What is the total kinetic energy of two protons in different reference frames?

AI Thread Summary
The discussion focuses on calculating the total kinetic energy of two protons moving at 0.5c in different reference frames. In frame S', the calculated kinetic energy is 290 MeV, but the initial attempt yielded 270 MeV using the classical formula 1/2 mv^2, which is incorrect for relativistic speeds. Participants emphasize the necessity of using the relativistic kinetic energy formula instead. The conversation highlights the confusion surrounding the application of classical mechanics in relativistic contexts. Accurate calculations require understanding relativistic effects on mass and energy.
PsychonautQQ
Messages
781
Reaction score
10

Homework Statement


In the reference frame S', two protons, each moving at .5c, approach each other head on. Calculate the total kinetic energy of the two protons in frame S', and calculate the total kinetic energy of the two protons as seen in reference frame S which is moving with one of the protons.






The Attempt at a Solution


So for the first part apparently the answer I'm looking for is 290MeV

v = .5c so γ = 1.1547

using 1/2 mv^2
m = 1.6726E-27
v = .5(3E8)

and then multiplying this times two for two protons and 1.15 from gamma, then converting that to eV by 1J = 6.24E18 eV I get 270MeV.. back of the book says 290 MeV.. what am I doing wrong ;-(?
 
Physics news on Phys.org
PsychonautQQ said:
using 1/2 mv^2

You'll need to use the relativistic formula for kinetic energy.
 
However we will not provide it. lol
 
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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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