2d absolutely elastic vector collision

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?

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.

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.

rock.freak667

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

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.

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

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.

dimension10

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

wolfblood

Nor in one direction?

dimension10

I don't get what you are saying?

Infinitum

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

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

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
Please. I get lost

dimension10

Oh, ok.

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 [itex]a^2-b^2 = (a+b)(a-b)[/itex]. You will get a similar term from the momentum equations which you can substitute to solve further.

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.

Infinitum

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

wolfblood

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

wolfblood

On wikipedia http://en.wikipedia.org/wiki/Elastic...ee-dimensional
there is part with thetas

but i couldn't change into equation with xA xB yA yB