in the world won't help if the physics isn't right.
I like the way you presented it, showing that the first scenario is a different system (because of the external force). I think the way I read the question, and the way that I said it may have been misleading into suggesting that the momentum in the first situation is the same as the second situation, when in fact, the momentum before the explosion is the same as after the explosion.

## 1. What is a gun-projectile elastic collision?

A gun-projectile elastic collision is a type of collision in which a gun fires a projectile (such as a bullet) and the projectile collides with a target or surface, bouncing off without any loss of kinetic energy.

## 2. How does the elasticity of the projectile affect the collision?

The elasticity of the projectile determines how much energy is conserved during the collision. In an elastic collision, the projectile will bounce off the target with the same velocity and kinetic energy as before the collision. In an inelastic collision, some of the energy is lost and the projectile may not bounce off with the same velocity.

## 3. What factors can influence the outcome of a gun-projectile elastic collision?

The outcome of a gun-projectile elastic collision can be influenced by factors such as the mass and velocity of the projectile, the angle of impact, and the elasticity of the materials involved. These factors can affect the amount of kinetic energy transferred during the collision.

## 4. How does the conservation of momentum apply to gun-projectile elastic collisions?

The conservation of momentum states that the total momentum of a system remains constant in the absence of external forces. In a gun-projectile elastic collision, the total momentum of the system (gun + projectile) before and after the collision is equal, as long as there are no external forces acting on the system.

## 5. Can a gun-projectile elastic collision be perfectly elastic?

Yes, a gun-projectile elastic collision can be perfectly elastic if there is no loss of kinetic energy during the collision. However, in real-world scenarios, some energy will always be lost due to factors such as air resistance and friction, making the collision technically inelastic.

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