Momentum Conservation in Inelastic Collisions: Explaining Newton's Laws

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Momentum is conserved in inelastic collisions because the total momentum before and after the collision remains constant, despite energy loss in the form of heat. The confusion arises from the distinction between momentum and energy; while momentum is a vector quantity that is conserved, kinetic energy is not conserved in inelastic collisions. Forces are not calculated directly during collisions because they depend on the time of contact, which is typically not provided in problems. The concept of "force conservation" does not exist; instead, momentum conservation applies in isolated systems without external forces. Understanding these principles clarifies the relationship between momentum, energy, and forces in physics.
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I'm preparing myself for my first physics exam and I have some general problems, I hope you great people will help me (you never let me down so far).

Why is momentum conserved in inelastic collision?
Is it because we count that "macro" momentum of two clay balls is transferred to "micro" momentum of it's particles ?

And I'm highly interested III Newton's law, it seems unbreakable (is there any exceptions), but I'm frustrated when I'm dealing with those collisions (both elastic and inelastic), we have this great talk about force in previous paragraphs of my physics textbook, and suddenly after it force is gone from all calculations, and all we calculate in theory and practice is speed and energy (talking about collisions).

Why can't I calculate force, I would love to, if III Newton's says (e.g. inelastic collision) every external force that acts on an object there is a force of equal magnitude but opposite direction which acts back on the object which exerted that external force, IF I understand correctly III law, those two balls acts on each other with same force, and I would like to know what's that force. All I have is change in energy, velocity or momentum of the balls, but F is = m * a , and a = dv/dt, I'm, always short in that dt, even if I wright Fdt=dp, again I have dp but no dt ? I don't get it, I can't calculate force involving collision.

+ I'm confused with acceleration, if we say that rotational motion is always accelerated (v changes), why is rotational motions so "stabile" (like the Moon rotating around the earth), I know how centrifugal force is making this possible, but is there any example of rectilinear accelerated motion that is so "stabile" and happening "forever". It confuses me because rotational and linear acceleration seems so deferent to me (I would gave them separate names).

OK, I've already asked too much ,
txn

p.s. why's speed of sound in air 1000km/h :)?
 
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The reason force isn't used in these collision problems is that force isn't conserved. Momentum is conserved as a vector (i.e. the directions add up too), and energy as a number (no directions). So you can set up and solve the equations. Then when you have the other numbers you can calculate the forces. Physics is like this all the way through; the number you want may not be the first number the equation gives. Willy Sutton said he robbed banks because "That's where the money is". The physics student has to go where the equation is.

In elastic collisions the micro strucures give back the momentum and energy (or almost all of it) to the macro scale, so the macro scale numbers add up. In inelastic collisions that doesn't happen, some important amount of energy gets lost (probably turned into heat and radiated) at the micro scale.
 
Tnx selfAdjoint, but I still don't understand something.
At inelastic collision mechanical energy isn't conserved, but we can still calculate it's “loose” (but we have absolutely NO information about Force-why). Excuse me, but I still haven't heard anything about concept of “Force conservation”, could you help me with that?
And I have trouble finding equations for inertial and non inertial reference systems on internet … google didn't help, maybe I need better keywords (+forced and sepulchral oscillations-translation from my antique dictionary ;)) ?

hey man thanks again for helping me :)
 
"why's speed of sound in air 1000km/h :)?"
I think, I got this one.

But still need help with those other questions …

Hey is that rotation of water in toilet consequence of Coriolis forces or what ?
 
It seems Coriolis has nothing to do with this effect (as much as i get it), but still i can't remeber some realtimelife example for this forces (except textbook one's)...
 
That's an urban myth. The coriolis force is proportional to the speed of the thing affected and the sine of the latitude. It turns the motion of a moving object (or stream) to the left in the Northern hemisphere and to the right in the southern hemisphere. For a speed as low as you find in a toilet bowl, the curvature would be undetctible. Conversely to get a curvature radius of under 50 cm the speed would have to be hundreds of km per hour. Your toilet doesn't get up that high, and the swirl of water is due to the direction the jets under the toilet rim are pointed.
 
jhirlo: " Excuse me, but I still haven't heard anything about concept of “Force conservation”, "

Yes, you have. Self Adjoint told you just before you posted that that there is no such thing as "Force Conservation". Force is not conserved (that's why we can use a lever to move a heavy object with less force). As long as there is no external force, momentum is conserved. Getting back to your original question, no, momentum is not transferred to its molecules. You may be thinking of conservation of energy: Kinetic energy is not conserved in an inelastic collision (basically, that's the definition of inelastic collision!) but if we take into account the increase in heat (the energy of the molecules) due to the collision, then total energy is conserved.

You cannot calculate the force between two objects when they collide, in a typical collision problem, because that would depend upon length of time they were in contact and you normally are not given that. You could calculate the "impulse"- that is "force times time" and measures the amount of momentum transferred from one object to another in a collision.
 
eh... i don't quite understand the concept of momentum. Can someone explain it? and how is it related to energy?
 
HallsofIvy, after your post thing seems much clearer for me. But I'm still confused with conserving momentum, you say: "Getting back to your original question, no, momentum is not transferred to its molecules",
then how is it conserved?

Let's say we have two clay bolls, one standing still with zero momentum p1=0, and another with some nonzero momentum p2 colliding with first one. Collision is inelastic and after it both balls are standing still deformed but now both with p'=0 momentum. There's definitely change in momentum: p1+p2 \neq 2p' (p2\neq
0) and dp\neq
0, so where's that momentum conservation?

When you say "but if we take into account the increase in heat (the energy of the molecules)" isn't that transfer of momentum to molecules ? are we just increasing their speed (v-factor from p=mv, or v from E=1/2mv^2) ?
 
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  • #10
I've just read, in example with cylinder rotated by rotating cord that it eventually rotates in fashion in which it has the largest angular momentum. Why's that, there's no explanation?

Hey and if you let body on it's own it'll rotate forever .And I thought that rotational motion is accelerated, and not stabile by I Newton's law.



Hey I'm sorry if I'm boring you, this is just too reveling for me, i fell like baby exploring new things (and I have to do it all in very short time) . Or maybe physics and I are just incompatible, and I have really bad, if they deserve to be called, books ...
 
  • #11
by selfAdjoint
The coriolis force is proportional to the speed of the thing affected and the sine of the latitude. It turns the motion of a moving object (or stream) to the left in the Northern hemisphere.
Just being picky here. But doesn't the coriolis force cause objects to move to the right in the northern hemisphere?
 
  • #12
Originally posted by dlgoff
Just being picky here. But doesn't the coriolis force cause objects to move to the right in the northern hemisphere?
It depends.
It’s not same if you are traveling from N to S or from S to N, of course all on northern hemisphere (it’s the same principle for southern :)).

hint: L-const.
 
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  • #13
It’s not same if you are traveling from N to S or from S to N, of course all on northern hemisphere (it’s the same principle for southern :)).
Well, the force is max at the pole and zero at the equator but the direction is still the same. (to the right in the northern hemisphere)

The force is the cross product of the angular velocity vector of the Earth with the linear velocity vector of the object. Isn't that right?

added by edit: oops. SelfAdjoint is right. It's to the left. I should have looked at my right-hand thumb.
 
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  • #14
Yes on northern hemisphere to west (I don’t use rules(right hand etc.), just imagining it).

But as I imagine it, if you’re traveling from S to N, on northern hemisphere, you’ll move to right (east), because you have grater (gained at south of N.hemisphere) velocity than Earth has at north of north hemisphere :).
 
  • #15
Originally posted by jhirlo
But I'm still confused with conserving momentum, you say: "Getting back to your original question, no, momentum is not transferred to its molecules",
then how is it conserved?
The momentum of a body is the momentum of its molecules.
Let's say we have two clay bolls, one standing still with zero momentum p1=0, and another with some nonzero momentum p2 colliding with first one. Collision is inelastic and after it both balls are standing still deformed but now both with p'=0 momentum. There's definitely change in momentum: p1+p2 \neq 2p' (p2\neq
0) and dp\neq
0, so where's that momentum conservation?
Your example is flawed. Assuming no other bodies are involved except the two clay balls, then the final momentum will equal the original momentum. (The two balls form an isolated system, so there are no external forces to change the total momentum.) Note that momentum is conserved in any collision, not just inelastic collisions.
 
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