Maximum and Minimum Kinetic Energies following decay

[paodealho]

Homework Statement

An unstable particle of mass M = m1 + m2 decays into two particles
of masses m1 and m2, releasing an amount of energy Q. Determine
the kinetic energies of the two particles in the CM frame. Given that
m1/m2 = 4, Q = 1 MeV, and that the unstable particle is moving in
the Lab with kinetic energy 2.25 MeV, find the maximum and minimum
Lab kinetic energies of the particle of mass m1.

Homework Equations

Conservation of Energy: $\frac{1}{2}$MV² = $\frac{1}{2}$m₁v₁² + $\frac{1}{2}$m₂v₂²

Conservation of Momentum: MV = m₁v₁ + m₂v₂

The Attempt at a Solution

In the center of mass frame, the kinetic energy and momentum are zero initially. The particle decays and releases energy Q which becomes the kinetic energy of the two particles following decay in this frame.

So Q = $\frac{1}{2}$m₁v₁² + $\frac{1}{2}$m₂v₂² (1)

and m₁v₁ = m₂v₂
v₁ = -m₂/m₁ * v₂ (2)

Substituting (2) into (1):
Q = $\frac{1}{2}$[m₁(-m₂/m₁ * v₂)² + m₂v₂²]

After some algebra ...

T₁ = $\frac{1}{2}$m₁v₁² = Q * m₂ / M

and

T₂ = $\frac{1}{2}$m₂v₂² = Q * m₁ / M

The first part is all fine and dandy. I'm struggling with the second portion. I don't think I'm to consider the rest mass of the particles resulting from the decay... If all of the available energy becomes kinetic energy for m₁ in the lab frame, wouldn't the maximum Kinetic energy just be 3.25 MeV for m₁ ? (the textbook indicates otherwise: 3.20 MeV)

Following that line of logic, the minimum would just be zero but something doesn't seem right about that.

My apologies for the essay. I've tried all manner of shenanigans (conservation law manipulation) to find the min/max kinetic energies but have not succeeded. Am I just over-thinking this?

Thanks!

 

[vanhees71]

Are you sure that you are supposed to use non-relativistic kinematics? From the problem setting I'd think so, because they say $M=m_1+m_2$, which is not correct in a relativistic setup.

In any case you have to rethink your energy-bilance equation! Note that in the decay an additional energy $Q$ is released.

 

[paodealho]

Hi,

Thanks for the reply!

Yes, I'm fairly certain that this is non-relativistic (classical mechanics course + problem description seem to indicate this).

With the additional energy Q released, shouldn't the equation for Energy just be:

$\frac{1}{2}$MV² + Q = $\frac{1}{2}$m₁v₁² + $\frac{1}{2}$m₂v₂² ?

So the total "available" energy for the decay products should be 3.25 MeV (KE of unstable particle + Q)?

I don't believe that the KE of either decay product can be zero due to conservation of momentum and the KE relations derived for the center of mass frame. If I'm looking for a minimum/maximum KE, it seems like I should be looking at a quadratic somewhere.