Proof on two dimensional elastic collision.

[mysqlpress]

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?

 

[Doc Al]

Can you prove the first result? (For equal masses.)

 

[mysqlpress]

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'

 

[Doc Al]

Good. Now redo it without canceling the masses. You'll get an expression for cos(x+y) that will depend on m - m1.

 

[mysqlpress]

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. :)

 

[44r0n0wnz]

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.

 

[44r0n0wnz]

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

 

[44r0n0wnz]

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

 

[Doc Al]

NvM. I got it. Law of cosines and sum difference formulas.

Good! (Sorry I didn't have time to respond earlier.)