2d absolutely elastic vector collision

[wolfblood]

Homework Statement

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]

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.

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

 

[dimension10]

Homework Statement

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?

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.

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

 

[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]

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.

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

 

[wolfblood]

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

Nor in one direction?

 

[dimension10]

Nor in one direction?

I don't get what you are saying?

 

[Infinitum]

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

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]

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

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

Oh, ok.

 

[Infinitum]

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

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.

 

[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]

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.

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_collision#Two-_and_three-dimensional
there is part with thetas

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

 

[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.