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For the diagram, how would I replace the sum or difference of two momentum vectors with an equivalent momentum vector? So, for example in the x-direction:

$$m_e u_e cos\beta = m_1 u_1 cos\beta + m_2 u_2 cos\beta \tag{1}$$

and in the y-direction,

$$m_e u_e sin\beta = m_1 u_1 sin\beta - m_2 u_2 sin\beta \tag{2}$$

where $m_e$ is the mass and $u_e$ is the velocity for the equivalent single vector. What I find confusing is if Eqns 1 and 2 are simplified further, then the following is obtained,

$$m_e u_e = m_1 u_1 + m_2 u_2 \tag{3}$$

$$m_e u_e = m_1 u_1 - m_2 u_2 \tag{4}$$

Obviously something is not correct. I'm not sure if I can make this equivalency but if possible, how would I correctly setup the equations to solve for $m_e$ and $u_e$?

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  • $\begingroup$ What is the angle $\beta$? Are both $m_1$ and $m_2$ traveling in the same direction? Why are you subtracting in equation (2)? A diagram of the problem would help clear up these confusions. $\endgroup$
    – Mark H
    Commented Sep 4, 2023 at 22:27
  • $\begingroup$ Diagram has been added. $\endgroup$
    – rdemyan
    Commented Sep 4, 2023 at 22:30
  • $\begingroup$ If $m_1$ and $m_2$ are different and $v_1$ and $v_2$ are different then there is no simplification that can be done other than $(m_1 v_1 + m_2 v_2) \cos \beta$ since you cannot factor out anything else. $\endgroup$
    – John Alexiou
    Commented Sep 5, 2023 at 16:00
  • $\begingroup$ Since my goal is to compare velocities before the collision with after the collision ($v_m$ - see comment below), then I think maybe what I should do is just define an average velocity before the collision. $\endgroup$
    – rdemyan
    Commented Sep 5, 2023 at 20:20

1 Answer 1

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The resultant vector does not have the same angle as the vectors being summed. Equation (1) should be $$m_eu_e\cos\alpha = m_1u_1\cos\beta + m_2u_2\cos\beta$$ and similarly for equation (2).

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  • $\begingroup$ What I posted was only the LHS of the momentum eqn. For Eqn. 1, the full equation is: $$m_1 u_1 cos\beta + m_2 u_2 cos\beta = (m_1 + m_2)v_m cos\alpha $$ What I was hoping to do was to combine the LHS using $m_eu_e$ and then compare $u_e$ with $v_m$. This is not an elastic collision. I'm trying to come up with a kind of coefficient of restitution but based on a single stream - hence the reason for wanting to combine the incoming streams. I was hoping to get an equivalent vector for the LHS which is prior to the collision. $\endgroup$
    – rdemyan
    Commented Sep 4, 2023 at 23:27
  • $\begingroup$ So, ultimately something like: $$m_e u_e cos\beta = (m_1 + m_2)v_m cos\alpha$$ So, I need to figure out what $m_e$ and $u_e$ are. $\endgroup$
    – rdemyan
    Commented Sep 4, 2023 at 23:34
  • $\begingroup$ @rdemyan I don't understand what it is you're trying to calculate. The sum of the two incoming momentum vectors (with angles $\pm\beta$) is the momentum vector after collision (with angle $\alpha$). I don't see any way, nor any point, in trying to combine the incoming vectors into one. $\endgroup$
    – Mark H
    Commented Sep 6, 2023 at 4:47
  • $\begingroup$ I am trying to combine the two terms on the left hand side into an equivalent single term. Is it possible to write the following for the x-direction momentum? $$m_1u_1cos\beta + m_2u_2cos\beta = (m_1+m_2)u_{ave}cos\beta$$ $\endgroup$
    – rdemyan
    Commented Sep 6, 2023 at 15:00
  • $\begingroup$ @rdemyan Unless the initial vectors are parallel, their sum will not be parallel to either vector. This means that the combined vectors cannot have the same angle as either of the initial vectors. The $\beta$ in the combined form can't mean anything because there's no vector with that angle except one of the initial vectors. The vector you're trying to calculate seems meaningless to me. The actual combined momentum vector is the momentum after the collision with angle $\alpha$. $\endgroup$
    – Mark H
    Commented Sep 7, 2023 at 10:03

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