# 2D elastic collision

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1. Jan 11, 2016

### Jam51

1. The problem statement, all variables and given/known data
A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball. If the first ball moves away with angle 30° to the original path, determine:

a. the speed of the first ball after the collision.

b. the speed and direction of the second ball after the collision.

2. Relevant equations
px = p'x (momentum conservation in the x direction)
v1x = v1'cosθ + v2'cosθ

piy = pfy (momentum conservation in the y direction)
0 = v1'sinθ + v2'sinθ

Ki = Kf (conservation of kinetic energy)
v1^2 = v1'^2 + v2'^2

3. The attempt at a solution
I've attempted solving it in a number of values, but am finding the math challenging and unable to isolate variables. First I attempted to eliminate theta:
v1x = v1'cosθ + v2'cosθ
v1x - v1'cosθ = v2'cosθ ---square--
v1x^2 - v1'^2cos^2θ = v2'^2cos^2θ

0 = v1'sinθ + v2'sinθ
-v1'^2sin^2θ = v2'^2sin^2

v1x^2 - v1'^2cos^2θ - v1'^2sin^2θ = v2'^2cos^2θ + v2'^2sin^2θ
v1x^2 - v1'^2(cos^2θ + sin^2θ) = v2'^2(cos^2θ + sin^2θ)
v1^2 - v1'^2 = v2'^2
(v1 - v1')(v1 + v1') = v2'^2

Arranging Energy Conservation, v1^2 = v1'^2 + v2'^2 :
v1^2- v1'^2 = v2'^2
(v1 - v1')(v1 + v1') = v2'^2

I feel like I've done something wrong up until this point because everything I get after just doesn't make sense to me. I'm not sure what to equate with what.

Help is greatly appreciated!

2. Jan 11, 2016

### Ray Vickson

These types of problems are typically much easier to do by first transforming from the original frame of reference (the lab frame) into the so-called center-of-mass (CM) [or center-of-momentum (COM)] frame, in which the total initial momentum = 0. For identical masses, that means that the initial velocities of balls 1 and 2 are exactly opposite, and their kinetic energies are equal in the CM frame. For a perfectly elastic collision, their individual kinetic energies remain the same after the collision, and the new velocities are exact opposites again. So, if you know the deflection angle of ball 1 in the lab frame, you can set up an equation to determine the deflection angle of ball 1 in the CM frame, so find the final (vector) velocities of balls 1 and 2 in the CM frame. Then you can transform those back into the lab frame.

OK, that method takes a bit more work, than the all-lab-frame method (the one you are using), but it is typically more straightforward.

See. eg., http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node52.html or
http://vallance.chem.ox.ac.uk/pdfs/Collisions.pdf

3. Jan 11, 2016

### WrongMan

yes i highly support the CM method, deduction are much more straightforward, and btw the angle on an elastic colision between v2' and v1' when particles have the same mass is 90º, allways you seemed to have figured that out anyway...

But the Cm method is defenitely the best, i dont think it takes more work. the lab frame method has a ton of formulas and deductions.
but yeah your way seems fine
this should be (v1x - v1'cosθ)^2 = v2'^2cos^2θ
then you do everything else you have done, add the sine and cosine equations, solve for v2' and then substitute that into the kinectic energy and solve for v1', and everything else is just a matter of substitution now i think.

but i highly recomend understanding and using the CM method, as @Ray Vickson said... it's so much simpler, in my opinion

4. Jan 11, 2016

### Jam51

Thank you for your reply. For some reason the prof omitted CM/COM from the course. However, if it makes it easier than I am going to give it a try because what I am doing is not getting me too far.

Thanks for the reference/examples as well!

5. Jan 11, 2016

### Jam51

Wicked, thanks WrongMan. I'm going to correct my math and see if that answers that. But then definitely exploring the CM method!

6. Jan 11, 2016

### Jam51

Right now I am not completely following the CM context - probably because my brain is too tired for new material. However, I tried the math again and it looks ugly, but I am hoping it's on the right track:

(v1 - v1'cosθ)^2 - v1'^2sin^2θ = v2'^2cos^2θ + v2'^2sin^2θ
v1^2 - 2v1v1'cosθ + v1'^2cos^2θ - v1'^2sin^2θ = v2'^2cos^2θ + v2'^2sin^2θ
v1^2 - 2v1v1'cosθ - v1'^2(cos^2θ + sin^2θ) = v2'^2(cos^2θ + sin^2θ)
v1^2 - 2v1v1'cosθ - v1'^2 = v2'^2

7. Jan 11, 2016

### Jam51

If v2^2 = v1^2 - 2v1v1'cosθ - v1'^2

and (v1 - v1')(v1 + v1') = v2'^2

Can I equate them and say:

v1^2 - 2v1v1'cosθ - v1'^2 = (v1 - v1')(v1 + v1')

???

I'm quite certain I will chuckle once I figure out the other method.

8. Jan 11, 2016

### Jam51

I may have done something wrong again because I have:

v1^2 - 2v1v1'cosθ - v1'^2 = (v1 - v1')(v1 + v1')
- 2v1v1'cosθ = 0

9. Jan 11, 2016

### WrongMan

yes, something is wrong, im trying to see what... might have to bust out the old pen and paper :D

10. Jan 11, 2016

### WrongMan

i think i've figured it out... source is way back in your formulas...

its 0 = - v1'sinθ + v2'sinθ
vertical velocity must cancel out

Last edited: Jan 11, 2016
11. Jan 11, 2016

### WrongMan

Ok now:
arrange equations so that all v1 coomponents are on the same side, and v2 are on the opposite.
so you get:
(x)=> (V1-V1'cos(θ))^2=V2'^2cos^2(θ) <=> V1^2 -2V1V1'cos(θ) +V1'^cos^2(θ)=V2'^2cos^2(θ)
(y)=> V1'^2sen^2(θ)=V2'^2sen^2(θ)
V1^2 - 2V1V1'cos(θ) + V1'^2cos^2(θ) V1'^2sen^2(θ)= V2'^2cos^2(θ)+V2'^2sen(θ)
solve for v2'
v1^2-2v1v1'cos(θ) +v1'^2 = v2'^2

now its okay, not sure were you messed up

ok your mistake was the one i stated above
that made you commit this one:
the sign on -v1'^2sin^2(θ) is not a "-"
and bellow you had +v1'^2 and a -v1^2 multiplying with cos and sin respectivly, and you should have felt something was wrong when applying the distributivity property, this mistake could have passed unoticed if you had chosen to use +v1 instead of -v1 bellow so good thing it happened

12. Jan 11, 2016

### ehild

Why did you want to eliminate theta, when it was given as 30°? substitute the values of sinθ and cosθ into your equations. They can be solved easily.
The problem did not state that it was an elastic collision.

13. Jan 11, 2016

### SammyS

Staff Emeritus
But it did state that it's elastic. (Easy to miss that.)

14. Jan 11, 2016

### ehild

Oh, I see. But the angles are different. The first equations are not correct. The angle of the first ball is given, but the angle of the second ball is unknown.

Last edited: Jan 11, 2016
15. Jan 11, 2016

### WrongMan

in elastic colisions if particles with the same mass the angle bettween final velocity vectors is 90º degrees is it not?

16. Jan 11, 2016

### ehild

Yes, but the OP used the same angle for both balls.

Last edited: Jan 11, 2016
17. Jan 11, 2016

### WrongMan

oh right, i don't know how i missed that... he can find it easly after finding v2' ...

18. Jan 11, 2016

### ehild

@Jam51: The velocities of the balls enclose different angles with the original direction. It is easy to prove what WrongMan said, that the angles add up to 90° when particles of equal mass collide elastically.

You have the following equations:
v1=u1 cosθ +u2 cosΦ
0=u1 sinθ-u2 sin Φ
u12+u22=v12
Square the first two equations, and add them.
v12=(u1 cosθ +u2 cosΦ)2+(u1 sinθ-u2 sin Φ)2 ---> v12=u12+u22+2 u1 u2 (cosθ cosΦ-sinθ sinΦ)
Because of conservation of energy, v12=u12+u22, so 2 u1 u2 (cosθ cosΦ-sinθ sinΦ) =0. But the term in the parentheses is equal to cos(θ+Φ) =0 so
θ+Φ=90°.

19. Jan 12, 2016

### Jam51

@ehild I'm following everything until "But the term in the parentheses is equal to cos(θ+Φ) =0 so θ+Φ=90°." I understand you're proving what was stated in previous posts, which would make Φ = 60°. However, I have no idea how you went from 2 u1 u2 (cosθ cosΦ-sinθ sinΦ) =0 to cos(θ+Φ) =0.

20. Jan 12, 2016

### WrongMan

that is a trig identitiy:
cos(x +/- y) = cos(x)cos(y) +/- sen(x)sen(y)

there is probably a way to deduce it but i dont know it...