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I know that usually collision with velocity collinear can be solved by simultaneous equations of both conservation of energy and linear momentum. But my question is when 2D velocity is encountered, we do only have a vector and a scalar (isn't it?) equations, how can we then find out the velocity (which is equivalent to four scalars) after collision?

But on the other hand, since they will collide, the relative velocity of two particles should be collinear. Then this can be solved by providing an equivalent 1D model without dealing with perpendicular component of velocity.

This seemed to be a contradiction that we've used 3 scalar eqns to find out 4 variables which is obviously impossible mathematically. But see this: $$ m_1\vec{v_1}^2-m_1\vec{v_1'}^2+m_2\vec{v_2}^2-m_2\vec{v_2'}^2=0 {(1)}\\ m_1(\vec{v_1}-\vec{v_1'})=m_2(\vec{v_2'}-\vec{v_2}){(2)}$$ After simple maths, you got ${(1)}/{(2)}$ is a vector equation, i.e. we've made a new equation with physical meaning unknown. Is there any physics concept behind this? Futhermore, for rigid bodies with their shape, can we also find out their angular velocity by providing conservation of angular momentum?

YiFei
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    Possible duplicate of Elastic collision between two circles – John Alexiou Dec 21 '15 at 15:46
  • yes, you can either use the two equations for momentum along the two different directions, or solve in a reference system where only one of the particles move.In my experience the first is usually used when solving collision problems. –  Dec 21 '15 at 15:47
  • you can use something called coefficient of restitution (e) which is kind of refers to the elasticity of the collision and is defined as relative velocity of seperation/relative velocity of approach. You actually get conservation of energy from this and conservation of momentum. If e=1, kinetic energy is conserved for 2-d collisions also.A google search would reveal further details. – Skawang Dec 21 '15 at 15:50
  • @brucesmitherson - and then, of course, we confuse students once we get to high energy physics scattering problems when we solve them all in the center-of-mass frame and have to translate back into the lab frame to get the observed scattering pattern. And I always found it easier to consider one object at rest, and then translate into the lab frame... – Jon Custer Dec 21 '15 at 16:05
  • This is a more general question that deserves a much better explanation than the "duplicate". This question is primarily asking about how a 4 degree of freedom system can be solved with 3 equations. And the answer is: it can't! it depends on the geometry of the impact point. The "duplicate" assumes circles which makes this dependence hide, and thus not answer the question. – Rick May 19 '16 at 15:29
  • In reality, it's a 6 degree of freedom system, as each object also has a rate of rotation about its COM. The conservation of momentum give you three (two axises and one rotation) and conservation of energy (or prescribed loss with a COR) gives you one more, but there are still two more equations that are needed to solve the system. These can be derived by looking at the impact point. If the question reopens I'll answer it. – Rick May 19 '16 at 15:38
  • @Rick If you feel that way, why not cast a re-open vote to submit the question back to the review queue? – dmckee --- ex-moderator kitten May 21 '16 at 16:10
  • @dmckee I would if I could. I don't have 3k yet. – Rick May 23 '16 at 10:07
  • @Rick I'm so happy that someone had really got what I'm really looking for, hope to see your elaborate answer:). So would you please edit my question if it really messes things up and confuses some people, thanks in advance. – YiFei May 27 '16 at 13:40

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