Conservation of linear momentum

In summary, the problem involves a positive pion at rest decaying into a positive muon and a neutrino with a measured kinetic energy of 4.1 MeV. Using both nonrelativistic and relativistic equations, the mass of the pion is found to be 139.52 MeV. The mass of the neutrino is assumed to be zero, and it is shown that it can only have kinetic energy in relativistic theory. Nonrelativistically, the neutrino has zero kinetic energy.
  • #1
davidpotts
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Homework Statement


A positive pion at rest decays to a positive muon and a neutrino. The kinetic energy of the muon has been measured to be T(muon) = 4.1 MeV. The mass of the muon is known from other experiments to be 105.7 MeV. Find the mass of the pion. Do this nonrelativistically, and then repeat your calculation relativistically.

Homework Equations


Nonrelativistic: T = p^2 / 2m
Relativistic: T = E - mc^2; E^2 = (pc)^2 + (mc^2)^2
p(pion) = p(muon) + p(neutrino) = 0

The Attempt at a Solution


Since I'm given T and m for the muon, I can find p(muon) from the above formulas, both nonrelativistically and relativistically. And by conservation of linear momentum, I know that p(neutrino) = -p(muon). But from here I'm stumped. I can't find out anything more about the neutrino, because there's no further data. And even if I could, I wouldn't know what to do with it. For example, are the masses of the muon and neutrino supposed to be simply added to find the mass of the pion? I feel like I'm not being given enough information to solve this.
 
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  • #2
The mass of the neutrino ought to be zero, so its Kinetic Energy is pc .
Then you can add all 3 Energies together (including mass Energy), to get the initial Energy.
 
  • #3
lightgrav said:
The mass of the neutrino ought to be zero, so its Kinetic Energy is pc .
Then you can add all 3 Energies together (including mass Energy), to get the initial Energy.

Is everybody just supposed to know that m(neutrino) = 0? Maybe so, I have very little formal physics training...

Anyway, so using the relativistic equations, we can get:

T(muon) = 4.1 = E(muon) - m(muon)c^2
= 4.1 = E(muon) - (105.7 / c^2)(c^2)
So E(muon) = 4.1 + 105.7 = 109.8 MeV

Now E^2 = (pc)^2 + (mc^2)^2
So 109.8^2 = (pc)^2 + 105.7^2
And pc(muon) = 29.72.

By conservation of linear momentum, p(muon) = p(neutrino).

So pc(neutrino) = 29.72.
And E(neutrino)^2 = (pc)^2 + (mc^2)^2
= 29.72^2 + (0)(c^2)^2
= 29.72^2
So E(neutrino) = 29.72 MeV

Now by conservation of energy, E before pion decay = E after = E(muon) + E(neutrino).
So E(pion) = 109.8 + 29.72 = 139.52 MeV

And since the pion is at rest, E(pion) = mc^2 = 139.52 MeV, which is the right answer.

What about the nonrelativistic case? Here we have T = p^2 / 2m.

So T(muon) = 4.1 = p^2 / (2)(105.7 / c^2)
So p(muon) = [(4.1)(2)(105.7 / c^2)]^.5
= 29.44 / c
Or pc(muon) = 29.44, which is very close to what we found in the relativistic case.

And is the idea that that's as far as we can go nonrelativistically? I don't see any nonrelativistic equations that allow you to have a nonzero momentum with a mass of zero. So it looks like a neutrino can't have kinetic energy in nonrelativistic theory. The neutrino is a relativistic animal. Is that right?
 

FAQ: Conservation of linear momentum

What is the law of conservation of linear momentum?

The law of conservation of linear momentum states that the total momentum of a closed system remains constant over time, meaning that the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction.

Why is the conservation of linear momentum important?

The conservation of linear momentum is important because it is a fundamental law of physics that helps us understand and predict the behavior of objects in motion. It also allows us to analyze and solve problems involving collisions and other interactions between objects.

Does the conservation of linear momentum apply to all types of motion?

Yes, the law of conservation of linear momentum applies to all types of motion, including both linear and rotational motion. However, it only applies to closed systems, meaning that there are no external forces acting on the system.

What are some real-world applications of the conservation of linear momentum?

The conservation of linear momentum has many real-world applications, including in the fields of engineering, sports, and transportation. For example, understanding the conservation of momentum is crucial in designing car safety features, analyzing the motion of projectiles, and predicting the trajectory of satellites in space.

Can the conservation of linear momentum be violated?

No, the conservation of linear momentum is a fundamental law of physics and cannot be violated. However, it may seem like it is being violated in certain situations, such as when an object appears to change direction without any external force acting on it. In these cases, there may be unseen forces at play, such as friction or air resistance.

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