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

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)

fckm · 12-08-2004, 02:55 PM

Are you sure this is true?
I think that they just exchange velocities in poportion to their masses.

edit: spelling

filtherton · 12-08-2004, 03:09 PM

It's true. At least its approximately true. If it weren't, billiards would be a very different sport. I don't know of a proof, though, just experience and intuition.

1010011010 · 12-08-2004, 04:04 PM

Think relatively. Rather than a cue ball striking a (stationary) 8-ball, you put them in a frame of reference where they're approaching each other with identical but mirrored trajectories. More or less, they'll swap trajectories for identical objects- basically like reflection off a plane tangent/normal to the point of contact... so, no, not always 90 degrees.

fckm · 12-08-2004, 04:36 PM

So let's assume that you have two velocities, v1, v2.
Perform a galilean transformation into the rest frame of v2. Object one collides with v1-v2.
Assuming both objects have the same mass:
object 1 has final velocity 0.
object two has final velocity v1-v2.
Reverse the transformation:
object 1 has velocity v2
object 2 has velocity v1

Thus, the objects are traveling at the same angle (with respect to the velocities) that they started with.

Karkaboosh1 · 12-10-2004, 08:19 AM

billiards arnt exactly the perfect example, since bll spin can change some things about it. energy does transfer, but instead of ball 1 having all energy and ball to having none (stationary), its split between the two depending on the angle. just need the 'mathmatical' proof that it works.

fckm · 12-10-2004, 06:37 PM

I think my proof proves that it doesn't work... The resulting angle is the same as the incident angle

Thermopyle · 12-13-2004, 11:48 AM

fckm is right, the resulting angle is the same as the incident angle. Anyway, if two balls hit each other with a slight angel the resulting angle will definitly not be 90 degrees.

jay_21 · 01-07-2005, 02:46 PM

How About this....
Make a vector triangle such that the length and directions of the sides give the
Initial KE(Kinetic Energy) 1st object K=1/2 mv^2
Final KE of 1st object K1=1/2 mv1^2
Final KE of 2nd object K2=1/2 mv2^2
By Law of conservation kinetic energy for elastic collision
K=K1+K2 ==> v2= v1^2 + v2^2 ==> By Pythagoras Th. Angle between v1 and v2 should be 90 degrees.
I think this works <==> angle not equal to 0 and 90 given that the second object is initially at rest and both are of same mass.
you could prove this too by taking final resultant momenta of the initial momentum and do lil'bit of trig... and this one would be it's special case......... but just to prove the final momenta are at right angles this should be enough.....

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)

12-08-2004, 02:55 PM	#2 (permalink)
fckm Insane Location: Ithaca, New York	Are you sure this is true? I think that they just exchange velocities in poportion to their masses. edit: spelling __________________ And if you say to me tomorrow, oh what fun it all would be. Then what's to stop us, pretty baby. But What Is And What Should Never Be.

12-08-2004, 04:04 PM	#4 (permalink)
1010011010 Psycho Location: Virginia Beach, VA	Think relatively. Rather than a cue ball striking a (stationary) 8-ball, you put them in a frame of reference where they're approaching each other with identical but mirrored trajectories. More or less, they'll swap trajectories for identical objects- basically like reflection off a plane tangent/normal to the point of contact... so, no, not always 90 degrees. __________________ *Simple Machines in Higher Dimensions*

12-08-2004, 04:36 PM	#5 (permalink)
fckm Insane Location: Ithaca, New York	So let's assume that you have two velocities, v1, v2. Perform a galilean transformation into the rest frame of v2. Object one collides with v1-v2. Assuming both objects have the same mass: object 1 has final velocity 0. object two has final velocity v1-v2. Reverse the transformation: object 1 has velocity v2 object 2 has velocity v1 Thus, the objects are traveling at the same angle (with respect to the velocities) that they started with. __________________ And if you say to me tomorrow, oh what fun it all would be. Then what's to stop us, pretty baby. But What Is And What Should Never Be.

12-10-2004, 06:37 PM	#7 (permalink)
fckm Insane Location: Ithaca, New York	I think my proof proves that it doesn't work... The resulting angle is the same as the incident angle __________________ And if you say to me tomorrow, oh what fun it all would be. Then what's to stop us, pretty baby. But What Is And What Should Never Be.

12-08-2004, 03:09 PM	#3 (permalink)
filtherton Junkie Location: In the land of ice and snow.	It's true. At least its approximately true. If it weren't, billiards would be a very different sport. I don't know of a proof, though, just experience and intuition.

12-10-2004, 08:19 AM	#6 (permalink)
Karkaboosh1 Upright	billiards arnt exactly the perfect example, since bll spin can change some things about it. energy does transfer, but instead of ball 1 having all energy and ball to having none (stationary), its split between the two depending on the angle. just need the 'mathmatical' proof that it works.

12-13-2004, 11:48 AM	#8 (permalink)
Thermopyle Crazy Location: Top of the World, Mom!	fckm is right, the resulting angle is the same as the incident angle. Anyway, if two balls hit each other with a slight angel the resulting angle will definitly not be 90 degrees. __________________ Live life like you're gonna die, beacause you're gonna! - William Shatner.

01-07-2005, 02:46 PM	#9 (permalink)
jay_21 Upright	How About this.... Make a vector triangle such that the length and directions of the sides give the Initial KE(Kinetic Energy) 1st object K=1/2 mv^2 Final KE of 1st object K1=1/2 mv1^2 Final KE of 2nd object K2=1/2 mv2^2 By Law of conservation kinetic energy for elastic collision K=K1+K2 ==> v2= v1^2 + v2^2 ==> By Pythagoras Th. Angle between v1 and v2 should be 90 degrees. I think this works <==> angle not equal to 0 and 90 given that the second object is initially at rest and both are of same mass. you could prove this too by taking final resultant momenta of the initial momentum and do lil'bit of trig... and this one would be it's special case......... but just to prove the final momenta are at right angles this should be enough.....