Inelastic collision: Momentum conserved, KE not- How?

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Inelastic collision: Momentum conserved, KE not-- How?

This is a question about the text of Hailiday/Resnick/Walker 8th ed. p217 - 219. It's not a homework problem, but just about understanding the text, so I hope I am posting to the correct location, if not, please advise. In the text they say that in an inelastic collision kinetic energy is not conserved (p.217)(this is the definition of inelastic), but also that total momentum IS conserved (p.218). I can't visualize how this can be.

I understand how in an inelastic collision kinetic energy (KE) is not conserved because some of the original KE goes into heat, deformation, etc. For example if you have a putty ball and it rolls in the +x direction into a mass on a frictionless surface (a one-dimensional completely inelastic collision) we can imagine that the putty ball squishes into a disk. So the KE is not conserved.

What I don't understand is how, despite this deformation/heat/etc. we can still say that momentum in the +x is conserved. Since no mass is lost in the collision, isn't there a "loss of the component of v in the +x direction," in that some of the original +x v gets used to deform the ball in the y,z directions? So our new object (ball+ mass) when it keeps moving in the +x direction must have less momentum.

Since this view is presumably wrong, could someone please explain how a decrease in (1/)mv^2 with m constant, requiring a decrease in v, can not lead to a decrease in mv? Thanks.
 

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  • #2
jtbell
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Indeed, the momentum of the ball decreases, just as its kinetic energy does. Conservation of momentum applies to the system of the ball plus the mass that it collides with. The mass gains an amount of momentum that is exactly equal to the amount lost by the ball.

This follows directly from Newton's Third Law of Motion. During the collision, the force the ball exerts on the mass is equal in magnitude and opposite in direction to the force the mass exerts on the ball. The two forces act for the same time period. Change in momentum = impulse = force x time. Therefore the change in momentum is also equal in magnitude and opposite in direction for the ball and the mass.
 
  • #3
marcusl
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Since this view is presumably wrong, could someone please explain how a decrease in (1/)mv^2 with m constant, requiring a decrease in v, can not lead to a decrease in mv? Thanks.
You can't just look at one component of v for one mass, because it's the sum of vector momenta that is conserved.

You might think of it this way: both total energy and momentum are conserved. Some of the energy is converted to heat (frictional deformation of the materials, sound, etc.), so some kinetic energy "disappears", but there is no such conversion of momentum into something else. Momentum must be conserved. Just like total E is conserved, although KE is not.

EDIT: sorry, Jtbell beat me to it!
 
  • #4
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Thank you very much for your very quick reply. I guess what has me confused is this: when the putty ball deforms, their is work done. This work is "coming from" the original v of the ball. But then this means there is less v in the +x direction to give to the mass. You said "the mass gains an amount of momentum that is exactly equal to the amount lost by the ball." But the momentum lost by the ball was transferred to the y,z directions (the deformation). So from this I have this wrong image that there is less mv to give to the +x direction. I do understand that the momentum of the whole system must be constant absent external forces, and that the vector sum of the deformation momentum is zero in the y,z plane. So it can't be taking any of the momentum from the x-direction.

Can you help me shake this image that there is less mv to transfer to the mass, because work is done in the y,z plane?

(edit) Marcusl thanks to you too-- i feel I'm almost getting it...
 
  • #5
nasu
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Don't forget that momentum is a vector. When you have two particles moving toward each other, the total momentum is actually the difference of the two momenta.
During the collision both particles will "loose momentum" but the difference can still be the same (and it is).
Example: For a plastic collision between two particles of the same mass and the same speed, initial momentum is zero and it stays zero after collision (they are at rest after collision, right?)
They both lost all momentum during collision.
 
  • #6
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"There is no such conversion of momentum into something else"-- okay-- I'm almost getting it. The putty ball hits, and its molecules move in the y,z plane= deformation, but this is not taking away any mv from the x direction. Even though work is being done to deform, which is (1/2) mv^2. This v of work is in y-z work, even though all the original energy was in the x-direction, x-work, and you get this y,z work from the x-work. So original KE from x-motion becomes KE of y,z motion, but this does not decrease the mv that is transferred to the mass in the x-direction.

(edit) Whew! Thanks to all of you. I admit I find this relationship between KE and mv slippery...but then one's vector and one isn't...
 
  • #7
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You are correct in saying that there is less velocity in the +x-direction, so it would seem like there should be less final momentum due to the fact that momentum is dependent on velocity. But this would be the wrong assumption because we did not take into consideration the other variable that momentum is dependent upon, namely, mass.

Before the collision, only one piece of mass is moving, after the collision, two pieces of mass are moving. So even though it moves slower (due to the fact that some KE was lost to heat, deformation, etc.), the fact that more mass is now moving makes up for exact amount of velocity lost - every time.

For example,

Before the collision, we have two pieces of mass, each weighing (5kg). One of them is stationary (v2 = 0 m/s) the other is moving (v1 = 10 m/s). So the total momentum before the collision is:

(momentum of object 1) + (momentum of object 2)
(5kg)(10m/s) + (5kg)(0m/s)
50 kg*m/s = total momentum before the collision

Now, after the inelastic collision, some of the energy gets converted to heat, lost in deformation, etc. As a result, their is going to be less velocity in the +x-direction, in fact, the final velocity of the two objects (since they are stuck together now, by definition an inelastic collision) is 5m/s. So, what is the final total momentum?

(momentum of new mass system)
(10kg)(5m/s)
50kg*m/s = total momentum after the collision

As you can see, even though it lost velocity, it gained moving-mass, which conserves momentum every time. This stems from the fact that the total energy in a closed system must be conserved before and after a collision: So, If some velocity is lost, then more mass must be moving in order to make up for it and keep the total energy in the closed system constant. Notice that the amount of energy added by the moving-mass is always exactly the same amount of energy lost in velocity by the single mass. This observation lead to the conclusion that momentum is always conserved.

I hope this helps!
 
  • #8
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I guess I should have highlighted on more thing:

I know what you're thinking, "if before the collision there is "x" amount of total energy, and if after the collision some of that energy gets spent on heat and deformation, then there should be less energy for velocity in order to make up for the added energy due to heat and deformation. To add moving mass to make up for this loss in velocity is to ignore the fact that energy was added to heat and deformation!"

Alright, good point, but such a thought has the premise that only energy in the form of moving-mass can be transferred, which is wrong.

In an inelastic collision, two objects must touch (by definition). When they touch, the energy gained in heat by object 2 is lost by object 1, and the energy gained in deformation by object 2 is lost by object 1 (how does something lose or gain deformation?! - that just simply means that it shapes into a more stable or less stable form, respectively), and the amount of energy lost in moving-mass by object 1 is gained in moving-mass by object two (which in this case is by the "sticking together" of mass 1 and 2).

The energy that was absorbed to increase heat was taken from the heat energy of the first object - it was not taken from the first object's moving-mass energy.

The energy absorbed to deform the second object was taken from the energy of the first object's form - it was not taken from the first object's moving-mass energy.

As you can see, total energy is still conserved, and so is momentum. Just remember one thing, Newton's laws are for general purposes, they are not universal laws that apply to every situation, they are not perfect. This becomes apparent when you start dealing with collisions or object that are approaching the speed of light, which is why Einstein made his equations to account for what Newton's equations couldn't.
 
  • #9
Doc Al
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As you can see, even though it lost velocity, it gained moving-mass, which conserves momentum every time. This stems from the fact that the total energy in a closed system must be conserved before and after a collision
I hope you are not implying that momentum conservation is a consequence of energy conservation.

Also: Realize that the thread you are responding to is several years old.
 

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