- #1

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- Thread starter Or Ozery
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- #1

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- #2

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- #3

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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...

- #4

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

- #5

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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.

- #6

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

[tex] u1 = \frac{v_1(m_1 - m_2) + 2m_2v_2}{m_1 + m_2} [/tex]

[tex] u2 = \frac{v_2(m_2 - m_1) + 2m_1v_1}{m_1 + m_2} [/tex]

Hope this is of some use.

- #7

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u1 and u2 depand on alpha as well...

Energy conservation:

[tex]m_{1}v_{1}^2 = m_{1}u_{1}^2 + m_{2}u_{2}^2[/tex]

Momentum conservation:

[tex]m_{1}v_{1}= m_{1}u_{1}\cos\alpha + m_{2}u_{2}\cos\beta[/tex]

[tex]m_{1}u_{1}\sin\alpha = m_{2}u_{2}\sin\beta[/tex]

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

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