Transition from Elastic to Inelastic collision

[Saptarshi Sarkar]

Homework Statement

A neutron moving with speed v (v<<c) collides head on with a H-atom kept at rest. The minimum K.E of the neutron for which inelastic collision takes place with the neutron and H-atom, both assumed to be of mass m, travel with velocities $v_1$ and $v_2$ respectively, after collision?

a) $T>2mv_1v_2$
b) $T>mv_1v_2$
c) $T>\frac {mv_1v_2} 2$
d) $T>\frac {mv_1v_2} 4$

Relevant Equations

By conservation of momentum :

$mv=mv_1+mv_2$

If the collision was elastic, we would have conservation of kinetic energy :

$\frac {mv^2} 2 = \frac {mv_1^2} 2 + \frac {mv_2^2} 2$

I know that if the collision was not elastic, some of the kinetic energy of the incident neutron wound be used up in some other process. But, I can't understand how I can figure out exactly how much. Even if I can calculate it, I don't know how to find the condition for the collision to go from elastic to inelastic.

 

[TSny]

Homework Statement:: A neutron moving with speed v (v<<c) collides head on with a H-atom kept at rest. The minimum K.E of the neutron for which inelastic collision takes place with the neutron and H-atom, both assumed to be of mass m, travel with velocities $v_1$ and $v_2$ respectively, after collision?

The grammar of the second sentence is confusing. Please check to make sure that you copied the question exactly. Is this a translation from another language?

 

[Saptarshi Sarkar]

The grammar of the second sentence is confusing. Please check to make sure that you copied the question exactly. Is this a translation from another language?

This is what the source book says, maybe it's a misprint.

It should be :

A neutron moving with speed v (v<<c) collides head on with a H-atom kept at rest. What is the minimum K.E of the neutron for which inelastic collision takes place?

Mass of both particles is assumed to be m. They travel with velocities $v_1$ and $v_2$ respectively after collision.

 

[kuruman]

I have verified that one of the answers is correct.
Hint: You know that momentum is conserved. This means that the velocity of the center of the mass does not change. Hence the kinetic energy $K_{cm}$ of the center of mass does not change whether the collision is elastic or not. The neutron must have enough energy to make this happen.

 

[PeroK]

Homework Statement:: A neutron moving with speed v (v<<c) collides head on with a H-atom kept at rest. The minimum K.E of the neutron for which inelastic collision takes place with the neutron and H-atom, both assumed to be of mass m, travel with velocities $v_1$ and $v_2$ respectively, after collision?

a) $T>2mv_1v_2$
b) $T>mv_1v_2$
c) $T>\frac {mv_1v_2} 2$
d) $T>\frac {mv_1v_2} 4$
Relevant Equations:: By conservation of momentum :

$mv=mv_1+mv_2$

If the collision was elastic, we would have conservation of kinetic energy :

$\frac {mv^2} 2 = \frac {mv_1^2} 2 + \frac {mv_2^2} 2$

I know that if the collision was not elastic, some of the kinetic energy of the incident neutron wound be used up in some other process. But, I can't understand how I can figure out exactly how much. Even if I can calculate it, I don't know how to find the condition for the collision to go from elastic to inelastic.

You seem to post quite a few questions that look like nonsense to me! Where do you find them? How can there be a "minimum" KE for a collision to be inelastic?

In an elastic collision, $v_1 = 0$. So, $v_1, v_2 \ne 0$ implies an inelastic collision. Then you might look for a relationship between $T$ and $mv_1v_2$. But that seems pointless to be honest.

The quantity you are being asked to calculate looks like the loss in KE.

That said, where does the KE in a collision between atomic particles go?

 

[neilparker62]

Does minimum KE correspond to the collision being perfectly inelastic ?

 

[kuruman]

Does minimum KE correspond to the collision being perfectly inelastic ?

I don't think so. I think that the assumption here is that $v_1$ and $v_2$ are given, they are non-zero and not necessarily equal. Given that assumption, one needs to find the minimum neutron energy that will provide the given particle speeds and leave enough kinetic energy for the CM.

 

[gneill]

If the KE of the neutron is less than the binding energy of deuterium, will the neutron stick? Just thought I'd ask .

 

[PeroK]

If the KE of the neutron is less than the binding energy of deuterium, will the neutron stick? Just thought I'd ask .

The velocities of both particles after the collision are given ($v_1, v_2$). I suspect the question setter doesn't know what they are talking about.

 

[haruspex]

If the KE of the neutron is less than the binding energy of deuterium, will the neutron stick? Just thought I'd ask .

Isn't that backwards? Binding energy is energy released by the binding of nucleons, not energy required to bind them.
On the other hand, a certain threshold has to be overcome to achieve nucleosynthesis...

 

[gneill]

Isn't that backwards? Binding energy is energy released by the binding of nucleons, not energy required to bind them.

Yeah, I was typing faster than I was thinking.

 

[haruspex]

The velocities of both particles after the collision are given ($v_1, v_2$). I suspect the question setter doesn't know what they are talking about.

Yes, that last part is weird . But it doesn’t say v1 and v2 are different

 

[Saptarshi Sarkar]

If the KE of the neutron is less than the binding energy of deuterium, will the neutron stick? Just thought I'd ask .

Could the neutron somehow change the hydrogen atom to be a deuterium one like what happens in nuclear reactors using light water as moderator? That would make the collision inelastic and I guess the minimum kinetic energy required will be that excitation energy.

 

[Saptarshi Sarkar]

You seem to post quite a few questions that look like nonsense to me! Where do you find them?

I am preparing for a test called IIT JAM in India and these are some of the questions I can't wrap my head around.

The quantity you are being asked to calculate looks like the loss in KE.

That said, where does the KE in a collision between atomic particles go?

After reading @gneill 's reply, I am guessing the KE went to change the Hydrogen atom into a Deuterium one.

 

[gneill]

After reading @gneill 's reply, I am guessing the KE went to change the Hydrogen atom into a Deuterium one.

I suspect that the question posers were not going in that direction. The deuterium idea was just an idle thought on my part. Sorry if I caused any confusion.

 

[PeroK]

I am preparing for a test called IIT JAM in India and these are some of the questions I can't wrap my head around.

The thing that is useful to know is for equal mass one-dimensional collisions you have only two possibilities. Note that these are non-relativistic collisions.

1) If energy is conserved, then the first particle stops and the second moves off with all the initial KE. This is the only solution that conserves momentum and energy.

Note that the first particle cannot rebound. You should confirm why this is.

2) If energy is not conserved, then both particles must move in the initial direction. And, by conservation of momentum we have $v_1 + v_2 = v$.

You can now calculate the KE before and after the collision and calculate the loss in KE. This is what I suggest you do to understand this problem. Then you should be able to answer the muddled question you've been given.

Note that the maximum possible loss in KE is where $v_1 = v_2$. This is called a totally inelastic collision.

Note, however, that inelastic collisions should not be relevant to non-relativistic atomic collisions. Consider, for example, the collision of air molecules with each other. Where would the dissipated energy go? Internal thermal energy of the air is the KE of air molecules! And sound is also the movement of air molecules!

Inelastic collisions are only relevant for larger objects, where KE energy can be lost to internal thermal energy and sound.

Note, finally, for relativistic collisions we have different formulae for energy and momentum and an elastic collision is redefined as one where rest mass is conserved and an inelastic collision as one where rest mass in not conserved. But that's a different ball game.

 

[Saptarshi Sarkar]

The thing that is useful to know is for equal mass one-dimensional collisions you have only two possibilities. Note that these are non-relativistic collisions.

1) If energy is conserved, then the first particle stops and the second moves off with all the initial KE. This is the only solution that conserves momentum and energy.

Note that the first particle cannot rebound. You should confirm why this is.

2) If energy is not conserved, then both particles must move in the initial direction. And, by conservation of momentum we have $v_1 + v_2 = v$.

You can now calculate the KE before and after the collision and calculate the loss in KE. This is what I suggest you do to understand this problem. Then you should be able to answer the muddled question you've been given.

Note that the maximum possible loss in KE is where $v_1 = v_2$. This is called a totally inelastic collision.

Note, however, that inelastic collisions should not be relevant to non-relativistic atomic collisions. Consider, for example, the collision of air molecules with each other. Where would the dissipated energy go? Internal thermal energy of the air is the KE of air molecules! And sound is also the movement of air molecules!

Inelastic collisions are only relevant for larger objects, where KE energy can be lost to internal thermal energy and sound.

Note, finally, for relativistic collisions, we have different formula for energy and momentum and an elastic collision is redefined as one where rest mass is conserved and an inelastic collision as one where rest mass in not conserved. But that's a different ball game.

Thanks! This was very informative!

 

[kuruman]

As I intimated in #4 and #7 the kinetic energy of the the center of mass plays a role here. The relevant momentum and energy conservation equations are$$v=v_1+v_2~~~~~~(1)$$ $$T=\frac{1}{2}m(v_1+v_2)^2=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2+Q~~~~~~(2)$$where Q is the energy lost in the collision. It is clear from equation (2) what $Q$ ought to be in terms of the mass and the final speeds. When Q = 0, the collision is elastic. That sets the floor for $Q$. For an inelastic collision Q must be greater than zero. Can Q increase indefinitely if the initial speed is fixed? The answer is no because at some point we violate the conservation equations. The maximum value of Q is reached when the initial energy $T$ is allocated to two pieces after the collision, one is the kinetic energy of the center of mass (a conserved quantity) and the other is Q.$$T_{min}=\frac{1}{2}(2m)V_{cm}^2+Q_{max}~~~~~~(3)$$So in what sense is the initial kinetic energy minimum? In the sense that $T$ cannot be less than the above value without violation of the conservation laws. The rest is algebra to get one of the answers provided.

 

[PeroK]

... an alternative analysis is:
$$v = v_1 + v_2$$
$$T = \frac 1 2 mv^2 = \frac 1 2 m (v_1 + v_2)^2 = T_1 + T_2 + mv_1v_2$$

So, we have $T > mv_1v_2$ and inequalities b), c) and d) all hold.

If we take the case $v_1 = v_2 = \frac v 2$, then we see that:
$$2mv_1v_2 = 2m(\frac v 2)(\frac v 2) = \frac 1 2 mv^2 = T$$
Which is counterexample to the strict inequality a), which holds in all other cases incidentally(!).

Then - insert some muddled question - b) is the answer!

Although, for any inelastic collision $T \ge 2mv_1v_2$ would, IMO, be the answer to a better question.