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As mentioned in every text book I own, the relative velocities of approach and separation are equal and opposite for (only) 1D collisions. For 2D collisions, only the component of velocities along the line of impact has to be considered. This makes sense since the (impulsive) force, ignoring friction, is always along the line of impact, and so must be the changes in momentum.


Here is the problem. In the derivation of this result, $\vec U_1-\vec U_2=-(\vec V_1- \vec V_2)$, at no point did I make the assumption that these velocities were 1 dimensional vectors. So shouldnt this formula apply to 2D and 3D collisions as well( without considering the velocities along the line of impact instead of the actual velocities)? Or is there some implicit assumption I already made in the Derivation that Restricts this?


thanks

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  • $\begingroup$ In 2-d, 3-d: we have to use the net velocities, which are vectors. The derivation uses scalar velocities, which is true only in 1-d.(In 1-d: the velocity vectors are scalars with proper signs) $\endgroup$ Aug 14 '20 at 12:49
  • $\begingroup$ For a collision in 2 or 3D, the direction of the impact force depends on the shape of the objects and the offset of the incoming velocities. All you can say other than conservation of energy and momentum is that the two speeds relative to the center of mass will be the same before and after the collision. $\endgroup$
    – R.W. Bird
    Aug 14 '20 at 13:11
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We know that the total momentum of the system is conserved in elastic collisions, and momentum is a vector. So if we have two particles with masses $m_1, m_2$ and velocities $\vec u_1, \vec u_2$ before the collision and $\vec v_1, \vec v_2$ after the collision then

$m_1 \vec u_1 + m_2 \vec u_2 = m_1 \vec v_1 + m_2 \vec v_2$

and this vector equation holds whether we are working in one, two or three dimensions.

We also know that the only force acting on the particles during the collision is along the line of impact, so any component of either particle's momentum (and hence velocity) that is perpendicular to the line of impact is unchanged by the collision. So it is often convenient to resolve momentum/velocity components along the line of impact, where we have the scalar equation

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where $u_1, u_2, v_1, v_2$ are now the scalar components of $\vec u_1, \vec u_2, \vec v_1, \vec v_2$ along the line of impact. Note that $u_1, u_2, v_1, v_2$ are signed - this can cause confusion if you use a convention that positive velocities are in one direction before the collision and in the opposite direction after.

If we know $m_1, m_2, \vec u_1, \vec u_2$ then this equation together with conservation of energy gives us enough information to find $\vec v_1$ and $\vec v_2$.

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  • $\begingroup$ So this is just a matter of convenience and if i were instead to equaTe the relative velocity of separation to that if separation, it would yield the same result? $\endgroup$ Aug 14 '20 at 11:35
  • $\begingroup$ @OVERWOOTCH Yes. As long as you use the same sense/direction for positive velocities for both objects and before and after the collision, then you will find that the vector and scalar equations are saying the same thing. $\endgroup$
    – gandalf61
    Aug 14 '20 at 12:53

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