# Why can we still assume momentum is conserved in an inelastic collision?

This is very similar to this question:

How can momentum but not energy be conserved in an inelastic collision?

However, I feel as though an important caveat was not resolved for me which is why I am asking it here.

We have that in a collision (namely, an inelastic one) energy is not necessarily conserved among the objects in the collision. A common explanation that makes sense to me is that the energy of the collision is converted into heat/sound energy in the outlying particles. However, if these outlying particles that were once stationary or moving more slowly begin to move more quickly (due to an increase in kinetic energy), wouldn't their momentum increase as well due to an increase in velocity?

One explanation I have for this is because particles will vibrate in such a way that the net velocity remains unchanged (i.e. there will be equal acceleration in all directions among the particles that heat up). However, I have no conclusive reason to believe that this is the case.

• That momentum is conserved in an inelastic collision is an observation from experiment rather than an assumption.
– user256444
Commented Jun 2, 2020 at 1:22

When two bodies make inelastic collision with each other, some of their total kinetic energy is lost in plastic deformation (i.e. can't be fully recovered) at the point of contact, heat or sound. This results in decrease of total kinetic energy.

On the other hand, linear momentum ($$m\vec v$$) is a vector quantity which has both magnitude and direction for addition to or subtraction from other linear momenta. Thus when two bodies make inelastic collision with each other, they separate each other in such a way that the vector sum of their linear momenta after collision remain the same even after change in their magnitudes and directions.

• This doesn't fully answer my question. When the kinetic energy is lost due to heat or sound, the energy is transferred to the particles conducting the heat of sound. Why isn't there an increase in momentum in these particles? Commented Jun 2, 2020 at 1:17
• The total momentum is conserved, momentum is a vector quantity so if one particle is going one way and another is going the opposite way, the total momentum is not increased because they cancel each other out.
– user256444
Commented Jun 2, 2020 at 1:23

1) Energy can be converted into different forms, momentum cannot 2) even if you lose the energy into heat, those photons do not have enough momentum to make a noticeable change in momentum but they have enough energy to notice the change.

Does any part of either object break off and leave the system? If not, there's nothing to carry away momentum. That is, there is no mechanism to change the momentum. The total momentum is conserved. That means that the averaging that you suggest has to work.

• But what about the particles surrounding the system that the heat is transferred to - isn't the momentum of these particles increasing? Commented Jun 2, 2020 at 0:55
• @somil If there are particles surrounding and interacting with the system, then it is not closed, and neither energy nor momentum is conserved. Either or both can be transferred out of the system. Commented Jun 2, 2020 at 1:43

If I have to explain in simple language I will say that:

In collisions between two isolated objects Newton's third law implies that momentum is always conserved. In collisions, it is assumed that the colliding objects interact for such a short time, that the impulse due to external forces is negligible. Thus the total momentum of the system just before the collision is the same as the total momentum just after the collision.

HOPE THIS HELPED YOU..

One explanation I have for this is because particles will vibrate in such a way that the net velocity remains unchanged

This is not so. It is net momentum that is unchanged, not velocity.

i.e. there will be equal acceleration in all directions among the particles that heat up

So there is not equal acceleration in all directions among the particles that heat up but equal forces. And this is what the third law of motion says. At least if I interpret "equal in all directions" as "in every direction the inner force acting in that direction is paired with another inner force with same strength and acting in opposite direction" so they cancel each other out.

Due to third law of motion, every interaction between two constituent particles comes in pair. So any force $$\vec{F}_{ij}$$ that starts acting in isolated system is paired by another force $$\vec{F}_{ji}=-\vec{F}_{ij}$$ with same strength but opposite direction. So the interaction between $$i$$th and $$j$$th particle produces two new change of inner momenta of particles $$d\vec{p}_{ij}=\vec{F}_{ij}dt$$ and $$d\vec{p}_{ji}=\vec{F}_{ji}dt=-\vec{F}_{ij}dt=-d\vec{p}_{ij}.$$ Thus the net change is zero.

PS.

...is converted into heat/sound energy in the outlying particles.

I am not sure what exactly "outlying" means, but for momentum conservation the system must be isolated. If it is isolated, then every particle can increase momentum and it will be paired with opposite increase of momentum of another particle. Once the system is not isolated the momentum conservation depends on the situation. If for example the collision produces sound waves in surrounding air, then momentum will be conserved only if the sound waves have zero net momenta. If they do not, like in the case where sound wave is more stronger in one direction that the other, the momentum will not be conserved.

• Doesn’t net momentum remaining unchanged imply that velocity is unchanged? Doesn’t equal incident forces imply equal acceleration components? Commented Jun 3, 2020 at 4:02
• @somil only if all the bodies have the same mass. In case of two particles system with masses $m_1$ and $m_2$ and momenta $\vec{p}_1$ and $\vec{p}_2$, the conservation of momentum implies $m_1\Delta\vec{v}_1=-m_2\Delta\vec{v}_2$, where $\vec{v}_1$ and $\vec{v}_2$ are velocities of the particles. For net velocity difference you will get $\Delta\vec{v}=\Delta\vec{v}_1+\Delta\vec{v}_2=(1-m_1/m_2)\Delta\vec{v}_1$ which is not zero in general. It is the same case for forces and accelerations. Commented Jun 3, 2020 at 8:31