## 2d absolutely elastic vector collision

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?

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 Recognitions: Homework Help 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.
 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.

Recognitions:
Homework Help

## 2d absolutely elastic vector collision

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

 Quote by 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?
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.

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

 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.

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

 Quote by dimension10 No, during elastic collisions, kinetic energy is conserved, not momentum.
Nor in one direction?

 Quote by wolfblood Nor in one direction?
I don't get what you are saying?

 Quote by dimension10 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.

 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

 Quote by Infinitum Both momentum and kinetic energy are conserved in elastic collisions.... In fact, momentum is conserved in both elastic and inelastic collisions.
Oh, ok.

 Quote by 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
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.

 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.

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

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

 Tags elastic collision, momentum, vector