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$K.E=\frac{1}{2}mv^2$ , $P=mv$ thus there is a relation between them. I really can't understand why there is no decrease in momentum when kinetic energy is decreased in an inelastic collision. I am only a high school passed student(ready for college). Should I leave this question until I learn "Lagrangian" and "Noether's Theorem"? Because I literally read most of the answers about these kinds of questions and spend almost half a day on it, BUT I do not get it. Can anyone please answer this question? please help me.

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    $\begingroup$ You really don't need Lagrangian to understand these. The Lagrangian is just another equivalent formulation to newton's laws. $\endgroup$
    – Babu
    Commented Oct 2, 2020 at 7:24
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    $\begingroup$ Also I think it will be good if you saw this stack: [physics.stackexchange.com/questions/92051/… collisions) $\endgroup$
    – Babu
    Commented Oct 2, 2020 at 8:13
  • $\begingroup$ It means that we cannot say that kinetic energy is always proportional to momentum, but K.E is directly proportional to energy right? $\endgroup$ Commented Oct 2, 2020 at 8:21
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    $\begingroup$ Yes kinetic energy is not a linear function of momentum but a quadratic function of it. ( you only get proportionality if you have a linear function) $\endgroup$
    – Babu
    Commented Oct 2, 2020 at 8:23

8 Answers 8

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$K.E=\frac{1}{2}mv^2$, $P=mv$ thus there is a relation between them. I really cant understand why there is no decrease in momentum when kinetic energy is decreased in inelastic collision.

It doesn't make sense to consider a single body during a collision (i.e. during interaction with a second body). You need to look at both bodies together. Therefore you need to consider the total momentum of both bodies. $$\vec{P}=m_1\vec{v}_1+m_2\vec{v}_2 \tag{1}$$ It is this quantity which is conserved during the collision.

Likewise you need to consider the total kinetic energy of both bodies. $$E_\text{kin}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$ It is this quantity which is conserved during an elastic collision.

So there are many possible ways how the velocities $\vec{v}_1$ and $\vec{v}_2$ can change to make the total kinetic energy $E_\text{kin}$ decrease while still preserving the total momentum $\vec{P}$.

Should i leave this question until i learn "Lagrangian" and "Noether's Theorem" ?

To understand momentum conservation (1) you don't need Lagrangian mechanics or Noether's theorem. Newton's mechanics is just enough.

According to Newton's third law (actio = reactio) you have $$\vec{F}_{2\to 1}=-\vec{F}_{1\to 2}$$ By applying Newton's second law ($\vec{F}=m\frac{\Delta \vec{v}}{\Delta t}$) to these two forces you further get $$m_1\frac{\Delta \vec{v}_1}{\Delta t}=-m_2\frac{\Delta \vec{v}_2}{\Delta t}$$

Rearranging this you get $$\frac{\Delta(m_1\vec{v}_1+m_2\vec{v}_2)}{\Delta t}=\vec{0}$$ and hence $$m_1\vec{v}_1+m_2\vec{v}_2=\text{const}$$ which is just the above mentioned conservation of total momentum.

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Just to add a bit of more mathematical flavour to the answers, here's why momentum can stay constant, yet kinetic energy may decrease. First we set up some ground rules.

You mention that momentum is $p=mv$ and kinetic energy is $\frac12mv^2$ and that there should be a relation between them. And yes, there is. Basic algebra tells you $K=\frac{p^2}{2m}$. Well and good.

Now, consider 2 bodies colliding, or rather just interacting. In this picture, the total momentum, at say time $t=0$ is $p_1+p_2$. As others have shown, Newton's third law guarantees that $p_1+p_2$ is constant at all times. If so, then $$\frac{\partial}{\partial t}(p_1+p_2)=0$$ where by $\frac{\partial}{\partial t}$ I mean the derivative with respect to time treating all other variables like space etc as constants. This mathematical statement contains the same information as the one written in words just before that. But this will come in handy.

Now, what about the total kinetic energy? At time $t=0$, it's $K_1+K_2=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}$. To make life simpler, we assume $m_1=\frac12=m_2$. This won't change much. Hence $K_1+K_2=p_1^2+p_2^2$. It would help us if we write this as $K_1+K_2=(p_1+p_2)^2-2p_1p_2$ Then, how does the kinetic energy vary in time? To find that, we take a time derivative of the kinetic energy, and get

$$\begin{align*}\frac{\partial}{\partial t}(K_1+K_2)&=\frac{\partial}{\partial t}(p_1^2+p_2^2)\\&=\frac{\partial}{\partial t}\{(p_1+p_2)^2-2p_1p_2\}\\&=2(p_1+p_2)\frac{\partial}{\partial t}(p_1+p_2)-2\frac{\partial}{\partial t}(p_1p_2)\\&=-2\frac{\partial}{\partial t}(p_1p_2)\end{align*}$$

where the last inequality follows because of our momentum conservation equation.

The leftover term is not necessarily 0. Check this by taking a collision with two balls which stick after impact and replace the partial derivative with differences.

An elastic collision is defined as a collision where this leftover term is $0$ and an inelastic collision as one where it's not.

As to where the lost kinetic energy goes, it goes into the internal energy of the bodies themselves, like vibrational motion, heat etc.

Also, this did not require Lagrange, right? There's the answer to this question as well.

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Actually there is no relationship between Kinetic Energy and Momentum (for a generalized system of particles). You can have a non-zero kinetic energy at the same time with momentum being zero.

You can see this from the following equations:

$$\mathcal E = \frac 12 \sum m_i v_i^2$$

$$\mathbf P = \sum m_i \mathbf v_i$$

Now if $v_i \neq 0$ then $\mathcal E$ will never be zero, whereas $\mathbf P$ can easily be zero.

But if you were talking about some special cases then yes there do exist some relationship between Kinetic Energy and Momentum (which you can easily derive yourself).

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Total energy and total momentum of two interacting particles are conserved. In an inelastic collision between them, the kinetic energy is not conserved, but total momentum is still conserved.

In order that the kinetic energy is not conserved some degree of freedom other than velocity must be present. The particle(s) should have internal energy, for example a temperature, vibration or rotation, dissociation - breaking up.

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Your misunderstanding is in writing $p=mv$ and forgetting that $v$ should really be a vector, i.e. the correct relation is $\vec p=m\vec v$. Momentum can be positive or negative. On the other hand, kinetic energy $mv^2/2$ is necessarily non-negative.

Imagine a system where $m_1=m_2=m$ and the two particles approach each other with the same speed but different direction, along the $x$-axis. To simplify: $v_2=-v_1$ in 1d. Both particles have the same (positive) kinetic energy $\frac{1}{2}mv^2$ but opposite momentum $p_2=-p_1$: the momentum has a direction (and thus a sign) but the kinetic energy does not (it is a scalar quantity). This sign is important as it indicates the direction of motion.

The net momentum before collision is $P=p_1+p_2=0$ since $p_2=-p_1$. The net kinetic energy is the sum of two positive terms so it positive and certainly non-$0$.

After the collision, many $v’_2=-v’_1$ will conserve the total momentum but not necessarily the total kinetic energy. Say: $v_1’=v_1/2$ and $v_2’=-v_1/2$ will do the trick. You can verify that the total kinetic is decreased by $4$ after the collision. Note that both momenta after collision have decreased in magnitude, i.e. $ p_1’< p_1$ in magnitude, and also $p_2’<p_2$ in magnitude, but the net momentum is still $0$ because one momentum is negative and the other is positive.

The point here is that, because momentum is a vector, it’s possible to combine momenta in more than one way to add to $0$ and conserve total momentum because momentum is a signed quantity.

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The question you’re asking is why is momentum always conserved even during collisions where kinetic energy is not conserved. Consider Newton’s third law of motion which states that during any interaction, the force one body exerts on another is equal to, but opposite in direction, to that exerted by the second body on the first, so that for the rate of change of total momentum

$$\frac{dP}{dt} = 0$$

so that for a two-body collision

$$m_1 v_1 + m_2 v_2 = constant$$ and

$$F = \frac{dp}{dt} $$

which is Newton’s second law. So momentum must be conserved.

Also, kinetic energy can be transferred to other forms of energy, such as heat and sound during collisions. This cannot happen for momentum in this fashion since momentum is not “a form of energy”.

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  • $\begingroup$ your equations will look a bit nicer if you encase them in two dollars on both sides. I have done it for you now. $\endgroup$
    – Babu
    Commented Oct 3, 2020 at 6:35
  • $\begingroup$ Thanks again Buraian. Very kind of you my friend! $\endgroup$
    – joseph h
    Commented Oct 3, 2020 at 7:06
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The relation between momentum and kinetic energy for a single particle:

Momentum is defined as:

$$\vec{p} = m \vec{v}$$

So, we can write velocity as:

$$ \frac{\vec{p}}{m} = \vec{v}$$

Kinetic energy is defined as:

$$ K = \frac{1}{2} mv^2$$

Using the previous equation,

$$ K = \frac{p^2}{2m}$$

So, it's very easy to see that it can be said that kinetic energy is a function of momentum and mass for a single particle.


In an inelastic collision, the momentum is conserved but the kinetic energy is not. You may think otherwise from the previous equation I wrote because it says that kinetic energy is a direct function of momentum.

When we have a collision what happens is the total momentum is split up between the different bodies such that if we summed momentum of every single body then the total momentum is the same.

The simple idea is that in an inelastic collision, the total momentum redistributes among the objects of collision in a way that when you take the difference of kinetic energy between the final and initial state, it gives a negative number. The lost energy goes into heat and vibrational energy.


Example:

Consider two clay balls body $A$ and $B$ with the same mass m. Each momentum velocity $v$ and collide head-on undergoing inelastic collision. Let velocity after collision be $v'$

The initial kinetic energy is:

$$ K = \frac{mv^2}{2} + \frac{mv^2}{2} = mv^2$$

Now after collision, they become one body with a velocity of zero (Refer). By momentum conservation for before and after collision:

$$ mv - mv = (m+m) v'$$

Hence,

$$ v'=0$$

Putting this into the kinetic energy post-collision is:

$$ K' = \frac{ (2m) (0)^2}{2} = 0$$

So we can see that the kinetic energy $ mv^2$ was lost completely. This energy went into deforming the clay and vibrational energies as Feynman has said in a quote that I put in the references (*).

The heart of the matter is that for an individual particle, we can relate it's kinetic energy and momentum but for a system of particles the two are not directly related.


Derivation the loss in energy:

Let two bodies $A$ and $B$ with mass $m_a$ and $m_b$ respectively. Let their initial momentums be $\vec{p_a} $ and $ \vec{p_b}$ and after collision in which they stick let their momentum be $\vec{p_{ab}}$ then their loss in kinetic energy.

$$K_{i} = \frac{ (\vec{p_a})^2}{2m_a} + \frac{ (\vec{p_b})^2}{2m_b}$$

The final kinetic energy is given as:

$$ K_{f} = \frac{ (\vec{p_a} +\vec{p_b})^2}{2(m_b +m_a)}$$

Note the momentum in the final must be equal to the initial momentum of $p_a +p_b$ due to the momentum conservation.

Hence,

$$ K_f = \frac{ (\vec{p_a})^2 + ( \vec{p_b})^2 + 2 \vec{p_a} \cdot \vec{p_b}}{2(m_b +m_a)}$$

Now consider the difference of kinetic energy between final and initial states:

$$ K_f - K_i = \frac{ \vec{p_a} \cdot \vec{p_b} }{m_a + m_b} - [ \frac{(m_a \vec{p_b})^2 + (m_b \vec{p_a})^2}{2(m_a + m_b)(m_a m_b)}]$$

$$ K_f - K_i =-\bigg[ \frac{(m_a \vec{p_b})^2 + (m_b \vec{p_a})^2 - 2m_a m_b \vec{p_a} \cdot \vec{p_b}}{2(m_a + m_b)(m_a m_b)} \bigg]$$

Or,

$$ K_f -K_i = - \bigg[ \frac{|m_a \vec{p_b} - m_b \vec{p_a}|^2}{2(m_a + m_b)(m_a m_b)} \bigg] $$

Since $m_a m_b$ is a strictly positive quantity, we can move it into the modulus :

$$ K_f - K_i =\frac{m_a m_b}{2(m_a +m_b)} (|\frac{\vec{p_b}}{m_b} - \frac{\vec{p_a}}{m_a}|)^2$$

Since we can switch terms in the square modulus,

$$ K_f -K_i = \frac{m_a m_b}{2(m_a +m_b)} (|\frac{\vec{p_a}}{m_a} - \frac{\vec{p_b}}{m_b}|)^2$$

Now, we can say that kinetic energy change has been reduced in the final state. This absolute value of the amount which is lost is taken as $K_{loss}$ and is given as:

$$ K_{loss} = \frac{m_a m_b}{2(m_a +m_b)} (|\frac{p_a}{m_a} - \frac{p_b}{m_b}|)^2$$

q.e.d


Further reading:

HC-Verma: Concept's of Physics

Feynman lectures (under energy and momentum of chapter-10)

To understand these ideas in more depth, see the answer by Ron Maimon here

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You've run into a mathematical issue, not a physical one.

Kinetic energy $K=\frac 12 mv^2$ and momentum $p=mv$ may be related, but not uniquely. They share not one but two parameters, both $m$ and $v$, and their relations to them are not the same (some are linear and others squared). Therefore they are not directly related.

Imagine doubling $p$.

  • Did you do that by doubling $m$? Then $K$ also doubles.
  • Or did you double $v$? then $K$ quadruples.
  • Or did you maybe change both $m$ and $v$? Maybe you increased $v$ a lot but also lowered $m$, so it together equated to a doubled $p$? Then maybe this combination causes some entirely different change in $K$.
  • And maybe even, if they are changed just right, $K$ doesn't change at all.

All this is possible if you find the right combination of changes in the parameters.

In short: When two properties are related via multiple parameters, then they are not necessarily directly related. A change in one does not directly or necessarily correspond to a change in the other. The change may be "absorbed" in the internal changes of those parameters.

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