Help relativity exam in in 5 hours

  • Thread starter Thread starter nemzy
  • Start date Start date
  • Tags Tags
    Exam Relativity
AI Thread Summary
To solve the kinetic energy of an electron with mass 9.11e-31 kg and momentum 4e-22, the formula E² = m₀²c⁴ + p²c² is essential. This equation allows the calculation of total energy, from which rest energy can be subtracted to find kinetic energy. The correct kinetic energy is 6.3e-14 J, but the challenge arises from the lack of velocity information. Understanding the relationship between momentum and energy in relativity is crucial for solving such problems. Memorizing this formula is recommended for anyone studying Special Relativity.
nemzy
Messages
124
Reaction score
0
urgent help relativity exam in in 5 hours

an electrion with mass 9.11e031 kg has momentum 4e-22. what is its kinetic energy

the answer is 6.3e-14 J but i have no idea how to get here

relativity kinetic energy = (y-1)mc^2 where y = 1/(square root of 1-(v^2/c^2)

momentum = ymu

thats as far as i can get , how the hell can u solve this type of problem if they don't give u the velocity of the electron?
 
Last edited:
Physics news on Phys.org
i got 8.78e-14

when plugging into that equation?
 
Use this
E^{2}=m_{0}^{2}c^{4}+p^{2}c^{2} to find the total energy and then from that simply subtract the rest energy...

Daniel.
 
Last edited:
thank u! i was stuck on that problem forever
 
U'd better memorize that formula.It's the most important in the Special Relativity,at least,not in Minkowski space.

Daniel.
 

Thread 'Errors and significant figures'

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...
View full post »
Thread 'A cylinder connected to a hanging mass'

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...
View full post »

Similar threads

Solution of basic special relativity problem
Replies
4
Views
1K
Potential Energy - 2D inelastic collision - ballistic pendulum
Replies
10
Views
2K
Body attached to a block by a spring, shaped like an inclined plane
Replies
1
Views
891
Confused about a conservation of energy problem
Replies
4
Views
2K
Challenging problem about an impact with a smooth frictionless surface
2
Replies
55
Views
5K
What is the energy conservation principle in a two masses and pulley problem?
Replies
33
Views
4K
Distribution of Energy when work is done on a system of 2 masses connected by a spring
Replies
17
Views
1K
Physics Exam Revision Help - Exam in 5 hours
Replies
17
Views
2K
Elastic Collision of Hydrogen and Carbon Atoms
Replies
9
Views
2K
Calculating the Travel Time of High-Energy Particles Across Our Galaxy
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top