How do you know if the rest mass changes or not?

AI Thread Summary
In a completely inelastic collision, the rest mass of the resulting composite object can exceed the sum of the individual rest masses due to energy transformations, such as heat. However, in other scenarios, the rest mass of particles remains constant before and after collisions, leading to confusion. Understanding whether rest mass changes requires knowledge of mathematical concepts like four-momentum and conservation laws. The term 'invariant mass' is often used when multiple particles are involved, while 'rest mass' typically refers to single particles. Clarifying these concepts is essential for grasping the principles of relativity and mechanics.
sspitz
Messages
18
Reaction score
0
My book says that if two objects collide completely inelastically, the rest mass of the composite is greater than the sum. I assume this happens because the objects heat up and this is counted in their energy.

However, in other problems the book assumes the rest mass of objects/particles is the same before and after collisions. Explanations are sparse.

How do you know if the rest mass changes or not?
 
Physics news on Phys.org
I am guessing this is a book on relativity? What mathematical methods have you been through so far? Have you learned how to calculate the four-momentum of a system yet? and how to get the rest mass from this? It would be best to understand it mathematically, but I could help with an intuitive explanation if you haven't been through the maths yet.
 
End of a book on mechanics. The next chapter is more formal. I would just like an intuitive grasp of what the rest mass is and how to set up simple conservation law equations such as in compton scattering.
 
hmm. I could write out a big paragraph explaining 4-momentum as well as I can, but I don't think the homework section is the right place for that kind of thing. But I can help with the confusion you had in your first post. Right, so the rest mass of two particles is not just the sum of their individual rest masses. I think this is where the confusion stemmed from.

Edit: Also, when several particles are involved, it is often called the 'invariant mass' of the system, and is usually just called 'rest mass' when only one particle is involved.
 
Last edited:
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 '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...
Back
Top