48
$\begingroup$

In inelastic collisions, kinetic energy changes, so the velocities of the objects also change.

So how is momentum conserved in inelastic collisions?

$\endgroup$

17 Answers 17

39
$\begingroup$

I think all of the existing answers miss the real difference between energy and momentum in an inelastic collision.

We know energy is always conserved and momentum is always conserved so how is it that there can be a difference in an inelastic collision?

It comes down to the fact that momentum is a vector and energy is a scalar.

Imagine for a moment there is a "low energy" ball traveling to the right. The individual molecules in that ball all have some energy and momentum associated with them: low energy ball traveling to the right

The momentum of this ball is the sum of the momentum vectors of each molecule in the ball. The net sum is a momentum pointing to the right. You can see the molecules in the ball are all relatively low energy because they have a short tail.

Now after a "simplified single ball" inelastic collision here is the same ball:

high energy ball traveling to the right

As you can see, each molecule now has a different momentum and energy but the sum of all of all of their momentums is still the same value to the right.

Even if the individual moment of every molecule in the ball is increased in the collision, the net sum of all of their momentum vectors doesn't have to increase.

Because energy isn't a vector, increasing the kinetic energy of molecules increases the total energy of the system.

This is why you can convert kinetic energy of the whole ball to other forms of energy (like heat) but you can't convert the net momentum of the ball to anything else.

$\endgroup$
  • 3
    $\begingroup$ The "vector" versus "scalar" argument lacks rigor. Can you back this up with some math? $\endgroup$ – Paul Feb 13 '16 at 20:46
  • 4
    $\begingroup$ @Paul: consider two identical particles, mass $m$: their center of mass moves with velocity $\vec v=(\vec v_1+\vec v_2)/2;$ their kinetic energy is $\frac12mv_1^2+\frac12mv_2^2.$ Of this energy, only $\frac12(2m)v^2=\frac14m\left(v_1^2+v_2^2+2v_1v_2\right)$ is externally visible as center-of-mass motion; the remaining $\frac14m(v_1-v_2)^2$ energy forms a "reservoir" for energy which cannot be observed in their center-of-mass; any situation with hidden reservoirs of stuff can violate conservation laws of that stuff. There is no analogous situation here for momentum as $(2m)v=mv_1+mv_2.$ $\endgroup$ – CR Drost Feb 19 '16 at 17:54
  • 4
    $\begingroup$ Downvoted. This answer is so very wrong. Momentum and energy are conserved quantities in classical mechanics. There's a lot more to energy than just kinetic energy, which means that kinetic energy is not necessarily a conserved quantity. $\endgroup$ – David Hammen Feb 19 '16 at 18:44
  • 10
    $\begingroup$ @DavidHammen my answer wasn't intended to be a comprehensive treatise on the subject but rather offer an intuitive understanding of how such a thing could be possible. I also mentioned other manifestations of energy in the last sentence. With that said, do you think the answer is technically wrong or just an incomplete simplification? $\endgroup$ – Brandon Enright Feb 19 '16 at 21:01
  • $\begingroup$ Wouldn't the same logic that says momentum is conserved (treating the system as a whole), also say that the energy of the system as a whole is conserved as well? $\endgroup$ – Yogi DMT Oct 26 '16 at 13:53
34
$\begingroup$

So how is momentum conserved in inelastic collisions?

It is a basic law of physics that momentum is always conserved - there is no known exception. Kinetic energy does not need to be conserved, because it can turn into other forms of energy - for example potential energy or internal/thermal energy ("heat"). Momentum can also turn into other form of momentum - momentum of the EM field - but the amount of momentum so transformed seems negligible in ordinary collisions of macroscopic bodies.

$\endgroup$
  • 1
    $\begingroup$ This and Eric Angle's answer are the best and most concise here. $\endgroup$ – WetSavannaAnimal Aug 11 '16 at 6:22
  • 3
    $\begingroup$ The OP asked the question "how" momentum was conserved. This answer doesn't really address that, it just restates the (unchallenged) premise of the question which is that momentum is of course conserved. $\endgroup$ – DaveInCaz Mar 28 '17 at 11:41
  • $\begingroup$ Hi Jan Lalinsky: Did you accidentally create two accounts and want to merge them? $\endgroup$ – Qmechanic Oct 4 '17 at 15:56
  • $\begingroup$ @Qmechanic, it is the unconfirmed account from the past that I no longer use. You may merge it with my main account. $\endgroup$ – Ján Lalinský Oct 4 '17 at 16:05
  • $\begingroup$ The mods cannot merge accounts, only the owner or the SE team. I leave it to you to contact the SE team. $\endgroup$ – Qmechanic Oct 4 '17 at 16:08
20
$\begingroup$

Energy and momentum are always conserved. Kinetic energy is not conserved in an inelastic collision, but that is because it is converted to another form of energy (heat, etc.). The sum of all types of energy (including kinetic) is the same before and after the collision.

$\endgroup$
  • 2
    $\begingroup$ This and Ján Lalinský's answer are the best and most concise here. $\endgroup$ – WetSavannaAnimal Aug 11 '16 at 6:22
  • 1
    $\begingroup$ If some kinetic energy is converted to heat, this must make the particles who absorb this energy vibrate with greater KE. Doesnt a faster moving particle have more momentum too? $\endgroup$ – Ubaid Hassan May 19 at 14:46
10
$\begingroup$

None of these answers really address the question; mostly they just reiterate physics principles that I suspect the poster already understands.

The question is saying, 'If kinetic energy changes in different types of collision, then the final velocities must be changing. If the final velocities are changing, the final momentum must be changing, but momentum is supposed to be conserved. How can this be?'

I had the same question this morning, which is how I ended up here. The key to understanding this is to realize that there is a range of final velocities that all conserve momentum, and that each point in this range represents a different amount of kinetic energy.

For example: a 1kg ball labeled $A$ moving at 3m/s strikes another 1kg ball $B$ that is at rest. Here are a few of the possible end velocities, with total kinetic energy for each case, and total momentum:

\begin{array}{|c|c|c|c|} \hline v_A & v_B & K_{AB} & p_{AB} \\ \hline \mathrm{m/s} & \mathrm{m/s} & \mathrm{J} & \mathrm{kg \cdot m/s} \\ \hline 0.0 & 3.0 & 0.0 + 4.5 = 4.5 & 0.0 + 3.0 = 3.0 \\ 0.5 & 2.5 & 0.125 + 3.125 = 3.25 & 0.5 + 2.5 = 3.0 \\ 1.0 & 2.0 & 0.5 + 2.0 = 2.5 & 1.0 + 2.0 = 3.0 \\ 1.5 & 1.5 & 1.125 + 1.125 = 2.25 & 1.5 + 1.5 = 3.0 \\ \hline \end{array} Notice that each of these represents the same amount of momentum (equal to the starting momentum) but they all yield different amounts of kinetic energy! This is why a certain amount of kinetic energy can be converted to thermal or other kinds of energy without decreasing total momentum. This is also why we need both conservation laws — momentum and energy — to solve certain types of problems; without both, we wouldn't be able constrain the answer to a particular point within the range of solutions (see above!) that conserve momentum.

$\endgroup$
6
$\begingroup$

Conservation of Momentum falls directly out of Newton's Laws.

Consider Newton's Third Law: $\sum \vec{F} = 0$

And Newton's Second Law: $\vec{F} = m \vec{a} = \frac{d\vec{p}}{dt}$

Combining these two laws we find: $\sum \frac{d\vec{p}}{dt} = 0$

This equation states that total momentum cannot change with respect to time. That is, the total momentum cannot change before or after the collision, irregardless of the type of collision. Thus momentum is always conserved.

$\endgroup$
6
$\begingroup$

Law of conservation of momentum is directly implied by Newton's laws of motion. Basically it is conserved even in inelastic collision because forces appear in pairs with equal magnitude and opposite direction as shown :
The two dark dots are two particles. The direction of arrows shows the direction of force and the length of the arrows shows their magnitude.
In all the physical phenomena the forces can be represented by the above mentioned image. For the inelastic collision this image also holds e.g. consider an inelastic collision as shown:
The block of mass $M$ is initially at rest and a bullet is moving towards it with a velocity $v_i$ and mass $m_w$. What happens during the collision? There appears a force pair of equal magnitude and in opposite direction. The force pair continuously varies in magnitude during the collision. The force pair is kinetic friction It continue to act until the relative velocity of bullet w.r.t the block becomes zero that is, both bullet and block acquires same velocity. These forces are represented as $\vec F_b$ and $\vec F_w$. $\vec F_b$ acts on the bullet towards left and $\vec F_w$ acts on block towards right.
By newton's third law $\vec F_w=-\vec F_b$
Change in momentum of bullet $= \Delta p_b = \int_{t_1}^{t_2}\vec F_b dt$
Change in momentum of the block$ = \Delta p_w = \int_{t_1}^{t_2}\vec F_w dt$ It is easy to recognize that since $\vec F_w=-\vec F_b$ the decrease in momentum of bullet appears as the increase in momentum of the block. Now the underlying fact is that if decrease in velocity of bullet causes a decrease in the momentum of bullet then at the same time the velocity of the block increases which causes the momentum of block to increase.
A different scenario can take place if the block is not stationary but is moving towards the bullet. Further let the initial momentum of the system is 0 what would happen after the collision? the bullet will sink in the block and the velocities of both block and bullet will become 0! That is, total kinetic energy of the system becomes 0!. We see K.E of the system may be changed (not the total energy) but momentum of the system cannot.
For a better explanation you should read these
1. http://www.feynmanlectures.caltech.edu/I_10.html
2. http://www.physicsclassroom.com/class/momentum/u4l2b.cfm


$\endgroup$
4
$\begingroup$

Eric Angle has it pretty much right. In an inelastic collision some of the kinetic energy is absorbed by the deformation of the material. For example, if two balls of putty collide and stick together, kinetic energy is absorbed by squishing the putty. In a second example, if you shoot a bullet at a log, some of the kinetic energy is absorbed by friction as the bullet passes into the wood. In both cases, some of the kinetic energy is turned into heat, so although energy is conserved, kinetic energy is not.

In an elastic collision, the objects bounce off each other. During the collision, the material momentarily deforms and absorbs some of the energy, but then bounce back like a spring, giving the energy back up. So in a elastic collision, kinetic energy is conserved.

$\endgroup$
0
$\begingroup$

Despite the inelasticity of the collision, the momentum will be conserved. The kinetic energy will change. Some work was done in the process of the "absorption" of energy during the inelastic collision and this will reduce the resultant kinetic energy.

Some of the other answers have derived the conservation principle from Newton's laws, but I think the more fundamental derivation was done by Emmy Noether, who discovered that our notion of invariance of physical laws under infinitesimal coordinate changes gives rise to conservation laws. Momentum is the conserved quantity that falls out of the symmetry with respect to spatial translation.

$\endgroup$
0
$\begingroup$

The conservation of momentum is simply a statement of Newton's third law of motion. During a collision the forces on the colliding bodies are always equal and opposite at each instant. These forces cannot be anything but equal and opposite at each instant during collision. Hence the impulses (force multiplied by time) on each body are equal and opposite at each instant and also for the entire duration of the collision. Impulses of the colliding bodies are nothing but changes in momentum of colliding bodies. Hence changes in momentum are always equal and opposite for colliding bodies. If the momentum of one body increases then the momentum of the other must decrease by the same magnitude. Therefore the momentum is always conserved.

On the other hand energy has no compulsion like increasing and decreasing by same amounts for the colliding bodies. Energy can increase or decrease for the colliding bodies in any amount depending on their internal make, material, deformation and collision angles. The energy has an option to change into some other form like sound or heat. Hence if the two bodies collide in a way that some energy changes from kinetic to something else or if the deformation of the bodies takes place in a way that they cannot recover fully then energy is not conserved. This option of changing into something else is not available to momentum due to Newton's third law of motion.

This is why momentum is always conserved but kinetic energy need not be conserved.

Further an elastic collision is defined in such a way that it's energy is taken to be conserved. Nothing like an elastic collision exists in nature. It is an ideal concept defined as such. Empirical measurements will always show that collisions are always inelastic

$\endgroup$
  • 3
    $\begingroup$ Dear sukhveer choudhary. It is often frown upon to post nearly identical answers to similar posts. In such cases, it is often better to just flag/comment about duplicate questions, so they can get closed. $\endgroup$ – Qmechanic May 12 '15 at 12:01
0
$\begingroup$

Who ever said the answer is "momentum is a vector and energy is a scalar" is correct, saying energy gets transformed into heat just kicks the can down the road, to "why can the KE get transformed?" I think the best answer starts by pretending that energy and momentum don't really exist (other than in our mind as a mathematically construction to help solve problems easier and quicker). What really exists are masses that interact by applying forces through time (impulses which are a vector) and/or by applying force over a distance (work is is the dot product of two vectors and therefore a scalar). Its the loss of the sign on the scalar that hurts our ability to the work. Example a internal spring causes two carts to separate. The fact the speed is increased for both carts increases the kinetic energy of system from zero to a positive number. Also heat is just internal kinetic energy (force times distance) in various random directions and again hard to track because of the loss of the sign when force and displacement are in the same direction. The fact that positive work is defined at when Force and motion are in the same direction means sometimes an upward force is positive at the same time a downward force is positive. This removes our ability to "follow" the discriminate force displacement interactions. this is my 1st post so sorry if it is crude.

$\endgroup$
  • 3
    $\begingroup$ I'm sorry but I fail to see how this answers the question... $\endgroup$ – AccidentalFourierTransform Feb 19 '16 at 17:32
  • $\begingroup$ Let me welcome you to Physics Stack Exchange! This is a question-and-answer site for people with specific conceptual questions about physics and their answers. I hope that you're able to stick around and contribute productively to our task of answering discussion questions. $\endgroup$ – CR Drost Feb 19 '16 at 18:05
  • $\begingroup$ With that said, if you're open to criticism: the question was posted years ago and already attracted some quality answers, one of which resolved the problem to the asker's satisfaction: you might want to turn your attentions to posts where you can contribute more! Your opening sentence also communicates a conversational feel among answers, as in, say, a Facebook thread--I prefer to treat Stack answers as independent cells. Finally your writing may strengthen, becoming more persuasive, if you lean less on insubstantial verbs like "is" in your answers. Welcome again, and happy answering! $\endgroup$ – CR Drost Feb 19 '16 at 18:13
0
$\begingroup$

The net mass after an inelastic collision changes.

We know that the momentum will still remain conserved.

Hence,when two bodies collide inelastically

$$(mv)=(m+M)u $$ $$=> u=\frac{m}{(M+m)}v$$

Now,finding the change in kinetic energy of the body with mass $m$

$\frac{1}2mv^2 -1/2(m){[m/(m+M)]v}^2$

and hence we can see that that the kinetic energy is not conserved

Using the work energy theorem, we can say that the lost kinetic energy must get converted into another form of energy. Generally this energy is some form of non-mechanical energy such as thermal energy.

$\endgroup$
  • $\begingroup$ Welcome on Physics SE and thank you for the answer :) Please refer to physics.stackexchange.com/help/notation/ for help in typesetting formulas. $\endgroup$ – Sanya Aug 11 '16 at 7:13
  • $\begingroup$ Net mass of the system remains constant. $\endgroup$ – Aniansh Nov 18 '16 at 14:10
0
$\begingroup$

Consider two particles of masses 'a' and 'b' , having variable velocities 'x' and 'y' respectively.

Now this is perfectly a math problem.

The two variables 'x' and 'y' are dependent on each other such that the term ax+by is always a constant. [By the conservation of linear momentum]

Now, how sure are you that the term (1/2)ax² + (1/2)by² (i.e, kinetic energy) is constant ?

It's not constant, right ?

$\endgroup$
0
$\begingroup$

The question is based on a mistaken assumption that the loss of total KE from a system of two or more particles implies a corresponding loss of overall momentum, which is not true.

For a single particle, a reduction of kinetic energy does imply a reduction in momentum- this is because a change to the quantity v2 necessarily involves a change in v.

For two or more interacting particles, a reduction in their combined KE does not necessarily result in a reduction of momentum. This is because it is always possible to find a set of numbers DVi that sum to zero even if the sum of their squares is non-zero.

Consider the trivial example of two particles of unit mass, one at rest the other moving with a speed of 4. Their combined KE is 8 units, and their combined momentum is 4 units. If they collide and coalesce (a totally inelastic collision) their speed will be 2 units. Their combined momentum will be 4 units, as before, and their combined KE will be 4 units, a reduction of 50%.

$\endgroup$
0
$\begingroup$

I think that there's a misunderstanding here, momentum is conserved, but also energy is conserved in the collition. The quantity that isnt conserved is the kinetic energy, and the reason it is not conserved in an inellastic collition is because the system needs to use some of the mechanical energy to deform the ball ( in other words work was done). Though the energy just got transformed into heat.

$\endgroup$
-1
$\begingroup$

Total momentum changes due to the influence of external forces over time, and during a collision the time taken is aproximately zero. So there is no net effect on total momentum. On the other hand kinetic energy is influenced by internal forces also which are substatial for a collision.

$\endgroup$
  • $\begingroup$ The total momentum can't be changed regardless of time because momentum is a vector. With energy (not a vector), you can convert kinetic energy of the objects into kinetic energy of the molecules in any direction (heat). With momentum though, the vector direction is conserved so even though the molecules gain momentum from their movement, the sum of all molecule momentums is still exactly the net momentum in the direction the object was traveling in the first place. $\endgroup$ – Brandon Enright Jan 4 '14 at 20:14
-1
$\begingroup$

I get that there are some geniuses of physics here answering this question. But I think the answer is simply that the kinetic energy was not conserved because work was done. So the kinetic energy is converted to work.

$\endgroup$
  • 1
    $\begingroup$ If work is done by a body on another body in the system, it doesn't necessarily imply that the energy of the system is being reduced. Total mechanical energy is only reduced when there are non conservative forces involved in the collision. $\endgroup$ – Aniansh Nov 18 '16 at 14:08
  • $\begingroup$ Why downvote the comment? My statement is correct. $\endgroup$ – Aniansh Mar 5 '18 at 14:00
-1
$\begingroup$

It is todo with the fact that kinetic energy depends on a non linear function of two velocities.

Consider the simple case of two 1kg balls in the PHET physics simulator.

This is a fascinating question. It would seem that if kinetic energy and momentum both depend on the same velocities they would both be conserved.

But the kinetic energy has a non linear dependence on velocity. This permits energy to be lost while the momentum remains conserved.

  • Conservation of momentum is a consequence of Newtons laws in a system isolated from external forces.

  • This example shows that it may not be to do with the scalar /vector argument mentioned above but instead as a result of the non linearity of the kinetic energy polynomial.


Ball 1 is moving to the right at 1 m/s

Ball 2 is at rest

Before the inelastic collision the energy of the system is $\color{blue}{0.5 \text{J} }$

$$\frac{1}{2}m v^2+\frac{1}{2}m v^2 = \frac{1}{2}(1) (1)^2+\frac{1}{2}(1) (0)^2$$

The total momentum of the system is $\color{red}{1\text{ kg m/s} }$ $$\frac{1}{2}m v^2+\frac{1}{2}m v^2 = (1) (1)+(1) (0)$$


enter image description here

After the inelastic collision the energy of the system is $\color{blue}{0.31 \text{J} }$ This is because Ball1 is now going at $\color{red}{0.25 \text{m/s}}$ and Ball2 is now going at $\color{red}{0.75 \text{m/s}}$

$$\frac{1}{2}m v^2+\frac{1}{2}m v^2 = \frac{1}{2}(1) (0.25)^2+\frac{1}{2}(1) (0.75)^2 = \color{blue}{0.31 }$$

But the total momentum of the system is still $\color{red}{1\text{ kg m/s} }$!!! $$\frac{1}{2}m v^2+\frac{1}{2}m v^2 = (1) (0.25)+(1) (0.75)= \color{red}{1}$$

enter image description here

Conclusion

Even though the velocities change in the inelastic collision, the momentum still adds up to 1, but the energy square law cannot follow. Mathematics permits the system to have the same momentum but a different kinetic energy after the collision.

enter image description here

The blue line is a line of constant momentum of 1 kg m/s. The orange circle is a constant kinetic energy of 0.5. The green circle is a constant kinetic energy of 0.31. Even though both the initial velocities (red point) and the final velocities (black point) lie on the blue line and correspond to the same momentum, they can lie on different kinetic energy circles.

$\endgroup$
  • $\begingroup$ The different power dependence tells you that a very limited set of imaginable outcomes conserve both momentum and bulk-kinetic energy. Fine. But you haven't explained why the observed outcomes that fail to conserve both are those than conserve momentum and not those that conserve bulk kinetic energy. $\endgroup$ – dmckee Jan 27 '18 at 21:31

Not the answer you're looking for? Browse other questions tagged or ask your own question.