Gun-Projectile Elastic Collision

[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]

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...

 

[scottdave]

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

 

[Calvin Pitts]

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?

 

[scottdave]

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

 

[berkeman]

Original thread for background... https://www.physicsforums.com/threads/gun-projectile-elastic-collisions.919690/

 

[Calvin Pitts]

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

 

[Hiero]

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.

 

[Calvin Pitts]

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).

 

[Hiero]

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.

 

[Calvin Pitts]

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]

 

[ehild]

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.

 

[Calvin Pitts]

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?

 

[Calvin Pitts]

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.

 

[ehild]

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.

 

[berkeman]

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

 

[Calvin Pitts]

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)

?

 

[ehild]

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)

 

[Calvin Pitts]

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.

 

[ehild]

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

 

[Calvin Pitts]

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.

 

[Calvin Pitts]

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

 

[ehild]

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.
Not bad for a first thread!

 

[Calvin Pitts]

The result is correct, although I can not follow your notations. Sooner or later you learn to type in the Maths.
Not bad for a first thread!

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. :(

 

[scottdave]

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.