Relativity-- Mass energy equivalence (1 Viewer)

[Roobs]

this questions got me confused:

"an electron travelling at 0.25c and of rest mass 9.11*10^-31 kg collides head on with a positron with the same mass but travelling at 0.5c. when matter and antimatter meet in this way they anihilate eachother to produce energy, how much energy is released"

the answers to the textbook (physics contexts 2) just use E=mc^2, with the rest masses of the particles eg:

E=2*9.11*10^-31*(3*10^8)^2
E=1.64*10^-12 J

but i know this book has some bullshit in their answers-- it seems illogical to me that the question specifies the velocities of the particles, but then infers that they make no difference to the amount of energy relesed. surely a collison at a higher velocity will release more energy.

i was thinking that kinetic energy KE=Mv*c^2-Mo*c^2 should be included as well, or using Mv in the e=mc^2 equation..or something

anyone clear this up for me?
cheers

 

[insert-username]

I think the speeds are there to throw you off - I'm not 100% sure on annihilation, but I don't think previous kinetic energy has any or much impact on the produced energy, since it's comparatively much smaller (i.e. you multiply by the speed of light squared for annihilation, compared to velocity squared and mass for KE). I could be wrong though.


I_F

 

[Roobs]

hmmm.. well the text i was doing theres questions from says that the standard formula for kinetic energy is wrong at relativistic speeds and that apparently einstein derived

KE=Mv*c^2-Mo*c^2

eh..i asked my physics teacher today, and he says that "he didnt really know" (douche!) but on the same token we went going to be asked such a question in the HSC....so meh...my best guess is still to use the dillated mass Mv in the calcualtion....but as i said thats just my best guess