# Maximum and Minimum Kinetic Energies following decay

1. Oct 26, 2013

### paodealho

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

Conservation of Momentum: MV = m1v1 + m2v2

3. 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!

2. Oct 26, 2013

### 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.

3. Oct 26, 2013

### paodealho

Hi,

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.