Momentum in a Head-on collision

  • #1
Rayan
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Misplaced Homework Thread
An electron and muon collide head-on, with energies 35 GeV and 50 GeV, the following reaction takes place:

$$ e^- + \mu^+ \rightarrow \nu_e + \nu_{\vec{\mu}} $$If the electron neutrino has energy of 25 GeV, and collides at angle 20 with respect to incoming electron direction, what is the muon neutrinos momentum?

Now I calculated the center of mass frame energy, which happened to be 84 GeV, and I was thinking that I can use energy conservation to determine muon neutrinos energy:

E = 60 GeV

and then since neutrinos mass is very small it can be neglected, so E ~ pc and therefore p~E/c .

But why is the angle between electron and electron neutrino given? Am I missing something here?
 
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  • #2
You calculated the magnitude of the momentum. The momentum is a vector and you should find its direction.

It's interesting that the problem statement gives you more information than required to solve the problem. Just giving the angle or the energy would be sufficient.
 
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  • #3
mfb said:
You calculated the magnitude of the momentum. The momentum is a vector and you should find its direction.

It's interesting that the problem statement gives you more information than required to solve the problem. Just giving the angle or the energy would be sufficient.
You are totally right! I did calculate the angle and got

$$\theta=10.8^{\circ}$$

with respect to the direction of incoming muon.

But I did indeed use the energy in my calculation! I can't really think of a way to determine the angle without using it?
 

Related to Momentum in a Head-on collision

What is momentum in a head-on collision?

Momentum in a head-on collision refers to the quantity of motion possessed by the colliding objects. It is the product of an object's mass and velocity. During the collision, the total momentum of the system (both objects) is conserved, meaning the total momentum before the collision is equal to the total momentum after the collision.

How is momentum conserved in a head-on collision?

Momentum is conserved in a head-on collision due to the principle of conservation of momentum, which states that in the absence of external forces, the total momentum of a closed system remains constant. This means that the sum of the momenta of the colliding objects before the collision is equal to the sum of their momenta after the collision.

What is the difference between elastic and inelastic head-on collisions?

In an elastic head-on collision, both momentum and kinetic energy are conserved. The colliding objects bounce off each other without any loss of kinetic energy. In an inelastic head-on collision, momentum is conserved, but kinetic energy is not. Some of the kinetic energy is converted into other forms of energy, such as heat or sound, and the objects may stick together after the collision.

How do you calculate the final velocities in a head-on collision?

To calculate the final velocities in a head-on collision, you need to apply the conservation of momentum and, if applicable, the conservation of kinetic energy. For elastic collisions, you can use the equations: \[ v1' = \frac{(m1 - m2)v1 + 2m2v2}{m1 + m2} \]\[ v2' = \frac{(m2 - m1)v2 + 2m1v1}{m1 + m2} \]For inelastic collisions, you use the conservation of momentum equation: \[ m1v1 + m2v2 = (m1 + m2)v' \]where \( v' \) is the common velocity after the collision if the objects stick together.

What factors affect the outcome of a head-on collision?

The outcome of a head-on collision is affected by several factors, including the masses of the colliding objects, their initial velocities, the nature of the collision (elastic or inelastic), and any external forces acting on the system. The material properties of the objects, such as their elasticity and structural integrity, also play a role in determining how they interact during the collision.

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