Question

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.30 ✕ 10−28 kg and that of the heavier fragment is 1.77 ✕ 10−27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment?

          An unstable particle at rest breaks up into two fragments
of unequal mass. The mass of the lighter
fragment is equal
to 2.30 ✕ 10−28 kg and that of
the heavier fragment
is 1.77 ✕ 10−27 kg. If the
lighter fragment has a speed of 0.893c after the
breakup, what is the speed of the heavier fragment?
        
Show more…

Added by Michelle C.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.30 ✕ 10−28 kg and that of the heavier fragment is 1.77 ✕ 10−27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment?
Close icon
Play audio
Feedback
Powered by NumerAI
Danielle Fairburn Jennifer Stoner
Ivan Kochetkov verified

Narayan Hari and 72 other subject Physics 101 Mechanics educators are ready to help you.

Ask a new question

*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.

*

Recommended Videos

-
an-unstable-particle-at-rest-breaks-into-two-fragments-of-unequal-mass-the-mass-of-the-lighter-fragm

An unstable particle at rest breaks into two fragments of unequal mass. The mass of the lighter fragment is $2.50 \times 10^{-28} \mathrm{~kg}$, and that of the heavier fragment is $1.67 \times 10^{-27} \mathrm{~kg}$. If the lighter fragment has a speed of $0.893 c$ after the breakup, what is the speed of the heavier fragment?

Fundamentals of Physics

an-unstable-particle-at-rest-spontaneously-breaks-into-two-fragments-of-unequal-mass-the-mass-of-the

An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is $2.50 \times 10^{-28} \mathrm{kg}$ , and that of the other is $1.67 \times$ $10^{-27} \mathrm{kg}$ . If the lighter fragment has a speed of 0.893$c$ after the breakup, what is the speed of the heavier fragment?

Physics for Scientists and Engineers with Modern Physics

an-unstable-particle-at-rest-breaks-into-two-fragments-of-unequal-mass-the-mass-of-the-lighter-fragm

An unstable particle at rest breaks into two fragments of unequal mass. The mass of the lighter fragment is $2.50 \times 10^{-28} \mathrm{~kg}$, and that of the heavier fragment is $1.67 \times 10^{-27} \mathrm{~kg}$. If the lighter fragment has a speed of $0.893 c$ after the breakup, what is the speed of the heavier fragment?

Fundamentals of Physics


*

Recommended Textbooks

-
University Physics with Modern Physics

University Physics with Modern Physics

Hugh D. Young 14th Edition
achievement 1,903 solutions
Physics: Principles with Applications

Physics: Principles with Applications

Douglas C. Giancoli 7th Edition
achievement 1,981 solutions
Fundamentals of Physics

Fundamentals of Physics

David Halliday, Robert Resnick , Jearl Walker 10th Edition
achievement 1,933 solutions

*

Transcript

-
00:01 To solve this problem, first i will just write the given data here.
00:05 Given data.
00:09 So here the value of m10 is given as 2 .30 multiplication 10 to the power minus 28 kg.
00:21 This is raised mass of lighter fragment and the value of m20 is given as 1 .77 multiplication 10 to the power minus 27 k.
00:32 The value of v1 is given as 0893c now from momentum conservation law i can write the expression for initial momentum is equal to initial momentum is equal to final momentum is equal to final momentum on further simplification i can write the expression as 0 is equal to m10 by under root 1 minus v1 square by c square multiplication v1 plus m 20 multiplication v2 by under root 1 minus v2 square by c square so simplifying it further this expression will become 0 is equal to 2 .30 multiplication 10 to the power minus 28, multiplication 0 .893c by under root 1 minus 0 .893c whole square by c square, plus 1 .77, multiplication 10 to the power minus 27, multiplication v2 by under root 1 minus v2 by under root 1 minus v2, by c squared.
02:08 So simplifying it further, the expression will become 0 is equal to 10 .14 multiplication 10 to the power minus 28 c plus 1 .77 multiplication 10 to the power minus 27 by under root 1 minus v square by c squared multiplication v2...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever