# Homework Help: Speed of Muons, calculate KE and Momentum

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1. Apr 12, 2015

### rlc

1. The problem statement, all variables and given/known data
The average lifetime of muons at rest is 2.20 μs. A laboratory measurement on muons traveling in a beam emerging from a particle accelerator yields an average muon lifetime of 14.718 μs.
a) What is the speed of the muons in the laboratory?
b) What is their kinetic energy? (MeV)
c) What is their momentum? (MeV/c) The mass of a muon is 207 times that of an electron.

2. Relevant equations
Part A: SQRT(1-(To/T)^2), where To is the average lifetime at rest and T is the second lifetime given in the problem.

Part B: KE= M[(1/sqrt(1-B^2))-1]
where: M=mass of muon= 105MeV
B= v/c, where v (in m/s) is the speed calculated in part A

Part C:

3. The attempt at a solution
I found Part A, have a question about B, and I don't understand how to calculate C.

Part A: SQRT(1-(2.2/14.718)^2)=0.98876c=0.98876(3E8)=2.966E8 m/s

Part B equation was given by one of my classmates, but I know I'm doing something wrong.
KE=(105 MeV)((1/SQRT(1-((2.966E8)/(3E8))^2)-1)=594.41 MeV
I kept getting this one wrong due to user error with those parentheses, but the online homework told me that number was correct, but changed it to 602 MeV?? Is that equation right, and the number accepted was in a range? Is there a way to be more exact in the equation?

Part C:
p=SQRT(KE*2*m)
If the mass of an electron is 9E-31 kg, then a muon is 207 times that, =12439 kg
p=SQRT(602*2*12439)=3869, which isn't right.
The homework asks for units of MeV/c...does that mean I divide by 3E8?
Is that even the correct equation to use?

2. Apr 12, 2015

### Orodruin

Staff Emeritus
You have used a non-relativistic expression for the relation between kinetic energyy and momentum, you need to use the relativistic expression for momentum, involving the particle velocity, mass, and gamma factor.

3. Apr 12, 2015

### rlc

p=(m)(v) / SQRT(1-(v^2)/(c^2))
Where would the kinetic energy be in the equation?

4. Apr 12, 2015

### unscientific

Energy would simply be:

$$E = \gamma m_0 c^2$$

where $m_0$ is the rest mass. (Try proving that relation from $E^2 = p^2c^2 + m_0^2c^4$ and $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$).

5. Apr 12, 2015

### rlc

Is this for calculating KE or momentum?

6. Apr 12, 2015

### Orodruin

Staff Emeritus
You do not need the kinetic energy to compute the momentum. It is sufficient with the mass and the velocity you computed in A.

7. Apr 12, 2015

### unscientific

Hint: If total energy = KE + rest mass energy = $\gamma m_0 c^2$, what is the KE?

With this KE, how do you work out the momentum using the total energy expression?

8. Apr 12, 2015

### unscientific

Orodruin is right. But maybe if you want to understand it more, derive an expression for total energy in terms of $\gamma$ and rest mass $m_0$. Then you will see that at any given speed there is a given $\gamma$, and the total energy (KE + rest mass energy) is simply a factor of $\gamma$ times rest mass energy.

9. Apr 12, 2015

### rlc

So would this equation end up working?
p=(m)(v) / SQRT(1-(v^2)/(c^2))

10. Apr 12, 2015

### Orodruin

Staff Emeritus
Also, your method on B is correct, you just have not used the same value for the mass as stated in the problem (which equates to 105.77 MeV rather than 105 MeV).

11. Apr 12, 2015

### Orodruin

Staff Emeritus
Yes. However, I suggest keeping the expression T/T0 for the gamma factor to avoid rounding errors.

12. Apr 12, 2015

### rlc

Oh, so would I use the average lifetime at rest value and the average accelerated lifetime value in this equation as well?

13. Apr 12, 2015

### rlc

p=[1/SQRT(1-(2.2/14.718)^2]*(mass)(velocity)
where mass=207*9E-31=12439
velocity is what I solved for in part a: 2.966E8

Did I miss something?

14. Apr 12, 2015

### Orodruin

Staff Emeritus
Now you inserted the expression for the velocity in terms of the times instead of the gamma factor. The gamma factor has a far simpler expression in T and T0...

15. Apr 12, 2015

### rlc

I'm sorry, I just really don't understand.

16. Apr 13, 2015

### rlc

Would T and T0 have the same values in the calculation for momentum as they did in the calculation for speed?
p=(m)(v) / SQRT(1-(T/T0)) ?

Last edited: Apr 13, 2015
17. Apr 13, 2015

### Orodruin

Staff Emeritus
What is the gamma factor $\gamma$ in terms of the times T and T0?

18. Apr 13, 2015

### rlc

Would that be: 1/SQRT(1-(T/T0)) ?

19. Apr 13, 2015

### Orodruin

Staff Emeritus
No. How did you solve part A? What were the quantities involved?

20. Apr 13, 2015

### rlc

SQRT(1-(To/T)^2)
Sorry, I kept forgetting the squared and I put them out of order. This worked for part a, but would this be 1/sqrt or just the sqrt?

21. Apr 13, 2015

### Orodruin

Staff Emeritus
Yes but how did you arrive at this formula? What is the gamma factor?

22. Apr 13, 2015