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A proton is traveling to the right at 2.5×10^7m/s . It has a head-on perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton.
What are the speeds of each after the collision?
3 answers
Ignoring relativistic effects...
mp*vp=mp*vp' + 12mpv'
vp=vp' + 12v' all from conservation of momentum.
Energy:
1/2 mp vp^2=1/2mp*vp'^2+ 1/2 12mp v'^2
vp^2=vp'^2 + 12 v'^2
solve for vp' in the first equation
vp'=vp-12v' put that in the second equation
vp^2=(vp-12v')^2+12v'^2
expand it, and solve the quadratic equation.
mp*vp=mp*vp' + 12mpv'
vp=vp' + 12v' all from conservation of momentum.
Energy:
1/2 mp vp^2=1/2mp*vp'^2+ 1/2 12mp v'^2
vp^2=vp'^2 + 12 v'^2
solve for vp' in the first equation
vp'=vp-12v' put that in the second equation
vp^2=(vp-12v')^2+12v'^2
expand it, and solve the quadratic equation.
just answer it jeez
0 answers