One type of decay seen in some subatomic particles occurs when a \'parent\' part

Question

One type of decay seen in some subatomic particles occurs when a 'parent' particle (mass M) spontaneously breaks up into two smaller 'daughter' particles (masses m1, m2). If the parent is initially at rest, then momentum is conserved during this decay as the daughters move in opposite directions with equal magnitudes of momentum: p1 = p2. Energy is also conserved: although the parent is not moving, it still has rest energy due to its mass (ER = Mc2), and all of this is used to create the two daughters: ER = E1 + E2,. But the daughters are both moving, so some of the parent's rest (mass) energy is converted to kinetic energy. The result is that the total mass of the daughters is smaller than the mass of the parent: m1 + m2 < M.

The question for this experiment: is more of the parent's mass M lost (converted to KE) when the daughters have the same speed, or when they have different speeds?

To design an experiment to answer this question, choose a parent particle mass, then chose speeds for the daughter particles. Apply the momentum conservation and energy conservation principles to determine the masses of the two daughter particles.

You should use the relativistic expressions for momentum and energy conservation as your starting point:

p1 = p2 1m1v1 = 2m2v2

E1 + E2 = ER 1m1c2 + 2m2 c2 = M c2

Requirements

Mandatory units:

energy: MeV

momentum: MeV/c

mass: MeV/c2

speed: c

Speed choices:

no speed can be > c (physically impossible - so far as we know!)

first decay: choose v1 = v2 > 0.90c for both daughters

second decay: choose any two speeds you like, so long as they average out to the speed used in the first decay

e.g., if you were choosing car speeds in mph, for the first case you could choose 100 mph for both, and in the second case you could choose 50 mph and 150 mph, which average out to 100 mph. Or you could choose 120 mph and 80 mph. And so forth.

After finding m1 and m2, verify explicitly (in your solution - show work!) that

p1 = p2

E1 + E2 = ER

Hints

For the parent particle, you can choose any mass from this list of baryon masses: https://en.wikipedia.org/wiki/List_of_baryons#Lists_of_baryons, or you can make up a mass, so long as it isn't any larger than the largest mass on that list.

When considering how to solve for the daughter masses, remember that after you choose v1 and v2, 1 and 2 can be found immediately, so m1 and m2 are the only truly unknown quantities in the two relativistic equations.

Never substitute a numerical value for "c". If you use the required units above, everything will work out fine

Explanation / Answer

when they have same speed:
let v1=v2=v=0.95*c

then gamma1=gamma2=1/sqrt(1-(v/c)^2)

=1/sqrt(1-0.95^2)

=3.2026

using conservation of momentum:

gamma1*m1*v1=gamma2*m2*v2

==>m1=m2

using energy conservation principle:

gamma1*m1*c^2+gamma2*m2*c^2=M*c^2

==>M=3.2026*(m1+m2)

==>m1+m2=M/3.2026

then lost mass=M-(m1+m2)

=0.68775*M

when they have separate speed:

let v1=0.96*c==>gamma1=1/sqrt(1-0.96^2)=3.5714

v2=0.94*c==>gamma2=1/sqrt(1-0.94^2)=2.9311

conserving momentum:

gamma1*m1*v1=gamma2*m2*v2

==>3.5714*m1*0.96*c=2.9311*m2*0.94*c

==>m2=1.24444*m1...(1)

conserving energy:

M*c^2=3.5714*m1*c^2+2.9311*m2*c^2

using equation 1,

M*c^2=3.5714*m1*c^2+2.9311*1.24444*m1*c^2=7.219*m1*c^2

==>M=7.219*m1...(2)

mass lost=M-(m1+m2)

=7.219*m1-m1-1.24444*m1

=4.9746*m1

=4.9746*M/7.219=0.6891*M

so mass lost is higher when they have separate speeds.

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