# Two dimensional elastic collision - unequal masses

Hi,

I need to find an expression for u1 and u2 using m1, m2, v1 and alpha.
See attached image for more details.

Or Ozery

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I can't open the image (yet), but I'm sure you have to find expressions for the initial and final energy and momentum and use energy, momentum conservation to solve for the two unknowns.

Yep

I know, I get three equations (1 from energy, and 2 from the momentum vector) with 3 variables, but they are very cumbersome - I can't find a solution...

Should be enough - in the mean time, want to describe the problem so we can look at it?

Consider the elastic collision of two particles with rest masses m1 and m2.
Particle 1 is moving with speed v1 and particle 2 is at rest.
We choose the coordinate system such that particle 1 is initially moving along the x axis.
The two vectors, initial and final velocity of particle 1, will define the x-y plane.
Because of conservation of momentum, the final velocity of particle 2 is also confined to the x-y plane.
After the collision particle 1 makes an angle alpha with the x axis and its velocity is u1(cos alpha; sin alpha).

Express u1 using v1, m1, m2 and alpha.

Right. So you do realize that Kinetic energy and momentum are both conserved. The velocity of the center of mass is also unchanegd.

Solving for the final velocities from using these principles... we get..

$$u1 = \frac{v_1(m_1 - m_2) + 2m_2v_2}{m_1 + m_2}$$

$$u2 = \frac{v_2(m_2 - m_1) + 2m_1v_1}{m_1 + m_2}$$

Hope this is of some use.

But where is alpha in your formulas?
u1 and u2 depand on alpha as well...

Energy conservation:
$$m_{1}v_{1}^2 = m_{1}u_{1}^2 + m_{2}u_{2}^2$$

Momentum conservation:
$$m_{1}v_{1}= m_{1}u_{1}\cos\alpha + m_{2}u_{2}\cos\beta$$
$$m_{1}u_{1}\sin\alpha = m_{2}u_{2}\sin\beta$$

Need to solve these equations (unknowns are u1, u2 and beta).