Proof on two dimensional elastic collision.

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Discussion Overview

The discussion revolves around the mathematical proof of the angles resulting from a two-dimensional elastic collision between two balls of equal mass. Participants explore the conditions under which the angle between the two masses after the collision is 90 degrees, as well as how this changes when the masses differ.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that for equal masses, the angle between the two masses after the collision is 90 degrees, while noting that if one mass is greater, the angle is less than 90 degrees, and if it is smaller, the angle is greater than 90 degrees.
  • Another participant requests a proof for the angle being 90 degrees for equal masses.
  • Several participants discuss the conservation of kinetic energy and momentum as part of the proof, presenting equations related to these principles.
  • A participant suggests redoing the proof without canceling the masses to derive an expression for cos(x+y) that depends on the mass difference.
  • One participant expresses confusion about the combination of equations and how the angle of 90 degrees is derived, seeking clarification on specific steps in the proof.
  • Another participant mentions using the law of cosines and sum difference formulas to relate the angles and states that cos(Ø1 + Ø2) equals zero, leading to the conclusion that the angles sum to 90 degrees.

Areas of Agreement / Disagreement

Participants generally agree on the principles of conservation of momentum and energy but do not reach a consensus on the clarity of the proof or the specific steps involved in deriving the angle relationships.

Contextual Notes

Some participants express uncertainty regarding the mathematical steps involved in combining equations and deriving the angle relationships, indicating that further clarification is needed on specific calculations.

mysqlpress
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When a ball with mass m collide with another ball with equal mass as m=> m1 at rest, the mathematical proof shows that after the collision, the angle between two masses would be 90'

I know that when m> m1 , angle <90'
when m<m1 , angle >90'

but there is no mathematical proof to show this result...
...
does anyone help me solve this problem?
 
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Can you prove the first result? (For equal masses.)
 
making use of KE conserved and momentum conserved...

mv=m1v1cosx+mv2cosy- (1)
v=v1cosx+v2cosy
0=m1v1sinx-mv2siny - (2)
0=v1sinx-v2siny
(1)^2+(2)^2

v^2 = (v1)^2+(v2)^2 +2v1v(cosxcosy-sinysinx)

KE conserved
mv^2 =m1v1^2+mv2^2
v^2=v1^2+v2^2
cos(x+y) = 90'
 
Good. Now redo it without canceling the masses. You'll get an expression for cos(x+y) that will depend on m - m1.
 
Doc Al said:
Good. Now redo it without canceling the masses. You'll get an expression for cos(x+y) that will depend on m - m1.
Thanks ! I've proven that. :)
 
Could you please describe the work in more detail? I understand the entire concept entirely, and even matched up with a lot of the math you did, but I didn't understand the combining you did in the middle and how the angle 90 just appeared out of nowhere.
 
mysqlpress said:
making use of KE conserved and momentum conserved...

mv=m1v1cosx+mv2cosy- (1)
v=v1cosx+v2cosy
0=m1v1sinx-mv2siny - (2)
0=v1sinx-v2siny
(1)^2+(2)^2

v^2 = (v1)^2+(v2)^2 +2v1v(cosxcosy-sinysinx)

KE conserved
mv^2 =m1v1^2+mv2^2
v^2=v1^2+v2^2
cos(x+y) = 90'

I understand everything except what is in bold. First, you are finding the momentum along the X. Then finding the momentum along the Y. Then somehow combine them in the equation for the conservation of energy. I have all of that already, but I don't understand the combining itself and don't understand the bolded areas
 
NvM. I got it. Law of cosines and sum difference formulas.

Ø1 + ø2 = 90,

so it is safe to assume that :

cos(Ø1 + ø2) = 0

since cos(90) = 0. Now, for the actual proof, start by using the formula for conservation of kinetic energy:

K1i + K2i = K1f + K2f.

After simplifying, you will get:

V1i^2 = V1f^2 + V2f^2.

According to law of cosines,

c^2 = a^2 + b^2 -2(ab)cosC.

CosC is The same value as cos(ø1 + ø2), so

c^2 = a^2 + b^2 -2(ab)cos
 
44r0n0wnz said:
NvM. I got it. Law of cosines and sum difference formulas.
Good! (Sorry I didn't have time to respond earlier.)
 

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