An unstable particle called the pion has a mean lifetime of 25 ns in its own rest frame. A beam of pions travels through the laboratory at a speed of $0.60 c .$ (a) What is the mean lifetime of the pions as measured in the laboratory frame? (b) How far does a pion travel (as measured by laboratory observers) during this time?

Short Answer

Expert verified
Answer: The mean lifetime of a pion in the laboratory frame is 31.25 ns, and it travels approximately 5,625 meters during this time.

Step by step solution

01

Find the mean lifetime in the laboratory frame

We are given the mean lifetime of a pion in its rest frame, \(t_0 = 25\textrm{ ns}\) and its speed, \(v = 0.60c\). To find the mean lifetime in the laboratory frame, we will use the time dilation formula: \(t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}\) where \(t\) is the time in the laboratory frame, \(t_0\) is the time in the pion's rest frame, \(v\) is the speed of the pion, and \(c\) is the speed of light. Plugging in the given values: \(t = \frac{25\textrm{ ns}}{\sqrt{1 - \frac{(0.60c)^2}{c^2}}}\)
02

Simplify the expression

Now, let's simplify the expression: \(t = \frac{25\textrm{ ns}}{\sqrt{1 - 0.36}}\) \(t = \frac{25\textrm{ ns}}{\sqrt{0.64}}\) \(t = \frac{25\textrm{ ns}}{0.8}\) \(t = 31.25\textrm{ ns}\) So, the mean lifetime of the pions as measured in the laboratory frame is \(31.25\textrm{ ns}\).
03

Calculate the distance traveled by the pions

Now that we have the mean lifetime of pions in the laboratory frame, we can calculate the distance they travel during this time. We use the formula: \(d = vt\) where \(d\) is the distance traveled, \(v\) is the speed of the pions, and \(t\) is the mean lifetime in the laboratory frame. Using the given speed and the calculated mean lifetime: \(d = (0.60c)(31.25\textrm{ ns})\)
04

Convert the time to distance using the speed of light

The speed of light, \(c\), is equal to approximately \(3.00\times10^8\textrm{ m/s}\), and there are \(10^9\textrm{ ns}\) in one second. Therefore, we can convert the time to distance using the speed of light: \(d = (0.60)(3.00\times10^8\textrm{ m/s})(31.25\textrm{ ns})\) \(d = (0.60)(3.00\times10^8\textrm{ m/s})(31.25\times10^{-9}\textrm{ s})\)
05

Calculate the distance

Finally, let's calculate the distance: \(d = (0.60)(3.00\times10^8\textrm{ m})(31.25\times10^{-9})\) \(d \approx 5.625\times10^3\textrm{ m}\) So, a pion travels approximately \(5.625\times10^3\textrm{ m}\), as measured by laboratory observers during its mean lifetime of \(31.25\textrm{ ns}\) in the laboratory frame.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A spaceship is traveling away from Earth at \(0.87 c .\) The astronauts report home by radio every 12 h (by their own clocks). (a) At what interval are the reports sent to Earth, according to Earth clocks? (b) At what interval are the reports received by Earth observers, according to their own clocks?
For a nonrelativistic particle of mass \(m,\) show that \(K=p^{2} /(2 m) .\) [Hint: Start with the nonrelativistic expressions for kinetic energy \(K\) and momentum \(p\).]
Muons are created by cosmic-ray collisions at an elevation \(h\) (as measured in Earth's frame of reference) above Earth's surface and travel downward with a constant speed of \(0.990 c .\) During a time interval of \(1.5 \mu \mathrm{s}\) in the rest frame of the muons, half of the muons present at the beginning of the interval decay. If one fourth of the original muons reach Earth before decaying, about how big is the height \(h ?\)
A constant force, acting for \(3.6 \times 10^{4} \mathrm{s}(10 \mathrm{h}),\) brings a spaceship of mass \(2200 \mathrm{kg}\) from rest to speed \(0.70 c\) (a) What is the magnitude of the force? (b) What is the initial acceleration of the spaceship? Comment on the magnitude of the answer.
Kurt is measuring the speed of light in an evacuated chamber aboard a spaceship traveling with a constant velocity of \(0.60 c\) with respect to Earth. The light is moving in the direction of motion of the spaceship. Siu- Ling is on Earth watching the experiment. With what speed does the light in the vacuum chamber travel, according to Siu-Ling's observations?
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free