# 2d absolutely elastic vector collision

1. May 30, 2012

### wolfblood

1. The problem statement, all variables and given/known data
I have mass of two points ma and mb than their vertical horizontal vector velocity parts before collision vxa0 vya0 vxb0 vyb0 and i need to calculate vector velocities after collision vxa1 vya1 vxb1 vyb1

I found something like the same vector componets of momemtums will exchange so ma.vxa0<=>mb.vxb0 and of course ma and mb is constant so i got this vxa=(mb.vxb)/ma. Am I right? If no, how to calculate them?

2. May 30, 2012

### rock.freak667

Were the two masses moving before or was one stationary?

With each component, the momentum will be conserved.

So you'd have in x-direction: momentum before impact = momentum after impact.

The same for the y-direction.

Since it is elastic, you know that KE before = KE after impact.

3. May 30, 2012

### wolfblood

Both was moving it's collision of two moving balls representated by circles in code but theoreticaly there can one stay on place if three balls colide in one thick (but i don't need it so complex) and I need right equations for counting that four velocities vxa1 vya1 vxb1 vyb1 from this 6 variables m1 m2 vxa0 vya0 vxb0 vyb0.

4. May 30, 2012

### rock.freak667

Based on the information you have such as values for m1,m2 and the velocities, you can form the equationjs and solve.

5. Jun 1, 2012

### dimension10

Apply conservation of mommentum. You get

$$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$$

Edit: Oops. I mean conservation of energy.

$$\left(m_1v_{1i}^2+m_2v_{2i}^2\right)=\left(m_1v_{1f}^2+m_2v_{2f}^2\right)$$

Last edited: Jun 1, 2012
6. Jun 1, 2012

### wolfblood

Fine from conservation of momentum in directions i can say that
vxA0*mA+vxB0*mB=vxA1*mA+vxB1*mB
vyA0*mA+vyB0*mB=vyA1*mA+vyB1*mB
from KE conservation I can say that
mA*(vxA0^2+vyA0^2)/2+mB*(vxB0^2+vyB0^2)/2=mA*(vxA^2+vyA^2)/2+mB*(vxB^2+vyB^2)/2
I have 3 equations with 4 unkowns what am i missing? Are these right?

Is this rigth? I found something like the same vector componets of momemtums will exchange so ma.vxa0<=>mb.vxb0 and of course ma and mb is constant so i got this vxa=(mb.vxb)/ma.

7. Jun 1, 2012

### dimension10

No, during elastic collisions, kinetic energy is conserved, not momentum.

8. Jun 1, 2012

### wolfblood

Nor in one direction?

9. Jun 1, 2012

### dimension10

I don't get what you are saying?

10. Jun 1, 2012

### Infinitum

:uhh:

Both momentum and kinetic energy are conserved in elastic collisions....

In fact, momentum is conserved in both elastic and inelastic collisions.

11. Jun 1, 2012

### wolfblood

Please can you simply tell me the equations? May be when i will see them i will understand. The four equations for
vxa=
vya=
vxb=
vyb=
with variables of
ma
mb
vxa0
vya0
vxb0
vyb0

12. Jun 1, 2012

### dimension10

Oh, ok.

13. Jun 1, 2012

### Infinitum

Yes, these are correct.

A little rearrangement in the equations will help. To start with, eliminate 1/2 from the kinetic energy equation, and then bring the terms with the same masses in each equation to the same side and use $a^2-b^2 = (a+b)(a-b)$. You will get a similar term from the momentum equations which you can substitute to solve further.

14. Jun 4, 2012

### wolfblood

I hava another question because i still have 3 equations with 4 unknowns could I do something like

mA*(vxA0^2+vyA0^2)/2+mB*(vxB0^2+vyB0^2)/2=mA*(vxA^2+vyA^2)/2+mB*(vxB^2+vyB^2)/2

mA*(vxA0^2+vyA0^2)+mB*(vxB0^2+vyB0)=mA*(vxA^2+vyA^2)+mB*(vxB^2+vyB^2)

fine

mA*(vxA0^2+vyA0^2)-mA*(vxA^2+vyA^2)=mB*(vxB^2+vyB^2)-mB*(vxB0^2+vyB0)

vxA0*mA-vxA1*mA=vxB1*mB-vxB0*mB
vyA0*mA-vyA1*mA=vyB1*mB-vyB0*mB
from it
mA*(vxA0-vxA1)=mb*(vxB1-vxB0)
mA*(vyA0-vyA1)=mb*(vyB1-vyB0)

but i still see 4 things i need to get vxA1 vyA1 vxB1 vyB1 and 3 equations plese which one is fourth?

Or can somebody solve this to get vxA1 vyA1 vxB1 vyB1 because i know one mathematical rule if i have n unkowns i need n equations.

15. Jun 4, 2012

### Infinitum

Let (vxA0-vxA1) = p be one variable, and (vxB1-vxB0)=q be another. Do you see how to solve this now?

16. Jul 3, 2012

### wolfblood

I don't see i think i am missing something about angles and vector redistribution :(

17. Jul 14, 2012

### wolfblood

18. Nov 30, 2012

### tpfitzpatrick

First: Momentum is always conserved.
Second: In elastic collisions, kinetic energy is also conserved.
When objects collide elastically in 2 dimensions, the geometry of the collision is also important because the impulse delivered to each object is along the lined connecting the two centers of mass.