K Energy & Momentum: Driving Force or Lost Heat?

[mark2142]

https://www.physicsforums.com/threa...y-in-inelastic-collisions.311037/post-2182192
If I understand correctly mathematically the momentum of the system remains unchanged but individual momentums decreases always. In an inelastic collision the momentum always decreases and so is the KE. KE is due to momentum of the body. Whenever this happens the lost energy goes into heat or thermal energy of the body.

When a force is applied internally the momentum changes and then so is KE. This lost KE goes to heat.
So is momentum driving the body forward or KE?

 

[anuttarasammyak]

When a force is applied internally the momentum changes and then so is KE. This lost KE goes to heat.
So is momentum driving the body forward or KE?

The equation of motion is
$$F=\frac{dp}{dt}=\frac{dp}{dE}\frac{dE}{dt}=\frac{\sqrt{2m}}{2\sqrt{E}}\frac{dE}{dt}$$
We can express force by dp/dt and by dE/dt. They are equivalent.

[EDIT] For 3D case
$$\mathbf{F}=\frac{d\mathbf{p}}{dt}$$
$$\frac{\mathbf{p}\cdot\mathbf{F}}{p}=\frac{\sqrt{2m}}{2\sqrt{E}}\frac{dE}{dt}$$

 

[PeroK]

https://www.physicsforums.com/threa...y-in-inelastic-collisions.311037/post-2182192
If I understand correctly mathematically the momentum of the system remains unchanged but individual momentums decreases always.

Momentum is a vector and the total momentum of the system remains constant. The magnitude of the momentum of an object may increase or decrease or stay the same in a collision.

In an inelastic collision the momentum always decreases and so is the KE.

No. There is always a loss of KE, but no momentum is lost.

KE is due to momentum of the body.

True, but KE is a scalar, whereas momentum is a vector.

So is momentum driving the body forward or KE?

Neither. By Newton's first law a body moves uniformly unless acted on by an unbalanced external force. Such a force changes the momentum and may or may not change the KE. E.g. a particle moving uniformly in a circle has continuously changing (direction of) momentum but constant KE.

 

[Delta2]

If I understand correctly mathematically the momentum of the system remains unchanged but individual momentums decreases always

The total momentum of a system remains unchanged (constant) both in magnitude and direction. I am not sure what exactly you mean by "individual momentum" but if the momentum of a subsystem decreases, then the momentum of another subsystem must increase so that the total momentum of the whole system remains unchanged.

 

[mark2142]

The magnitude of the momentum of an object may increase or decrease or stay the same in a collision.

Ok. I thought both body looses momentum because both looses velocity. So are you saying that the velocities are exchanged between the bodies in other words momentum (mass is constant) but the KE decreases and gets converted to heat? But KE is associated to velocity? I don't get it.
Also $KE=\frac{p^2}{2m}$. So if momentums are exchanged so has to KE?

 

[PeroK]

Ok. I thought both body looses momentum because both looses velocity. So are you saying that the velocities are exchanged between the bodies in other words momentum (mass is constant) but the KE decreases and gets converted to heat? But KE is associated to velocity? I don't get it.
Also $KE=\frac{p^2}{2m}$. So if momentums are exchanged so has to KE?

You need to write out 100 times:

Momentum is a vector
Momentum is a vector
Momentum is a vector ...

 

[PeroK]

Ps there is always a frame of reference where the total momentum is zero. It's called the centre of momentum frame. In that frame you cannot lose momentum.

What you are talking about is the magnitude of momentum of each particle/object. But, the magnitude of the momentum is not the same thing as momentum

Momentum is a vector.

 

[anuttarasammyak]

Let us see a system made of N>>1 similar particles of mass m and its total momentum is zero which is conserved during the process of course. Let the particles do elastic collision. The conservation laws are
$$\sum_{i=1}^N \mathbf{p_i}=0$$
$$\sum_{i=1}^N \mathbf{p_i}^2=2mE=const.$$
where elastic collisions take place.
State 1:
$$\mathbf{p}_1=-\mathbf{p}_2\ where\ |\mathbf{p}_1|=|\mathbf{p}_2|=\sqrt{mE}$$
and
$$p_3=p_4=p_5=...=p_N=0$$
Only two particles have non zero momentum.
State 2:
$$p_1=p_2=p_3=...=p_N=\sqrt{\frac{2mE}{N}}$$
All the particle have equal magnitude of momentum where they cancel as vectors. There are a lot of other cases. After many collisions we may feel that both state 1 and state 2 are NOT NATURAL distributions of momentum/energy among N particles. With some postulates we can get statistical solution of the distribution.

This example may give you a suggestion on your concern of KE decrease, e.g. starting from initial state 1, the two particles would lose energy and momentum in magnitude by collisions thus the state of the system transfers to others e.g. state 2.

 

[mark2142]

Ps there is always a frame of reference where the total momentum is zero. It's called the center of momentum frame. In that frame you cannot lose momentum.

I get it. The momentum of the COM (mass) cannot change without the external force. So even though the sum of the momentums can't change the individual momentum can? And so the KE of the particles loses into heat energy?

 

[PeroK]

I get it. The momentum of the COM cannot change without the external force. So even though the sum of the momentums can't change the individual momentum can? And so the KE of the particles loses into heat energy?

The individual particles may individually lose or gain KE even in an elastic collision. But, the total (vector) momentum remains the same.

One classic example of an elastic collision is a particle of mass $m$ with velocity $\vec v$ hitting another particle of mass $m$ at rest. The first particle stops and the second particle moves off with the original velocity $\vec v$. All KE is transferred from one particle to another.

If we look at that collision from the centre of momentum frame, then the particles are initially moving towards each other with velocities $\pm \vec v/2$. And, after the collision, each particle has retained all it KE and reversed its direction of motion.

In the inelastic version, each particle would reverse its direction but lose some KE. The total momentum remains zero, but the speed of each particle would be less after the collision than it was before.

 

[mark2142]

The total momentum remains zero,

In which? In the inelastic version of this

One classic example of an elastic collision is a particle of mass with velocity hitting another particle of mass at rest.

 

[PeroK]

In which? In the inelastic version of this

Always, elastic and inelastic. Momentum is always conserved (in a closed system) as a consequence of Newton's third law.

KE is only conserved when no energy is lost to other forms of energy (like sound or internal heat energy).

 

[mark2142]

Always, elastic and inelastic. Momentum is always conserved (in a closed system) as a consequence of Newton's third law.

KE is only conserved when no energy is lost to other forms of energy (like sound or internal heat energy).

Thank you for sparing some time for me. I get it now.