Speed of Pi Meson - Relativistic question

AI Thread Summary
The discussion focuses on calculating the speed of a pi meson created in a particle accelerator, which has a rest lifetime of 2.60 x 10^-8 seconds. It travels 120 meters before decaying, leading to the conclusion that its speed is 0.998c. Participants suggest using the time dilation formula to find the lifetime in the lab frame and solving for speed by eliminating the unknowns. The approach of assuming the speed of light for initial calculations is deemed acceptable due to minimal error. The conversation emphasizes the importance of correctly applying relativistic equations to determine the meson's velocity.
ZedCar
Messages
353
Reaction score
1
After being created in a high-energy particle particle accelerator, a pi meson at rest has an average lifetime of 2.60 x 10^-8s. Travelling at a speed very close to the speed of light, a pi meson travels a distance of 120m before decaying. How fast is it moving?

Answer: 0.998c


Could anyone give me an ideas as to how to start this, or which equation(s) I should be using?

Thank you
 
Physics news on Phys.org
First calculate the pion life time in the lab frame. It's ok to use v = c here.

Then all you need is time dilatation and how the gamma factor depends on v.
 
ZedCar said:
Could anyone give me an ideas as to how to start this, or which equation(s) I should be using?
Hint: You are given ΔT0. What would the lifetime be in the lab frame? Write it in terms of the unknown speed v. (Do not assume v = c!) You'll then set up an equation and solve for the speed.
 
Doc Al said:
Hint: You are given ΔT0. What would the lifetime be in the lab frame? Write it in terms of the unknown speed v. (Do not assume v = c!) You'll then set up an equation and solve for the speed.

I would use the time dilation formula to obtain the lifetime in the lab frame.

Δt = γΔt0

Δt = γ(2.60x10^-8)
So I would have two unknowns, Δt and the v value in γ.

Is this correct so far? I solve this for v, which means the expression will have an unknown of Δt contained in it.
 
yes you're correct ... i just solved it this way and it gives the correct answer ...
 
ZedCar said:
Is this correct so far? I solve this for v, which means the expression will have an unknown of Δt contained in it.
So far, so good. Now combine this with the basic equation for velocity, which will eliminate Δt. Then you can solve for v, which will be the only unknown.

FYI, the method suggested by niklaus, of calculating Δt by taking v = c, is perfectly fine for this problem. The error introduced by that short cut is too small to matter.

But you should have no problem solving for v without needing any approximations.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top