0
$\begingroup$

If two bodies $M_1$ and $M_2$ collide inelastically, the net external force on the system $(M_1+M_2)$ is $0$, so total momentum of the system is conserved. If we ignore the change in their mass due to deformation,then the total Kinetic energy of the system should remain same right ?
But, since some of the individual kinetic energies of the bodies is lost, then that would mean that total kinetic energy would not remain same... I'm a bit confused over this ..

(Note that I have considered a simple case of collision where the only external force on each body is their normal force to each other, other than the gravitational force and the normal from ground..)

$\endgroup$
4
  • 1
    $\begingroup$ we include gravity? Momentum is not conserved in an external gravitational field. $\endgroup$
    – JEB
    Feb 13, 2023 at 5:54
  • 1
    $\begingroup$ Kinetic energy is lost in any inelastic collision $\endgroup$
    – Bob D
    Feb 13, 2023 at 17:26
  • $\begingroup$ @JEB momentum of the system is conserved in external gravitational field only if gravity can do work on the system, here in this case I have mentioned the only external force on each ball is normal from each other, consider them moving on a smooth horizontal ground.. and therefore there is no resultant external force on the system..also, gravity and normal on both bodies from the ground are always perpendicular to the line of motion of the bodies and thus they do no work on the bodies.. $\endgroup$ Feb 14, 2023 at 1:01
  • $\begingroup$ @VenuGopalDash that's on way to look at it. Momentum is conserved if the system is invariant under translations, which is not generally true in a gravitational field. It's not clear from the question that the system is confined to 2 dimensions, in which gravity and normal forces don't matter. $\endgroup$
    – JEB
    Feb 14, 2023 at 17:33

4 Answers 4

2
$\begingroup$

In Newtonian Physics the total mass is the same before and after the collision. The total energy is also conserved, but the total energy includes many other types of energy besides kinetic energy. In an elastic collision the kinetic energy is conserved as well as the total energy. We use the word inelastic for collisions where the kinetic energy is not conserved because some of the kinetic energy is changed to other forms of energy.

The main other form is usually heat, though energy can be taken up by changing internal structures when materials are permanently deformed, by sound (though this is usually quite small), and by other things. When some kinetic energy is changed to other forms of energy the total energy remains the same, obeying the Law of Conservation of Energy.

$\endgroup$
2
$\begingroup$

By right the total energy in the system should remain the same in the context of Newtonian physics, according to the law of conservation of energy, $K_1+U_1=K_2+U_2$, where $K_1$ and $K_2$ are the initial and final kinetic energies respectively, $U_1$ and $U_2$ are the initial and final potential energies respectively. Simply put the total energy at the start of the experiment would be equal considering that it is a closed system (which in this case it is as it was stated to be an elastic collision).

If there are any loss of energy measured at the end, it is likely through a loss of energy through other factors, such as heat et cetera.

$\endgroup$
1
  • 1
    $\begingroup$ Put braces i.e., the "$" symbol at the ends of your equations to render them in Latex format. See the edits to you answer. $\endgroup$
    – joseph h
    Feb 13, 2023 at 8:16
0
$\begingroup$

Inelastic collision in itself means there is no kinetic energy conserved between two bodies. Most collisions are considered inelastic in macroscopic scale because they lose some energy as heat , sound etc. Elastic collision refers to the kinetic energy being conserved most probably in molecules because the energy they lose are negligible.

$\endgroup$
0
$\begingroup$

The answer depends on what the definition of "the system" is, and under which physical laws we're analyzing the problem.

In the simplest case, the system is two (inertial) masses (no gravity), obeying Newtonian mechanics. There is no loss of generality by analyzing the problem in the center-of-mass frame:

$$ \vec p_1 = -\vec p_2 $$

defines the momentum, and the total kinetic energy is:

$$ T = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} = p \frac{m_1 + m_2}{2m_1m_2} $$

where $p = |\vec p_1| = |\vec p_2|$.

After any collision:

$$ \vec p'_1 = -\vec p'_2 $$

and

$$ T' = p' \frac{m_1 + m_2}{2m_1m_2} $$

where $p' = |\vec p'_1| = |\vec p'_2|$.

If the collision is inelastic:

$$ p' < p $$

so clearly kinetic energy is lost:

$$ T' < T $$

Here our masses were just abstractions: structureless matter, perhaps point masses.

More realistically, one can consider them as macroscopic masses made up of microscopic constituents. In this case, the missing energy

$$ \Delta T = T-T' $$

goes into heat and deformation. The heat is just kinetic energy of the microscopic constituents...but now our system isn't two masses, it's made up of $O(N_A)$ objects. (One may also consider energy loss to sound waves, but that just add as few more $N_A$ degrees-of-freedom.) Assuming these objects are bound into masses, the deformation can store/release potential energy due to whatever binds them together.

A way to gain insight into that process is to consider two (abstract) masses, one with an ideal spring attached, in 1 dimension. When they collide, the spring is compressed and locked at the point of maximum compression ($x$), a totally inelastic collision. If you work this out using $F=ma$, you should find that:

$$ \Delta T = \frac 1 2 kx^2 $$

Thus, the missing kinetic energy is entirely potential energy.

Were the collision elastic, then the ideal spring would push the masses apart, returning all the potential energy back to kinetic energy. Of course, any realistic spring would have some inertia, and overshoot the release, and be left with some is oscillating energy: there is just no way to deliver all the energy back to the masses that leaves the spring motionless at its equilibrium position.

(If you work that out relativistically, you'll find that:

$$ \Delta M = \frac{\frac 1 2 kx^2}{c^2} $$

is required by 4-momentum conservation).

If you then generalize the 1 spring to $N_A$ intermolecular bonds, you'll find there is absolutely no way have all the internal vibrational modes excited under deformation line up and coherently dump all their energy at once back to the two masses: in reality, they are going to "ring".

There is simply no way to construct an elastic collision out of macroscopic objects. (Gravitation scattering of planets notwithstanding...though gravitational radiation will ruin that, too).

Then: you play the infinity card. Assume the masses are infinitely rigid, and bounce with no deformation. That will recover an elastic collision; however, it violates special relativity (see: Born Rigidity). An infinitely rigid object transmits the collision faster than light, and is ruled out.

The only way to save "perfect" elasticity is to bring in quantum mechanics. If the masses are coherent quantum objects (say, a proton and a $\pi^+$), then they can elastically scatter...even if there's an intermediate state such as a $\Delta^{++}$. Here the delta baryon decays and leaves a proton and pion, which are indistinguishable, from the initial proton and pion..even if they swapped quarks.

Whether macroscopic quantum objects can scatter elastically is another question, for a condensed matter expert.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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