Impact force of Rigid objects and no displacement

  • #51
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 lets 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
Drakkith
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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.

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.

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 lets 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
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
Drakkith
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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
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
Drakkith
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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
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
Drakkith
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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
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
Drakkith
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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
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
Drakkith
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This is nonsense. Myself and others have explained to you multiple times why your scenarios can't work. Thread locked.
 

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