How can momentum be conserved when energy is lost in an inelastic collision?

[mahela007]

The kinetic energy that an object possesses is dependant on it's speed according to the equation E=1/2 mv^2

So wouldn't a loss of energy mean that the object has slowed down and lost some momentum?

[Bignose]

Firstly, welcome to the forum

 

Secondly, your statement need a slight amendment: "The kinetic energy that an object possesses in a certain reference frame is dependant (sic) on it's speed according..." The "in a certain reference frame" is a very important qualifier because kinetic energy is very highly observer dependent.

 

Thirdly, you have to remember that a momentum conservation does have sources and sinks -- specifically forces. When an inelastic collision occurs, there are forces that have to be accounted to deform the two particles that are colliding. This deformation is both a momentum sink and an energy sink, and if the amount if momentum lost is a constant, you can use a coefficient of restitution formulation to get exact numbers for the momentum and energy losses.

[mahela007]

Thank you for your reply.

As I'm still at somewhat of an introductory level of studying physics I'm having a bit of trouble understanding the terms you have used. For example I don't know what the following mean.

Reference frame

Observer dependent

Source and Sinks

[Bob_for_short]

The kinetic energy that an object possesses is dependant on it's speed according to the equation E=1/2 mv^2

So wouldn't a loss of energy mean that the object has slowed down and lost some momentum?

 

Yes, in interactions the particle energy and momentum (as well as the angular momentum) are not conserved but exchanged. The total energy, momentum and the angular momentum of a system may conserve (transferred from one subsystem to another with no change of the sum).

 

Bob.

[Bignose]

Thank you for your reply.

As I'm still at somewhat of an introductory level of studying physics I'm having a bit of trouble understanding the terms you have used. For example I don't know what the following mean.

Reference frame

Observer dependent

Source and Sinks

 

Reference frame is the point of view you take.

 

Consider a universe with only 3 balls in it (each with mass of 1 kg). All three are moving. Now, consider a point of view (or observer or reference frame) wherein you travel with ball A. From ball A's point of view, ball B moves at 1 m/s, and ball C moves at 10 m/s. Because we are travelling with ball A, it appears that ball A is not moving at all (0 m/s). So, from the reference frame of moving with A, it appears that the universe has 101(=10^2 + 1^2) J of kinetic energy.

 

Now, change to a reference frame where B is stationary -- A is now moving at -1m/s and C is now moving at 9 m/s. In this frame, it appears that the universe has 82(=1^2 + 9^2) J of kinetic energy.

 

Finally, in the reference frame were C is stationary, A is moving at -10 m/s and B is moving at -9 m/s. In this frame, the universe has 181(=10^2 + 9^2) J of kinetic energy.

 

So, which one is right? They all are. Each is right from it's own frame. This is why kinetic energy is very, very observer dependent, very, very reference frame dependent. So, it is important to point out what reference frame you are working in when doing these kinds of problems.

 

It is also a very important result to understand that there is no such thing as a "preferred" reference frame. That any and all reference frames are equally valid and the laws of physics are necessarily reference frame independent.

 

Sources and sinks are part of the conservation equations.

 

If I had a box (assume it is perfectly rigid and non-deformable), conservation of mass shows that the amount of stuff that goes into the box must be equal to the amount of stuff that comes out.

 

Conservation of momentum and energy is similar -- if you draw a box around fluxes in and out, what comes in must come out. However, unlike mass, it is easy to imagine sources or sinks for momentum and energy. A source for momentum may be something like a pressure gradient in a fluid that acts like a source for momentum in the fluid mechanics equations. In this case, more momentum may "come out" than went in because the pressure force accelerated the fluid. A sink for momentum may be that same fluid being pumped vertically and the gravity force will act on the fluid decelerating it.

 

A sink for energy is something like a chemical reaction that may be either exothermic (source for energy) or endothermic (sink for energy).

 

Similarly, what happens in inelastic collisions is that the inelasticity acts as a sink for momentum and energy. The two inelastic objects that collide deform and fail to rebound and that momentum and energy is effectively lost. It isn't "gone", it gets turned into heat and noise and whatnot, but it isn't affecting the movement of the objects anymore.

Edited May 3, 2009 by Bignose

[J.C.MacSwell]

Consider a completely inelastic collision between two objects. They collide and stick together. Total momentum (a vector quantity) is conserved, yet total kinetic energy (a scalar quantity) is lost no matter what inertial frame you use for measurement.

[mahela007]

But Bignose says that there are momentum sinks.(Thanks for your explanation on reference frame) In that case how will momentum be conserved?

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I think I'd find it easier to understand via an example:

Imagine an elastic collision between a hard metal ball and a ball of putty. On collision some of the kinetic energy of the metal ball would be used to deform the putty. So how could the momentum be conserved if some energy was lost in other forms?

Edited May 4, 2009 by mahela007
Consecutive posts merged.

[swansont]

But Bignose says that there are momentum sinks.(Thanks for your explanation on reference frame) In that case how will momentum be conserved?

 

Forces are momentum source/sinks. Newton's second law can be written as F = dp/dt, so if there are no net external forces, momentum will be conserved.

 

In a collision there are no momentum sinks, because all the forces are internal.

[Severian]

We can put it a little more mathematically for you if you like. It is not the energy that is conserved in an eleastic collision, but the total energy.

 

The total energy is [math]E = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2[/math].

 

The total momentum on the other hand (sticking to one dimesnion for simplicity) is [math]p = m_1 v_1 +m_2 v_2[/math].

 

Now, in an inelastic collision although [math]p[/math] stays the same, [math]v_1[/math] and [math]v_2[/math] can individually change. Since the energy is the sum of the squares, it can change too.

[J.C.MacSwell]

But Bignose says that there are momentum sinks.(Thanks for your explanation on reference frame) In that case how will momentum be conserved?

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Consecutive posts merged

I think I'd find it easier to understand via an example:

Imagine an elastic collision between a hard metal ball and a ball of putty. On collision some of the kinetic energy of the metal ball would be used to deform the putty. So how could the momentum be conserved if some energy was lost in other forms?

 

Any momentum gained by one is lost by the other. Both might be "slowed down", each may have lost speed in the direction they were headed, but when you add their momentums together, it will add up to the same total momentum (same total mass movement in the same direction)

[Bignose]

In a broader sense, the sources and sinks are part of the "conservation". The full conservation equation looks like:

 

[math]\frac{\partial \rho\mathbf{v}}{\partial t} +\mathbf{v}\cdot \rho \nabla\mathbf{v} = \nabla^2 \mathbf{T} + \mathbf{f}[/math]

 

where v is the velocity of the object

[math]\rho[/math] is the density

T is the stress tensor

and f is the body forces.

 

This equation simply says that the rate of change of momentum = rate of momentum advecting in - rate of momentum advecting out + rate of momentum diffusing in - rate of momentum diffusing out + forces acting.

 

Those forces are included in the conservation equation, and simple show how much the momentum changes per unit time.

 

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Consider a simple example: You hold a ball perfectly still in your palm with your arm outstretched while standing. The ball is still and hence has no velocity. A time t=0, you drop the ball. At t>0, the ball is moving -- it has gained momentum.

 

The force of gravity acted on the ball and the ball gained momentum.

 

So, external forces are part of the momentum conservation as well.

 

A opposite example is true, too. You putt a golf ball, with the instant of impact being t=0. As the ball rolls over the ground, it slows down and at some t>0, the ball will come to a standstill. In this case at t=0, the ball definitely had momentum, and then some time later it is still and of course has no momentum. The force of friction acts on the ball to lower its momentum down to zero.

 

Forces are part of the momentum conservation as well.

[mahela007]

Thanks for all your replies. I think I understand why momentum is conserved now.

[J.C.MacSwell]

 

Those forces are included in the conservation equation, and simple show how much the momentum changes per unit time.

 

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Consider a simple example: You hold a ball perfectly still in your palm with your arm outstretched while standing. The ball is still and hence has no velocity. A time t=0, you drop the ball. At t>0, the ball is moving -- it has gained momentum.

 

The force of gravity acted on the ball and the ball gained momentum.

So, external forces are part of the momentum conservation as well.

 

A opposite example is true, too. You putt a golf ball, with the instant of impact being t=0. As the ball rolls over the ground, it slows down and at some t>0, the ball will come to a standstill. In this case at t=0, the ball definitely had momentum, and then some time later it is still and of course has no momentum. The force of friction acts on the ball to lower its momentum down to zero.

 

Forces are part of the momentum conservation as well.

 

True, but in each case if you look at the whole system, including the ball and the Earth, the forces are internal and in each case momentum is conserved at all times. Any momentum gained or lost by the ball is respectively lost or gained by the Earth.

[Bignose]

J.C.MacSwell,

 

sure, but in the case of a typical ball dropped on the Earth -- how much does the Earth move? A trillionth of a meter? The amount of velocity and movement and force on the earth is so small it is for all purposes zero. Technically non-zero, but effectively zero.

[J.C.MacSwell]

J.C.MacSwell,

 

sure, but in the case of a typical ball dropped on the Earth -- how much does the Earth move? A trillionth of a meter? The amount of velocity and movement and force on the earth is so small it is for all purposes zero. Technically non-zero, but effectively zero.

 

The point is with regard to the momentum. The change in the Earths momentum is the same and opposite of that of the ball.

 

Momentum is conserved. Using an example where one mass is so large relative to the other can be misleading with regard to that point. The force on the Earth and the resulting change in it's momentum, due to the interaction with the ball, may seem insignificant - but it is exactly equal but opposite to the force and resulting change in momentum of the much smaller ball.

 

You may understand that, but I think it should be kept clear to the OP.

[swansont]

J.C.MacSwell,

 

sure, but in the case of a typical ball dropped on the Earth -- how much does the Earth move? A trillionth of a meter? The amount of velocity and movement and force on the earth is so small it is for all purposes zero. Technically non-zero, but effectively zero.

 

That's why you can treat such a situation as the ball experiencing an external force — modeling the earth as an unmoving, effectively infinite mass is a reasonable approximation of reality.

[J.C.MacSwell]

That's why you can treat such a situation as the ball experiencing an external force — modeling the earth as an unmoving, effectively infinite mass is a reasonable approximation of reality.

 

Extremely reasonable...and that is why it can be confusing with regard to conservation of momentum, which is a fairly simple concept.

 

Changes of reference frames, observer dependency, momentum sources or sinks, and external forces can be useful ideas or tools to simplify getting a solution for a particular situation, but IMO they cloud the issue when trying to understand conservation of momentum in the most basic way.

[Bignose]

But, J.C., doing the complete momentum conservation isn't the way the principle is applied very often at all.

 

When water in a pipe goes around an elbow, no one takes into account the momentum of the fluid that pushes into the pipe wall, which then pushes against the brackets and hardware that hold the pipe, which then push against the building, which then pushes against the earth. Etc. etc. This is not how anyone solves this problem.

 

They solve the problem with the Navier-Stokes equations, which are an expression for the conservation of momentum in a Newtonian fluid, and include the gravity and pressure forces as part of the equation.

 

The real applications of the conservation of momentum include forces as sources and sinks.

[mahela007]

Going back to my initial question.. could you describe what happen to the kinetic energy when a solid metal ball collides with a ball of putty and the putty deforms?

[swansont]

Going back to my initial question.. could you describe what happen to the kinetic energy when a solid metal ball collides with a ball of putty and the putty deforms?

 

The energy goes into other forms. The putty will heat up, there is work done in deforming it, and any sound you hear contains a small amount of energy. Total energy is conserved.

[Bignose]

mahela,

 

you may be interested in post#10 of this thread: http://www.scienceforums.net/forum/showthread.php?t=35615&highlight=restitution

 

I wrote out all the equations for a collision with a coefficient of restitution -- and the energy term involving that coefficient is in there. The rest of the thread is a lot of the OP making a bunch of mistakes and spouting nonsense, but it may be helpful for you to read the replies to the garbage, too.

[J.C.MacSwell]

But, J.C., doing the complete momentum conservation isn't the way the principle is applied very often at all.

 

Who is claiming that it is?

 

All I have been doing is pointing out where the momentum went in the examples in answer to "How is momentum conserved?"

 

When water in a pipe goes around an elbow, no one takes into account the momentum of the fluid that pushes into the pipe wall, which then pushes against the brackets and hardware that hold the pipe, which then push against the building, which then pushes against the earth. Etc. etc. This is not how anyone solves this problem.

 

What problem? They certainly take the momentum of the fluid into account when choosing and placing the brackets. Do you think I'm suggesting everything should be calculated for every application?

 

 

They solve the problem with the Navier-Stokes equations, which are an expression for the conservation of momentum in a Newtonian fluid, and include the gravity and pressure forces as part of the equation.

 

Who solves any problem related to water flow around an elbow with Navier Stokes Equations? Flow around an elbow is virtually always turbulent. Navier Stokes Equations for this application would be unsolvable, or with enough simplifying assumptions unreliable, for any practical application where the answer wasn't already known.

 

Understanding of basic principals are essential when making simplifying assumptions and choosing your equations.

 

The real applications of the conservation of momentum include forces as sources and sinks.

 

If sinks and sources are fully understood they may help explain how momentum is conserved. If not they might hide it.

 

 

 

Tell me if the following is not misleading with regard to momentum :

Similarly, what happens in inelastic collisions is that the inelasticity acts as a sink for momentum and energy. The two inelastic objects that collide deform and fail to rebound and that momentum and energy is effectively lost. It isn't "gone", it gets turned into heat and noise and whatnot, but it isn't affecting the movement of the objects anymore.

 

I think this describes where the energy goes quite well.

All I was attempting to do was clarify what seemed unnecessarily complicated and somewhat ambiguous with regard to momentum.

Edited May 7, 2009 by J.C.MacSwell

[swansont]

If sinks and sources are fully understood they may help explain how momentum is conserved. If not they might hide it.

 

I agree. Often the best approach is to define the system such that you minimize the sources and sinks. Keep it simple.

[mahela007]

thanks!