We know that in an inelastic collision that total momentum of the system before collision equals the total momentum after collision. But total kinetic energy before collision is not equal to total kinetic energy after collision.

How is possible given that the formula of momentum is $mv$ and the formula of kinetic energy is $\frac{1}{2} mv^2$; both are dependent on mass and velocity.

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    $\begingroup$ The premise of the question is incorrect, if we consider the momentum and kinetic energy of all particles in the collision. Heat is kinetic energy... $\endgroup$
    – DJohnM
    Aug 27 '14 at 17:38

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

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    $\begingroup$ 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

Here are two separate ways to address the issue you bring up. One is more mathematical---comparing the relations $mv$ and $\frac{1}{2}mv^2$. The other has more to do with force and energy, which I'm calling physical.


Let's imagine two objects that are moving in the same direction collide with each other. Just to keep things simple, let's also imagine that they move in the same direction after the collision. (This can always be set up, so you don't lose anything by assuming it.)

Before and after the collision, the quantity

$$p_\text{tot} \equiv m_1v_1 + m_2v_2 \tag{1}$$

is unchanged. The speeds may have changed from before & after the collision, but you can plug in either set (either the initial speeds or the final speeds) that sum won't change.

Now what can be said about the quantity

$$2K_\text{tot} \equiv m_1v_1^2 + m_2v_2^2?\tag{2}$$

(I moved the $\frac{1}{2}$ to the other side; hope that's okay with you. Just makes the expression look more similar.) Well, not much really. They are both composed of the same quantities, but they aren't necessarily the same because there's no mathematical way to manipulate Eqn. 1 to make it look like Eqn. 2. Try it, you won't be able to. Here's what I mean. I can multiply $p_\text{tot}$ by $v_{1f}$ (that's object 1's final velocity) and end up with an invented quantity I'm calling $Q$:

$$Q\equiv p_\text{tot}v_{1f} = m_1v_1v_{1f} + m_2v_2v_{1f}. \tag{3}$$

Now that quantity is the same before and after the collision. How do I know? Because $p_\text{tot}$ is the same, so $p_\text{tot}$ multiplied by the same number $v_{1f}$ must also be the same.

That's what I mean when I said you can't manipulate $p_\text{tot}$ to make it look like kinetic energy. So there's no reason kinetic energy should be the same before and after the collision.


The momentum of a system of objects is the same before and after the collision if the net impulse on the system is zero:

$$\int F_\text{net}\,dt = \Delta p$$

That is Newton's 2nd law, but written in a different form than you might have seen.

So now we know when and "why" momentum is constant. What about kinetic energy? That's actually harder. The governing equation is

$$\sum_i \vec F_i\cdot d\vec s = \Delta K + \Delta U + \Delta E_\text{thermal} + \cdots$$

In other words, the sum of the external works on your system equals the change in total energy, but that doesn't tell you anything about the kinetic energy. Energy can change forms. So if kinetic energy is lost in some collision, it went into potential, thermal, etc.


Let's take an example with simple numbers :



This can represent the momentum conservation. Now look at the sum of squares :



The sum is not conserved because the momentum that was transferred changed differently the result of the squares. In a word, kinetic energy doesn't change linearly with speed (which is obvious since it's a square).


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