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The statement that baffles me:

During most of the collisions, part of the kinetic energy evolve as heat, nevertheless momentum is still conserved.

Ok, the statement may be true. But what confuses me is that: Is there no relation between the momentum and kinetic energy? Are they not linked with each other? Both concerns with velocity. If one decreases, generally it would effect the other. I know something is wrong in my understanding. But I am unable to visualize the above statement. Please help me explaining this.

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marked as duplicate by ACuriousMind, Brandon Enright, John Rennie, Danu, David Z Nov 9 '14 at 11:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Momentum $\vec p = m\vec v$ is conserved. Kinetic energy $E = \frac{1}{2}mv^2$ isn't. What exactly about that is unclear to you? $\endgroup$ – ACuriousMind Nov 8 '14 at 19:58
  • $\begingroup$ Conservation of momentum happens because of lack of external forces during the collision. The $\int F_{\rm external} \,{\rm d}t=0$. $\endgroup$ – ja72 Nov 9 '14 at 4:19
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    $\begingroup$ possible duplicate of How can momentum but not energy be conserved in an inelastic collision? $\endgroup$ – Brandon Enright Nov 9 '14 at 5:26
  • $\begingroup$ @Brandon Enright: Thanks sir. You have brought before me this question. Thanks again. $\endgroup$ – user36790 Nov 9 '14 at 6:01
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Possibly your confusion arises from not considering that KE is a scalar, whilst momentum is a vector.

Yes, there is of course a connection between KE and momentum: $K = p^2/2m$ (for non-relativistic bodies).

Thus, for two equal-mass particles heading directly towards each other at equal speeds $v$, their velocities are $\pm {\bf v}$ and their momenta $\pm m{\bf v}$. The total momentum before the collision is zero, whilst the total kinetic energy is $mv^2$. i.e. whilst the momenta add in vector fashion, the scalar nature of kinetic energy means you just sum the magnitudes of each one.

If the particles stick together on collision, the momentum will still be zero (as it must), whilst the total kinetic energy will also be zero (the difference must appear as energy in other forms).

In other words, whilst there is a connection between the individual momentum and kinetic energy of each particle. There is no relationship between the total momentum and the total kinetic energy of all the particles.

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  • $\begingroup$ It's worth noting that $K=p^2/2m$ only holds for rigid objects, or any collection of particles that all move together, but not for systems of objects with nonzero relative velocities. $\endgroup$ – David Z Nov 9 '14 at 11:04
  • $\begingroup$ @David Z Precisely. There is no relationship between the KE and total momentum of a group of "particles". $\endgroup$ – Rob Jeffries Nov 9 '14 at 12:06
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I'm 27 and since I was about 15 I had the same doubt you do. Only a couple of years ago I realized why momentum is always conserved in a collision, whereas the same is not enforced for energy. (They must have told me this at some point -- or points --, but I guess sometimes I just don't pay much attention)

First of all, let's make it empirically clear that there is no hard and fast constraint on how energy will behave in a collision. Imagine different materials involved in a head-on collision (billiard balls, basketballs, you know the drill) and we're done with that.

However, conservation of momentum is much, much more straightforward. Let's recapitulate the basics: momentum is $\vec p_0 = m\vec v_0$. When a force $\vec F$ acts on our body for $\Delta t$ seconds, the new momentum will be $\vec p = \vec p_0 + \vec F\Delta t$.

$\vec F\Delta t$ is called impulse and it even has the same units as momentum (that's why we can sum them): $[m][v]=kg\cdot\frac{m}{s}=\frac{kg\cdot m}{s^2}\cdot s=[F][\Delta t]$.

So, what happens during a collision is that, during that brief amount of contact time, two objects exert force on one another. According to Newton's Third Law, the two forces are of equal magnitude and opposite directions. This is true in every instant of that brief amount of time (would be true if it was a large amount of time too).

So, for every infinitesimal amount of time $\mathrm dt$ during contact, if object $A$'s momentum gains the impulse $\vec F\mathrm dt$, then it is apparent that object $B$'s momentum will be decreased of $-\vec F\mathrm dt$! And, in the end of the collision, the variation of momentum in $A$ will be the same as in $B$ (but reversed). After all, both objects suffered the exact same force $F$ (but reversed), for the exact same amount of time!

We can state that as (if $\vec F$ is the force acted upon object $A$): $$\Delta\vec p_A = \int_0^T\vec F(t)\mathrm dt$$ $$\Delta\vec p_B = \int_0^T-\vec F(t)\mathrm dt$$ $$\Rightarrow\Delta\vec p_A+\Delta\vec p_B=\vec0$$

SUMMING UP: momentum conservation is an obvious consequence of Newton's Third Law, while energy conservation, although it exists, develops in more subtle and elusive ways, like conversion from one kind to another (in collisions, usually it goes from kinetic to thermal).

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Is there no relation between the momentum and kinetic energy? Are they not linked with each other?

No, they're not. Conservation of momentum only says that $$m_1 v_{1_-} + m_2 v_{2_-} = m_1 v_{1_+} + m_2 v_{2_+}$$ I used subscripts 1 and 2 denote the particles involved in the collision and the subscripts - and + denote the pre- and post-collision conditions. There is nothing in the above that dictates that $$m_1 {v_{1_-}}^{\!2} + m_2 {v_{2_-}}^{\!2} = m_1 {v_{1_+}}^{\!2} + m_2 {v_{2_+}}^{\!2}$$ which is what would be required to conserve kinetic energy. There is no such thing as a law of conservation of kinetic energy. There are many ways in which momentum can be conserved but kinetic energy is not conserved. For example, consider a bullet shot into a block of wood, which is a common way to measure muzzle velocity. This is a purely inelastic collision. Kinetic energy is not conserved in this case; it's not even close.

Kinetic energy is only conserved in purely elastic collisions. Purely elastic collisions are a fiction. There is no such thing as a purely elastic collision except at the atomic scale.

Even momentum is only approximately conserved in "real world" collisions. For example, when two cars collide, the collision event is not quite instantaneous. Friction between tires (and possibly other parts of the car) and road takes place during the brief timespan during which the collision event occurs. This means that the combined momentum of the two cars just before they collided and their combined momentum just after the dust settles are not quite equal to one another. The pre- and post- combined momenta are however very nearly equal because the collision event is very short in duration.

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Suppose total kinetic energy were conserved (i.e. if you can ignore potential energy changes), then because it's a nonlinear fuunction of the velocities, the sum of the kinetic energies of the particles that make up a composite body does not equal the kinetic energy of the center the center of mass of the object.

In practice, what this means is that you cannot describe the macroscopic world in terms of only macroscopic variables. So, even if you are interested in macroscopic balls that collide with each other and you want to describe that in terms of the relevant macroscopic varables like the mass velocities and center or mass of the balls, then this will not work, because energy will leak into the microscopic degrees of freedom and you need to keep that into account. Since there are so many microscopic degrees of freedom (each ball is made out of a large number of atoms), this looks like an impossible task. However precisely because there are so many degrees of freedom, a statistical approach is feasible leading to a thermodynamic description (heat, temperature etc. which are actually statistical concepts).

Now, you can actually consider an ideal elastic collision where no energy is dissipated and derive that momentum is conserved. This works as follows. Conservation of kinetic implies that:

$$\sum_{i}m_i {\bf v}_i^2 = \sum_{i}m_i {\bf u}_i^2 $$

where the $v_i$ are the initial velocities and the $u_i$ are the final velocities. Now, this is relative to some reference frame. Suppose that I am in another frame of reference that is moving with velocity $\bf{w}$ in that reference frame. Then in terms of the velocities in my frame, conservation of kinetic energy yields the equation:

$$\sum_{i}m_i ({\bf v}_i-{\bf w})^2 = \sum_{i}m_i ({\bf u}_i-{\bf w})^2 $$

If you then expand the squares inside the summations:

$$({\bf v-w})^2 = {\bf v}^2 - 2 {\bf v\cdot w} + {\bf w}^2$$

then you get back conservation of energy in the first frame, conservation of momentum in the direction that I'm moving in and conservation of mass (which we already assumed by taking the initial and final masses to be equal). Since my velocity $\bf w$ is arbitrary, we have derived conservation of momentum from conservation of kinetic energy.

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Individual momenta are not conserved, only the total momentum is. It is not entirely determined with the kinetic energy, there are some additional "degrees of freedom" or "dimensions", loosely speaking.

The total energy is conserved too, you just have to consider a potential energy of interaction during (even elastic) collision.

Heat released in a body is an increase of kinetic and potential energy of its constituents.

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