Impact force of Rigid objects and no displacement

In summary: Force and kinetic energy aren't directly related to each other in this manner. You don't even need to know the kinetic energy of an object to figure out the force applied during a collision. Indeed, it doesn't even help you in many cases since it isn't a conserved quantity (total energy is, but not kinetic energy). Also note that neither momentum nor kinetic energy are affected by mass.
  • #36
normal_force said:
I would ask you, how would YOU solve it.

You can't solve it under the conditions you've specified because they are unphysical. The answer, if we attempted to use real physical laws to find it, would be meaningless. A bullet cannot strike a target and transfer all of its kinetic energy to that target unless the collision is totally inelastic. But then the bullet and the target MUST deform and most of the KE will be lost as heat and as work done during the deformation process.

There is no idealized scenario which will change this and you can't handwave any of this away. If you use perfectly rigid objects then you can't have an inelastic collision and your bullet WILL rebound.

Nothing else is physically possible.
 
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  • #37
Drakkith said:
You can't solve it under the conditions you've specified because they are unphysical. The answer, if we attempted to use real physical laws to find it, would be meaningless. A bullet cannot strike a target and transfer all of its kinetic energy to that target unless the collision is totally inelastic. But then the bullet and the target MUST deform and most of the KE will be lost as heat and as work done during the deformation process.

There is no idealized scenario which will change this and you can't handwave any of this away. If you use perfectly rigid objects then you can't have an inelastic collision and your bullet WILL rebound.

Nothing else is physically possible.
Okay, that is true...but I've changed the situation if you have noticed...the bullet DOES deform but not by a whole lot but it doesn't penetrate its target, it may deform it a little...but WE are saying...like how a hammer and nail work...

Like, how a baseball will do work on a baseball glove...displacing it...

Like this right here...



at 0:48 you can just skip to

and ignore the political stuff obviously...
 
  • #38
normal_force said:
Okay, that is true...but I've changed the situation if you have noticed...the bullet DOES deform but not by a whole lot but it doesn't penetrate its target, it may deform it a little...

Okay. So now I think you'd need to know how the bullet responds to stress (stiffness? elasticity?), its velocity, and the mass of both the bullet and target in order to find out how long the collision takes and what the average force is.
 
  • #39
Drakkith said:
Okay. So now I think you'd need to know how the bullet responds to stress (stiffness? elasticity?), its velocity, and the mass of both the bullet and target in order to find out how long the collision takes and what the average force is.
Now we are getting some where, we can apply those, I just need the equation. I also assume you saw the video?
 
  • #40
normal_force said:
Now we are getting some where, we can apply those, I just need the equation.

Wish I could help you. I don't know how to figure out and apply the elasticity/stiffness/whatever to a bullet impact.

normal_force said:
I also assume you saw the video?

I did. What about it?
 
  • #41
normal_force said:
I just need the equation.
You need numerical simulations for such deformation problems.
 
  • #42
normal_force said:

at 0:48 you can just skip to

The money shot is at 0.53. The bullet is deformed.
 
  • #43
Drakkith said:
Wish I could help you. I don't know how to figure out and apply the elasticity/stiffness/whatever to a bullet impact.
I did. What about it?
hmph, there is an equation...but...also, that video shows energy transfer and such...

however, I was speaking with other people another website and they were talking about this, they said some interesting stuff that can help to this.
 
  • #44
A.T. said:
You need numerical simulations for such deformation problems.

which would makes sense but I am trying to find that equation...
 
  • #45
jbriggs444 said:
The money shot is at 0.53. The bullet is deformed.

like I've corrected...the bullet is deformed, true...but energy is transferred...
 
  • #46
The energy that is transferred is a function of the force applied and the distance over which that force is applied. The distance over which the force is applied is not available from the givens of the problem (mass of bullet, mass of bullet cap, muzzle velocity of bullet). So it is difficult to use work-energy to determine the energy or velocity of the bullet+cap.

An analysis based on momentum allows you to determine the velocity of the bullet+cap immediately. That, in turn, allows you to calculate its energy.
 
  • #47
jbriggs444 said:
The energy that is transferred is a function of the force applied and the distance over which that force is applied. The distance over which the force is applied is not available from the givens of the problem (mass of bullet, mass of bullet cap, muzzle velocity of bullet). So it is difficult to use work-energy to determine the energy or velocity of the bullet+cap.

An analysis based on momentum allows you to determine the velocity of the bullet+cap immediately. That, in turn, allows you to calculate its energy.

jbriggs444 said:
The energy that is transferred is a function of the force applied and the distance over which that force is applied. The distance over which the force is applied is not available from the givens of the problem (mass of bullet, mass of bullet cap, muzzle velocity of bullet). So it is difficult to use work-energy to determine the energy or velocity of the bullet+cap.

An analysis based on momentum allows you to determine the velocity of the bullet+cap immediately. That, in turn, allows you to calculate its energy.
Indeed but it isn't very hard to calculate...
 
  • #48
normal_force said:
Indeed but it isn't very hard to calculate...
Are you participating in this thread in order to learn something or in order to promote a private theory?

Edit: This thread is going nowhere and a good bit of the volume is a result of my postings. I'm out.
 
Last edited:
  • #49
jbriggs444 said:
Are you participating in this thread in order to learn something or in order to promote a private theory?

Im learning something, that is what I am trying to do.
 
  • #50
normal_force said:
Indeed but it isn't very hard to calculate...

It may help to start with a very simple problem. If a 10 kg ball moving at 10 m/s has a perfectly inelastic collision with a stationary 100 kg ball, do you know how to find the velocity of the two balls after the collision?
 
  • #51
Drakkith said:
It may help to start with a very simple problem. If a 10 kg ball moving at 10 m/s has a perfectly inelastic collision with a stationary 100 kg ball, do you know how to find the velocity of the two balls after the collision?

Momentum transfer after 10kg ball hit is 100 kg-m/s, the 100kg ball's velocity is ~.0909m/s or 1m/s.

okay, well...Imma just come straight. The idea is like a baseball hitting a glove or a hammer hitting a nail or as I just did, a thrown at a nail. I forgot to add Mechanical energy(because I forgot).

TMEi + Wext = TMEf ; if we have really tough objects or psuedo-rigid

If I throw a ball at a catchers glove, the catchers glove exerts 500 Newtons at opposite to my incoming ball which I threw

The ball has 30 joules, without to much calculations let's say the glove is displaced .045 meters, we will say this is our final displacement; KE=TMEi

(30Joules)+(-500N's•.045 meters) = TMEf= 7.5 Joules, the 7.5 joules is expended into deformation and heat,etc
 
  • #52
normal_force said:
Momentum transfer after 10kg ball hit is 100 kg-m/s, the 100kg ball's velocity is ~.0909m/s or 1m/s.

Pretty much. The momentum before the collision is 100 kg*m/s. So when the two collide and "stick together", the momentum after the collision still has to be 100 kg*m/s. So now it's a mass of 110 kg. P=MV, so V=P/M, or V=100/110 = 0.909 m/s.

But notice that the kinetic energy is initially 500 joules. After the collision it is only 45.4 joules. So less than one-tenth of the K.E. is left over. And note that this is on a large mass that is completely free to move under the force applied by the first ball.

A near-rigid pair of objects actually fairs a little better. Assuming a nearly elastic collision: momentum before collision is 100 kg*m/s. The first ball rebounds off of the first. The velocity of the 10 kg ball ends up being -8.18 m/s and the velocity of the 100 kg ball is 1.818 m/s. KE of the 100 kg ball is now 165 joules and the KE of the 10 kg ball is 335 joules. You actually transfer more kinetic energy to the 2nd object with an elastic collision than you do with an inelastic one.

normal_force said:
okay, well...Imma just come straight. The idea is like a baseball hitting a glove or a hammer hitting a nail or as I just did, a thrown at a nail. I forgot to add Mechanical energy(because I forgot).

Those situations aren't simple collisions, so they can't be modeled as such. There are other forces acting on the glove/nail/hammer that have to be taken into account.

normal_force said:
If I throw a ball at a catchers glove, the catchers glove exerts 500 Newtons at opposite to my incoming ball which I threw

The ball has 30 joules, without to much calculations let's say the glove is displaced .045 meters, we will say this is our final displacement; KE=TMEi

(30Joules)+(-500N's•.045 meters) = TMEf= 7.5 Joules, the 7.5 joules is expended into deformation and heat,etc

I don't even know why you're posting this example. Unless you know a realistic force and displacement, then this is utterly meaningless.
 
  • #53
Drakkith said:
Pretty much. The momentum before the collision is 100 kg*m/s. So when the two collide and "stick together", the momentum after the collision still has to be 100 kg*m/s. So now it's a mass of 110 kg. P=MV, so V=P/M, or V=100/110 = 0.909 m/s.

But notice that the kinetic energy is initially 500 joules. After the collision it is only 45.4 joules. So less than one-tenth of the K.E. is left over. And note that this is on a large mass that is completely free to move under the force applied by the first ball.

A near-rigid pair of objects actually fairs a little better. Assuming a nearly elastic collision: momentum before collision is 100 kg*m/s. The first ball rebounds off of the first. The velocity of the 10 kg ball ends up being -8.18 m/s and the velocity of the 100 kg ball is 1.818 m/s. KE of the 100 kg ball is now 165 joules and the KE of the 10 kg ball is 335 joules. You actually transfer more kinetic energy to the 2nd object with an elastic collision than you do with an inelastic one.
Those situations aren't simple collisions, so they can't be modeled as such. There are other forces acting on the glove/nail/hammer that have to be taken into account.
I don't even know why you're posting this example. Unless you know a realistic force and displacement, then this is utterly meaningless.
T

There is a a loss of energy, thus inelastic collision...

however, example of that? Think, Mechanical energy-> work...think baseball and catchers glove, hammer and nail, skidding car.
 
  • #54
normal_force said:
There is a a loss of energy, thus inelastic collision...

however, example of that? Think, Mechanical energy-> work...think baseball and catchers glove, hammer and nail, skidding car.

I'm sorry I can't understand what you're trying to say. Can you re-iterate what you wanted from this thread in the first place? After all this talk I'm having trouble figuring out how best to try to help you.
 
  • #55
Drakkith said:
I'm sorry I can't understand what you're trying to say. Can you re-iterate what you wanted from this thread in the first place? After all this talk I'm having trouble figuring out how best to try to help you.
They are mechanical energy examples, how KE->ME->force, the energy is conserved a bit...some lost in deformation...but these are times were momentum is less relevant then KE, this is W=∆KE stands true in a solid body external sense.
 
  • #56
normal_force said:
They are mechanical energy examples, how KE->ME->force, the energy is conserved a bit...some lost in deformation...but these are times were momentum is less relevant then KE, this is W=∆KE stands true in a solid body external sense.

Are we talking about a simple collision, like a bullet hitting a target, or about something more complicated?
 
  • #57
Drakkith said:
Are we talking about a simple collision, like a bullet hitting a target, or about something more complicated?
We are talking about psuedo-rigid collisions, where the bullet is deformed slightly but is able to apply force, doing work.
 
  • #58
normal_force said:
We are talking about psuedo-rigid collisions, where the bullet is deformed slightly but is able to apply force, doing work.

This is nothing like your examples. This would obey the simple examples I gave above about collisions.
 
  • #59
Drakkith said:
This is nothing like your examples. This would obey the simple examples I gave above about collisions.

for a psuedo-rigid object with slight deformation and no rebound, its literally just like a hammer and nail, Its KE->ME->Force•distance->work. The object losses some energy in deformation, however in this example, much like bowling pins and bowling ball, baseball and glove, hammer and nail. The psuedo-rigid bullet, this bullet is not made of typical materials, yet it deforms despite it being very tough. The bullet is allowed to apply its force, as it slows down it exerts force, bullet to target and target to bullet. IT does work, and work gives KE as work is energy transfer->W=∆KE, W=1/2m•vf^2 - 1/2m•vi^2...this is a case of where the bullet doesn't penetrate but deforms a little yet retaining enough KE->Mechanical energy to still do work.

In typical bullets, they deform, shatter, rebound, deflects, etc...thus energy is converted into other forms, like how the force of impact causes a typical lead core, copper jacketed bullet, they penetrate or shatter...thus they do not do work, they penetrate so they do work on material, the impact force of a bullet Favg=Ke/d

http://www.physicsclassroom.com/class/energy/Lesson-1/Mechanical-Energy

Found this one, gives a pretty good run down, even things about KE and such.
 
  • #60
normal_force said:
for a psuedo-rigid object with slight deformation and no rebound

This isn't possible and therefore the rest of your post isn't applicable. It's not even possible in a thought experiment because is violates several physical laws.
 
  • #61
Drakkith said:
This isn't possible and therefore the rest of your post isn't applicable.

that is not true, its abides by the laws of energy but it hasn't been done, no object has proven tough enough to withstand that force...now the material is possible...
However, my post relates to hammer and nail, that is KE doing work, this is a staple fact of the laws of physics...
 
  • #62
This is nonsense. Myself and others have explained to you multiple times why your scenarios can't work. Thread locked.
 

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