Tilted Forum Project Discussion Community - View Single Post - Proof that elastic collisions always form 90 degrees?

Karkaboosh1 · 12-08-2004, 01:34 PM

kind of stuck here, i know that the proof stands correct, but just need to get the math of it down on paper...

when 2 objects collide and do not stick together, let say 2 billiard balls, assuming that the angle of collision is not 0 (direct/strait hit) or 90 degrees (miss), they will form a 90 degree angle between them. IE the degree between ray A (from white cue ball) and ray B (um, 8 ball?), at the origin of collision C 90 degree's exactly.

equasion taht might help (havnt looked into this much... just cant get my though proccesses so to say to get going)

mv1(i) = mv1(f) cos ө(1) + mv2(f) cos ө(2) <-- energy conversion along x-axis

0 = mv1(f) sin ө(1) - mv2(f) sin ө(2) <-- energy conversion along y-axis

(i) = initial
(f) = final
m = mass
v = velocity
1 = first object (cue ball)
2 = second object (8 ball)

12-08-2004, 01:34 PM	#1 (permalink)
Karkaboosh1 Upright	Proof that elastic collisions always form 90 degrees? kind of stuck here, i know that the proof stands correct, but just need to get the math of it down on paper... when 2 objects collide and do not stick together, let say 2 billiard balls, assuming that the angle of collision is not 0 (direct/strait hit) or 90 degrees (miss), they will form a 90 degree angle between them. IE the degree between ray A (from white cue ball) and ray B (um, 8 ball?), at the origin of collision C 90 degree's exactly. equasion taht might help (havnt looked into this much... just cant get my though proccesses so to say to get going) mv1(i) = mv1(f) cos ө(1) + mv2(f) cos ө(2) <-- energy conversion along x-axis 0 = mv1(f) sin ө(1) - mv2(f) sin ө(2) <-- energy conversion along y-axis (i) = initial (f) = final m = mass v = velocity 1 = first object (cue ball) 2 = second object (8 ball)