How Does a Muon's Travel Illustrate Relativity Principles?

  • Thread starter Thread starter EgpYo
  • Start date Start date
  • Tags Tags
    Muons Relativity
AI Thread Summary
The discussion focuses on the calculations related to a muon's travel and its implications for the theory of relativity. When moving at 0.99c, the muon's travel distance is analyzed using both Newtonian mechanics and relativistic principles. Initial calculations suggest a distance of 653.4 meters under Newtonian mechanics, while relativistic effects indicate a shorter distance of 463.32 meters. The discrepancies between these distances highlight the importance of relativity, as they demonstrate how time dilation affects the muon's observed lifetime. Overall, the discussion emphasizes the critical role of relativity in understanding particle behavior at high velocities.
EgpYo
Messages
25
Reaction score
0

Homework Statement


A muon has a lifetime of 2.0x10^-6 seconds when at rest, after which time it decays into other particles
a) Ignore any effects of relativity discussed in this section. If the muon was moving at 0.99c, how far would it travel before decaying into other particles, according to Newtonian mechanics?
b) How long would the muon last, according to an observer in Earth's frame of reference?
c)How far would the muon actually travel, when viewed moving at 0.99c?
d) Compare the two distances travelled. Explain why this type of evidence is excellent support for the theory of relativity.

Homework Equations



tm = ts/ sqrt(1-v^2/c^2)

The Attempt at a Solution



a) (0.99)(3x10^8) = v
v = 2.97x10^8

v=dt
d = (2.97x10^8)(2.2x19^-6)
d = 653.4 m

b) tm = ts/ (1-v^2/c^2)
tm = 2.2x10^-6/0.141
tm = 0.000015602 s

c) d = vt
= (2.97x19^-6)(1.56x10^-6)
= 463.32 m

d) ?
 
Physics news on Phys.org
What are you calculating in (b) ? The expression is not your relevant formula and the calculation is way off !
Fix that and then c and d will be a lot more doable.
 
EgpYo said:
b) tm = ts/ (1-v^2/c^2)
tm = 2.2x10^-6/0.141
tm = 0.000015602 s

Ok, couple typos there. You forgot the square-root sign. And it might be helpful if you converted your tm into scientific notation.

EgpYo said:
c) d = vt
= (2.97x19^-6)(1.56x10^-6)
= 463.32 m

The (2.97x19^-6) looks like you intended to write (2.97x10^8) but your fingers got away from you.
Check the (1.56x10^-6) .
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top